AQA GCSE Mathematics Paper 1Non-calculator (Higher Tier) — every question, answered
AQA GCSE Mathematics (8300) Higher Tier Paper 1 is the non-calculator paper. We analysed every sitting we could obtain the real question paper and mark scheme for: June 2018, June 2019, June 2022 and June 2023 (June 2020 and June 2021 do not exist as normal exam sittings, since GCSE exams were cancelled in both years because of the pandemic). Unlike a science paper with a handful of long discursive questions, a maths paper is around 25 to 30 short questions worth 1 to 5 marks each, covering a huge spread of separate skills. Below is what each recurring skill has actually asked across the four sittings we have, with a complete worked answer written to the mark scheme for each one, every paragraph explained.
Questions © AQA, quoted for analysis. Diagrams and figures described or recreated in our own words, not reproduced verbatim. Mark scheme content translated into plain English, not copied. PrepWise is independent and not endorsed by AQA.
The paper is a template
Same 22 question types, every sitting“All four sittings we have full papers for test standard form directly, most often a conversion or a division that must end in a specific form (standard form or an ordinary number).”
Every sitting we have tests standard form at least once, and June 2019 and June 2023 both test it twice in the same paper.
1×asked“Put these numbers in order from smallest to largest: 8 times 10 to the power negative 4, 4 times 10 to the power negative 2, 6 times 10 to the power negative 4, 0.07”
“Put these numbers in order from smallest to largest: 8 times 10 to the power negative 4, 4 times 10 to the power negative 2, 6 times 10 to the power negative 4, 0.07”
What it’s really asking
Convert every value to an ordinary decimal so the sizes can be compared directly.
The full worked answer — June 2018
Converting each value to an ordinary decimal: 8 times 10 to the power negative 4 equals 0.0008, 4 times 10 to the power negative 2 equals 0.04, 6 times 10 to the power negative 4 equals 0.0006, and 0.07 is already an ordinary decimal.
Ordering these four decimals from smallest to largest gives 0.0006, then 0.0008, then 0.04, then 0.07, so the smallest is 6 times 10 to the power negative 4 and the largest is 0.07.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise standard form questionsWhat the mark scheme rewarded
- One correct conversion to a comparable form, such as 0.08 times 10 to the power negative 2 or 0.0008 (1 mark)
- The fully correct order given with no clearly incorrect working shown (1 mark)
- Convert every value the same way, either all to ordinary decimals or all to a matching power of 10, before comparing them
- A negative power of 10 makes a number smaller, not larger, so 8 times 10 to the power negative 4 is far smaller than 0.07
Traps examiners saw
- Comparing only the number in front, the 8 or the 4, and ignoring the power of 10 attached to it
- Mixing up which direction the decimal point moves for a negative index, placing a value in completely the wrong position
Full-mark self-check 0 of 3
1×asked“Work out (3 times 10 to the power 5) divided by (4 times 10 to the power 3). Give your answer as an ordinary number.”
“Work out (3 times 10 to the power 5) divided by (4 times 10 to the power 3). Give your answer as an ordinary number.”
What it’s really asking
Divide the number parts and the powers of 10 separately, then convert the result out of standard form.
The full worked answer — June 2019
Dividing the two standard form numbers: (3 times 10 to the power 5) divided by (4 times 10 to the power 3) equals (3 divided by 4) times (10 to the power 5 divided by 10 to the power 3), which is 0.75 times 10 squared.
0.75 times 10 squared equals 0.75 times 100, which equals 75, so the answer as an ordinary number is 75.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise standard form questionsWhat the mark scheme rewarded
- A correct method that splits the calculation into the number parts and the powers of 10, such as reaching 0.75 times 10 squared, or a correct unsimplified equivalent like 300000 divided by 4000 (1 mark)
- The single correct final answer, 75, as an ordinary number and not left in standard form (1 mark)
- Divide the number parts and the powers of 10 separately rather than converting both numbers to ordinary form first, since that risks a slip with the extra zeros
- If the answer on the line is given in standard form as well as 75, make clear which one is the final choice, since an unclear answer can lose the accuracy mark
Traps examiners saw
- Stopping at the standard form result, 0.75 times 10 squared, without converting it into the ordinary number the question specifically asks for
- Losing track of a zero when converting the powers of 10 back to an ordinary number, especially with a value like 0.75 times 100
Full-mark self-check 0 of 3
1×asked“Work out 80 000 000 divided by 200. Give your answer in standard form.”
“Work out 80 000 000 divided by 200. Give your answer in standard form.”
What it’s really asking
Do the division as an ordinary number first, then convert the result into a times 10 to the power n form.
The full worked answer — June 2022
80 000 000 divided by 200 is the same as 80 000 000 divided by 2, then divided by 100, which is 40 000 000 divided by 100, which equals 400 000.
Writing 400 000 in standard form: the first non-zero digit is 4, with five more digits after it, so 400 000 equals 4 times 10 to the power 5.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise standard form questionsWhat the mark scheme rewarded
- Any correct value seen from the division, whether left as an ordinary number such as 400000 or partly converted, and then correctly converted to standard form (up to 2 marks)
- 400000 alone on the answer line, with no standard form given, scores only 1 mark since the question specifically asks for standard form
- It is often quicker on a non-calculator paper to divide by 2 then by 100 rather than dividing by 200 directly
- Marks are not awarded for the correct answer from working that is clearly incorrect, so keep every division step visible
Traps examiners saw
- Giving the correct ordinary number, 400000, but forgetting to convert it into standard form as the question specifically asks
- Writing the standard form with the wrong power of 10, such as 4 times 10 to the power 4 instead of 10 to the power 5, by miscounting the digits
Full-mark self-check 0 of 3
1×asked“b times 10 to the power n is the number 7200 written in standard form. Work out b times 10 to the power negative n. Write your answer as an ordinary number.”
“b times 10 to the power n is the number 7200 written in standard form. Work out b times 10 to the power negative n. Write your answer as an ordinary number.”
What it’s really asking
Find b and n by writing 7200 in standard form first, then switch the sign of the power before converting back to an ordinary number.
The full worked answer — June 2023
7200 written in standard form is 7.2 times 10 to the power 3, so b equals 7.2 and n equals 3.
b times 10 to the power negative n equals 7.2 times 10 to the power negative 3, which equals 7.2 divided by 1000, which equals 0.0072.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise standard form questionsWhat the mark scheme rewarded
- The correct value 0.0072, or the correctly identified b times 10 to the power negative n before conversion, such as 7.2 times 10 to the power negative 3 (1 mark for identifying b and n correctly, 1 further mark for the final ordinary number)
- Identify b and n from the given number first by writing it in standard form yourself, rather than guessing which digits they refer to
- Switching the sign of the power reverses whether the value grows or shrinks, so double check the direction of the decimal point move
Traps examiners saw
- Using the original power, 10 to the power 3, instead of switching its sign to 10 to the power negative 3 as the question asks
- Losing a zero when converting 7.2 times 10 to the power negative 3 back to an ordinary decimal, giving 0.072 instead of 0.0072
Full-mark self-check 0 of 3
The method for every Q12 (Jun18) / Q5 (Jun19) / Q5 (Jun22) / Q12 (Jun23) — same every sittingMark bands, steps, timing
What this question type rewards
The topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
- Writing every number as a times 10 to the power n with 1 is less than or equal to a which is less than 10
- Dealing with the number parts and the powers of 10 separately when multiplying or dividing
- Reading the question carefully: standard form or an ordinary number, and giving the answer in that exact form
The steps
- Write each number in the calculation as a times 10 to the power n
- Multiply or divide the number parts, then add or subtract the powers of 10 separately
- Convert back to an ordinary number by moving the decimal point the number of places the power shows
- Check the final answer is in the exact form the question asks for
Which of these numbers is written in standard form?
Standard form conversions and calculations come up in every sitting we have, worth 1 to 2 marks each. Practise converting confidently between standard form and ordinary numbers without a calculator.
Practise standard form questions“Three of the four sittings we have full papers for test the laws of indices directly, and June 2019 tests it twice in the same paper.”
Every sitting from June 2019 onward tests combining powers of the same base, sometimes to reach a single power and sometimes to solve for an unknown index.
1×asked“Work out the value of (3 to the power 12 divided by 3 to the power 5) divided by (3 squared times 3)”
“Work out the value of (3 to the power 12 divided by 3 to the power 5) divided by (3 squared times 3)”
What it’s really asking
Simplify each bracket to a single power of 3, then divide those two powers.
The full worked answer — June 2019
3 to the power 12 divided by 3 to the power 5 equals 3 to the power 7, and 3 squared times 3 equals 3 cubed, since the lone 3 is 3 to the power 1.
3 to the power 7 divided by 3 cubed equals 3 to the power 4, which equals 81.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise index law questionsWhat the mark scheme rewarded
- Rewriting every term as a power of 3 by combining the division and multiplication inside each bracket into a single power (1 mark)
- Reaching a single power in the form 3 to the power n, using any correct combination of the indices (1 mark)
- The single correct evaluated answer, 81 (1 mark)
- Deal with each bracket separately first, combining it into a single power of 3, before dividing the two brackets by each other
- Remember a lone number like the 3 in the second bracket is still 3 to the power 1, and must be added to the other index in that bracket
Traps examiners saw
- Forgetting that the lone 3 in 3 squared times 3 is a power of 1, and missing it out when combining the bracket into a single power
- Subtracting the indices for the multiplication inside a bracket instead of adding them
Full-mark self-check 0 of 3
1×asked“Simplify 8 to the power 4 divided by 32 to the power two fifths. Give your answer in the form 2 to the power m where m is an integer.”
“Simplify 8 to the power 4 divided by 32 to the power two fifths. Give your answer in the form 2 to the power m where m is an integer.”
What it’s really asking
Rewrite both 8 and 32 as powers of 2, since the final answer must be a single power of 2.
The full worked answer — June 2019
8 equals 2 cubed, so 8 to the power 4 equals 2 to the power 12. Also 32 equals 2 to the power 5, so 32 to the power two fifths equals 2 squared.
2 to the power 12 divided by 2 squared equals 2 to the power 10, so m equals 10.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise index law questionsWhat the mark scheme rewarded
- Correctly converting 8 to a power of 2, reaching 2 to the power 12 for the numerator (1 mark)
- Correctly converting the fractional index on 32, reaching 2 squared for the denominator (1 mark)
- The single correct final value, m equals 10 (1 mark)
- Recognise 8 and 32 as small powers of 2 instantly, since converting to a common base is the only way to combine them
- A fractional index m over n means take the nth root first, then raise the result to the power m
Traps examiners saw
- Trying to combine 8 and 32 without first converting both to powers of 2, since they cannot be combined with different bases
- Misapplying the fractional index on 32, for example taking the wrong root or forgetting to raise to the power 2 afterwards
Full-mark self-check 0 of 3
1×asked“Simplify 8 times 2 to the power 6 times 2 to the power 4. Give your answer as a power of 2.”
“Simplify 8 times 2 to the power 6 times 2 to the power 4. Give your answer as a power of 2.”
What it’s really asking
Convert the 8 into a power of 2 so every term shares the same base, then add the indices.
The full worked answer — June 2022
8 equals 2 cubed, so the calculation becomes 2 cubed times 2 to the power 6 times 2 to the power 4.
Adding the three indices, 3 plus 6 plus 4, gives 2 to the power 13.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise index law questionsWhat the mark scheme rewarded
- A correct method that combines the three terms into a single power of 2, such as 2 cubed times 2 to the power 6 times 2 to the power 4, or the evaluated total 8192 (up to 2 marks)
- The correct final power, 2 to the power 13, or 8192 with correct working, alone on the answer line (both marks)
- Spot that 8 is not already written as a power of 2 and convert it first, since every term must share the same base before the indices can be added
- Adding three indices together is no different from adding two, just add all three powers in one step
Traps examiners saw
- Leaving the 8 unconverted and only combining the two powers of 2 that are already written that way, losing the contribution from 8
- Multiplying the indices together instead of adding them when combining powers of the same base
Full-mark self-check 0 of 3
1×asked“2 to the power a times 3 times 5 squared equals 600. Work out the value of a. You must show your working.”
“2 to the power a times 3 times 5 squared equals 600. Work out the value of a. You must show your working.”
What it’s really asking
Evaluate the known factors first to isolate 2 to the power a, then recognise the result as a power of 2.
The full worked answer — June 2023
3 times 5 squared equals 3 times 25, which equals 75, so 2 to the power a times 75 equals 600.
2 to the power a equals 600 divided by 75, which equals 8. Since 8 equals 2 cubed, a equals 3.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise index law questionsWhat the mark scheme rewarded
- Evaluating the known factors of the equation, 3 times 5 squared, correctly, or an equivalent method to isolate 2 to the power a (1 mark)
- Correct method to divide out those known factors from 600, isolating 2 to the power a (1 mark)
- The single correct value, a equals 3, with working shown (1 mark, since this question asks for working to be shown)
- Evaluate every part of the equation that does not involve the unknown index first, so the unknown term stands alone
- Recognise the result of the division, 8, as a small power of 2 immediately to read off the index
Traps examiners saw
- Trying to take a log or root instead of simply recognising 8 as 2 cubed, which wastes time on a non-calculator paper
- Not showing any working, since the question specifically asks for it and marks are not usually awarded without it
Full-mark self-check 0 of 3
The method for every Q8, Q23 (Jun19) / Q6b (Jun22) / Q7 (Jun23) — same every sittingMark bands, steps, timing
What this question type rewards
The topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
- Writing every number in the calculation as a power of the same base before combining anything
- Adding indices when multiplying and subtracting indices when dividing
- Converting a fractional index into a root and a power
The steps
- Rewrite every number in the calculation as a power of the same base
- Multiply powers of the same base by adding the indices, divide by subtracting them
- A fractional index m over n means take the nth root, then raise to the power m
- Simplify to a single power and check it matches the form the question asks for
Which of these is equivalent to a³ × a⁵?
The laws of indices come up in every sitting we have from June 2019 onward, worth 2 to 3 marks. Practise combining and solving powers without reaching for a calculator.
