WJEC GCSE Mathematics and Numeracy (Double Award) Revision
Explore WJEC GCSE Mathematics and Numeracy (Double Award) revision topics. Check the official specification for complete requirements for your exam year and route.
Course details
Assessment Choose a tier separately for each unit. Unit 1: calculator, 30%; Unit 2: non-calculator, 30%; Unit 3: calculator, 40%. Foundation times are 90, 90 and 105 minutes; Higher times are 105, 105 and 120 minutes. This guide is for the new Double Award, not the legacy resit.
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Reviewed course guidance
Revision points, with the explanation beside them
Open a point to read what to know, how to use it in an assessment, a plain-language explanation and a self-check. Answers stay hidden until you choose to reveal them.
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Unit 2 tier
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Unit 3 tier
- Foundation tier
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Topic group
Number
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §1.1.1 foundation; Unit 2 §1.1.1 foundation; Unit 3 §1.1.1 foundation. Printed pages 11: read and write whole numbers of any magnitude expressed in figures or words
Open revision guidance
What to know
- Whole numbers consist of digits 0 through 9 positioned to represent powers of ten. Understanding magnitude requires reading digits from left to right, where the leftmost digit holds the highest place value. For example, the number 450,231 consists of 4 hundred thousands, 5 ten thousands, 0 thousands, 2 hundreds, 3 tens, and 1 unit. Writing these in words follows specific conventions where periods like thousands and millions are separated by commas to aid readability, such as writing four hundred fifty thousand, two hundred thirty one.
- When converting written words to numeric figures, identify the largest denomination mentioned first. If presented with 'two million, seventy five thousand, and twelve', interpret 'two million' as 2,000,000. 'Seventy five thousand' occupies the thousands block as 075, and 'twelve' occupies the final three digits as 012. Combining these, the numerical representation is 2,075,012. This structure ensures that place holders are filled correctly with zeros, preventing errors in magnitude interpretation regardless of whether the input is in descriptive text or standard digit form.
What to do in an assessment
- To practise reading numbers, isolate each grouping of three digits separated by a comma. Read each triad as a standalone number and append the period name, such as million, thousand, or units. For example, in 8,409,005, read eight (million), four hundred nine (thousand), and five (units). When writing figures from text, create blank slots for each place value to ensure precision. If given 'six hundred thousand and eight', write six, zero, zero, zero, zero, eight, resulting in 600,008. Verify the total count of digits matches the expected scale.
- Perform a self-check by converting 1,005,432 into words and vice versa. The conversion yields 'one million, five thousand, four hundred thirty two'. Repeat this for a different magnitude like 99,004 to ensure proficiency with varied lengths. Always check that trailing or empty placeholders are correctly assigned zero digits. Consistent practice of this systematic approach minimises errors when working with large datasets or financial figures in mathematical problems.
In plain language
Reading and writing large numbers means breaking them into groups of three and assigning the right place value to each, like hundreds, thousands, or millions.
Self-check
Write the number 402,050 in words.
Show the answer
Four hundred two thousand and fifty.
View diagrams
Place-value table for 50.063
Key and description
50.063 = 5 tens + 0 ones + 0 tenths + 6 hundredths + 3 thousandths.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §1.1.2 foundation; Unit 2 §1.1.2 foundation; Unit 3 §1.1.2 foundation. Printed pages 11: round whole numbers to the nearest 10, 100, 1000, etc.
Open revision guidance
What to know
- Rounding involves approximating a number to a simpler value based on specific place values like the nearest 10, 100, or 1000. To round to the nearest 10, examine the units digit. If the digit is 5 or greater, round up the tens digit by 1 and set the units digit to 0. If it is less than 5, keep the tens digit the same and set the units digit to 0. This logic extends to other powers of ten. For rounding 3,462 to the nearest 100, look at the tens column, which is 6. Because 6 is 5 or greater, round up the hundreds digit (4 becomes 5) and set all digits to the right to zero, resulting in 3,500.
- Understanding the boundary condition is essential for accurate rounding. The halfway point, typically represented by a 5, dictates that one must round up to ensure a consistent standard. When rounding 7,500 to the nearest 1,000, observe the hundreds digit, which is 5. According to the rounding rule, this requires rounding up the thousands digit (7 becomes 8) and zeroing the remainder, giving 8,000. Mastery of this skill allows for quick estimation in mental arithmetic and provides a systematic way to manage numerical precision in various mathematical contexts, from financial accounting to statistical summary.
What to do in an assessment
- Identify the target place value for rounding and underline the digit to its immediate right. For rounding 12,389 to the nearest 1,000, the thousands place is 2 and the hundreds digit to its right is 3. Since 3 is less than 5, keep the thousands digit as 2 and change all digits to the right to zero, yielding 12,000. If the digit to the right had been 5 or greater, the thousands digit would have increased to 3, resulting in 13,000. Apply this same mechanical step to every rounding challenge.
- Check your work by drawing a number line. If you are rounding 56 to the nearest 10, mark 50 and 60 on the line. Since 56 is closer to 60, verify the result matches your calculated rounding. This visual check confirms the accuracy of your application of the rounding rules, especially for numbers close to the halfway point. Consistent practice builds intuition for where a number sits relative to its nearest multiples, ensuring that rounding tasks are performed reliably and without confusion.
In plain language
Rounding makes numbers easier to work with by finding the nearest 10, 100, or 1000. If the next digit is 5 or more, round up; if it is less than 5, round down.
Self-check
Round 8,472 to the nearest 100.
Show the answer
8,500.
View diagrams
Rounding and integer number lines
Key and description
56 lies between 50 and 60, so to the nearest ten it rounds to 60. The second line marks the integers −5, −2, 0 and 3.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §1.1.3 foundation; Unit 2 §1.1.3 foundation; Unit 3 §1.1.3 foundation. Printed pages 11: understand place value of whole numbers and those written in decimal form
Open revision guidance
What to know
- Place value dictates the weight of a digit based on its position relative to the decimal point. To the left, each position represents increasing powers of ten: units, tens, hundreds, thousands, and so on. To the right, each position represents decreasing fractions of ten: tenths (1/10), hundredths (1/100), and thousandths (1/1000). For example, in the number 12.458, the '1' is in the tens place, '2' is in the units place, '4' is in the tenths place, '5' is in the hundredths place, and '8' is in the thousandths place. The decimal point acts as an anchor for these values.
- Understanding this system is critical for comparing decimal values. A common error is assuming that a longer decimal string is larger, but 0.125 is actually smaller than 0.2 because the first decimal place (tenths) in 0.2 is higher than in 0.125. When ordering decimals, align the numbers by their decimal point and compare each column from left to right. This ensures that the most significant digits are compared first, correctly identifying which value possesses greater magnitude. This method remains consistent whether dealing with simple currency values or complex scientific measurements.
What to do in an assessment
- To analyse place value, create a table with columns for each position (thousands, hundreds, tens, units, decimal point, tenths, hundredths, thousandths). Place your number into the table aligning digits to the decimal point to visualise the value of each digit. For 50.063, place 5 in tens, 0 in units, 0 in tenths, 6 in hundredths, and 3 in thousandths. This grid approach prevents errors when reading or writing numbers with zeros in significant positions, clarifying that the 0 in the tenths column is a place holder, not absence of value.
- Order a set of decimals by normalising their length with zeros. For example, to compare 0.5, 0.49, and 0.512, rewrite them as 0.500, 0.490, and 0.512. Now, compare the columns left to right: the tenths column shows 0.512 is the largest (5), followed by 0.500 (5), and 0.490 (4). Order them as 0.49, 0.5, 0.512. This padding technique provides a reliable, logical framework for sorting any list of decimal figures correctly every time.
In plain language
Place value shows what each digit is worth. To the right of the decimal point, numbers get smaller (tenths, hundredths, thousandths). Compare numbers by lining up the decimal points.
Self-check
Which is larger: 0.8 or 0.795?
Show the answer
0.8.
View diagrams
Place-value table for 50.063
Key and description
50.063 = 5 tens + 0 ones + 0 tenths + 6 hundredths + 3 thousandths.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §1.1.4 foundation; Unit 2 §1.1.4 foundation; Unit 3 §1.1.4 foundation. Printed pages 11: round decimals to the nearest whole number or a given number of decimal places
Open revision guidance
What to know
- Rounding decimals involves trimming a value to a specified number of decimal places (dp) or to the nearest whole number. The rule remains consistent with whole numbers: identify the digit immediately after your target position. If that following digit is 5 or greater, round the target digit up by one. If it is less than 5, leave the target digit unchanged. For example, rounding 4.672 to 1 decimal place (dp) requires looking at the second digit, 7. Because 7 is 5 or greater, the 6 in the tenths place increases to 7, resulting in 4.7.
- When rounding to the nearest whole number, the target is the units digit, and the 'following' digit is the first digit after the decimal point. For instance, in 12.83, the units digit is 2. The digit to its right is 8. Since 8 is 5 or greater, the 2 rounds up to 3, yielding 13. Similarly, rounding 0.045 to 2 decimal places looks at the thousandths digit, 5. The digit 4 in the hundredths place increases to 5, resulting in 0.05. Mastery of this logic ensures that precision is maintained consistently, preventing overestimation or underestimation in various measurements.
What to do in an assessment
- To round 3.14159 to 3 decimal places, locate the third digit after the decimal point, which is 1. Observe the digit immediately following it, which is 5. Since 5 is 5 or greater, increase the target 1 to 2. The final result is 3.142. By focusing specifically on the digit one place to the right, you remove all ambiguity from the rounding process. Always check that your resulting number maintains the requested number of decimal places.
- For self-check, round 7.996 to 2 decimal places. Locating the second decimal place (9) and the digit following it (6), you must round up the 9 to 10, which carries over to the tenths place, turning 7.99 into 8.00. Writing the two zeros is important to show that the rounding was performed to the precision of 2 decimal places. This demonstrates precision and adherence to the standard conventions for decimal rounding tasks.
In plain language
Rounding decimals means changing them to a specific number of places. Look at the digit after your target; if it is 5 or more, add one. If less than 5, keep it the same.
Self-check
Round 5.438 to 2 decimal places.
Show the answer
5.44.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §1.1.5 foundation; Unit 2 §1.1.5 foundation; Unit 3 §1.1.5 foundation. Printed pages 12: round numbers to a given number of significant figures
Open revision guidance
What to know
- Significant figures (sf) provide a measure of precision for a number regardless of its magnitude. The first significant figure is the first non-zero digit when reading from left to right. For example, in 0.00405, the first significant figure is 4. Subsequent zeros count if they are between non-zero digits, but leading zeros never count. To round to a given number of significant figures, identify the target digit (the one at the specified count) and use the next digit to decide whether to round up, just as with decimal places.
- Rounding 45,720 to 2 significant figures requires finding the first two digits: 4 and 5. The next digit is 7, which is 5 or greater, so the 5 rounds up to 6. To maintain the magnitude of the number, the remaining places are filled with zeros, resulting in 46,000. For decimals, rounding 0.00284 to 2 significant figures targets the 2 and 8. The next digit is 4, so the 8 remains unchanged, yielding 0.0028. This system ensures that numbers maintain their scale while providing the desired level of accuracy required for the calculation.
What to do in an assessment
- To round 0.05062 to 2 significant figures, ignore leading zeros and find the first significant figure (5). The second is 0 (it is between 5 and 6, so it counts). The digit following the second sf is 6. Because 6 is 5 or greater, the second sf (0) becomes 1. The result is 0.051. If you were asked for 3 significant figures, you would look at the digit after the third sf (2), which is less than 5, so the result remains 0.0506.
- Check your understanding by rounding 1,008 to 2 significant figures. The first sf is 1, the second is 0. The next digit is 0. Therefore, the result is 1,000. If rounding to 3 sf, the third figure is 0, the next is 8, so the 0 becomes 1, giving 1,010. By systematically identifying the target sf, you ensure precision and consistency in reporting data, a critical step for solving complex multi-stage numerical problems where error propagation must be strictly controlled.
In plain language
Significant figures measure precision. They are the first non-zero digits. To round, identify the target digit and use the one to its right to decide if you round up.
Self-check
Round 0.00762 to 2 significant figures.
Show the answer
0.0076.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §1.1.6 foundation; Unit 2 §1.1.6 foundation; Unit 3 §1.1.6 foundation. Printed pages 12: understand, use and order directed numbers
Open revision guidance
What to know
- Directed numbers, also known as integers, include positive numbers, negative numbers, and zero. On a number line, positive numbers extend to the right of zero, while negative numbers extend to the left. The value of a number decreases as you move further left. For instance, -10 is smaller than -2. When comparing directed numbers, remember that any positive number is always greater than any negative number. Understanding this ordering is essential for correctly placing numbers in a sequence or performing basic operations like finding the difference between two temperatures.
- Applying operations to directed numbers requires following specific sign rules. When adding a negative number, it is equivalent to subtracting the positive version, e.g., 5 + (-3) = 2. Subtracting a negative number is equivalent to adding the positive version, e.g., 5 - (-3) = 8. Multiplication and division follow the same logic: multiplying two numbers with the same sign results in a positive outcome, while different signs result in a negative outcome. For example, -4 multiplied by 3 equals -12, and -4 multiplied by -3 equals 12.
What to do in an assessment
- Practice ordering directed numbers by visualising a vertical thermometer. Temperatures like -5, -2, 0, and 3 are ordered from bottom to top. To order these, identify the most negative as the smallest (-5), then move towards zero (-2), then the positive values (0, 3). This physical representation clarifies that -5 is further from zero on the negative side, confirming its smaller value compared to -2. Write these lists carefully to avoid confusing sign placement.
- Solve expressions involving directed numbers by processing one sign pair at a time. For 10 - (-5) + (-2), first change 10 - (-5) to 10 + 5 = 15. Then add the (-2), which simplifies to 15 - 2 = 13. By breaking down multi-step expressions into basic binary operations, you reduce the risk of arithmetic errors. Check the final sign of each result against the multiplication/division rules: negative times positive is negative, negative times negative is positive.
In plain language
Directed numbers are positive and negative numbers. Negative numbers are to the left of zero. Subtracting a negative is the same as adding a positive.
Self-check
Evaluate 7 - (-4) + (-3).
Show the answer
8.
View diagrams
Rounding and integer number lines
Key and description
56 lies between 50 and 60, so to the nearest ten it rounds to 60. The second line marks the integers −5, −2, 0 and 3.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §1.1.7 foundation; Unit 3 §1.1.7 foundation. Printed pages 12: decide whether to round up or down, as appropriate, in a problem
Open revision guidance
What to know
- Rounding in real-world problems often requires a strategic decision to round up or down depending on the desired outcome. Standard rounding (round to nearest) is not always appropriate when constraints are present. For example, if you are organising transport for a group of people and the capacity of a bus is 50, but you have 102 people, you must round up to the nearest whole bus (102 / 50 = 2.04), requiring 3 buses to ensure everyone is transported. This is known as 'rounding up' to satisfy the constraint of full coverage.
- Conversely, rounding down is necessary when you have a limiting budget or fixed capacity. If you have 50 pounds and each item costs 7 pounds, you must calculate 50 / 7 = 7.14. You cannot buy the 8th item because you lack the funds, so you must round down to 7 items regardless of the decimal value. Understanding these scenarios is critical for accuracy in applied mathematics. Contextual awareness is the primary driver for choosing the rounding direction, ensuring solutions are practical, feasible, and compliant with real-world limitations.
What to do in an assessment
- When faced with a word problem, first define what your result represents: people, currency, or materials. If the result involves splitting items that cannot be divided, like buses or people, you almost always round up. For a calculation requiring 4.1 taxis for 21 people, you cannot book 0.1 of a taxi, so round up to 5 taxis. Write a brief statement justifying why you rounded up or down, confirming that your answer is a whole number that satisfies the physical requirements of the scenario.
- For problems involving funds or resources, use division then check the remainder. If you are calculating how many full packs of 6 tiles you can buy with 38 tiles needed, 38 / 6 = 6.33. You need to buy 7 packs to cover the full requirement, rounding up. If the budget constraint is the focus, such as buying as many as possible with 40 pounds where items cost 6 pounds, 40 / 6 = 6.66. You can only purchase 6 items, so round down. These distinctions are vital for real-world mathematical success.
In plain language
In real-world problems, rounding up or down depends on the situation. If you need full coverage, always round up. If you are limited by budget, round down.
Self-check
If you need to transport 145 students and each coach holds 50, how many coaches are needed?
Show the answer
3 coaches.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §1.1.8 foundation; Unit 2 §1.1.8 foundation; Unit 3 §1.1.8 foundation. Printed pages 12: check methods and solutions using appropriate strategies
Open revision guidance
What to know
- Checking mathematical work is essential for accuracy in GCSE exams. One primary strategy is to use inverse operations to verify results. If a problem involves addition, such as calculating the sum 145 + 236 = 381, you can check this by performing the inverse operation: 381 - 236. If the result is 145, your original sum is likely correct. This logic extends to multiplication and division; to verify 12 * 15 = 180, perform 180 / 15, which must equal 12. These inverse checks are vital for identifying simple arithmetic errors that often occur during pressurized testing environments.
- Estimation provides a second layer of verification for complex multi step calculations. By rounding each component of a calculation to one significant figure, you can determine if your final answer is reasonable. For instance, if you are asked to calculate (49 * 21) / 4.8, you can estimate this by computing (50 * 20) / 5. The calculation becomes 1000 / 5, which equals 200. If your calculated answer significantly deviates from 200, such as 20 or 2000, you immediately know that a decimal point error or order of operations mistake has occurred. Using these strategies together ensures your work remains robust throughout the assessment.
What to do in an assessment
- To demonstrate the inverse checking method, solve the equation 567 - 289. After calculating 567 - 289 = 278, verify this by performing the addition 278 + 289. The result of 278 + 289 is 567, which matches the original number, confirming the subtraction was accurate. If you were solving an algebraic equation like 3x = 42, first divide 42 by 3 to find x = 14. To check this, substitute 14 back into the equation: 3 * 14 = 42. Since 42 = 42, the solution is verified as correct.
- Apply the estimation check to the problem: Calculate 8.1 * 42.4 divided by 1.9. First, round the numbers: 8 * 40 divided by 2. This simplifies to 320 divided by 2, which equals 160. Now perform the actual calculation: 8.1 * 42.4 = 343.44. Then, 343.44 divided by 1.9 is 180.758. The estimate 160 is close to 180, indicating the calculation is likely correct in terms of magnitude. Always perform the full calculation first, then perform the estimate to verify if the scale of the answer is logical.
- Checks apply beyond arithmetic. For a triangle with angles 45°, 60° and a proposed third angle 80°, their sum is 185°, not 180°, so at least one value is wrong. For a frequency table, add every frequency and compare with the sample size. For a graph intersection, substitute the read-off coordinates into both equations, allowing for the accuracy of the graph.
In plain language
Check your answers by doing the opposite sum or by rounding numbers to check if the answer makes sense.
Self-check
Calculate 412 + 159 and check your answer using subtraction.
Show the answer
571; check: 571 - 159 = 412.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §1.1.9 foundation. Printed pages 12: estimate solutions to numerical calculations by approximating the numbers in the calculations
Open revision guidance
What to know
- Estimation is a critical mathematical skill used to find approximate solutions to calculations without performing precise arithmetic. The primary technique involves rounding each number in a complex calculation to its nearest sensible value, typically one significant figure or a round number that makes mental arithmetic straightforward. This allows you to quickly assess the size of an answer and detect major errors during examination conditions. By simplifying numbers to powers of ten or simple multiples, complex operations such as (39 * 51) / 9.8 can be approximated as (40 * 50) / 10, which yields 200. This approximation provides a reliable yardstick to ensure your final calculated answer is of the correct order of magnitude.
- Beyond simple multiplication and division, estimation is useful for checking more complicated scenarios involving roots or additions of large numbers. For example, to estimate the square root of 83, note that 81 is a perfect square (9*9) and 100 is a perfect square (10*10). Since 83 is much closer to 81, the estimate should be just over 9. When dealing with addition, rounding to the nearest ten or hundred helps prevent catastrophic errors where thousands might be mistaken for hundreds. Mastering these approximation techniques ensures that you are consistently performing valid mathematical checks before finalising your work on a paper.
What to do in an assessment
- To practise estimation, consider the calculation 192 * 4.9. To estimate this, round 192 to 200 and 4.9 to 5. The calculation becomes 200 * 5, which results in 1000. When performing the actual calculation, 192 * 4.9 equals 940.8. The estimate 1000 is very close to 940.8, which confirms that your calculation is likely correct. This process should be applied consistently to all numerical problems involving multiplication to ensure that your results align with basic expectation.
- Consider the calculation (508 + 291) / 3.9. Use estimation to predict the answer. Round 508 to 500 and 291 to 300, giving 800. Then round 3.9 to 4. The estimation is 800 / 4 = 200. Now perform the full calculation: 508 + 291 = 799. Next, 799 / 3.9 equals approximately 204.87. Since 204.87 is very close to your estimate of 200, you have verified the validity of your result through approximation. Using this strategy regularly prevents mistakes that are caused by simple calculator entry errors.
- Estimate before accepting an answer: a rectangle roughly 20 m by 30 m has area about 600 m², so an answer near 60 m² suggests a misplaced decimal. In 100 tosses of a fair coin, about 50 heads is reasonable; 500 heads is impossible because it exceeds the total tosses. A plausible result is a check, not proof that all the working is correct.
In plain language
Round the numbers in your sum to the nearest simple number so you can do the math in your head and see if your final answer is roughly correct.
Self-check
Estimate the value of 203 * 4.1 using rounding.
Show the answer
Estimate: 200 * 4 = 800 (Actual: 832.3).
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §1.2.1 foundation; Unit 2 §1.2.1 foundation; Unit 3 §1.2.1 foundation. Printed pages 12: the common properties of numbers, including knowledge of odd, even, integers, multiples, factors, primes
Open revision guidance
What to know
- Understanding the fundamental classification of integers is essential for number theory. Integers consist of all whole numbers, including negative numbers, zero, and positive values. Even numbers are integers divisible by 2 with no remainder, generally represented by 2n, while odd numbers are those that result in a remainder of 1 when divided by 2, represented by 2n+1. This binary classification defines the parity of any integer on the number line.
- Further properties include factors, multiples, and prime numbers. A factor is an integer that divides another integer exactly without leaving a remainder; for example, the factors of 12 are 1, 2, 3, 4, 6, and 12. A multiple is the product of an integer and any other integer; for example, 24 is a multiple of 6. A prime number is a natural number greater than 1 that has exactly two distinct positive divisors: 1 and itself. These properties form the basis for analysing numeric relationships and solving complex arithmetic problems efficiently.
What to do in an assessment
- To identify properties, start with prime verification. To check if 13 is prime, attempt to divide it by prime numbers up to the square root of 13. Since 2 and 3 do not divide 13 perfectly, it is confirmed as a prime number. Conversely, to find factors, list pairs that multiply to the target. For 20, test: 1x20, 2x10, and 4x5. The set of factors is {1, 2, 4, 5, 10, 20}. This systematic approach ensures all divisors are captured.
- For multiples, identify the sequence of products. The first five multiples of 7 are calculated by 7*1=7, 7*2=14, 7*3=21, 7*4=28, and 7*5=35. When classifying parity, observe the last digit of the number. If the final digit is 0, 2, 4, 6, or 8, the number is even; otherwise, it is odd. This simple visual check allows for immediate classification of large integers without performing the full division operation.
In plain language
This lesson explains how numbers are grouped, such as whether they are even or odd, and how to find their factors, multiples, and prime numbers.
Self-check
Identify if 17 is prime and list all factors of 16.
Show the answer
17 is prime. Factors of 16 are 1, 2, 4, 8, 16.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §1.2.2 foundation; Unit 2 §1.2.2 foundation; Unit 3 §1.2.2 foundation. Printed pages 12: the meaning of the terms square, square root, cube and cube root
Open revision guidance
What to know
- Squaring a number means multiplying it by itself, denoted by the exponent 2 (x^2). For example, 5 squared (5^2) is 5 * 5 = 25. The inverse operation, finding the square root, asks what number multiplied by itself results in the original value. For example, the square root of 25 (√25) is 5. This relationship is critical in geometric applications like area, where the area of a square with side length s is s squared.
- Cubing a number involves multiplying it by itself three times, denoted by the exponent 3 (x^3). For example, 3 cubed (3^3) is 3 * 3 * 3 = 27. The cube root of a number is the value that, when multiplied by itself three times, yields the original number. For example, the cube root of 27 (∛27) is 3. These operations are essential for understanding volumes and more advanced algebraic relationships, serving as the building blocks for power-based calculations.
What to do in an assessment
- When calculating squares or cubes, ensure the exponent is applied correctly to the base. For 4^2, calculate 4 * 4 = 16. For 4^3, perform 4 * 4 * 4, which is 16 * 4 = 64. When working with roots on a calculator, use the specific keys for square root (√) or the function for cube root (usually ∛x or x^(1/3)). To solve √81, input 81 then the root key to retrieve 9, because 9 * 9 = 81.
- To solve more complex problems, combine these steps. If asked for the value of 2^3 + √16, first resolve the powers and roots separately. Calculate 2^3 = 2 * 2 * 2 = 8. Then calculate √16 = 4. Add the results: 8 + 4 = 12. This method of breaking down expressions follows standard order of operations, ensuring that exponential and radical terms are evaluated prior to performing addition or subtraction.
In plain language
This lesson teaches you how to square and cube numbers, and how to find their roots, which is useful for geometry and calculating volumes.
Self-check
Calculate 6 squared and the cube root of 64.
Show the answer
6 squared is 36; the cube root of 64 is 4.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier or Unit 3 tier · Unit 2 §1.2.3 foundation; Unit 3 §1.2.3 foundation. Printed pages 12: the meaning of the term reciprocal
Open revision guidance
What to know
- A reciprocal of a number is defined as 1 divided by that number. For any non-zero number x, its reciprocal is 1/x. This concept is fundamental in understanding multiplicative inverses. For instance, the reciprocal of 5 is 1/5. If the number is a fraction like a/b, its reciprocal is found by flipping the numerator and denominator, resulting in b/a. Therefore, the reciprocal of 2/3 is 3/2.
- Understanding reciprocals is vital for algebraic manipulation and solving equations. When a number is multiplied by its reciprocal, the result is always 1 (e.g., 5 * 1/5 = 1). This is why they are often called multiplicative inverses. Note that zero does not have a reciprocal, as division by zero is undefined in standard arithmetic. Recognising how to convert integers, decimals, and fractions into their reciprocals is a necessary skill for advanced mathematical operations.
What to do in an assessment
- To find the reciprocal of an integer, write it as a fraction over 1 and invert it. For the number 8, write it as 8/1; its reciprocal is therefore 1/8. To express this as a decimal, divide 1 by 8, which equals 0.125. For a fraction, such as 7/4, simply swap the numbers to get 4/7. Always verify that the product of the number and its reciprocal equals 1 to ensure the calculation is accurate.
- When dealing with mixed numbers, convert them to improper fractions first. For 1 1/2, convert to 3/2. The reciprocal is then 2/3. For decimal values like 0.4, convert to a fraction first: 0.4 is 4/10, or 2/5. The reciprocal is then 5/2, which is 2.5. This conversion process avoids errors that often occur when attempting to calculate the reciprocal of a decimal directly using division.
In plain language
This lesson explains that a reciprocal is what you get when you turn a number into a fraction like 1/x, which is helpful for algebra.
Self-check
What is the reciprocal of 4, and what is the reciprocal of 2/5?
Show the answer
The reciprocal of 4 is 1/4 (0.25); the reciprocal of 2/5 is 5/2 (2.5).
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §1.2.4 foundation. Printed pages 13: express numbers as the product of their prime factors in index form
Open revision guidance
What to know
- Every composite integer can be expressed uniquely as a product of prime numbers. This is known as prime factor decomposition. To write this in index form, we use exponents to represent repeated prime factors. For example, the number 72 can be broken down: 72 = 8 * 9 = 2 * 2 * 2 * 3 * 3. In index form, this is written as 2^3 * 3^2. This notation is a compact and efficient way to describe the numerical structure of integers.
- The process of index form notation ensures that prime bases are grouped together, making it easier to see the structure of a number. This is crucial for comparing numbers, simplifying fractions, and finding common factors. By representing a number like 100 as 2^2 * 5^2, one can instantly identify the prime components that contribute to the value, which assists in solving more complex arithmetic and algebraic problems later in the curriculum.
What to do in an assessment
- To find the prime factors, use a factor tree. Start with any pair of factors for the given number. For 60, choose 6 * 10. Since 6 = 2 * 3 and 10 = 2 * 5, the factors are 2, 2, 3, and 5. Collect them: 2, 2, 3, 5. Write the final result in index form by counting the occurrences of each prime. Since 2 appears twice, write 2^2. Thus, the prime factorization of 60 is 2^2 * 3^1 * 5^1 (or simply 2^2 * 3 * 5).
- Repeat this method for larger numbers. For 120, break it down: 120 = 12 * 10 = (3 * 4) * (2 * 5) = 3 * (2 * 2) * 2 * 5. Collecting the prime factors yields 2 * 2 * 2 * 3 * 5. Converting this to index form results in 2^3 * 3^1 * 5^1. This systematic reduction ensures that no prime factor is missed. Always perform a quick multiplication check at the end to confirm the product matches the original number.
In plain language
This lesson teaches you how to break a number down into its prime components and write it using powers, which helps with advanced math.
Self-check
Write 40 as a product of its prime factors in index form.
Show the answer
40 = 2 * 2 * 2 * 5 = 2^3 * 5^1.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §1.2.5 foundation. Printed pages 13: find the least common multiple (LCM) and highest common factor (HCF) using prime factor decomposition or other appropriate methods
Open revision guidance
What to know
- The Highest Common Factor (HCF) of two or more numbers is the largest positive integer that divides all of the given numbers without a remainder. Conversely, the Least Common Multiple (LCM) is the smallest positive integer that is a multiple of each of the given numbers. Prime factorization is a powerful tool to determine both; by breaking numbers down into prime factors, one can compare their composition to find these values systematically.
- To find the HCF, identify the lowest power of all common prime factors. To find the LCM, identify the highest power of every prime factor present in any of the numbers. These metrics are essential for simplifying expressions, solving proportions, and finding common denominators. Mastering these techniques using prime decomposition allows for a more reliable approach to number problems compared to simple listing, especially for large integers where manual listing is inefficient and prone to error.
What to do in an assessment
- To find HCF and LCM of 24 and 36, first find prime factors. 24 = 2^3 * 3^1. 36 = 2^2 * 3^2. For HCF, take the common prime factors with the lowest power: 2^2 (common to both) and 3^1 (common to both). HCF = 2^2 * 3^1 = 4 * 3 = 12. For LCM, take every prime factor with the highest power: 2^3 (from 24) and 3^2 (from 36). LCM = 2^3 * 3^2 = 8 * 9 = 72.
- Verify calculations by checking if the results make sense. Since 12 is the largest number that divides both 24 and 36, and 72 is the smallest number divisible by both, the calculation is logically sound. For larger numbers or three numbers, apply the same steps: write all as prime products, then select the minimum powers for HCF and maximum powers for LCM across the full set of factors found in all numbers.
In plain language
This lesson shows how to find the largest factor two numbers share (HCF) and the smallest number they both divide into (LCM) using prime factorization.
Self-check
Find the HCF and LCM of 12 and 18.
Show the answer
HCF is 6; LCM is 36.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §1.2.6 foundation. Printed pages 13: use prime factor decomposition to help solve other numerical problems, including links to square numbers
Open revision guidance
What to know
- Prime factor decomposition serves as a diagnostic tool for identifying square numbers and roots. A number is a perfect square if, and only if, all the exponents in its prime factorization are even. For example, 36 = 2^2 * 3^2; since both exponents (2) are even, 36 is a perfect square. This property allows for quick verification of whether a number has an integer root without manually testing every candidate divisor.
- Beyond square numbers, prime decomposition helps in solving problems involving divisibility and simplification. If you need to know if 1008 is divisible by 7, decomposing it reveals 2^4 * 3^2 * 7^1; the presence of the 7^1 term confirms divisibility. This method connects abstract prime theory to practical numerical tasks, providing a robust logic for manipulating complex products and ratios that appear in various GCSE math applications.
What to do in an assessment
- To check if 144 is a perfect square, calculate its prime factors: 144 = 12 * 12 = (2^2 * 3) * (2^2 * 3) = 2^4 * 3^2. Observe the exponents: 4 and 2. Because both are even, 144 is a perfect square. To find its square root, halve the exponents: 2^(4/2) * 3^(2/2) = 2^2 * 3^1 = 4 * 3 = 12. This verifies √144 = 12 accurately through prime factors.
- For further application, check if 200 is a perfect square. 200 = 2 * 100 = 2 * 10 * 10 = 2 * (2 * 5) * (2 * 5) = 2^3 * 5^2. The exponent of 2 is 3 (odd). Since not all exponents are even, 200 cannot be a perfect square. This process provides a definitive test for square numbers and simplifies related problems, such as reducing complex surds to their simplest form by extracting perfect squares from roots.
In plain language
This lesson teaches you to use prime factors to quickly check if a number is a perfect square or to simplify roots.
Self-check
Is 81 a perfect square using prime factorization?
Show the answer
Yes. 81 = 3^4. Since 4 is even, it is a perfect square; √81 = 3^(4/2) = 3^2 = 9.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §1.3.1 foundation; Unit 2 §1.3.1 foundation; Unit 3 §1.3.1 foundation. Printed pages 13: the notation for positive integral indices
Open revision guidance
What to know
- Positive integral indices represent repeated multiplication. When we see a base number like 5 raised to a power of 3, written as 5^3, it means the base 5 is multiplied by itself 3 times. Thus, 5^3 equals 5 * 5 * 5, which results in 125. The base is the number being multiplied, and the exponent or index, in this case 3, indicates how many times that base is used as a factor. This notation provides a concise way to express large numbers that involve recurring factors.
- For variables, the same principle applies. If we have x^4, this represents x * x * x * x. It is important to distinguish this from multiplication involving the index itself. For instance, 2^4 is 16, whereas 2 * 4 is 8. These indices follow specific algebraic rules, such as a^m * a^n = a^(m+n) for multiplication of terms with the same base, where you add the exponents, or (a^m)^n = a^(mn) when raising a power to another power, where you multiply the exponents.
What to do in an assessment
- To evaluate expressions with positive indices, identify the base and the exponent clearly. For 4^3, write it out as 4 * 4 * 4. First, calculate 4 * 4 = 16. Then, multiply that result by the final 4, which is 16 * 4 = 64. For algebraic simplification using index rules, take 3^2 * 3^4. Since the bases are identical, add the exponents to get 3^(2+4) = 3^6. Finally, evaluate 3^6 by calculating 3 * 3 * 3 * 3 * 3 * 3 = 729.
- Practice converting expanded forms into index notation. If you are given 7 * 7 * 7 * 7 * 7, identify that 7 is the base and it appears 5 times, so the index notation is 7^5. When dealing with mixed bases, group like terms before applying rules. For example, in 2^2 * 3^2 * 2^3, rewrite it as (2^2 * 2^3) * 3^2. Use the addition rule to get 2^(2+3) * 3^2, resulting in 2^5 * 3^2. Calculating these gives 32 * 9 = 288.
In plain language
Positive indices are just a shorthand for multiplying a number by itself a specific number of times.
Self-check
Simplify and calculate 2^3 * 2^2.
Show the answer
2^5 = 32
- Higher
Scope: Higher · Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §1.3.2 higher; Unit 2 §1.3.2 higher; Unit 3 §1.3.2 higher. Printed pages 13: the notation for zero and negative indices
Open revision guidance
What to know
- Indices extend beyond positive integers to include zero and negative values. The Zero Index Rule states that any non-zero number raised to the power of zero equals 1. For example, 7^0 = 1, and (1/2)^0 = 1. This arises because dividing a power by itself, like a^n / a^n, equals 1, and according to the index subtraction rule, this is a^(n-n) = a^0. Understanding this is essential for simplifying complex algebraic expressions where terms might cancel out entirely.
- Negative Indices are defined by the reciprocal. The rule is that a^(-n) = 1 / (a^n). For instance, 5^(-2) is equivalent to 1 / (5^2), which simplifies to 1/25. This concept is vital for scientific notation and calculus. When dealing with fractions, (a/b)^(-n) is equivalent to (b/a)^n. Recognising these patterns allows for the conversion of complex fractional indices into more manageable forms or vice versa, ensuring accurate manipulation of mathematical expressions.
What to do in an assessment
- To work with zero indices, always replace any base (other than zero) raised to power 0 with 1. If you have 3x^0, this simplifies to 3 * 1, which is 3. For negative indices, convert them to their positive form before calculating. For example, simplify 2^(-3). Write this as 1 / 2^3. Since 2^3 is 8, the result is 1/8 or 0.125. Always ensure the base is positive before applying the reciprocal rule unless the context dictates otherwise.
- For more complex calculations, such as 4^(-2) * 4^3, use the standard index addition rule: 4^(-2+3) = 4^1 = 4. Alternatively, express as (1 / 4^2) * 4^3 = (1 / 16) * 64. Since 64 / 16 equals 4, both methods confirm the result. When dealing with negative indices in fractions, such as (2/3)^(-2), invert the fraction to (3/2) and change the index to positive, giving (3/2)^2 = 9/4 = 2.25. This step-by-step approach ensures precision.
In plain language
A zero index turns any number into 1, while a negative index tells you to flip the number into a fraction.
Self-check
Simplify 3^(-2) and 5^0.
Show the answer
1/9 and 1
- Higher
Scope: Higher · Higher tier in either Unit 2 tier · Unit 2 §1.3.3 higher. Printed pages 13: the notation for fractional indices
Open revision guidance
What to know
- Fractional indices introduce the relationship between exponents and roots. A fractional index of 1/n signifies the nth root of the base. Specifically, a^(1/2) is equivalent to the square root of a, written as √a, and a^(1/3) is the cube root, written as ∛a. This notation is powerful because it allows us to express radical operations using the standard laws of indices. For any fraction m/n as an index, the numerator m represents a power, and the denominator n represents the root, expressed as a^(m/n) = (root n of a)^m.
- When calculating with fractional indices, ensure you understand the order of operations. Usually, it is easier to find the root first before applying the power. For example, 8^(2/3) can be calculated as the cube root of 8 squared, or (8^(1/3))^2. Calculating the cube root of 8 first gives 2, and then squaring that result gives 2^2 = 4. If you had squared 8 first to get 64 and then took the cube root, you would get 4, illustrating that the sequence of root and power operations leads to the same result.
What to do in an assessment
- To evaluate 16^(3/2), break the problem into parts. The denominator 2 indicates a square root, and the numerator 3 indicates a power. Write it as (√16)^3. First, calculate √16 = 4. Then, raise this result to the power of 3, so 4^3 = 4 * 4 * 4 = 64. For another example, 27^(2/3), the denominator 3 indicates a cube root. Find ∛27, which is 3 because 3 * 3 * 3 = 27. Then, square the result to get 3^2 = 9. This systematic approach avoids large number errors.
- Practice converting radical forms into fractional index notation. For instance, the square root of 5 is 5^(1/2). The cube root of 10 squared is 10^(2/3). When using a calculator, ensure you use parentheses for the fractional exponent, such as entering 16^(3/2) as 16^(3/2). If you do not use parentheses, the calculator might perform 16^3 then divide by 2, leading to an incorrect answer of 2048 instead of 64. Always verify the exponent is calculated as a single block.
In plain language
Fractional indices are a way to show roots and powers, where the bottom number is the root and the top number is the power.
Self-check
Calculate 8^(4/3).
Show the answer
16
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §1.3.4 foundation. Printed pages 13: use the rules of indices to perform calculations with numbers written in index form for positive integral indices
Open revision guidance
What to know
- The laws of indices allow for rapid calculation and simplification of numerical expressions without needing full expansion. The three primary rules for positive integral indices are multiplication, division, and power of a power. When multiplying terms with the same base, such as x^a * x^b, we sum the exponents to get x^(a+b). When dividing, such as x^a / x^b, we subtract the exponents to get x^(a-b). These rules only apply when the bases are identical, which is a common area for error.
- The third rule, the power of a power rule, involves raising a base that is already exponentiated to another power, such as (x^a)^b. In this scenario, we multiply the exponents to result in x^(a*b). These laws form the foundation for all further work in algebra, including scientific notation and polynomial manipulation. Mastery of these rules simplifies tasks involving large numbers by reducing them to manageable operations with smaller values before computing the final result.
What to do in an assessment
- To simplify expressions like 2^3 * 2^4, use the addition rule: 2^(3+4) = 2^7. Compute 2^7, which is 128. For a division problem, simplify 5^6 / 5^2. Using the subtraction rule, 5^(6-2) = 5^4. Compute 5^4 = 5 * 5 * 5 * 5 = 625. If an expression includes a power of a power, such as (3^2)^3, use the multiplication rule: 3^(2*3) = 3^6. Calculate 3^6 = 729. These steps keep calculations clean and logical.
- When facing a mix of operations, such as (2^3 * 2^2) / 2^4, tackle the numerator first. The numerator 2^3 * 2^2 simplifies to 2^5 = 32. Then divide by the denominator 2^4 = 16. The resulting division is 32 / 16 = 2. Alternatively, use the index laws directly: 2^5 / 2^4 = 2^(5-4) = 2^1 = 2. Direct index law usage is often faster and less prone to arithmetic error than evaluating each term first. Always check the bases match before starting.
In plain language
Use index laws to add powers when multiplying, subtract powers when dividing, and multiply powers when raising to a new power.
Self-check
Simplify (4^2 * 4^3) / 4^4.
Show the answer
4^1 = 4
- Higher
Scope: Higher · Higher tier in either Unit 2 tier · Unit 2 §1.3.5 higher. Printed pages 14: use the rules of indices to perform calculations with numbers written in index form for positive, negative and fractional indices
Open revision guidance
What to know
- For the same base, multiply by adding indices, divide by subtracting indices, and raise a power to another power by multiplying indices. A negative index means a reciprocal: x^(−n)=1/x^n for x≠0.
- For positive x, x^(m/n)=(ⁿ√x)^m. State the domain when using even roots or denominators. For example, √(x⁵) and (√x)⁵ agree for x≥0, but a reciprocal of either requires x>0. Use brackets to show exactly what is under a root.
What to do in an assessment
- 9^(1/2)×9^(−1)=9^(1/2−1)=9^(−1/2)=1/√9=1/3. Directly, √9×(1/9)=3/9=1/3, giving an independent check.
- For x>0, (x²×x^(−4))/x^(1/2)=x^(2−4−1/2)=x^(−5/2)=1/√(x⁵). The whole x⁵ is inside the square root. Also, (x^(1/2)×x²)/x^(−3)=x^(1/2+2−(−3))=x^(11/2): subtracting a negative adds its magnitude.
- Keep numerical coefficients separate from powers. (6x³×2x^(−1))/(3x^(1/2))=(12/3)x^(3−1−1/2)=4x^(3/2), for x>0.
- Include coefficients when applying index laws: (12x⁵)/(3x⁻¹) = 4x⁶ for x ≠ 0. For a positive x, (4x²)^(1/2) = 2x. For 1/√(x⁵), write x^(−5/2), explicitly putting the whole x⁵ inside the square root.
In plain language
Combine coefficients and indices separately, show the extent of a root with brackets, and check that denominators and even roots are defined.
Self-check
For x>0, simplify x³×x^(−1)×x^(1/2).
Show the answer
x^(3−1+1/2)=x^(5/2).
- Higher
Scope: Higher · Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §1.3.6 higher; Unit 2 §1.3.6 higher; Unit 3 §1.3.6 higher. Printed pages 14: convert ordinary numbers into and out of standard form
Open revision guidance
What to know
- Standard form is a systematic method for writing extremely large or small numbers using powers of 10. A number in standard form is always written as A * 10^n, where A is a decimal such that 1 ≤ A < 10, and n is an integer. For example, 4500 becomes 4.5 * 10^3 because the decimal point moves three places to the left. Conversely, 0.00032 becomes 3.2 * 10^(-4) because the decimal point moves four places to the right.
- Converting out of standard form requires reversing this process. If you have 2.7 * 10^4, the positive exponent tells you to move the decimal point four places to the right, yielding 27000. If the exponent is negative, such as 5.1 * 10^(-3), move the decimal point three places to the left, resulting in 0.0051. This format is critical in scientific fields as it provides a clear scale for comparison and ensures accuracy when handling scientific data and calculations across widely varying magnitudes.
What to do in an assessment
- To convert 670,000 to standard form, place the decimal after the first digit to get 6.7. Count how many places you moved the decimal point from the original end of the number to this new position. Moving it 5 places to the left means 6.7 * 10^5. To convert 0.000089, move the decimal to get 8.9. Since you moved it 5 places to the right, the index is -5, giving 8.9 * 10^(-5). Accuracy in counting the decimal jumps is essential.
- To convert 3.14 * 10^3 into an ordinary number, recognise that 10^3 means 1000. So, 3.14 * 1000 = 3140. For 7.5 * 10^(-2), recognise that 10^(-2) means 1/100 or 0.01. So, 7.5 * 0.01 = 0.075. Always verify the result by checking if it matches the magnitude of the original scientific form. If the exponent is positive, the number must be large; if negative, the number must be a fraction of 1.
In plain language
Standard form uses A * 10^n to write very large or small numbers, where A is between 1 and 10 and n is an integer.
Self-check
Convert 0.00045 to standard form.
Show the answer
4.5 * 10^(-4)
- Higher
Scope: Higher · Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §1.3.7 higher; Unit 2 §1.3.7 higher; Unit 3 §1.3.7 higher. Printed pages 14: use numbers written in standard form
Open revision guidance
What to know
- Standard form allows for efficient multiplication and division of very large or small numbers. When multiplying, you multiply the A coefficients and add the exponents of the 10s. For example, (2 * 10^3) * (3 * 10^2) equals (2 * 3) * 10^(3+2), which is 6 * 10^5. This avoids having to convert everything to long-form numbers first, minimising the chance of error with extra zeros. This property makes it standard practice in scientific calculations where magnitude is key.
- When dividing, you divide the A coefficients and subtract the exponents of the 10s. For (8 * 10^6) / (2 * 10^2), you compute (8 / 2) * 10^(6-2), resulting in 4 * 10^4. Sometimes, the result of the division or multiplication will not be in the form 1 ≤ A < 10. In such cases, a final adjustment step is required to move the decimal point and compensate by updating the power of 10. Mastering this adjustment is vital for maintaining the strict standard form format.
What to do in an assessment
- To compute (5 * 10^4) * (3 * 10^3), multiply 5 and 3 to get 15. Then add exponents to get 10^7, resulting in 15 * 10^7. This is not in standard form because 15 is not less than 10. Adjust it by dividing 15 by 10 to get 1.5 and increasing the exponent by 1. The correct standard form is 1.5 * 10^8. Always look for this final step when multiplying coefficients.
- To compute (4 * 10^5) / (8 * 10^2), divide the coefficients: 4 / 8 = 0.5. Subtract the exponents: 10^(5-2) = 10^3. This gives 0.5 * 10^3. Since 0.5 is less than 1, you must adjust the coefficient by multiplying by 10 to get 5.0, then decreasing the exponent by 1 to get 10^2. The final answer is 5.0 * 10^2. Using this adjustment logic ensures all answers meet the standard requirements.
In plain language
When multiplying in standard form, multiply the coefficients and add the powers; when dividing, divide the coefficients and subtract the powers, then adjust to keep the number between 1 and 10.
Self-check
Calculate (2 * 10^3) * (6 * 10^4).
Show the answer
1.2 * 10^8
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §1.4.1 foundation; Unit 2 §1.4.1 foundation; Unit 3 §1.4.1 foundation. Printed pages 14: how to find equivalent fractions
Open revision guidance
What to know
- Equivalent fractions represent the same portion of a whole despite having different numerators and denominators. They are identified by multiplying or dividing both the numerator and the denominator by the same non-zero integer, which preserves the overall value of the fraction.
- For example, if you take the fraction 2/5, you can find an equivalent fraction by multiplying the numerator and denominator by 3, resulting in 6/15. This works because multiplying by 3/3 is equivalent to multiplying by 1, which does not change the magnitude of the quantity.
What to do in an assessment
- To convert a fraction to an equivalent one, perform the same arithmetic operation on both parts. Consider 4/7; to find an equivalent fraction with a denominator of 21, observe that 7 multiplied by 3 equals 21. Therefore, multiply the numerator 4 by 3 to get 12, yielding 12/21.
- To verify equivalence for 8/12, divide both by their highest common factor, which is 4. Calculating 8 divided by 4 gives 2, and 12 divided by 4 gives 3. Thus, 8/12 is equivalent to 2/3. Ensure all operations are applied consistently to the entire fractional value.
In plain language
Equivalent fractions are different ways of writing the same size piece. You find them by multiplying or dividing the top and bottom numbers by the same number.
Self-check
Find an equivalent fraction for 3/4 with a denominator of 20.
Show the answer
15/20. Working: 20 divided by 4 is 5. Multiply 3 by 5 to get 15.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §1.4.2 foundation; Unit 2 §1.4.2 foundation; Unit 3 §1.4.2 foundation. Printed pages 14: the equivalences between fractions, decimals and percentages
Open revision guidance
What to know
- Fractions, decimals, and percentages are three different ways to express the same rational number. A percentage represents a number out of 100, while a decimal is a base-10 representation. Converting between them requires understanding their common denominator relationship.