Practise index law questions“Three of the four sittings we have full papers for test converting a recurring decimal to a fraction, with June 2022 asking for the difference of two recurring decimals rather than a single conversion.”
Every sitting from June 2019 onward tests this skill, using the multiply-by-a-power-of-10-and-subtract method every time.
1×asked“Circle the fraction that is equivalent to 0.1 recurring”
“Circle the fraction that is equivalent to 0.1 recurring”
What it’s really asking
Recognise that a single repeating digit after the decimal point corresponds to a fraction with 9 in the denominator.
The full worked answer — June 2019
0.1 recurring means 0.1111 repeating forever. One ninth equals 0.1111 repeating forever, since dividing 1 by 9 gives a single repeating 1.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise recurring decimal questionsWhat the mark scheme rewarded
- The single correct option circled, one ninth (1 mark, point marked with no working required)
- Learn the standard result that a single repeating digit d after the decimal point always equals d over 9
- Test an option quickly by dividing its numerator by its denominator to check it genuinely produces the repeating pattern shown
Traps examiners saw
- Confusing a recurring decimal notation with a terminating one, for example mistaking 0.1 recurring for exactly 0.1
- Picking a fraction that looks similar, such as one tenth, which is a terminating decimal and not a match for 0.1 recurring
Full-mark self-check 0 of 2
1×asked“Work out 0.68 recurring on the 8, minus 0.45 recurring on the 5. Give your answer as a fraction in its simplest form.”
“Work out 0.68 recurring on the 8, minus 0.45 recurring on the 5. Give your answer as a fraction in its simplest form.”
What it’s really asking
Convert each recurring decimal to a fraction with the same method, then subtract and simplify.
The full worked answer — June 2022
For 0.68 recurring on the 8: let y equal 0.6888 repeating. 100y equals 68.888 repeating, and 10y equals 6.888 repeating, so 100y minus 10y equals 62, giving y equals 62 over 90.
For 0.45 recurring on the 5: the same method gives z equals 41 over 90.
62 over 90 minus 41 over 90 equals 21 over 90, which simplifies to 7 over 30.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise recurring decimal questionsWhat the mark scheme rewarded
- Multiplying the first recurring decimal by a power of 10 and subtracting to eliminate the recurring digit (1 mark)
- Multiplying the second recurring decimal by a power of 10 and subtracting to eliminate its recurring digit (1 mark)
- The correct unsimplified fraction for each recurring decimal, 62 over 90 and 41 over 90 (1 mark)
- The correct subtraction of the two fractions, reaching 21 over 90 (1 mark)
- The final answer simplified fully to 7 over 30, follow-through available for a correctly simplified version of an earlier slip (1 mark)
- Convert both recurring decimals to fractions using exactly the same multiply-and-subtract method before attempting to subtract them
- The two fractions do not need a common denominator before subtracting if they already share one, such as both having 90 as their denominator here
Traps examiners saw
- Subtracting the two recurring decimals directly as decimals instead of converting each to a fraction first, which the question specifically asks for
- Simplifying only one of the two fractions before subtracting, leading to a mismatched denominator
Full-mark self-check 0 of 3
1×asked“Write 0.13 recurring on the 3 as a fraction in its simplest form.”
“Write 0.13 recurring on the 3 as a fraction in its simplest form.”
What it’s really asking
Multiply by powers of 10 that line up the repeating 3, not the 1, since only the 3 recurs.
The full worked answer — June 2023
Let x equal 0.1333 repeating. 10x equals 1.333 repeating and 100x equals 13.333 repeating, so 100x minus 10x equals 12, giving 90x equals 12.
x equals 12 over 90, which simplifies to 2 over 15.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise recurring decimal questionsWhat the mark scheme rewarded
- Denoting the recurring decimal by a letter and multiplying by a power of 10, such as 10x equals 1.333 repeating (1 mark)
- Multiplying by a second power of 10 and subtracting to eliminate the recurring digits, reaching an equation such as 90x equals 12 (1 mark)
- The correct fraction, fully simplified to 2 over 15 (1 mark)
- Only the digit under the recurring dot actually repeats here, the leading 1 does not, so the powers of 10 must be chosen to line up the 3s specifically
- Always give the final fraction in its simplest form, since an unsimplified correct fraction can lose the last mark
Traps examiners saw
- Assuming the whole decimal repeats and multiplying by the wrong powers of 10, which fails to fully cancel the recurring part
- Leaving the answer as 12 over 90 without simplifying it fully to 2 over 15
Full-mark self-check 0 of 3
The method for every Q19 (Jun19) / Q22 (Jun22) / Q23 (Jun23) — same every sittingMark bands, steps, timing
What this question type rewards
The topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
- Multiplying the recurring decimal by the correct power of 10 so the recurring part lines up
- Subtracting to eliminate the recurring digits and leave a whole number equation
- Simplifying the resulting fraction fully
The steps
- Let x equal the recurring decimal
- Multiply x by 10, 100, or another power of 10 so the recurring digits line up exactly with the original
- Subtract the two equations to cancel the recurring part completely
- Solve for x as a fraction, then simplify it fully
Which of these fractions gives a recurring decimal when you divide?
Converting a recurring decimal to a fraction comes up in every sitting we have from June 2019 onward. Practise the multiply-and-subtract method until it is automatic.
Practise recurring decimal questions“Every sitting we have full papers for opens with at least one quick non-calculator arithmetic question involving fractions, decimals, or both.”
June 2023 alone opens with four separate 1-mark arithmetic questions in a row, and every other sitting has at least one similar question early in the paper.
1×asked“Work out the value of (1 and two thirds) squared. Circle your answer.”
“Work out the value of (1 and two thirds) squared. Circle your answer.”
What it’s really asking
Convert the mixed number to an improper fraction first, then square the numerator and denominator separately.
The full worked answer — June 2019
1 and two thirds equals five thirds. Squaring gives five thirds squared equals 25 over 9, which equals 2 and seven ninths.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise fraction and decimal arithmeticWhat the mark scheme rewarded
- The single correct answer circled, 2 and seven ninths, or its equivalent improper fraction 25 over 9 (1 mark, point marked)
- Always convert a mixed number to an improper fraction before squaring it, since squaring the whole number and fraction parts separately gives a completely wrong result
- Square the numerator and denominator of the improper fraction separately, then convert back to a mixed number if needed
Traps examiners saw
- Squaring the whole number 1 and the fraction two thirds separately instead of converting to five thirds first
- Selecting the improper fraction option, 25 over 9, when the answer options are written as mixed numbers, or the reverse
Full-mark self-check 0 of 3
1×asked“In each box, write a fraction less than 1 to make a correct calculation with a product of three tenths. In each box, write a decimal less than 1 to make a correct calculation with a product of 0.06.”
“In each box, write a fraction less than 1 to make a correct calculation with a product of three tenths. In each box, write a decimal less than 1 to make a correct calculation with a product of 0.06.”
What it’s really asking
Work backwards from the target product to find two values less than 1 that multiply to give it.
The full worked answer — June 2022
For the fraction part: three fifths times one half equals three tenths, and both three fifths and one half are less than 1.
For the decimal part: 0.3 times 0.2 equals 0.06, and both 0.3 and 0.2 are less than 1.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise fraction and decimal arithmeticWhat the mark scheme rewarded
- Two fractions, each less than 1, with a product of three tenths (1 mark, any correct pair such as three fifths and one half)
- Two decimals, each less than 1, with a product of 0.06 (1 mark, any correct pair such as 0.3 and 0.2)
- Work backwards from the target product by picking one simple factor first, then dividing the target by it to find the second
- Both values must genuinely be less than 1, so check a chosen pair does not include a whole number or a value of exactly 1
Traps examiners saw
- Writing one of the two values as exactly 1, or as a value greater than 1, which breaks the 'less than 1' condition even if the product is correct
- Using decimals inside a fraction answer, such as 0.6 over 1, which is not accepted as two proper fractions
Full-mark self-check 0 of 3
1×asked“Work out 0.7 times 0.5. Work out five sixths divided by 3. Work out 27 divided by 0.6.”
“Work out 0.7 times 0.5. Work out five sixths divided by 3. Work out 27 divided by 0.6.”
What it’s really asking
Apply place value for the decimal parts, and multiply by the reciprocal for the fraction division.
The full worked answer — June 2023
0.7 times 0.5: ignoring the decimal points, 7 times 5 equals 35, and since both numbers have one decimal place, the answer has two decimal places, giving 0.35.
Five sixths divided by 3 equals five sixths times one third, which equals five eighteenths.
27 divided by 0.6 equals 270 divided by 6, which equals 45.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise fraction and decimal arithmeticWhat the mark scheme rewarded
- 0.35, or an equivalent fraction such as 7 over 20, on the answer line for the first part (1 mark)
- 5 over 18, or a correct equivalent such as 10 over 36 or the recurring decimal 0.2777 repeating, for the second part (1 mark)
- 45 for the third part, with 270 divided by 6 accepted as working but not as an unprocessed final answer (1 mark)
- Mark the value on the answer line for each part; if it is left blank the working is checked instead, so always write a final answer even under time pressure
- If more than one form of an answer is given, such as both a fraction and a decimal, every form given must be correct for the mark to be awarded
Traps examiners saw
- Leaving an unprocessed calculation such as 270 divided by 6 as the final answer instead of evaluating it fully to 45
- Giving inconsistent forms across working and the answer line, for example correct working but a different, incorrect value copied onto the answer line
Full-mark self-check 0 of 3
1×asked“Work out the value of (three halves) squared. Give your answer as a mixed number.”
“Work out the value of (three halves) squared. Give your answer as a mixed number.”
What it’s really asking
Square the numerator and denominator separately, then convert the improper fraction to a mixed number.
The full worked answer — June 2023
Three halves squared equals 9 over 4, which converts to the mixed number 2 and one quarter.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise fraction and decimal arithmeticWhat the mark scheme rewarded
- The single correct answer, 2 and one quarter, given as a mixed number exactly as the question asks (1 mark, point marked)
- Square the numerator and denominator of the fraction separately, three squared over 2 squared, before converting the improper fraction to a mixed number
- Converting to a mixed number is the last step, not the first, since squaring a mixed number directly does not work
Traps examiners saw
- Leaving the answer as the improper fraction 9 over 4 instead of converting it to the mixed number the question specifically asks for
- Squaring only the numerator, 3, and forgetting to also square the denominator, 2
Full-mark self-check 0 of 3
The method for every Q2 (Jun19) / Q9 (Jun22) / Q1, Q3 (Jun23) — same every sittingMark bands, steps, timing
What this question type rewards
The topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
- Converting mixed numbers to improper fractions before multiplying or dividing
- Using place value to multiply or divide decimals without a calculator
- Recognising that dividing by a fraction means multiplying by its reciprocal
The steps
- Convert any mixed number into an improper fraction first
- For decimals, ignore the decimal points, do the calculation with whole numbers, then place the decimal point back using place value
- For dividing by a fraction, multiply by its reciprocal instead
- Simplify or convert the final answer to the form the question asks for
Which of these fractions is the largest? ⅔ ¾ ⅗ ⅝
Quick non-calculator fraction and decimal arithmetic opens nearly every paper we have. Practise these until they take seconds, not minutes, to free up time for the harder questions.
Practise fraction and decimal arithmetic“Three of the four sittings we have full papers for test a ratio problem set in a real context, always requiring at least two stages of working.”
Every one of these questions starts by finding the value of one part of a ratio, then uses that value to work out a further change.
1×asked“A different pattern is made using 20 straight lines and 16 arcs. The straight lines and arcs are made from metal. 20 straight lines cost 12 pounds. Cost of one straight line to cost of one arc is 2 to 3. Work out the total cost of the metal in the pattern.”
“A different pattern is made using 20 straight lines and 16 arcs. The straight lines and arcs are made from metal. 20 straight lines cost 12 pounds. Cost of one straight line to cost of one arc is 2 to 3. Work out the total cost of the metal in the pattern.”
What it’s really asking
Find the cost of one line first, use the ratio to find the cost of one arc, then total both.
The full worked answer — June 2018
One straight line costs 12 pounds divided by 20, which equals 0.60 pounds. Since the cost ratio of one line to one arc is 2 to 3, one arc costs 0.60 pounds times three halves, which equals 0.90 pounds.
The total cost is 20 times 0.60 pounds plus 16 times 0.90 pounds, which is 12 pounds plus 14.40 pounds, giving 26.40 pounds.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise ratio problemsWhat the mark scheme rewarded
- Finding the unit cost of one straight line, 0.60 pounds, or an equivalent correct step (1 mark)
- Using the given ratio correctly to find the unit cost of one arc, 0.90 pounds, or an equivalent dependent step (1 mark)
- The correct final total, 26.40 pounds, in correct money notation (1 mark)
- Find the value of one item first, here the cost of a single line, before scaling up to the cost of an arc using the ratio
- Give money answers in full correct notation, such as 26.40 rather than 26.4, since a missing zero can cost the final mark
Traps examiners saw
- Applying the 2 to 3 ratio the wrong way round, scaling the line's cost down instead of up to find the arc's cost
- Writing the final answer as 26.4 instead of 26.40, losing the mark for correct money notation
Full-mark self-check 0 of 3
The method for every Q14b (Jun18) / Q11 (Jun19) / Q6 (Jun23) — same every sittingMark bands, steps, timing
What this question type rewards
The topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
- Finding the value of one part of the ratio before using it in any further step
- Working with the correct fraction of each person's own share when amounts are redistributed
- Keeping units and money notation correct throughout
The steps
- Add the parts of the ratio to find the total number of parts, or use a given amount to find the value of one part
- Work out each named quantity from the ratio
- Apply any further change (a gift, a redistribution, or a unit cost) using the values just found
- Check the final answer is in the units and form the question asks for
A recipe uses flour and sugar in the ratio 5:2. What fraction of the mixture is flour?
Multi-stage ratio problems come up in three of the four sittings we have, worth 3 to 4 marks. Practise finding the value of one part before tackling the rest of the question.
Practise ratio problems“Three of the four sittings we have full papers for test a percentage increase directly, either as a calculation or as a multiple choice question.”
Every one of these questions uses the multiplier method, converting a percentage increase into a single decimal multiplier.
1×asked“The value of y is 20 percent more than the value of x. Circle the ratio x to y.”
“The value of y is 20 percent more than the value of x. Circle the ratio x to y.”
What it’s really asking
Write y as a multiple of x using the multiplier, then simplify the ratio of the two.
The full worked answer — June 2018
20 percent more means y equals 1.2 times x, so x to y is the same as 1 to 1.2, which simplifies to 5 to 6.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise percentage increase questionsWhat the mark scheme rewarded
- The single correct ratio circled, 5 to 6 (1 mark, point marked with no working required)
- Convert the percentage increase into a decimal multiplier, giving y equals 1.2 times x, before writing the ratio
- Simplify a ratio containing a decimal by multiplying both sides by the same value until whole numbers appear, here multiplying 1 to 1.2 by 5
Traps examiners saw
- Writing the ratio the wrong way round, giving y to x instead of the x to y the question specifically asks for
- Using 0.2 as the multiplier instead of the full 1.2, since only the increase itself is 20 percent, not the new total
Full-mark self-check 0 of 3
1×asked“Three friends arrive at a party. Their arrival increases the number of people at the party by 20 percent. In total, how many people are now at the party?”
“Three friends arrive at a party. Their arrival increases the number of people at the party by 20 percent. In total, how many people are now at the party?”
What it’s really asking
The 3 extra people represent 20 percent of the original total, so find the original total first.
The full worked answer — June 2019
If 3 people is 20 percent of the original number, the original number is 3 divided by 0.2, which equals 15.
The total number now at the party is 15 plus 3, which equals 18.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise percentage increase questionsWhat the mark scheme rewarded
- A correct method to find the original number, 3 divided by 0.2, which equals 15 (1 mark)
- The correct final total, 18 (1 mark)
- Treat the increase itself, the 3 new arrivals, as a known fraction of the unknown original total, then divide to find that original
- Read the question carefully to see whether it wants the original total, the increase, or the new total, since all three appear here
Traps examiners saw
- Treating the 3 new arrivals as 20 percent of the new total rather than the original total, which gives the wrong original number
- Stopping at the original total, 15, without adding the 3 new arrivals to reach the final total the question actually asks for
Full-mark self-check 0 of 3
1×asked“Circle the calculation that increases 240 by 7.5 percent”
“Circle the calculation that increases 240 by 7.5 percent”
What it’s really asking
Convert 7.5 percent to a decimal and add it to 1 to find the correct multiplier.
The full worked answer — June 2022
7.5 percent as a decimal is 0.075, so increasing by 7.5 percent means multiplying by 1 plus 0.075, which is 1.075, giving the calculation 240 times 1.075.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise percentage increase questionsWhat the mark scheme rewarded
- The single correct calculation circled, 240 times 1.075 (1 mark, point marked with no working required)
- Build the multiplier by adding the percentage as a decimal to 1, never leaving it as the percentage alone
- Check the decimal conversion of 7.5 percent carefully, since a single misplaced decimal place changes which option is correct
Traps examiners saw
- Misplacing the decimal point when converting 7.5 percent to 0.075, for example writing 0.75 instead
- Selecting an option using 0.075 alone as the multiplier instead of 1.075, which would find the increase itself rather than the new total
Full-mark self-check 0 of 3
The method for every Q16 (Jun18) / Q7 (Jun19) / Q3 (Jun22) — same every sittingMark bands, steps, timing
What this question type rewards
The topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
- Converting a percentage increase into a single decimal multiplier, such as 20 percent more meaning multiply by 1.2
- Working backwards from a known increase to find the original value
- Keeping the multiplier and the original value clearly separate in the working
The steps
- Convert the percentage increase into a decimal multiplier by adding it to 1
- Multiply the original value by this multiplier to find the increased value
- If the increase itself is known, divide it by the percentage as a decimal to find the original value
- Check whether the question wants the new value, the original value, or a ratio between them
What multiplier is used to increase a value by 15%?
Percentage increase questions come up in three of the four sittings we have. Practise building the correct multiplier every time, forwards and backwards.
Practise percentage increase questions“Three of the four sittings we have full papers for test a linear inequality, ranging from a one-step solve to a graphical region bounded by three inequalities.”
Every sitting we checked treats an inequality exactly like an equation for the algebra, only keeping the inequality sign throughout.
1×asked“Solve 5 times (x plus 3) is less than 60”
“Solve 5 times (x plus 3) is less than 60”
What it’s really asking
Expand the bracket and solve exactly like an equation, keeping the inequality sign.
The full worked answer — June 2018
Expanding the bracket gives 5x plus 15 is less than 60, so 5x is less than 45.
Dividing both sides by 5 gives x is less than 9.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise linear inequality questionsWhat the mark scheme rewarded
- Correct expansion of the bracket, 5x plus 15 is less than 60, or equivalent correct rearrangement (1 mark)
- The correct final inequality, x is less than 9, with the sign in the right direction (1 mark)
- Expand the bracket exactly as with an equation, then collect terms before dividing to isolate x
- Dividing by a positive number, 5 here, never changes the direction of the inequality sign
Traps examiners saw
- Writing the final answer with an incorrect sign, such as x is greater than 9, when the original inequality was strictly less than
- Expanding the bracket incorrectly, for example forgetting to multiply the 3 by 5 as well as the x
Full-mark self-check 0 of 3
1×asked“On the grid, identify the region represented by x is greater than 3, and y is greater than 1, and x plus y is less than or equal to 7. Label the region R.”
“On the grid, identify the region represented by x is greater than 3, and y is greater than 1, and x plus y is less than or equal to 7. Label the region R.”
What it’s really asking
What the sources actually showed — June 2022
A blank set of axes on which the student draws the three boundary lines x equals 3, y equals 1, and x plus y equals 7, then labels the enclosed triangular region.
The full worked answer — June 2022
The line x equals 3 is drawn as a dashed vertical line, since x is greater than 3 is a strict inequality, and y equals 1 is drawn as a dashed horizontal line for the same reason.
The line x plus y equals 7 is drawn as a solid line, since x plus y is less than or equal to 7 includes the line itself, running from around (3, 4) to (6, 1).
The triangular region enclosed between these three lines, where x is greater than 3 and y is greater than 1 and their sum is at most 7, is labelled R.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise linear inequality questionsWhat the mark scheme rewarded
- Correct dashed lines drawn for the two strict inequalities, x equals 3 and y equals 1 (1 mark)
- Correct solid line drawn for the non-strict inequality, x plus y equals 7 (1 mark)
- The correct region identified, by labelling it R or shading it consistently (1 mark)
- Use a dashed line for any strict inequality, greater than or less than, and a solid line for greater than or equal to, or less than or equal to
- Each line only needs to be drawn far enough to clearly show the boundary of the triangular region, not across the entire grid
Traps examiners saw
- Using a solid line for x equals 3 or y equals 1, which changes the meaning of a strict inequality to include the boundary itself
- Shading or labelling the wrong one of the several regions the three lines create, rather than the one where all three inequalities hold together
Full-mark self-check 0 of 3
1×asked“Solve 2x is less than 26”
“Solve 2x is less than 26”
What it’s really asking
Divide both sides by 2, keeping the inequality sign the same direction.
The full worked answer — June 2023
Dividing both sides by 2 gives x is less than 13.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise linear inequality questionsWhat the mark scheme rewarded
- The correct final inequality, x is less than 13, with the sign in the right direction (1 mark, point marked)
- Dividing both sides by a positive number never changes the direction of the inequality sign
- Check the final inequality can be written the other way round too, 13 is greater than x, without changing its meaning
Traps examiners saw
- Writing an equation, x equals 13, instead of keeping the inequality sign the question asks for
- Giving the value of x with the wrong inequality sign, such as x is greater than 13
Full-mark self-check 0 of 3
The method for every Q5 (Jun18) / Q23 (Jun22) / Q2 (Jun23) — same every sittingMark bands, steps, timing
What this question type rewards
The topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
- Expanding brackets and collecting terms exactly as with an equation
- Keeping the inequality sign the same direction throughout, unless multiplying or dividing by a negative number
- For a graphical region, drawing each boundary line correctly and identifying the overlap
The steps
- Expand any brackets first
- Collect terms on each side exactly as you would for an equation
- Divide to leave the unknown on its own, keeping the inequality sign the same direction
- For a region, draw each line and shade or label the overlap that satisfies every inequality at once
Which of the following correctly describes how to represent x > 3 on a number line?
Linear inequalities come up in three of the four sittings we have, from a one-step solve to a full graphical region. Practise keeping the sign right in both directions.
Practise linear inequality questions“Two of the four sittings we have full papers for test simultaneous equations, and June 2023 tests it twice, once as pure algebra and once as a word problem about balancing weights.”
Every version uses elimination or substitution to remove one unknown before solving for the other.
1×asked“Solve the simultaneous equations 2x plus 3y equals 5p and y equals 2x plus p, where p is a constant. Give your answers in terms of p in their simplest form.”
“Solve the simultaneous equations 2x plus 3y equals 5p and y equals 2x plus p, where p is a constant. Give your answers in terms of p in their simplest form.”
What it’s really asking
Substitute the expression for y into the first equation, then solve for x and y in terms of p.
The full worked answer — June 2019
Substituting y equals 2x plus p into the first equation gives 2x plus 3 times (2x plus p) equals 5p, which expands to 2x plus 6x plus 3p equals 5p.
Collecting terms gives 8x equals 2p, so x equals p over 4.
Substituting x equals p over 4 back into y equals 2x plus p gives y equals p over 2 plus p, which simplifies to three halves of p.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise simultaneous equationsWhat the mark scheme rewarded
- Substituting the given expression for y into the first equation to reach a single equation in x and p (1 mark)
- Collecting terms correctly to solve for x in terms of p, reaching x equals p over 4 (1 mark)
- Correct simplified terms for both x and p, with p not omitted from either (up to 2 further marks, awarded together)
- Since y is already given in terms of x and p, substitution is faster here than trying to eliminate an unknown by scaling both equations
- Keep p treated as a constant throughout the algebra, exactly like any other letter, until both x and y are found in terms of it
Traps examiners saw
- Losing the p term partway through the algebra, giving an answer for x or y with no p in it at all
- Substituting back into the more complicated original equation instead of the simpler one, increasing the chance of an arithmetic slip
Full-mark self-check 0 of 3
1×asked“Solve the simultaneous equations 2x minus 5y equals 13 and 3x plus 4y equals 8”
“Solve the simultaneous equations 2x minus 5y equals 13 and 3x plus 4y equals 8”
What it’s really asking
Scale both equations so one unknown has matching coefficients, then eliminate it by adding or subtracting.
The full worked answer — June 2023
Multiplying the first equation by 4 and the second by 5 gives 8x minus 20y equals 52, and 15x plus 20y equals 40.
Adding the two equations eliminates y: 23x equals 92, so x equals 4.
Substituting x equals 4 into 2x minus 5y equals 13 gives 8 minus 5y equals 13, so y equals negative 1.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise simultaneous equationsWhat the mark scheme rewarded
- Scaling both equations so one unknown, here y, has matching or opposite coefficients (1 mark)
- Adding the two scaled equations correctly to eliminate y, reaching 23x equals 92 (1 mark)
- The correct value of x, from correctly solving the eliminated equation (1 mark)
- The correct value of y, from substituting back into one of the original equations (1 mark)
- Choose which unknown to eliminate based on which coefficients are quickest to make match, here y needed multiplying by 4 and 5
- Substitute the found value back into one of the ORIGINAL equations, not a scaled or rearranged version, to reduce the risk of a repeated error
Traps examiners saw
- Scaling only one of the two equations instead of both, so the coefficients of the chosen unknown never actually match
- Adding when the signs require subtracting, or the reverse, introducing a sign error that changes both final answers
Full-mark self-check 0 of 3
1×asked“K, L and M are weights. Both of two given balance scales balance exactly. How many M weights are needed to balance one L weight?”
“K, L and M are weights. Both of two given balance scales balance exactly. How many M weights are needed to balance one L weight?”
What it’s really asking
Translate each balanced scale into an equation, then combine the two equations to compare L and M directly.
What the sources actually showed — June 2023
The first scale balances 3 K weights against 4 L weights. The second scale balances 1 K weight against 1 L weight together with 2 M weights, giving the two equations 3K equals 4L and K equals L plus 2M.
The full worked answer — June 2023
The two balanced scales give 3K equals 4L and K equals L plus 2M.
Substituting K equals L plus 2M into 3K equals 4L gives 3 times (L plus 2M) equals 4L, which expands to 3L plus 6M equals 4L, so 6M equals L.
Since L equals 6M, six M weights are needed to balance one L weight.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise simultaneous equationsWhat the mark scheme rewarded
- Correctly writing both balanced scales as equations, 3K equals 4L and K equals L plus 2M (1 mark)
- Correctly substituting to eliminate K, reaching an equation relating only L and M (1 mark)
- The correct final relationship, L equals 6M, and the final answer, six (1 mark)
- Translate each balanced scale into an equation before attempting any algebra, since the whole question depends on both being correct
- Substitute to eliminate the letter that appears in both equations, K here, since the question only wants a relationship between L and M
Traps examiners saw
- Setting up one of the two balance equations incorrectly, for example reversing which side has more weights
- Stopping once L is written in terms of M without converting the relationship into the specific number of M weights the question asks for
Full-mark self-check 0 of 3
The method for every Q21 (Jun19) / Q16, Q20 (Jun23) — same every sittingMark bands, steps, timing
What this question type rewards
The topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
- Multiplying one or both equations so that one unknown has matching or opposite coefficients
- Adding or subtracting the equations correctly to eliminate that unknown
- Substituting back to find the second unknown
The steps
- Multiply one or both equations so the coefficient of one unknown matches (or is the negative of) the other
- Add or subtract the equations to eliminate that unknown
- Solve the resulting equation for the remaining unknown
- Substitute back into either original equation to find the first unknown
Which method is most efficient for solving the following simultaneous equations? 5x + 2y = 14 3x + 2y = 10
Simultaneous equations come up in two of the four sittings we have, sometimes as pure algebra and sometimes hidden inside a word problem. Practise spotting both.