- For instance, 0.25 is equivalent to 25/100, which simplifies to 1/4 or 25 percent. Understanding these connections is essential for comparing values in different formats or solving problems that involve mixed numeric types.
What to do in an assessment
- To convert a percentage like 40 percent to a fraction, write it as 40/100. Simplify by dividing by 20 to get 2/5. To convert this to a decimal, divide 2 by 5 using long division or a calculator, which equals 0.4.
- To convert a decimal like 0.75 to a percentage, multiply by 100 to get 75 percent. To convert it to a fraction, write 75/100 and simplify by dividing by 25 to obtain 3/4. These methods allow for consistent translation between numeric forms.
In plain language
Fractions, decimals and percentages are three ways of writing the same number. A percentage means a number out of 100, so dividing a percentage by 100 gives its decimal form.
Self-check
Convert 0.2 to a percentage and a simplified fraction.
Show the answer
20% and 1/5. Working: 0.2 times 100 is 20%. 0.2 equals 20/100, which simplifies to 1/5.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §1.4.3 foundation; Unit 2 §1.4.3 foundation; Unit 3 §1.4.3 foundation. Printed pages 14: convert numbers from one form into another
Open revision guidance
What to know
- Number conversion allows you to move fluently between fractions, decimals, and percentages to suit specific mathematical tasks. This is primarily performed using place value principles and division operations to ensure the underlying quantity remains constant during the conversion process.
- For example, converting a fraction to a decimal is best achieved by dividing the numerator by the denominator. When dealing with fractions that result in repeating decimals, such as 1/3, the value is recognised as 0.333 recurring, which helps in identifying patterns for rounding or further calculations.
What to do in an assessment
- To convert a decimal to a fraction, such as 0.625, write it over 1000 to get 625/1000. Divide both by the greatest common factor, 125, to yield 5/8. This systematic reduction ensures the fraction is in its simplest form.
- To convert a percentage like 150 percent to a decimal, divide by 100 to get 1.5. To convert 1.5 to a fraction, write 150/100, which reduces to 3/2 or 1 and 1/2. Use these division and multiplication steps to accurately translate values.
In plain language
You can convert numbers from one form to another to make them easier to use. Use division to turn fractions into decimals, and multiply by 100 to get percentages.
Self-check
Convert 0.8 to a fraction and percentage.
Show the answer
4/5 and 80%. Working: 0.8 is 80/100, which simplifies to 4/5. 0.8 times 100 is 80%.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §1.4.4 foundation. Printed pages 14: order and compare whole numbers, decimals, fractions and percentages
Open revision guidance
What to know
- Comparing whole numbers, decimals, fractions, and percentages requires converting them into a common format, usually decimals. By ensuring all values are in the same representation, you can easily use place value to determine their order from least to greatest or vice versa.
- For example, when ordering 0.5, 3/4, and 40 percent, convert them all: 0.5 is 0.50, 3/4 is 0.75, and 40 percent is 0.40. Sorting the decimal equivalents 0.40, 0.50, and 0.75 shows the original order is 40 percent, 0.5, 3/4.
What to do in an assessment
- To compare 0.6, 2/3, and 65 percent, convert all to decimals. 0.6 is 0.60. 2 divided by 3 is approximately 0.666. 65 percent is 0.65. Comparing 0.60, 0.66, and 0.65, we find the order is 0.6 < 65% < 2/3.
- Use a number line to visualise these values. By plotting the decimal forms, you can see their relative distances from zero, which verifies your numeric comparison and helps identify outliers or clusters within your data set.
In plain language
To compare numbers, convert them all to decimals. You can then easily see which is the largest or smallest by looking at the numbers.
Self-check
Order 0.7, 3/5, and 75% from least to greatest.
Show the answer
3/5, 0.7, 75%. Working: 3/5=0.6, 0.7=0.7, 75%=0.75. 0.6 < 0.7 < 0.75.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §1.4.5 foundation; Unit 2 §1.4.5 foundation; Unit 3 §1.4.5 foundation. Printed pages 14: simplify fractions
Open revision guidance
What to know
- Simplifying a fraction means reducing it to its lowest terms where the numerator and denominator share no common factors other than 1. This is achieved by dividing both parts of the fraction by their highest common factor (HCF).
- For example, in the fraction 12/18, the factors of 12 are 1, 2, 3, 4, 6, 12 and the factors of 18 are 1, 2, 3, 6, 9, 18. The highest common factor is 6. Dividing both 12 and 18 by 6 results in the simplest fraction, 2/3.
What to do in an assessment
- To simplify a fraction, first list the factors of the numerator and the denominator. For 20/30, factors of 20 are 1, 2, 4, 5, 10, 20. Factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30. The highest common factor is 10.
- Next, divide the numerator 20 by 10 to get 2, and the denominator 30 by 10 to get 3. The simplified fraction is 2/3. This process ensures the representation is as concise as possible while maintaining the same original value.
In plain language
Simplifying fractions means making the numbers as small as possible. You divide the top and bottom by the largest possible number that divides into both.
Self-check
Simplify the fraction 15/25.
Show the answer
3/5. Working: The highest common factor of 15 and 25 is 5. 15 divided by 5 is 3, and 25 divided by 5 is 5.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §1.4.6 foundation; Unit 2 §1.4.6 foundation; Unit 3 §1.4.6 foundation. Printed pages 14: express one number as a fraction or percentage of another
Open revision guidance
What to know
- To express one quantity as a fraction of another, place the target value as the numerator and the total value as the denominator. This ratio can then be simplified, or if required, converted to a percentage by multiplying the resulting decimal by 100.
- Consider a class of 20 students where 5 are wearing hats. As a fraction, this is 5/20. To express this as a percentage, simplify the fraction to 1/4, which is equivalent to 0.25, and then multiply by 100 to get 25 percent.
What to do in an assessment
- To calculate what percentage 8 is of 40, write the fraction 8/40. Simplify the fraction by dividing both by 8, which gives 1/5. To turn this into a percentage, perform the division 1 divided by 5 to get 0.2, then multiply by 100 to get 20 percent.
- Alternatively, express 12 meters as a fraction of 60 meters. Write 12/60, which simplifies to 1/5. Use these ratios to describe relative proportions in datasets accurately, ensuring units match before creating the fraction.
In plain language
To show how much one number represents in terms of another, write it as a fraction. You can then turn that into a percentage by multiplying the decimal by 100.
Self-check
Express 4cm as a fraction and percentage of 20cm.
Show the answer
1/5 and 20%. Working: 4/20 simplifies to 1/5. 4 divided by 20 is 0.2, which is 20%.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §1.4.7 foundation; Unit 2 §1.4.7 foundation; Unit 3 §1.4.7 foundation. Printed pages 14: find a fraction or percentage of a quantity
Open revision guidance
What to know
- Finding a portion of a quantity involves multiplying the whole amount by the fractional or decimal equivalent of the percentage. This is a fundamental operation used in scaling measurements or calculating parts of a total set of items.
- For example, to find 3/4 of 80kg, multiply 80 by 3/4. This is done by dividing 80 by the denominator (4) to get 20, then multiplying by the numerator (3) to arrive at 60kg.
What to do in an assessment
- To find 20 percent of 50m, convert 20 percent to the decimal 0.2. Multiply 50 by 0.2 to get 10m. Alternatively, use the fraction method: 20/100 is 1/5. Calculate 50 divided by 5 to get 10m.
- For another example, find 7/10 of 40 units. Divide 40 by 10 to get 4, then multiply by 7 to get 28 units. These operations allow you to partition amounts systematically in any unit, such as length or mass.
In plain language
To find a portion of something, multiply the whole amount by the fraction or decimal. For example, 50% means dividing by 2.
Self-check
Find 25% of 40kg.
Show the answer
10kg. Working: 25% is 1/4. 40 divided by 4 is 10.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §1.4.8 foundation; Unit 3 §1.4.8 foundation. Printed pages 15: calculate fractional and percentage changes (increase and decrease)
Open revision guidance
What to know
- To calculate a percentage change, you must determine the absolute difference between the original value and the new value. This difference is then divided by the original starting quantity to find the fractional change. Converting this fraction into a percentage involves multiplying the result by 100. This process works for both increases and decreases, where an increase results in a positive percentage and a decrease results in a negative percentage value.
- When calculating a percentage increase, start with the original amount and add the calculated percentage of that original amount. For a decrease, subtract the percentage value from the original. For example, if an original price of 80 dollars increases by 25 percent, calculate 25 percent of 80, which is 20, then add 20 to 80 to get 100. If that same 80 dollars decreases by 10 percent, calculate 10 percent of 80, which is 8, then subtract 8 from 80 to get 72.
What to do in an assessment
- To compute a percentage change, use the formula: (Difference / Original) * 100. Consider a stock price that moves from 200 dollars to 240 dollars. The difference is 240 minus 200, which is 40. Divide 40 by the original 200 to get 0.2. Finally, multiply 0.2 by 100 to yield a 20 percent increase. This step-by-step approach ensures accurate identification of growth or decline relative to the starting point.
- For a percentage decrease, consider a population changing from 500 animals to 400 animals. Subtract 400 from 500 to find a reduction of 100. Divide the reduction of 100 by the original 500 to get 0.2. Multiply 0.2 by 100 to show a 20 percent decrease. Always ensure the original value is the denominator to maintain the correct reference frame for the proportional change calculation.
In plain language
Find how much something changed, then divide that change by the starting amount. Multiply by 100 to get the percentage.
Self-check
An item costing 150 pounds increases to 180 pounds. Calculate the percentage increase.
Show the answer
The difference is 30. Divide 30 by 150 to get 0.2. Multiply by 100 for a 20 percent increase.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §1.4.9 foundation; Unit 3 §1.4.9 foundation. Printed pages 15: understand and use multipliers
Open revision guidance
What to know
- A multiplier is a decimal factor used to represent a percentage change in a single operation. To create a multiplier for an increase, add the percentage as a decimal to 1. For instance, a 15 percent increase corresponds to a multiplier of 1.15 because 100 percent plus 15 percent equals 115 percent, which is 1.15 as a decimal. This simplifies multi-step calculations by allowing you to multiply the base value directly by this factor.
- For a percentage decrease, subtract the percentage from 100 percent and convert to a decimal. A 15 percent decrease corresponds to 85 percent, which equals a multiplier of 0.85. Multiplying any quantity by this value effectively reduces the number by the specified amount in one step. This method is highly efficient for recurring percentage problems where repeated scaling is necessary.
What to do in an assessment
- Apply multipliers by identifying the change percentage first. If an initial value of 400 kilograms increases by 12 percent, calculate the multiplier: 1 + 0.12 = 1.12. Multiply 400 by 1.12 to get 448 kilograms. This approach replaces the separate steps of calculating the percentage and adding it back, reducing errors during complex tasks.
- Use the decrease multiplier for cost reductions. If a 60 pound fee decreases by 30 percent, the multiplier is 1 - 0.30 = 0.70. Multiply 60 by 0.70 to reach a new price of 42 pounds. Ensure the multiplier is always checked against the direction of change, where an increase is greater than 1 and a decrease is less than 1.
In plain language
A multiplier is a number you multiply by to change a total. Add the percent to 100 for growth, or subtract from 100 for reduction, then change to a decimal.
Self-check
Use a multiplier to increase 250 meters by 8 percent.
Show the answer
The multiplier is 1.08. 250 multiplied by 1.08 equals 270 meters.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier · Unit 1 §1.4.10 foundation. Printed pages 15: solve problems with repeated proportional changes
Open revision guidance
What to know
- Repeated proportional changes involve applying a multiplier multiple times over successive periods. When a quantity changes by a fixed percentage for several intervals, such as compound interest or depreciation, you apply the same multiplier repeatedly. The formula for this is Original Amount multiplied by (Multiplier raised to the power of the number of periods). This captures the effect of the percentage change applying to the new total at each step.
- It is important to differentiate between simple linear addition and compound growth. If you are calculating the value of a 1000 pound investment with a 5 percent annual interest rate over 3 years, you do not just multiply 5 percent by 3 years. Instead, you use the multiplier 1.05 and calculate 1000 * (1.05^3). This accounts for the interest earning interest over time, leading to a higher total outcome.
What to do in an assessment
- To solve these problems, define the base, the multiplier, and the exponent. Consider 500 liters of a chemical reducing by 10 percent per day over 4 days. The multiplier is 0.9. The calculation is 500 * (0.9^4). Calculate 0.9^4 = 0.6561. Then multiply by 500 to get 328.05 liters remaining after the 4 day period.
- For investments, use the growth multiplier. If you invest 2000 pounds at 4 percent per year for 2 years, the multiplier is 1.04. The equation is 2000 * (1.04^2). First, square 1.04 to get 1.0816. Then, 2000 * 1.0816 equals 2163.20 pounds. Keep precision in the multiplier until the final calculation to avoid rounding errors.
In plain language
When a value changes by a percentage many times, use a power. Multiply the start value by the multiplier raised to the number of times it changes.
Self-check
A population of 1000 bacteria increases by 10 percent every hour for 3 hours. Calculate the final population.
Show the answer
1000 * (1.1^3) = 1000 * 1.331 = 1331 bacteria.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier · Unit 1 §1.4.11 foundation. Printed pages 15: calculate using ratios in a variety of situations
Open revision guidance
What to know
- Ratios express the relationship between two or more quantities, showing how they relate to the whole. A ratio of 3:2 means that for every 3 parts of the first item, there are 2 parts of the second. To work with these, first calculate the total number of parts by adding the components of the ratio together. This total represents the whole, allowing you to find the value of a single part by dividing the total quantity by the total number of parts.
- Once you know the value of one part, you can multiply this by the individual ratio components to find the specific amounts for each category. For example, if you have 50 kilograms to split in a 3:2 ratio, the total parts equal 5. Dividing 50 by 5 gives 10 kilograms per part. Multiplying 10 by 3 gives 30 kilograms, and multiplying 10 by 2 gives 20 kilograms, satisfying the ratio.
What to do in an assessment
- To share an amount in a given ratio, list the parts, add them to get the total, then find the multiplier. If 120 pounds is shared in a 5:3 ratio, the parts total 8. Divide 120 by 8 to get 15. The first share is 5 * 15 = 75 pounds, and the second share is 3 * 15 = 45 pounds. Check the sum: 75 + 45 = 120.
- Use ratios to simplify and scale. If a recipe uses a 4:1 ratio of flour to sugar and you have 200 grams of sugar, the sugar is the 1 part. Multiply 200 by 4 to get 800 grams of flour. Always identify which part of the ratio refers to the known quantity before scaling the other components.
In plain language
Add the ratio parts together to find the total. Divide the amount by the total to find the value of one part. Multiply that value by each ratio number.
Self-check
Divide 60 centimeters in a ratio of 2:1.
Show the answer
Total parts = 3. 60 / 3 = 20. Parts are 2 * 20 = 40 and 1 * 20 = 20.
- Higher
Scope: Higher · Higher tier in either Unit 1 tier · Unit 1 §1.4.12 higher. Printed pages 15: solve numerical problems involving direct and inverse proportion
Open revision guidance
What to know
- Direct proportion exists when two quantities increase or decrease together at a constant rate, represented by y = kx, where k is a constant. For example, if 5 pens cost 10 pounds, 1 pen costs 2 pounds. This constant k is the unit cost. Inverse proportion occurs when one value increases as the other decreases, represented by y = k / x, where xy = k. If 2 workers build a wall in 6 hours, 4 workers would take 3 hours, because the total worker-hours remains constant at 12.
- Understanding the nature of the variables is crucial. In direct proportion, the ratio y/x is always constant. In inverse proportion, the product xy is always constant. These relationships allow for algebraic solutions when parts of the data are missing, provided the constant of proportionality k can be determined from known pairs of values.
What to do in an assessment
- For direct proportion, find k first. If y = 15 when x = 3, then 15 = k * 3, so k = 5. Now, you can predict y for any x, such as if x = 10, y = 5 * 10 = 50. Use coordinates like (3, 15) and (10, 50) to visualise the linear growth passing through the origin.
- For inverse proportion, find k by multiplying. If y = 4 when x = 5, then k = 4 * 5 = 20. With this constant, you can find missing values. If x = 10, then 10 * y = 20, so y = 2. Use this to solve real-world timing problems where effort or speed impacts completion duration.
In plain language
Direct proportion means both values grow together by a constant. Inverse proportion means one grows while the other shrinks, but their product stays constant.
Self-check
If y is directly proportional to x, and y = 20 when x = 4, find y when x = 7.
Show the answer
k = 20 / 4 = 5. When x = 7, y = 5 * 7 = 35.
- Higher
Scope: Higher · Higher tier in either Unit 1 tier · Unit 1 §1.4.13 higher. Printed pages 15: find the original quantity given the result of a proportional change
Open revision guidance
What to know
- Finding an original quantity is the reverse of applying a percentage change. When you have the final value after an increase or decrease, you must reverse the multiplier to return to the base (100 percent) value. If an item costs 115 pounds after a 15 percent increase, the final price is 115 percent of the original. To find the original, divide the final value by the multiplier (1.15 in this case).
- This method prevents the common error of simply subtracting the percentage of the final value. For example, if a price is reduced by 20 percent to 80 pounds, the original is not 100 pounds (which would be 20 percent of 100, leaving 80). Rather, 80 pounds represents 80 percent of the original. Dividing 80 by 0.80 returns the original 100 pounds. This mathematical logic applies consistently to all proportional reverse-change problems.
What to do in an assessment
- Identify the multiplier and divide. If a total is 250 after a 25 percent increase, the multiplier is 1.25. Set up the equation: Original * 1.25 = 250. Divide 250 by 1.25 to get 200. Always check if the final value is higher (for increase) or lower (for decrease) than the expected original base.
- For decreases, if a population is 360 after a 10 percent decrease, the multiplier is 0.90. Set up the equation: Original * 0.90 = 360. Divide 360 by 0.90 to get 400. Use these steps to solve for starting values in financial reports, inventory counts, or percentage loss scenarios.
In plain language
To get the start value back, divide the final amount by the multiplier you used to change it. Do not just take the percent off the end result.
Self-check
An item costs 90 pounds after a 10 percent discount. What was the original price?
Show the answer
The multiplier is 0.90. 90 / 0.90 = 100 pounds.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §1.5.1 foundation; Unit 2 §1.5.1 foundation; Unit 3 §1.5.1 foundation. Printed pages 15: understand and use number operations and the relationships between them, including inverse operations and the hierarchy of operations
Open revision guidance
What to know
- Number operations involve addition, subtraction, multiplication, and division. Relationships between these operations are fundamental: subtraction is the inverse of addition, and division is the inverse of multiplication. For example, if 15 + 7 = 22, then 22 - 7 = 15. These relationships allow us to solve for unknowns in equations.
- The hierarchy of operations defines the sequence for evaluating expressions. We follow the order: Brackets, Orders (indices and roots), Division and Multiplication (left to right), and Addition and Subtraction (left to right). Following this order ensures consistent results for mathematical expressions across all calculations.
What to do in an assessment
- Apply the hierarchy to evaluate expressions. For 10 + 4 * (3 + 2), solve brackets first: 10 + 4 * 5. Then multiply: 10 + 20. Finally add: 30. This ensures the correct structure is followed every time.
- Use inverse operations to solve for missing variables. If x / 5 = 4, multiply by 5 to find x: 5 * 4 = 20. Similarly, if x - 12 = 8, add 12 to find x: 8 + 12 = 20. Each operation has an exact inverse path.
In plain language
This lesson teaches you the correct order to solve math problems so you always get the right answer and explains how to undo math operations to find missing numbers.
Self-check
Calculate 20 - 4 * (3 + 1) using the hierarchy of operations.
Show the answer
First add inside brackets: 3 + 1 = 4. Then multiply: 4 * 4 = 16. Finally subtract: 20 - 16 = 4.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §1.5.2 foundation; Unit 2 §1.5.2 foundation; Unit 3 §1.5.2 foundation. Printed pages 15: add, subtract, multiply and divide whole numbers, including large whole numbers
Open revision guidance
What to know
- Performing arithmetic on large whole numbers requires a systematic approach. Addition and subtraction use column methods to manage place value. Multiplication involves partitioning or formal long multiplication to manage digits, while division uses the short or long division method to determine how many times one number fits into another.
- The magnitude of large numbers does not change the fundamental properties of arithmetic. For example, 1,200 * 30 can be solved by calculating 12 * 3 = 36 and appending the three zeros to get 36,000. These principles apply to all whole numbers regardless of scale.
What to do in an assessment
- Calculate 1,450 + 2,675 using column addition. Align digits by place value: 1450 + 2675. 0+5=5, 5+7=12 (carry 1), 4+6+1=11 (carry 1), 1+2+1=4. The sum is 4,125.
- Calculate 5,600 / 20. Divide both by 10 to simplify: 560 / 2 = 280. Using inverse logic, check if 280 * 20 = 5,600. It is correct.
In plain language
This lesson explains how to add, subtract, multiply, and divide large numbers safely by keeping columns aligned and using simple place-value shortcuts.
Self-check
Calculate 3,500 - 1,280.
Show the answer
Using column subtraction: 3500 - 1280 = 2220.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §1.5.3 foundation; Unit 2 §1.5.3 foundation; Unit 3 §1.5.3 foundation. Printed pages 15: add, subtract, multiply and divide decimals, fractions and negative numbers
Open revision guidance
What to know
- Working with decimals requires aligning decimal points for addition and subtraction. For multiplication, multiply as whole numbers then place the decimal point based on total decimal places in the factors. Dividing by decimals involves shifting the point to make the divisor a whole number.
- For negative numbers, apply specific sign rules. When multiplying or dividing, two same signs produce a positive result, and different signs produce a negative result. Fractions require common denominators for addition and subtraction, while multiplication uses direct numerator and denominator product.
What to do in an assessment
- Solve -4.5 * 2.2. First, multiply 45 * 22 = 990. Because one is negative, the result is negative. With two total decimal places, we get -9.90.
- Solve 3/4 - 1/5. Convert to a common denominator of 20: 15/20 - 4/20 = 11/20.
In plain language
This lesson covers rules for decimals, negatives, and fractions, ensuring you handle signs and decimal points correctly during calculations.
Self-check
Calculate -2.5 * -4.
Show the answer
Two negative signs multiplied result in a positive: 2.5 * 4 = 10.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §1.5.4 foundation. Printed pages 15: understand and use operations written as number machines
Open revision guidance
What to know
- Number machines represent functions visually. An input enters, undergoes an operation, and produces an output. Understanding these helps in identifying function composition, where the output of one machine becomes the input of another.
- Inverse number machines are used to backtrack from output to input. By reversing the operation performed at each stage (e.g., changing + to - or * to /), you can solve for unknown values systematically.
What to do in an assessment
- Use a machine where input x enters, is multiplied by 3, then 5 is added to produce output 20. To solve for x, reverse: Start at 20, subtract 5 to get 15. Divide by 3 to get 5. So x = 5.
- Design a machine for the expression 2 * (x - 4). The first stage subtracts 4 from x. The second stage multiplies by 2. If input is 10, then (10-4)*2 = 12.
In plain language
This lesson shows how to use diagrams called number machines to follow the steps of a calculation and reverse them to find missing inputs.
Self-check
A machine multiplies by 4 then subtracts 2. If the output is 14, what is the input?
Show the answer
Reverse: 14 + 2 = 16. Then 16 / 4 = 4. The input is 4.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §1.5.5 foundation; Unit 3 §1.5.5 foundation. Printed pages 16: use a calculator efficiently and effectively, including: order of operations addition, subtraction, multiplication and division square, cube and other powers square root and cube root brackets other appropriate functions
Open revision guidance
What to know
- A calculator follows the order of operations, so brackets determine which part is calculated first. Know the keys for addition, subtraction, multiplication, division, powers, square roots and cube roots. Negative signs and subtraction can use different keys.
- Keep the full displayed value or use memory/Ans during intermediate steps, then round only the final result as requested. Estimate the answer first so an incorrect key sequence is easier to spot.
What to do in an assessment
- Enter (8+4)÷3 to obtain 4. Without the brackets, 8+4÷3 gives 9.333… because division happens first. Check the screen shows the intended expression.
- Calculate √81 + 2³ = 9 + 8 = 17. For ∛64, use the cube-root key or 64^(1/3) to obtain 4. To find 5⁴, use the power key and obtain 625. Check where a root or power entry ends before adding the next term.
In plain language
Use brackets and the correct power/root keys, check the entered expression, and postpone rounding.
Self-check
Calculate (7²−√25)÷4.
Show the answer
(49−5)÷4 = 44÷4 = 11.
- Higher
Scope: Higher · Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §1.5.5 higher; Unit 3 §1.5.5 higher. Printed pages 16: standard form
Open revision guidance
What to know
- Standard form is a × 10ⁿ with 1 ≤ |a| < 10 for a non-zero number. Calculators commonly use an EXP or ×10ˣ key and may display an exponent with an E, for example 3.2E−4 means 3.2 × 10⁻⁴, not 3.2 − 4.
- When using the exponent key, enter the coefficient then the exponent without entering another 10. Use the negative-number key for a negative exponent. Read the whole displayed answer and convert it into the requested form.
What to do in an assessment
- Calculate (3 × 10⁵)(2 × 10⁻³). Enter both standard-form numbers and multiply. The result is 600 = 6 × 10². Check independently: coefficients multiply to 6 and exponents add to 2.
- For (8 × 10⁷)/(2 × 10³), the result is 4 × 10⁴ = 40000. For 3.2E−4, move the decimal point four places left to obtain 0.00032.
In plain language
The exponent-entry key represents a power of ten. Check the displayed exponent, especially when it is negative.
Self-check
A calculator displays 5.6E−3. Write this as an ordinary decimal.
Show the answer
5.6 × 10⁻³ = 0.0056.
- Higher
Scope: Higher · Higher tier in either Unit 3 tier · Unit 3 §1.5.6 higher. Printed pages 16: use the trigonometric functions on a calculator efficiently and effectively
Open revision guidance
What to know
- Trigonometric functions (sine, cosine, tangent) are vital for calculating lengths and angles in right-angled triangles. Ensure the calculator is set to 'DEG' mode for degrees. The functions return ratios based on the angle inputs, mapping angles to specific coordinate properties on a circle.
- Inverse trigonometric functions (sin^-1, cos^-1, tan^-1) allow calculation of angles from side ratios. For example, if tan(A) = 1, then A = tan^-1(1) = 45 degrees. Understanding these requires identifying sides as Opposite, Adjacent, and Hypotenuse relative to the angle.
What to do in an assessment
- Calculate sin(30). Ensure calculator is in DEG mode. Input sin, then 30, then =. Result is 0.5.
- Find angle A if cos(A) = 0.5. Press SHIFT or 2nd function, then cos, input 0.5, then =. Result is 60 degrees.
In plain language
This lesson teaches how to use your calculator for sine, cosine, and tangent functions to solve triangle problems.
Self-check
Find angle B if tan(B) = 1.0.
Show the answer
Use inverse tangent: tan^-1(1.0) = 45 degrees.
View diagrams
Right-angled triangle with labelled sides
Key and description
For angle θ, the opposite side is 3, the adjacent side is 4 and the hypotenuse is 5.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §1.6.1 foundation; Unit 3 §1.6.1 foundation. Printed pages 16: round an answer to a reasonable degree of accuracy in light of the context
Open revision guidance
What to know
- Rounding is the process of adjusting a number to a simpler or more convenient value, usually by selecting a fixed number of decimal places or significant figures. Choosing the degree of accuracy depends on the context of the problem, such as financial calculations requiring two decimal places for currency or physical measurements needing to match the precision of the tools used. When a value is rounded, digits to the right of the target position determine whether to round up or stay the same.
- For example, if you are given the value 47.638 kilograms, rounding to one decimal place requires looking at the second decimal digit, which is 3. Since 3 is less than 5, you keep the first decimal digit as it is, resulting in 47.6 kilograms. Conversely, if the value were 47.658, the digit 5 would prompt an increase to 47.7 kilograms. This practice ensures that calculations remain manageable without losing essential information necessary for the problem context.
What to do in an assessment
- To round 124.567 to two decimal places, identify the second decimal place, which is 6. Check the digit immediately to its right, which is 7. Because 7 is 5 or greater, round the 6 up to 7, resulting in 124.57. Always perform the rounding step at the very end of a calculation to ensure the highest level of precision is maintained throughout your working.
- Consider a context where you calculate the area of a circle with a radius of 5 cm. Using pi as 3.14159, the calculation gives 78.5398... cm squared. Rounding this to a reasonable degree of accuracy, such as one decimal place, yields 78.5 cm squared. This result acknowledges that the precision of the final answer should reflect the input data. Use the calculator memory functions to store intermediate results before applying final rounding.
In plain language
Rounding makes numbers easier to handle by simplifying them to a specific point. Always check the next digit to decide whether to round up or stay the same, and remember that context dictates how precise you need to be.
Self-check
Round 0.8924 to two decimal places.
Show the answer
0.89. The third decimal digit is 2, which is less than 5, so the second digit remains 9.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier · Unit 1 §1.6.2 foundation. Printed pages 16: recognise that measurement is approximate and that a measurement expressed to a given unit is in possible error of half a unit
Open revision guidance
What to know
- Measurements in mathematics and science are never perfectly exact because they depend on the limitations of the instruments used. Every measurement has an inherent degree of uncertainty. When a value is stated to a specific unit, such as 15 cm measured to the nearest centimetre, it implies that the true value lies somewhere between the values that would round to 15. The error is defined as half of the smallest unit of measurement used.
- If a length is measured as 15 cm, the smallest unit is 1 cm. The error is half of this, which is 0.5 cm. This means the actual measurement could be as low as 14.5 cm or as high as 15.5 cm. This concept is fundamental when assessing the reliability of data in geometry or physics experiments, as it dictates the possible range of error for any reported value.
What to do in an assessment
- To find the range of a measurement, identify the precision of the unit. For a mass measured as 20 kg to the nearest kg, the unit is 1 kg. Divide this by 2 to find the error, which is 0.5 kg. The lower boundary is 20 minus 0.5, equal to 19.5 kg, and the upper boundary is 20 plus 0.5, equal to 20.5 kg. Always state the possible range as 19.5 kg <= mass < 20.5 kg.
- If a time is recorded as 120 seconds to the nearest 10 seconds, the smallest unit is 10 seconds. The error is 10 divided by 2, which is 5 seconds. The range is 120 minus 5 (115 seconds) to 120 plus 5 (125 seconds). When applying this to problem solving, use these bounds to determine the minimum and maximum possible outcomes for area, perimeter, or total distance.
In plain language
Every measurement is approximate. If you measure something to the nearest unit, the true value could be half a unit above or below that reading. This half-unit value creates the range for your measurement.
Self-check
A distance is measured as 8 metres to the nearest metre. What is the potential range of the true distance?
Show the answer
The error is 0.5 metres. The range is 7.5 metres <= distance < 8.5 metres.
- Higher
Scope: Higher · Higher tier in either Unit 1 tier · Unit 1 §1.6.3 higher. Printed pages 16: determine the upper and lower bounds of numbers expressed to a given degree of accuracy
Open revision guidance
What to know
- Upper and lower bounds define the interval in which a true value must reside given a measurement's stated precision. When a value is recorded as x to a specific degree of accuracy, the lower bound is the minimum value that would round up to x, and the upper bound is the value just below the point where it would round up to the next interval. These bounds are critical for understanding the potential variance in any calculated result derived from measured data.
- For example, if a measurement of 10.4 cm is rounded to one decimal place, the range of possible values is 10.35 <= x < 10.45. The value 10.35 is the lower bound, as any value smaller would round to 10.3. The value 10.45 is the upper bound, as 10.45 and anything above it would round to 10.5. These boundaries define the range of certainty for the measurement.
What to do in an assessment
- To determine bounds for a number like 420 rounded to the nearest 10, first find the error: 10 divided by 2 is 5. The lower bound is 420 minus 5, which is 415. The upper bound is 420 plus 5, which is 425. Represent this mathematically as 415 <= value < 425. This ensures that any input falling within this range is consistent with the reported value.
- Consider the coordinate point (5, 8) where both values are given to the nearest integer. The bounds for the x-coordinate are 4.5 <= x < 5.5, and for the y-coordinate are 7.5 <= y < 8.5. If calculating a distance or slope between points, always use these bounded intervals to establish the worst-case and best-case scenarios. This method avoids premature rounding and keeps calculations accurate by acknowledging the hidden range of the input.
In plain language
Bounds are the lowest and highest values a measurement could actually be based on its rounding. They help you understand the full range of possibilities for your data.
Self-check
What are the bounds for 75 rounded to the nearest 5?
Show the answer
The error is 5/2 = 2.5. The lower bound is 72.5 and the upper bound is 77.5. Thus, 72.5 <= value < 77.5.
- Higher
Scope: Higher · Higher tier in either Unit 1 tier · Unit 1 §1.6.4 higher. Printed pages 17: calculate the upper and lower bounds in the addition, subtraction, multiplication and division of numbers expressed to a given degree of accuracy
Open revision guidance
What to know
- Rounded measurements represent intervals. A length of 10 cm rounded to the nearest centimetre means 9.5 ≤ L < 10.5 cm. Its lower bound is 9.5 and upper bound is 10.5, but the upper endpoint is not included. An upper or lower bound need not be a value the calculation actually attains.
- For positive measurements, a product has lower bound lower × lower and upper bound upper × upper. A quotient has lower bound numerator-lower ÷ denominator-upper and upper bound numerator-upper ÷ denominator-lower. For a difference a − b, use a-lower − b-upper for its lower bound and a-upper − b-lower for its upper bound. These rules for products and quotients assume positive inputs; check signs before using them.
What to do in an assessment
- A rectangle has measured length 10 cm and width 5 cm, each to the nearest centimetre. Its area has lower bound 9.5 × 4.5 = 42.75 cm² and upper bound 10.5 × 5.5 = 57.75 cm². The lower endpoints are possible but the upper endpoints are not, so 42.75 ≤ area < 57.75 cm².
- A distance of 100 m and time of 10 s are each rounded to the nearest whole unit. The speed's lower bound is 99.5/10.5 ≈ 9.476 m/s and upper bound is 100.5/9.5 ≈ 10.579 m/s. Keep the exact fractions when reporting the bounds: both speed endpoints are excluded because each requires one excluded upper input.
In plain language
Use the extreme possible inputs, choosing the correct extremes for the operation. Distinguish a bound from an endpoint that can actually occur.
Self-check
a = 10 and b = 2 are rounded to the nearest whole number. Find the bounds of a − b and decide whether they are included.
Show the answer
9.5 ≤ a < 10.5 and 1.5 ≤ b < 2.5. Lower bound = 9.5 − 2.5 = 7; upper bound = 10.5 − 1.5 = 9. Neither endpoint can occur, so 7 < a − b < 9.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §1.6.5 foundation; Unit 3 §1.6.5 foundation. Printed pages 17: premature rounding in problems involving multiple steps may affect the accuracy of the final answer
Open revision guidance
What to know
- Premature rounding occurs when you round intermediate values during a multi-step calculation. This practice introduces small errors at each step that aggregate, leading to a final result that deviates significantly from the true mathematical value. For example, if a calculation involves multiplying a value by one-third (0.333...) and you round to 0.3 before proceeding, the final answer will be inaccurate. It is essential to keep all digits in your calculator memory until the final step.
- When solving complex problems involving multiple operations, the compounding effect of rounding errors can turn a correct method into a flawed result. By retaining exact fractions or keeping as many decimal places as your calculator allows throughout your intermediate steps, you maintain the integrity of the math. Final rounding should only be performed once, at the very end, to provide an answer that is as close to the theoretical value as possible.
What to do in an assessment
- Suppose you must calculate (100 / 3) * 7. Correct method: Keep 100/3 as a fraction or use 33.3333333 on your calculator, then multiply by 7 to get 233.333... Rounding at the end to two decimal places yields 233.33. Incorrect method: 100/3 = 33.3. Then 33.3 * 7 = 233.1. The difference between 233.33 and 233.1 shows the inaccuracy introduced by rounding mid-step.
- Always check your display settings to ensure the calculator is not prematurely showing rounded values. Use the ANS or memory keys on your device to store the result of each sub-calculation. If a problem requires multiple steps, such as calculating the surface area of a complex shape by finding individual parts, sum the exact expressions or unrounded decimals, and apply the final requested rounding only to that total. This ensures precision is preserved consistently.
In plain language
Never round numbers until the very end of your math problem. If you round numbers in the middle of your steps, the small errors will add up and make your final answer wrong.
Self-check
If you need to calculate 200 divided by 7 then multiplied by 3, which approach is more accurate: rounding 200/7 to 28.6 first, or calculating (200/7)*3 entirely without intermediate rounding?
Show the answer
Calculating (200/7)*3 entirely is more accurate. 28.6*3=85.8, whereas (200/7)*3 = 85.714... (rounded to 85.71).
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier · Unit 1 §1.7.1 foundation. Printed pages 17: interpret and use mathematical information presented in written or visual form, including infographics, schedules, tables, timetables, calendars and charts
Open revision guidance
What to know
- Mathematical information is frequently presented via visual formats such as tables, calendars, or charts, which require specific techniques for extraction. To interpret a table, one must first identify the axes or headers, which define the variables provided. For example, in a bus timetable, columns often represent stops while rows represent journey times. Always scan for the unit of measure, such as minutes or meters, before performing any calculations.
- Infographics use scaled imagery to communicate quantitative data. To analyse these, identify the key and scale factor. If one icon represents 500 units, the total is the count of icons multiplied by the scale. In line charts, read the horizontal x-axis for the independent variable and the vertical y-axis for the dependent variable. Ensure that coordinates are tracked precisely across grid lines to maintain accuracy in data reporting.
What to do in an assessment
- Consider a train timetable where the 08:00 departure arrives at 09:15, and the 08:30 departure arrives at 09:45. Calculate the duration for both: 9:15 minus 8:00 equals 75 minutes, and 9:45 minus 8:30 equals 75 minutes. By isolating these values, we determine that the journey duration is consistent regardless of the departure time, providing a practical application for reading scheduled transport data.
- For a bar chart showing fruit sales where apples = 120, bananas = 80, and cherries = 40, sum the categories: 120 plus 80 plus 40 equals 240. Represent this in a table to practise organising visual data. Ensure every numerical value corresponds to the labelled axis. Validate the sum by cross-referencing the total count provided in the chart description to confirm accuracy in your data interpretation process.
In plain language
This lesson teaches how to extract numbers and meanings from tables, charts, and schedules by carefully reading titles, labels, and units of measurement.
Self-check
If a table shows a bus departing at 10:15 and arriving at 11:05, how many minutes is the journey?
Show the answer
50 minutes. Working: 11:05 minus 10:15 is 11:05 minus 10:05 minus 10 minutes, which results in 50 minutes total.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier · Unit 1 §1.7.2 foundation. Printed pages 17: create plans and schedules
Open revision guidance
What to know
- Creating a plan or schedule requires organising tasks chronologically or logically within a set timeframe. To build an effective schedule, start by listing all necessary components or events. Next, assign a specific duration to each task and order them sequentially. Use a timeline or grid format to ensure that no activities overlap unless they can be performed simultaneously. Identifying constraints, such as start and end times, is crucial for logical arrangement.
- When planning a project with specific steps, calculate total duration by summing the individual task times. For example, if Task A takes 20 minutes, Task B takes 45 minutes, and Task C takes 15 minutes, the total time is 80 minutes. Factor in transition or 'slack' time between tasks. Proper documentation in a clear table helps in communicating the schedule effectively to others, ensuring that milestones are met within the allotted constraints.
What to do in an assessment
- Draft a study schedule. Task 1: Maths (30 minutes), Task 2: Break (10 minutes), Task 3: English (40 minutes). Total time = 30 + 10 + 40 = 80 minutes. If the start time is 09:00, the schedule is: Maths 09:00 to 09:30, Break 09:30 to 09:40, English 09:40 to 10:20. Use a tabular format to clearly list these start and end times for each activity.
- Design a weekly calendar event plan. If a task requires 3 hours on Monday and 2.5 hours on Tuesday, the total commitment is 5.5 hours. Represent this in a weekly grid where each cell represents a 30-minute slot. Use specific coordinates or blocks to shade the occupied time. This visual method helps in identifying free capacity and avoiding scheduling conflicts within a 168-hour weekly cycle.
In plain language
This lesson explains how to organise tasks into logical schedules by calculating individual times, adding them together, and mapping them onto a timeline or calendar.
Self-check
You have a task starting at 14:00 that takes 90 minutes. What time does it end?
Show the answer
15:30. Working: 90 minutes is 1 hour and 30 minutes. 14:00 plus 1 hour 30 minutes equals 15:30.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier · Unit 1 §1.7.3 foundation; Unit 2 §1.7.3 foundation. Printed pages 17: use, interpret and produce Venn diagrams
Open revision guidance
What to know
- A Venn diagram is a tool for visualising relationships between sets of objects using overlapping circles within a universal set. Each circle represents a category, and the overlapping region represents the intersection, containing elements that belong to both categories. Elements outside the circles but within the bounding box represent items that belong to neither category. Mastering these diagrams involves correctly placing elements based on their shared attributes.
- To construct a Venn diagram, define your universal set (e.g., numbers 1 to 10) and your two criteria (e.g., Even numbers and Multiples of 3). The even numbers are {2, 4, 6, 8, 10} and multiples of 3 are {3, 6, 9}. The intersection is {6}. Place the numbers accordingly: {2, 4, 8, 10} in the non-overlapping part of 'Even', {3, 9} in the non-overlapping part of 'Multiples of 3', and {6} in the middle overlap. Numbers {1, 5, 7} are outside.
What to do in an assessment
- Practice identifying intersections. Given Set A = {1, 2, 3, 4} and Set B = {3, 4, 5, 6}, the intersection is {3, 4}. Place these two elements in the centre overlapping zone. Place 1 and 2 exclusively in the circle for Set A. Place 5 and 6 exclusively in the circle for Set B. This spatial mapping confirms that 3 and 4 satisfy both conditions simultaneously.
- Calculate probabilities using Venn diagrams. If 10 people like Tea and 15 like Coffee, with 5 liking both, draw the circles. The 'Tea only' region has 10 minus 5 = 5 people. The 'Coffee only' region has 15 minus 5 = 10 people. The total number of people is 5 (Tea only) + 5 (Both) + 10 (Coffee only) = 20 unique individuals. Verify the counts to ensure they match the dataset correctly.
In plain language
This lesson explains how to use Venn diagrams to sort items into groups, identify shared traits, and organise overlapping categories using circles and a universal bounding box.
Self-check
In a set of 10 numbers {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, how many numbers are in the intersection of 'Odd numbers' and 'Prime numbers'?
Show the answer
3. The odd numbers are {1, 3, 5, 7, 9} and prime numbers are {2, 3, 5, 7}. The shared numbers are {3, 5, 7}.
View diagrams
Venn diagram for two events
Key and description
Constructed group of 30 pupils: n(A) = 18, n(B) = 14, n(A ∩ B) = 8. A only = 18 − 8 = 10; B only = 14 − 8 = 6; neither = 30 − (10 + 8 + 6) = 6. P(A ∪ B) = 24/30 = 4/5; P(A ∩ B) = 8/30 = 4/15.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §1.8.1 foundation; Unit 2 §1.8.1 foundation; Unit 3 §1.8.1 foundation. Printed pages 18: carry out calculations involving knowledge of money; pounds (£) and pence
Open revision guidance
What to know
- Money calculations require careful attention to decimal notation where the decimal point separates pounds (£) from pence. Since there are 100 pence in one pound, any amount in pence must be expressed as a two-digit decimal after the point (e.g., 5p is £0.05, not £0.5). When performing arithmetic, align the decimal points vertically to ensure accurate addition and subtraction of monetary values.
- Multiplication and division involving currency follow standard decimal rules, but results must always be rounded to the nearest penny unless otherwise specified. For example, if a calculation results in £12.567, it should be rounded to £12.57 for payment purposes. Contextual awareness is vital, as money problems often involve calculating remainders or total costs from unit prices, requiring consistent unit conversion before performing operations.
What to do in an assessment
- To calculate the total cost of multiple items, multiply the quantity by the unit price. Suppose you buy 4 notebooks at £1.75 each and 3 pens at £0.85 each. Calculate: (4 * 1.75) + (3 * 0.85) = 7.00 + 2.55 = £9.55. Ensure all decimals are tracked: 4 * 1.75 is 7.00, and 3 * 0.85 is 2.55. Adding these sums together yields the final total of £9.55 exactly.
- To divide a total cost among a group, divide the total by the number of people. If a bill is £45.30 and is shared equally among 6 people, calculate 45.30 / 6 = 7.55. Each person pays £7.55. Check by multiplying the result: 7.55 * 6 = 45.30. Always verify the alignment of the decimal point during long division to avoid place value errors that would shift the pence into the pounds column.
In plain language
Money is written in pounds and pence with a decimal point. There are 100 pence in a pound. When working with money, use decimals, remember to round to two decimal places, and always check that your decimal points are lined up when adding or subtracting.
Self-check
A customer buys 5 items costing £2.40 each and 2 items costing £1.15 each. What is the total cost?
Show the answer
Total = (5 * 2.40) + (2 * 1.15) = 12.00 + 2.30 = £14.30.
- Core
Open revision guidance
What to know
- Personal finance management requires understanding the relationship between income and expenditure. Gross income is the total salary earned before deductions like Income Tax and National Insurance. Net pay is the amount received after these statutory deductions. Budgeting involves tracking regular costs such as utility bills, mobile phone contracts, and loan repayments to ensure total spending does not exceed total income. Understanding profit and loss is also key for enterprise: profit is the difference between revenue (money coming in) and costs (money spent).
- Calculations in finance often involve percentage changes, such as price increases or sales discounts. For example, to find a 15% discount on a £50 item, calculate 0.15 * 50 = £7.50, then subtract: 50 - 7.50 = £42.50. VAT (Value Added Tax) is typically calculated by adding a percentage to the net price. If VAT is 20%, multiply the price by 1.20. Exchange rates allow for travel calculations: if the exchange rate is £1 to 1.20 Euros, £100 converts to 100 * 1.20 = 120 Euros.
What to do in an assessment
- To calculate monthly net pay, assume a gross monthly salary of £2,000. If Income Tax is 20% and National Insurance is 10%, calculate deductions: Tax = 0.20 * 2000 = £400. NI = 0.10 * 2000 = £200. Total deductions = £600. Net pay = 2000 - 600 = £1,400. This logic applies to all financial deductions where percentages are applied to a base value.
- For VAT and price comparisons, consider a product costing £120 before 20% VAT. To find the total price, calculate 120 * 1.20 = £144. To compare two items, use 'best buy' analysis: if Item A is 500g for £2.00 and Item B is 800g for £3.00, find price per 100g. Item A: 2.00 / 5 = £0.40 per 100g. Item B: 3.00 / 8 = £0.375 per 100g. Item B is the better buy.
In plain language
Personal finance covers managing money including salaries, taxes, and daily costs. It uses percentages to calculate things like tax, discounts, VAT, and currency exchange. Budgeting is about balancing your money, and unit pricing helps find the best value when shopping.
Self-check
A shirt costs £40 before a 15% discount. What is the final price?
Show the answer
Discount = 0.15 * 40 = £6. Final price = 40 - 6 = £34.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier · Unit 1 §1.8.3 foundation. Printed pages 19: recognise the difference between simple and compound interest and be able to perform calculations with both, using efficient calculation methods and including the use of multipliers
Open revision guidance
What to know
- Simple interest is calculated only on the original principal amount. The interest earned each period is constant. The formula is Interest = Principal * Rate * Time. For example, if you invest £1,000 at 5% simple interest per year for 3 years, the interest is 1000 * 0.05 * 3 = £150. Total value = £1,150. Simple interest does not compound, meaning your interest does not earn further interest.
- Compound interest is calculated on the principal plus any interest accumulated from previous periods. This is represented by the formula Total = Principal * (1 + Rate)^Time. For a £1,000 investment at 5% compound interest for 3 years, calculate 1000 * (1.05)^3 = 1000 * 1.157625 = £1,157.63. Compound interest results in higher returns over time because the base amount grows every period.
What to do in an assessment
- To calculate simple interest on £2,000 at 4% for 5 years: Step 1, identify values: P=2000, R=0.04, T=5. Step 2, multiply: 2000 * 0.04 * 5 = 400. The interest is £400. Total = 2000 + 400 = £2,400.
- To calculate compound interest on £2,000 at 4% for 5 years: Step 1, use the multiplier (1 + 0.04) = 1.04. Step 2, apply exponent: 1.04^5 = 1.2166529. Step 3, multiply by principal: 2000 * 1.2166529 = 2433.3058. Rounded, this is £2,433.31. Note the difference: compound interest provides £33.31 more than simple interest.
In plain language
Simple interest is based on the original amount only. Compound interest is based on the original amount plus any interest already earned. Compound interest grows faster because you earn interest on your interest.
Self-check
Calculate compound interest for £500 at 3% for 2 years.
Show the answer
500 * (1.03)^2 = 500 * 1.0609 = £530.45.
- Higher
Scope: Higher · Higher tier in either Unit 1 tier · Unit 1 §1.8.4 higher. Printed pages 19: perform calculations involving multiple rates
Open revision guidance
What to know
- Complex financial scenarios often require applying multiple rates sequentially or across different segments. Examples include graduated tax brackets or tiered pricing structures. For instance, if an tax system applies 0% to the first £10,000, 20% to the next £30,000, and 40% above that, you must split the income into chunks and apply the specific rate to each chunk separately before summing the totals.
- When handling multiple rates, precision is essential. Identify the threshold for each rate. Calculate the taxable amount in each bracket, apply the specific percentage, and ensure the summation is complete. This method ensures that the progression of rates is correctly mapped against the actual value being assessed, avoiding errors associated with applying a single average rate across all segments.
What to do in an assessment
- Scenario: Earn £50,000. Tax rates: £0-10k (0%), £10k-40k (20%), Over £40k (40%). Split the income: First £10,000 * 0 = £0. Next £30,000 (from 10k to 40k) * 0.20 = £6,000. Remaining £10,000 (from 40k to 50k) * 0.40 = £4,000. Total tax = 0 + 6000 + 4000 = £10,000.
- Another example: Tiered utility usage. First 100 units at £0.20, next 200 units at £0.15. Usage = 250 units. Split: (100 * 0.20) + (150 * 0.15) = 20 + 22.50 = £42.50. Ensure the logic holds for each segment independently before adding.
In plain language
Some calculations use different rates for different amounts. For tax or utility bills, you break the total amount into chunks, calculate the cost for each chunk using its specific rate, and add those costs together.
Self-check
Calculate tax on £25,000 income. Rates: 0% on first £10,000, 20% on the remainder.
Show the answer
(10,000 * 0) + (15,000 * 0.20) = 0 + 3,000 = £3,000.
- Higher
Scope: Higher · Higher tier in either Unit 1 tier · Unit 1 §1.8.5 higher. Printed pages 19: calculate, use and apply Annual Equivalent Rate (AER) when comparing financial products
Open revision guidance
What to know
- AER is a standard measure used to compare the interest rates of different savings accounts or loans. Because interest can be compounded over different periods (e.g., monthly, quarterly, or annually), the nominal rate often misleads. AER standardises this by calculating the actual yearly return on an investment or cost of a loan, accounting for the effect of compounding throughout a single year.
- To use AER for comparing products, ensure all products are converted to an annual equivalent. For an account paying r% interest compounded n times per year, AER is calculated using the formula: AER = (1 + (r/n))^n - 1. This value represents the effective percentage gain over one full year, allowing direct comparison between two different financial products regardless of how often their interest is applied.
What to do in an assessment
- Compare Account A (5% annual interest) with Account B (4.9% compounded monthly). Step 1: For A, AER = 5%. Step 2: For B, r=0.049, n=12. AER = (1 + (0.049/12))^12 - 1. Step 3: (1 + 0.0040833)^12 - 1 = 1.05011 - 1 = 0.05011 or 5.011%. Conclusion: Account B is better.
- Always use the decimal form of the percentage rate in the formula. Convert n based on frequency: monthly=12, quarterly=4, daily=365. After calculating, convert back to a percentage by multiplying by 100.
In plain language
AER helps compare financial products by showing what the interest rate would be if it were paid once a year. It accounts for compounding, so you can see which account truly pays more interest even if they have different compounding frequencies.
Self-check
What is the AER of an account with 6% annual rate compounded half-yearly (n=2)?
Show the answer
(1 + (0.06/2))^2 - 1 = (1.03)^2 - 1 = 1.0609 - 1 = 0.0609 or 6.09%.
- Higher
Scope: Higher · Higher tier in either Unit 1 tier · Unit 1 §1.8.6 higher. Printed pages 19: calculate, use and apply Annual Percentage Rate (APR) when comparing financial products, including mortgages
Open revision guidance
What to know
- APR is a standardised annual percentage measure of the cost of credit, incorporating interest and relevant compulsory charges. Use a stated APR when comparing borrowing offers, rather than comparing only their headline interest rates. Repayment timing, term and charges matter.
- Adding a one-off fee percentage to an annual interest rate is not a general method for calculating a regulated APR, especially on a repaying loan or mortgage. A simple first-year cost calculation can illustrate why fees matter, but it must be labelled with its assumptions and not presented as the actual APR.
What to do in an assessment
- Two otherwise comparable loans advertise APRs of 8.9% and 10.2%. The first has the lower standardised annual cost of credit. Check that the amounts, terms and repayment arrangements are comparable, and read any conditions rather than assuming a lower introductory rate is a lower total cost.
- In a simplified one-year model, £10000 remains outstanding for the whole year, interest is 4% of that amount, and an additional £200 fee is payable. Interest = £400 and total stated first-year charges = £600. Charges/original principal = 6%, but this is not automatically the regulated APR. If a second offer has 3% interest and no fee under the same model, its first-year charges are £300, so it costs £300 less in that particular model.
In plain language
APR helps compare the whole cost of borrowing. Do not confuse a simplified fee calculation with a genuine APR quotation.
Self-check
In a simplified one-year loan model, £5000 remains outstanding throughout, interest is 5% of that amount and an extra £100 fee is charged. What are the first-year charges? Does that alone determine the regulated APR?
Show the answer
Interest = £5000×0.05 = £250; charges = £250+£100 = £350. No: a regulated APR calculation also depends on the timing and repayment arrangements, so do not label £350/£5000 = 7% as the actual APR without those assumptions.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §1.9.1 foundation. Printed pages 19: that recurring decimals are exact fractions, and that some exact fractions are recurring decimals
Open revision guidance
What to know
- A decimal is recurring if a pattern of digits repeats infinitely, such as 0.333... which is the fraction 1/3. Every rational number, which can be expressed as a ratio of two integers p/q, will manifest as either a terminating decimal or a recurring decimal in base 10.
- Conversely, any recurring decimal can be expressed as a fraction. If we take 0.444..., we can let x = 0.444.... Then 10x = 4.444.... Subtracting the original equation (x = 0.444...) from 10x gives 9x = 4.0, leading to x = 4/9. This proves that these decimals are indeed exact fractional values.
What to do in an assessment
- To convert a recurring decimal like 0.7272... to a fraction, set x = 0.7272.... Multiply by 100 because two digits repeat, giving 100x = 72.7272.... Subtracting x leaves 99x = 72. Divide by 99 to get x = 72/99. Simplify by dividing the numerator and denominator by 9, resulting in 8/11.
- Check the result by performing long division: 8 divided by 11. 11 into 80 goes 7 times with remainder 3; 11 into 30 goes 2 times with remainder 8. The cycle 72 repeats. This confirms 8/11 equals 0.7272.... Always ensure the multiplier matches the length of the recurring cycle to isolate the repeating part perfectly.
In plain language
Recurring decimals are simply another way to write fractions where the division never ends. You can turn them back into fractions using algebra by multiplying by 10 or 100 to shift the decimals and then subtracting the original value.
Self-check
Convert 0.1818... into its simplest fractional form.
Show the answer
Let x = 0.1818.... Then 100x = 18.1818.... 100x minus x equals 99x = 18. x = 18/99. Dividing both by 9 gives 2/11.
- Higher
Scope: Higher · Higher tier in either Unit 2 tier · Unit 2 §1.9.2 higher. Printed pages 20: convert recurring decimals to fractional form
Open revision guidance
What to know
- Converting recurring decimals to fractions involves identifying the repeating part and using algebraic manipulation to eliminate the decimal portion. This process relies on the fact that multiplying by a power of 10 shifts the decimal point, allowing for cancellation when subtracting the original number.
- For example, in 0.515151..., the cycle is 51. Let x = 0.515151.... Multiplying by 100 yields 100x = 51.515151.... By subtracting the original x from 100x, we isolate 99x = 51. This transformation proves that all pure recurring decimals are rational numbers representable as exact fractions.
- When the recurring part does not start immediately, such as 0.1666..., we must use a different multiplier. Let x = 0.1666.... Then 10x = 1.666... and 100x = 16.666.... Subtracting 10x from 100x gives 90x = 15, so x = 15/90, which simplifies to 1/6.
What to do in an assessment
- Convert 0.2333... to a fraction. Let x = 0.2333.... Multiply by 10 to shift non-repeating part: 10x = 2.333.... Multiply by 100 to shift one repeating digit: 100x = 23.333.... Now, subtract: 100x - 10x = (23.333...) - (2.333...). This gives 90x = 21. Therefore, x = 21/90. Simplifying by 3 gives 7/30.
- Check the work by verifying the division 7 divided by 30. 30 into 70 is 2 with remainder 10; 30 into 100 is 3 with remainder 10; the pattern 3 repeats. The fraction 7/30 is exact. Always look for the smallest repeating block to simplify the algebra efficiently.
In plain language
If a decimal repeats a pattern, it can be written as a fraction. By multiplying the decimal by powers of 10, you can line up the repeating parts and subtract them to find the hidden fractional value.
Self-check
Convert 0.41666... to a fraction.
Show the answer
Let x = 0.41666.... 10x = 4.1666.... 100x = 41.666.... Subtract 10x from 100x: 90x = 37.5. Since we need an integer fraction, multiply by 10 to get 900x = 375. x = 375/900 = 5/12.
- Higher
Scope: Higher · Higher tier in either Unit 2 tier · Unit 2 §1.9.3 higher. Printed pages 20: distinguish between rational and irrational numbers
Open revision guidance
What to know
- Rational numbers are numbers that can be written as a fraction p/q where p and q are integers and q is not zero. This includes all terminating decimals like 0.75 (3/4) and recurring decimals like 0.333... (1/3).
- Irrational numbers cannot be expressed as simple fractions. Their decimal expansions are infinite and never repeat. Familiar examples include pi and the square roots of non-square numbers, such as root 2. Distinguishing them requires checking if the decimal pattern repeats or terminates.
What to do in an assessment
- Classify 0.121212... as rational or irrational. Since the block 12 repeats infinitely, this is a recurring decimal. Based on the rules of algebra, we can express it as a fraction (12/99), so it is rational. Write the classification clearly and justify it by showing the fraction conversion.
- Classify the value of square root 5. The square root of any non-square integer is irrational. Its decimal expansion (approx 2.2360679...) does not terminate or repeat. Therefore, it cannot be written as a fraction. Label the categories for data sets like {0.5, 0.333..., root 7, 1/4} and explain each choice.
In plain language
Rational numbers are numbers that can be turned into a fraction. Irrational numbers are numbers like pi or root 2 that have infinite, non-repeating decimals and cannot be written as a simple fraction.
Self-check
Is 0.875 rational or irrational?
Show the answer
It is rational because it terminates, so it can be written as 875/1000, which simplifies to 7/8.
- Higher
Scope: Higher · Higher tier in either Unit 2 tier or Unit 3 tier · Unit 2 §1.9.4 higher; Unit 3 §1.9.4 higher. Printed pages 20: manipulate and simplify numerical expressions involving surds
Open revision guidance
What to know
- Surds are the square roots of non-square numbers. To simplify a surd like root 20, find the largest square factor. Since 20 = 4 * 5 and the square root of 4 is 2, root 20 simplifies to 2 * root 5.
- When adding or subtracting surds, they behave like algebraic variables. You can only combine terms with the same surd part. For example, 3 * root 2 + 5 * root 2 equals 8 * root 2. However, 3 * root 2 + 5 * root 3 cannot be simplified further as the surd parts are distinct.
What to do in an assessment
- Simplify root 72. Identify square factors: 72 = 36 * 2. Rewrite as sqrt(36 * 2) = sqrt(36) * sqrt(2) = 6 * root 2. This is the simplest form because 2 has no square factors other than 1.
- Perform the expression 2 * root 18 + 5 * root 8. Simplify each: 2 * (3 * root 2) + 5 * (2 * root 2). Calculate the products: 6 * root 2 + 10 * root 2. Sum the results: 16 * root 2. Always simplify the individual surds first before combining like terms to ensure accuracy.
In plain language
Surds are roots that do not result in a whole number. You simplify them by taking out any square numbers hidden inside the root and combining similar terms just like you would with algebra letters.
Self-check
Simplify 3 * root 27 + 2 * root 12.
Show the answer
3 * (3 * root 3) + 2 * (2 * root 3) = 9 * root 3 + 4 * root 3 = 13 * root 3.
- Higher
Scope: Higher · Higher tier in either Unit 2 tier · Unit 2 §1.9.5 higher. Printed pages 20: manipulate and simplify more complex numerical expressions involving surds, including multiplying expressions containing surds and simplifying fractions containing surds by division of common factors
Open revision guidance
What to know
- More complex surd manipulation involves multiplying expressions containing multiple terms, similar to expanding binomials (a + b)(c + d). For instance, (1 + root 2)(3 - root 2) expands to 1*3 - 1*root 2 + 3*root 2 - root 2*root 2.
- Simplifying fractions containing surds often requires dividing by common factors. If an expression like (4 + 2 * root 3) / 2 is given, divide every term by the common factor 2 to get 2 + root 3. This division technique simplifies expressions significantly.
What to do in an assessment
- Expand (2 + root 5)(3 - root 5). Multiply terms: 2*3 - 2*root 5 + 3*root 5 - root 5*root 5. Combine: 6 + root 5 - 5. Result: 1 + root 5. Show each step of the expansion to ensure signs are handled correctly during the multiplication.
- Simplify (6 + 9 * root 2) / 3. Divide each term by 3: (6/3) + (9 * root 2 / 3) = 2 + 3 * root 2. Use this approach for any polynomial-style surd expression where all coefficients are divisible by a common integer.
In plain language
You can multiply and divide complex surd expressions just like you do with algebraic brackets. Use FOIL (First, Outer, Inner, Last) to multiply brackets and divide each part by the denominator to simplify fractions.
Self-check
Expand and simplify (2 + root 3)(2 - root 3).
Show the answer
4 - 2*root 3 + 2*root 3 - 3 = 1.
- Higher
Scope: Higher · Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §1.9.6 higher; Unit 3 §1.9.6 higher. Printed pages 20: use pi in exact calculations
Open revision guidance
What to know
- Using pi in exact calculations means keeping the symbol instead of substituting 3.14 or a calculator approximation. This maintains mathematical precision. For a circle with a radius r = 5 cm, the circumference is 2 * pi * r = 10 * pi cm, and the area is pi * r^2 = 25 * pi cm^2.
- Exact values are critical in geometry to avoid rounding errors that propagate through subsequent steps. For example, a sphere with radius r = 3 m has a surface area of 4 * pi * r^2. Substituting r = 3 gives 4 * pi * 9 = 36 * pi m^2. By keeping pi, the result is exact rather than an estimate.
What to do in an assessment
- To compute the area of a sector with radius 6 cm and angle 60 degrees, use the formula (theta / 360) * pi * r^2. Calculation: (60 / 360) * pi * 6^2 = (1/6) * pi * 36. Multiply: 1/6 * 36 = 6. The result is 6 * pi cm^2. Do not convert to a decimal unless specifically requested.
- When solving problems involving composite shapes, maintain the pi symbol throughout. If you have a cylinder with radius 2 m and height 10 m, volume is pi * r^2 * h = pi * 2^2 * 10 = 40 * pi m^3. Ensure all units of length are squared or cubed appropriately, as the precision of pi is preserved alongside the unit dimensions.
In plain language
Exact calculations with pi mean leaving the symbol pi in your answer rather than using a long decimal. This is more accurate for geometry problems involving circles, spheres, or cylinders.
Self-check
Find the exact circumference of a circle with a diameter of 14 cm.
Show the answer
Circumference = pi * d. Since d = 14 cm, the circumference is 14 * pi cm.
- Higher
Scope: Higher · Higher tier in either Unit 2 tier or Unit 3 tier · Unit 2 §1.9.7 higher; Unit 3 §1.9.7 higher. Printed pages 20: use surds in exact calculations
Open revision guidance
What to know
- Exact calculations with surds involve performing arithmetic operations while maintaining the surd form. This ensures precision is not lost to decimals. For example, if a square has a side length of root 7 m, the area is (root 7)^2 = 7 m^2.
- When determining perimeters of rectangles with sides root 3 cm and root 12 cm, keep the surds exact. The perimeter is 2 * (root 3 + root 12). Since root 12 = 2 * root 3, the perimeter is 2 * (root 3 + 2 * root 3) = 2 * (3 * root 3) = 6 * root 3 cm.
What to do in an assessment
- Calculate the area of a rectangle with sides 5 cm and root 8 cm. Area = 5 * root 8. Simplify root 8 to 2 * root 2. Area = 5 * 2 * root 2 = 10 * root 2 cm^2. Keep the answer in this form to be exact.
- Calculate the volume of a rectangular prism with sides root 2 m, root 3 m, and root 6 m. Volume = root 2 * root 3 * root 6 = root (2 * 3 * 6) = root 36 = 6 m^3. Working with surds often reveals exact integer results through multiplication.
In plain language
Exact calculations with surds keep the roots in the answer. This is more precise than decimals for calculations involving area or volume with square roots.
Self-check
Find the area of a square with side length 3 * root 5 cm.
Show the answer
Area = (3 * root 5)^2 = 9 * 5 = 45 cm^2.
Topic group
Algebra
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §2.1.1 foundation; Unit 2 §2.1.1 foundation; Unit 3 §2.1.1 foundation. Printed pages 21: understand the basic conventions of algebra
Open revision guidance
What to know
- Algebraic notation uses letters called variables to represent unknown or changeable values. In the expression 5x, the 5 is the coefficient and x is the variable. When multiplying variables, we write them side by side without multiplication signs, so x multiplied by y is written as xy. This convention removes ambiguity between the letter x and the multiplication symbol.
- Mathematical operations follow an order similar to arithmetic. Powers are written as superscripts, such as x squared (x^2), and division is often represented using fraction bars, such as a/b. When coefficients are 1, they are usually omitted; for example, 1x is written simply as x. Similarly, indices of 1 are omitted, so x^1 is written as x.
What to do in an assessment
- To simplify expressions, apply the convention of order and omission. For the expression 3 multiplied by a, followed by b, write it as 3ab. For a multiplied by a multiplied by a, use the exponent notation to write a^3. If you have 4 times x divided by 2, simplify the coefficients: 4/2 equals 2, resulting in 2x.
- When converting words to algebra, translate operations carefully. The phrase three more than x is written as x + 3. The phrase double the sum of x and y is written as 2(x + y). Always ensure variables are grouped correctly using brackets when addition or subtraction occurs before multiplication.
In plain language
Algebra is a shorthand language for mathematics. By following specific rules for writing letters and numbers together, we can represent complex problems simply.
Self-check
Write the algebraic expression for 5 multiplied by y, all divided by 10, then add z.
Show the answer
y/2 + z. Working: (5 * y) / 10 = 5y/10 = y/2; adding z gives y/2 + z.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §2.1.2 foundation; Unit 2 §2.1.2 foundation; Unit 3 §2.1.2 foundation. Printed pages 21: substitute positive and negative whole numbers, fractions and decimals into simple formulae and expressions written in words or in symbols
Open revision guidance
What to know
- Substitution involves replacing variables in an expression with specific numerical values. If you are given a formula, such as A = lb (Area equals length times breadth), and provided with l = 5 and b = 3, you replace the letters with these numbers. Always use brackets when substituting negative numbers to ensure the signs are handled correctly during multiplication.
- Expressions can contain fractions and decimals. For a formula like P = 2(l + w), if l = 2.5 and w = 1.2, you first perform the addition within the brackets: 2.5 + 1.2 = 3.7. Then multiply by the coefficient: 2 * 3.7 = 7.4. Precision is vital when working with these types of numbers to ensure the final output is accurate.
What to do in an assessment
- To substitute correctly, write out the formula clearly and replace each letter with its assigned value. For an expression like 3x + y where x = -2 and y = 4, calculate: 3 * (-2) = -6, then add 4. The result is -2. Always show the substitution step before performing the arithmetic to avoid errors.
- When working with fractional inputs, treat them as regular numbers. For the expression 2a + b where a = 1/2 and b = 1/4, calculate 2 * (1/2) = 1. Then add 1/4 to get 1.25 or 5/4. Ensure that the units are consistent if the variables represent physical quantities in a specific formula.
In plain language
Substitution is swapping a letter for a known number. Once swapped, you just perform the sum using that number instead of the letter.
Self-check
Evaluate 4x - 2y given x = 3 and y = -1.
Show the answer
14. Working: 4(3) - 2(-1) = 12 - (-2) = 12 + 2 = 14.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §2.1.3 foundation; Unit 2 §2.1.3 foundation; Unit 3 §2.1.3 foundation. Printed pages 21: recognise the definitions of the terms equation, expression and formula and be able to distinguish between them
Open revision guidance
What to know
- An expression is a collection of terms and operators without an equals sign, such as 3x + 2. It represents a value that can change based on the variable. An equation contains an equals sign and states that two expressions are equivalent, such as 3x + 2 = 11, which allows us to find the value of x.
- A formula describes a relationship between two or more quantities, usually representing a real-world concept like Area = length * width. Unlike an equation which might have one specific solution, a formula acts as a general rule where variables are replaced to calculate a result for different inputs.
What to do in an assessment
- To identify these, look for the presence of an equals sign and the purpose of the statement. If you see '3a - 5', it is an expression because it has no equals sign. If you see 'y = 3a - 5', it is a formula because it defines one quantity in terms of others. If you see '3a - 5 = 10', it is an equation because it asks for a specific value of a.
- Practice distinguishing them by writing your own examples. Create an expression using variables (e.g., 2b + 7), transform it into an equation by adding a result (e.g., 2b + 7 = 15), and express it as a formula (e.g., V = 2b + 7). This exercise reinforces the structural differences and the logical purpose of each mathematical form.
In plain language
Expressions are like phrases without a conclusion. Equations are questions looking for a specific answer. Formulae are rules that show how things relate.
Self-check
Identify if 5 = 2x + 1 is an expression, equation, or formula.
Show the answer
Equation. It contains an equals sign and relates to a specific solution for x.
- Higher
Scope: Higher · Higher tier in either Unit 2 tier or Unit 3 tier · Unit 2 §2.1.4 higher; Unit 3 §2.1.4 higher. Printed pages 21: recognise the definition of the term identity and be able to distinguish between identities, equations, expressions and formulae
Open revision guidance
What to know
- An identity is a mathematical statement that is true for all possible values of the variables involved. It is denoted by the triple bar symbol (≡). For instance, 2(x + 3) ≡ 2x + 6 is an identity because, regardless of whether x is 1, 10, or -5, both sides will always yield the same result. The expansion or simplification of terms demonstrates the equivalence.
- Unlike an equation, which is only true for specific values of a variable, an identity is universally true. In 2x + 5 = 11, the statement is only true if x = 3. In 2(x + 1) ≡ 2x + 2, the statement is true for any value of x. Recognising this difference helps in simplifying complex algebraic proofs and determining if expressions are identical.
What to do in an assessment
- To test for an identity, substitute different values into both sides of the expression. For example, test x(x + 1) ≡ x^2 + x. If x = 2, then 2(2 + 1) = 6 and 2^2 + 2 = 6. If x = -3, then -3(-3 + 1) = 6 and (-3)^2 + (-3) = 9 - 3 = 6. Since they match for both, it is likely an identity.
- When simplifying, perform algebraic manipulations to see if the left side can be transformed into the right side. For example, to prove x^2 + 2x + 1 ≡ (x + 1)^2, expand the right side: (x + 1)(x + 1) = x^2 + x + x + 1 = x^2 + 2x + 1. Since both sides are now identical, the statement is an identity.
In plain language
An identity is true for any number you choose. An equation is only true for certain specific numbers.
Self-check
Is 3(x - 2) ≡ 3x - 6 an identity or an equation? Why?
Show the answer
Identity. It is true for any value of x because expanding 3(x - 2) results in 3x - 6.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §2.1.5 foundation; Unit 2 §2.1.5 foundation; Unit 3 §2.1.5 foundation. Printed pages 22: form and simplify expressions
Open revision guidance
What to know
- Forming expressions requires translating descriptive text into symbols. If you have a number x, adding five becomes x + 5. If you take a number y and multiply it by 3 then subtract 2, it becomes 3y - 2. Always define your variables clearly if they are not provided, using letters like x, y, or n.
- Simplifying expressions involves grouping together terms that have the same variables and powers. In an expression like 4x + 3y + 2x + y, you identify the 'like' terms (terms with x and terms with y). Adding 4x and 2x results in 6x, and adding 3y and y results in 4y, giving a simplified total of 6x + 4y.
What to do in an assessment
- To form an expression, identify the variables and the operations. For the problem 'the sum of a and b multiplied by 4', you write 4(a + b). If the problem is 'the product of a and b', you write ab. Always place numbers before variables and use alphabetical order for multiple variables where applicable.
- To simplify, scan the expression for common variable types. For 5a + 2b - a + 3b: group the a terms (5a - a = 4a) and the b terms (2b + 3b = 5b). The final simplified form is 4a + 5b. Ensure all signs (plus or minus) remain attached to the term that follows them.
In plain language
Creating expressions means turning word problems into maths language. Simplifying means making those expressions as short and tidy as possible.
Self-check
Form and simplify the expression: A number n is multiplied by 2, then added to the result of 3 multiplied by n.
Show the answer
5n. Working: (n * 2) + (3 * n) = 2n + 3n = 5n.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §2.1.6 foundation; Unit 2 §2.1.6 foundation; Unit 3 §2.1.6 foundation. Printed pages 22: collect like terms
Open revision guidance
What to know
- Collecting like terms is a process of simplifying an algebraic expression by grouping terms that have identical variables and powers. For example, in 3x^2 + 5x + 2x^2 + x, the terms 3x^2 and 2x^2 are 'like' because they both have the variable x squared, whereas 5x and x are like terms because they both have the variable x.
- When collecting, always treat the sign in front of the term as part of the term. In the expression 5a - 3b + 2a + 4b, the subtraction sign belongs to the 3b. When you move terms, they must carry their signs with them. This ensures that the result remains mathematically correct.
What to do in an assessment
- To collect like terms, reorder the expression to place similar terms next to each other. Given 7y + 3x - 2y + 5x, rearrange it to (3x + 5x) + (7y - 2y). Now perform the arithmetic: 8x + 5y. Ensure that you do not mix different powers; for example, x and x^2 cannot be combined.
- Check for hidden coefficients. If you see an expression like 3a + b - 2a + 4b, remember that 'b' is the same as '1b'. So, the calculation for the b terms is 1b + 4b = 5b. The calculation for the a terms is 3a - 2a = 1a, which is just written as a. Final answer: a + 5b.
In plain language
Like terms are items that match. You group them together to count how many you have of each, just like counting apples and oranges separately.
Self-check
Simplify: 6p + 4q - 2p + q.
Show the answer
4p + 5q. Working: (6p - 2p) + (4q + q) = 4p + 5q.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §2.1.7 foundation; Unit 2 §2.1.7 foundation; Unit 3 §2.1.7 foundation. Printed pages 22: expand expressions – single bracket
Open revision guidance
What to know
- Expanding a single bracket involves multiplying the term outside the bracket by every term inside the bracket. For an expression like 3(x + 5), the 3 must be multiplied by x and also by 5. The result is 3x + 15. This process uses the distributive law, where multiplication is distributed over addition.
- When dealing with variables and negative signs, caution is required. For -2(x - 4), you multiply -2 by x to get -2x, and -2 by -4 to get +8. The result is -2x + 8. Signs change according to standard multiplication rules: two negatives make a positive, while a negative and a positive result in a negative.
What to do in an assessment
- To expand, draw arrows to indicate the multiplication from the outside term to each inside term. For 4(2a - 3b), multiply 4 by 2a (giving 8a) and 4 by -3b (giving -12b). The final expanded expression is 8a - 12b. Always ensure all terms inside the bracket are included in the process.
- Handle multi-term expansions carefully. For 2(a + b + 3), distribute the 2 to all three terms: 2 * a = 2a, 2 * b = 2b, and 2 * 3 = 6. Combine these into 2a + 2b + 6. If variables share coefficients, simplify the result further by collecting terms if necessary.
In plain language
Expanding a bracket is like sharing a gift. Everything inside the bracket must be multiplied by the number waiting outside.
Self-check
Expand 5(2x - 3).
Show the answer
10x - 15. Working: 5 * 2x = 10x; 5 * -3 = -15.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier or Unit 3 tier · Unit 2 §2.1.8 foundation; Unit 3 §2.1.8 foundation. Printed pages 22: multiply and divide terms by applying rules of indices
Open revision guidance
What to know
- To multiply terms with the same base, you add the exponents. For example, x^3 multiplied by x^4 equals x^(3+4), which is x^7. This rule applies because x^3 is x*x*x and x^4 is x*x*x*x, totaling seven instances of x.
- To divide terms with the same base, you subtract the exponents. For example, x^9 divided by x^3 equals x^(9-3), which is x^6. This is because nine factors of x in the numerator cancel out three factors in the denominator, leaving six factors remaining.
What to do in an assessment
- To compute 5y^2 * 4y^3, first multiply the coefficients 5 * 4 = 20, then apply the addition rule for indices to the y terms, resulting in y^(2+3) = y^5. The final expression is 20y^5.
- To simplify 12a^7 / 3a^2, divide the numerical coefficients 12 / 3 = 4, then apply the subtraction rule for the indices: a^(7-2) = a^5. The complete simplified term is 4a^5.
In plain language
When multiplying variables with identical bases, add their powers. When dividing, subtract the power of the divisor from the power of the dividend.
Self-check
Simplify 3b^4 * 2b^5 / 6b^2.
Show the answer
3 * 2 = 6, and b^(4+5) = b^9. Then 6b^9 / 6b^2 = 1b^(9-2) = b^7.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier or Unit 3 tier · Unit 2 §2.1.9 foundation; Unit 3 §2.1.9 foundation. Printed pages 22: simplify algebraic fractions, including the addition and subtraction of fractions with constant terms as denominators
Open revision guidance
What to know
- Simplifying algebraic fractions requires identifying common factors between the numerator and denominator. By cancelling these factors, the expression is reduced to its simplest form without changing its value.
- When adding or subtracting algebraic fractions, it is necessary to establish a common denominator. For fractions like a/x + b/x, the sum is simply (a+b)/x, assuming x is a non-zero constant term.
What to do in an assessment
- To simplify 15x / 5, divide the numerator 15x by the constant 5 to get 3x. For (4x+8) / 2, factorise the numerator to get 2(2x+4) / 2, which simplifies to 2x+4.
- To add 2x/7 + 3x/7, combine the numerators since the denominators are equal, resulting in (2x+3x) / 7 = 5x/7. For subtraction, treat it similarly: 5y/4 - y/4 = (5y-y) / 4 = 4y/4 = y.
In plain language
Cancel common factors between the top and bottom of a fraction. To add or subtract, make sure the bottom numbers are the same, then add or subtract the top.
Self-check
Simplify 9x/3 + 4x/2.
Show the answer
9x/3 = 3x. 4x/2 = 2x. 3x + 2x = 5x.
- Higher
Scope: Higher · Higher tier in either Unit 2 tier or Unit 3 tier · Unit 2 §2.1.10 higher; Unit 3 §2.1.10 higher. Printed pages 22: simplify more complex algebraic fractions, including the addition and subtraction of fractions with linear expressions as denominators
Open revision guidance
What to know
- More complex fractions involve linear expressions in the denominator. To simplify, we often factorise both numerator and denominator to cancel common linear terms. For example, (x^2 + 5x + 6) / (x+2) simplifies by factorising the numerator into (x+2)(x+3), allowing (x+2) to cancel.
- For addition or subtraction, finding a lowest common multiple (LCM) of linear denominators is essential. When adding 1/(x+1) + 2/(x-3), multiply the terms to reach a common denominator of (x+1)(x-3).
What to do in an assessment
- To simplify (x^2 - 9) / (x-3), factorise x^2-9 into (x-3)(x+3). Thus, (x-3)(x+3) / (x-3) = x+3. This operation requires x not to equal 3.
- To compute 1/(x+2) + 1/(x+3), use the common denominator (x+2)(x+3). The numerator becomes 1*(x+3) + 1*(x+2), resulting in (x+3 + x+2) / (x+2)(x+3) = (2x+5) / (x^2+5x+6).
In plain language
Factorise expressions to cancel terms. To combine complex fractions, use the common denominator formed by multiplying the linear expressions together.
Self-check
Simplify 1/(x+1) - 1/(x+2).
Show the answer
(1*(x+2) - 1*(x+1)) / ((x+1)(x+2)) = (x+2-x-1) / (x^2+3x+2) = 1 / (x^2+3x+2).
- Higher
Scope: Higher · Higher tier in either Unit 2 tier or Unit 3 tier · Unit 2 §2.1.11 higher; Unit 3 §2.1.11 higher. Printed pages 22: expand two linear expressions in one or two variables
Open revision guidance
What to know
- Expanding two linear expressions involves multiplying every term in the first bracket by every term in the second bracket. Using the FOIL method (First, Outer, Inner, Last), we ensure every component is accounted for.
- When dealing with two variables, the same logic applies. For example, (x + a)(y + b) expands to xy + xb + ay + ab. Each term is distributed systematically to maintain the integrity of the expression.
What to do in an assessment
- To expand (x+4)(x+5), calculate: First (x*x = x^2), Outer (x*5 = 5x), Inner (4*x = 4x), and Last (4*5 = 20). Combining like terms, x^2 + (5x+4x) + 20 = x^2 + 9x + 20.
- To expand (x+2)(y+3), multiply as follows: x*y = xy, x*3 = 3x, 2*y = 2y, and 2*3 = 6. The expanded form is xy + 3x + 2y + 6. No terms here can be combined further as they are distinct.
In plain language
Multiply every part of the first bracket by every part of the second. Add up the pieces to get your final expression.
Self-check
Expand (x+3)(x-2).
Show the answer
x^2 - 2x + 3x - 6 = x^2 + x - 6.
- Higher
Scope: Higher · Higher tier in either Unit 2 tier or Unit 3 tier · Unit 2 §2.1.12 higher; Unit 3 §2.1.12 higher. Printed pages 22: expand two expressions in one variable, where one is linear and the other is quadratic
Open revision guidance
What to know
- Expanding a linear expression with a quadratic expression follows the distributive law. If you have (ax + b)(cx^2 + dx + e), you must multiply each of the two terms in the first bracket by each of the three terms in the second bracket.
- This produces six initial terms which must then be simplified by collecting like powers of x. This expansion is foundational for more advanced algebraic manipulation and polynomial arithmetic.
What to do in an assessment
- To expand (x+2)(x^2 + 3x + 4), distribute the x: x(x^2 + 3x + 4) = x^3 + 3x^2 + 4x. Then distribute the 2: 2(x^2 + 3x + 4) = 2x^2 + 6x + 8.
- Now, sum these results: x^3 + (3x^2 + 2x^2) + (4x + 6x) + 8 = x^3 + 5x^2 + 10x + 8. Always ensure each power of x is accounted for during the summation.
In plain language
Multiply every term in the small bracket by every term in the large bracket. Combine terms with the same powers of x to finish.
Self-check
Expand (x+1)(x^2 + 2x + 3).
Show the answer
x(x^2 + 2x + 3) = x^3 + 2x^2 + 3x. 1(x^2 + 2x + 3) = x^2 + 2x + 3. Sum: x^3 + 3x^2 + 5x + 3.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §2.1.13 foundation. Printed pages 22: factorise linear or quadratic expressions that have at least one common factor
Open revision guidance
What to know
- Factorising is the reverse of expanding. To factorise, identify the Highest Common Factor (HCF) of all terms in the expression. The HCF is then pulled outside the bracket, leaving the remaining components inside.
- This method applies to linear and quadratic expressions. For example, in 3x + 6, the HCF is 3, giving 3(x+2). In x^2 + 5x, the HCF is x, giving x(x+5). Always check if further factorisation is possible.
What to do in an assessment
- To factorise 12y + 18, find the HCF of 12 and 18, which is 6. Divide both terms by 6 to get 2y + 3. Write this as 6(2y+3).
- To factorise 4x^2 + 8x, identify the HCF of 4x^2 and 8x, which is 4x. Dividing each term yields x + 2. The result is 4x(x+2).
In plain language
Find the largest number or variable that divides evenly into all parts of the expression. Pull it out and place the rest inside brackets.
Self-check
Factorise 10x^2 - 15x.
Show the answer
The HCF is 5x. 5x(2x - 3).
- Higher
Scope: Higher · Higher tier in either Unit 2 tier · Unit 2 §2.1.14 higher. Printed pages 22: factorise more complex expressions by the extraction of common factors
Open revision guidance
What to know
- More complex factorisation involves extracting common factors from expressions that might contain multiple variables or higher-degree terms. The process remains the identification of the highest common factor, but requires careful attention to coefficients and variables present in each term.
- Sometimes you must extract a term that includes both a number and a variable. This is critical for simplifying expressions in higher-level algebra and preparing for solving equations.
What to do in an assessment
- To factorise 6x^2y + 9xy^2, identify the HCF of 6 and 9 (which is 3) and the HCF of x^2y and xy^2 (which is xy). The combined HCF is 3xy.
- Divide 6x^2y by 3xy to get 2x. Divide 9xy^2 by 3xy to get 3y. The factored form is 3xy(2x + 3y).
In plain language
Identify the largest shared number and the lowest power of each variable shared across all parts. Divide everything by this factor and keep the remainder in brackets.
Self-check
Factorise 16a^2b + 8ab^2.
Show the answer
The HCF is 8ab. 8ab(2a + b).
- Higher
Scope: Higher · Higher tier in either Unit 2 tier · Unit 2 §2.1.15 higher. Printed pages 23: factorise quadratic expressions of the form x 2 + ax + b and ax 2 + bx + c, including the difference of two squares
Open revision guidance
What to know
- To factorise x² + ax + b, find two numbers whose product is b and sum is a. For x² + 7x + 10, the numbers are 5 and 2, so x² + 7x + 10 = (x + 5)(x + 2). Expand to check: x² + 2x + 5x + 10.
- For a non-monic quadratic, split the middle term using factors of the product of the leading coefficient and constant. For 2x² + 7x + 3, 2 × 3 = 6 and 6 + 1 = 7. Then 2x² + 6x + x + 3 = 2x(x + 3) + (x + 3) = (2x + 1)(x + 3). The difference of squares is a² − b² = (a − b)(a + b).
What to do in an assessment
- Factorise 3x² + 10x + 8. The product 3 × 8 is 24; use 6 and 4, whose sum is 10. Then 3x² + 6x + 4x + 8 = 3x(x + 2) + 4(x + 2) = (3x + 4)(x + 2). Expansion gives 3x² + 6x + 4x + 8, confirming the result.
- 25x² − 49 = (5x)² − 7² = (5x − 7)(5x + 7). For 4x² − 100, first remove the common factor: 4(x² − 25) = 4(x − 5)(x + 5). Always look for a common factor before applying the difference-of-squares identity.
In plain language
Factorising reverses expansion. Preserve the square signs, choose the two numbers carefully, and expand your answer to check it.
Self-check
Factorise 2x² + 5x + 3.
Show the answer
(2x + 3)(x + 1), because expansion gives 2x² + 2x + 3x + 3 = 2x² + 5x + 3.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §2.1.16 foundation. Printed pages 23: change the subject of a formula when the subject appears in one term
Open revision guidance
What to know
- To change the subject of a formula, perform inverse operations on both sides to isolate the desired variable. If the subject appears in one term, identify the operations applied to the variable and reverse them in the correct order. For example, in the formula y = 3x + 5, we want x as the subject. First, subtract 5 from both sides: y - 5 = 3x. Then divide by 3: (y - 5) / 3 = x. This procedure ensures the balance of the equation is maintained at every step of the algebraic manipulation process.
- Consider the formula A = πr^2. To make r the subject, divide by π to get A / π = r^2. Then take the square root of both sides, resulting in r = sqrt(A / π). Always apply the inverse operation to both sides simultaneously. For equations like v = u + at, to isolate t, subtract u from both sides: v - u = at. Next, divide both sides by a, resulting in t = (v - u) / a. The use of brackets is essential if the numerator contains multiple terms that need to be divided by a single variable.
What to do in an assessment
- Change the subject of P = 2(l + w) to l. Step 1: Divide by 2 to get P / 2 = l + w. Step 2: Subtract w to get P / 2 - w = l. Alternatively, expand first: P = 2l + 2w. Subtract 2w: P - 2w = 2l. Divide by 2: (P - 2w) / 2 = l. Both methods produce mathematically equivalent results. Use the one that is more intuitive for the specific structure of the given formula.
- Change the subject of F = ma to m. Step 1: The equation is F = ma. Since m is multiplied by a, divide both sides by a. Step 2: F / a = m. Thus, m = F / a. For the formula E = mc^2, change the subject to m. Step 1: Divide by c^2. Step 2: E / c^2 = m. The variable m is now the subject of the formula. Ensure all terms are accounted for during each transformation stage.
In plain language
Changing the subject means rewriting a formula so that one letter is on its own on one side of the equals sign. Use inverse operations like adding/subtracting or multiplying/dividing to move other terms to the other side until only your chosen letter remains.
Self-check
Change the subject of v = u + at to a.
Show the answer
a = (v - u) / t. Working: Subtract u: v - u = at. Divide by t: (v - u) / t = a.
- Higher
Scope: Higher · Higher tier in either Unit 2 tier · Unit 2 §2.1.17 higher. Printed pages 23: change the subject of a formula when the subject appears in more than one term
Open revision guidance
What to know
- When the subject appears in more than one term, you must gather all terms containing that variable on one side of the equation. For the formula y = mx + nx, factorise the right side: y = x(m + n). Then, divide by the expression in the bracket to isolate x: x = y / (m + n). This technique is essential when the subject is trapped within different parts of an algebraic expression. Always ensure that every term containing the desired variable is moved, while terms without it are moved to the opposite side.
- Consider the formula v = u + at. If we need to make t the subject, subtract u: v - u = at. Then divide by a: t = (v - u) / a. If the formula is more complex, such as y = (ax + b) / (cx + d), clear the fraction first. Multiply by (cx + d): y(cx + d) = ax + b. Expand the bracket: ycx + yd = ax + b. Group x terms: ycx - ax = b - yd. Factorise x: x(yc - a) = b - yd. Finally, divide: x = (b - yd) / (yc - a). This covers standard rational formula rearrangements.
What to do in an assessment
- Rearrange y = ax + b to make x the subject. Step 1: y - b = ax. Step 2: Divide by a to isolate x, giving x = (y - b) / a. For the formula y = xz + xk, make x the subject. Step 1: Notice x is common to both terms. Factorise: y = x(z + k). Step 2: Divide both sides by (z + k). Result: x = y / (z + k). This method works for any linear algebraic structure where common factors can be extracted.
- Rearrange T = (x + 3) / (x - 2) to make x the subject. Step 1: Multiply by (x - 2): T(x - 2) = x + 3. Step 2: Expand: Tx - 2T = x + 3. Step 3: Gather x terms: Tx - x = 3 + 2T. Step 4: Factorise x: x(T - 1) = 3 + 2T. Step 5: Divide: x = (3 + 2T) / (T - 1). Check: If T = 4, x = (3 + 8) / (4 - 1) = 11 / 3. Testing in original formula: (11/3 + 3) / (11/3 - 2) = (20/3) / (5/3) = 4. The result is verified.
In plain language
If the letter you want is in more than one place, get all those terms onto one side of the equals sign. Use factorisation to turn multiple instances into a single product, then divide to get the letter on its own.
Self-check
Rearrange y = ax + bx + c to make x the subject.
Show the answer
x = (y - c) / (a + b). Working: Subtract c: y - c = ax + bx. Factorise x: y - c = x(a + b). Divide by (a + b): x = (y - c) / (a + b).
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §2.2.1 foundation; Unit 2 §2.2.1 foundation; Unit 3 §2.2.1 foundation. Printed pages 23: form, manipulate and solve linear and other simple equations with whole number and fractional coefficients
Open revision guidance
What to know
- Linear equations involve terms with variables raised to the power of one. The goal is to isolate the unknown variable, such as x, by performing inverse operations on both sides of the equal sign. When equations contain fractional coefficients, such as (1/4)x = 5, we multiply both sides by the denominator to clear the fraction, resulting in x = 20. This maintains equality because whatever operation is applied to one side must be applied to the other.
- For equations with multiple steps, such as 3x + 4 = 19, we first undo addition or subtraction, then undo multiplication or division. Subtracting 4 from both sides yields 3x = 15. Then, dividing by 3 gives x = 5. These principles apply to both whole numbers and fractions. The order of operations is effectively reversed to peel away the constants and coefficients attached to the variable until the variable is isolated.
What to do in an assessment
- To solve 2x + 7 = 15, subtract 7 from both sides: 2x = 8. Then, divide by 2: x = 4. For fractional coefficients like (2/3)x = 10, multiply by the denominator 3 to get 2x = 30. Then divide by the numerator 2 to get x = 15. Always check your work by substituting the result back into the original equation to ensure the left-hand side equals the right-hand side.
- Consider an equation with a fraction: (x/5) - 2 = 3. First, add 2 to both sides to get (x/5) = 5. Next, multiply both sides by 5, resulting in x = 25. For verification, plug 25 back into the original: (25/5) - 2 equals 5 - 2, which is 3. This confirms the solution is correct. Practice these steps with varying coefficients to ensure proficiency in balancing equations.
In plain language
An equation is like a balance scale; to find the mystery number, perform the same math on both sides until the variable stands alone.
Self-check
Solve for x: (x/4) + 3 = 8.
Show the answer
Subtract 3 from both sides to get x/4 = 5. Multiply by 4 to find x = 20.
- Higher
Scope: Higher · Higher tier in either Unit 2 tier or Unit 3 tier · Unit 2 §2.2.2 higher; Unit 3 §2.2.2 higher. Printed pages 23: form, manipulate and solve more complex linear equations, including equations with more than one fractional term
Open revision guidance
What to know
- More complex linear equations often involve multiple fractional terms, such as (x+3)/2 + (x-1)/4 = 5. To solve these, you must find a common denominator for all fractions involved. In this example, 4 is the common denominator. Multiplying the entire equation by the common denominator eliminates all fractions, allowing you to treat the equation as a standard linear equation. This process simplifies the expression by converting fractional parts into integer components.
- When variable terms appear in the numerator, ensure you group the terms correctly using parentheses before expanding. For instance, in (x+3)/2, the whole (x+3) is divided by 2. When multiplying by the common denominator, the 2 cancels out leaving 2(x+3). Applying this systematically avoids errors when coefficients are distributed across the numerator. Mastering this technique is essential for solving equations where variables are nested within complex fractional structures.
What to do in an assessment
- To solve (x+3)/2 + (x-1)/3 = 4, multiply the entire equation by the common denominator 6. This gives 3(x+3) + 2(x-1) = 24. Expand the brackets: 3x + 9 + 2x - 2 = 24. Combine like terms: 5x + 7 = 24. Subtract 7 from both sides: 5x = 17. Divide by 5 to find x = 3.4. Always verify by substituting x = 3.4 into the initial fractional expression.
- Consider the equation (2x+1)/5 - (x-2)/2 = 1. Multiply by the common denominator 10: 2(2x+1) - 5(x-2) = 10. Distribute the multipliers: 4x + 2 - 5x + 10 = 10. Simplify: -x + 12 = 10. Subtract 12: -x = -2, so x = 2. Check: (2(2)+1)/5 - (2-2)/2 = 5/5 - 0 = 1. The solution holds. Use these algebraic manipulation steps consistently for any linear equation involving multiple fractional terms.
In plain language
Clear away complex fractions by multiplying every term by the common denominator, then solve the resulting simpler equation using standard balancing methods.
Self-check
Solve for x: (x+1)/3 + (x-2)/4 = 2.
Show the answer
Multiply by 12: 4(x+1) + 3(x-2) = 24. Expand: 4x+4+3x-6=24. Simplify: 7x-2=24. Result: 7x=26, so x = 26/7 or approximately 3.71.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §2.2.3 foundation. Printed pages 23: form, manipulate and solve simple linear inequalities with whole number and fractional coefficients
Open revision guidance
What to know
- Linear inequalities represent ranges of values rather than a single point. They use symbols like < (less than), > (greater than), ≤ (less than or equal to), and ≥ (greater than or equal to). The rules for manipulating inequalities are similar to equations, with one crucial exception: if you multiply or divide by a negative number, you must reverse the inequality sign. This ensures the range of valid solutions remains accurate.
- When solving, isolate the variable just as you would in an equation. For example, in 3x - 5 < 10, add 5 to both sides to get 3x < 15, then divide by 3 to get x < 5. This tells us that any number less than 5 is a valid solution. For inequalities with fractional coefficients, follow the same clearing process used in linear equations, while remaining vigilant about the sign-reversal rule during multiplication or division by negatives.
What to do in an assessment
- To solve 4 - 2x ≥ 8, subtract 4 from both sides to get -2x ≥ 4. Divide both sides by -2. Because we divided by a negative, the sign changes: x ≤ -2. Check this by picking a value in the range, such as -3. Plugging -3 into the original: 4 - 2(-3) = 4 + 6 = 10, and 10 ≥ 8, which is true. Pick a value outside, like 0: 4 - 2(0) = 4, and 4 ≥ 8 is false. The range x ≤ -2 is correct.
- For an inequality with a fraction, such as (x/3) + 1 > 2, subtract 1 to get (x/3) > 1. Multiply both sides by 3 to get x > 3. Since we multiplied by a positive number, the inequality sign does not flip. Test with x = 4: (4/3) + 1 = 1.33 + 1 = 2.33, which is greater than 2. This satisfies the inequality. Ensure you always apply inverse operations correctly and watch the sign.
In plain language
Inequalities show a range of numbers. Treat them like equations, but remember to flip the arrow sign whenever you multiply or divide by a negative number.
Self-check
Solve for x: 6 - 3x < 12.
Show the answer
Subtract 6 to get -3x < 6. Divide by -3 and flip the sign: x > -2.