Practise simultaneous equations“Two of the four sittings we have full papers for test simplifying an algebraic fraction, always requiring a full factorisation before any cancelling.”
Every version needs the numerator and denominator factorised first; cancelling terms without factorising first scores nothing.
1×asked“Simplify fully (4x minus 8x squared) over (12x minus 6)”
“Simplify fully (4x minus 8x squared) over (12x minus 6)”
What it’s really asking
Factorise the numerator and denominator so a shared bracket appears in both, then cancel it.
The full worked answer — June 2019
The numerator 4x minus 8x squared factorises to 4x times (1 minus 2x), and the denominator 12x minus 6 factorises to 6 times (2x minus 1).
Since (1 minus 2x) is the negative of (2x minus 1), the fraction becomes negative 4x over 6, which simplifies to negative two thirds of x.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise algebraic fraction questionsWhat the mark scheme rewarded
- Correct factorisation of the numerator, 4x times (1 minus 2x), or an equivalent (1 mark)
- Correct factorisation of the denominator, 6 times (2x minus 1), or a correct cancelling of a common factor of 2 throughout (1 mark)
- The fully simplified final fraction, negative two thirds of x (1 mark)
- Factorise numerator and denominator completely before attempting to cancel anything between them
- Spot a bracket and its negative, such as (1 minus 2x) and (2x minus 1), which cancel to give a factor of negative 1
Traps examiners saw
- Cancelling an x from one term of the numerator with an x in the denominator without factorising first, which is not a valid cancellation
- Missing the sign change when cancelling (1 minus 2x) with (2x minus 1), losing the negative in the final answer
Full-mark self-check 0 of 3
1×asked“Simplify fully 6 over a minus 11 over 4a. Simplify fully (y squared minus 3y) times (y squared plus 10y plus 21) over (y squared minus 9).”
“Simplify fully 6 over a minus 11 over 4a. Simplify fully (y squared minus 3y) times (y squared plus 10y plus 21) over (y squared minus 9).”
What it’s really asking
For the first part, use a over 4a as the common denominator; for the second, factorise every part before cancelling.
The full worked answer — June 2022
For 6 over a minus 11 over 4a: writing both fractions over the common denominator 4a gives 24 over 4a minus 11 over 4a, which equals 13 over 4a.
For the product: y squared minus 3y factorises to y times (y minus 3), y squared plus 10y plus 21 factorises to (y plus 7) times (y plus 3), and y squared minus 9 factorises to (y plus 3) times (y minus 3).
Cancelling the shared factors of (y minus 3) and (y plus 3) leaves y times (y plus 7), which equals y squared plus 7y.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise algebraic fraction questionsWhat the mark scheme rewarded
- For the first part, converting 6 over a to the common denominator 4a correctly (1 mark)
- For the first part, the correct simplified answer, 13 over 4a (1 mark)
- For the second part, all three quadratics correctly factorised: y times (y minus 3), (y plus 7) times (y plus 3), and (y plus 3) times (y minus 3) (1 mark each, up to 3 marks)
- For the second part, the shared factors cancelled correctly to reach the fully simplified answer, y squared plus 7y (1 mark)
- Use the smaller denominator, a, times the extra factor needed, 4, to reach the common denominator 4a quickly, rather than multiplying both denominators together
- Spot y squared minus 9 as a difference of two squares immediately, since it factorises to (y plus 3) times (y minus 3) without needing the quadratic formula
Traps examiners saw
- Using a times 4a as the common denominator instead of the smaller, correct common denominator 4a itself
- Missing that y squared minus 9 is a difference of two squares and failing to factorise the denominator at all
Full-mark self-check 0 of 3
The method for every Q16 (Jun19) / Q24 (Jun22) — same every sittingMark bands, steps, timing
What this question type rewards
The topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
- Factorising the numerator and denominator fully before cancelling anything
- Cancelling a common factor, not a common term, between numerator and denominator
- Recognising a difference of two squares or a common quadratic factor across two fractions being multiplied
The steps
- Factorise the numerator fully
- Factorise the denominator fully
- Cancel any factor that appears in both the numerator and the denominator
- Check the simplified fraction cannot be reduced any further
Simplify 6x²/3x
Simplifying algebraic fractions comes up in two of the four sittings we have, worth 3 marks or more. Practise factorising fully before you cancel anything.
Practise algebraic fraction questions“Three of the four sittings we have full papers for use a given identity or a comparison of coefficients, always testing whether the identity is applied rather than just recognised.”
Each version gives the student a true identity or equivalence and asks them to use it, not just state it.
1×asked“Here is an identity: x squared minus y squared is identically equal to (x plus y) times (x minus y). Use the identity to work out the value of 193 squared minus 7 squared. You must show your working. Factorise 100a squared minus 81b squared.”
“Here is an identity: x squared minus y squared is identically equal to (x plus y) times (x minus y). Use the identity to work out the value of 193 squared minus 7 squared. You must show your working. Factorise 100a squared minus 81b squared.”
What it’s really asking
Match 193 and 7 to x and y in the given identity to avoid squaring large numbers directly, then apply the same pattern to factorise.
The full worked answer — June 2019
Matching 193 to x and 7 to y in the identity gives 193 squared minus 7 squared equals (193 plus 7) times (193 minus 7), which is 200 times 186.
200 times 186 equals 37 200.
Since 100a squared is (10a) squared and 81b squared is (9b) squared, the same identity gives 100a squared minus 81b squared equals (10a plus 9b) times (10a minus 9b).
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise algebraic identity questionsWhat the mark scheme rewarded
- Correctly matching 193 to x and 7 to y to reach 200 times 186 (1 mark)
- The correct final value, 37200, from correct working using the identity (1 mark)
- Recognising 100a squared and 81b squared as perfect squares and correctly factorising using the identical pattern (1 mark)
- Use the exact identity the question gives rather than attempting to square 193 directly, which is both slower and riskier without a calculator
- Recognise a term is a perfect square by checking whether its coefficient and power are both perfect squares, such as 100a squared being (10a) squared
Traps examiners saw
- Squaring 193 directly instead of using the given identity, wasting time and risking an arithmetic slip on a non-calculator paper
- Factorising the second expression from scratch instead of recognising it follows the exact same difference of two squares pattern just used
Full-mark self-check 0 of 3
1×asked“12x cubed plus 7x squared plus 3x minus 10 is identically equal to 2 times (ax cubed plus x squared plus 2x minus 5) plus x times (bx plus c). Work out the values of a, b and c.”
“12x cubed plus 7x squared plus 3x minus 10 is identically equal to 2 times (ax cubed plus x squared plus 2x minus 5) plus x times (bx plus c). Work out the values of a, b and c.”
What it’s really asking
Expand the right hand side fully, then match each power of x to the corresponding term on the left.
The full worked answer — June 2022
Expanding the right hand side gives 2ax cubed plus 2x squared plus 4x minus 10, plus bx squared plus cx, which combines to 2ax cubed plus (2 plus b)x squared plus (4 plus c)x minus 10.
Comparing the x cubed terms, 2a equals 12, so a equals 6.
Comparing the x squared terms, 2 plus b equals 7, so b equals 5, and comparing the x terms, 4 plus c equals 3, so c equals negative 1.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise algebraic identity questionsWhat the mark scheme rewarded
- Fully expanding the right hand side and combining like terms, or directly comparing the x cubed coefficients to find a equals 6 (1 mark)
- Comparing the x squared and x terms to form correct equations for b and c (1 mark)
- The correct final values, a equals 6, b equals 5 and c equals negative 1, all three needed for full marks (1 mark)
- Expand the entire right hand side fully before comparing anything, since the identity holds true for every value of x only once both sides are fully expanded
- Compare coefficients power by power, starting with the highest power of x, since each comparison gives one equation for one unknown
Traps examiners saw
- Stating only one or two of the three constants correctly with no working shown, which only earns partial credit rather than the full marks
- Making an expansion error in one term, such as missing the bx squared or cx contribution from the second bracket
Full-mark self-check 0 of 3
1×asked“Here is a number machine: multiply by a, then add b. Show that when the input increases by 2, the output increases by 2a.”
“Here is a number machine: multiply by a, then add b. Show that when the input increases by 2, the output increases by 2a.”
What it’s really asking
Write the output for an input of x, then for an input of x plus 2, and compare the two expressions.
The full worked answer — June 2023
For an input of x, the machine gives an output of ax plus b. For an input of x plus 2, the output is a times (x plus 2) plus b, which equals ax plus 2a plus b.
The difference between the two outputs is (ax plus 2a plus b) minus (ax plus b), which equals 2a, proving the output increases by 2a.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise algebraic identity questionsWhat the mark scheme rewarded
- Writing correct general expressions for the output at an input of x and at an input of x plus 2, in terms of a, b and x (1 mark)
- Correctly subtracting the two expressions to reach exactly 2a, completing the proof (1 mark)
- Use general algebra, a letter such as x, rather than substituting a specific number, since only general algebra proves the result for every possible input
- Written explanations without algebra do not score marks on a 'show that' question, even if the reasoning in words is correct
Traps examiners saw
- Checking the result with a single specific number instead of general algebra, which does not constitute a proof and does not score full marks
- Making a sign or expansion error when multiplying a by (x plus 2), losing the 2a term needed for the proof
Full-mark self-check 0 of 3
The method for every Q18 (Jun19) / Q18 (Jun22) / Q13a (Jun23) — same every sittingMark bands, steps, timing
What this question type rewards
The topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
- Substituting the given identity directly into the numbers or expression in the question
- Expanding fully and comparing matching terms on both sides
- Showing every algebraic step so the working can be checked, not just the final result
The steps
- Identify exactly how the given identity maps onto the specific numbers or expression in the question
- Substitute or expand fully
- Compare coefficients of matching powers on both sides if constants are being found
- State every unknown value clearly at the end
Which expression represents an even number for all integer values of n?
Using a given identity or comparing coefficients comes up in three of the four sittings we have. Practise applying the exact structure the question gives you.
Practise algebraic identity questions“Three of the four sittings we have full papers for test a sequence beyond the standard nth term, using an unusual rule such as adding or multiplying the previous two terms.”
Each version gives a rule for generating the sequence and asks the student to apply it forwards or use given terms to find unknown constants.
1×asked“A linear sequence starts a plus 2b, a plus 6b, a plus 10b, and so on. The 2nd term has value 8. The 5th term has value 44. Work out the values of a and b.”
“A linear sequence starts a plus 2b, a plus 6b, a plus 10b, and so on. The 2nd term has value 8. The 5th term has value 44. Work out the values of a and b.”
What it’s really asking
Use the constant difference of 4b between terms to write the 5th term, then solve two equations for a and b.
The full worked answer — June 2018
Since each term increases by 4b, the 5th term is a plus 2b plus 4 times 4b, which equals a plus 18b.
This gives two equations: a plus 6b equals 8, and a plus 18b equals 44. Subtracting the first from the second gives 12b equals 36, so b equals 3.
Substituting b equals 3 into a plus 6b equals 8 gives a plus 18 equals 8, so a equals negative 10.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise sequence questionsWhat the mark scheme rewarded
- Correctly extending the sequence's constant difference of 4b to write the 5th term as a plus 18b (1 mark)
- Setting up correct simultaneous equations from the two given term values (1 mark)
- The correct value of b, found first by eliminating a (1 mark)
- The correct value of a, found by substituting back (1 mark)
- Use the sequence's constant algebraic difference, 4b here, to write any further term without needing to know a or b individually first
- Eliminate one unknown by subtracting one equation from the other, since both equations share the same a term
Traps examiners saw
- Extending the sequence by the wrong number of steps, for example treating the given terms as 3 steps apart instead of the correct number
- Solving for only one of the two unknown constants and forgetting to substitute back to find the second
Full-mark self-check 0 of 3
1×asked“The next term of a sequence is made by adding the previous two terms. Which of these sequences follows this rule: negative 9, 2, negative 7, negative 5, negative 12; or 0, negative 3, negative 3, 0, negative 3?”
“The next term of a sequence is made by adding the previous two terms. Which of these sequences follows this rule: negative 9, 2, negative 7, negative 5, negative 12; or 0, negative 3, negative 3, 0, negative 3?”
What it’s really asking
Check each pair of consecutive terms against the stated addition rule to see which whole sequence is consistent.
The full worked answer — June 2019
Checking negative 9, 2, negative 7, negative 5, negative 12: negative 9 plus 2 equals negative 7, then 2 plus negative 7 equals negative 5, then negative 7 plus negative 5 equals negative 12. Every term matches the rule.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise sequence questionsWhat the mark scheme rewarded
- The single correct sequence identified, with every consecutive pair genuinely checked against the given rule (1 mark, point marked)
- Test the rule against every consecutive pair in a sequence, not just the first, since a sequence can start correctly and then break the rule later
- Work through the whole sequence systematically rather than guessing which option looks right
Traps examiners saw
- Checking only the first pair of terms and assuming the whole sequence follows the rule without testing every remaining pair
- Applying the rule in the wrong order, for example adding the two terms after the ones intended by the rule
Full-mark self-check 0 of 2
1×asked“The first three terms of a sequence are x, y, xy. The sequence is continued by multiplying the previous two terms. Circle the 5th term. The 8th term of the sequence is x to the power 8 times y to the power 13, and the value of this term is negative. What does this mean about the values of x and y?”
“The first three terms of a sequence are x, y, xy. The sequence is continued by multiplying the previous two terms. Circle the 5th term. The 8th term of the sequence is x to the power 8 times y to the power 13, and the value of this term is negative. What does this mean about the values of x and y?”
What it’s really asking
Multiply consecutive terms to build up the powers of x and y, then reason about which sign combinations make the final term negative.