- Higher
Scope: Higher · Higher tier in either Unit 2 tier · Unit 2 §2.2.4 higher. Printed pages 23: form, manipulate and solve linear inequalities where the variable appears on both sides of the inequality or where two separate inequalities are written as a double inequality
Open revision guidance
What to know
- Advanced inequalities involve variables appearing on both sides of the inequality sign or double inequalities, such as 2 < 3x - 1 < 8. For variables on both sides, such as 5x + 2 > 2x + 11, the approach is to gather all variable terms on one side and all constant terms on the other. Subtracting 2x from both sides gives 3x + 2 > 11, which simplifies to 3x > 9, resulting in x > 3. The same logic of sign reversal applies if dividing by a negative.
- Double inequalities, written as a <= bx + c <= d, represent the range of x that satisfies both conditions simultaneously. To solve, perform the same operation across all three sections of the inequality. For 2 < 3x - 1 < 8, add 1 to all parts: 3 < 3x < 9. Then, divide by 3: 1 < x < 3. This indicates that x must be greater than 1 and less than 3. This technique keeps the logical boundaries of the solution intact.
What to do in an assessment
- To solve 7x - 4 ≥ 3x + 8, subtract 3x from both sides: 4x - 4 ≥ 8. Add 4 to both sides: 4x ≥ 12. Divide by 4: x ≥ 3. Verify: If x = 4, 7(4)-4 = 24 and 3(4)+8 = 20. Since 24 ≥ 20, the solution x ≥ 3 is consistent. Always perform steps on both sides of the inequality to maintain the balance of the expression.
- Solve the double inequality: -5 ≤ 2x + 1 < 7. Subtract 1 from all parts: -6 ≤ 2x < 6. Divide all parts by 2: -3 ≤ x < 3. This means x is in the closed-open interval [-3, 3). For validation, test x = 0: 2(0)+1 = 1, which fits within the original boundary -5 ≤ 1 < 7. The logic remains sound when operations are applied to all components simultaneously.
In plain language
For complex inequalities, isolate the variable by grouping it on one side or perform operations on all parts of a double inequality to find the range.
Self-check
Solve for x: 3 < 2x + 5 < 11.
Show the answer
Subtract 5 from all sides: -2 < 2x < 6. Divide by 2: -1 < x < 3.
- Higher
Scope: Higher · Higher tier in either Unit 2 tier · Unit 2 §2.2.5 higher. Printed pages 24: form, manipulate and solve by factorisation, quadratic equations of the form x 2 + bx + c = 0 or ax 2 + bx + c = 0
Open revision guidance
What to know
- Quadratic equations take the form x² + bx + c = 0 or ax² + bx + c = 0. Solving by factorisation involves finding two binomials (x + p)(x + q) = 0 such that their product expands to the original quadratic. This relies on the property that if the product of two factors is zero, then at least one factor must be zero. We look for two numbers that multiply to c and add to b.
- For equations where the coefficient a is not 1, such as 2x² + 7x + 3 = 0, we split the middle term. We need two numbers that multiply to 2*3=6 and add to 7. These are 6 and 1. We rewrite the middle term: 2x² + 6x + 1x + 3 = 0. Factor by grouping: 2x(x + 3) + 1(x + 3) = 0. This gives (2x + 1)(x + 3) = 0. Solving gives x = -0.5 and x = -3.
What to do in an assessment
- To solve x² + 5x + 6 = 0, find two numbers that multiply to 6 and add to 5. These are 2 and 3. Write as (x + 2)(x + 3) = 0. Set each bracket to zero: x + 2 = 0 so x = -2, or x + 3 = 0 so x = -3. Check by putting -2 back: (-2)² + 5(-2) + 6 = 4 - 10 + 6 = 0. This confirms the roots are correct.
- Consider 2x² + 5x - 3 = 0. Multiply a and c: 2 * -3 = -6. Find two numbers that multiply to -6 and add to 5. These are 6 and -1. Rewrite: 2x² + 6x - x - 3 = 0. Grouping: 2x(x + 3) - 1(x + 3) = 0. Factors: (2x - 1)(x + 3) = 0. Solutions: x = 0.5 and x = -3. Verify: 2(0.5)² + 5(0.5) - 3 = 2(0.25) + 2.5 - 3 = 0.5 + 2.5 - 3 = 0. Accurate.
- For the monic equation x² + 5x + 6 = 0, factorise to (x+2)(x+3) = 0. A product is zero only if at least one factor is zero, so x = −2 or x = −3. Substitute each root into the original equation to check.
In plain language
To solve quadratics, break the equation into two parts that multiply to zero. If the x-squared term has a number, split the middle term first.
Self-check
Solve by factorising: x² + 7x + 10 = 0.
Show the answer
(x + 2)(x + 5) = 0. So x = -2 or x = -5.
- Higher
Scope: Higher · Higher tier in either Unit 2 tier · Unit 2 §2.2.6 higher. Printed pages 24: form, manipulate and solve two simultaneous linear equations with whole number coefficients by algebraic methods
Open revision guidance
What to know
- Simultaneous equations involve finding the common values for two or more variables that satisfy multiple equations at once. For two linear equations with whole number coefficients, we use algebraic methods like substitution or elimination. Elimination works by multiplying one or both equations by a constant so that the coefficients of one variable match in magnitude. Adding or subtracting the resulting equations then eliminates that variable, leaving an equation with only one unknown.
- For example, given 2x + 3y = 12 and x - y = 1. We can multiply the second equation by 3 to get 3x - 3y = 3. Now, we have equations with +3y and -3y. Adding these together cancels the y variable, leaving 5x = 15. Solving for x gives 3, then substitute 3 back into any original equation to find y. This method provides a clear, logical path to identifying the intersection point of two lines in a coordinate plane.
What to do in an assessment
- Solve 3x + 2y = 16 and x - 2y = 0. Since the y coefficients are +2 and -2, add the equations directly: 4x = 16, which means x = 4. Substitute x = 4 into x - 2y = 0: 4 - 2y = 0, so 2y = 4, and y = 2. The solution is (4, 2). Check: 3(4) + 2(2) = 12 + 4 = 16. Correct.
- Consider 2x + y = 7 and 3x + 2y = 12. Multiply the first by 2: 4x + 2y = 14. Subtract the second equation from this: (4x - 3x) + (2y - 2y) = 14 - 12. This leaves x = 2. Substitute x = 2 into 2x + y = 7: 4 + y = 7, so y = 3. Solution (2, 3). Check in original: 3(2) + 2(3) = 6 + 6 = 12. Verified.
In plain language
To solve simultaneous equations, make one set of coefficients match, then add or subtract the equations to remove one variable and solve the remaining part.
Self-check
Solve 2x + y = 10 and x - y = 2.
Show the answer
Add the equations: 3x = 12, so x = 4. Substitute: 4 - y = 2, so y = 2. Solution: (4, 2).
- Higher
Scope: Higher · Higher tier in either Unit 2 tier or Unit 3 tier · Unit 2 §2.2.7 higher; Unit 3 §2.2.7 higher. Printed pages 24: solve equations involving fractions with linear denominators leading to quadratic or linear equations
Open revision guidance
What to know
- Equations involving fractions with linear denominators, such as 1/(x+1) + 2/(x-2) = 3, require algebraic manipulation to solve. The process involves identifying a common denominator to clear the fractions. Here, (x+1)(x-2) is the common denominator. Multiplying the entire equation by this product converts the expression into a standard quadratic or linear equation. It is crucial to multiply each term properly to avoid missing factors.
- After clearing fractions, expand the resulting terms. If a quadratic form arises, such as ax² + bx + c = 0, solve using factorisation or the quadratic formula. These equations can lead to multiple solutions. Always check for values that make denominators zero, as these are invalid in the context of the original equation, such as x = -1 or x = 2 in the above example. Systematic expansion ensures no terms are overlooked during the algebraic conversion.
What to do in an assessment
- Solve 1/x + 1/(x+1) = 2. Common denominator: x(x+1). Multiply through: (x+1) + x = 2x(x+1). Simplify: 2x + 1 = 2x² + 2x. Subtract 2x from both sides: 1 = 2x². Solve: x² = 0.5, so x = ±√(0.5) ≈ ±0.707. Always perform the multiplication step carefully to clear denominators.
- Consider 2/(x-3) = 1/x. Multiply by common denominator x(x-3): 2x = 1(x-3). This simplifies to 2x = x - 3. Subtract x: x = -3. Verify: 2/(-3-3) = 2/-6 = -1/3. And 1/(-3) = -1/3. The solution x = -3 is correct. Practise this method to convert complex fractional equations into manageable linear or quadratic forms.
In plain language
Multiply the whole equation by the common denominator to remove fractions, then solve the remaining standard equation while checking for impossible answers where the denominator becomes zero.
Self-check
Solve for x: 3/x = 1/(x-2).
Show the answer
Multiply by x(x-2): 3(x-2) = x. Expand: 3x - 6 = x. 2x = 6, so x = 3.
- Higher
Scope: Higher · Higher tier in either Unit 3 tier · Unit 3 §2.2.8 higher. Printed pages 24: form, manipulate and solve by formula, quadratic equations of the form x 2 + bx + c = 0, or ax 2 + bx + c = 0
Open revision guidance
What to know
- Quadratic equations take the form ax^2 + bx + c = 0. The quadratic formula x = (-b ± sqrt(b^2 - 4ac)) / 2a is a universal method for finding the roots of these equations when factorization is not straightforward. The term b^2 - 4ac is called the discriminant. If the discriminant is positive, there are two distinct real roots. If it equals zero, there is one repeated root. If it is negative, there are no real roots. This formula is derived from completing the square.
- For example, consider the equation 2x^2 + 5x - 3 = 0. Here, a = 2, b = 5, and c = -3. First, calculate the discriminant: (5)^2 - 4(2)(-3) = 25 + 24 = 49. Because 49 is a positive square number, we expect two rational roots. Substitute into the formula: x = (-5 ± sqrt(49)) / (2 * 2). This yields x = (-5 + 7) / 4 = 0.5 and x = (-5 - 7) / 4 = -3. These roots solve the equation perfectly.
What to do in an assessment
- To solve equations like x^2 - 6x + 8 = 0, first identify the coefficients a = 1, b = -6, c = 8. Calculate the discriminant: (-6)^2 - 4(1)(8) = 36 - 32 = 4. Apply the formula: x = (6 ± sqrt(4)) / 2. This results in x = (6 + 2) / 2 = 4 and x = (6 - 2) / 2 = 2. Verify by substituting x = 4: (4)^2 - 6(4) + 8 = 16 - 24 + 8 = 0.
- When handling the form ax^2 + bx + c = 0 where a does not equal 1, ensure you account for the leading coefficient in the denominator. For 3x^2 + 7x + 2 = 0, use a = 3, b = 7, c = 2. The discriminant is 49 - 4(3)(2) = 49 - 24 = 25. The roots are x = (-7 ± 5) / 6. This gives x = -2/6 = -1/3 and x = -12/6 = -2. Always double check your arithmetic with negative signs when evaluating b^2 - 4ac.
In plain language
Solving quadratic equations involves finding the values of x that make the equation equal zero using a standard formula.
Self-check
Solve 4x^2 - 12x + 5 = 0 using the quadratic formula.
Show the answer
a=4, b=-12, c=5. Discriminant is (-12)^2 - 4(4)(5) = 144 - 80 = 64. x = (12 ± 8) / 8. Solutions are x = 2.5 and x = 0.5.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 3 tier · Unit 3 §2.2.9 foundation. Printed pages 24: solve a range of cubic equations by trial and improvement methods, justifying the accuracy of the solution
Open revision guidance
What to know
- Trial and improvement is a numerical method used to find solutions for complex equations like x^3 + x - 10 = 0, where direct algebraic methods are unavailable or difficult. You start by identifying two integers that bound the solution. By substituting these into the equation, you observe where the result changes from positive to negative, indicating a root exists between those two integers. This iterative process refines the estimate systematically.
- To improve accuracy, we look at midpoints. If testing x = 2 gives a result of 2^3 + 2 - 10 = 0 (too small) and x = 3 gives 3^3 + 3 - 10 = 20 (too large), the root is between 2 and 3. We try x = 2.1, 2.2, etc. As we move closer, the value of the function approaches zero. This method requires clear record-keeping of every step taken to ensure the final decimal value is verified.
What to do in an assessment
- Let us solve x^3 + 4x - 20 = 0. Try integers: x=2 gives 8+8-20 = -4. Try x=3 gives 27+12-20 = 19. The root is between 2 and 3. Now try midpoints: try x=2.3. (2.3)^3 + 4(2.3) - 20 = 12.167 + 9.2 - 20 = 1.367 (too high). The root is between 2.2 and 2.3. Try x=2.25. (2.25)^3 + 4(2.25) - 20 = 11.39 + 9 - 20 = 0.39 (still high). Keep refining until the required precision is reached.
- Justifying the solution involves showing the values for x just above and just below the target value. If 2.24 produces a negative result and 2.25 produces a positive result, we conclude the root lies in this interval. This confirms that the solution is accurate to a specific decimal place. Always maintain a table format to document trials clearly: Column A for x, Column B for x^3 + 4x - 20, and Column C for Status (Low/High).
- For x³+x = 10 with a positive root, trials give x=1.8 → 7.632, x=1.85 → 8.182125, x=1.9 → 8.759, x=1.95 → 9.364875, x=2 → 10. To settle rounding for a non-exact root, test the midpoint between adjacent rounded values, not just their endpoints, and keep the sign of each comparison with the target.
In plain language
Trial and improvement is a step by step approach to finding approximate answers to equations by testing numbers.
Self-check
Use trial and improvement to find the root of x^3 + x = 7 to 1 decimal place.
Show the answer
x=1 gives 2, x=2 gives 10. Root is between 1 and 2. Testing 1.8 gives 7.632 (high), 1.7 gives 6.613 (low). 1.75 gives 7.109 (high). The answer is 1.7.
- Higher
Scope: Higher · Higher tier in either Unit 3 tier · Unit 3 §2.2.10 higher. Printed pages 24: construct and use equations that describe direct and inverse proportion
Open revision guidance
What to know
- Proportion describes the relationship between two variables. Direct proportion means that as one variable increases, the other increases at a constant rate, represented as y = kx, where k is the constant of proportionality. For example, if y is directly proportional to x and y=10 when x=2, then 10 = k(2), so k=5. The equation is y = 5x. With this, we can predict values: if x=4, then y=5(4)=20.
- Inverse proportion means that as one variable increases, the other decreases, represented as y = k/x. Here, the product of the variables remains constant (xy = k). For instance, if y is inversely proportional to x and y=4 when x=3, then 4 = k/3, implying k=12. The equation is y = 12/x. If we need to find y when x=6, we calculate y = 12/6 = 2. These relationships are fundamental for modelling physical phenomena like speed, time, and distance.
What to do in an assessment
- To construct equations, read the problem to determine the type of variation. If a quantity 'y' is directly proportional to the square of 'x', write y = kx^2. Given coordinates (2, 20), substitute to find k: 20 = k(2^2) = 4k, so k=5. The finished model is y = 5x^2. Using this, calculate new coordinates; at x=3, y = 5(3^2) = 45. Point is (3, 45). Always state the value of k clearly as part of your solution process.
- For inverse square variation, y = k/x^2. If y=2 when x=4, then 2 = k/16, so k=32. The equation is y = 32/x^2. Test it by finding y at x=8: y = 32/64 = 0.5. These calculations allow you to map out curves on a graph. Ensure you use appropriate units if provided in the data, such as meters per second or kilograms, and maintain consistent dimensionality throughout your algebraic manipulations.
In plain language
Proportion tells us how two values change in relation to each other, either by growing together or one growing as the other shrinks.
Self-check
If y is inversely proportional to x, and y=6 when x=2, find y when x=4.
Show the answer
y = k/x. 6 = k/2, so k=12. Equation is y = 12/x. When x=4, y = 12/4 = 3.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §2.3.1 foundation. Printed pages 25: recognise, describe and continue patterns in number
Open revision guidance
What to know
- A number sequence is an ordered set of numbers defined by a specific rule. Recognising these patterns involves looking for the difference between successive terms. In arithmetic sequences, the change is constant; for example, in the set 4, 7, 10, 13, the difference is +3. Identifying this constant difference allows you to predict future terms by applying the same operation to the last known number.
- Patterns can also be geometric or multiplicative, where each term is generated by multiplying the previous term by a fixed number. For instance, in 2, 6, 18, 54, the rule is to multiply by 3. Understanding whether a sequence grows through addition or multiplication is vital. By observing the gaps, you define the growth logic which acts as the foundation for all sequence analysis, whether the progression is linear or non-linear.
What to do in an assessment
- To continue a sequence like 5, 11, 17, 23, first calculate the difference between terms: 11 - 5 = 6 and 17 - 11 = 6. Since the difference is a consistent +6, add 6 to the final term. Thus, 23 + 6 = 29, and 29 + 6 = 35. Always verify the difference across at least three terms to ensure the pattern is stable before calculating subsequent values.
- For multiplicative sequences such as 1, 4, 16, 64, calculate the ratio between terms: 4 divided by 1 is 4, and 16 divided by 4 is 4. Since the multiplier is 4, multiply the last term 64 by 4 to get 256. Continue by multiplying 256 by 4 to yield 1024. Writing down the intermediate steps for each jump prevents calculation errors when dealing with larger numbers or rapidly increasing sequences.
In plain language
Number patterns are lists where each new number follows a set rule. You find the rule by checking how much you add or multiply to get from one number to the next.
Self-check
Find the next two terms in the sequence: 7, 12, 17, 22.
Show the answer
The rule is to add 5 each time. 22 + 5 = 27, and 27 + 5 = 32. The terms are 27 and 32.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §2.3.2 foundation. Printed pages 25: describe, in words and symbols, the rule for the next term of a sequence
Open revision guidance
What to know
- Describing a sequence rule involves identifying the relationship between the position of a term (n) and the value of the term itself. A word-based description explains the logic, such as 'add 4 to the previous term'. A symbolic rule provides a formula where 'n' represents the position in the sequence (e.g., 1st, 2nd, 3rd term). This allows for the calculation of any term without listing all preceding values.
- Rules for linear sequences usually take the form an + b, where 'a' is the constant difference between terms and 'b' is the value that adjusts the first term. If a sequence is 5, 8, 11, the difference is 3, so the rule starts with 3n. Since the first term (n=1) is 5, we need to add 2 to 3(1). The resulting rule is 3n + 2. Symbols provide a precise mathematical shorthand for complex verbal descriptions.
What to do in an assessment
- To describe the rule for the sequence 7, 10, 13, 16, first identify the difference between terms: 10 - 7 = 3. In words, the rule is to start with 7 and add 3 for each step. In symbols, use the structure 3n + b. For n = 1, 3(1) + b = 7, which means b = 4. The rule is 3n + 4. Always substitute n = 2 (3(2)+4 = 10) to confirm the formula aligns with the given sequence.
- When describing a sequence that decreases, like 20, 17, 14, 11, note the negative difference: 17 - 20 = -3. The rule in words is to start with 20 and subtract 3 each time. In symbols, the rule is -3n + b. For n = 1, -3(1) + b = 20, so b = 23. The symbolic rule is -3n + 23. Testing with n = 3: -3(3) + 23 = -9 + 23 = 14, which confirms the rule is correct.
In plain language
You can explain sequence rules using words or math symbols. A word rule tells you what to do to the last number, while a symbol rule lets you find any number in the list instantly using its position number (n).
Self-check
Find the symbolic rule for the sequence: 4, 9, 14, 19.
Show the answer
The difference is 5, so the rule starts with 5n. For n = 1, 5(1) + b = 4, so b = -1. The rule is 5n - 1.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §2.3.3 foundation. Printed pages 25: generate linear and non-linear sequences given the nth term rule
Open revision guidance
What to know
- Generating a sequence means calculating individual terms using a provided nth term rule. Given a rule such as 2n + 3, you substitute values of n starting from n = 1 to find the terms of the sequence. This process transforms a static formula into a dynamic list of values. It is a fundamental skill that connects algebraic rules to numerical progressions, helping to visualise how variables dictate growth.
- For non-linear sequences, the rule will involve n raised to a power, such as n^2 + 1. In these cases, the difference between consecutive terms is not constant, which signifies quadratic or higher-order growth. By substituting integer values for n, you compute the sequence step-by-step. Ensuring accurate arithmetic during these substitutions is critical, as a single error in one term affects the perceived pattern of the entire sequence.
What to do in an assessment
- To generate the first four terms of the rule 4n - 2, substitute n = 1, 2, 3, 4 sequentially: For n=1: 4(1) - 2 = 2. For n=2: 4(2) - 2 = 6. For n=3: 4(3) - 2 = 10. For n=4: 4(4) - 2 = 14. The sequence is 2, 6, 10, 14. Always use brackets during substitution to maintain clarity and avoid calculation errors, especially when the rule includes negative coefficients.
- For a non-linear rule like n^2 + n, generate the first three terms as follows: For n=1: (1)^2 + 1 = 1 + 1 = 2. For n=2: (2)^2 + 2 = 4 + 2 = 6. For n=3: (3)^2 + 3 = 9 + 3 = 12. The sequence is 2, 6, 12. The differences here are 4 and 6, showing the non-constant change characteristic of quadratic rules. Double-check squares before adding.
In plain language
Generating a sequence means finding the list of numbers created by a math rule. You put the position number (n=1, 2, 3...) into the rule and solve it for each step.
Self-check
Generate the first three terms of the sequence given by 3n^2 - 1.
Show the answer
n=1: 3(1)-1=2; n=2: 3(4)-1=11; n=3: 3(9)-1=26. The terms are 2, 11, 26.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §2.3.4 foundation. Printed pages 25: find the nth term of a sequence, given numerically or diagrammatically, where the rule is linear
Open revision guidance
What to know
- Finding the nth term rule for a linear sequence involves two steps: identifying the common difference and finding the adjustment constant. The common difference represents the 'a' in the standard form an + b. Because it is a linear sequence, the difference between any two consecutive terms remains constant. This constant difference acts as the multiplier for the position variable n.
- Once the difference 'a' is determined, you must find 'b', the value needed to reach the first term when n = 1. By setting up the equation a(1) + b = First Term, you solve for b. This algebraic approach works for any linear sequence, whether presented as a list of numbers or a diagrammatic pattern of objects like squares or circles. Mastery of this allows for the prediction of distant terms in a sequence without manual counting.
What to do in an assessment
- Given the sequence 8, 13, 18, 23, find the difference: 13 - 8 = 5. So, the rule is 5n + b. Substitute n = 1 to find b: 5(1) + b = 8, which gives 5 + b = 8, so b = 3. The final rule is 5n + 3. Always check this rule against another term, such as n = 3: 5(3) + 3 = 15 + 3 = 18. This matches the third term in the sequence, confirming the rule is accurate.
- For a diagrammatic pattern where Figure 1 has 4 dots, Figure 2 has 7 dots, and Figure 3 has 10 dots: The sequence is 4, 7, 10. The difference is 3. Start with 3n + b. For n = 1, 3(1) + b = 4, so 3 + b = 4, b = 1. The nth term rule is 3n + 1. Verify with n = 2: 3(2) + 1 = 7 dots, which matches Figure 2.
In plain language
To find the rule for a linear sequence, find the constant gap between numbers (this is the number next to n) and then find the correction needed to make it match the first number when n is 1.
Self-check
Find the nth term for the sequence: 5, 8, 11, 14.
Show the answer
The gap is 3, so it starts with 3n. For n=1, 3(1)+b=5, so b=2. The rule is 3n + 2.
- Higher
Scope: Higher · Higher tier in either Unit 2 tier · Unit 2 §2.3.5 higher. Printed pages 25: find the nth term of a sequence, given numerically or diagrammatically, where the rule is quadratic
Open revision guidance
What to know
- A quadratic sequence follows an nth term rule in the form an^2 + bn + c. Unlike linear sequences where the first difference is constant, quadratic sequences have a constant second difference. To identify the rule, first find the differences between terms. Then, calculate the differences between those differences (the second differences). The coefficient 'a' of n^2 is always half of the second difference.
- Once 'a' is identified, you subtract the sequence generated by an^2 from the original sequence. The remaining numbers form a new linear sequence, which you solve for the 'bn + c' part of the rule. This systematic decomposition allows you to build the complex quadratic formula from the observed numbers. Accuracy in calculating the second differences is vital, as any miscalculation will cascade into the linear part of the expression.
What to do in an assessment
- For the sequence 4, 11, 20, 31: Differences are 7, 9, 11. Second differences are 9-7=2 and 11-9=2. Since the second difference is 2, a = 2/2 = 1. The rule starts with 1n^2. Subtract 1n^2 from the sequence: n=1: 1^2=1 (4-1=3); n=2: 2^2=4 (11-4=7); n=3: 3^2=9 (20-9=11). The new sequence is 3, 7, 11, which has a difference of 4. So bn+c is 4n-1. The rule is n^2 + 4n - 1.
- Check the rule n^2 + 4n - 1 for n=4: (4)^2 + 4(4) - 1 = 16 + 16 - 1 = 31. This matches the fourth term in the sequence. By isolating the quadratic portion first, we simplify the problem into a standard linear sequence task, ensuring robust results for all terms within the quadratic pattern.
In plain language
Quadratic sequences have a changing gap that also grows at a steady rate. You find the rule by checking the 'second difference', which helps define the n^2 part of the formula, then solving for the remaining linear part.
Self-check
Find the nth term for 3, 8, 15, 24.
Show the answer
Diffs: 5, 7, 9. 2nd diff: 2. a=1. Seq minus n^2: 3-1=2, 8-4=4, 15-9=6. Linear is 2n. Rule is n^2 + 2n.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §2.4.1 foundation. Printed pages 25: use coordinates in 4 quadrants
Open revision guidance
What to know
- The Cartesian coordinate plane is divided into four quadrants by the horizontal x-axis and the vertical y-axis, which intersect at the origin (0, 0). Coordinates are expressed as ordered pairs (x, y), where the x-value indicates the horizontal displacement from the y-axis, and the y-value indicates the vertical displacement from the x-axis. Moving right increases x, and moving up increases y.
- In Quadrant I, both x and y are positive. In Quadrant II, x is negative but y is positive. In Quadrant III, both x and y are negative. In Quadrant IV, x is positive but y is negative. Locating points accurately is essential for graphical work, ensuring that signed values are plotted correctly relative to the origin.
What to do in an assessment
- To plot a coordinate, start at the origin (0, 0). For a point (x, y), move x units horizontally (right if positive, left if negative) then move y units vertically (up if positive, down if negative). For example, to plot A(-2, 3), move 2 units left from the origin and 3 units up.
- To identify a point's coordinates, determine how many units along the horizontal axis you must move to align with the point, then count the units required to reach it vertically. For a point in the third quadrant, say B(-4, -1), confirm the position is 4 units left of the y-axis and 1 unit below the x-axis.
In plain language
A coordinate system is like a grid map. You use pairs of numbers (x, y) to find a specific spot by moving sideways first and then up or down.
Self-check
Plot the point P(3, -2) and state its quadrant.
Show the answer
Start at (0,0), move 3 units right, then 2 units down. It is in Quadrant IV.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §2.4.2 foundation. Printed pages 25: draw, interpret, recognise and sketch the graphs of x = a, y = b, y = ax + b
Open revision guidance
What to know
- Equations of the form x = a represent vertical lines passing through a on the x-axis, where all points on the line have the same x-coordinate. Equations of the form y = b represent horizontal lines passing through b on the y-axis, where all points have the same y-coordinate. These lines are special cases of linear functions where the gradient is either undefined or zero.
- Graphs of the form y = ax + b are linear. Drawing these involves creating a table of values for x, calculating corresponding y-values, and plotting them. For example, y = 2x + 1: if x = 0, y = 1; if x = 1, y = 3; if x = 2, y = 5. Connecting these points (0, 1), (1, 3), and (2, 5) produces a straight line.
What to do in an assessment
- To sketch x = a, draw a vertical line at the mark 'a' on the x-axis. To sketch y = b, draw a horizontal line at the mark 'b' on the y-axis. Ensure the line extends beyond the plotted points to represent an infinite line.
- To sketch y = ax + b, select three x-values (e.g., -1, 0, 1), substitute them into the equation to find y, and plot these points. Use a ruler to draw a straight line through them. If x = 2 and the equation is y = 3x - 2, substitute 2 to find y = 3(2) - 2 = 4. The point (2, 4) must lie on the line.
In plain language
Vertical and horizontal lines are simple straight lines. For more complex lines (y = ax + b), pick a few x numbers, find their y matches, and draw the line that passes through them.
Self-check
Sketch the line y = 2x - 1. What is the y-value when x = 3?
Show the answer
Calculate y = 2(3) - 1 = 5. The line passes through (0,-1), (1,1), and (2,3).
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §2.4.3 foundation. Printed pages 25: know and use the form y = mx + c to represent a straight line where m is the gradient of the line, and c is the value of the y-intercept
Open revision guidance
What to know
- The equation y = mx + c is the standard form for a straight line where m represents the gradient (steepness) and c represents the y-intercept (the point where the line crosses the y-axis). The gradient m is defined as the change in y divided by the change in x. A positive m means the line slopes upward to the right, while a negative m means it slopes downward to the right.
- The y-intercept c is found at coordinate (0, c). By knowing these two values, one can define any straight line uniquely. If a line passes through (1, 5) and (3, 11), the gradient is (11 - 5) / (3 - 1) = 6 / 2 = 3. Using y = 3x + c and point (1, 5), we solve 5 = 3(1) + c, giving c = 2. Thus, the equation is y = 3x + 2.
What to do in an assessment
- To identify m and c from an equation, ensure it is written in the form y = mx + c. For example, in y = -4x + 7, m = -4 and c = 7. If given 2y = 4x + 6, divide the entire equation by 2 to get y = 2x + 3, so m = 2 and c = 3.
- To find the equation from a graph, identify the y-intercept (0, c) to determine c. Then pick another point on the line (x1, y1) and calculate the gradient using m = (y1 - c) / (x1 - 0). Once m and c are known, write the full equation y = mx + c.
In plain language
This equation tells you how steep the line is (m) and where it hits the vertical y-axis (c). It is the easiest way to write the rule for a straight line.
Self-check
Find m and c for the line 3y = 9x - 12.
Show the answer
Divide by 3: y = 3x - 4. So, m = 3 and c = -4.
View diagrams
Two simultaneous linear equations
Key and description
Constructed example: x is the number of £1 items and y is the number of £2 items. There are 5 items, so x + y = 5 and y = 5 − x. There is one more £2 item than £1 items, so y = x + 1. The lines meet at (2, 3): 2 one-pound items and 3 two-pound items. The cost relation is C = x + 2y, giving C = 2 + 6 = £8.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §2.4.4 foundation. Printed pages 25: draw and interpret quadratic graphs of the form y = ax 2 + bx + c, and draw the line y = k in order to solve ax 2 + bx + c = k
Open revision guidance
What to know
- A quadratic graph has the form y = ax² + bx + c. Unlike linear graphs, these are curved (parabolas). If a is positive, the graph has a minimum point and looks like a 'U' shape. If a is negative, it has a maximum point and looks like an inverted 'U'. The y-intercept is simply the value c, where the curve crosses the y-axis (when x = 0).
- Solving ax² + bx + c = k involves finding the x-coordinates where the quadratic curve intersects the horizontal line y = k. By drawing both the quadratic curve and the line y = k on the same axes, the intersection points reveal the solutions. If y = x² - 4, and we solve x² - 4 = 0, we look for where the curve hits the x-axis.
What to do in an assessment
- To draw y = ax² + bx + c, build a table of x values (e.g., from -3 to 3), calculate the corresponding y values by substituting into the quadratic, and plot the points. For example, if y = x² - 2: if x = 1, y = 1² - 2 = -1; if x = 2, y = 2² - 2 = 2. Smoothly connect the points.
- To solve ax² + bx + c = k, first draw the curve y = ax² + bx + c. Then, draw the straight line y = k across the grid. The points where the curve and the line cross provide the x-values that satisfy the equation. If the line is y = 3 and the curve is y = x², the solutions are found where x² = 3, which is approximately ±1.73.
- In the labelled example y = x²−2x−3 = (x−3)(x+1), the roots are −1 and 3, the y-intercept is −3, and the turning point is (1,−4). The graph opens upwards. The function is negative between its roots, zero at the roots and positive outside them. These are interpretations of the completed graph, not just plotting instructions.
In plain language
Quadratic graphs are U-shaped curves. To solve an equation like ax² + bx + c = k, you draw the curve and a flat line at height k. Where they meet, the x-values are your answers.
Self-check
For y = x² - 1, calculate y when x = 2.
Show the answer
y = 2² - 1 = 4 - 1 = 3. Point is (2, 3).
View diagrams
Roots and turning point of a quadratic
Key and description
y = x² − 2x − 3 = (x − 3)(x + 1). Roots: (−1, 0) and (3, 0). Completing the square gives (x − 1)² − 4, so the turning point is (1, −4).
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Higher
Scope: Higher · Higher tier in either Unit 2 tier · Unit 2 §2.4.5 higher. Printed pages 26: use straight line graphs to locate regions given by inequalities
Open revision guidance
What to know
- Straight line graphs define boundaries on a coordinate plane. An inequality like y > 2x + 1 indicates a region rather than just a line. The line y = 2x + 1 acts as a boundary; points above the line satisfy the 'greater than' condition. If the inequality includes 'or equal to' (≥ or ≤), the boundary line is solid; if it is strict (> or <), the line is dashed.
- To identify a region, consider a test point not on the line, such as (0, 0). If (0, 0) satisfies the inequality, the region containing the origin is the solution. For instance, testing 0 > 2(0) + 1 results in 0 > 1, which is false. Therefore, the region that does not include the origin is the correct solution space for y > 2x + 1.
What to do in an assessment
- First, draw the boundary line by plotting y = mx + c. Use a dashed line for strict inequalities and a solid line for inclusive ones. For 3x + y < 4, draw the line y = -3x + 4 as a dashed line.
- Second, determine which side to shade. Choose a test point like (1, 1). If 3(1) + 1 < 4, which is 4 < 4, it is false. So, (1, 1) is not in the region. Shade the side of the line that does not contain (1, 1). Always ensure the region is clearly demarcated by the axes or other given boundaries if specified.
In plain language
An inequality shows a whole area on a graph instead of just a line. You draw the boundary line and then shade the part of the graph that makes the statement true.
Self-check
For y < x, is the point (2, 1) in the shaded region?
Show the answer
Yes. 1 < 2 is true. The point (2, 1) is in the valid region.
- Higher
Scope: Higher · Higher tier in either Unit 2 tier · Unit 2 §2.4.6 higher. Printed pages 26: identify the equations of lines parallel or perpendicular to a given line
Open revision guidance
What to know
- Lines are parallel if they have the same gradient (m). For example, y = 2x + 3 and y = 2x - 5 are parallel because both have m = 2. Regardless of the y-intercept, parallel lines will never intersect. This is a fundamental property of linear equations that share the same slope in the Cartesian plane.
- Lines are perpendicular if the product of their gradients is -1. If a line has gradient m, the perpendicular line has a gradient of -1/m (the negative reciprocal). For instance, if a line is y = 3x + 1, a perpendicular line would have a gradient of -1/3. The equation could be y = -1/3x + 4. Perpendicular lines meet at a right angle.
What to do in an assessment
- To find a parallel line equation, keep the m value the same. If you need a line parallel to y = 4x + 2 passing through (0, 5), the gradient remains 4 and the new intercept is 5, so y = 4x + 5.
- To find a perpendicular line, find the negative reciprocal of the gradient. For line y = -2x + 1, the gradient is -2, so the perpendicular gradient is -1/(-2) = 0.5. To find the full equation, use y = 0.5x + c and substitute a known point. If it must pass through (2, 3), 3 = 0.5(2) + c gives 3 = 1 + c, so c = 2. The equation is y = 0.5x + 2.
In plain language
Parallel lines have the same steepness (gradient) so they never touch. Perpendicular lines meet at a 90-degree angle; their steepness values are negative reciprocals of each other (like 2 and -1/2).
Self-check
What is the gradient of a line perpendicular to y = 5x - 3?
Show the answer
The negative reciprocal of 5 is -1/5.
- Higher
Scope: Higher · Higher tier in either Unit 2 tier · Unit 2 §2.4.7 higher. Printed pages 26: form, manipulate and solve two simultaneous linear equations with whole number coefficients by graphical methods
Open revision guidance
What to know
- Simultaneous equations are two or more algebraic equations that share the same variables. Graphically, solving these means finding the coordinate (x, y) where the two lines intersect. Since both equations must be true at the same time, the intersection point is the unique solution where x and y values satisfy both original linear equations.
- To solve 2x + y = 6 and x - y = 3, we rearrange them into y = mx + c form: y = -2x + 6 and y = x - 3. Plotting these two lines, we observe they cross at a point. Calculating the intersection shows the values of x and y that solve the system. This method visually confirms the solution for consistent linear equations.
What to do in an assessment
- Rearrange both equations into y = mx + c format. For 2x + y = 6, subtract 2x to get y = -2x + 6. For x - y = 3, add y then subtract 3 to get y = x - 3.
- Plot both lines on the same set of axes. The intersection point (x, y) is the answer. If they cross at (3, 0), then x = 3 and y = 0. Check the result by substituting these values into the original equations: 2(3) + 0 = 6 and 3 - 0 = 3. Both are true, so (3, 0) is correct.
- Form the equations from the situation before drawing. Let x be the number of £1 items and y the number of £2 items. Five items gives x+y=5; one more £2 item than £1 item gives y=x+1. The plotted lines y=5−x and y=x+1 meet at (2,3), meaning two £1 items and three £2 items. Their cost is 2+2×3 = £8.
In plain language
To solve two equations at once, draw them on the same graph. The point where the two lines cross gives the values for x and y that work for both equations.
Self-check
Solve y = x + 1 and y = -x + 3 graphically.
Show the answer
They intersect at (1, 2). Check: 2 = 1+1 and 2 = -1+3. Both true.
View diagrams
Two simultaneous linear equations
Key and description
Constructed example: x is the number of £1 items and y is the number of £2 items. There are 5 items, so x + y = 5 and y = 5 − x. There is one more £2 item than £1 items, so y = x + 1. The lines meet at (2, 3): 2 one-pound items and 3 two-pound items. The cost relation is C = x + 2y, giving C = 2 + 6 = £8.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Higher
Scope: Higher · Higher tier in either Unit 2 tier · Unit 2 §2.4.8 higher. Printed pages 26: draw, interpret, recognise and sketch the graphs of y = ax 2 + b, y = (ax + b)(cx + d), y = a/x, y = ax 3
Open revision guidance
What to know
- Quadratic graphs follow the form y = ax^2 + b or y = (ax + b)(cx + d). For y = 2x^2 + 3, the graph is a parabola opening upwards with the vertex at (0, 3). For factored forms like y = (x - 2)(x + 4), the roots occur where y = 0, specifically at x = 2 and x = -4. These roots define the x-intercepts of the parabolic curve.
- Reciprocal graphs take the form y = a/x. When a is positive, the branches occupy the first and third quadrants. For y = 6/x, as x approaches zero from the right, y increases towards infinity. Cubic graphs y = ax^3 represent smooth curves passing through the origin. For y = 2x^3, when x = 1, y = 2; when x = -1, y = -2, showcasing point symmetry about the origin (0, 0).
What to do in an assessment
- To sketch y = 2x^2 - 8, calculate key points. Let x = 0, then y = -8. Set y = 0 to find roots: 2x^2 = 8, x^2 = 4, so x = 2 and x = -2. Plot points (-2, 0), (2, 0), and (0, -8). Draw a smooth parabola connecting these points through the vertex at (0, -8).
- For the reciprocal graph y = 4/x, generate a table of values: x = 1, y = 4; x = 2, y = 2; x = 4, y = 1; x = -1, y = -4; x = -2, y = -2; x = -4, y = -1. Plot these six coordinate pairs on a Cartesian plane. Draw two continuous curves that approach the x-axis and y-axis as asymptotes without ever touching them.
In plain language
We learn to draw and interpret different types of curves. Quadratics create U-shaped parabolas, reciprocals create split curves that get close to axes but never touch them, and cubics form S-shaped curves.
Self-check
For the graph y = x^2 - 9, what are the roots?
Show the answer
Set x^2 - 9 = 0. Then x^2 = 9, so x = 3 and x = -3.
- Higher
Scope: Higher · Higher tier in either Unit 2 tier · Unit 2 §2.4.9 higher. Printed pages 26: draw and interpret graphs of the form y = ax 3 + b and y = ax 3 + bx 2 + cx + d
Open revision guidance
What to know
- Cubic functions take the general form y = ax^3 + b or the expanded polynomial form y = ax^3 + bx^2 + cx + d. The term ax^3 dictates the long-term behaviour of the graph. When a is positive, the graph starts in the bottom-left and trends towards the top-right, crossing the x-axis at least once.
- The constants b, c, and d influence the vertical position, curvature, and intercepts. For example, in y = x^3 - x, the graph crosses the x-axis at x = 0, x = 1, and x = -1. Unlike quadratic curves, cubic curves can exhibit two turning points, a local maximum and a local minimum, creating a characteristic S-bend shape.
What to do in an assessment
- To draw y = x^3 - 2x^2, first find the roots by factoring: x^2(x - 2) = 0. The roots are x = 0 (repeated) and x = 2. Create a coordinate table: x = -1, y = -3; x = 0, y = 0; x = 1, y = -1; x = 2, y = 0; x = 3, y = 9. Connect these points with a smooth, continuous line ensuring the curve touches the x-axis at the origin.
- For y = x^3 + 1, identify the y-intercept at (0, 1) and the root at x = -1. Evaluate points: x = -2, y = -7; x = 0, y = 1; x = 1, y = 2. Plotting these confirms the curve's steep growth for positive x values and steep descent for negative x values. Ensure the inflection point is drawn clearly.
In plain language
Cubic graphs involve x^3 terms. They create smooth curves that can wiggle. They are drawn by plotting several points to determine the shape, especially focusing on where they cross the x-axis (roots).
Self-check
For y = x^3 - 8, find the root.
Show the answer
Set x^3 - 8 = 0, so x^3 = 8. The cube root of 8 is 2, so x = 2.
- Higher
Scope: Higher · Higher tier in either Unit 2 tier · Unit 2 §2.4.10 higher. Printed pages 26: use a graphical method to solve ax 2 + bx + c = dx + e and ax 3 + bx 2 + cx + d = ex + f
Open revision guidance
What to know
- Graphical solutions for equations like ax^2 + bx + c = dx + e involve identifying the intersection points of the two separate curves y = ax^2 + bx + c and y = dx + e. The x-coordinates of these points are the solutions to the equation. A quadratic and a linear graph can intersect at zero, one, or two points.
- Similarly, for ax^3 + bx^2 + cx + d = ex + f, we plot the cubic y = ax^3 + bx^2 + cx + d and the line y = ex + f. The intersection points provide the values of x that satisfy the equality. If the line is tangent to the curve, there is a repeated solution, meaning the graphs touch at exactly one specific coordinate point.
What to do in an assessment
- Solve x^2 - 4 = x - 2 graphically. Plot y = x^2 - 4 (parabola) and y = x - 2 (line). For the parabola, points are (-2, 0), (0, -4), (2, 0). For the line, points are (0, -2) and (2, 0). The graphs intersect at (2, 0) and (-1, -3). The solutions are x = 2 and x = -1.
- Solve x^3 = 2x + 1 graphically. Plot y = x^3 and y = 2x + 1. For y = x^3, points: (-1, -1), (0, 0), (1, 1), (2, 8). For y = 2x + 1, points: (-1, -1), (0, 1), (1, 3). The point (-1, -1) is an intersection. Using a graph, another intersection exists near x = 1.6.
In plain language
To solve an equation graphically, you draw two separate graphs and see where they cross. The x-value at each crossing point is the answer to the equation.
Self-check
If the intersection of x^2 and x + 2 is (2, 4) and (-1, 1), what are the solutions?
Show the answer
The solutions are the x-coordinates: x = 2 and x = -1.
- Higher
Scope: Higher · Higher tier in either Unit 2 tier · Unit 2 §2.4.11 higher. Printed pages 26: draw and interpret graphs when y is given implicitly in terms of x
Open revision guidance
What to know
- Implicit graphs occur when y is not isolated, often in the form f(x, y) = 0. A classic example is the circle equation x^2 + y^2 = r^2, where r is the radius. For x^2 + y^2 = 25, the graph represents a circle centred at (0, 0) with a radius of 5 units.
- Unlike standard functions where every x yields one y, implicit graphs may have multiple y-values for a single x-value. For instance, in x^2 + y^2 = 25, if x = 3, y^2 = 16, resulting in y = 4 or y = -4. Drawing these requires checking both positive and negative roots for y across the defined range of x.
What to do in an assessment
- Draw the graph of x^2 + y^2 = 16. Recognise this as a circle centred at origin (0, 0) with radius = 4. Plot points on axes: (4, 0), (-4, 0), (0, 4), (0, -4). Use a compass or careful sketching to create a circle passing through these four coordinate points.
- To sketch x^2 + y^2 = 9, identify the radius as 3. Evaluate points where x=0, y=+/-3 and y=0, x=+/-3. Additionally, for x=2, y^2 = 9 - 4 = 5, so y = +/-2.24. Plot the points (0, 3), (0, -3), (3, 0), (-3, 0), (2, 2.24), and (2, -2.24) to construct the circular shape.
In plain language
Implicit graphs show relationships where x and y are mixed together. An example is a circle, where the x-values and y-values squared add up to a fixed number. They can have multiple y-values for one x-value.
Self-check
What is the radius of x^2 + y^2 = 100?
Show the answer
The equation is x^2 + y^2 = r^2. Since r^2 = 100, r = 10 units.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier · Unit 1 §2.5.1 foundation. Printed pages 27: draw and interpret the following graphs: Conversion graphs Travel graphs Other graphs that describe real-life situations.
Open revision guidance
What to know
- Conversion graphs represent the direct relationship between two different units. For example, to convert miles to kilometres, we use the conversion factor where 5 miles equals 8 kilometres. A conversion graph is a straight-line graph passing through the origin, which allows us to find values by reading across from one axis and down to another. If 10 miles is 16 kilometres, the point (10, 16) lies on the line starting at (0, 0).
- Travel graphs, often called distance-time graphs, display how distance from a starting point changes over time. The slope of the line represents speed, calculated as change in distance divided by change in time. For instance, if a car travels 60 kilometres in 1.5 hours, its average speed is 60 divided by 1.5, resulting in 40 kilometres per hour. Stationary periods are represented by horizontal lines where the distance remains constant.
What to do in an assessment
- To construct a conversion graph, create a table of values. If 1 British Pound equals 1.25 Euros, mark (0, 0), (10, 12.5), and (20, 25) on a coordinate plane with Pounds on the x-axis and Euros on the y-axis. Draw a straight line through these points. To use it, find 15 on the x-axis, move vertically to the line, then horizontally to the y-axis to read approximately 18.75 Euros.
- To interpret a travel graph, identify segments. Suppose a person walks 4 kilometres in 1 hour (segment A), rests for 0.5 hours (segment B, horizontal), and walks back 4 kilometres in 1 hour (segment C). The gradient of segment A is 4/1 = 4 km/h. The gradient of segment C is -4/1 = -4 km/h, representing the return journey speed. The total distance covered is 8 kilometres over 2.5 hours.
In plain language
Graphs help us visualise relationships like currency conversion or journey progress by mapping values as coordinates on a grid and using slopes to explain rates of change.
Self-check
A car travels at a constant speed of 30 metres per second for 10 seconds. What is the total distance covered?
Show the answer
Distance = speed multiplied by time. 30 metres per second multiplied by 10 seconds equals 300 metres.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier · Unit 1 §2.5.2 foundation. Printed pages 27: interpret graphical representation used in the media, recognising that some graphs may be misleading
Open revision guidance
What to know
- Graphs in the media are designed to convey information quickly, but they can be misleading. A common trick is manipulating the scale of the vertical axis. If an axis does not start at zero, small differences between data points appear much larger than they truly are. This exaggerates the visual impact of data trends, potentially causing viewers to draw incorrect conclusions about the scale of change.
- Another method used to mislead involves the use of 3D objects to represent 1D data. When an object like a bar is replaced with a 3D volume, doubling the height of the object also multiplies its width and depth. Consequently, the volume increases by a factor of 8 rather than 2, making the data appear vastly more significant than the actual values suggest. Always verify the scale, labels, and baseline of every chart.
What to do in an assessment
- When analysing media graphs, first check the vertical axis. If a bar chart comparing sales shows 100 units versus 110 units but starts the y-axis at 90, the difference looks huge, but it is only a 10 percent increase. Examine the scale increments. Are they uniform? If the intervals between markers are inconsistent, the representation is distorted. Always look for the actual numbers written on the graph.
- Check for visual inconsistencies. If a pictogram uses icons of different sizes to show growth, be aware that area or volume changes exponentially compared to height. For a data set where Year 1 is 5 units and Year 2 is 10 units, a graphic representing Year 2 that is twice as tall and twice as wide actually looks four times larger. Always rely on the raw numerical data rather than the visual size.
In plain language
Media graphs can sometimes distort facts by changing scales or using misleading visuals. Always look at the axis numbers and base values rather than just the visual shape.
Self-check
A bar chart shows two bars. Bar A is 2cm high representing 50 visitors. Bar B is 4cm high representing 55 visitors. Why might this chart be considered misleading?