The full worked answer — June 2022
Term 3 is x times y, term 4 is y times xy, which is x times y squared, and term 5 is xy times xy squared, which is x squared times y cubed.
Since x to the power 8 has an even power, it is positive or zero whatever the sign of x, so x could be positive or negative. Since y to the power 13 has an odd power, it keeps the same sign as y, so for the whole term to be negative, y must be negative.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise sequence questionsWhat the mark scheme rewarded
- Correctly applying the multiply-the-previous-two-terms rule to build up to the 5th term, x squared times y cubed (1 mark)
- Correct reasoning that an even power, such as x to the power 8, cannot make a term negative on its own, so x could be either sign (1 mark)
- Correct reasoning that an odd power, such as y to the power 13, keeps the sign of the base, so y specifically must be negative (1 mark)
- Apply the term-to-term rule carefully step by step, since each new term depends on both of the previous two terms, not just the one immediately before it
- Reason about even and odd powers separately: an even power is always positive or zero, while an odd power always keeps the same sign as its base
Traps examiners saw
- Concluding that x must be negative from the negative final term, without checking that an even power on x cannot be responsible for the negative sign
- Making an error building up the powers term by term, for example adding the powers instead of multiplying the actual terms together
Full-mark self-check 0 of 3
The method for every Q20 (Jun18) / Q12 (Jun19) / Q19 (Jun22) — same every sittingMark bands, steps, timing
What this question type rewards
The topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
- Applying the stated rule correctly to generate further terms
- Setting up and solving equations from two known terms when constants are unknown
- Reasoning about which values (positive, negative, either) are possible from the rule's structure
The steps
- Read the rule carefully: is it term-to-term (using previous terms) or position-to-term (using the term number)?
- Generate any further terms needed using the rule exactly as stated
- If constants are unknown, use two known terms to set up and solve simultaneous equations
- Check any reasoning about sign (positive or negative) against every case the rule allows
What is the common difference of the arithmetic sequence below? 4, 11, 18, 25, 32, ...
Sequence questions come up in three of the four sittings we have, often with an unusual rule rather than a standard nth term. Practise reading the rule carefully every time.
Practise sequence questions“Three of the four sittings we have full papers for use the completed square form directly, either to solve an equation, express an expression, or find a turning point.”
Every version relies on the same fact: the turning point of y equals (x plus a) squared plus b is at the point (negative a, b).
1×asked“Without expanding any brackets, show how to work out the exact solutions of 9 times (x plus 3) squared equals 4. Give the solutions.”
“Without expanding any brackets, show how to work out the exact solutions of 9 times (x plus 3) squared equals 4. Give the solutions.”
What it’s really asking
Divide to isolate the squared bracket, square root both sides, then solve for x without ever expanding the bracket.
The full worked answer — June 2018
Dividing both sides by 9 gives (x plus 3) squared equals four ninths, so x plus 3 equals plus or minus two thirds.
Subtracting 3 from both sides gives x equals negative 3 plus two thirds, or x equals negative 3 minus two thirds, which are negative seven thirds and negative eleven thirds.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise completing the square questionsWhat the mark scheme rewarded
- Correctly isolating the squared bracket by dividing by 9, reaching (x plus 3) squared equals four ninths (1 mark)
- Correctly square rooting both sides, keeping both the positive and negative root, x plus 3 equals plus or minus two thirds (1 mark)
- Both correct exact solutions, negative seven thirds and negative eleven thirds (1 mark)
- Isolate the squared bracket completely before square rooting, since square rooting too early leads to an incorrect expression
- Always keep both the positive and negative root when square rooting an equation, since the question asks for exact solutions, plural
Traps examiners saw
- Expanding the bracket first despite the question specifically saying not to, which loses the method mark for the intended approach
- Giving only one solution instead of both, forgetting the negative root entirely
Full-mark self-check 0 of 3
1×asked“Here are sketches of two graphs. The graph of y equals x squared minus 1 is translated 3 units to the left to give graph A. The equation of graph A can be written in the form y equals x squared plus bx plus c. Work out the values of b and c. The graph of y equals x squared minus 1 is reflected in the x-axis to give graph B. Work out the equation of graph B.”
“Here are sketches of two graphs. The graph of y equals x squared minus 1 is translated 3 units to the left to give graph A. The equation of graph A can be written in the form y equals x squared plus bx plus c. Work out the values of b and c. The graph of y equals x squared minus 1 is reflected in the x-axis to give graph B. Work out the equation of graph B.”
What it’s really asking
Replace x with (x plus 3) for the translation, and negate the whole expression for the reflection.
What the sources actually showed — June 2018
A sketch showing the original curve y equals x squared minus 1 and graph A, its image after a translation of 3 units to the left.
The full worked answer — June 2018
Translating 3 units left replaces x with (x plus 3), giving graph A as y equals (x plus 3) squared minus 1, which expands to x squared plus 6x plus 9 minus 1, or x squared plus 6x plus 8.
Reflecting y equals x squared minus 1 in the x-axis negates the whole expression, giving graph B as y equals 1 minus x squared.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise completing the square questionsWhat the mark scheme rewarded
- Correctly substituting (x plus 3) for x for the translation, and expanding fully to reach x squared plus 6x plus 8, giving b equals 6 and c equals 8 (1 mark each for b and c, so 2 marks)
- Correctly negating the whole expression for the reflection, reaching y equals 1 minus x squared (1 mark)
- A translation of 3 units to the left replaces every x in the original equation with (x plus 3), never (x minus 3)
- A reflection in the x-axis negates the whole expression, including any constant term, not just the x squared part
Traps examiners saw
- Substituting (x minus 3) instead of (x plus 3) for a translation to the LEFT, which is a very common direction mix-up
- Negating only the x squared term for the reflection and forgetting to also negate the constant, minus 1
Full-mark self-check 0 of 3
1×asked“The turning point of the graph y equals (x plus a) squared plus b has x-coordinate negative 2. The point (3, 1) is another point on the graph. Work out the y-coordinate of the turning point.”
“The turning point of the graph y equals (x plus a) squared plus b has x-coordinate negative 2. The point (3, 1) is another point on the graph. Work out the y-coordinate of the turning point.”
What it’s really asking
Find a from the turning point's x-coordinate, then substitute the known point to find b.
The full worked answer — June 2019
The turning point of y equals (x plus a) squared plus b is at x equals negative a, so negative a equals negative 2, giving a equals 2.
Substituting the point (3, 1) and a equals 2 into the equation gives 1 equals (3 plus 2) squared plus b, which is 1 equals 25 plus b, so b equals negative 24.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise completing the square questionsWhat the mark scheme rewarded
- Correctly reading the turning point's x-coordinate as negative a, giving a equals 2 (1 mark)
- Correctly substituting the known point and the value of a into the full equation (1 mark)
- The correct value of b, negative 24, which is also the y-coordinate of the turning point (1 mark)
- Learn as a fact that the turning point of y equals (x plus a) squared plus b is always at (negative a, b), rather than re-deriving it each time
- Substitute a known point on the graph, not the turning point itself, to find the remaining unknown constant, b
Traps examiners saw
- Reading the turning point's x-coordinate directly as a instead of negative a, giving the wrong sign for a
- Substituting the coordinates the wrong way round, using 3 for y and 1 for x instead of the reverse
Full-mark self-check 0 of 3
1×asked“Express x squared minus 6x minus 15 in the form (x minus a) squared minus b where a and b are integers.”
“Express x squared minus 6x minus 15 in the form (x minus a) squared minus b where a and b are integers.”
What it’s really asking
Halve the coefficient of x to find a, then adjust the constant to find b.
The full worked answer — June 2023
Halving the coefficient of x, negative 6, gives negative 3, so the expression starts as (x minus 3) squared, which expands to x squared minus 6x plus 9.
Since the original expression has minus 15 but (x minus 3) squared already contributes plus 9, subtracting 9 and a further 15 gives (x minus 3) squared minus 24, so a equals 3 and b equals 24.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise completing the square questionsWhat the mark scheme rewarded
- Correctly halving the coefficient of x, negative 6, to reach (x minus 3) squared as the starting point (1 mark, condoning a equals 3 if the correct constant follows)
- The correct final expression, (x minus 3) squared minus 24, with a equals 3 and b equals 24 (1 further mark)
- Halve the coefficient of x to find the value inside the bracket, then square that value to find what must be subtracted to correct the constant
- Check the sign of a matches the form given in the question, here (x minus a) squared, so a is positive 3, not negative 3
Traps examiners saw
- Forgetting to subtract the extra 9 introduced by squaring (x minus 3), leaving the wrong constant term
- Giving a with the wrong sign for the form asked, writing (x plus 3) squared instead of (x minus 3) squared
Full-mark self-check 0 of 3
The method for every Q27b, Q29 (Jun18) / Q26 (Jun19) / Q21 (Jun23) — same every sittingMark bands, steps, timing
What this question type rewards
The topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
- Halving the coefficient of x to find the value inside the bracket
- Adjusting the constant term correctly after completing the square
- Reading the turning point directly from the completed square form
The steps
- Halve the coefficient of x to find the value that goes inside the bracket
- Square that value and subtract it to correct the constant term
- For a turning point, read the coordinates directly as (negative a, b) from y equals (x plus a) squared plus b
- For an equation already in completed square form, isolate the bracket and square root both sides
To write x² + 10x + 3 in the form (x + p)² + q, what is the value of p?
Completing the square and quadratic turning points come up in three of the four sittings we have. Practise reading the turning point straight from the completed square form.
Practise completing the square questions“Two of the four sittings we have full papers for combine a given angle ratio with a standard angle fact, rather than testing the angle fact on its own.”
Both questions use the ratio to write every angle as a multiple of one unknown, then use the angle sum to solve for it.
1×asked“Here is a quadrilateral. Angle a equals 90 degrees, and a to b is 5 to 3. x to y is 1 to 3. Show that b equals x.”
“Here is a quadrilateral. Angle a equals 90 degrees, and a to b is 5 to 3. x to y is 1 to 3. Show that b equals x.”
What it’s really asking
Find b from the first ratio, find x from the second ratio and the quadrilateral's angle sum, then compare the two values.
What the sources actually showed — June 2019
A quadrilateral with one angle labelled a, a second labelled b, and the remaining two angles labelled x and y, drawn not to scale.
The full worked answer — June 2019
Since a to b is 5 to 3 and a equals 90 degrees, one part of the ratio is 90 divided by 5, which is 18 degrees, so b equals 3 times 18, which is 54 degrees.
The four angles of the quadrilateral sum to 360 degrees, so x plus y equals 360 minus 90 minus 54, which is 216 degrees. Since x to y is 1 to 3, one part is 216 divided by 4, which is 54 degrees, so x equals 54 degrees.
Since b equals 54 degrees and x equals 54 degrees, b equals x, as required.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise angle fact questionsWhat the mark scheme rewarded
- Using the given ratio and the known value of a to find b, 54 degrees (1 mark)
- Using the quadrilateral's angle sum of 360 degrees together with the second ratio to find x, also 54 degrees (1 mark)
- Explicitly stating that b equals x, completing the show that proof (1 mark)
- Find the value of one part of each ratio first, using whichever angle is already known, before tackling the rest of the shape
- Know the angle sum of a quadrilateral, 360 degrees, cold, since it is needed here without any diagram measurements given
Traps examiners saw
- Using the wrong angle sum, such as 180 degrees, for a four-sided shape instead of the correct 360 degrees
- Stopping once b and x are both found without explicitly stating the comparison the question specifically asks to show
Full-mark self-check 0 of 3
1×asked“ABC, BD and BE are straight lines. Angle EBD equals 5 times angle ABE. Angle DBC equals 3 times angle ABE. Work out the size of angle EBD.”
“ABC, BD and BE are straight lines. Angle EBD equals 5 times angle ABE. Angle DBC equals 3 times angle ABE. Work out the size of angle EBD.”
What it’s really asking
Write every angle as a multiple of angle ABE, then use the straight line's 180 degree sum to solve for it.
What the sources actually showed — June 2023
A straight line ABC with a point B on it, from which two further straight lines BD and BE are drawn, splitting the angle on the straight line at B into three parts: angle ABE, angle EBD, and angle DBC.
The full worked answer — June 2023
Letting angle ABE equal x, angle EBD equals 5x and angle DBC equals 3x. Since ABC is a straight line, x plus 5x plus 3x equals 180 degrees, giving 9x equals 180.
Solving gives x equals 20 degrees, so angle EBD equals 5 times 20, which is 100 degrees.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise angle fact questionsWhat the mark scheme rewarded
- Writing every angle as a multiple of angle ABE, using the given ratios 5x and 3x (1 mark)
- Forming a correct equation using the straight line angle sum of 180 degrees, x plus 5x plus 3x equals 180 (1 mark)
- The correct final answer, angle EBD equals 100 degrees, found by substituting back (1 mark)
- Write every angle in the diagram as a multiple of the same single unknown before forming any equation
- Know the angle sum on a straight line, 180 degrees, cold, since the whole equation depends on it
Traps examiners saw
- Using the wrong angle sum, such as 360 degrees, for angles that actually lie on a single straight line
- Solving correctly for the unknown x but giving x itself as the final answer instead of substituting back to find angle EBD specifically
Full-mark self-check 0 of 3
The method for every Q14 (Jun19) / Q4 (Jun23) — same every sittingMark bands, steps, timing
What this question type rewards
The topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
- Writing every angle in the ratio as a multiple of one unknown letter
- Using the correct angle fact (angles on a straight line sum to 180, angles in a quadrilateral sum to 360) to form an equation
- Solving the equation and substituting back to find the specific angle asked for
The steps
- Let one part of the ratio equal an unknown letter, and write every other angle as a multiple of it
- Use the relevant angle fact to write an equation involving all the angles
- Solve the equation for the unknown letter
- Substitute back to find the specific angle the question asks for
An angle measures 135°. What type of angle is this?
Angle ratio problems come up in two of the four sittings we have, always combining a ratio with a standard angle fact. Practise writing every angle in terms of one unknown first.
Practise angle fact questions“All four sittings we have full papers for include at least one area or surface area question built from circles or 3D shapes, and June 2023 tests it twice.”
Every version keeps pi symbolic throughout the working, since a decimal approximation for pi is never needed or wanted on a non-calculator paper.
1×asked“Here is a tunnel for a toy train. AD is a semicircular arc of radius 10cm. BC is a semicircular arc of radius 7cm. The length of the tunnel is 30cm. Work out the total area of all six faces of the tunnel. Give your answer in terms of pi.”
“Here is a tunnel for a toy train. AD is a semicircular arc of radius 10cm. BC is a semicircular arc of radius 7cm. The length of the tunnel is 30cm. Work out the total area of all six faces of the tunnel. Give your answer in terms of pi.”
What it’s really asking
Identify all six faces (two semicircular annuli, two curved surfaces, two flat rectangular strips) and add their areas.
What the sources actually showed — June 2018
A 3D tunnel shape whose cross-section is the region between two semicircular arcs, an outer arc of radius 10cm and an inner arc of radius 7cm, extruded to a length of 30cm, so that the front and back faces are semicircular rings and the shape has an outer and an inner curved surface plus two flat rectangular strips where the arcs meet the ground.
The full worked answer — June 2018
Each of the two end faces is a semicircular ring, with area (1 half) times (pi times 10 squared minus pi times 7 squared), which is (1 half) times 51 pi, or 25.5 pi. The two end faces together give 51 pi.
The outer curved surface is (1 half) times 2 times pi times 10 times 30, which is 300 pi, and the inner curved surface is (1 half) times 2 times pi times 7 times 30, which is 210 pi.
The two flat rectangular strips where the arcs meet the ground each measure (10 minus 7) by 30, giving 2 times 3 times 30, which is 180. Adding every face together gives 51 pi plus 300 pi plus 210 pi plus 180, which is 561 pi plus 180.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise area and surface area questionsWhat the mark scheme rewarded
- Correctly finding the area of the two end faces, the difference of two semicircles, 51 pi (1 mark)
- Correctly finding the outer curved surface, 300 pi (1 mark)
- Correctly finding the inner curved surface, 210 pi (1 mark)
- Correctly finding the two flat rectangular strips, 180, using no pi since they are not circular (1 mark)
- The correct final total, 561 pi plus 180, combining every one of the six faces (1 mark)
- Break a composite 3D shape into every distinct face before finding any area, since forgetting a single face is the most common way to lose marks here
- Keep the circular faces (with pi) and the flat rectangular faces (without pi) clearly separate in the final total, since only some faces contain pi
Traps examiners saw
- Forgetting one of the six faces entirely, most often one of the two flat rectangular strips where the arcs meet the ground
- Using a decimal approximation for pi partway through, which is never needed and risks losing accuracy on this non-calculator question
Full-mark self-check 0 of 3
1×asked“A shaded semicircle is inside a circle. The radius of the circle is 10cm. The diameter of the semicircle is 8cm. How many times bigger is the unshaded area than the shaded area?”
“A shaded semicircle is inside a circle. The radius of the circle is 10cm. The diameter of the semicircle is 8cm. How many times bigger is the unshaded area than the shaded area?”
What it’s really asking
Find the full circle's area and the semicircle's area, subtract to find the unshaded area, then divide.
What the sources actually showed — June 2019
A circle of radius 10cm with a shaded semicircle of diameter 8cm drawn inside it, touching the edge of the larger circle, with the remaining region left unshaded.
The full worked answer — June 2019
The full circle has area pi times 10 squared, which is 100 pi. The shaded semicircle has radius 4cm, so its area is (1 half) times pi times 4 squared, which is 8 pi.
The unshaded area is 100 pi minus 8 pi, which is 92 pi, so the unshaded area is 92 pi divided by 8 pi, which is 11.5 times bigger than the shaded area.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise area and surface area questionsWhat the mark scheme rewarded
- Correctly finding the area of the full circle, 100 pi (1 mark)
- Correctly finding the area of the shaded semicircle, using its radius of 4cm, 8 pi (1 mark)
- Correctly subtracting to find the unshaded area, 92 pi (1 mark)
- The correct final ratio, 11.5, from dividing the unshaded area by the shaded area (1 mark)
- Read the semicircle's diameter carefully and halve it to find its radius, since the area formula needs the radius, not the diameter
- Cancel pi completely when dividing two areas that both contain it, leaving a pure number ratio, exactly what the question asks for
Traps examiners saw
- Using the given diameter, 8cm, directly as the radius in the area formula instead of halving it first
- Leaving pi in the final answer instead of cancelling it, since the question asks how many times bigger, a pure number
Full-mark self-check 0 of 3
1×asked“Here are two shapes, P and Q. How many times bigger is the area of P than the area of Q? You must show your working.”
“Here are two shapes, P and Q. How many times bigger is the area of P than the area of Q? You must show your working.”
What it’s really asking
Find each sector's area as a fraction of its full circle, then divide the two sector areas.
What the sources actually showed — June 2022
Shape P is a sector covering three quarters of a full circle of radius 20, and shape Q is a sector covering one third of a full circle of radius 15.
The full worked answer — June 2022
Shape P's area is three quarters of pi times 20 squared, which is three quarters of 400 pi, giving 300 pi. Shape Q's area is one third of pi times 15 squared, which is one third of 225 pi, giving 75 pi.
P is 300 pi divided by 75 pi, which equals 4 times bigger than Q.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise area and surface area questionsWhat the mark scheme rewarded
- Correctly finding shape P's area as three quarters of a full circle of radius 20, 300 pi (1 mark)
- Correctly finding shape Q's area as one third of a full circle of radius 15, 75 pi (1 mark)
- The correct final ratio, 4, from dividing the two sector areas (1 mark)
- Find each sector's full circle area first, then multiply by the fraction of the circle the sector actually covers
- Cancel pi completely when dividing the two sector areas, since the question asks how many times bigger, a pure number
Traps examiners saw
- Using the wrong radius for each sector, mixing up shape P's radius of 20 with shape Q's radius of 15
- Forgetting to apply the fraction of the circle each sector covers, and using the full circle's area instead of the sector's area
Full-mark self-check 0 of 3
1×asked“A solid hemisphere has radius x. A solid cylinder has radius 3x and height x. Surface area of a sphere equals 4 pi r squared, where r is the radius. Work out the ratio total surface area of the hemisphere to total surface area of the cylinder. Give your answer in its simplest form. You must show your working.”
“A solid hemisphere has radius x. A solid cylinder has radius 3x and height x. Surface area of a sphere equals 4 pi r squared, where r is the radius. Work out the ratio total surface area of the hemisphere to total surface area of the cylinder. Give your answer in its simplest form. You must show your working.”
What it’s really asking
Write the total surface area of each solid in terms of x, then simplify the ratio between them.
What the sources actually showed — June 2023
A solid hemisphere of radius x is shown alongside a solid cylinder of radius 3x and height x, with the given formula for the surface area of a full sphere provided.
The full worked answer — June 2023
The hemisphere's curved surface is half of 4 pi x squared, which is 2 pi x squared, plus its flat circular face, pi x squared, giving a total of 3 pi x squared.
The cylinder's two flat circular ends each have area pi times (3x) squared, which is 9 pi x squared, giving 18 pi x squared for both, and its curved surface is 2 pi times 3x times x, which is 6 pi x squared, giving a total of 24 pi x squared.
The ratio is 3 pi x squared to 24 pi x squared, which cancels pi x squared completely to give 3 to 24, or 1 to 8 in its simplest form.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise area and surface area questionsWhat the mark scheme rewarded
- Correctly finding the hemisphere's total surface area in terms of x, 3 pi x squared, combining its curved and flat parts (1 mark)
- Correctly finding the cylinder's total surface area in terms of x, 24 pi x squared, combining its two flat ends and curved side (1 mark)
- The correct simplified final ratio, 1 to 8, from cancelling pi x squared from both sides (1 mark)
- Use the given formula for a full sphere's surface area, then halve it for the curved part of a hemisphere, remembering to also add its flat circular base
- Cancel the common factor of pi x squared from both sides of the ratio before simplifying to whole numbers
Traps examiners saw
- Forgetting the hemisphere's flat circular base, giving only its curved surface as the total, which is one whole term short
- Using the wrong radius for the cylinder's flat ends, forgetting they use 3x, not x, since the cylinder's radius is three times the hemisphere's
Full-mark self-check 0 of 3
1×asked“A circle, centre O, has circumference 20 pi cm. Q is a point on the circle. OPQR is a square. Perimeter of the square to circumference of the circle equals the square root of a, to pi, where a is an integer. Work out the value of a. You must show your working.”
“A circle, centre O, has circumference 20 pi cm. Q is a point on the circle. OPQR is a square. Perimeter of the square to circumference of the circle equals the square root of a, to pi, where a is an integer. Work out the value of a. You must show your working.”
What it’s really asking
Find the circle's radius from its circumference, use it as the square's diagonal via Pythagoras, then compare perimeters.
What the sources actually showed — June 2023
A circle centre O with circumference 20 pi cm. OPQR is a square with O at the centre and Q on the circle, so that OQ, the diagonal from O to Q, equals the circle's radius.
The full worked answer — June 2023
Since the circumference is 20 pi, the radius is 10cm, and OQ, the diagonal of the square, equals this radius, 10cm.
By Pythagoras on the square's diagonal, side squared plus side squared equals 10 squared, so 2 times side squared equals 100, giving side squared equals 50, so the side length is the square root of 50, which is 5 times the square root of 2.
The perimeter of the square is 4 times 5 times the square root of 2, which is 20 times the square root of 2, so the ratio to the circumference, 20 pi, is the square root of 2 to pi, giving a equals 2.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise area and surface area questionsWhat the mark scheme rewarded
- Correctly finding the radius from the circumference, 10cm, and identifying OQ as the square's diagonal (1 mark)
- Correctly applying Pythagoras to the square's diagonal to find the side length, 5 times the square root of 2 (1 mark)
- Correctly finding the perimeter and forming the ratio with the circumference (1 mark)
- The correct final value, a equals 2 (1 mark)
- Use circumference equals 2 pi r to find the radius before anything else, since every later step depends on it
- Recognise that O and Q, being opposite corners of square OPQR, means OQ is the square's diagonal, not one of its sides
Traps examiners saw
- Treating OQ as a side of the square instead of correctly identifying it as the diagonal, which changes every later step
- Leaving the side length as the square root of 50 instead of simplifying the surd fully to 5 times the square root of 2
Full-mark self-check 0 of 3
The method for every Q18 (Jun18) / Q9 (Jun19) / Q11 (Jun22) / Q17, Q26 (Jun23) — same every sittingMark bands, steps, timing
What this question type rewards
The topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
- Keeping pi symbolic throughout rather than substituting a decimal approximation
- Breaking a composite shape into simple areas (full circles, semicircles, sectors, rectangles) before combining them
- Simplifying a ratio of two areas that both contain pi by cancelling it
The steps
- Identify every simple shape (circle, semicircle, sector, rectangle) that makes up the composite shape
- Write the area or surface area of each simple shape, keeping pi as a symbol
- Add or subtract the areas as the question requires
- For a ratio, cancel any common factor of pi and simplify to whole numbers
A circle has radius r. Which pair of formulas gives the correct circumference C and area A?
Area and surface area questions involving circles and 3D shapes come up in every sitting we have. Practise keeping pi symbolic and breaking composite shapes into simple parts.
Practise area and surface area questions“Two of the four sittings we have full papers for use the perpendicular gradient relationship to solve a coordinate geometry problem, one involving a circle's tangent and one involving a right-angled triangle.”
Both questions rely on the same fact: perpendicular lines have gradients that multiply to give negative 1.
1×asked“P (4, 8) is a point on a circle, centre O. The tangent at P intersects the axes at points A and B. Show that the gradient of the tangent is negative one half.”
“P (4, 8) is a point on a circle, centre O. The tangent at P intersects the axes at points A and B. Show that the gradient of the tangent is negative one half.”
What it’s really asking
Find the gradient of the radius OP, then use the fact that a tangent is perpendicular to the radius at that point.
The full worked answer — June 2019
The gradient of the radius OP, from (0,0) to (4,8), is 8 over 4, which equals 2.
Since the tangent is perpendicular to the radius, its gradient is the negative reciprocal of 2, which is negative one half, as required.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise coordinate geometry questionsWhat the mark scheme rewarded
- Correctly finding the gradient of the radius OP, 2 (1 mark)
- Correctly applying the negative reciprocal relationship to reach the required gradient, negative one half (1 mark)
- Know that a tangent to a circle is always perpendicular to the radius at the exact point where it touches the circle
- Perpendicular gradients always multiply to give negative 1, so the negative reciprocal of 2 is negative one half, not one half
Traps examiners saw
- Finding the gradient of the tangent directly rather than using the radius, which is far harder without more information
- Taking the reciprocal of the radius's gradient without also changing its sign, giving one half instead of negative one half
Full-mark self-check 0 of 3
1×asked“Points P, Q and R (8, 22) form a triangle. PQ is a horizontal line, with P on the y-axis. Angle PRQ is a right angle. The gradient of PR is 2. Work out the coordinates of Q.”
“Points P, Q and R (8, 22) form a triangle. PQ is a horizontal line, with P on the y-axis. Angle PRQ is a right angle. The gradient of PR is 2. Work out the coordinates of Q.”
What it’s really asking
Find P from the gradient of PR, then use the perpendicular gradient of RQ and the horizontal condition to find Q.
The full worked answer — June 2023
The line PR has gradient 2 and passes through R (8, 22), so its equation is y equals 2x plus 6. Since P is on the y-axis, its x-coordinate is 0, giving P equals (0, 6).
Since angle PRQ is a right angle, RQ is perpendicular to PR, so its gradient is the negative reciprocal of 2, which is negative one half.
Since PQ is horizontal, Q has the same y-coordinate as P, which is 6. Using RQ's equation, y minus 22 equals negative one half times (x minus 8), and substituting y equals 6 gives negative 16 equals negative one half times (x minus 8), so x equals 40, giving Q equals (40, 6).