Show the answer
The chart doubles the height for a very small numerical increase of 5 visitors. Because the axis likely does not start at 0, it creates a false sense of a large difference.
- Higher
Scope: Higher · Higher tier in either Unit 3 tier · Unit 3 §2.5.3 higher. Printed pages 27: construct and use tangents to curves to estimate rates of change for non-linear functions, and use appropriate compound measures to express results, including finding velocity in distance-time graphs and acceleration in velocity-time graphs
Open revision guidance
What to know
- The gradient of a tangent gives an instantaneous rate of change at its point of contact with a curve. A secant through two points on the curve gives the average rate across that interval instead. Draw a tangent that follows the curve's local direction; it is not necessarily a line that never crosses the curve elsewhere.
- On a distance–time graph the tangent's gradient is instantaneous speed. On a displacement–time graph it is velocity, and on a velocity–time graph it is acceleration. Use the axis labels and units to identify the rate.
What to do in an assessment
- A tangent at t=5 s on a distance–time curve passes through (4,10) and (6,20), with distance in metres. Use these two points on the tangent, not two arbitrary curve points: gradient = (20−10)/(6−4) = 5 m/s.
- For v=t², a tangent at (2,4) has equation v=4t−4. Two points on this tangent are (1,0) and (3,8), so instantaneous acceleration = (8−0)/(3−1) = 4 m/s². By contrast, the secant through curve points (2,4) and (4,16) gives average acceleration (16−4)/(4−2) = 6 m/s² across that interval.
In plain language
For a rate at one instant, read two clear points on the tangent. For an average rate, use two points on the curve.
Self-check
A tangent on a velocity–time graph passes through (2 s,6 m/s) and (5 s,12 m/s). What acceleration does it represent?
Show the answer
(12−6)/(5−2) = 6/3 = 2 m/s².
- Higher
Scope: Higher · Higher tier in either Unit 3 tier · Unit 3 §2.5.4 higher. Printed pages 27: interpret the meaning of the area under a graph, including the area under velocity-time graphs and graphs in other practical contexts
Open revision guidance
What to know
- Area under a rate graph gives the accumulated quantity: flow in litres per second × time in seconds gives litres. For a velocity–time graph, signed area gives displacement. If velocity is non-negative throughout, this also equals distance travelled. For negative velocity, add the magnitudes of areas to find total distance instead.
- Split a piecewise-straight graph into rectangles, triangles and trapeziums. Rectangle area = base × height; triangle area = ½ × base × height; trapezium area = ½ × (parallel sides added) × perpendicular separation. Check the units by multiplying the units on the two axes.
What to do in an assessment
- Velocity rises uniformly from 0 to 10 m/s in 5 s. The graph forms a triangle with base 5 s and height 10 m/s. Displacement = ½ × 5 × 10 = 25 m. Velocity stays non-negative, so distance is also 25 m.
- Velocity decreases uniformly from 8 to 4 m/s over 4 s. The area is a trapezium: ½ × (8 + 4) × 4 = 24 m. A pump delivers 3 litres/s for 20 s: the rectangle under its rate–time graph has area 3 × 20 = 60 litres.
In plain language
The area combines a rate with how long it operates. Decide which quantity and units the area represents before calculating.
Self-check
A vehicle travels at +10 m/s for 6 s, then −5 m/s for 2 s. Find its displacement and total distance.
Show the answer
Displacement = 10 × 6 + (−5) × 2 = 50 m. Total distance = 60 + 10 = 70 m.
View diagrams
Area under a velocity-time graph
Key and description
Velocity rises uniformly from 0 to 10 m/s over 5 s. The shaded triangle has base 5 s and height 10 m/s: displacement = ½ × 5 × 10 = 25 m. Because velocity is non-negative, distance travelled is also 25 m.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Higher
Scope: Higher · Higher tier in either Unit 3 tier · Unit 3 §2.5.5 higher. Printed pages 27: use the trapezium rule to estimate the area under a curve
Open revision guidance
What to know
- Divide an interval into equal-width strips. Replace each piece of the curve by a straight segment, so the area below it is a trapezium. Add these areas to estimate the total. The result is approximate, not the exact curved area.
- For n strips of width h and ordinates y₀, y₁, …, yₙ, estimate area by ½h[y₀ + 2(y₁ + … + yₙ₋₁) + yₙ]. The two end values appear once; each interior value appears twice. Smaller strips generally improve the estimate for a smooth curve.
What to do in an assessment
- For y = x² from x = 0 to 2 using four strips, h = 0.5. Values at x = 0, 0.5, 1, 1.5, 2 are 0, 0.25, 1, 2.25, 4. The estimate is 0.25[0 + 2(0.25 + 1 + 2.25) + 4] = 0.25 × 11 = 2.75 square units.
- Using eight strips, h = 0.25 and the interior values are 0.0625, 0.25, 0.5625, 1, 1.5625, 2.25, 3.0625. Their sum is 8.75. The estimate is 0.125[0 + 2 × 8.75 + 4] = 2.6875 square units. This is closer to the exact area 8/3 ≈ 2.6667 than the four-strip estimate.
In plain language
Calculate every ordinate, double the interior ones, and keep both endpoints once.
Self-check
Estimate the area under y = x² from x = 0 to 1 using two strips.
Show the answer
h = 0.5; values are 0, 0.25, 1. Estimate = 0.25[0 + 2 × 0.25 + 1] = 0.375 square units.
View diagrams
Four trapezium strips under y = x²
Key and description
From x=0 to 2 the width is h=0.5 and ordinates are 0, 0.25, 1, 2.25, 4. The straight tops of the four strips approximate the curve. Total estimated area = ½ × 0.5[0 + 2(0.25+1+2.25) + 4] = 2.75 square units.
Labelled constructed examples. Each diagram’s key identifies its own data and method.
Topic group
Geometry and measures
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §3.1.1 foundation; Unit 2 §3.1.1 foundation; Unit 3 §3.1.1 foundation. Printed pages 28: Geometric terms, including: point, line and plane horizontal, vertical, diagonal midpoint parallel and perpendicular clockwise and anticlockwise turns acute, obtuse, reflex, right angle, straight angle, full turn exterior, interior angles faces, edges and vertices
Open revision guidance
What to know
- Geometric properties are defined by specific relationships between points, lines, and planes. A point represents a precise location, while a line extends infinitely in two directions. A plane is a flat, two-dimensional surface. Angles define the rotation between two intersecting rays, measured in degrees. For example, a right angle is exactly 90 degrees, an acute angle is less than 90 degrees, and an obtuse angle is between 90 and 180 degrees. Reflex angles measure between 180 and 360 degrees.
- Spatial relationships are described using terms such as parallel, where lines never meet, and perpendicular, where lines meet at 90 degrees. A midpoint divides a line segment into two equal lengths. Turning involves rotation, where a clockwise turn proceeds in the direction of clock hands and anticlockwise in the opposite. Three-dimensional objects have faces (flat surfaces), edges (lines where faces meet), and vertices (points where edges meet). Interior angles reside within a shape's boundary, whereas exterior angles are formed by extending a side.
What to do in an assessment
- To identify a midpoint, calculate the average of the coordinates. Given points A at (2, 4) and B at (6, 8), the midpoint M is ((2+6)/2, (4+8)/2) = (4, 6). The distance from A to M is square root of ((4-2) squared + (6-4) squared) = square root of 8, which is approximately 2.83 units. This demonstrates the application of coordinate geometry to find the middle point of a segment located on a Cartesian plane.
- To measure angles in a shape, use a protractor to find the degree of rotation. For a triangle with vertices at (0, 0), (4, 0), and (0, 3), the interior angle at (0, 0) is 90 degrees. Clockwise movement from (0, 3) around the origin by 90 degrees maps to (3, 0). The interior angles must sum to 180 degrees. A full turn represents 360 degrees, which equals four right angles.
In plain language
This lesson covers the basic vocabulary and properties used to describe geometry, such as lines, angles, and the parts of 3D objects like faces and vertices.
Self-check
What is the midpoint of a line segment connecting (1, 2) and (5, 6)?
Show the answer
The midpoint is (3, 4) calculated as ((1+5)/2, (2+6)/2).
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §3.1.2 foundation; Unit 2 §3.1.2 foundation; Unit 3 §3.1.2 foundation. Printed pages 29: Vocabulary and essential properties of 2-D shapes, including: triangles – scalene, isosceles, equilateral, right-angled quadrilaterals – square, rectangle, parallelogram, rhombus, kite, trapezium polygons – including pentagon, hexagon, octagon, regular and irregular circles – radius, diameter, tangent, circumference, chord, arc, sector, segment.
Open revision guidance
What to know
- Two-dimensional shapes are categorised by their sides, angles, and rotational symmetry. Triangles are classified as scalene (no equal sides), isosceles (two equal sides), equilateral (three equal sides), or right-angled. Quadrilaterals include squares, rectangles, parallelograms, rhombi, kites, and trapeziums, each possessing specific properties regarding parallel sides and angle sums. Polygons are multi-sided figures; regular polygons have all equal sides and angles, while irregular polygons do not.
- Circles require specific vocabulary: the radius is the distance from the centre to the edge, the diameter is twice the radius, the circumference is the perimeter, and a chord connects two points on the circumference. Tangents touch the circle at exactly one point, while arcs are segments of the circumference. Sectors are slice-like regions bounded by two radii and an arc, and segments are regions bounded by a chord and an arc. These definitions are essential for calculating areas and perimeters.
What to do in an assessment
- To classify a triangle with sides 3 cm, 4 cm, and 5 cm, apply the Pythagorean theorem: 3 squared + 4 squared = 9 + 16 = 25. Since 5 squared = 25, this confirms a right-angled triangle. Calculating the perimeter is 3 + 4 + 5 = 12 cm. For a circle with a radius of 7 cm, the diameter is 2 times 7 = 14 cm, and the circumference is 2 times pi times 7 = 43.98 cm, using pi as 3.14159.
- For a quadrilateral with vertices A(0,0), B(4,0), C(6,3), and D(2,3), evaluate parallel sides. The slope of AB is 0. The slope of DC is (3-3)/(6-2) = 0. Since slopes match, AB is parallel to DC. The slope of AD is (3-0)/(2-0) = 1.5. The slope of BC is (3-0)/(6-4) = 1.5. Since both pairs of opposite sides are parallel, this is a parallelogram. Area is base (4 cm) times height (3 cm) = 12 square cm.
In plain language
This lesson details the names and unique geometric properties of different 2-dimensional shapes including triangles, quadrilaterals, polygons, and circles.
Self-check
What type of quadrilateral has vertices (0,0), (4,0), (5,2), and (1,2)?
Show the answer
It is a parallelogram because opposite sides are parallel with slopes of 0 and 2.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §3.1.3 foundation; Unit 2 §3.1.3 foundation; Unit 3 §3.1.3 foundation. Printed pages 29: Vocabulary and essential properties of 3-D shapes including cube, cuboid, cylinder, prism, pyramid, cone, sphere, tetrahedron
Open revision guidance
What to know
- Three-dimensional shapes exist in space and possess volume, surface area, and specific geometric features like faces, edges, and vertices. A cube has six square faces, eight vertices, and twelve equal edges. A cuboid has rectangular faces. A cylinder features two circular faces and a curved surface. Prisms have a consistent cross-section throughout their length, while pyramids taper to a single apex point. Cones have one circular base and a curved surface meeting at a point, spheres are perfectly round, and tetrahedra are triangular-based pyramids.
- Understanding these shapes involves analysing their nets, which are 2D patterns that fold into 3D objects, and calculating their properties using dimensions. For instance, the volume of a cuboid is found by multiplying its length, width, and height. The orientation and properties of these shapes are foundational for solving spatial problems. Each shape has unique symmetry properties that define how they can be rotated or reflected without changing their appearance.
What to do in an assessment
- Calculate the volume of a cuboid with length 5 cm, width 3 cm, and height 4 cm. Volume = length times width times height = 5 times 3 times 4 = 60 cubic cm. To find the surface area, calculate the area of all faces: 2 times (5 times 3) + 2 times (5 times 4) + 2 times (3 times 4) = 30 + 40 + 24 = 94 square cm.
- For a cylinder with radius 2 cm and height 10 cm, the volume is pi times radius squared times height = 3.14159 times 4 times 10 = 125.66 cubic cm. The surface area consists of two circular bases and a rectangular side surface. Total area = (2 times pi times radius squared) + (2 times pi times radius times height) = (2 times 3.14159 times 4) + (2 times 3.14159 times 2 times 10) = 25.13 + 125.66 = 150.79 square cm.
In plain language
This lesson explains the features of 3D objects such as cubes, cylinders, and pyramids, and how to use their dimensions to calculate volume and surface area.
Self-check
What is the volume of a cube with an edge length of 3 cm?
Show the answer
The volume is 3 cubed = 27 cubic cm.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §3.2.1 foundation; Unit 3 §3.2.1 foundation. Printed pages 30: measure and accurately draw: a straight line a circle or arc of a circle an angle of any size.
Open revision guidance
What to know
- To measure and draw a straight line, use a transparent ruler. Align the zero mark at the start point and read the scale to the desired length in centimeters or millimeters. For example, to draw a 7.5 cm line, mark a point at 0, place the ruler edge precisely against the paper, and mark another point at 7.5. Connect these with a firm pencil stroke.
- Angles are measured using a protractor. Place the centre of the protractor on the vertex and align the baseline with one ray. Read the scale starting from zero to the point where the second ray intersects the arc. When drawing an angle of 55 degrees, mark the vertex, draw a baseline ray, align the protractor centre, mark 55 degrees on the scale, and draw the second ray through that mark.
What to do in an assessment
- To draw a circle with radius 4 cm, set your compasses to 4 cm by measuring against a ruler. Place the compass point on the paper at a centre point, keep the compass fixed, and rotate the pencil end around to form the perimeter. For an arc, follow the same steps but only draw the required portion of the circumference.
- Practice measuring an angle between two segments AB and AC. Place the protractor centre at A. Align 0 degrees with AB. If the angle is 110 degrees, locate 110 on the inner or outer scale as appropriate. Draw the ray AC. Verify accuracy by checking if the shape reflects the intended measurement, ensuring the pencil line is thin for precision.
In plain language
This lesson covers using basic tools to create and measure lines, shapes, and angles accurately.
Self-check
Draw a line of 5.2 cm and an angle of 40 degrees. What is the length and angle?
Show the answer
The line measures 5.2 cm with a ruler. The angle measures 40 degrees using a protractor.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 3 tier · Unit 3 §3.2.2 foundation. Printed pages 30: accurately draw, using a ruler and a protractor: an angle bisector a perpendicular line bisector 2-D shapes given side lengths and, if appropriate, angles (compasses will be required to draw triangles when three side lengths are known) the locus of a point which moves such that it satisfies certain conditions, including: a given distance from a fixed point or line (compasses will be required) equidistant from two fixed points or lines.
Open revision guidance
What to know
- A perpendicular bisector passes through the midpoint of a segment at 90°. Every point on it is equally far from the two endpoints. Use the same compass radius, greater than half the segment length, to draw arcs from each endpoint above and below the segment; join their intersections.
- To bisect an angle, draw an arc centred on its vertex cutting both arms. From those two intersections, draw intersecting arcs of equal radius. Join their intersection to the original vertex. For a perpendicular at a point on a line, mark equal distances on the line either side of the point and bisect that new segment. Leave construction arcs visible.
What to do in an assessment
- Construct the perpendicular bisector of PQ with P(0,0) and Q(4,0), using equal compass openings greater than 2 units. The two arc intersections have x = 2 but their y-values depend on the compass opening. Joining them gives x = 2. If the opening is √8 units, the intersections specifically are (2,2) and (2,−2).
- For a 60° angle at A, draw a circle centred at A and mark B on it. With the same compass opening draw a circle centred at B and choose an intersection C. Triangle ABC is equilateral, so ∠BAC = 60°. Bisect that angle to construct 30°. Do not set a compass radius by guessing the final intersection coordinates.
In plain language
Equal-radius arcs enforce equal distances. Their crossings determine the construction rather than being arbitrary points.
Self-check
Why must the compass radius exceed half the length of a segment when constructing its perpendicular bisector?
Show the answer
The equal-radius circles must meet at two distinct points on opposite sides of the segment. A smaller radius gives no intersection; exactly half gives only the midpoint.
View diagrams
Compass-and-straightedge constructions
Key and description
Perpendicular bisector: equal-radius arcs from A and B meet at C and D; CD crosses AB at its midpoint M at 90°. Angle bisector: an arc from O cuts both arms at P and Q; equal-radius arcs from P and Q meet at R; OR bisects the angle.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Core
Scope: Core · Foundation tier, Higher tier in either Unit 3 tier · Unit 3 §3.2.3 foundation. Printed pages 30: solve problems involving intersecting loci in two dimensions – this will include the identification of regions that satisfy certain conditions
Open revision guidance
What to know
- Intersecting loci problems involve finding specific regions that satisfy multiple conditions simultaneously. For example, you might be asked to shade the region where a point is less than 3 cm from point A(0,0) AND closer to line L1 than line L2. These problems are solved by sketching both individual loci on the same set of axes and identifying the overlapping area.
- Conditions often involve inequalities. A region 'less than distance d' is the interior of a circle or a space between lines. The intersection is the set of points belonging to both logical sets. Always verify boundary conditions; if the condition is 'less than', the boundary is usually excluded, which is represented by a dashed line on your sketch.
What to do in an assessment
- Consider points P(0,0) and Q(4,0). Identify the region where points are within 2 cm of P and within 3 cm of Q. First, draw a circle of radius 2 centred at (0,0). Second, draw a circle of radius 3 centred at (4,0). The region satisfying both conditions is the overlap of these two circular discs.
- Calculate the boundary of the intersection. The circles have equations x^2 + y^2 = 4 and (x-4)^2 + y^2 = 9. Solve for intersection points: x^2 - (x-4)^2 = 4 - 9, which simplifies to 8x - 16 = -5, so 8x = 11, x = 1.375. Substitute x back to find y: y^2 = 4 - (1.375)^2 = 4 - 1.890625 = 2.109375, so y = +/- 1.45. The overlap is defined between these coordinates.
In plain language
This lesson teaches how to find areas on a map or graph that satisfy two or more distance rules at the same time.
Self-check
If a point must be within 2 units of (0,0) and within 2 units of (2,0), does the point (1, 1.5) satisfy this?
Show the answer
Distance to (0,0) is sqrt(1^2+1.5^2) = 1.8 < 2. Distance to (2,0) is sqrt((1-2)^2+1.5^2) = 1.8 < 2. Yes, it satisfies both.
View diagrams
Intersection of a line and circular locus
Key and description
A = (0, 0), B = (4, 0). The line locus x = 2 intersects the circle centre A, radius 3 at (2, √5) and (2, −√5). The shaded solution is inside the circle and to the right of x = 2: x > 2.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §3.3.1 foundation; Unit 3 §3.3.1 foundation. Printed pages 31: read and interpret scales
Open revision guidance
What to know
- To read and interpret linear or non-linear scales, one must first identify the value represented by each interval on the scale. By subtracting two consecutive marked major values (e.g., 10 and 20) and dividing by the number of smaller subdivisions between them, you determine the unit value per graduation. For a ruler or speedometer, ensure the unit of measurement is consistent with the context provided, such as millimeters, centimeters, or kilometers per hour.
- When dealing with dials or gauge instruments, pay attention to the starting zero position. If a needle points between 40 and 50 on a gauge with four equal subdivisions, each subdivision equals 2.5 units. Therefore, a reading at the second mark is 40 + (2 * 2.5) = 45 units. Accurate interpretation relies on careful observation of the scale starting point and the incremental value of each tick mark, ensuring units are clearly appended to the final scalar result.
What to do in an assessment
- Examine a linear temperature scale where 0 degrees Celsius and 10 degrees Celsius have five subdivisions between them. Calculate the value of one subdivision: (10 - 0) / 5 = 2 degrees Celsius per mark. If the liquid in the thermometer rests at the third mark, the reading is 0 + (3 * 2) = 6 degrees Celsius. Always perform the division check to avoid assuming every scale is marked in single units.
- Consider a weight scale marked at 0 kg, 5 kg, and 10 kg with ten small intervals between 0 and 5. Calculate: (5 - 0) / 10 = 0.5 kg per interval. If the pointer rests on the 7th mark, the total weight is 7 * 0.5 = 3.5 kg. Apply this logic to any non-standard gauge by checking the total range (max - min) and dividing by the count of gaps between marks.
In plain language
Reading a scale is about finding out what each small dash is worth. You find the gap between two numbers on the scale, count how many little lines are between them, and divide the gap by that number.
Self-check
A scale goes from 0 to 20 cm with 8 equal intervals. What is the value of 5 intervals?
Show the answer
Each interval is 2.5 cm (20 / 8 = 2.5). Therefore, 5 intervals equals 5 * 2.5 = 12.5 cm.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §3.3.2 foundation; Unit 3 §3.3.2 foundation. Printed pages 31: use and interpret maps
Open revision guidance
What to know
- Maps utilise scales to represent real-world distances on a manageable surface. A ratio scale, such as 1:50000, indicates that 1 unit on the map equals 50,000 of the same units in reality. To convert map distance to real distance, multiply the measured distance in centimeters by the scale factor. If the map scale is 1 cm to 2 km, a 5 cm line on the map represents 10 km in the real world (5 * 2 = 10).
- When interpreting map data, one must account for units of measurement carefully. Convert map centimeters to meters or kilometers by dividing by 100 for meters or 100,000 for kilometers. For example, a scale of 1:100000 means 1 cm represents 1 km. Always measure using a straight edge and align precisely with the scale provided on the map legend to ensure accuracy.
What to do in an assessment
- Identify the scale factor on a given map, for instance 1:25000. Measure a path between two coordinates, e.g., Point A (x=2, y=3) and Point B (x=6, y=6). The map distance is calculated using the distance formula: sqrt((6-2)^2 + (6-3)^2) = sqrt(4^2 + 3^2) = 5 cm. Convert this to real life: 5 cm * 25000 = 125,000 cm. Divide by 100,000 to get 1.25 km.
- Practice converting ratios to verbal scales. Given a ratio of 1:10000, express this as '1 cm represents 100 meters' by dividing 10,000 cm by 100. Apply this to determine that a 4.5 cm line on the map corresponds to 450 meters (4.5 * 100). Use consistent units throughout to avoid errors in conversion, especially when switching between centimeters, meters, and kilometers.
In plain language
Maps use scales to show big areas on a small piece of paper. If the map says 1:10000, it means that every centimeter on the map equals 10,000 centimeters in real life. You measure with a ruler and then multiply by the scale number.
Self-check
The map scale is 1:5000. How far is a 3 cm line on the map in meters?
Show the answer
3 * 5000 = 15,000 cm. Since 100 cm = 1 meter, 15000 / 100 = 150 meters.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §3.3.3 foundation; Unit 3 §3.3.3 foundation. Printed pages 31: interpret and produce scale drawings; scales may be written in the form 1 cm represents 5 m, or 1:500
Open revision guidance
What to know
- Scale drawings require a constant relationship between the drawing and the object. Scales are presented as verbal statements (1 cm represents 5 m) or ratios (1:500). To convert between these, determine the number of centimeters in the target unit. If 1 cm represents 5 m, the ratio is 1:500 because 5 m = 500 cm. Consistent conversion is essential for accuracy.
- To produce a drawing, define a scale that fits your paper, measure the object, and divide by the scale factor. For a 10 m by 20 m building, a scale of 1:200 means drawing sides of 5 cm (1000 cm / 200 = 5 cm) and 10 cm (2000 cm / 200 = 10 cm). Ensure that angles remain identical to the original object during the drawing process.
What to do in an assessment
- Create a scale drawing for a rectangular room measuring 4 m by 6 m using a scale of 1:100. Convert actual lengths to cm: 4 m = 400 cm, 6 m = 600 cm. Divide by 100 to get drawing lengths: 400 / 100 = 4 cm and 600 / 100 = 6 cm. Draw the rectangle with sides of 4 cm and 6 cm. Label all sides clearly with original dimensions.
- Interpret a drawing with scale 1 cm represents 2 m. If a wall in the drawing is 3.5 cm long, find the real length: 3.5 cm * 2 m/cm = 7 m. Use this method for compound shapes by breaking them into rectangles or triangles. If a shape has a side of 2.5 cm, multiply by the scale 2 to get the actual 5 m measurement.
In plain language
A scale drawing is a mini version of a real object. You keep all shapes the same but shrink the size using a scale. If the scale is 1:100, everything real is 100 times bigger than on the paper.
Self-check
A drawing uses a scale of 1:200. A wall on the drawing is 5 cm. How long is the real wall in meters?
Show the answer
5 * 200 = 1000 cm. Since 100 cm = 1 m, the real wall is 10 meters long.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §3.3.4 foundation; Unit 3 §3.3.4 foundation. Printed pages 31: understand 3-figure bearings and use this knowledge to interpret and draw bearings
Open revision guidance
What to know
- Bearings are used to navigate or locate positions accurately. A 3-figure bearing is measured clockwise from North and is always expressed as three digits (e.g., 045 degrees instead of 45 degrees). If a direction is exactly East, the bearing is 090 degrees. If it is South-West, the bearing is 225 degrees, calculated by adding 180 degrees to the South bearing of 180 degrees.
- To draw a bearing, draw a North line at your starting point. Use a protractor to measure the angle clockwise from North. Connect the points to form the direction line. When interpreting a bearing, recognise that North is 000 degrees/360 degrees, East is 090 degrees, South is 180 degrees, and West is 270 degrees. Every point is fixed relative to North.
What to do in an assessment
- Practice calculating bearings. Point A is North of Point B. If you are at B and move towards A, the bearing is 000 degrees. If you are at A and move towards B, the bearing is 180 degrees. If you move at a 45 degree angle North-East, the bearing is 045 degrees. Use a protractor to mark 45 degrees clockwise from a vertical line representing North.
- Solve for reverse bearings. If Point B is at a bearing of 060 degrees from A, find the bearing of A from B. Draw North lines at both points. The angle at A is 60 degrees. The alternate interior angle at B makes a 60 degree angle with the South line, so the bearing of A from B is 180 + 60 = 240 degrees. Always sketch North lines to visualise the geometry.
In plain language
Bearings tell you which direction to go, measured clockwise from North. We always write them as three numbers, like 030 degrees, to be clear. North is always 000.
Self-check
What is the bearing of a point that is directly West?
Show the answer
West is 90 degrees left of North, which is 360 - 90 = 270 degrees.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 3 tier · Unit 3 §3.3.5 foundation. Printed pages 31: interpret plans and elevations of 3-D shapes
Open revision guidance
What to know
- Plans and elevations provide a 2-D representation of a 3-D object. The 'plan' is the view from directly above. 'Front elevation' is the view from the front, and 'side elevation' is the view from the side. Each view is drawn as a flat shape. For example, a cylinder's plan is a circle, its front elevation is a rectangle, and its side elevation is a rectangle.
- When looking at a 3-D shape like a cuboid measuring 5 cm by 3 cm by 2 cm, the plan is a 5 cm by 3 cm rectangle. The front elevation is a 5 cm by 2 cm rectangle, and the side elevation is a 3 cm by 2 cm rectangle. By combining these three perspectives, you can reconstruct the 3-D shape's dimensions and spatial orientation.
What to do in an assessment
- Draw the elevations for a block stack. Imagine a shape built of 3 cubes. From the front, you see 3 squares. From the side, you see 1 square. From the top (plan), you see 2 squares. Sketch these grids using a scale of 1 square = 1 cm. Ensure the heights align across the elevations; for example, the height on the front elevation must match the height on the side elevation.
- Analyse the given elevations to determine the original shape. Front: 4 cm by 3 cm rectangle. Side: 4 cm by 2 cm rectangle. Plan: 3 cm by 2 cm rectangle. This describes a cuboid of dimensions 4 cm, 3 cm, and 2 cm. Always use isometric paper or squared paper to ensure parallel lines stay parallel when sketching these multi-view drawings.
In plain language
Plans and elevations are just the view of an object from different sides. A plan is from above, a front elevation is from the front, and a side elevation is from the side.
Self-check
If the front elevation of a shape is a 5 cm by 4 cm rectangle and the side is 2 cm by 4 cm, what is the height of the shape?
Show the answer
The height is 4 cm, as it appears in both the front and side elevations.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 3 tier · Unit 3 §3.3.6 foundation. Printed pages 31: interpret 2-D representation of 3-D shapes on isometric paper
Open revision guidance
What to know
- Isometric paper consists of a grid of equilateral triangles arranged to represent 3-D shapes. Lines are drawn at 30-degree angles to the horizontal to provide depth. A 1 cm edge on a cube is represented by a line spanning two dots or the side of an isometric triangle. This allows for a clear, non-distorted perspective of 3-D objects.
- When representing a 3-D object like a 2cm x 2cm x 2cm cube, every edge is drawn to the same scale. The vertical lines remain vertical, but the 'horizontal' edges are drawn at 30-degree angles along the grid lines. This technique helps in visualising how different faces connect in space without the need for complex perspective drawing.
What to do in an assessment
- Draw a 3-D cuboid of dimensions 3 cm by 2 cm by 4 cm. Start with a corner vertex. Draw the front-facing vertical edge as 4 cm (4 units down). From the bottom, draw the width (3 units) and length (2 units) along the isometric grid lines. Connect all corners to complete the cuboid. Each side must follow the grid lines exactly.
- Interpret a drawing on isometric paper. Identify a shape with a base of 3 units by 2 units and a height of 1 unit. By counting the grid units along each axis (vertical, left 30 degrees, right 30 degrees), you can define the volume or surface area. If one unit = 1 cm, the volume is 3 * 2 * 1 = 6 cubic centimeters.
In plain language
Isometric paper has triangles that help you draw 3-D shapes like cubes so they look real. You use the grid lines to show the height, width, and depth of the shape.
Self-check
A shape on isometric paper has 2 units height, 2 units width, and 2 units depth. If each unit is 1 cm, what is the shape?
Show the answer
It is a cube with sides of 2 cm.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 3 tier · Unit 3 §3.3.7 foundation. Printed pages 31: interpret nets of 3-D shapes
Open revision guidance
What to know
- A net is a 2-D shape that can be folded to form a 3-D object. When interpreting a net, imagine how the edges will meet to form vertices and faces. For a cube, the net consists of 6 squares arranged in a cross or T-shape. Each square edge must match an adjacent square edge in the final folded form.
- Common nets include the cube (6 squares), triangular prism (2 triangles and 3 rectangles), and square-based pyramid (1 square and 4 triangles). To verify a net, ensure that all faces are present and that they are arranged such that they can fold correctly without overlapping or leaving gaps.
What to do in an assessment
- Design a net for a cuboid measuring 3 cm by 2 cm by 1 cm. You need 6 rectangles: 2 of 3x2, 2 of 3x1, and 2 of 2x1. Arrange them on a grid: start with the 3x2 base, attach the 3x1 rectangles to the sides, the 2x1 rectangles to the ends, and the top 3x2 rectangle to one of the side pieces. Check the folding mentally to ensure all sides close.
- Identify the shape from a net consisting of a central triangle surrounded by three identical triangles. This net folds into a tetrahedron. Measure the sides of the triangles in the net. If each side is 4 cm, the total surface area of the folded tetrahedron is 4 * (sqrt(3)/4 * 4^2) = 27.71 square centimeters.
In plain language
A net is a flat pattern that you can fold into a 3-D shape. If you cut out the net and fold the lines, the faces will meet to create the object.
Self-check
A net of a cube is made of 6 squares. If each square has a side of 3 cm, what is the surface area of the cube?
Show the answer
One square has area 3 * 3 = 9 sq cm. 6 squares have total area 6 * 9 = 54 sq cm.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §3.4.1 foundation; Unit 2 §3.4.1 foundation; Unit 3 §3.4.1 foundation. Printed pages 31: recall and use the following angle properties: Sum of angles at a point Sum of angles on a straight line Opposite angles at a vertex Alternate, corresponding and interior angles within parallel lines Sum of angles in a triangle
Open revision guidance
What to know
- Angles around a point sum to 360 degrees, while angles on a straight line sum to 180 degrees. These properties allow for solving unknown angles in geometric figures by creating simple linear equations. For example, if three angles at a point are x, 2x, and 90 degrees, then x + 2x + 90 = 360, leading to 3x = 270 and x = 90 degrees.
- Parallel lines intersected by a transversal create specific angle relationships. Alternate angles are equal, corresponding angles are equal, and interior angles sum to 180 degrees. When two lines intersect, opposite angles at the vertex are identical. In a triangle, the interior angles always sum to 180 degrees, allowing calculation of a missing angle if two others are known, such as 180 - (50 + 60) = 70 degrees.
What to do in an assessment
- To find a missing angle on a straight line, subtract the sum of known angles from 180. Given a line with angles 45 degrees and y, calculate y = 180 - 45 = 135 degrees. For parallel lines, identify the Z, F, or C shapes. In a Z shape with a 75 degree angle at one corner, the alternate angle is 75 degrees.
- Apply the triangle sum property to a set of angles (a, b, c). If a=40 and b=80, calculate c = 180 - 40 - 80 = 60 degrees. Use vertical opposite angles: if two lines cross and one angle is 110 degrees, the angle directly opposite is also 110 degrees. Use these facts together to solve complex diagrams by isolating triangles or intersecting lines.
- Angles around a point total 360°. If three adjacent angles around a point are 95°, 125° and x, then x = 360−95−125 = 140°. Check that all sectors together complete one turn, with no overlap or gap.
In plain language
This topic teaches how to find missing angles using standard geometric rules like line sums, triangle sums, and parallel line patterns.
Self-check
In a triangle with angles 2x, 3x, and 40 degrees, find x.
Show the answer
2x + 3x + 40 = 180 becomes 5x = 140, so x = 28 degrees.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier or Unit 3 tier · Unit 2 §3.4.2 foundation; Unit 3 §3.4.2 foundation. Printed pages 32: recall and use the following angle properties: Angle properties of right-angled, isosceles and equilateral triangles The exterior angle of a triangle is equal to the sum of the interior angles at the other two vertices Sum of angles in a quadrilateral Angle properties of special quadrilaterals, including rectangles, parallelograms and kites
Open revision guidance
What to know
- Triangles have specific properties based on their type. In an isosceles triangle, two sides and two base angles are equal. In an equilateral triangle, all three sides and angles are 60 degrees. The exterior angle of any triangle equals the sum of the two opposite interior angles. This helps solve problems where a triangle side is extended to create an exterior angle, such as an exterior angle of 120 degrees created by interior angles of 50 and 70 degrees.
- Quadrilaterals, including rectangles, parallelograms, and kites, have specific angle rules. The sum of interior angles in any quadrilateral is 360 degrees. In a parallelogram, opposite angles are equal and adjacent angles sum to 180 degrees. For a kite, one pair of opposite angles are equal. Knowing these shapes allows for the calculation of missing angles using basic algebraic substitution when variables represent unknown values.
What to do in an assessment
- To find a missing angle in an isosceles triangle with a vertex angle of 40 degrees, calculate the base angles as (180 - 40) / 2 = 70 degrees. To calculate the exterior angle of a triangle with interior angles 55 and 65, add them to get 120 degrees. Ensure these rules are applied strictly to the identified shape properties during calculation.
- Calculate unknown angles in a parallelogram by setting up equations. If one angle is 2x and the adjacent angle is x + 30, use the property 2x + x + 30 = 180 to solve for x, resulting in 3x = 150 and x = 50. Therefore, the angles are 100 degrees and 80 degrees. Verify sums against the 360 degree total for the quadrilateral.
In plain language
Learn how to use properties of specific triangles and four-sided shapes like parallelograms to calculate unknown interior and exterior angles.
Self-check
In a parallelogram, one angle is 3x and the adjacent is x + 20. Find the value of x.
Show the answer
3x + x + 20 = 180 means 4x = 160, so x = 40.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §3.4.3 foundation. Printed pages 32: Polygons: Regular and irregular polygons Sum of exterior angles of a polygon Sum of an interior and exterior angle of a polygon Sum of interior angles of a polygon
Open revision guidance
What to know
- A polygon is a closed shape with straight sides. Regular polygons have equal sides and equal angles. The sum of the exterior angles of any convex polygon is always 360 degrees, regardless of the number of sides. The interior angle and exterior angle at any vertex of a polygon sum to 180 degrees because they sit on a straight line. This fundamental relationship allows for calculating one if the other is known.
- To find the sum of the interior angles of a polygon with n sides, use the formula (n - 2) * 180 degrees. This works because any polygon can be split into (n - 2) triangles. For a regular polygon, you can divide the total interior sum by the number of sides to find the value of each individual interior angle. For example, a pentagon has 5 sides, so (5 - 2) * 180 = 540 degrees total interior sum, and each interior angle is 540 / 5 = 108 degrees.
What to do in an assessment
- Use the exterior angle property to determine the number of sides of a regular polygon. If an exterior angle is 36 degrees, then the number of sides n = 360 / 36 = 10. To find the interior angle of this decagon, calculate 180 - 36 = 144 degrees.
- Calculate the sum of interior angles for an octagon (8 sides) as (8 - 2) * 180 = 1080 degrees. To find the value of one interior angle of a regular hexagon (6 sides), use (6 - 2) * 180 = 720 degrees total, then divide by 6 sides to get 120 degrees per interior angle. Ensure units are recorded in degrees.
- Exterior angles, taking one consistent turn at each vertex of a convex polygon, sum to 360°. A regular pentagon has each exterior angle 360/5 = 72°, hence each interior angle 180−72 = 108°. Its interior-angle sum is (5−2)×180 = 540°, agreeing with 5×108.
In plain language
Understand how to find the total sum of angles inside any polygon and how to calculate individual angles for regular shapes using side counts.
Self-check
What is the interior angle of a regular polygon with 9 sides?
Show the answer
(9 - 2) * 180 / 9 = 7 * 180 / 9 = 1260 / 9 = 140 degrees.
- Higher
Open revision guidance
What to know
- Circle theorems describe specific geometric relationships within a circle. Key properties include: the tangent at any point is perpendicular (90 degrees) to the radius; an angle at the centre is twice the angle at the circumference; the angle in a semicircle is 90 degrees; angles in the same segment are equal; opposite angles of a cyclic quadrilateral sum to 180 degrees; and the alternate segment theorem states the angle between a tangent and chord equals the angle in the alternate segment.
- These theorems are derived from circle geometry principles. For example, the tangent-radius perpendicularity holds because the shortest distance from a point to a line is the perpendicular segment. Cyclic quadrilaterals have vertices all touching the circle circumference. Using these, one can find missing angles in complex circle diagrams by identifying the relevant theorem for the given arc, tangent, or chord configuration.
What to do in an assessment
- To apply the centre/circumference theorem: if an arc subtends 40 degrees at the circumference, the angle at the centre is 80 degrees. For a cyclic quadrilateral with one angle of 105 degrees, calculate the opposite angle as 180 - 105 = 75 degrees. Always label vertices (e.g., A, B, C) to track angles clearly.
- Identify segments to use the 'same segment' theorem. If angle ACB = 30 degrees and angle ADB = 30 degrees, they are equal. For tangents from an external point P meeting the circle at A and B, remember PA = PB. Combine these rules to solve geometry problems involving coordinate points on a circle, such as (0, 5) and (5, 0) on a circle of radius 5, calculating slopes to verify perpendicularity.
In plain language
This lesson covers the rules governing angles and lines within circles, such as tangent properties and segment relationships, essential for solving geometry problems.
Self-check
A cyclic quadrilateral has an interior angle of 85 degrees. What is the opposite angle?
Show the answer
180 - 85 = 95 degrees.
View diagrams
Tangent and radius at a point of contact
Key and description
The radius to the point of contact is perpendicular to the tangent.
Angle in a semicircle: labelled proof
Key and description
AB is a diameter and O is the centre. C lies on the semicircle. OA = OC = OB, so the two small triangles are isosceles. Put α at A and between CA and CO; put β at B and between CB and CO. The angle sum of ABC gives 2(α + β) = 180°, hence ∠ACB = α + β = 90°.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Higher
Scope: Higher · Higher tier in either Unit 2 tier · Unit 2 §3.4.5 higher. Printed pages 33: construct geometric proofs using angle properties and facts, including circle theorems
Open revision guidance
What to know
- A proof is a sequence of justified steps, not a measurement or a drawing that looks correct. Name the angles and lines, state a relevant theorem for every step, and show that the conclusion follows from the givens.
- In a cyclic quadrilateral ABCD, ∠ABC + ∠ADC = 180°. If CB is extended beyond B to E, ∠ABC + ∠ABE = 180° because C, B and E are collinear. Subtracting the common angle gives ∠ABE = ∠ADC: an exterior angle equals the opposite interior angle.
What to do in an assessment
- Prove that the angle in a semicircle is a right angle. Let AB be a diameter, O its midpoint and C another point on the circle. Join OC. Put ∠OAC = α and ∠OBC = β. Since OA = OC and OB = OC, the isosceles triangles give ∠ACO = α and ∠OCB = β. Triangle ABC has angle sum α + β + (α + β) = 180°, so α + β = 90° and ∠ACB = 90°.
- Prove that a radius is perpendicular to a tangent at its contact point T. Every point of the tangent other than T lies outside the circle and is farther from centre O than the radius OT. Thus OT is the shortest distance from O to that line. The shortest distance from a point to a line is perpendicular, so OT meets the tangent at 90°. State the outside-circle assumption rather than merely reading a right-angle mark.
In plain language
Start with what is given, use a named rule at each step, and do not assume the fact you are trying to prove.
Self-check
In cyclic quadrilateral ABCD, AB is extended beyond B to E. Prove that ∠CBE equals ∠CDA.
Show the answer
∠ABC + ∠CBE = 180° (straight line ABE). ∠ABC + ∠CDA = 180° (opposite angles of a cyclic quadrilateral). Subtract ∠ABC to obtain ∠CBE = ∠CDA.
View diagrams
Angle in a semicircle: labelled proof
Key and description
AB is a diameter and O is the centre. C lies on the semicircle. OA = OC = OB, so the two small triangles are isosceles. Put α at A and between CA and CO; put β at B and between CB and CO. The angle sum of ABC gives 2(α + β) = 180°, hence ∠ACB = α + β = 90°.
Tangent and radius at a point of contact
Key and description
The radius to the point of contact is perpendicular to the tangent.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §3.5.1 foundation; Unit 2 §3.5.1 foundation; Unit 3 §3.5.1 foundation. Printed pages 34: Time: Notation for 12- and 24-hour clock Seconds in a minute, minutes in an hour, hours in a day, days in a week and months in a year
Open revision guidance
What to know
- Time notation uses the 12-hour and 24-hour clock systems. The 12-hour clock relies on am and pm designations to distinguish between morning and afternoon, while the 24-hour clock uses a four-digit format ranging from 00:00 to 23:59. For example, 2:30 pm in the 12-hour clock corresponds to 14:30 in the 24-hour clock format.
- Standard time units are foundational for measurements. There are 60 seconds in one minute and 60 minutes in one hour. Moving to larger scales, there are 24 hours in a single day, 7 days in one week, and varying days in months, though for calculations, we assume a standard year contains 365 days or 12 months.
What to do in an assessment
- To convert 12-hour times to 24-hour times, add 12 to any hour value from 1 pm to 11 pm. For 12 am, use 00:00. To convert 24-hour times back, subtract 12 from hours greater than 12. For instance, convert 16:45 to 12-hour format: 16 minus 12 equals 4, so the time is 4:45 pm.
- Perform calculations involving durations. If an event starts at 08:45 and ends at 10:15, calculate the duration by counting to the next hour (15 minutes to 09:00) then adding the remaining 1 hour and 15 minutes. Total time is 1 hour and 30 minutes, which is 90 minutes.
In plain language
Understanding how to read and write time in different formats is essential for scheduling and calculating durations between events using standard day and month cycles.
Self-check
Convert 19:20 to 12-hour time and calculate the duration between 19:20 and 21:05.
Show the answer
19:20 is 7:20 pm. Duration is 1 hour 45 minutes or 105 minutes.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §3.5.2 foundation; Unit 2 §3.5.2 foundation; Unit 3 §3.5.2 foundation. Printed pages 34: Metric Units: Standard metric units for length, mass and capacity and the relationships between them
Open revision guidance
What to know
- Metric units provide a standardised system for measuring length, mass, and capacity. Basic units include the metre (m) for length, the gram (g) or kilogram (kg) for mass, and the litre (l) or millilitre (ml) for capacity. Understanding these prefixes is key: kilo means 1000 times, and milli means 1/1000th.
- The relationships are consistent: 1 kilometre = 1000 metres, 1 metre = 100 centimetres, 1 centimetre = 10 millimetres. For mass, 1 kilogram = 1000 grams. For capacity, 1 litre = 1000 millilitres. These scales ensure that measurements can be converted precisely by multiplying or dividing by factors of 10, 100, or 1000.
What to do in an assessment
- Convert between units using multiplication or division. To convert 2.5 kilograms to grams, multiply by 1000: 2.5 * 1000 = 2500 grams. To convert 450 centimetres to metres, divide by 100: 450 / 100 = 4.5 metres. Apply these consistently across all metric dimensions.
- Work with combined measurement problems. If a container holds 1.5 litres of water and you add 250 millilitres, first convert units to be identical. Convert 1.5 litres to 1500 millilitres, then add 250 millilitres to get 1750 millilitres or 1.75 litres. This ensures accuracy in volumetric addition.
In plain language
The metric system uses powers of ten to relate units, making it simple to scale measurements for length, mass, and capacity by adjusting decimals.
Self-check
Convert 3.2 metres to millimetres and 750 grams to kilograms.
Show the answer
3.2 metres = 3200 millimetres. 750 grams = 0.75 kilograms.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 2 tier or Unit 3 tier · Unit 1 §3.5.3 foundation; Unit 2 §3.5.3 foundation; Unit 3 §3.5.3 foundation. Printed pages 34: carry out calculations involving time
Open revision guidance
What to know
- Time calculations differ from decimal math because they are base-60. When adding or subtracting time, once the minutes reach 60, they must be converted to an additional hour. For example, 40 minutes plus 30 minutes is 70 minutes, which is 1 hour and 10 minutes.
- When calculating across hours, treat the hour and minute components separately. Subtract or add minutes first, handling any borrowing or carrying across the 60-minute threshold. When calculating duration between two different times, use a timeline or count up to the nearest hour to ensure accuracy.
What to do in an assessment
- To add 1 hour 50 minutes to 2 hours 30 minutes, add the hours (1 + 2 = 3) and minutes (50 + 30 = 80). Since 80 minutes equals 1 hour and 20 minutes, combine these to get 4 hours and 20 minutes total.
- To subtract, e.g., 45 minutes from 3:15, note that you cannot subtract 45 from 15. Borrow 1 hour (60 minutes) from the 3, making it 2 hours and (60 + 15) = 75 minutes. Now, 75 - 45 = 30. The result is 2:30.
In plain language
Working with time requires managing the 60-minute cycle, ensuring that totals exceeding 60 minutes are converted back into hours correctly.
Self-check
Add 2 hours 40 minutes to 3 hours 30 minutes. Subtract 50 minutes from 4:20.
Show the answer
Addition: 6 hours 10 minutes. Subtraction: 3:30.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §3.5.4 foundation; Unit 3 §3.5.4 foundation. Printed pages 34: convert between units of time
Open revision guidance
What to know
- Converting time involves using factors based on the relationships between units. There are 60 seconds in a minute and 60 minutes in an hour. When converting smaller units to larger ones, divide by these factors. When converting larger units to smaller ones, multiply by these factors.
- For example, to convert 120 minutes to hours, we divide 120 by 60 to get 2 hours. To convert 3 hours to minutes, we multiply 3 by 60 to get 180 minutes. These conversions are linear and apply across all base time calculations.
What to do in an assessment
- Perform unit conversion using the scale factor. To convert 450 minutes into hours: 450 / 60 = 7.5 hours. Since 0.5 hours is 30 minutes, this is 7 hours and 30 minutes. Always check the final unit requested to ensure the conversion is correct.
- Handle conversion involving seconds. To convert 2 minutes 15 seconds into total seconds: (2 * 60) + 15. First, 120 + 15 = 135 seconds. Conversely, for 200 seconds to minutes: 200 / 60 = 3 remainder 20, which is 3 minutes 20 seconds.
In plain language
Time conversion is a matter of multiplying or dividing by 60, depending on whether you are moving to a larger or smaller unit of time.
Self-check
Convert 300 minutes to hours. Convert 4 minutes 20 seconds to seconds.
Show the answer
300 minutes = 5 hours. 4 minutes 20 seconds = 260 seconds.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier · Unit 1 §3.5.5 foundation. Printed pages 34: carry out calculations involving different time zones
Open revision guidance
What to know
- Time zones exist because the Earth rotates. The world is divided into zones relative to Coordinated Universal Time (UTC). Locations east of the prime meridian are generally ahead of UTC, while those to the west are behind. Calculations require finding the difference in hours between two specific locations.
- When calculating time zone differences, use the offset provided. If City A is UTC+1 and City B is UTC-5, the difference is 6 hours. When moving from City B to City A, you add 6 hours. When moving from City A to City B, you subtract 6 hours from the local time.
What to do in an assessment
- Calculate arrival times. If a flight leaves City A (UTC+2) at 09:00 and takes 5 hours to reach City B (UTC-1), first determine the time difference: 2 - (-1) = 3 hours ahead. To find local arrival time, convert departure to City B time (09:00 - 3 = 06:00) then add flight duration (06:00 + 5 = 11:00).
- Verify the direction of the adjustment. If crossing zones where the time is later, add the difference. If crossing zones where the time is earlier, subtract the difference. Ensure the date is consistent, particularly when the calculation crosses midnight, which changes the day.
In plain language
Time zones are offsets from a central time reference; calculate differences by comparing offsets and adjusting your starting time accordingly.
Self-check
If it is 10:00 in London (UTC+0) and the time difference to New York is 5 hours behind, what is the time in New York?
Show the answer
10:00 - 5 hours = 05:00.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §3.5.6 foundation; Unit 3 §3.5.6 foundation. Printed pages 34: make sensible estimates of metric measurements in everyday situations, recognising the appropriateness of units in different contexts
Open revision guidance
What to know
- Estimation requires recognising the physical scale of standard metric units. For example, a standard door is roughly 2 metres high; a paperclip is roughly 1 centimetre wide. Understanding these real-world benchmarks helps determine which units are appropriate for different contexts.
- Choose units based on size. Use millimetres for small items like a coin thickness, centimetres for books, metres for furniture or room dimensions, and kilometres for travel distances. For mass, grams are for items held in one hand, while kilograms are for heavier items like humans or appliances.
What to do in an assessment
- Practice estimation by comparing objects. To estimate the length of a desk, identify an object of known length, such as an A4 sheet (approx 30 cm) and see how many fit along the desk edge. If 4 sheets fit, estimate 4 * 30 = 120 cm or 1.2 metres.
- Select the correct unit for the context. If measuring the height of a house, metres is appropriate (e.g., 6 metres). If measuring the capacity of a medicine spoon, millilitres is appropriate (e.g., 5 millilitres). Avoid units that lead to excessively large or small numbers when a more standard unit exists.
In plain language
Use real-world object benchmarks to estimate lengths, masses, and capacities, ensuring that the unit chosen is practical for the size of the item measured.
Self-check
Which unit is best for a spoon of cough syrup: millilitres, litres, or grams? Estimate the length of a car.
Show the answer
Millilitres. A car is typically 4 to 5 metres long.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier · Unit 1 §3.5.7 foundation. Printed pages 35: convert between the following metric and Imperial units: km - miles; cm, m - inches, feet; kg - lb; grams – ounces; litres - pints, gallons - the appropriate metric to imperial equivalences will be given to learners
Open revision guidance
What to know
- Converting between metric and imperial systems relies on approximate equivalences. Since these systems have different origins, conversions are rarely exact and usually provided in exam questions. Common equivalences include 1 mile being approximately 1.6 kilometres, 1 inch equalling 2.54 centimetres, and 1 kilogram being approximately 2.2 pounds (lb).
- Capacity conversions are also common, such as 1 gallon being approximately 4.5 litres or 1 pint equalling roughly 568 millilitres. These values are constants you apply to solve problems when given in the question prompt.
What to do in an assessment
- Use the provided conversion factor to solve word problems. If 1 inch = 2.5 cm, how many centimetres are in 12 inches? Calculate 12 * 2.5 = 30 cm. If 1 kg = 2.2 lb, how many kg is 11 lb? Calculate 11 / 2.2 = 5 kg. Always verify if you need to multiply or divide based on the target unit.
- Apply conversion to compound units or travel. If travelling 50 miles and knowing 1 mile = 1.6 km, calculate 50 * 1.6 = 80 km. If fuel consumption is 5 miles per litre, and you convert to kilometres, check the context of whether conversion occurs before or after the calculation.
In plain language
When converting between metric and imperial units, use the provided conversion factors to scale the measurements appropriately, knowing that these are approximate values.
Self-check
Given 1 inch = 2.5 cm and 1 kg = 2.2 lb, convert 20 inches to centimetres and 4.4 lb to kilograms.
Show the answer
20 inches = 50 cm. 4.4 lb = 2 kg.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §3.5.8 foundation; Unit 3 §3.5.8 foundation. Printed pages 35: recall and use compound measures for speed and fuel consumption. Units include: m/s, km/h, mph and mpg
Open revision guidance
What to know
- Compound measures involve combining two different physical quantities to describe a rate. Speed is defined as distance divided by time. If a vehicle travels 150 km in 2 hours, the speed is 150 km / 2 h = 75 km/h. When calculating, always ensure that the time units match the speed unit provided. For example, to find distance using m/s, time must be in seconds: if an object moves at 10 m/s for 60 seconds, the distance is 10 m/s * 60 s = 600 m.
- Fuel consumption is often measured in miles per gallon (mpg). This represents the distance a vehicle can travel per unit of fuel consumed. If a car consumes 5 gallons of fuel to travel 200 miles, the consumption is 200 miles / 5 gallons = 40 mpg. To determine the fuel needed for a trip, rearrange the formula: Fuel = Distance / Consumption. For a 300 mile trip at 40 mpg, the fuel required is 300 / 40 = 7.5 gallons.
What to do in an assessment
- To calculate speed, use the formula S = D / T. Imagine a cyclist travelling 45 kilometers in 3 hours. Divide the distance (45 km) by the time (3 h) to find the speed: 45 / 3 = 15 km/h. To calculate the distance from speed and time, use D = S * T. If a train travels at 80 km/h for 2.5 hours, the distance is 80 * 2.5 = 200 km. Always confirm that units are consistent throughout your calculations before solving.
- To determine fuel usage, use the relationship F = D / C. Suppose a car travels 480 miles with an efficiency of 32 mpg. First, perform the division 480 / 32. The result is 15 gallons of fuel used. If you need to find how far a vehicle can travel with a specific amount of fuel, use D = C * F. If you have 12 gallons of fuel and the car gets 35 mpg, the distance potential is 35 * 12 = 420 miles. Double check your arithmetic by multiplying the result back to ensure it matches the original totals.
In plain language
Compound measures link two units like distance and time to form a new measurement like speed. You find speed by dividing the distance travelled by the time taken, and fuel efficiency by dividing the distance by the amount of fuel used.
Self-check
A car travels 240 miles using 8 gallons of fuel. What is its fuel consumption in mpg?
Show the answer
240 miles / 8 gallons = 30 mpg. The calculation is 240 divided by 8.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 3 tier · Unit 3 §3.5.9 foundation. Printed pages 35: recall and use other compound measures, including density, population density and flow rates. Units include kg/m 3 , g/cm 3 , population per km 2 , m 3 per hour, litres per second
Open revision guidance
What to know
- Density is defined as mass per unit volume. It indicates how tightly packed the matter is within an object. The formula is Density = Mass / Volume. For instance, if a block of metal has a mass of 500 g and occupies a volume of 25 cm cubed, its density is 500 / 25 = 20 g/cm cubed. Units are commonly kg/m cubed or g/cm cubed. When dealing with fluid flow rates, the principle remains similar: Flow Rate = Volume / Time. If 120 meters cubed of water flows through a pipe in 2 hours, the flow rate is 60 m cubed per hour.
- Population density measures the concentration of people in a given area. The formula is Population Density = Total Population / Area. If a town with an area of 50 km squared has a population of 10,000 people, the population density is 10,000 / 50 = 200 people per km squared. This helps compare population levels across different geographic regions regardless of their total size. Always verify that units match the formula's requirements before dividing.
What to do in an assessment
- To find density, divide mass by volume. A piece of iron has a mass of 15.7 kg and a volume of 0.002 m cubed. The density is 15.7 / 0.002 = 7850 kg/m cubed. If you need to find the mass, multiply density by volume: Mass = Density * Volume. For a block with a volume of 0.5 m cubed and density of 7000 kg/m cubed, the mass is 7000 * 0.5 = 3500 kg. When measuring flow rates, if a pump moves 20 litres per second, to find the volume moved in 60 seconds, calculate 20 * 60 = 1200 litres.
- To calculate population density, divide the total number of people by the area. A region has a population of 45,000 people spread over 150 km squared. The calculation is 45,000 / 150 = 300 people per km squared. To find the population of an area, use Population = Density * Area. If a region has a density of 50 people per km squared and covers an area of 1200 km squared, the total population is 50 * 1200 = 60,000 people. Always list units clearly in your final answer to ensure correctness.
In plain language
These measures describe rates like mass per volume, people per area, or volume per time. You find them by dividing the total amount by the total space or time.
Self-check
A village has 1200 people living in an area of 3 km squared. What is the population density?
Show the answer
1200 people / 3 km squared = 400 people per km squared.
- Higher
Scope: Higher · Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §3.5.10 higher; Unit 3 §3.5.10 higher. Printed pages 35: convert between units of area and volume
Open revision guidance
What to know
- When converting units of area, you must square the conversion factor. For example, since 1 m = 100 cm, a square of 1 m by 1 m has an area of 10,000 cm squared (100 * 100 = 10,000). Always apply the square to the scale factor, not just the base units. If you are converting from cm squared to m squared, divide by 10,000. For instance, 50,000 cm squared is equivalent to 50,000 / 10,000 = 5 m squared. Understanding this relationship prevents errors when scaling dimensions.
- Volume conversion requires cubing the conversion factor. Because 1 m = 100 cm, 1 cubic meter is 100 * 100 * 100 = 1,000,000 cm cubed. To convert from m cubed to cm cubed, multiply by 1,000,000. Conversely, to convert cm cubed to m cubed, divide by 1,000,000. Consider a container with a volume of 2,500,000 cm cubed. To convert this to cubic meters, divide by 1,000,000 to get 2.5 m cubed. This principle applies to all linear scale factors regardless of the units involved.
What to do in an assessment
- To convert area, identify the linear scale factor and square it. Let us convert 4.5 meters squared to centimeters squared. The scale factor for length is 100. The squared scale factor is 100 * 100 = 10,000. Multiply the area by the squared factor: 4.5 * 10,000 = 45,000 cm squared. If you need to convert 750,000 mm squared to m squared, identify the factor 1 m = 1000 mm. The square factor is 1,000,000. Divide the area: 750,000 / 1,000,000 = 0.75 m squared.
- To convert volume, cube the linear scale factor. Convert 0.003 m cubed to cm cubed. The linear factor is 100, so the cubic factor is 100 * 100 * 100 = 1,000,000. Multiply: 0.003 * 1,000,000 = 3000 cm cubed. Conversely, convert 450,000,000 mm cubed to meters cubed. The linear factor is 1000, so the cubic factor is 1,000,000,000 (1000 cubed). Divide the volume: 450,000,000 / 1,000,000,000 = 0.45 m cubed. Always verify the exponent matches the dimension of the units.
In plain language
To convert area, square the conversion factor for length. To convert volume, cube the conversion factor for length. Multiply when moving to smaller units and divide when moving to larger units.
Self-check
Convert 12,000,000 cm cubed into meters cubed.
Show the answer
12,000,000 / 1,000,000 = 12 m cubed. The cubic conversion factor is 1,000,000.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §3.6.1 foundation; Unit 3 §3.6.1 foundation. Printed pages 36: 2-D shapes: Estimate of the area of an irregular shape drawn on a square grid Perimeter and area of a square, rectangle, triangle, parallelogram, trapezium, circle, semicircle and a composite shape
Open revision guidance
What to know
- To find the area of an irregular shape on a grid, count the number of full squares completely inside the boundary and then estimate the contribution of partial squares. A partial square covering more than half a unit is generally counted as one, while those covering less are ignored. Summing these provides a reasonable approximation of the total area in square units.
- For standard polygons, the area is calculated using specific geometric formulas. A rectangle with sides 5 cm and 3 cm has an area of 5 times 3, equalling 15 square cm. A triangle with base 6 cm and height 4 cm uses the formula 0.5 times base times height, resulting in 12 square cm. A trapezium with parallel sides 4 cm and 6 cm and height 3 cm uses the formula 0.5 times the sum of parallel sides times height, which is 0.5 times 10 times 3, equalling 15 square cm.
What to do in an assessment
- Calculate the perimeter and area of a composite shape made of a rectangle (8 cm by 4 cm) and a triangle (base 8 cm, height 3 cm) attached to the top. Perimeter involves summing the external edges: 8 plus 4 plus 4 plus 5 (hypotenuse) plus 5, totalling 26 cm. Area is the sum of the two parts: 32 square cm for the rectangle and 12 square cm for the triangle, for a total of 44 square cm.
- To find the area of a circle with radius 7 cm, use pi times radius squared. Using pi as 3.14, the calculation is 3.14 times 49, resulting in 153.86 square cm. For a semicircle, divide this result by 2 to get 76.93 square cm. Always ensure units are squared for area and linear for perimeter measurements.
In plain language
Measuring 2D shapes involves counting squares for irregular outlines or using specific math formulas for regular ones. By breaking complex shapes into simpler ones like rectangles or triangles, you can add their individual areas together to find the total size.
Self-check
Calculate the area of a rectangle 6 cm by 4 cm joined to a triangle with base 6 cm and height 3 cm.
Show the answer
Area = (6 times 4) + (0.5 times 6 times 3) = 24 + 9 = 33 square cm.
- Higher
Scope: Higher · Higher tier in either Unit 3 tier · Unit 3 §3.6.2 higher. Printed pages 36: 2-D shapes: Length of a circular arc Area of a sector and a segment
Open revision guidance
What to know
- An arc is a portion of a circle's circumference. Its length is calculated using the fraction of the circle represented by the central angle in degrees: (angle/360) times 2 times pi times radius. For a circle with radius 10 cm and an angle of 90 degrees, the arc length is (90/360) times 2 times 3.14 times 10 = 0.25 times 62.8 = 15.7 cm.
- A sector is a pie-shaped slice of a circle, while a segment is the area between an arc and a chord. Sector area is calculated as (angle/360) times pi times radius squared. Using the same circle, area is 0.25 times 3.14 times 100 = 78.5 square cm. To find the segment area, subtract the area of the triangular part (formed by the two radii and the chord) from the sector area.
What to do in an assessment
- Consider a sector with radius 6 cm and angle 60 degrees. First, find sector area: (60/360) times 3.14 times 36 = 1/6 times 113.04 = 18.84 square cm. Next, calculate the triangular area using 0.5 times r squared times sin(theta). Using 60 degrees: 0.5 times 36 times sin(60) = 18 times 0.866 = 15.588 square cm.
- Subtract the triangle area from the sector area to isolate the segment area: 18.84 - 15.588 = 3.252 square cm. Ensure the calculator is set to degree mode for trigonometric operations. These methods allow for precise measurement of curved geometric segments which are essential for engineering and design applications involving rotational symmetry.
In plain language
These calculations measure parts of a circle. Arcs are the edges, sectors are the slices like pizza, and segments are the smaller areas cut off by a straight line between two points on the circle's edge.
Self-check
Find the area of a sector with radius 4 cm and angle 90 degrees (pi = 3.14).
Show the answer
Area = (90/360) times 3.14 times 16 = 0.25 times 50.24 = 12.56 square cm.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §3.6.3 foundation; Unit 3 §3.6.3 foundation. Printed pages 36: 3-D shapes: Surface area, cross- sectional area, volume and capacity of a cube, cuboid, prism, and a composite solid Cross-sectional area, volume and capacity of a cylinder
Open revision guidance
What to know
- For a prism with a constant cross-section, volume = cross-sectional area × length. A cuboid is a rectangular prism; a cylinder has a circular cross-section, so its volume is πr²h. The height h is perpendicular to the base. These formulas do not apply unchanged to a cone or pyramid.
- Surface area is the sum of the areas of all exposed faces. A cuboid of dimensions l,w,h has surface area 2(lw+lh+wh); a cube of side a has surface area 6a². On a composite solid, do not count internal joined faces. Capacity is an internal volume: 1 cm³ = 1 ml, 1000 cm³ = 1 litre, and 1 m³ = 1000 litres.
What to do in an assessment
- A cuboid measures 5 cm × 4 cm × 3 cm. Volume = 5×4×3 = 60 cm³. Surface area = 2(5×4 + 5×3 + 4×3) = 94 cm². A triangular prism with cross-section base 4 cm and perpendicular height 5 cm has cross-sectional area ½×4×5 = 10 cm². With length 6 cm, its volume is 60 cm³.
- A cylinder of radius 2 cm and height 10 cm has cross-sectional area 4π cm² and volume 40π ≈ 125.66 cm³, or about 125.66 ml. For two non-overlapping cuboids joined together, add their volumes. If they have dimensions 4×3×2 and 2×3×2 cm and share a 3×2 face, volume = 24+12 = 36 cm³; exposed surface area = 52+32−2×6 = 72 cm².
In plain language
Work out a prism's cross-section first. Count only exposed faces for surface area and convert cubic units carefully.
Self-check
A rectangular tank has internal dimensions 50 cm × 40 cm × 30 cm. What is its capacity in litres?
Show the answer
Volume = 50×40×30 = 60000 cm³. Divide by 1000 to get 60 litres.
- Higher
Scope: Higher · Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §3.6.3 higher; Unit 3 §3.6.3 higher. Printed pages 36: Surface area of a cylinder Surface area, volume and capacity of a sphere, cone, pyramid and a compound solid
Open revision guidance
What to know
- A closed cylinder has surface area 2πr² + 2πrh. A sphere has surface area 4πr² and volume (4/3)πr³. A cone has curved surface area πrl, total surface area πrl + πr², and volume (1/3)πr²h. In a cone, l is slant height and h is perpendicular height.
- A pyramid has volume (1/3) × base area × perpendicular height. Its total surface area is the base plus all triangular face areas, using each face's perpendicular height within that face. That is generally a slant height, not the pyramid's vertical height. Compound solids can require adding volumes or subtracting a removed part; exposed surface areas must exclude internal faces.
What to do in an assessment
- A closed cylinder of radius 3 cm and height 5 cm has surface area 2π×9 + 2π×3×5 = 48π ≈ 150.80 cm². A sphere of radius 3 cm has volume (4/3)π×27 = 36π ≈ 113.10 cm³ and area 36π ≈ 113.10 cm²: the equal numerical expressions here have different units.
- A cone has radius 4 cm and vertical height 3 cm. By Pythagoras, slant height = √(4²+3²) = 5 cm. Volume = (1/3)π×16×3 = 16π ≈ 50.27 cm³; total area = π×4×5 + π×16 = 36π ≈ 113.10 cm². A square pyramid with base side 6 cm and vertical height 4 cm has face slant height √(3²+4²) = 5 cm, volume 36×4/3 = 48 cm³, and area 36 + 4(½×6×5) = 96 cm².
- A hemisphere of radius 3 cm is placed on a cylinder of the same radius and height 5 cm. Volume = 45π + 18π = 63π ≈ 197.92 cm³. The external area includes the cylinder's curved surface, its bottom and the hemisphere's curved surface, but not their shared circular face: 30π + 9π + 18π = 57π ≈ 179.07 cm².
In plain language
Choose the formula for the solid, distinguish vertical and slant heights, and omit hidden joined faces.
Self-check
A square-based pyramid has base side 10 cm and vertical height 12 cm. Find its volume and total surface area.
Show the answer
Volume = (1/3)×100×12 = 400 cm³. Face slant height = √(5²+12²) = 13 cm. Total surface area = 100 + 4(½×10×13) = 360 cm².
View diagrams
Vertical and face slant heights
Key and description
A square pyramid has base side 6 cm and vertical height 4 cm. Join the apex to the midpoint of a base side: this face slant height is √(4² + 3²) = 5 cm. Volume = ⅓ × 36 × 4 = 48 cm³; total area = 36 + 4 × ½ × 6 × 5 = 96 cm². The perspective drawing is schematic.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Core
Scope: Core · Foundation tier, Higher tier in either Unit 3 tier · Unit 3 §3.7.1 foundation. Printed pages 37: use Pythagoras’ theorem in 2-D, including reverse problems
Open revision guidance
What to know
- Pythagoras theorem states that for a right-angled triangle with legs a and b and hypotenuse c, the relationship holds that a squared plus b squared equals c squared. This rule allows for the calculation of missing side lengths in two dimensions whenever two sides are known, or verifying if a triangle is right-angled using its side lengths.
- Reverse problems involve finding a leg length given the hypotenuse and the other leg. By rearranging the formula to a squared equals c squared minus b squared, you can calculate the unknown side. Ensure all dimensions are in consistent units, such as centimeters or meters, before performing the calculation to avoid errors in the final result.
What to do in an assessment
- To find the hypotenuse of a triangle with legs 3cm and 4cm, calculate 3 squared plus 4 squared which is 9 plus 16, resulting in 25. Taking the square root of 25 yields 5cm as the hypotenuse. Always perform the squaring step first and the square root operation as the final step in the sequence.
- For a reverse problem where the hypotenuse is 13cm and one leg is 5cm, calculate 13 squared minus 5 squared, which is 169 minus 25, resulting in 144. The square root of 144 is 12cm. This confirms the second leg is 12cm, providing a complete solution for the triangle dimensions.
In plain language
Use the relationship between sides of a right-angled triangle to find missing lengths using squares and square roots.
Self-check
A right triangle has a hypotenuse of 10m and one leg of 6m. What is the length of the other leg?
Show the answer
8m. Calculated by square root of (10 squared minus 6 squared) which is square root of (100 minus 36) equals square root of 64.
View diagrams
Right-angled triangle with labelled sides
Key and description
For angle θ, the opposite side is 3, the adjacent side is 4 and the hypotenuse is 5.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Higher
Scope: Higher · Higher tier in either Unit 3 tier · Unit 3 §3.7.2 higher. Printed pages 37: use Pythagoras’ theorem in 3-D, including reverse problems
Open revision guidance
What to know
- In three dimensions, Pythagoras theorem is applied to determine the space diagonal within a cuboid or prism. For a cuboid with dimensions length L, width W, and height H, the space diagonal D is found by the equation D squared equals L squared plus W squared plus H squared.
- Reverse problems in 3D require finding one missing dimension when the space diagonal and two other dimensions are known. By subtracting the squares of known sides from the square of the diagonal, the missing dimension is found. Always verify the orientation of the shape to ensure the correct sides are being squared.
What to do in an assessment
- Consider a cuboid with length 4m, width 3m, and height 12m. The space diagonal D squared is 4 squared plus 3 squared plus 12 squared, which is 16 plus 9 plus 144 equals 169. Thus, D is the square root of 169, which is 13m. Always sum the squares of all three dimensions before taking the square root.
- To solve for a missing height where the diagonal is 15cm, length is 9cm, and width is 8cm, rearrange to H squared equals 15 squared minus 9 squared minus 8 squared. This equals 225 minus 81 minus 64, resulting in 80. Therefore, H equals the square root of 80, which is approximately 8.94cm.
In plain language
Find the longest internal line in a 3D box by squaring all three sides, adding them, and taking the square root.
Self-check
A cuboid has length 2cm, width 2cm, and height 1cm. Find the space diagonal.
Show the answer
3cm. Calculated by square root of (2 squared plus 2 squared plus 1 squared) equals square root of (4 plus 4 plus 1) equals square root of 9.
- Higher
Scope: Higher · Higher tier in either Unit 3 tier · Unit 3 §3.7.3 higher. Printed pages 37: use trigonometric relationships in right-angled triangles to solve problems, including those involving bearings and angles of elevation and depression
Open revision guidance
What to know
- Relative to the chosen angle θ, sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse and tan θ = opposite/adjacent. The hypotenuse is opposite the right angle. Set the calculator to degrees. To find an angle, use the corresponding inverse function.
- Bearings are three-figure angles measured clockwise from North, for example 053°. An angle of elevation is measured upwards from a horizontal line; depression is downwards from a horizontal line. In a height problem include any stated eye height rather than assuming the observer is at ground level.
What to do in an assessment
- A ladder 5 m long makes 60° with level ground. It is the hypotenuse and the height is opposite: height = 5 sin 60° ≈ 4.33 m. An observer's eyes are 1.5 m above ground, 20 m horizontally from a tower, with elevation 40°. Height above the eyes = 20 tan 40° ≈ 16.78 m, so tower height ≈ 18.28 m.
- A boat finishes 3 km north and 4 km east of its starting point. Draw North vertically and East horizontally. The angle clockwise from North has opposite side 4 and adjacent side 3, so θ = tan⁻¹(4/3) ≈ 53.13°. Its bearing is 053° to the nearest degree and its distance is √(3² + 4²) = 5 km. From the destination back to the start, add 180°: 233°.
In plain language
Draw and label the right triangle, then choose the ratio using the sides relative to the required angle.
Self-check
A point is 8 km north and 6 km east of A. Find its bearing from A to the nearest degree.
Show the answer
θ = tan⁻¹(6/8) ≈ 36.87° clockwise from North, so the three-figure bearing is 037°.
View diagrams
Right-angled triangle with labelled sides
Key and description
For angle θ, the opposite side is 3, the adjacent side is 4 and the hypotenuse is 5.
Bearing and component distances
Key and description
Travel 10 km on a bearing of 030° (30° clockwise from north). North component = 10 cos 30° = 5√3 km (about 8.66 km); east component = 10 sin 30° = 5 km.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Higher
Scope: Higher · Higher tier in either Unit 3 tier · Unit 3 §3.7.4 higher. Printed pages 37: calculate a side or an angle of a right-angled triangle in 2-D and 3-D
Open revision guidance
What to know
- Calculating sides and angles in 3D often involves identifying hidden 2D right-angled triangles within the shape. For example, a rectangular prism might require finding the diagonal of a base before calculating the angle of the space diagonal using trigonometry.
- When solving for angles, use the inverse trigonometric functions on your calculator, denoted as sin-1, cos-1, or tan-1. These functions take the ratio of sides and return the corresponding angle in degrees. Ensure consistent identification of the hypotenuse, opposite, and adjacent sides relative to the angle being calculated.
What to do in an assessment
- To find the angle of elevation in 3D for a pyramid with square base side 10m and height 12m, first find the half-diagonal of the base. Half-diagonal is square root of (5 squared plus 5 squared) which is 7.07m. Now use Tan angle equals Height / half-diagonal: Tan angle equals 12 / 7.07 equals 1.697.
- Calculate the inverse tangent of 1.697. Tan-1(1.697) equals 59.5 degrees. This angle represents the tilt of the edge toward the centre. Always visualise the 2D plane cutting through the 3D object to define the necessary triangle clearly.
In plain language
Find angles and sides in complex 3D shapes by isolating the correct 2D right-angled triangles inside them.
Self-check
In a cube of side 5cm, what is the angle between a diagonal of the base and the space diagonal of the cube?
Show the answer
35.3 degrees. Calculated using Tan angle equals 5 / (5 * square root of 2), which is 1 / 1.414 equals 0.707. Tan-1(0.707) equals 35.3 degrees.
- Higher
Scope: Higher · Higher tier in either Unit 3 tier · Unit 3 §3.7.5 higher. Printed pages 37: extend trigonometry to angles of any size
Open revision guidance
What to know
- Trigonometric functions are extended beyond 90 degrees using the unit circle. Angles are measured from the positive x-axis. Sine is related to the y-coordinate, Cosine to the x-coordinate, and Tangent is the ratio y/x. Symmetry properties allow finding values for obtuse and reflex angles.
- The four quadrants dictate the signs of the functions. In the first quadrant, all are positive. In the second, only Sine is positive. In the third, only Tangent is positive. In the fourth, only Cosine is positive. This is remembered as CAST. For an obtuse angle x, Sin(x) equals Sin(180-x).
What to do in an assessment
- To find Cosine 150 degrees, identify it is in the second quadrant where Cosine is negative. The reference angle is 180 minus 150 which is 30 degrees. Thus, Cos 150 equals -Cos 30. Since Cos 30 is 0.866, Cos 150 equals -0.866.
- For Tangent 210 degrees, this is in the third quadrant where Tangent is positive. The reference angle is 210 minus 180 which is 30 degrees. Therefore, Tan 210 equals Tan 30. Since Tan 30 is approximately 0.577, Tan 210 equals 0.577. Always use the reference angle relative to the x-axis.
In plain language
Use the unit circle and quadrant rules to find values for angles greater than 90 degrees.
Self-check
What is the value of Sine 120 degrees?
Show the answer
0.866. Since it is in the second quadrant, Sine 120 equals Sine 60, which is approximately 0.866.
- Higher
Scope: Higher · Higher tier in either Unit 3 tier · Unit 3 §3.7.6 higher. Printed pages 37: apply knowledge of trigonometry with angles of any size to the solution of problems in 2-D or 3-D, including appropriate use of the sine rule and cosine rule
Open revision guidance
What to know
- Label sides a,b,c opposite angles A,B,C. The sine rule a/sin A = b/sin B = c/sin C is useful when a matching side–angle pair is known. The cosine rule c² = a²+b²−2ab cos C finds a side from two sides and their included angle, or an angle from three sides.
- Keep unrounded intermediate values. In a side-side-angle sine-rule problem, inverse sine may give an acute angle even when an obtuse alternative is possible. Consider both θ and 180°−θ, and reject a candidate only using the other information or the angle sum. Using the cosine rule with three known sides avoids that ambiguity.
What to do in an assessment
- For a=10 cm, b=12 cm and included C=60°, c² = 100+144−240×0.5 = 124, so c=√124 ≈ 11.14 cm. To find B use cos B = (a²+c²−b²)/(2ac) = (100+124−144)/(20√124). Thus B ≈ 68.95° and A ≈ 51.05°. The sine rule agrees when the correct angle is chosen.
- For a=5, b=6 and C=70°, c² = 25+36−60 cos70° ≈ 40.47879. Hence c ≈ 6.36, not 6.39. Check that the side opposite the largest angle is the longest side.
- For a=10, A=30° and b=12, sin B = 12 sin30°/10 = 0.6. B can be about 36.87° or 143.13°. The remaining angle is about 113.13° or 6.87°, respectively, so both triangles are possible unless more information selects one.
In plain language
Use matching opposite pairs for the sine rule and the included angle for the cosine rule. Keep full precision and check possible obtuse angles.
Self-check
Sides a=5, b=6 include C=70°. Find c to two decimal places.
Show the answer
c=√(5²+6²−2×5×6 cos70°) ≈ √40.47879 ≈ 6.36.
- Higher
Scope: Higher · Higher tier in either Unit 3 tier · Unit 3 §3.7.7 higher. Printed pages 37: use the formula: area of a triangle = ½absinC
Open revision guidance
What to know
- The area of a triangle can be calculated without the perpendicular height if two sides and the included angle are known. The formula is Area equals 0.5 * a * b * Sin C, where a and b are sides and C is the angle between them.
- This formula is particularly useful for finding the area of non-right-angled triangles or for calculating areas within 3D coordinate geometry problems. Ensure that the angle used is the one explicitly contained between the two specified side lengths.
What to do in an assessment
- Calculate the area of a triangle with side a=8m, side b=10m, and included angle 45 degrees. Using the formula, Area equals 0.5 * 8 * 10 * Sin 45. Sin 45 is approximately 0.707.
- Multiply 0.5 * 8 * 10 which is 40. Then, 40 * 0.707 equals 28.28 square meters. This result provides the total enclosed space of the triangle. Always ensure units are squared as this represents an area calculation.
In plain language
Find the area of any triangle using two sides and the angle between them multiplied by 0.5 and sine.
Self-check
A triangle has sides 6cm and 9cm with an included angle of 30 degrees. Find the area.
Show the answer
13.5 square centimeters. Calculated by 0.5 * 6 * 9 * Sin 30 equals 0.5 * 54 * 0.5 equals 13.5.
- Higher
Scope: Higher · Higher tier in either Unit 3 tier · Unit 3 §3.7.8 higher. Printed pages 37: sketch, understand the behaviour of, and use the graphs of trigonometric functions
Open revision guidance
What to know
- The graph of y = sin(x) is a periodic wave that begins at the origin (0,0), reaching a maximum height of 1 at 90 degrees, crossing the x-axis at 180 degrees, hitting a minimum of -1 at 270 degrees, and returning to the x-axis at 360 degrees. This function repeats its values every 360 degrees, which is defined as its period. The amplitude of this standard wave is 1, measured as the distance from the centre line to the peak. Understanding these points allows you to sketch the curve accurately by plotting coordinates such as (0, 0), (90, 1), (180, 0), (270, -1), and (360, 0).
- The graph of y = cos(x) behaves similarly as a periodic wave, but it starts at its maximum value of 1 when x = 0 degrees. The function crosses the x-axis at 90 degrees, reaches a minimum value of -1 at 180 degrees, crosses the x-axis again at 270 degrees, and returns to 1 at 360 degrees. By memorising these key coordinates like (0, 1), (90, 0), (180, -1), (270, 0), and (360, 1), you can produce a consistent sketch. Because both sine and cosine functions oscillate between 1 and -1, they are foundational for modelling periodic real-world phenomena like tide heights or sound waves.
What to do in an assessment
- To sketch y = sin(x) for the interval 0 to 360 degrees, first draw axes with degrees on the horizontal x-axis and values on the vertical y-axis. Mark the amplitude as 1 and -1. Calculate the y-value for x = 45 degrees using a calculator: sin(45) equals approximately 0.707. Since sin(45) is positive, you place a point halfway between the origin and 90 degrees at 0.707. Continue this process for 135 degrees, where sin(135) also equals 0.707. Join the points (0,0), (45, 0.707), (90,1), (135, 0.707), (180,0), (225, -0.707), (270,-1), (315, -0.707), and (360, 0) with a smooth curve.
- To graph y = cos(x) over the same interval, plot your points starting at (0, 1). Calculate the value at 60 degrees: cos(60) equals 0.5. Plot (60, 0.5). Next, find cos(120), which equals -0.5. Plot (120, -0.5). By calculating these intermediate steps, you ensure the wave transition through the x-axis at 90 degrees is accurate. Connect (0,1), (60, 0.5), (90,0), (120, -0.5), (180, -1), (240, -0.5), (270, 0), (300, 0.5), and (360, 1). Notice that the curve is symmetrical about the y-axis, providing a visual verification of the cosine identity.
In plain language
Trigonometric graphs are repeating waves that show how sine and cosine values change as an angle increases. Sine starts at zero and rises, while cosine starts at one and falls. Both waves move smoothly between 1 and -1 and repeat their cycle every 360 degrees.
Self-check
For the function y = sin(x), what is the value of y when x = 270 degrees?
Show the answer
y = -1. Working: From the definition of the sine wave, the graph reaches its minimum value of -1 at 270 degrees.
View diagrams
Sine and cosine over a full turn
Key and description
Solid curve: y=sin x; dashed curve: y=cos x, in degrees. At 0°,90°,180°,270°,360°, sine is 0,1,0,−1,0 and cosine is 1,0,−1,0,1. Both have amplitude 1 and period 360°.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §3.8.1 foundation. Printed pages 38: find coordinates identified by given geometrical information including: the midpoint of a line the fourth vertex of a parallelogram location determined by distance from a given point and angle made with a given line.
Open revision guidance
What to know
- For endpoints (x₁,y₁) and (x₂,y₂), the midpoint is ((x₁+x₂)/2, (y₁+y₂)/2). Average the coordinates separately. This produces a point halfway along the segment.
- For a parallelogram ABCD with vertices named consecutively round its boundary, the displacement A to D equals B to C, and A to B equals D to C. Equivalently, both diagonals have the same midpoint. Naming the vertex order matters: the same three points can give different fourth points if the order changes.
What to do in an assessment
- The midpoint of A(2,4) and B(6,10) is ((2+6)/2, (4+10)/2) = (4,7). Check that the displacement from A to the midpoint is (2,3), the same as the displacement from the midpoint to B.
- In consecutive parallelogram ABCD, A(1,1), B(4,2), C(5,5). B to C is (1,3), so D = (1,1)+(1,3) = (2,4). Check the diagonal midpoints: AC has midpoint (3,3) and BD also has midpoint (3,3).
- To locate a point 5 units from A(1,2) at 90° anticlockwise from the positive horizontal direction, draw that direction and move 5 units upwards: the point is (1,7). If AB has displacement (3,1) and C(2,4), a parallel segment CD with the same length and direction has D = (2+3,4+1) = (5,5). Reversing the direction would give a different endpoint.
In plain language
Average coordinates for a midpoint; use equal displacements or bisecting diagonals for a parallelogram.
Self-check
What is the midpoint of (0,0) and (10,8)?
Show the answer
((0+10)/2, (0+8)/2) = (5,4).
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §3.8.2 foundation. Printed pages 38: describe and draw shapes with line symmetry
Open revision guidance
What to know
- Line symmetry, or reflectional symmetry, occurs when a shape can be folded along a line such that both halves match perfectly. A shape might have one, multiple, or no lines of symmetry. For instance, an isosceles triangle has one line of symmetry, whereas a regular pentagon has five.
- To describe line symmetry, identify the number of lines and their orientation, such as vertical, horizontal, or diagonal. Drawing these shapes requires plotting key vertices and ensuring that the reflected points are equidistant from the symmetry line on the opposite side.
What to do in an assessment
- Example: Consider a square with vertices at (0,0), (2,0), (2,2), and (0,2). This shape has four lines of symmetry: the vertical line x=1, horizontal line y=1, and diagonal lines y=x and y=2-x. To draw this, plot the vertices and use a dashed line to indicate the axes of symmetry passing through the centre at (1,1).
- Example: Consider a kite with vertices (0,1), (2,0), (0,-3), and (-2,0). This shape has one vertical line of symmetry along the y-axis (x=0). By drawing the line of symmetry, you confirm that for every point (x, y) on the right, there is a corresponding (-x, y) on the left, maintaining the shape's property.
In plain language
A line of symmetry is a line that divides a shape into two halves that are perfect reflections of each other.
Self-check
How many lines of symmetry does a rectangle that is not a square have?
Show the answer
It has two lines of symmetry: one vertical through the centre and one horizontal through the centre.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §3.8.3 foundation. Printed pages 38: draw lines of symmetry on a shape
Open revision guidance
What to know
- Drawing lines of symmetry requires precision to ensure the split sections match correctly when folded. For complex shapes, consider the geometric centre and the path between matching vertices. A line of symmetry always bisects the distance between corresponding points.
- When working on a grid, you can use the coordinate system to verify the line. For instance, if a shape is defined by coordinates, the line of symmetry is often a grid line or a diagonal path (y=x or y=-x). Identifying these helps in maintaining accurate geometry.
What to do in an assessment
- Example: Draw the lines of symmetry for a regular hexagon with centre (0,0) and vertices at (2,0), (1, 1.73), (-1, 1.73), (-2,0), (-1, -1.73), and (1, -1.73). The lines of symmetry include the x-axis, y-axis, and lines through opposite vertices such as y=1.73x and y=-1.73x. Use a ruler to mark these lines clearly through the centre point (0,0).
- Example: For an isosceles trapezoid with coordinates (1,1), (3,1), (4,3), and (0,3), identify the centre. The line of symmetry is x=2. Drawing this vertical line shows it passes through the midpoints of the parallel sides, confirming the reflectional balance between the left and right sections of the trapezoid.
In plain language
Creating lines of symmetry means finding the path that divides a shape into two parts that are a reflection of each other.
Self-check
If a shape is a circle, how many lines of symmetry can be drawn?
Show the answer
An infinite number, as any line passing through the centre is a line of symmetry.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §3.8.4 foundation. Printed pages 38: understand order of rotational symmetry and describe and draw shapes with rotational symmetry
Open revision guidance
What to know
- The order of rotational symmetry counts how many times a shape matches its original position during a full 360° turn, including the final full turn but not counting the starting position again. Order 1 means only the full turn matches.
- A regular n-sided polygon has order n, with matching positions every 360°/n. A non-square rectangle has order 2, whereas a square has order 4. The rotation must use the shape's correct centre.
What to do in an assessment
- A rectangle with vertices (0,0),(4,0),(4,2),(0,2) has centre (2,1). A 180° rotation swaps (0,0) with (4,2) and (4,0) with (0,2). It also matches at 360°, giving order 2.
- An exact equilateral triangle has vertices (0,√3),(−1,0),(1,0) and centre (0,√3/3). A 120° clockwise rotation sends the top vertex to the lower-right vertex. The whole triangle matches at 120°, 240° and 360°, so its order is 3.
In plain language
Rotate about the centre and count all matching positions in one complete turn.
Self-check
What is the order of rotational symmetry of a regular pentagon, and what is its smallest positive matching rotation?
Show the answer
Order 5; smallest rotation = 360°/5 = 72°.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §3.8.5 foundation. Printed pages 38: describe and draw the following transformations: reflection in a given line rotation of 90 º or 180 º , clockwise or anticlockwise, using a centre of rotation translation with direction and distance of horizontal and vertical movement enlargement with positive integer or a positive, fractional scale factor, and a centre of enlargement
Open revision guidance
What to know
- Transformations involve mapping shapes to new positions through reflection, rotation, translation, or enlargement. Reflection uses a line of symmetry; rotation requires a centre, angle, and direction; translation uses a vector (horizontal, vertical); and enlargement uses a scale factor and a centre point.
- Enlargement changes the size of the shape. If the scale factor k is positive, the shape expands or shrinks from the centre of enlargement. For example, a scale factor of 2 doubles all distances from the centre to the vertices. A fractional scale factor (e.g., 0.5) results in a smaller shape.
What to do in an assessment
- Example (Rotation): Rotate the point A(1, 1) by 90 degrees clockwise about centre (0, 0). The rule for 90 clockwise is (x, y) -> (y, -x). So A(1, 1) becomes (1, -1).
- Example (Enlargement): Enlarge the triangle with vertices A(1, 1), B(2, 1), and C(1, 2) by a scale factor of 3 from the centre (0, 0). The new vertices are (1*3, 1*3) = (3, 3), (2*3, 1*3) = (6, 3), and (1*3, 2*3) = (3, 6).
- A fractional enlargement can shrink a shape: with centre (0,0), factor ½ maps (2,2), (4,2), (2,4) to (1,1), (2,1), (1,2). The centre can be outside, inside or on the boundary of a shape. To describe an enlargement from two drawings, compare corresponding lengths for its scale factor; if a centre is required, extend the lines joining corresponding vertices to find it. Questions can also ask for an enlargement without specifying a centre, as clarified in the September 2026 guidance.
In plain language
Transformations move shapes using reflection, rotation, translation, or enlargement.
Self-check
Translate the point (2, 3) by the vector (4, -1). What is the new coordinate?
Show the answer
(6, 2).
- Higher
Scope: Higher · Higher tier in either Unit 2 tier · Unit 2 §3.8.5 higher. Printed pages 39: enlargement with a negative scale factor and centre of enlargement.
Open revision guidance
What to know
- For centre C and scale factor k, map each point P to P′ = C + k(P−C). A negative k places the image on the opposite side of the centre from the object. Lengths multiply by |k| and corresponding angles remain equal.
- The straight line joining each point and its image passes through the centre. For k = −1, distances from the centre stay equal and the result is equivalent to a 180° rotation about the centre. A negative fractional factor also reduces the size.
What to do in an assessment
- With centre (0,0) and k = −2, multiply both coordinates by −2. Triangle (1,1), (2,1), (1,2) maps to (−2,−2), (−4,−2), (−2,−4). Draw both triangles and join corresponding vertices through the origin.
- With centre C(1,1), point P(5,3) has displacement (4,2). For k = −½, multiply that displacement to get (−2,−1), then add the centre: P′ = (−1,0). Multiplying the original coordinates directly would incorrectly assume the centre was the origin.
- A fractional enlargement can shrink a shape: with centre (0,0), factor ½ maps (2,2), (4,2), (2,4) to (1,1), (2,1), (1,2). The centre can be outside, inside or on the boundary of a shape. To describe an enlargement from two drawings, compare corresponding lengths for its scale factor; if a centre is required, extend the lines joining corresponding vertices to find it. Questions can also ask for an enlargement without specifying a centre, as clarified in the September 2026 guidance.
In plain language
Start at the centre, reverse the direction, and scale the distance by the magnitude of the factor.
Self-check
Enlarge P(3,−2) by factor −3 about (0,0).
Show the answer
P′ = (−9,6), on the opposite side of the origin and three times as far from it.
View diagrams
Negative enlargement from centre O
Key and description
Original triangle: (1, 1), (3, 1), (1, 2). Enlargement centre O, scale factor −2 gives (−2, −2), (−6, −2), (−2, −4). Join O to each vertex and extend through O: the image is on the opposite side and twice as far away. This is a negative enlargement, not a reflection in a mirror line.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Higher
Scope: Higher · Higher tier in either Unit 2 tier · Unit 2 §3.8.6 higher. Printed pages 39: describe and draw two successive transformations
Open revision guidance
What to know
- Successive transformations occur when one shape undergoes two distinct movements in sequence. For example, a rotation followed by a translation results in a final position. It is critical to perform the transformations in the specified order, as the intermediate shape serves as the input for the second step.
- When describing two successive transformations, one might sometimes express the outcome as a single equivalent transformation, such as a single rotation or reflection. This requires tracking the movement of individual vertices through the coordinate grid to find the final mapping.
What to do in an assessment
- Example: Apply a rotation of 90 degrees clockwise about (0,0) to point P(2, 1), then a translation by vector (3, 2). First, (2, 1) rotated 90 clockwise becomes (1, -2). Then, applying translation (3, 2): (1+3, -2+2) = (4, 0).
- Example: Apply an enlargement of factor 2 about (0,0) to a square with vertices (1, 1), (2, 1), (2, 2), (1, 2). The new vertices are (2, 2), (4, 2), (4, 4), (2, 4). Then, reflect these points in the line x=0. The final vertices are (-2, 2), (-4, 2), (-4, 4), (-2, 4).
In plain language
Consecutive transformations occur when a shape undergoes two different movements one after the other.
Self-check
If you rotate (1, 0) by 90 degrees counter-clockwise about (0,0) and then translate by (0, 1), where does the point end up?
Show the answer
Rotating (1, 0) 90 deg CCW gives (0, 1). Translating (0, 1) by (0, 1) results in (0, 2).
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 3 tier · Unit 3 §3.9.1 foundation. Printed pages 39: identify congruent and similar shapes
Open revision guidance
What to know
- Congruent shapes are defined as figures that are identical in both shape and size. Two triangles, for example, are congruent if all three sides correspond exactly in length and all three internal angles match. If you can move, rotate, or reflect one shape to perfectly overlay another, they are congruent. This relationship ensures that area, perimeter, and side lengths remain invariant across the shapes.
- Similar shapes possess the same internal angles but differ in scale. They maintain the same relative proportions between corresponding sides. For instance, a rectangle with sides 2 cm and 4 cm is similar to one with sides 4 cm and 8 cm because the ratio of corresponding sides is constant at 2. Similarity does not imply congruence unless the scale factor is exactly 1.
What to do in an assessment
- To identify congruent shapes, compare the internal angles and side lengths systematically. Given Triangle A with sides (3 cm, 4 cm, 5 cm) and Triangle B with sides (3 cm, 4 cm, 5 cm), the sides match perfectly under SSS. If Triangle C has sides (6 cm, 8 cm, 10 cm), it is similar to A and B because 6/3 = 8/4 = 10/5 = 2. These ratios define the scale factor.
- Practise comparing geometric sets. Consider a square with vertices at (0,0), (2,0), (2,2), and (0,2). A second square with vertices at (0,0), (4,0), (4,4), and (0,4) is similar but not congruent. The side length ratio is 2, confirming it is an enlargement. Always check that all corresponding angles are equal to verify similarity.
In plain language
Congruent means exactly the same shape and size. Similar means the same shape but a different size, like a photo that has been zoomed in.
Self-check
Are two triangles with side lengths (5, 12, 13) and (10, 24, 26) congruent or similar?
Show the answer
They are similar. The side ratios are 10/5 = 2, 24/12 = 2, and 26/13 = 2. Since the ratio is constant but not 1, they are similar, not congruent.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 3 tier · Unit 3 §3.9.2 foundation. Printed pages 39: use the knowledge that, for two similar 2-D or 3-D shapes, one is an enlargement of the other
Open revision guidance
What to know
- An enlargement occurs when a shape is scaled by a factor to create a similar shape. Whether in 2-D or 3-D, if one shape is an enlargement of the other, every linear dimension is multiplied by a common scale factor. This process increases the size of the shape while preserving the geometric integrity, ensuring all angles remain identical.
- In a 3-D context, such as a cuboid with dimensions (l, w, h), an enlargement by a scale factor of k results in new dimensions (kl, kw, kh). The spatial orientation and relative proportions are maintained, which is why the two shapes are considered similar. It is essential to recognise that enlargement is a transformation that changes length while maintaining the shape definition.
What to do in an assessment
- Identify the scale factor by dividing a side of the larger shape by the corresponding side of the smaller shape. For a 2-D rectangle with sides 4 cm and 6 cm being enlarged to 8 cm and 12 cm, calculate the scale factor: 8/4 = 2 and 12/6 = 2. Since both ratios yield 2, the shape is an enlargement by a scale factor of 2.
- Apply enlargement to coordinates. If a triangle has vertices at A(1,1), B(2,1), and C(1,2), an enlargement with scale factor 3 from the origin moves the vertices to A'(3,3), B'(6,3), and C'(3,6). Verify the similarity by checking that side AB has length 1 and A'B' has length 3, confirming the scale factor 3 applies correctly to all distances.
In plain language
When a shape is enlarged, all its sides grow by the same ratio. This keeps the shape the same but makes it larger.
Self-check
If a square with side length 5 cm is enlarged to a side length of 15 cm, what is the scale factor?
Show the answer
The scale factor is 3, because 15 cm / 5 cm = 3.
- Higher
Scope: Higher · Higher tier in either Unit 3 tier · Unit 3 §3.9.3 higher. Printed pages 39: use the knowledge that, in similar shapes, corresponding dimensions are in the same ratio
Open revision guidance
What to know
- Similar shapes have one fixed ratio between corresponding lengths. If the scale factor from a small shape to a larger one is k, multiply every corresponding linear dimension by k, including heights, diagonals and perimeters.