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise coordinate geometry questionsWhat the mark scheme rewarded
- Forming the equation of PR and correctly finding P, using that P lies on the y-axis (1 mark)
- Correctly applying the negative reciprocal to find RQ's gradient, negative one half (1 mark)
- Correctly using the horizontal condition to fix Q's y-coordinate as matching P's (1 mark)
- Correctly forming and solving the equation of RQ for Q's x-coordinate (1 mark)
- The correct final coordinates, Q equals (40, 6) (1 mark)
- A point on an axis has one coordinate equal to zero, use this to find P's coordinates quickly from the equation of PR
- A horizontal line means two points share the same y-coordinate, not the same x-coordinate, a common source of confusion
Traps examiners saw
- Applying the perpendicular gradient relationship at the wrong point, forgetting the right angle is specifically at R, between RP and RQ
- Confusing which coordinate stays fixed for the horizontal line PQ, using matching x-coordinates instead of matching y-coordinates
Full-mark self-check 0 of 4
The method for every Q25 (Jun19) / Q24 (Jun23) — same every sittingMark bands, steps, timing
What this question type rewards
The topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
- Finding the gradient of a known line before working out the gradient of a line perpendicular to it
- Using the negative reciprocal relationship between perpendicular gradients
- Forming and solving the equation of a line to find where it crosses an axis or meets another line
The steps
- Find the gradient of any line whose coordinates are known
- For a perpendicular line, take the negative reciprocal of that gradient
- Use the point and the new gradient to form the equation of the perpendicular line
- Solve for any coordinates the question asks for, using the equation formed
Which of the following is true about two parallel straight lines?
Perpendicular gradient problems come up in two of the four sittings we have. Practise the negative reciprocal relationship until it is instant.
Practise coordinate geometry questions“Every one of the four sittings we have full papers for ends its trigonometry content with an exact-value question, always the final or near-final question on the paper.”
Every version requires the exact surd values of sine, cosine and tangent at 30, 45 and 60 degrees rather than a calculator or decimal approximation.
1×asked“Show that the value of cos 30 degrees times tan 60 degrees plus sin 30 degrees is an integer.”
“Show that the value of cos 30 degrees times tan 60 degrees plus sin 30 degrees is an integer.”
What it’s really asking
Substitute the exact surd values for cos 30, tan 60 and sin 30, then simplify to show the surds cancel out completely.
The full worked answer — June 2018
Cos 30 degrees equals the square root of 3 over 2, tan 60 degrees equals the square root of 3, and sin 30 degrees equals one half.
Substituting gives (the square root of 3 over 2) times the square root of 3, plus one half, which equals three halves plus one half, since the square root of 3 times the square root of 3 is 3.
Three halves plus one half equals 2, which is an integer, as required.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise exact trig value questionsWhat the mark scheme rewarded
- All three exact values stated correctly, since this question requires every step to be shown for full marks (1 mark)
- Correctly multiplying the surds together, using that the square root of 3 times itself equals 3 (1 mark)
- The correct final integer, 2, reached with every step shown (1 mark)
- This particular question demands every step is shown, since the mark scheme explicitly requires it for the full 3 marks
- A surd multiplied by itself always cancels to the whole number underneath the root, here the square root of 3 times the square root of 3 equals exactly 3
Traps examiners saw
- Giving only the final integer, 2, with no working shown, which does not score full marks even though the answer is correct
- Mixing up cos 30 and sin 30, both of which appear in this expression alongside tan 60, since one is a half and the other involves a surd
Full-mark self-check 0 of 3
1×asked“Angle x is acute. Cos x equals sin 60 degrees times tan 30 degrees. Work out the size of angle x. You must show your working.”
“Angle x is acute. Cos x equals sin 60 degrees times tan 30 degrees. Work out the size of angle x. You must show your working.”
What it’s really asking
Substitute the exact values, simplify the product, then recognise the result as another exact cosine value.
The full worked answer — June 2019
Sin 60 degrees equals the square root of 3 over 2, and tan 30 degrees equals one over the square root of 3.
Multiplying gives (the square root of 3 over 2) times (one over the square root of 3), which equals one half, since the square roots of 3 cancel completely.
Since cos x equals one half, and cos 60 degrees equals one half, x equals 60 degrees.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise exact trig value questionsWhat the mark scheme rewarded
- Both exact values needed stated correctly, sin 60 degrees and tan 30 degrees (1 mark)
- Correctly multiplying and simplifying the surds to reach one half, cancelling the matching roots of 3 (1 mark)
- The correct final angle, x equals 60 degrees, from recognising the simplified value as an exact cosine (1 mark)
- Write tan 30 degrees as one over the square root of 3, since it is often more useful in this form for cancelling than as a decimal
- After simplifying to a value like one half, check the exact values table for which angle and which ratio gives exactly that result
Traps examiners saw
- Leaving a surd in the denominator uncancelled, missing that the two square roots of 3 in the product cancel completely
- Confusing cos 60 degrees with sin 60 degrees, both of which are common exact values but equal different things, one half and the square root of 3 over 2
Full-mark self-check 0 of 3
1×asked“Work out the value of (cos 30 degrees times sin 45 degrees times tan 60 degrees) squared”
“Work out the value of (cos 30 degrees times sin 45 degrees times tan 60 degrees) squared”
What it’s really asking
Substitute all three exact values, multiply, then square the result and simplify.
The full worked answer — June 2022
Cos 30 degrees equals the square root of 3 over 2, sin 45 degrees equals the square root of 2 over 2, and tan 60 degrees equals the square root of 3.
Multiplying the three values gives (the square root of 3 over 2) times (the square root of 2 over 2) times the square root of 3, which equals 3 times the square root of 2, over 4.
Squaring gives (3 times the square root of 2, over 4) squared, which equals 9 times 2 over 16, which equals 18 over 16, or 9 over 8, which is 1 and one eighth.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise exact trig value questionsWhat the mark scheme rewarded
- All three exact values needed stated correctly, cos 30 degrees, sin 45 degrees and tan 60 degrees (1 mark)
- Correctly multiplying the three values together, simplifying the product of surds (1 mark)
- Correctly squaring the simplified expression (1 mark)
- The correct final value, 9 over 8, or the equivalent mixed number 1 and one eighth (1 mark)
- Simplify the product of three surds first before squaring, since squaring the unsimplified expression is far more error-prone
- Remember that squaring the square root of 2 gives exactly 2, and squaring the square root of 3 gives exactly 3, cancelling both roots
Traps examiners saw
- Squaring each of the three exact values individually instead of squaring the whole simplified product at the end
- Losing a factor when squaring a fraction, forgetting to square both the numerator and the denominator
Full-mark self-check 0 of 4
1×asked“Show that (4 sin 30 degrees minus tan 45 degrees) over (2 cos 30 degrees) can be written as tan x, where x is an acute angle.”
“Show that (4 sin 30 degrees minus tan 45 degrees) over (2 cos 30 degrees) can be written as tan x, where x is an acute angle.”
What it’s really asking
Substitute the exact values into the numerator and denominator separately, then simplify the resulting fraction to a known tangent value.
The full worked answer — June 2023
Sin 30 degrees equals one half, tan 45 degrees equals 1, and cos 30 degrees equals the square root of 3 over 2, so the numerator is 4 times one half minus 1, which equals 1, and the denominator is 2 times the square root of 3 over 2, which equals the square root of 3.
This gives the fraction one over the square root of 3, which is the exact value of tan 30 degrees.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise exact trig value questionsWhat the mark scheme rewarded
- All three exact values needed stated correctly, sin 30 degrees, tan 45 degrees and cos 30 degrees (1 mark)
- Correctly substituting into the numerator and simplifying it, reaching 1 (1 mark)
- Correctly substituting into the denominator and simplifying it, reaching the square root of 3 (1 mark)
- Recognising the final fraction as the exact value of tan 30 degrees, completing the show that (1 mark)
- Simplify the numerator and denominator completely separately before combining them into a single fraction
- After reaching a simplified fraction, check the exact values table to identify which tangent value it matches
Traps examiners saw
- Combining the numerator and denominator too early, before either has been fully simplified, which makes errors harder to spot
- Confusing tan 30 degrees with tan 60 degrees, which are reciprocals of each other, one over the square root of 3 and the square root of 3
Full-mark self-check 0 of 3
The method for every Q30 (Jun18) / Q27 (Jun19) / Q28 (Jun22) / Q25 (Jun23) — same every sittingMark bands, steps, timing
What this question type rewards
The topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
- Recalling the exact values of sine, cosine and tangent at 0, 30, 45, 60 and 90 degrees
- Substituting these exact surd values and simplifying algebraically
- Recognising the simplified result as another exact trigonometric value where the question asks for an angle
The steps
- Recall or derive the exact value of every trig ratio in the expression
- Substitute the exact values in place of the trig ratios
- Simplify the resulting surd expression fully
- If an angle is asked for, recognise the simplified value as a known exact trig ratio
What is the exact value of sin 30°?
An exact trig values question closes the trigonometry content in every single sitting we have. Practise the exact values table until it is instant recall.
Practise exact trig value questions“Two of the four sittings we have full papers for test a reciprocal-shaped curve, one by asking the student to draw and read it, and one by asking the student to criticise an incorrect sketch.”
Both questions depend on the same underlying shape: a reciprocal curve never touches either axis and has two separate branches.
1×asked“The number of items, n, made in 1 hour by a machine is given by n equals 60 over t, where t is the time in minutes the machine takes to make one item. On the grid, draw the graph of n equals 60 over t for values of t from 1 to 4. The machine takes 3 minutes 30 seconds to make one item. Use your graph to estimate the value of n.”
“The number of items, n, made in 1 hour by a machine is given by n equals 60 over t, where t is the time in minutes the machine takes to make one item. On the grid, draw the graph of n equals 60 over t for values of t from 1 to 4. The machine takes 3 minutes 30 seconds to make one item. Use your graph to estimate the value of n.”
What it’s really asking
Plot accurate points for the reciprocal relationship, join them smoothly, then read the value at t equals 3.5.
What the sources actually showed — June 2019
A blank grid with axes for t from 1 to 4 and n on the vertical axis, for the student to plot and draw the curve n equals 60 over t.
The full worked answer — June 2019
Calculating n at t equals 1, 2, 3 and 4 gives n equals 60, 30, 20 and 15, so the four points (1, 60), (2, 30), (3, 20) and (4, 15) are plotted and joined with a smooth curve.
Reading from the curve at t equals 3.5, n is a little more than 15 and a little less than 20, close to the exact value of 60 divided by 3.5, which is approximately 17.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise reciprocal graph questionsWhat the mark scheme rewarded
- All four points correctly calculated and plotted from the given formula (1 mark)
- A correct smooth curve drawn through the four points (1 mark)
- A correct method to read the graph at t equals 3.5 (1 mark)
- The correct estimate for n, read consistently from the drawn curve (1 mark)
- Calculate every point accurately from the given formula before plotting, since incorrect points cannot be joined into a correct curve
- Read an estimate from the actual drawn curve rather than recalculating the exact value, since the question specifically asks for an estimate using the graph
Traps examiners saw
- Joining the plotted points with straight line segments instead of a single smooth curve
- Reading the estimate at the wrong value on the horizontal axis, for example at t equals 3 instead of the required 3.5
Full-mark self-check 0 of 3
1×asked“Erika tries to sketch the graph y equals 1 over x with x not equal to 0. Make two different criticisms of her sketch.”
“Erika tries to sketch the graph y equals 1 over x with x not equal to 0. Make two different criticisms of her sketch.”
What it’s really asking
Compare the incorrect sketch against the two key properties of a genuine reciprocal graph: it never touches either axis, and its branches sit in opposite quadrants.
What the sources actually showed — June 2023
A sketch showing a curve in the top right region that incorrectly touches the y-axis, with a matching curve appearing in the top left region instead of the bottom left, so both branches sit above the x-axis.
The full worked answer — June 2023
The first criticism is that the sketch shows the curve touching the y-axis, but a genuine reciprocal graph gets closer and closer to the y-axis without ever reaching it, since x cannot equal 0.
The second criticism is that the branch for negative x values is drawn in the wrong place, above the x-axis, when it should be below the x-axis, since 1 divided by a negative number is always negative.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise reciprocal graph questionsWhat the mark scheme rewarded
- A valid, specific criticism referencing the graph incorrectly touching the y-axis (1 mark)
- A valid, specific criticism referencing the branch for negative x values being drawn in the wrong quadrant (1 mark)
- A genuine reciprocal graph gets closer and closer to both axes without ever touching them, since x cannot equal 0 and y can never equal 0 either
- Dividing 1 by a negative number always gives a negative result, which fixes exactly which quadrant the negative branch must sit in
Traps examiners saw
- Giving a vague criticism such as 'it's wrong' instead of a specific, correct reference to the axis or the quadrant that is actually incorrect
- Criticising a feature that is not actually wrong in the sketch, which does not score credit even if another part of the answer is correct
Full-mark self-check 0 of 3
The method for every Q10 (Jun19) / Q9 (Jun23) — same every sittingMark bands, steps, timing
What this question type rewards
The topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
- Plotting or checking points that genuinely satisfy the reciprocal relationship
- Recognising that a reciprocal curve never crosses or touches either axis
- Recognising that the two branches sit in opposite quadrants when the constant is positive
The steps
- Calculate a handful of accurate coordinate pairs from the given reciprocal relationship
- Plot these points and join them with a smooth curve, never a straight line
- Check the curve gets closer to both axes without ever touching them
- For negative input values, check the curve appears in the correct opposite quadrant
The graph of y = 1/x has an asymptote along the x-axis. What does this mean?
Reciprocal graphs come up in two of the four sittings we have, either to draw or to critique. Practise the two properties every genuine reciprocal graph must have.
Practise reciprocal graph questions“Two of the four sittings we have full papers for build a tree diagram problem with two stages, both requiring the branches to be completed before any calculation.”
Both questions test the same two-stage combined probability skill: multiply along a branch, and add across separate winning routes.