- Identify corresponding dimensions before forming a ratio. Base-small/base-large = height-small/height-large. This is a linear scale relationship; area and volume use different powers of the scale factor.
What to do in an assessment
- A small right triangle has base 3 cm and perpendicular height 4 cm. A similar triangle has corresponding base 9 cm. Scale factor = 9/3 = 3, so the corresponding height is 4×3 = 12 cm. Check: base ratio and height ratio are both 3.
- A trapezium has parallel bases 2 cm and 4 cm and perpendicular height 3 cm. A similar trapezium has shorter base 6 cm. Scale factor = 6/2 = 3, so its longer base is 12 cm and height is 9 cm. Each corresponding length ratio agrees.
In plain language
Match like dimensions, find the scale factor and apply it to every corresponding length.
Self-check
A rectangle has sides 4 cm and 6 cm. A similar rectangle has shorter side 10 cm. What is its longer side?
Show the answer
Scale factor = 10/4 = 2.5. Longer side = 6×2.5 = 15 cm.
- Higher
Scope: Higher · Higher tier in either Unit 3 tier · Unit 3 §3.9.4 higher. Printed pages 39: use the knowledge that, in similar and congruent shapes, corresponding angles are equal
Open revision guidance
What to know
- Similar shapes have equal corresponding angles and corresponding lengths in one fixed ratio. Congruent shapes have the same shape and size, so corresponding angles and lengths are equal. Identify which vertex matches which before comparing angles.
- For triangles, matching corresponding angles is enough to establish similarity (AA, with the third angle following from a 180° sum). This converse is not true for arbitrary polygons: two rectangles can both have four right angles but different length-to-width ratios.
What to do in an assessment
- Triangle ABC is similar to DEF with A matching D, B matching E and C matching F. If A = 40° and B = 60°, then C = 180° − 40° − 60° = 80°. Therefore D = 40°, E = 60° and F = 80°, regardless of the scale factor.
- An equilateral triangle with side 3 cm and another with side 6 cm have angles 60°, 60°, 60° and are similar, with length scale factor 2, but not congruent. A 2-by-3 rectangle and a 2-by-5 rectangle have matching right angles but side ratios 3/2 and 5/2, so they are not similar.
In plain language
Match corresponding corners carefully. Equal angles imply triangle similarity, but other polygons also need proportional corresponding sides.
Self-check
Two triangles both have angles 40°, 60° and 80°. Are they necessarily similar? Are they necessarily congruent?
Show the answer
They are necessarily similar by equal corresponding angles. They need not be congruent because their side lengths may differ by a scale factor other than 1.
- Higher
Scope: Higher · Higher tier in either Unit 3 tier · Unit 3 §3.9.5 higher. Printed pages 39: use the relationships between the ratios of lengths, areas, volumes and capacities of similar shapes
Open revision guidance
What to know
- When comparing similar shapes, the ratios of their dimensions change based on the power of the dimension being measured. If the scale factor for length is k, then the ratio of their surface areas is k^2, and the ratio of their volumes or capacities is k^3. This is essential for scaling complex 3-D objects where volume increases much faster than surface area.
- For example, a cube with side 1 cm has area 6 cm^2 and volume 1 cm^3. If scaled by k=3, the new cube has side 3 cm, area 6 * 3^2 = 54 cm^2, and volume 1 * 3^3 = 27 cm^3. This exponential growth relationship is vital for understanding physical scaling, such as packaging design or fluid dynamics within similar vessels.
What to do in an assessment
- Determine the scale factor first. If the length of Shape A is 2 cm and Shape B is 4 cm, k = 4/2 = 2. Use k^2 for area and k^3 for volume. If the area of Shape A is 5 cm^2, the area of Shape B is 5 * (2^2) = 5 * 4 = 20 cm^2. If the volume of Shape A is 10 cm^3, the volume of Shape B is 10 * (2^3) = 10 * 8 = 80 cm^3.
- Apply this to capacity. If a jug with height 10 cm holds 500 ml, a similar jug with height 20 cm has a scale factor of 2. The volume capacity ratio is 2^3 = 8. Therefore, the larger jug holds 500 ml * 8 = 4000 ml. Always ensure you are cubing the linear scale factor, not the original volume, to find the new capacity.
In plain language
When you enlarge a shape, its length increases by a factor, its area increases by the squared factor, and its volume increases by the cubed factor.
Self-check
A sphere has a radius of 2 cm and volume 33.5 cm^3. If the radius is doubled (scale factor 2), what is the new volume?
Show the answer
The scale factor k=2. Volume increases by k^3 = 8. New volume = 33.5 * 8 = 268 cm^3.
- Higher
Scope: Higher · Higher tier in either Unit 3 tier · Unit 3 §3.9.6 higher. Printed pages 39: understand and use the following conditions to prove the congruence of triangles using formal arguments: side-side-side (SSS) side-angle-side (SAS) angle-angle-side (AAS) right angle-hypotenuse- side (RHS).
Open revision guidance
What to know
- To prove two triangles are congruent, you must establish one of four criteria: Side-Side-Side (SSS), where all three sides match; Side-Angle-Side (SAS), where two sides and the included angle match; Angle-Angle-Side (AAS), where two angles and a non-included side match; or Right angle-Hypotenuse-Side (RHS), which applies specifically to right-angled triangles.
- Formal proofs require stating the condition used and identifying the matching elements. For instance, in an RHS proof, you must explicitly state that the angle is 90 degrees, the hypotenuses are equal, and one other pair of sides is equal. This structure provides a logically sound basis for claiming congruence without needing to measure every single feature of the shapes.
What to do in an assessment
- Apply SSS: Triangle 1 has sides (3,4,5), Triangle 2 has (3,4,5). Since all sides are equal, they are congruent. Apply SAS: Triangle 1 has side 5, angle 30, side 6. Triangle 2 has side 5, angle 30, side 6. Since the angle is 'included' between the sides, they are congruent. Provide these identifiers clearly in any proof.
- Use RHS: Triangle 1 has hypotenuse 10, leg 6, angle 90. Triangle 2 has hypotenuse 10, leg 6, angle 90. Because the hypotenuse and one leg match, they are congruent by RHS. Document each step by citing the specific measurements that satisfy the chosen criteria, such as 'AB = DE = 5 cm' or 'angle B = angle E = 90 degrees'.
In plain language
You can prove that two triangles are exactly the same (congruent) by using specific rules about their sides and angles, such as SSS, SAS, AAS, or RHS.
Self-check
Two right-angled triangles have hypotenuse 13 cm and one leg 5 cm. Are they congruent?
Show the answer
Yes, by RHS (Right angle-Hypotenuse-Side). The third side would be 12 cm by Pythagoras, confirming congruence.
Topic group
Statistics and probability
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §4.1.1 foundation; Unit 3 §4.1.1 foundation. Printed pages 40: understand and use the statistical problem solving process: specify the problem/planning; collect, process and represent data; interpret and discuss results, including limitations of data and anomalies
Open revision guidance
What to know
- The statistical cycle consists of a logical flow of steps starting with specifying the problem, planning data collection, collecting the data, processing and representing it through graphs or tables, and finally interpreting the results to answer the original question. It is essential to understand that this process is cyclical because the findings often lead to new questions or a need to refine the original research objectives.
- Anomalies or outliers are data points that do not fit the general pattern and must be identified during the processing stage. One must always discuss the limitations of the data, such as whether the sample was large enough to be representative or if the measurement techniques used were biased. A dataset example for testing would be measuring the heights of 5 students: 150cm, 155cm, 160cm, 158cm, and 280cm; the 280cm value is an anomaly requiring investigation.
What to do in an assessment
- To process data, first define your research question, such as 'Does the average time spent studying per week exceed 10 hours?' Then, collect primary data from a set group using a tally chart or survey. Represent this using a bar chart where the x-axis shows study intervals (0-5, 5-10, 10-15 hours) and the y-axis shows the frequency. For a sample of 20 students, if the frequencies are 4, 8, and 8, you would calculate the mean by multiplying the midpoints (2.5, 7.5, 12.5) by frequencies.
- Interpret the results by comparing your calculated average to the initial hypothesis. If the mean is 8.25 hours, explain that the hypothesis is not supported. Discuss limitations: perhaps the data collection only included one class, which is not representative of the whole school population. Always verify if any extreme values were included in the calculation and state their effect on the final mean or range.
In plain language
The statistical cycle is a way to solve real-world problems by planning, collecting, calculating, and then explaining what the data means, while being honest about any errors or missing information.
Self-check
If you have data points 2, 3, 3, 4, 25, what is the anomaly and why does it change the mean?
Show the answer
The anomaly is 25. The mean including 25 is 37/5 = 7.4. The mean excluding the anomaly is 12/4 = 3. The anomaly pulls the average far from the typical data values.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §4.1.2 foundation; Unit 3 §4.1.2 foundation. Printed pages 40: specify and test hypotheses, taking account of the limitations of the data available
Open revision guidance
What to know
- Hypothesis testing begins with a statement that can be proved or disproved by examining data. A hypothesis is a claim about a population parameter or a relationship between variables. For example, a hypothesis might be 'Students who drink water perform better on tests.' To test this, you must specify the null hypothesis (no effect) and the alternative hypothesis (a real effect), while acknowledging that data availability often limits the scope of your investigation.
- When testing hypotheses, consider the limitations of the data, such as sample size, bias in data collection, or measurement errors. If your data set consists of only 5 test scores, your ability to make a valid claim is weak. Always evaluate if the conclusion is based on enough evidence. For instance, comparing the test results of two groups: Group A (5 students: 60, 65, 70, 75, 80) and Group B (5 students: 85, 90, 95, 100, 105). The means are 70 and 95, showing a clear difference.
What to do in an assessment
- First, set your hypothesis: 'Students studying at night achieve higher scores than those studying in the morning.' Next, collect data by recording scores for two distinct groups. Ensure the groups are of sufficient size to mitigate random variation. Use a table to organise your data. Calculate the average score for each group and compare the outcomes. If the night group averages 72 and the morning group averages 70, note that the difference is small and might be due to chance.
- Critique your results. If you only surveyed 10 students, state this as a limitation. If one student in the night group scored 0 due to illness, note this as an anomaly that skewed your result. Conclude by either accepting or rejecting the hypothesis based on your quantitative findings. Ensure your explanation includes why the data was or was not sufficient, considering the sample size and any potential bias present in the selection process.
In plain language
Hypothesis testing means making a statement about something and checking it using data, while considering whether the data is sufficient or if there are errors in your information.
Self-check
If Group X has a mean of 50 and Group Y has a mean of 52 based on 3 people each, is the hypothesis that Group Y is better supported?
Show the answer
Not strongly. Because the sample size (3) is very small, the difference of 2 could easily be caused by chance variation rather than a real performance difference.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §4.1.3 foundation; Unit 3 §4.1.3 foundation. Printed pages 40: design and criticise questions for a questionnaire, including notions of fairness and bias
Open revision guidance
What to know
- Designing a questionnaire requires careful attention to the clarity and neutrality of the questions. A good question avoids bias, which is when a question leads the respondent toward a specific answer. For example, 'Don't you think our local park is excellent?' is a leading question. A better, fairer question would be 'How would you rate the local park?' with a neutral scale like 'Very Good, Good, Average, Poor, Very Poor'.
- When designing questionnaires, consider the target audience and ensure options cover all possible answers. Overlapping ranges in numeric options must be avoided. For instance, if asking for age, use ranges such as '0-10, 11-20, 21-30'. If you use '0-10' and '10-20', a 10-year-old would not know which box to tick. Fairness implies that the questionnaire should not pressure participants or exclude relevant perspectives by limiting choices.
What to do in an assessment
- To design a survey, start by listing the specific information needed. Construct questions that are short, clear, and unambiguous. Use open-ended questions for detailed feedback and closed-ended questions for statistical analysis. For a survey on transportation, use: 'How many days per week do you cycle to school?' with options: '0, 1-2, 3-4, 5'. This set covers all possibilities without overlap. Pilot the questionnaire with a small group to identify any confusing wording.
- Critique your survey for potential bias. If asking about satisfaction with cafeteria food, avoid asking 'Is the food tasty?' and instead use 'How would you rate the taste of the cafeteria food?' Provide a balanced scale from 1 (Very Dissatisfied) to 5 (Very Satisfied). Ensure anonymity if the topic is sensitive to encourage honest responses. Check that your question does not assume the respondent performs an activity, such as 'When did you stop smoking?' (if they never smoked, this is biased).
In plain language
Designing a questionnaire means creating clear and unbiased questions, ensuring that the answers provide information without leading respondents to a specific answer.
Self-check
Identify the issue: 'Why is the school uniform so uncomfortable?'
Show the answer
This is a leading question because it assumes the respondent finds the uniform uncomfortable, which introduces bias into the data.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §4.1.4 foundation; Unit 3 §4.1.4 foundation. Printed pages 40: consider the effect of sample size and other factors that affect the reliability of data and conclusions drawn
Open revision guidance
What to know
- Reliability refers to the consistency and trustworthiness of data. Sample size is a critical factor; a larger, randomly selected sample is generally more reliable than a small or non-representative one. If you sample 5 people to represent a town of 10,000, your results will likely be unreliable because the sample fails to capture the diversity of the population. Other factors affecting reliability include how the data was measured and whether the environment was controlled.
- Consider the effect of bias, such as asking questions to only one age group when the survey topic affects everyone. Reliability is also threatened by human error during collection. For example, if recording the height of participants (150cm, 155cm, 160cm), an error where one person is recorded as 1500cm instead of 150cm will drastically lower the reliability of the mean. Always re-check data for such outliers before drawing conclusions.
What to do in an assessment
- To improve reliability, first increase your sample size. If investigating the preference for ice cream flavors, surveying 100 people is more reliable than surveying 10. Next, use random sampling to ensure every member of the population has an equal chance of being selected. Avoid convenience sampling, such as only surveying your friends, as this introduces bias. If you gather data on commute times (e.g., 20 mins, 25 mins, 22 mins), ensure the measuring instrument (a clock) is accurate.
- When analysing the reliability of conclusions, compare data across different sub-groups. If one group reports significantly different outcomes, investigate if it's due to a real trend or an issue with the sample selection. Document all methods used to collect the data to allow for replication. If your findings show that 80% of 10 people prefer tea, but 55% of 100 people prefer tea, acknowledge that the larger sample size provides a more accurate representation of the population's preference.
In plain language
Reliability is about how consistent data is. Larger samples and random selections make your conclusions more reliable and accurate.
Self-check
Which is more reliable: a survey of 50 randomly chosen people or a survey of 10 friends?
Show the answer
The survey of 50 randomly chosen people is more reliable because it has a larger sample size and is less likely to be influenced by the specific bias of a small friend group.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 3 tier · Unit 3 §4.1.5 foundation. Printed pages 41: understand and use tallying methods
Open revision guidance
What to know
- A tally chart records one mark for each observation, normally grouping marks into fives to aid counting. Convert the tallies into frequencies, then check that the total frequency equals the number of observations.
- Categories must be clear and not overlap. For grouped numerical data, define boundaries so that every possible observation goes into exactly one class, for example 0 ≤ x < 10 and 10 ≤ x < 20.
What to do in an assessment
- For scores 5, 8, 5, 9, 8, 5, 5, 9, 8, 5, tally each score in turn. Score 5 occurs five times, 8 occurs three times and 9 occurs twice. The frequency table is 5:5, 8:3, 9:2. The total is 5 + 3 + 2 = 10, matching the ten observations.
- Of twelve vehicles, six are red, four blue and two green. Record these three categories with frequencies 6, 4 and 2. Their total is 12. If totals disagree, return to the original observations rather than altering a frequency to force the total.
In plain language
Count each observation once and verify the frequencies against the original data.
Self-check
Make a frequency table for 2, 3, 2, 4, 2, 3.
Show the answer
Value 2 has frequency 3, value 3 frequency 2, and value 4 frequency 1. Total = 3 + 2 + 1 = 6.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 3 tier · Unit 3 §4.1.6 foundation. Printed pages 41: understand and use frequency tables
Open revision guidance
What to know
- A frequency table is a structured tabular representation that maps individual data values or ranges to the total count (frequency) of their occurrences within a dataset. By consolidating raw, unorganized data into a summary format, frequency tables enable analysts to identify patterns, such as the mode (the most frequent value), and to perform further statistical analysis like calculating the mean or median. This is particularly useful for managing large datasets where simple lists would become cumbersome or impossible to interpret visually.
- The layout of a frequency table requires two primary columns: the data category or interval and the corresponding frequency. For discrete data, such as shoe sizes (3, 4, 3, 5, 4, 3), the table lists sizes in the first column and the number of times each appears in the second column. For example, if size 3 appears three times, the frequency is 3. This organisation transforms disjointed observations into a clear, usable format that serves as the foundation for more advanced visualizations like bar charts or histograms, ensuring data integrity is maintained throughout the aggregation process.
What to do in an assessment
- To construct a frequency table, identify your dataset and group the values. Let us use a dataset of student ages: 12, 13, 12, 14, 13, 12, 12, 14, 13, 12. First, identify the unique ages: 12, 13, and 14. Set up a table with two columns: 'Age (Years)' and 'Frequency'. Go through the list and count each occurrence carefully. For age 12, count: 12, 12, 12, 12, 12. That is 5 occurrences. For age 13, count: 13, 13, 13. That is 3 occurrences. For age 14, count: 14, 14. That is 2 occurrences. Ensure your total sum (5 + 3 + 2 = 10) matches the original count of observations.
- After populating the table, you can calculate total frequencies or interpret the distribution. For example, by checking the frequency column, you can immediately state that the modal age is 12 because it has the highest frequency of 5. This method is mathematically robust because it accounts for every individual data point, leaving no information unaccounted for. When dealing with continuous data, you would group the data into classes like '10-12', '13-15', and '16-18', ensuring that these intervals are mutually exclusive and collectively exhaustive to prevent data overlapping or loss.
In plain language
A frequency table is a grid that shows how many times each item appears in a list. It helps you see which items are most common without needing to look through a messy list. It organises information so you can answer questions about the data quickly.
Self-check
Using the dataset of scores: 2, 4, 2, 3, 2. What is the frequency of score 2?
Show the answer
The frequency of score 2 is 3, as it appears three times in the dataset.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 3 tier · Unit 3 §4.1.7 foundation. Printed pages 41: sort, classify and tabulate qualitative (categorical) data, discrete or continuous quantitative data
Open revision guidance
What to know
- Sorting and classifying data is a fundamental skill that involves categorising information based on its specific nature, such as qualitative (categorical) or quantitative (numerical). Qualitative data describes attributes or characteristics that are not inherently numerical, such as eye colour or favourite pet. Conversely, quantitative data is measurable and can be further subdivided into discrete data, which consist of distinct values (like the number of people in a room), or continuous data, which can take any value within a range (like weight or temperature measured in kilograms or Celsius).
- Effective data management begins by correctly identifying the type of information being handled. If you are tabulating the number of siblings in families, you are working with discrete quantitative data because the counts must be whole numbers. If you are tabulating the height of plants in centimeters, you are dealing with continuous quantitative data because height can fall anywhere on a continuum (e.g., 10.5 cm or 10.52 cm). Correct classification dictates the choice of appropriate statistical tools and visual representations, ensuring that the final analysis accurately reflects the underlying data reality.
What to do in an assessment
- To sort data effectively, follow a logical categorisation process. Imagine a dataset: (Blue, 15, 2.3 kg, Green, 18, 4.1 kg, Blue, 12, 1.2 kg). First, extract the categories. The colours are qualitative (Blue, Green). The integers (15, 18, 12) represent discrete data (perhaps counts of items). The decimals (2.3, 4.1, 1.2) with units represent continuous data (mass in kilograms). Create three separate tables for each type to organise them. Use the tallying method to count the frequency of each category within these groups.
- For the quantitative discrete data (15, 18, 12), create a table with columns 'Value' and 'Frequency'. Input the values and their frequencies (each is 1 here, totaling 3). For continuous data (2.3 kg, 4.1 kg, 1.2 kg), if the dataset was larger, you would define intervals such as '0.0 - 2.0 kg' and '2.1 - 5.0 kg'. By sorting these, you can calculate the arithmetic mean of discrete items or determine the range of the continuous variables. This rigorous approach prevents the error of treating categorical identifiers as numerical values, which would invalidate any statistical outcome.
In plain language
Sorting data means putting things into groups based on what they are. You can have names or colours (qualitative), numbers you count (discrete), or numbers you measure like height (continuous). Sorting helps you choose the right way to show your information.
Self-check
If you are measuring the time in seconds it takes for a runner to finish a race, is this discrete or continuous data?
Show the answer
This is continuous data, because time is a measurement that can take any value within a range and is not limited to whole numbers.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §4.1.8 foundation; Unit 3 §4.1.8 foundation. Printed pages 41: group discrete or continuous data into class intervals of equal or unequal widths
Open revision guidance
What to know
- Grouping data into intervals is essential when dealing with large datasets or continuous variables where individual values vary too much to list separately. An interval (or class) covers a range of values, such as 10 ≤ x < 20. The width of the interval is the difference between the upper and lower boundaries. Equal width intervals are preferred for easier comparison across groups, though unequal intervals can be used if some ranges are more frequent than others.
- When creating class intervals, ensure every data point fits into exactly one category and that no gaps exist. For example, if you have heights of students: 140cm, 145cm, 150cm, 155cm, 160cm, you could group them into intervals: 140-149, 150-159, and 160+. If a student is 149.5cm, you must decide how to handle decimals. Use '140 ≤ h < 150' to avoid ambiguity.
What to do in an assessment
- To group data, first identify the range of your data set (maximum minus minimum). Decide on the number of groups you want (usually between 5 and 10). Divide the total range by the number of groups to estimate the interval width. For instance, with ages ranging from 5 to 55, a range of 50, using 5 intervals would give a width of 10. Your intervals would be: 5-14, 15-24, 25-34, 35-44, 45-55.
- Apply the intervals to your dataset: {7, 12, 18, 22, 28, 33, 40, 50}. Tally the counts: 5-14 (7, 12), 15-24 (18, 22), 25-34 (28, 33), 35-44 (40), 45-55 (50). This results in frequencies: 2, 2, 2, 1, 1. Ensure the intervals are mutually exclusive (e.g., using ≤ and <) so that a value like 15 does not fall into two different classes. Label the intervals clearly on a table.
In plain language
Grouping data means creating intervals to organise many numbers into clear categories, which makes the data easier to read and interpret.
Self-check
If you have data points 10, 15, 20, 25 and use intervals 10-19 and 20-29, how many fall in each?
Show the answer
10-19 contains 10 and 15 (frequency 2). 20-29 contains 20 and 25 (frequency 2).
- Higher
Scope: Higher · Higher tier in either Unit 1 tier · Unit 1 §4.1.9 higher. Printed pages 41: specify the data needed and consider potential sampling methods
Open revision guidance
What to know
- Specifying data needs requires clarity on what information will answer the research question. One must determine the population of interest and decide on the sampling frame. A population could be 'all students in the school', while the sample is 'one class per year group'. The method chosen to collect data depends on resources, time, and the required precision. Planning involves anticipating potential obstacles, such as non-response or difficulty accessing data.
- Different sampling methods provide varying levels of accuracy. Random sampling gives every individual an equal chance, reducing bias, but may require a comprehensive list of the population. Other methods include stratified sampling, which ensures sub-groups are represented proportionally, or systematic sampling, which involves selecting every nth person. For a population of 100 students, picking every 10th student is a systematic approach to reach a sample size of 10.
What to do in an assessment
- To specify data needs, first define the hypothesis: 'Does regular exercise improve concentration scores?' To collect data, define the sample group: 50 students chosen at random from a school roster of 500. Design a simple measurement tool: a 10-minute concentration test. Ensure the data collected captures the necessary variables (exercise hours per week and test scores). Consider the potential for bias if only athletes are sampled.
- Critique the sampling strategy. If you choose 50 students from the first 50 on the roster, you create alphabetical bias. Instead, use a random number generator to select 50 indices from 1 to 500. If you need representation from all grades, use stratified sampling: calculate the proportion of students in each grade (e.g., 20% in Year 7) and select that same percentage from the random sample. Document why this method is chosen to defend your results.
In plain language
Identifying data needs and sampling methods means planning carefully to obtain accurate and representative information for your research.
Self-check
If you need 10% of a school with 1000 students, how many do you select for a random sample?
Show the answer
You select 100 students (1000 multiplied by 0.10).
- Higher
Scope: Higher · Higher tier in either Unit 1 tier · Unit 1 §4.1.10 higher. Printed pages 41: understand, describe and use different sampling techniques – i.e. random, systematic and stratified sampling
Open revision guidance
What to know
- Sampling techniques are methods used to select a subset of a population to make inferences. Random sampling ensures every member has an equal probability of selection, eliminating systematic bias. Systematic sampling involves a regular interval, such as picking every 5th item. While simple, it can be biased if the list has a hidden pattern. For example, if you pick every 5th house on a street and every 5th house is a corner house, your sample will not be representative.
- Stratified sampling divides the population into strata (groups) based on characteristics like age, gender, or grade, then selects a proportionate random sample from each. If a school has 400 boys and 600 girls (1000 total), and you need a sample of 100, you must choose 40 boys and 60 girls to maintain the population proportion. This ensures that the diversity within the population is captured accurately, making conclusions much more reliable.
What to do in an assessment
- To use random sampling, assign each member a unique number and use a random number generator to pick them. For a population of 200, generate numbers between 1 and 200. To use systematic sampling, determine your interval (k = Population Size / Sample Size). For a population of 200 and a sample of 20, k = 10. Select a random starting point between 1 and 10, then pick every 10th member thereafter.
- To use stratified sampling, first define your strata. E.g., for a population of 100 workers, 50 managers and 50 laborers, to get a sample of 20, calculate the 50/50 split. Select 10 managers at random from their group and 10 laborers at random from their group. This technique is vital when variables within groups differ significantly. Compare results between different sampling methods to see if the chosen technique affected the final outcome and why.
In plain language
Sampling techniques are ways of selecting a smaller group of people to represent a whole population, ensuring that the results are fair and reliable.
Self-check
You have 600 students in Grade 1, 400 in Grade 2. To get a sample of 100, how many from each grade?
Show the answer
Grade 1: (600/1000) * 100 = 60 students. Grade 2: (400/1000) * 100 = 40 students.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §4.2.1 foundation; Unit 3 §4.2.1 foundation. Printed pages 41: construct and interpret pictograms, bar charts and pie charts for qualitative data and for discrete quantitative data
Open revision guidance
What to know
- Data presentation involves visualising qualitative categories or discrete numerical values. Pictograms use symbols to represent frequency, requiring a clear key, such as one star symbol equals 5 items. Bar charts utilise rectangular bars of uniform width where height corresponds to frequency, ideal for comparing distinct groups.
- Pie charts display proportions of a whole by dividing a circle into sectors. Each sector angle is calculated by finding the fraction of the total frequency and multiplying by 360 degrees. For a dataset of 20 items with 5 items in a category, the angle is (5/20) multiplied by 360, equaling 90 degrees.
What to do in an assessment
- To construct a bar chart for categorical data (e.g., Car colors: Red 10, Blue 5, Green 5), draw a horizontal axis with category labels and a vertical axis for frequency. Plot bars of width 1 cm separated by gaps. Ensure the vertical scale is consistent and clearly marked.
- To draw a pie chart for the same data, sum the frequencies to 20. Calculate each angle: Red (10/20)*360=180 degrees, Blue (5/20)*360=90 degrees, and Green (5/20)*360=90 degrees. Use a protractor to mark these angles from a common centre point on a circle, ensuring the sum is 360 degrees.
- To construct a pictogram, choose a key, name the categories, and align the symbols from a common baseline. For red 6, blue 4 and green 2, a key of one circle representing two vehicles gives red ●●●, blue ●● and green ●. If a category has 3 vehicles, show one full circle and half a circle, with the same key.
In plain language
This lesson teaches how to draw and read pictograms, bar charts, and pie charts so that you can show information clearly and compare different groups of data using shapes and angles.
Self-check
If a survey shows 12 people like apples, 8 like bananas, and 4 like pears, what is the angle for bananas in a pie chart?
Show the answer
Total = 24. Banana angle = (8/24) * 360 = 120 degrees.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §4.2.2 foundation; Unit 3 §4.2.2 foundation. Printed pages 41: construct and interpret vertical line diagrams for discrete data
Open revision guidance
What to know
- Vertical line diagrams are used specifically for discrete quantitative data. They are similar to bar charts but use simple vertical lines instead of rectangles to represent frequency. This focuses the eye on individual values on the horizontal axis rather than continuous ranges, which is essential for counts like shoe size or number of siblings.
- The vertical axis represents the frequency count, while the horizontal axis represents the discrete variable. Each line is drawn to a height corresponding to the data point's frequency. Because the data is discrete, the lines are singular, and there is no implication of values existing between the plotted integers.
What to do in an assessment
- To construct a diagram for data (e.g., Number of pets: 0 pets = 3 people, 1 pet = 6 people, 2 pets = 2 people), label the horizontal axis 'Number of Pets' and the vertical axis 'Frequency'. Place ticks at 0, 1, and 2 on the bottom.
- Draw a vertical line at 0 up to 3 units height, at 1 up to 6 units height, and at 2 up to 2 units height. Ensure the scale on the vertical axis is consistent, for example, 1 cm per unit of frequency, to allow accurate interpretation of the relative differences in counts.
In plain language
Learn how to use vertical line diagrams to show numerical data that has discrete values, using simple lines instead of bars.
Self-check
Data: 5 children have 0 books, 7 have 1 book, 3 have 2 books. How high is the line for 1 book?
Show the answer
The line for 1 book should reach a height of 7 units on the vertical frequency axis.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §4.2.3 foundation; Unit 3 §4.2.3 foundation. Printed pages 42: construct line graphs for the values of a variable at different points in time; understand that intermediate values in a line graph may or may not have meaning
Open revision guidance
What to know
- Line graphs are essential for tracking changes in a variable over time. Points are plotted at specific time intervals and joined by straight line segments. This visualises trends, such as growth or decline, across a period. It is crucial to distinguish between meaningful intermediate values and situations where points exist only at discrete intervals.
- When interpreting line graphs, understand that intermediate points on the line may represent estimates if the trend is smooth, or they may have no physical reality if the data is purely episodic. For example, temperature over 24 hours can be interpolated, but the number of cars sold per month cannot logically be interpreted as a partial car sold in the middle of the month.
What to do in an assessment
- Construct a line graph for a dataset (Time: 08:00, 10:00, 12:00; Temperature: 10C, 15C, 12C). Plot (08:00, 10), (10:00, 15), and (12:00, 12). Connect these points with straight lines.
- To interpret the graph, identify the slope between points. Between 08:00 and 10:00, the temperature rose by 5 degrees over 2 hours, a rate of 2.5 degrees per hour. Observe the peak at 10:00 as the maximum temperature point.
In plain language
Line graphs show how things change over time. You plot points for different times and connect them with lines to see the trend.
Self-check
If at 12:00 temperature is 12C and at 14:00 it is 18C, what is the rate of increase?
Show the answer
Increase = 6 degrees over 2 hours, so 3 degrees per hour.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 3 tier · Unit 3 §4.2.4 foundation. Printed pages 42: construct scatter diagrams for data on paired variables
Open revision guidance
What to know
- Scatter diagrams represent paired data variables on a Cartesian plane to identify relationships. Each data entry is a coordinate (x, y), where 'x' is the independent variable and 'y' is the dependent variable. Plotting these allows visual inspection of how one variable relates to the other.
- Unlike graphs, scatter diagrams do not join points with lines. The resulting cloud of points shows the strength and direction of the association. For example, plotting hours studied (x) against test scores (y) helps determine if more study time generally correlates with higher grades.
What to do in an assessment
- Given data points (1, 2), (2, 4), (3, 5), (4, 7), (5, 9), set up a grid with x-axis 0 to 6 and y-axis 0 to 10. Mark the points accurately at their exact coordinates.
- Ensure the scales for both axes are consistent. Check for outliers which may deviate significantly from the general grouping of points, as these can impact the interpretation of the dataset.
- For data pairs (1,2), (2,4), (3,5), (4,8), (5,9), label the horizontal and vertical variables and plot five points, without joining them. They show positive correlation. The displayed dashed trend line y ≈ 1.8x+0.2 is an estimate. At x=3.5 it predicts y≈6.5; this is interpolation, not proof of causation.
In plain language
Scatter diagrams show the relationship between two variables by plotting points on a graph, without connecting the points with lines.
Self-check
If you have points (1, 10) and (2, 20), what does a scatter diagram show?
Show the answer
A linear positive relationship where y increases as x increases.
View diagrams
Scatter diagram and estimated trend
Key and description
Constructed data: (1, 2), (2, 4), (3, 5), (4, 8), (5, 9). These show positive correlation, not proof of causation. The dashed line y ≈ 1.8x + 0.2 is an estimate, not an exact model.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Core
Scope: Core · Foundation tier, Higher tier in either Unit 3 tier · Unit 3 §4.2.5 foundation. Printed pages 42: draw ‘by eye’ a line of ‘best fit’ and understand and interpret what this represents. In questions where the mean point has been given, calculated or plotted, candidates will be expected to draw the line of ‘best fit’ through that point
Open revision guidance
What to know
- A line of best fit is a straight line drawn through a scatter diagram that represents the overall trend of the data. It does not need to pass through every point but should follow the general direction of the trend, with roughly an equal number of points above and below the line.
- When a mean point is provided, the line of best fit must pass through it. The mean point is calculated by finding the mean of all x-coordinates and the mean of all y-coordinates. This point serves as an anchor for the line, ensuring it reflects the central tendency of the paired variables.
What to do in an assessment
- With points (1, 2), (2, 4), (3, 5), (4, 7), (5, 9), first calculate the mean: x-mean = 15/5 = 3, y-mean = 27/5 = 5.4. Plot the point (3, 5.4).
- Place a ruler over the scatter plot. Align it to pass through the calculated mean point (3, 5.4) and balance it so the spread of points above and below the line is visually consistent. Draw the line carefully from edge to edge of the chart area.
In plain language
A line of best fit is a straight line that you draw through data on a scatter diagram to show the general trend.
Self-check
If the mean point is (2, 3), must the line pass through it?
Show the answer
Yes, a line of best fit must pass through the mean point when it is provided or calculated.
View diagrams
Scatter diagram and estimated trend
Key and description
Constructed data: (1, 2), (2, 4), (3, 5), (4, 8), (5, 9). These show positive correlation, not proof of causation. The dashed line y ≈ 1.8x + 0.2 is an estimate, not an exact model.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Core
Scope: Core · Foundation tier, Higher tier in either Unit 3 tier · Unit 3 §4.2.6 foundation. Printed pages 42: interpret and draw conclusions from scatter diagrams; use terms such as positive correlation, negative correlation, little or no correlation
Open revision guidance
What to know
- Scatter diagrams allow us to define correlation, which is the relationship between variables. Positive correlation exists when both variables increase together. Negative correlation is observed when one variable increases while the other decreases. Little or no correlation suggests no predictable pattern between the two variables.
- Interpret the strength of the relationship by how closely points cluster around the line of best fit. A tight cluster implies a strong correlation, while a loose scattering of points indicates a weak correlation. Identifying these patterns helps in understanding data behaviour even if variables are not perfectly aligned.
What to do in an assessment
- Observe a plot where points move from bottom-left to top-right (e.g., Temperature 5C, Sales 100; Temperature 20C, Sales 500). This indicates positive correlation.
- Observe a plot where points move from top-left to bottom-right (e.g., Speed 100km/h, Time 1h; Speed 50km/h, Time 2h). This indicates negative correlation. Describe the relationship clearly using these specific terms.
In plain language
Scatter diagrams show the type of relationship between two things: positive (both increase), negative (one increases and the other decreases), or no correlation.
Self-check
If points cluster tightly moving downwards to the right, what is this?
Show the answer
Strong negative correlation.
View diagrams
Scatter diagram and estimated trend
Key and description
Constructed data: (1, 2), (2, 4), (3, 5), (4, 8), (5, 9). These show positive correlation, not proof of causation. The dashed line y ≈ 1.8x + 0.2 is an estimate, not an exact model.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Core
Scope: Core · Foundation tier, Higher tier in either Unit 3 tier · Unit 3 §4.2.7 foundation. Printed pages 42: appreciate that correlation does not imply causality
Open revision guidance
What to know
- It is a fundamental statistical error to assume that correlation implies causality. Just because two variables show a statistical relationship (a line of best fit can be drawn), it does not mean that one variable directly causes the other to change.
- Extraneous factors often influence two correlated variables. For example, ice cream sales and sunscreen sales both increase in summer. While they show a positive correlation, ice cream sales do not cause sunscreen sales; both are caused by the higher temperature in summer.
What to do in an assessment
- When analysing data, look for underlying reasons for a trend. If a scatter graph shows that as ice cream consumption increases, so do sunburn cases, identify that both are linked to increased solar exposure.
- State clearly that correlation measures the strength of a relationship but does not provide evidence of cause. Always phrase conclusions carefully, distinguishing between 'associated with' and 'caused by'.
In plain language
Correlation means that data is linked, but it does not mean that one thing causes the other. Other factors often affect both things.
Self-check
If shark attacks increase as ice cream sales increase, does ice cream cause shark attacks?
Show the answer
No. Both increase due to warmer weather encouraging swimming, which is a third variable.
View diagrams
Scatter diagram and estimated trend
Key and description
Constructed data: (1, 2), (2, 4), (3, 5), (4, 8), (5, 9). These show positive correlation, not proof of causation. The dashed line y ≈ 1.8x + 0.2 is an estimate, not an exact model.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Core
Scope: Core · Foundation tier, Higher tier in either Unit 3 tier · Unit 3 §4.2.8 foundation. Printed pages 42: find the mean, median, mode and range of a list of values
Open revision guidance
What to know
- A data set consists of numerical observations. The mean is calculated by summing all values and dividing by the total count. The median is the central value of an ordered sequence. If the number of values is even, the median is the average of the two middle terms. The mode is the most frequently occurring value in the set. The range measures the spread of data by calculating the difference between the maximum and minimum values.
- Consider the data set of test scores: 5, 8, 8, 12, 17. The sum is 50. With 5 values, the mean is 50/5 = 10. The ordered set has 8 as the middle value, so the median is 8. The value 8 appears twice, making it the mode. The range is 17 - 5 = 12. These metrics describe central tendency and dispersion, providing a foundational summary of the distribution characteristics of numerical populations.
What to do in an assessment
- To find these statistics for a new set: 3, 4, 4, 7, 12. First, sort the set: 3, 4, 4, 7, 12. Count is 5. Sum: 3+4+4+7+12 = 30. Mean: 30 / 5 = 6. Median: The third position is 4. Mode: 4 appears twice, so 4 is the mode. Range: 12 - 3 = 9. Record each result clearly using appropriate units if provided in context.
- For an even set: 2, 5, 5, 9, 10, 11. Count is 6. Sum: 42. Mean: 42 / 6 = 7. Median: Average of 3rd and 4th terms (5 and 9) is (5+9)/2 = 7. Mode: 5. Range: 11 - 2 = 9. Always verify the sorted order before selecting the median to ensure accuracy in the central positioning of the data.
In plain language
We use the mean, median, and mode to find the middle of data, and the range to see how far apart the highest and lowest numbers are.
Self-check
Given 2, 3, 3, 8, 14, find the mean, median, mode, and range.
Show the answer
Mean = 6; Median = 3; Mode = 3; Range = 12. Working: Sum=30, Count=5, Mean=30/5=6. Median is 3rd term. Range=14-2=12.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 3 tier · Unit 3 §4.2.9 foundation. Printed pages 42: construct and interpret grouped frequency diagrams and frequency polygons
Open revision guidance
What to know
- Grouped frequency diagrams display data organised into intervals, showing how many items fall within each class. These diagrams help visualise large data sets. Frequency polygons are constructed by plotting the frequency against the midpoints of each class interval and connecting these points with straight lines. These polygons are useful for comparing distributions across different groups or identifying trends over ranges.
- Example: Heights (cm) grouped as 150-160, 160-170, 170-180 with frequencies 5, 12, 8. Midpoints are 155, 165, 175. Plotting (155, 5), (165, 12), and (175, 8) and joining with lines creates the frequency polygon. These visual tools summarise large amounts of information efficiently for quick interpretation of where data concentration occurs within specific intervals.
What to do in an assessment
- To construct a frequency polygon: 1. Calculate the midpoint for each interval (sum boundaries and divide by 2). 2. Plot the frequency on the y-axis against the midpoint on the x-axis. 3. Connect each point with a ruler to form the polygon. Ensure labels are included for axes such as 'Height (cm)' and 'Frequency'.
- For interpretation, observe the peak of the polygon to identify the modal class. For groups 10-20 (freq 4), 20-30 (freq 10), 30-40 (freq 6), the midpoints are 15, 25, 35. The peak at (25, 10) confirms the modal group is 20-30. Verify all arithmetic for midpoints to ensure correct plot placement.
In plain language
We use frequency diagrams and polygons to see how data is distributed over ranges of numbers.
Self-check
Data: 0-10 (freq 2), 10-20 (freq 8), 20-30 (freq 3). Calculate midpoints and identify the modal class.
Show the answer
Midpoints: 5, 15, 25. Modal class: 10-20. Working: (0+10)/2=5, (10+20)/2=15, (20+30)/2=25. Frequency 8 is the highest.
View diagrams
Frequency polygon from grouped data
Key and description
Classes and frequencies: [0, 10): 2; [10, 20): 6; [20, 30): 8; [30, 40): 4. Plot the class midpoints: (5, 2), (15, 6), (25, 8), (35, 4), then join consecutive points.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Core
Scope: Core · Foundation tier, Higher tier in either Unit 3 tier · Unit 3 §4.2.10 foundation. Printed pages 43: find the mean, median, mode and range for a discrete (ungrouped) frequency distribution
Open revision guidance
What to know
- For exact values x with frequencies f, the mean is Σfx/Σf. Multiply each value by its frequency, sum those products and divide by the total number of observations. The mode is the value with the greatest frequency.
- Locate the median in the ordered observations using cumulative frequencies. With an odd total N, use observation (N+1)/2. With an even total, average observations N/2 and N/2+1. A halfway rank such as 6.5 means averaging the sixth and seventh observations, not looking for a fractional observation.
What to do in an assessment
- Values 2 and 3 have frequencies 3 and 2. Total = 5; Σfx = 2×3 + 3×2 = 12, so mean = 12/5 = 2.4. The ordered observations are 2,2,2,3,3. The third observation is 2, so median = 2; mode = 2.
- Values 1,2,3 have frequencies 4,6,2. Total = 12; Σfx = 4 + 12 + 6 = 22, so mean = 22/12 ≈ 1.833. The sixth and seventh observations both lie among the six 2s, so median = (2+2)/2 = 2. The mode is 2 because its frequency is largest. The range is 3−1 = 2, using the observed values, not their frequencies.
In plain language
Frequency tells you how often a value occurs. Count repeated observations when finding the mean and the median.
Self-check
Values 1 and 2 have frequencies 2 and 3. Find their mean and median.
Show the answer
Mean = (1×2 + 2×3)/5 = 8/5 = 1.6. The ordered list is 1,1,2,2,2, so the third observation and median are 2.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 3 tier · Unit 3 §4.2.11 foundation. Printed pages 43: find an estimate for the mean of a grouped frequency distribution
Open revision guidance
What to know
- When data is grouped into classes, the exact values are unknown. Therefore, we use the midpoint of each class interval to represent the typical value of that group. The estimated mean is the sum of (midpoint multiplied by frequency) divided by the total frequency. This approach provides a reliable approximation for central tendency in grouped data where individual values are not provided.
- Example: 10-20 (freq 5), 20-30 (freq 3). Midpoints: 15, 25. Calculation: (15 * 5) + (25 * 3) = 75 + 75 = 150. Total freq: 8. Estimated mean: 150 / 8 = 18.75. This estimate relies on the assumption that data points are distributed evenly across the intervals, which serves as an acceptable approximation for statistical analysis in GCSE maths contexts.
What to do in an assessment
- Step 1: Calculate the midpoint for every interval. Step 2: Multiply each midpoint by its corresponding frequency. Step 3: Add these products to find the total sum. Step 4: Divide the total sum by the sum of frequencies. Ensure you show all intermediate steps: list columns for interval, midpoint, frequency, and f*midpoint.
- Example application: 0-10 (freq 10), 10-20 (freq 20). Midpoints: 5, 15. Calculations: 5*10 = 50; 15*20 = 300. Total sum = 350. Total freq = 30. Estimate: 350 / 30 = 11.67 (rounded). Precision is key in stating units correctly or rounding to appropriate decimals as specified by the task requirements.
In plain language
Since we do not know the exact values, we use the midpoint of the group to estimate the mean.
Self-check
Group 0-5 (freq 2), 5-10 (freq 2). Find the estimated mean.
Show the answer
Estimated mean = 5. Working: Midpoints 2.5 and 7.5. (2.5*2 + 7.5*2) / 4 = (5 + 15) / 4 = 20 / 4 = 5.
- Higher
Scope: Higher · Higher tier in either Unit 3 tier · Unit 3 §4.2.12 higher. Printed pages 43: find an estimate for the median of a grouped frequency distribution
Open revision guidance
What to know
- To estimate the median in grouped data, we use linear interpolation. First, determine the position of the median: (N/2), where N is the total frequency. Locate the class interval containing this position using cumulative frequency. Then, apply the interpolation formula: L + [ ( (N/2 - CF_prev) / f_median ) * width ], where L is the lower boundary, CF_prev is cumulative frequency before the class, f_median is the frequency of the median class, and width is the class size.
- Consider 0-10 (f=5), 10-20 (f=10). N=15. Median pos=7.5. The class 10-20 contains the 7.5th item. Lower limit L=10, CF_prev=5, f_median=10, width=10. Median = 10 + [((7.5 - 5) / 10) * 10] = 10 + 2.5 = 12.5. This provides a specific numerical estimate within the interval.
What to do in an assessment
- 1. Calculate cumulative frequencies. 2. Identify the median class (the interval containing the N/2 value). 3. Identify L (start of that class), CF_prev (total frequency before the class), f_class (frequency of that class), and class width (difference between boundaries). 4. Use the formula to calculate. Ensure all algebraic steps are clear.
- Application: Data 20-30 (f=4), 30-40 (f=6). N=10. Median pos=5. Median class is 30-40. L=30, CF_prev=4, f=6, width=10. Median = 30 + ((5 - 4)/6) * 10 = 30 + 1.67 = 31.67. This result represents the estimated median value based on linear interpolation assumptions.
In plain language
We use a formula to find a detailed estimate of the median within a group, assuming that the data is evenly spread.
Self-check
Constructed data have classes 0 ≤ x < 10 (frequency 2) and 10 ≤ x < 20 (frequency 2). Using N/2 and linear interpolation, estimate the median.
Show the answer
The estimated median is 10. N = 4, so use N/2 = 2 on the cumulative-frequency axis. This lies at the boundary between the classes: the point (10, 2). Interpolating within the first class gives 0 + (2/2) × 10 = 10. This is a grouped-data estimate, not the exact median of the unknown individual values.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 3 tier · Unit 3 §4.2.13 foundation. Printed pages 43: determine the modal category for qualitative data and modal class for grouped data
Open revision guidance
What to know
- The modal category applies to qualitative data, where the mode is simply the category with the highest frequency. The modal class applies to grouped quantitative data. It is defined as the interval or class that has the highest frequency in a given distribution. Unlike the mean or median, the modal class is descriptive of the most populated range in a set.
- Example: Colours of cars are Red (10), Blue (5), Green (2). The modal category is Red because it has the highest count. For grouped data: 0-10 (freq 5), 10-20 (freq 15), 20-30 (freq 8). The modal class is 10-20 as it has the maximum frequency (15). This helps identify where the most frequent observations lie.
What to do in an assessment
- 1. Review the frequency table provided. 2. Compare the frequencies of all categories or intervals. 3. Select the category or class that corresponds to the highest numerical value in the frequency column. If two classes share the same highest frequency, it may be a bimodal distribution.
- Example: Ages 0-10 (f=2), 10-20 (f=10), 20-30 (f=10), 30-40 (f=3). Here, 10-20 and 20-30 both have a frequency of 10. The set is bimodal. Always justify your choice by pointing to the frequency count. Clear labelling is required to identify the mode effectively.
In plain language
The modal class is the group that has the largest number of items in it.
Self-check
Data: Red (f=3), Blue (f=7), Green (f=4). Identify the modal category.
Show the answer
Blue. It has the highest frequency of 7.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 3 tier · Unit 3 §4.2.14 foundation. Printed pages 43: determine the group containing the median for grouped data
Open revision guidance
What to know
- Determining the group containing the median is the first step before calculating an exact estimate. The median position is found by N/2, where N is the total frequency. By calculating the cumulative frequency, you can identify which class interval contains the value at the (N/2)th position. This defines the range where the true median lies.
- Example: N=20. Median position = 10. Cumulative frequencies: 0-10 (f=5, CF=5), 10-20 (f=10, CF=15). The 10th value falls within the 10-20 interval because the cumulative frequency reaches 5 before this class and 15 after it. Thus, the 10-20 interval contains the median.
What to do in an assessment
- 1. Sum the frequencies to find N. 2. Calculate N/2. 3. Create a cumulative frequency column by adding successive frequencies. 4. Find the first interval where the cumulative frequency is equal to or greater than N/2. This specific interval is the group containing the median.
- Example: Class 0-5 (f=4), 5-10 (f=6), 10-15 (f=2). Total N=12. Median pos = 6. CF: 4, 10, 12. Since CF reaches 10 at the 5-10 class, and 4 < 6 <= 10, the median lies in the 5-10 group. Always verify the running total to avoid errors.
In plain language
We use cumulative frequency to see which group contains the middle value.
Self-check
Data: 0-10 (f=3), 10-20 (f=4). N=7. Find the group containing the median.
Show the answer
10-20. N/2=3.5. Cumulative freq: 3, 7. 3.5 falls in the second interval.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 3 tier · Unit 3 §4.2.15 foundation. Printed pages 43: select, calculate and estimate appropriate measures of central tendency (i.e. the mean, median or the mode)
Open revision guidance
What to know
- Measures of central tendency summarise a dataset with a single value that represents the centre or typical value. The mean is calculated by summing all observations and dividing by the count. For a dataset of test scores (4, 7, 7, 10, 12), the mean is (4+7+7+10+12)/5 = 40/5 = 8.
- The median is the middle value when data is ordered numerically. If the number of data points is even, it is the average of the two middle values. The mode is the most frequently occurring value in the dataset, while the mean is often used for general averages but can be influenced by extreme outliers.
What to do in an assessment
- To estimate the mean from a frequency table, multiply each value (or midpoint of a group) by its frequency, sum these products, and divide by the total frequency. For scores (midpoints) 2, 5, 8 with frequencies 3, 4, 3, total sum = (2*3)+(5*4)+(8*3) = 6+20+24 = 50, and total frequency = 10. Estimated mean = 50/10 = 5.
- Identify the median position for n data points using (n+1)/2. For n=10, the median is between the 5th and 6th values. If these are 5 and 6, the median is (5+6)/2 = 5.5. Use the mode for categorical data as it does not require numerical ordering.
In plain language
Central tendency tells us the middle of our data. Use the mean for calculations, median for central ranking, and mode for popularity.
Self-check
Calculate the mean of 2, 3, 3, 8.
Show the answer
16/4 = 4.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 3 tier · Unit 3 §4.2.16 foundation. Printed pages 43: compare data distributions using one measure of central tendency and/or one measure of spread
Open revision guidance
What to know
- To compare distributions, calculate one measure of central tendency (mean or median) and one measure of spread (range or IQR). For Group A (10, 12, 14, 16, 18) and Group B (10, 10, 14, 18, 18), Group A has a mean of 14 and range of 8. Group B has a mean of 14 and range of 8.
- Comparison requires evaluating both the average and the consistency of data. Even if measures are identical, the distribution shape differs. Group A is clustered closely, while Group B is spread toward the extremes.
What to do in an assessment
- Compare the mean or median to decide which group performed better on average. Then use the range to assess the consistency. If Group A range is 8 and Group B range is 10, Group A is more consistent.
- Apply the logic to specific sets. For dataset X: 1, 2, 3, 4, 5 (Mean 3, Range 4) vs dataset Y: 1, 3, 3, 3, 5 (Mean 3, Range 4). Here, despite identical measures, dataset X is uniform, while Y is concentrated at the centre.
In plain language
When comparing two sets of numbers, look at their average to see which is higher and their range to see which is more predictable.
Self-check
Compare the range of {2, 4, 6} and {1, 4, 7}.
Show the answer
Range A = 4; Range B = 6. Set A is more consistent.
- Higher
Scope: Higher · Higher tier in either Unit 3 tier · Unit 3 §4.2.17 higher. Printed pages 43: select, calculate and estimate appropriate measures of spread, including the range and interquartile range applied to discrete, grouped and continuous data
Open revision guidance
What to know
- Spread measures describe how data varies. The range is the difference between the maximum and minimum. The Interquartile Range (IQR) focuses on the middle 50 percent of data, specifically Q3 minus Q1. This makes it resistant to extreme outliers.
- For discrete data: {2, 5, 5, 8, 10, 12, 15}. Median = 8. Q1 is the median of the lower half {2, 5, 5} which is 5. Q3 is the median of the upper half {10, 12, 15} which is 12. IQR = 12 - 5 = 7.
What to do in an assessment
- For grouped data, identify the cumulative frequency thresholds. If total frequency is 40, Q1 is at the 10th value and Q3 is at the 30th value. Use linear interpolation if the exact value falls within a class interval.
- Calculate IQR for continuous data sets using the frequency density if provided. Always state your units clearly, e.g., if measuring height in cm, the IQR should also be in cm.
In plain language
Measures of spread show how far numbers are from the centre. IQR is the difference between the 75th and 25th percentiles.
Self-check
Find the IQR of {1, 3, 4, 5, 7, 8, 9}.
Show the answer
Q1=3, Q3=8, IQR=5.
- Higher
Scope: Higher · Higher tier in either Unit 3 tier · Unit 3 §4.2.18 higher. Printed pages 44: construct and interpret cumulative frequency tables and diagrams, including estimating the median, interquartile range and other percentages
Open revision guidance
What to know
- Cumulative frequency is a running total of frequencies. It is used to plot a cumulative frequency diagram (an ogive). The y-axis represents cumulative frequency, and the x-axis represents the data values. These diagrams allow you to estimate the median at 50 percent of the total frequency and the IQR using the 25th and 75th percentiles.
- For a frequency table (0-10, 10-20, 20-30 with frequencies 5, 10, 5), the cumulative frequencies are 5, 15, 20. Total frequency is 20. Median is at 10th value, Q1 at 5th, Q3 at 15th.
What to do in an assessment
- To estimate the median: Find 50 percent of the total frequency (10). Draw a horizontal line from 10 on the y-axis to the curve, then drop down to the x-axis. If the x-axis point is 15, the median is 15 units.
- To calculate IQR: Find the x-values for the 25 percent and 75 percent points. For a total frequency of 20, 25 percent is 5 and 75 percent is 15. Read x-values at cumulative frequencies 5 and 15, then subtract them.
In plain language
Use cumulative frequency tables and graphs to see where the middle and quartiles of your data sit by adding up frequencies step by step.
Self-check
For a total frequency of 40, what cumulative value represents the median?
Show the answer
20.
View diagrams
Cumulative-frequency curve and quartiles
Key and description
Frequencies 2, 6, 8, 4 give cumulative points (0, 0), (10, 2), (20, 8), (30, 16), (40, 20). N = 20. Interpolating on the graph: Q₁ at N/4 = 5 is 15; median at N/2 = 10 is 22.5; Q₃ at 3N/4 = 15 is 28.75. Do not use (N + 1)/2.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Higher
Scope: Higher · Higher tier in either Unit 3 tier · Unit 3 §4.2.19 higher. Printed pages 44: interpret and use box-and- whisker diagrams to compare distributions
Open revision guidance
What to know
- A box plot locates the minimum, lower quartile Q1, median, upper quartile Q3 and maximum on one scale. The box spans Q1 to Q3, so its length is the interquartile range Q3−Q1. A line within it marks the median; whiskers reach the stated extremes. Here we use the ordinary five-number diagram, not a separately defined outlier convention.
- Compare medians to discuss typical values and IQRs to compare the spread of the middle half. Compare ranges separately because extreme values affect the range more directly. Use the same scale and units, and give comparisons in the problem’s context.
What to do in an assessment
- For 10, 12, 15, 18, 20, 22, 25 kg, the median is 18 kg. Exclude it before finding the medians of the two halves: Q1=12 kg and Q3=22 kg. On a labelled scale draw a box from 12 to 22, a median line at 18 and whiskers to 10 and 25. IQR=22−12=10 kg; range=25−10=15 kg.
- Using that same median-of-halves convention, C={1,3,4,6,9} has median 4, Q1=(1+3)/2=2 and Q3=(6+9)/2=7.5. Its IQR is 5.5 and range is 8. D={4,5,5,6,8} has median 5, Q1=(4+5)/2=4.5 and Q3=(6+8)/2=7. Its IQR is 2.5 and range is 4. D has a higher median but less spread. State the quartile convention if it is not supplied.
- A supplied five-number summary avoids any rank convention: for minimum 4, Q1=8, median 12, Q3=16 and maximum 20, draw the box from 8 to 16 with its median at 12 and whiskers to 4 and 20.
In plain language
Read the typical value from the median line, the middle-half spread from the box and the full spread from the whiskers.
Self-check
For minimum 4, Q1=8, median 12, Q3=16 and maximum 20, find the IQR and range.
Show the answer
IQR=16−8=8 units; range=20−4=16 units.
View diagrams
Box plot and five-number summary
Key and description
Constructed summary: minimum 2, lower quartile 5, median 8, upper quartile 12, maximum 16. Range = 16 − 2 = 14; interquartile range = 12 − 5 = 7. The box spans the middle half of the distribution, not half its numerical range.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Higher
Scope: Higher · Higher tier in either Unit 3 tier · Unit 3 §4.2.20 higher. Printed pages 44: construct and interpret histograms with unequal class widths, including calculating the median and other percentages of the distribution
Open revision guidance
What to know
- In a histogram with unequal class widths, the y-axis represents 'frequency density' rather than frequency. Frequency density is calculated by dividing the frequency of the class by the class width. The area of each bar is proportional to the frequency. Because classes have different widths, bars are not uniform; a tall, narrow bar signifies a high concentration of data in a small range, while a wide, short bar signifies a lower concentration over a larger range. This method accurately represents data density.
- To find the median or specific percentiles within a histogram, one must calculate the cumulative area. The total area of all bars represents the total frequency. For example, if total frequency is 100, the median is the value where 50 units of area are accumulated. One calculates the area of each bar: (width * frequency density). If the first bar (width 5, density 4) covers 20 units, and the second bar (width 10, density 5) covers 50 units, the median must lie within the second bar. Linear interpolation is then used to find the exact value.
What to do in an assessment
- Construct a histogram for this data (time in minutes): [0-10, freq 20], [10-25, freq 30], [25-30, freq 25]. First, calculate frequency densities: 20/10=2, 30/15=2, 25/5=5. Plot class intervals on the x-axis and frequency density on the y-axis. The first bar has height 2 and width 10 (area 20). The second has height 2 and width 15 (area 30). The third has height 5 and width 5 (area 25). The total area is 75.
- To find the median (where area is 37.5), note that the first bar covers 20 units. We need 17.5 more units from the second bar. The second bar has width 15 and density 2, total area 30. The distance into this bar is 17.5 / 2 = 8.75. Adding this to the start of the second interval (10), the median is 18.75 minutes. This process ensures accurate analysis when categories are not balanced in size.
In plain language
When groups in a histogram differ in size, we must use frequency density (height of the bar) to ensure that the area of the bar corresponds to the actual frequency. To find the median, we calculate the area of the bars until we reach half the total.
Self-check
If a bar has a width of 10 and frequency density of 3, what is the frequency represented by that bar?
Show the answer
Frequency = Width * Frequency Density = 10 * 3 = 30.
View diagrams
Histogram with unequal class widths
Key and description
The vertical scale is frequency density. Bar area, not bar height alone, represents frequency.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §4.2.21 foundation; Unit 3 §4.2.21 foundation. Printed pages 44: recognise that graphs may be misleading
Open revision guidance
What to know
- Graphs can be misleading if the axes are truncated or not labelled correctly. For instance, a bar chart starting at 50 instead of 0 exaggerates the difference between bars of height 52 and 54.
- Another issue involves using inconsistent scales or 3D images that distort proportions. A data set showing profit: Year 1 = £1000, Year 2 = £1100, might look like a huge increase if the y-axis is compressed.
What to do in an assessment
- Always inspect the axes of a graph. Check if the vertical axis begins at zero or if it breaks. If the scale is non-linear, it misrepresents growth rates.
- Look for consistent intervals. If a histogram shows intervals 0-10, 10-20, 20-50, the last interval is disproportionately large, creating a false visual impression of density.
In plain language
Be careful with graphs that look dramatic. They might be designed to hide information or overstate a small change.
Self-check
If a y-axis starts at 1000 and goes to 1010, what is the risk?
Show the answer
It exaggerates small differences between values.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §4.2.22 foundation; Unit 3 §4.2.22 foundation. Printed pages 44: look at data to find patterns and exceptions
Open revision guidance
What to know
- Analysing data involves looking for trends and anomalies. A pattern might show a consistent increase, like a straight line on a scatter plot of coordinates (1, 10), (2, 20), (3, 30).
- Exceptions (outliers) are data points that do not follow the pattern. In a set of temperature readings: 15, 16, 17, 85, 18, 19, the value 85 is an anomaly that significantly affects the mean but does not represent the trend.
What to do in an assessment
- Identify correlations. Use a scatter plot to plot points, for example: X-axis represents study hours, Y-axis represents score. If points cluster near a line, a pattern exists.
- Exclude or note outliers when drawing conclusions. Always explain why an outlier might exist, such as a recording error or a genuine unusual event.
In plain language
Look for the general trend in data, but keep an eye out for strange numbers that stand out from the rest.
Self-check
In the set {10, 12, 11, 13, 90, 12}, identify the outlier.
Show the answer
90, because it is much higher than others.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Unit 1 §4.2.23 foundation; Unit 3 §4.2.23 foundation. Printed pages 44: draw inferences and conclusions from summary measures and data representations, relate results back to the original problem
Open revision guidance
What to know
- Drawing inferences means making an informed judgement based on data analysis. For a car fuel efficiency study, if the mean distance is 40 miles per gallon (mpg), we conclude the car is efficient for standard driving.
- Relating results back to the original problem is vital. If we analysed test scores of 50 students, we must conclude whether the intervention (e.g., a new textbook) was effective based on the comparison of before and after measures.
What to do in an assessment
- Base conclusions on specific statistics. Use the range and mean to support claims. For example: 'The mean increased from 60 to 70 units, suggesting the method works'.
- Check the context. A high average score might mask the fact that some students failed significantly; always examine the spread as well as the average.
In plain language
Use your findings to answer the original question. Ensure your conclusion is supported by the numbers you calculated.
Self-check
If test scores rose from 50 to 60, what can you infer?
Show the answer
Performance improved by 10 units on average.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §4.3.1 foundation. Printed pages 45: the vocabulary of probability, including notions of uncertainty and risk
Open revision guidance
What to know
- Probability is the numerical measure of how likely an event is to occur, representing the degree of uncertainty associated with a future event. Uncertainty refers to outcomes that are not known in advance, while risk specifically involves scenarios where an outcome might lead to a negative consequence or loss within a controlled experiment.
- In a standard experiment, such as rolling a fair 6-sided die, the sample space is the set of all possible outcomes: {1, 2, 3, 4, 5, 6}. Vocabulary is essential to distinguish between theoretical events and empirical observations, ensuring that mathematical communication about chance is precise and understood in a quantitative context.
What to do in an assessment
- To quantify uncertainty, identify all possible outcomes of an event. For example, if a bag contains 3 red, 2 blue, and 5 green marbles, there are 10 total outcomes. The risk of picking a blue marble is the count of blue outcomes (2) divided by the total (10), simplified to 1/5.
- Practice evaluating language used in news or reports. If a weather forecast states a 20 percent chance of rain, translate this into a frequency of 20 outcomes out of 100 theoretical days with identical conditions. By mapping real-world terminology to numerical values, you ensure that abstract uncertainty is transformed into measurable probability.
In plain language
Probability uses specific words and numbers to describe how likely something is to happen, helping us measure risk and uncertainty in experiments.
Self-check
In an experiment with 20 total outcomes where 5 outcomes are defined as 'risk', what is the count of the remaining non-risk outcomes?
Show the answer
15 outcomes (20 minus 5).
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §4.3.2 foundation. Printed pages 45: the meaning of the terms 'fair', 'an even chance', 'certain', 'likely', 'unlikely' and 'impossible'
Open revision guidance
What to know
- Probability terms describe the likelihood of events on a qualitative scale. 'Impossible' events have a probability of 0, meaning they never happen. An 'even chance' describes a probability of 0.5 or 50 percent, such as flipping a fair coin where heads and tails are equally weighted.
- Terms like 'likely' and 'unlikely' define events that are more or less probable than an even chance, respectively. 'Certain' events are guaranteed to happen, assigned a probability of 1. Identifying these states requires understanding the total number of outcomes versus the subset of outcomes that satisfy the specific condition being described.
What to do in an assessment
- To categorise an event, list all possible outcomes. If you have a bag with 10 white balls, picking a white ball is 'certain' because all 10/10 outcomes match. If you pick a red ball from that same bag, it is 'impossible' because 0/10 outcomes match.
- Compare probabilities to 0.5. If an event has 7 successful outcomes out of 10 total, the fraction is 7/10 or 0.7. Since 0.7 is greater than 0.5, we classify this event as 'likely' to occur. Always calculate the numerical value first before applying the qualitative label.
In plain language
We use specific terms like certain, likely, even chance, unlikely, and impossible to describe how probable an outcome is on a scale.
Self-check
If you roll a 6-sided die, is rolling a 7 'likely', 'unlikely', or 'impossible'?
Show the answer
Impossible, because there is no face numbered 7 on a standard 6-sided die.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §4.3.3 foundation. Printed pages 45: that the probability scale extends from 0 to 1
Open revision guidance
What to know
- The probability scale is a linear interval strictly bounded between 0 and 1 inclusive. A value of 0 indicates an impossible event, and a value of 1 indicates a certain event. Any outcome that is neither impossible nor certain must fall within the open interval (0, 1).
- This scale provides the framework for all probability calculations. Because the scale is restricted to the interval [0, 1], probabilities cannot be less than 0 or greater than 1. When working with complex problems, ensuring your final result stays within these boundaries is a necessary step for verification.
What to do in an assessment
- To visualise the scale, draw a line segment 10 cm long. Mark 0 at the left end and 1 at the right. Place markers for 0.25, 0.5, and 0.75. For an event with 3 wins out of 4 trials, calculate 3/4 = 0.75 and place it on your line.
- When converting data to this scale, always check that your inputs are valid. If you have an experimental result that suggests a probability of 1.2, you have performed a calculation error, as 1.2 exceeds the maximum of 1. Re-evaluate your counts to ensure the subset does not exceed the total sample size.
- A labelled probability scale is 0 (impossible) — 0.25 — 0.5 (even chance) — 0.75 — 1 (certain), with equally spaced marks. Place P=¾ at the 0.75 mark, not at 3 or 4. The endpoints 0 and 1 are included.
In plain language
The probability scale goes from 0 (impossible) to 1 (certain). Every possible event must be represented as a number between these two values.
Self-check
Is a probability of 1.5 valid on the probability scale?
Show the answer
No, it is invalid because the scale is strictly limited to 0 through 1.
View diagrams
Labelled probability scale
Key and description
The equally spaced marks are 0, ¼, ½, ¾ and 1. Zero means impossible, ½ means an even chance, and 1 means certain. Both endpoints are included. The highlighted point represents probability ¾.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §4.3.4 foundation. Printed pages 45: that probabilities may be expressed as fractions, decimals or percentages
Open revision guidance
What to know
- Probabilities can be represented in three equivalent forms: fractions, decimals, and percentages. A fraction such as 1/4 represents the ratio of target outcomes to total outcomes. Converting this to a decimal is achieved by division: 1 divided by 4 equals 0.25.
- To express the decimal as a percentage, multiply by 100 and add the percent sign. Thus, 0.25 times 100 equals 25 percent. These conversions are fundamental because they allow for the comparison of data sets that may initially appear in different units.
What to do in an assessment
- To convert a probability of 2/5 to a decimal and percentage, first divide 2 by 5 to get 0.4. Then, multiply 0.4 by 100 to get 40 percent. Always perform the division carefully to ensure the precision of the decimal representation.
- For data sets, if you observe 12 hits in 20 attempts, the fraction is 12/20. Simplify this by dividing both numerator and denominator by 4 to get 3/5. Then, convert 3/5 to 0.6 and 60 percent. Keeping a consistent method for these conversions prevents errors when summarising experimental results.
In plain language
Probabilities are written as fractions, decimals, or percentages. You can easily switch between them by dividing or multiplying.
Self-check
What is 3/4 as a decimal and a percentage?
Show the answer
0.75 and 75 percent.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §4.3.5 foundation. Printed pages 45: that the total probability of all the possible outcomes of an experiment is 1
Open revision guidance
What to know
- The sum of the probabilities of all mutually exclusive and exhaustive outcomes in an experiment is always equal to 1. If an experiment has outcomes A, B, and C, then P(A) + P(B) + P(C) = 1.
- This property is useful for finding the probability of an event when you know the probabilities of all other outcomes. For example, if P(A) = 0.2 and P(B) = 0.5, then P(C) must be 1 - (0.2 + 0.5) = 0.3. This principle is fundamental for checking the consistency of probability models.
What to do in an assessment
- List all outcomes for an experiment. Assign each outcome a probability value. Sum these values. If the sum is 1, the model is complete. If you are missing one outcome, subtract the sum of known probabilities from 1 to identify the missing probability.
- Apply this to a dataset of 100 spins of a wheel where there are 3 colors: Red, Blue, and Green. If Red happens 30 times and Blue 40 times, we have 70/100 or 0.7 accounted for. The probability of Green must be 1 - 0.7 = 0.3. Always verify that your subset sums do not exceed 1.
In plain language
If you add up the probabilities of every possible outcome in an experiment, the total must always be 1.
Self-check
In an experiment with 3 outcomes, if P(A) = 0.1 and P(B) = 0.4, what is P(C)?
Show the answer
0.5 (Calculated as 1 - 0.5 = 0.5).
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §4.3.6 foundation. Printed pages 45: calculate theoretical probabilities based on equally likely outcomes
Open revision guidance
What to know
- Theoretical probability is defined as the number of favourable outcomes divided by the total number of equally likely outcomes. The formula is P(Event) = (Number of favorable outcomes) / (Total number of possible outcomes).
- This method assumes each outcome in the sample space has the exact same likelihood of occurring. For example, a fair die has 6 faces. If you want to roll a number greater than 4, the favourable outcomes are 5 and 6. This gives 2 favourable outcomes out of 6, resulting in 2/6, which simplifies to 1/3.
What to do in an assessment
- To find a theoretical probability, start by listing the sample space. For a bag containing 3 red, 4 blue, and 3 yellow marbles (total 10), finding the probability of picking red involves identifying 3 favorable outcomes. Divide 3 by 10 to get 0.3 or 30 percent.
- Use coordinate points for geometry problems. If a grid has 10 points labelled (1,1) to (2,5), and you select a point at random, the probability of selecting a point with an x-coordinate of 2 is determined by counting how many points have x=2 (which are (2,1), (2,2), (2,3), (2,4), (2,5)) and dividing by 10. Thus, 5/10 = 0.5.
In plain language
Calculate theoretical probability by dividing the number of results you want by the total number of all possible results.
Self-check
If a spinner has 8 equal sections numbered 1-8, what is the probability of landing on a number less than 3?
Show the answer
2/8, which simplifies to 1/4 (0.25).
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §4.3.7 foundation. Printed pages 45: estimate the probability of an event as the proportion of times it has occurred – link to experimental evidence and relative frequency
Open revision guidance
What to know
- Experimental probability, or relative frequency, is the ratio of the number of times an event occurs in an actual series of trials to the total number of trials. P(Event) = (Number of occurrences) / (Total trials).
- Unlike theoretical probability, experimental probability uses actual data collected through observation. As the number of trials increases, the experimental probability typically becomes a better estimate of the theoretical probability of an event.
What to do in an assessment
- Conduct an experiment by flipping a coin 50 times. Record the results in a frequency table. If heads appears 28 times, the relative frequency is 28/50. Simplify this fraction to 14/25, which equals 0.56 or 56 percent.
- Analyse the dataset: Suppose a die was rolled 100 times, and the results for rolling a 6 were as follows: 18 times. The experimental probability of rolling a 6 is 18/100 = 0.18. Compare this to the theoretical probability (1/6 = 0.167) to observe the variance. Keep records of your raw counts to demonstrate the experimental process clearly.
In plain language
Experimental probability is based on real results from doing an experiment. You find it by dividing the number of successful outcomes by the total number of trials.
Self-check
In 40 trials, a specific event occurred 10 times. What is the experimental probability?
Show the answer
10/40, which simplifies to 1/4 (0.25 or 25 percent).
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §4.3.8 foundation. Printed pages 46: draw and interpret a graphical representation of relative frequency against the number of trials and understand that the long- term stability of relative frequency is expected
Open revision guidance
What to know
- Relative frequency is the ratio of the number of times an event occurs to the total number of trials conducted. For an experiment, such as rolling a biased die, let X be the outcome of rolling a 6. If we perform 50 trials and observe 12 successes, the relative frequency is 12/50, which equals 0.24. As the number of trials increases, the relative frequency typically converges toward a stable value representing the experimental probability of the event.
- Long term stability refers to the observation that as we increase the number of trials indefinitely, the fluctuations in relative frequency decrease and the value settles around the underlying probability. For example, in 100 trials, the frequency might be 0.22, and in 1000 trials, it might be 0.201. Graphing these values with the number of trials on the x-axis and the relative frequency on the y-axis shows a curve that flattens out, illustrating how empirical evidence supports the Law of Large Numbers.
What to do in an assessment
- To draw a relative frequency graph, create a table with two columns: Trial Count (n) and Relative Frequency (RF). Consider a simulation of tossing a coin where the first 10 trials show 6 heads (RF 0.6), 20 trials show 11 heads (RF 0.55), 50 trials show 24 heads (RF 0.48), and 100 trials show 49 heads (RF 0.49). Plot these points: (10, 0.6), (20, 0.55), (50, 0.48), (100, 0.49). Draw a smooth curve through these points and note how the line approaches the theoretical probability of 0.5.
- When interpreting the graph, identify the region where the curve becomes nearly horizontal. This stabilization point provides an estimate for the event probability. In our dataset, the final points hover near 0.49. To calculate the intermediate growth, verify each RF value by dividing the cumulative success count by the trial count. For instance, at 50 trials, 24/50 equals 0.48, and at 100 trials, 49/100 equals 0.49. Always use clearly labelled axes on your graph, noting the Trial Count on the horizontal axis and Relative Frequency on the vertical axis.
In plain language
Relative frequency is how often something happens compared to how many times you tried. If you keep trying more and more, the result usually settles down to a steady number.
Self-check
An experiment has 40 successes in 200 trials. Calculate the relative frequency. If the theoretical probability is 0.25, is this result close?
Show the answer
The relative frequency is 40/200 = 0.2. This is close to 0.25 but lower, indicating that further trials might bring it closer to the theoretical value.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §4.3.9 foundation. Printed pages 46: compare an estimated probability from experimental results with a theoretical probability
Open revision guidance
What to know
- Experimental probability is derived from actual results of an experiment, calculated as (Number of successful outcomes / Total number of trials). Theoretical probability is based on mathematical expectation, calculated as (Number of favorable outcomes / Total number of possible outcomes) assuming each outcome is equally likely. Comparing these two helps determine if a process is fair or biased.
- When comparing, find the difference between the experimental probability (P_exp) and the theoretical probability (P_theo). A small difference, such as |P_exp - P_theo| < 0.05, suggests the experimental results align well with expectations, while a large discrepancy suggests potential bias in the mechanism or an insufficient number of trials to reach stability.
What to do in an assessment
- Example: A fair six-sided die is rolled 60 times. Theoretically, the probability of rolling a 1 is 1/6 (approx 0.167). If the experiment results in 8 occurrences of a 1, the experimental probability is 8/60, which is approximately 0.133. Calculate the difference: |0.133 - 0.167| = 0.034. Since 0.034 is small, the result is considered consistent with theoretical expectations.
- When performing this analysis, always calculate the exact fractions first before converting to decimals for comparison. Use a calculator for division, ensuring at least three decimal places for accuracy. List your steps clearly: define the theoretical outcome, count the observed successes, calculate P_exp, find the magnitude of the difference, and state whether the values are reasonably close given the trial size.
In plain language
Experimental probability is what you get from testing, while theoretical probability is what math says should happen. Comparing them shows if your test results make sense.
Self-check
A fair coin has a theoretical probability of 0.5 for heads. In 200 flips, you get 90 heads. Calculate the difference between experimental and theoretical probability.
Show the answer
Experimental probability = 90/200 = 0.45. Difference = |0.45 - 0.5| = 0.05.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §4.3.10 foundation. Printed pages 46: understand and use the expected number of successes of an event when an experiment is repeated, and events are equally likely
Open revision guidance
What to know
- The expected number of successes of an event is calculated by multiplying the total number of repetitions (trials) by the probability of the event occurring in a single trial. For an event with probability p, and n trials, the expected number (E) is E = n * p. This applies when each trial is independent and the probability of success remains constant across all trials.
- If an event has several equally likely outcomes, the probability p is defined as (Number of target outcomes) / (Total number of outcomes). For example, if you roll a die 120 times, the probability of getting a 4 is 1/6. The expected number of times a 4 appears is E = 120 * (1/6) = 20. This does not guarantee that you will get exactly 20 successes, but it represents the long-term average.
What to do in an assessment
- To determine the expected number of successes, first identify the total trials n. Next, establish the probability p of the specific outcome occurring once. Use the formula E = n * p. For instance, if you flip a coin 500 times, the probability p is 0.5. Calculate E = 500 * 0.5 = 250. You should expect 250 heads. If you are provided with multiple outcomes, ensure the sum of probabilities is 1.0 before calculating E.
- When calculating for multiple trials, always present your intermediate steps. For example, if you pick a card from a deck of 52 cards 26 times, and you seek a Heart (p = 1/4), calculate: 26 * 0.25 = 6.5. Since you cannot have half a success, interpret this as 6 or 7. For school work, report the result as the decimal E = 6.5 if asked, as this represents the exact mathematical expectation.
In plain language
To find the expected number of successes, multiply the number of times you do the trial by the chance of it happening once.
Self-check
A spinner has 4 equal sections: Red, Blue, Green, Yellow. You spin it 80 times. How many times would you expect it to land on Red?
Show the answer
Probability p = 1/4 = 0.25. Expected successes = 80 * 0.25 = 20.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §4.4.1 foundation. Printed pages 46: that if A and B are mutually exclusive events, then the probability of A or B occurring is P(A) + P(B)
Open revision guidance
What to know
- Mutually exclusive events are two or more outcomes that cannot happen at the same time. For example, if you roll a single fair six-sided die, the event of rolling a 2 and the event of rolling a 5 are mutually exclusive because you cannot roll both numbers with one die roll. Mathematically, the probability of either A or B occurring is the sum of their individual probabilities, expressed as P(A or B) = P(A) + P(B).
- This principle applies to various scenarios where outcomes are partitioned. If you pick a marble from a bag containing 5 red, 3 blue, and 2 green marbles, the events of picking a red marble or picking a blue marble are mutually exclusive. Since the events do not overlap, we calculate the combined probability simply by adding the likelihood of each distinct outcome together to find the total probability of one of these events occurring.
What to do in an assessment
- To solve problems involving mutually exclusive events, first identify the individual probabilities. Suppose a box contains 10 cards numbered 1 to 10. The probability of choosing a card with a 3 is P(3) = 1/10. The probability of choosing a card with a 7 is P(7) = 1/10. Because these are mutually exclusive, P(3 or 7) = 1/10 + 1/10 = 2/10, which simplifies to 1/5.
- Apply this to a probability distribution table where the sum of all probabilities equals 1. If P(Win) = 0.4 and P(Draw) = 0.3, then P(Loss) = 1 - (0.4 + 0.3) = 0.3. Since these events cannot happen simultaneously, the probability of (Win or Draw) is 0.4 + 0.3 = 0.7. Always verify that the sum of the probabilities of all possible exclusive outcomes adds up to exactly 1.
In plain language
Mutually exclusive means two things cannot happen at the same time. To find the chance of one or the other happening, just add their individual chances together.
Self-check
A bag has 4 red, 3 green, and 3 yellow counters. What is the probability of picking a red OR a green counter?
Show the answer
P(Red) = 4/10, P(Green) = 3/10. Since they are mutually exclusive, P(Red or Green) = 4/10 + 3/10 = 7/10 or 0.7.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §4.4.2 foundation. Printed pages 46: that if A and B are independent events, then the probability of A and B occurring is P(A) x P(B)
Open revision guidance
What to know
- Independent events are occurrences where the outcome of the first event does not affect the probability of the second event. For instance, flipping a coin twice: the result of the first flip (Heads or Tails) has no influence on the result of the second flip. The probability of both events A and B occurring is found by multiplying their individual probabilities: P(A and B) = P(A) x P(B).
- Understanding independence is crucial for calculating complex probabilities involving sequential trials. If you roll a die and then spin a spinner, the outcome of the die roll is independent of the spinner's result. Because the events are independent, we multiply the chance of each specific outcome to find the probability of the specific combined intersection of those outcomes occurring in sequence.
What to do in an assessment
- Calculate independent events by isolating the individual probability of each trial. Consider a coin (P(Heads) = 0.5) and a standard six-sided die (P(6) = 1/6). The probability of getting Heads AND a 6 in a single combined experiment is 0.5 x 1/6 = 0.5/6 = 1/12. Always verify the events are truly independent before multiplying.
- Use a table for two events. If event A has probabilities {0.2, 0.8} and event B has {0.5, 0.5}, the four combined outcomes occur with probabilities 0.1, 0.1, 0.4, and 0.4. This multiplication rule persists regardless of the number of trials as long as each trial remains independent of the others. Ensure that all fractions are simplified correctly during your calculation process to maintain accuracy.
In plain language
Independent events mean the first one does not change the result of the second one. To find the chance of both happening, multiply the two individual chances together.
Self-check
A coin is tossed and a die is rolled. What is the probability of getting Tails and a 4?
Show the answer
P(Tails) = 1/2, P(4) = 1/6. Since they are independent, P(Tails and 4) = 1/2 * 1/6 = 1/12.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §4.4.3 foundation. Printed pages 47: identify all the outcomes of a combination of experiments, including lists, sample space diagrams, tree diagrams and Venn diagrams
Open revision guidance
What to know
- To understand combinations of experiments, we use visual tools to track every possible outcome. A sample space diagram lists all outcomes of two events on a grid, such as the sum of two dice rolls, where coordinates (1,1) through (6,6) represent every result. Tree diagrams show sequential steps as branches, multiplying along paths for dependent or independent events. Venn diagrams illustrate overlaps between sets, allowing us to subtract common outcomes to find unique probabilities.
- Lists are the simplest method for smaller sample spaces, while tree diagrams excel at showing sequences of trials. Sample space grids are superior for two-dimensional outcomes, like the position of two moving objects on a coordinate plane, for instance at points A(1,2) and B(3,4). By organising outcomes systematically, we ensure no possibility is missed, which is the foundational step for calculating accurate probabilities.
What to do in an assessment
- Construct a sample space diagram for two dice. Create a grid with rows 1 to 6 and columns 1 to 6. Each cell represents a coordinate pair (die1, die2). There are 36 possible outcomes. To find the probability of a sum of 4, identify coordinates (1,3), (2,2), and (3,1). This gives 3/36 = 1/12.
- Draw a tree diagram for picking two counters without replacement from a bag of 3 red and 2 blue. Start at a point, draw two branches: Red (3/5) and Blue (2/5). From each end, draw another set of branches adjusted for the remaining counters. If the first was Red, the second branches are Red (2/4) and Blue (2/4). Multiply along the branches to find the probability of Red then Blue: (3/5) x (2/4) = 6/20 = 3/10.
- For two coins, label the sample-space rows first coin H,T and columns second coin H,T. The four cells are HH, HT in the H row and TH, TT in the T row. If both fair tosses are independent, all four cells are equally likely.
- For a group of 30 pupils with 18 in A, 14 in B and 8 in both, put 8 in the overlap first. Then A-only is 18−8 = 10, B-only is 14−8 = 6 and neither is 30−(10+8+6) = 6. The displayed Venn diagram records these four non-overlapping regions.
In plain language
To find the chance of complex outcomes, use diagrams to list all possibilities. Grids work well for two dice, and tree diagrams help track sequential steps like picking cards or counters.
Self-check
Using a sample space grid for two coins, what is the probability of getting at least one Head?
Show the answer
Outcomes: HH, HT, TH, TT. 3 out of 4 have at least one Head. P(at least one Head) = 3/4.
View diagrams
Sample space for two coin tosses
Key and description
Rows identify the first toss and columns the second. Cells are HH, HT, TH, TT (H=head, T=tail). For two independent fair tosses each cell has probability ¼. Three cells contain at least one head, giving probability ¾.
Venn diagram for two events
Key and description
Constructed group of 30 pupils: n(A) = 18, n(B) = 14, n(A ∩ B) = 8. A only = 18 − 8 = 10; B only = 14 − 8 = 6; neither = 30 − (10 + 8 + 6) = 6. P(A ∪ B) = 24/30 = 4/5; P(A ∩ B) = 8/30 = 4/15.
Labelled constructed examples. Each diagram’s key identifies its own data and method. - Core
Scope: Core · Foundation tier, Higher tier in either Unit 2 tier · Unit 2 §4.4.4 foundation. Printed pages 47: recognise when the addition of probabilities for mutually exclusive events and the multiplication of probabilities for two independent events is needed
Open revision guidance
What to know
- The fundamental decision in probability is whether to add or multiply. You add probabilities when you have mutually exclusive events where you need the chance of 'this OR that' occurring. This assumes the events cannot happen simultaneously, such as picking a single card that is either an Ace or a King from a deck. You are essentially increasing the number of favourable outcomes in your numerator by summing the specific events.
- You multiply probabilities when dealing with independent events where you need the chance of 'this AND that' occurring. This applies when the outcomes are sequential or simultaneous in different experiments, such as flipping a coin and rolling a die. Multiplication is required because the second event's outcome space is scaled by the probability of the first event's success, creating a joint probability space.
What to do in an assessment
- To decide, ask: can both happen? If no, add. If the experiment has separate stages or trials, ask: do I need both to happen? If yes, multiply. For example, if you pick one ball from a bag, the chance of Red or Blue is additive. If you pick a ball, replace it, and pick again, the chance of Red then Blue is multiplicative.
- Consider a scenario with a biased spinner (P(A)=0.6, P(B)=0.4). If you spin twice, the probability of getting the same result (A and A or B and B) involves both operations: (0.6 x 0.6) + (0.4 x 0.4) = 0.36 + 0.16 = 0.52. Here, you multiply for the individual sequences and add because the sequences themselves are mutually exclusive paths.
In plain language
Add when you need one result OR another from exclusive options. Multiply when you need one result AND another from independent trials.
Self-check
A box has 2 red and 3 blue balls. If you pick one, what is the probability it is red or blue? If you pick one, replace it, and pick another, what is the probability both are red?
Show the answer
P(Red or Blue) = 2/5 + 3/5 = 1. P(Red and Red) = 2/5 * 2/5 = 4/25.
- Higher
Scope: Higher · Higher tier in either Unit 2 tier · Unit 2 §4.4.5 higher. Printed pages 47: recognise when problems involve three independent events, and be able to calculate the required probabilities
Open revision guidance
What to know
- Calculations involving three independent events follow the same multiplication rule as two events. Because each event occurs independently, the joint probability of events A, B, and C occurring is the product of their individual probabilities: P(A and B and C) = P(A) x P(B) x P(C). This rule holds true regardless of the type of trials, whether they are coin tosses, rolling dice, or independent selections from different containers.
- In high-tier applications, you might be required to solve for complex combinations, such as the probability of getting at least one success in three trials. This often involves calculating the complement: P(at least one success) = 1 - P(no successes). Since each trial is independent, P(no success in three trials) = P(Fail) x P(Fail) x P(Fail). By isolating the 'failure' path, you can efficiently derive the outcome for the 'at least one' scenario.
What to do in an assessment
- To find the probability of three specific independent events, multiply the three probabilities directly. Suppose you flip a coin (P=0.5), roll a die (P=1/6), and choose a counter from a bag of 4, where one is red (P=0.25). The probability of getting Heads, a 6, and a red counter is 0.5 x 1/6 x 0.25 = 0.5/24 = 1/48.
- For problems involving 'at least one', define your probability of success (s) and failure (f=1-s). For three trials, P(at least one) = 1 - (f x f x f). If a machine has a 10% (0.1) failure rate per day, the probability of at least one failure in three days is 1 - (0.9 x 0.9 x 0.9) = 1 - 0.729 = 0.271. Always ensure your decimals or fractions are handled precisely.
In plain language
When three events are independent, multiply all three probabilities together. To find the chance of 'at least one', it is easier to find the chance of none happening and subtract that from 1.
Self-check
A coin is tossed 3 times. What is the probability of getting 3 Heads?
Show the answer
P(Head) = 1/2. P(H and H and H) = 1/2 * 1/2 * 1/2 = 1/8.
- Higher
Scope: Higher · Higher tier in either Unit 2 tier · Unit 2 §4.4.6 higher. Printed pages 47: recognise when problems involve two or three dependent events, and be able to calculate the required probabilities, including sampling without replacement
Open revision guidance
What to know
- Dependent events occur when the result of the first event changes the probability of subsequent events. A primary example is sampling without replacement: if you pick an item from a set and do not put it back, the total count and the specific category count for the next pick have changed. This implies that the probabilities of later events are conditional on the outcomes of earlier events, necessitating a update to the fraction denominator and potentially the numerator.
- For three dependent events, you continue this logic sequentially. If you start with 10 items (3 Red, 7 Blue) and pick three times without replacement, the probability of picking Red then Red then Blue is calculated by adjusting the pool size each time: (3/10) x (2/9) x (7/8). This demonstrates that the outcome space for each subsequent trial is directly affected by previous selections, which is central to high-tier probability theory.
What to do in an assessment
- When dealing with dependent events, track the state of the pool after each step. Suppose you have 5 red and 5 blue marbles. To pick two red marbles without replacement: 1st pick P(Red) = 5/10. Because one red is removed, 2nd pick P(Red) = 4/9. The joint probability is (5/10) x (4/9) = 20/90 = 2/9.
- Apply this to coordinate-based sampling. If you have a set of labelled points A(1,1), B(2,2), C(3,3), and D(4,4) in a bag, picking two points without replacement involves 4 options for the first, then 3 for the second. The total number of outcomes is 4 x 3 = 12. The probability of picking A then B is 1/4 x 1/3 = 1/12. Ensure you decrement both the numerator and denominator appropriately for each step.
In plain language
Dependent events mean the first pick changes what is left for the second pick. Always reduce the number of total items, and the count of the specific item if it was picked, before calculating the next step.
Self-check
A bag has 3 white and 2 black balls. If you pick 2 balls without replacement, what is the probability of picking white then white?
Show the answer
1st pick P(W) = 3/5. 2nd pick P(W) = 2/4. P(W and W) = 3/5 * 2/4 = 6/20 = 3/10.
Topic group
Assessment and supplied formulas
- Core
Open revision guidance
What to know
- This is the new Mathematics and Numeracy Double Award, first awarded in November 2026. Each unit has its own Foundation or Higher entry; do not assume all three tiers are the same. Unit 1 is Financial Mathematics and Other Applications of Numeracy: 30%, calculator allowed, with Higher 1 h 45 min/80 marks and Foundation 1 h 30 min/65 marks. Unit 2 is Non-calculator Mathematics: 30%, with the same times and marks. Unit 3 is Calculator Mathematics: 40%, Higher 2 h/90 marks or Foundation 1 h 45 min/75 marks.
- Unit 1 uses personal and real-world applications. Units 2 and 3 include contextual and context-free problems. The assessment objectives cover using techniques, reasoning and communicating mathematically, and interpreting and solving problems. Show methods, interpret the units and context, and check whether the answer is sensible. The specification's formula lists appear at the front of Units 1 and 3 only; they are limited lists, not a promise that every needed formula is provided.
What to do in an assessment
- For a shopping problem, identify quantities and relevant prices, choose a method, calculate with consistent units, then state what your result means. If three identical items cost £7.20, one costs £7.20÷3 = £2.40. Checking 3×£2.40 = £7.20 supports the result; merely writing 2.4 without context is less clear.
- Practise Unit 2 without a calculator. For Units 1 and 3 use an approved calculator: no symbolic algebra, symbolic differentiation or integration, internet access, translation, or retrievable stored text/formulas. Check your centre's instructions. For the new award, Units 1 and 2 are available from summer 2026 and Unit 3 from November 2026. In 2027, both summer and November series are part of the specification's schedule.
- Ask the centre about cash-in and resits. For pre-16 candidates the terminal rule is 30%, each individual unit can be resat once before a fresh start, and a terminal attempt takes precedence even when the previous mark was better. For post-16 candidates there is no terminal or fresh-start rule, and the better of only the two most recent unit attempts is used. These are the specification rules, not advice that an individual is eligible for an entry.
In plain language
Confirm your unit tiers and exam arrangements with your centre, then practise explaining methods as well as obtaining answers.
Self-check
A candidate is entered for Unit 1 Higher, Unit 2 Foundation and Unit 3 Higher. Is that combination allowed, and which unit must be practised without a calculator?
Show the answer
Yes: tiers can differ across units. Unit 2 is non-calculator. Unit 1 is worth 30%, Unit 2 30%, and Unit 3 40%.
- Core
Scope: Core · Foundation tier, Higher tier in either Unit 1 tier or Unit 3 tier · Formula list Unit 1 Foundation p53; Formula list Unit 1 Higher p54 common formulas; Formula list Unit 3 Foundation p55; Formula list Unit 3 Higher p56 common formulas
Open revision guidance
What to know
- For Units 1 and 3, both tiers' supplied lists give trapezium area = ½(a+b)h, where a and b are the parallel sides and h their perpendicular separation. Do not use a sloping side as h.
- They also give uniform-cross-section volume = cross-sectional area × length, for example a prism or cylinder. Calculate the cross-sectional area first and use a length perpendicular to it. These formulas still require correct substitutions, units and interpretation.
What to do in an assessment
- A trapezium has parallel sides 4 cm and 8 cm, with perpendicular height 3 cm. Its area is ½(4+8)×3 = 18 cm².
- A prism has cross-sectional area 12 cm² and length 5 cm. Its volume is 12×5 = 60 cm³. The area unit multiplied by a length unit gives a volume unit.
In plain language
Identify what each letter represents in the diagram, then substitute consistent measurements.
Self-check
A trapezium's parallel sides are 5 m and 7 m; its perpendicular height is 4 m. Find its area.
Show the answer
½(5+7)×4 = 24 m².
- Higher
Scope: Higher · Higher tier in either Unit 1 tier · Formula list Unit 1 Higher p54 additional formulas
Open revision guidance
What to know
- Unit 1 Higher additionally supplies sphere volume (4/3)πr³ and surface area 4πr²; cone volume (1/3)πr²h and curved surface area πrl. Here r is radius, h perpendicular height, and l cone slant height. A cone's total area also includes its circular base.
- The supplied Annual Equivalent Rate formula is AER = (1+i/n)ⁿ−1 as a decimal, where i is the nominal annual interest rate as a decimal and n is the number of compounding periods per year. Convert a percentage rate to a decimal before substitution, and convert the final decimal back to a percentage if asked.
What to do in an assessment
- A sphere with radius 2 cm has volume (4/3)π×8 = 32π/3 ≈ 33.51 cm³ and surface area 4π×4 = 16π ≈ 50.27 cm².
- For nominal interest 6% per year compounded monthly, i=0.06 and n=12. AER = (1+0.06/12)¹²−1 ≈ 0.061678 = 6.168%. This is greater than 6% because interest itself earns interest.
In plain language
The list provides the formula, but you still need to choose the right measurements and interpret the rate.
Self-check
For nominal 4% interest compounded annually, what AER does the formula give?
Show the answer
i=0.04, n=1: (1+0.04)¹−1 = 0.04 = 4%.
- Higher
Scope: Higher · Higher tier in either Unit 3 tier · Formula list Unit 3 Higher p56 additional formulas
Open revision guidance
What to know
- Unit 3 Higher supplies sphere volume (4/3)πr³ and area 4πr²; cone volume (1/3)πr²h and curved area πrl. Distinguish radius from diameter and perpendicular height from slant height.
- For a triangle, sides a,b,c are opposite angles A,B,C. The supplied sine rule is a/sin A = b/sin B = c/sin C; cosine rule a² = b²+c²−2bc cos A; and area = ½ab sin C, where C is the included angle. The quadratic formula gives x = [−b ± √(b²−4ac)]/(2a) for ax²+bx+c=0 with a≠0.
What to do in an assessment
- For sides 5 and 7 with included angle 60°, area = ½×5×7×sin60° ≈ 15.16 square units. The opposite third side has square 5²+7²−2×5×7×cos60° = 39, so its length is √39 ≈ 6.245.
- For x²−3x−1=0, a=1, b=−3, c=−1. The discriminant is (−3)²−4×1×(−1)=13. The roots are (3+√13)/2 ≈ 3.303 and (3−√13)/2 ≈ −0.303. Include both signs; the whole numerator is divided by 2a.
In plain language
Match sides and opposite angles consistently. In a quadratic, substitute signed coefficients and retain both possible roots.
Self-check
Use the quadratic formula for x²−5x+6=0.
Show the answer
Discriminant = 25−24=1. x=(5±1)/2, giving x=3 or x=2.
Previous revision ticks
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Earlier topic checklist
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Mathematics and Numeracy (Double Award) topic list
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These ticks keep your earlier topic checklist. They do not count towards the new reviewed requirements above. Its numbering may be internal; use the linked references above for the exact reviewed source requirements.
Foundation tier topics
Higher tier topics
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