1×asked“Anna plays a game with an ordinary, fair dice. If she rolls 1 she wins. If she rolls 2 or 3 she loses. If she rolls 4, 5 or 6 she rolls again. When she has to roll again, if she rolls an odd number she wins, if she rolls an even number she loses. Complete the tree diagram with the four missing probabilities. Is Anna more likely to win or to lose? You must work out the probability that she wins.”
“Anna plays a game with an ordinary, fair dice. If she rolls 1 she wins. If she rolls 2 or 3 she loses. If she rolls 4, 5 or 6 she rolls again. When she has to roll again, if she rolls an odd number she wins, if she rolls an even number she loses. Complete the tree diagram with the four missing probabilities. Is Anna more likely to win or to lose? You must work out the probability that she wins.”
What it’s really asking
Fill in the branch probabilities from the rules given, then add the probability of winning outright to the probability of rolling again and then winning.
What the sources actually showed — June 2019
A tree diagram with a first set of branches for Win, Lose, and Roll Again on the first roll, and a second set of branches for Odd (win) and Even (lose) following only from Roll Again.
The full worked answer — June 2019
On the first roll, P(win) is one sixth, P(lose) is two sixths, or one third, and P(roll again) is three sixths, or one half. On the second roll, P(odd) is one half and P(even) is one half.
Anna can win in two ways: rolling a 1 outright, with probability one sixth, or rolling again and then getting an odd number, with probability one half times one half, which is one quarter.
Adding these gives P(win) equals one sixth plus one quarter, which is two twelfths plus three twelfths, or five twelfths, so P(lose) is seven twelfths. Since seven twelfths is greater than five twelfths, Anna is more likely to lose.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise tree diagram questionsWhat the mark scheme rewarded
- All four missing branch probabilities filled in correctly, checking each set sums to 1 (up to 2 marks)
- Correctly multiplying along the roll-again-then-win branch, one half times one half (1 mark)
- Correctly adding the two separate winning routes together, one sixth plus one quarter (1 mark)
- The correct final comparison and conclusion, that Anna is more likely to lose (1 mark, with follow-through available from an earlier slip)
- Multiply along a single branch for 'and', here rolling again AND then getting an odd number, and add across separate branches for 'or'
- Do not forget the outright win route (rolling a 1 first) when finding the total probability of winning, since there are two separate ways to win
Traps examiners saw
- Missing the outright win on the first roll and only counting the roll-again-then-win route, which misses one sixth of the total winning probability
- Comparing the two totals incorrectly, or forgetting the question specifically asks which outcome is more likely, not just for the probability of winning
Full-mark self-check 0 of 3
1×asked“In a game, two bags, A and B, contain cards, each marked Yes or No. Bag A has 3 Yes and 2 No cards. Bag B has 1 Yes and 9 No cards. A player picks one card at random from each bag, and the cards are put back. Complete the tree diagram. To win a prize, a player must pick two cards marked Yes. 450 people each play the game once. How many people are expected to win a prize?”
“In a game, two bags, A and B, contain cards, each marked Yes or No. Bag A has 3 Yes and 2 No cards. Bag B has 1 Yes and 9 No cards. A player picks one card at random from each bag, and the cards are put back. Complete the tree diagram. To win a prize, a player must pick two cards marked Yes. 450 people each play the game once. How many people are expected to win a prize?”
What it’s really asking
Complete each bag's Yes and No probabilities, multiply them for the combined win probability, then scale up to 450 people.
What the sources actually showed — June 2022
A tree diagram with one set of branches for Bag A (Yes or No) and a second, independent set of branches for Bag B (Yes or No), used to find the probability of picking Yes from both bags.
The full worked answer — June 2022
For Bag A, P(Yes) is three fifths and P(No) is two fifths. For Bag B, P(Yes) is one tenth and P(No) is nine tenths.
The probability of winning, picking Yes from both bags, is three fifths times one tenth, which equals three fiftieths.
Out of 450 people, the expected number of winners is three fiftieths times 450, which equals 27.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise tree diagram questionsWhat the mark scheme rewarded
- Both branches of Bag A correctly completed from the given card counts (1 mark)
- Both branches of Bag B correctly completed from the given card counts (1 mark)
- Correctly multiplying the two independent probabilities together, three fiftieths (1 mark)
- Correctly scaling the combined probability up to 450 people (1 mark)
- The correct final expected number of winners, 27 (1 mark)
- Complete each bag's branches directly from the given card counts, checking that Yes and No sum to 1 for each bag separately
- Multiply two independent probabilities together for 'and', since both cards must be Yes for a player to win
Traps examiners saw
- Adding the two bags' probabilities instead of multiplying them, which is only correct for 'or', not 'and'
- Scaling by the wrong number, forgetting to multiply by the full 450 people rather than a different given value
Full-mark self-check 0 of 3
The method for every Q6 (Jun19) / Q12 (Jun22) — same every sittingMark bands, steps, timing
What this question type rewards
The topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
- Completing every missing probability on the tree diagram, remembering that branches from one point sum to 1
- Multiplying probabilities along a single branch to find the probability of that sequence of outcomes
- Adding the probabilities of every separate route that leads to the outcome asked for
The steps
- Fill in every missing branch, using the fact that probabilities from the same point sum to 1
- Identify every route through the tree that leads to the outcome asked for
- Multiply the probabilities along each route
- Add the probabilities of all the routes together for the final combined probability
A fair coin is flipped twice. In a tree diagram, what must the probabilities on the branches from the same point always add up to?
Tree diagram problems come up in two of the four sittings we have, worth 3 marks or more. Practise multiplying along branches and adding across separate winning routes.
Practise tree diagram questions“Three of the four sittings we have full papers for use a Venn diagram, testing set notation, spotting errors in one, or reading probabilities from one.”
Every version depends on correctly identifying which region or regions a piece of set notation or a probability actually refers to.
1×asked“Which of these represents the shaded region on a Venn diagram of sets A and B? Circle your answer: A intersect B complement, B complement, A union B complement, A complement union B complement.”
“Which of these represents the shaded region on a Venn diagram of sets A and B? Circle your answer: A intersect B complement, B complement, A union B complement, A complement union B complement.”
What it’s really asking
Work out which combination of union, intersection and complement correctly describes the shaded area shown.
What the sources actually showed — June 2018
A Venn diagram showing two overlapping circles, A and B, inside a rectangle representing the whole set, with one region of the diagram shaded.
The full worked answer — June 2018
The shaded region covers every part of the diagram except the part of B that lies outside A, which is exactly the region described by A union B complement, everything in A together with everything outside B.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise Venn diagram questionsWhat the mark scheme rewarded
- The single correct option circled, A union B complement (1 mark, point marked with no working required)
- Build up the correct notation piece by piece: identify what is shaded inside A, what is shaded outside B, then combine with the right connective
- Union means everything in at least one of the two named sets, so 'A union B complement' includes all of A plus everything outside B
Traps examiners saw
- Confusing union with intersection, choosing an option that only covers the overlap rather than everything in at least one region
- Missing that a complement is involved, choosing an option that describes the shaded region without accounting for the part outside B
Full-mark self-check 0 of 3
1×asked“In a group of 98 students, 25 study both Art and French, 10 study Art but do not study French, and 41 study French. Joel draws a Venn diagram to represent this. Make two criticisms of his diagram.”
“In a group of 98 students, 25 study both Art and French, 10 study Art but do not study French, and 41 study French. Joel draws a Venn diagram to represent this. Make two criticisms of his diagram.”
What it’s really asking
Check whether the circles are labelled and whether the numbers shown genuinely add up to the correct total of 98.
What the sources actually showed — June 2022
Two overlapping circles with numbers written in each region but no labels naming which circle is Art and which is French, and with the numbers in the regions adding to 99 rather than the correct total of 98.
The full worked answer — June 2022
The first criticism is that neither circle is labelled, so it is not clear which circle represents Art and which represents French.
The second criticism is that the numbers shown in the diagram add up to 99, not 98, since French-only is 41 minus 25, which is 16, and 10 plus 25 plus 16 gives 51 students in the circles, meaning the number outside both circles should be 98 minus 51, which is 47, not 48 as drawn.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise Venn diagram questionsWhat the mark scheme rewarded
- A valid criticism referring to one or both circles not being labelled (1 mark)
- A valid criticism referring to the numbers not adding to the correct total of 98 (1 mark)
- Always check every circle on a Venn diagram is labelled, since a reader cannot otherwise tell which set each circle represents
- Add up every region shown on the diagram and compare the total against the number actually given in the question
Traps examiners saw
- Giving a vague criticism such as 'the numbers are wrong' instead of identifying exactly which number is wrong and what it should be
- Using an incorrect value as evidence for a criticism, such as claiming the total should be 100 instead of the correct 98
Full-mark self-check 0 of 3
1×asked“The Venn diagram represents 100 items. Write down P(A intersect B). Work out P(A complement). Work out P(A union B).”
“The Venn diagram represents 100 items. Write down P(A intersect B). Work out P(A complement). Work out P(A union B).”
What it’s really asking
Identify which regions each piece of notation refers to, then divide the count in those regions by 100.
What the sources actually showed — June 2023
A Venn diagram with two overlapping circles, A and B, over a total of 100 items: 28 in A only, 13 in the overlap of A and B, 48 in B only, and 11 outside both circles, values consistent with the mark scheme's confirmed answers of 13 over 100, 59 over 100 and 89 over 100.
The full worked answer — June 2023
P(A intersect B) is the number of items in the overlap of both circles, 13, out of 100, which is 13 over 100.
P(A complement) is everything outside circle A, which is the 48 in B only plus the 11 outside both, giving 59, so P(A complement) is 59 over 100.
P(A union B) is everything in either circle, which is 28 plus 13 plus 48, giving 89, so P(A union B) is 89 over 100.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise Venn diagram questionsWhat the mark scheme rewarded
- The correct probability for A intersect B, 13 over 100 (1 mark)
- The correct probability for A complement, 59 over 100 (1 mark)
- The correct probability for A union B, 89 over 100 (1 mark)
- Read each probability directly as the count in the relevant region divided by the total of 100 items
- A complement means everything outside the named set, so A complement combines the B-only region with the region outside both circles
Traps examiners saw
- Confusing intersection with union, reading off only the overlap when the question actually wants everything in at least one circle
- Forgetting to include the region outside both circles when finding a complement, undercounting the total
Full-mark self-check 0 of 3
The method for every Q22 (Jun18) / Q7 (Jun22) / Q11 (Jun23) — same every sittingMark bands, steps, timing
What this question type rewards
The topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
- Correctly matching set notation such as intersection, union and complement to the regions they describe
- Checking that every region in a Venn diagram is labelled and that the total across all regions matches the stated total
- Reading a probability directly as a count of items in a region divided by the total number of items
The steps
- Identify exactly which region or regions the set notation, or the question, refers to
- For a critique, check every circle is labelled and that all the region totals add up to the correct overall total
- For a probability, count the items in the relevant region and divide by the total number of items
- Simplify any resulting fraction fully
In a Venn diagram with two events A and B, which symbol represents the region where BOTH events occur at the same time?
Venn diagram questions come up in three of the four sittings we have, testing notation, critique and probability. Practise matching each symbol to the exact region it means.
Practise Venn diagram questions“Three of the four sittings we have full papers for build a question around a cumulative frequency diagram, either constructing one from a frequency table or reading estimates from a given one.”
Every version needs the running total built up correctly before any estimate or comparison can be trusted.
1×asked“Type A batteries and type B batteries were tested. The cumulative frequency diagram shows information about the battery life of type A. Estimate the interquartile range for type A. Estimate the number of type A batteries that had a battery life of more than 1600 hours. The box plot shows information about the battery life of type B. On average, which type had the greater battery life? State how you chose your answer, using data from both diagrams.”
“Type A batteries and type B batteries were tested. The cumulative frequency diagram shows information about the battery life of type A. Estimate the interquartile range for type A. Estimate the number of type A batteries that had a battery life of more than 1600 hours. The box plot shows information about the battery life of type B. On average, which type had the greater battery life? State how you chose your answer, using data from both diagrams.”
What it’s really asking
Read the lower and upper quartiles from the curve to find the interquartile range, then compare the two types' medians.
What the sources actually showed — June 2018
A cumulative frequency diagram for type A battery life, with readings confirmed against the mark scheme as a lower quartile of approximately 1100 hours and an upper quartile of approximately 1400 hours, alongside a box plot for type B battery life showing a median of 1300 hours.
The full worked answer — June 2018
Reading from the cumulative frequency curve, the lower quartile for type A is approximately 1100 hours and the upper quartile is approximately 1400 hours, giving an interquartile range of 1400 minus 1100, which is 300 hours.
Reading across from 1600 hours on the horizontal axis up to the curve and across to the cumulative frequency axis shows that only a small number of batteries, around 4, lasted longer than 1600 hours.
Type A's median, read from the cumulative frequency curve, is around 1260 hours, while type B's median, read from the box plot, is 1300 hours, which is greater, so type B had the greater average battery life.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise cumulative frequency questionsWhat the mark scheme rewarded
- Correctly reading both the lower and upper quartiles from the curve to find the interquartile range, 300 hours (1 mark)
- Correctly reading the cumulative frequency at 1600 hours and using it to estimate the number of batteries lasting longer (1 mark)
- Correctly reading and comparing both medians, one from each diagram, to reach the correct final comparison (up to 3 further marks across all three parts)
- Read both quartiles from the same curve consistently before subtracting, since the interquartile range needs both values together
- When a question asks for a comparison using data from two different diagrams, take one clear reading from each and state both before concluding
Traps examiners saw
- Reading only one quartile and treating it as the interquartile range, forgetting the range needs a subtraction of two values
- Comparing the wrong statistic, such as the range instead of the median, when the question specifically asks about average battery life
Full-mark self-check 0 of 3
1×asked“Here is information about the test marks of 120 students, given in a frequency table with class widths of 10 marks. Complete the cumulative frequency table. Draw a cumulative frequency graph. Students who scored 15 marks or fewer take another test. Use your graph to estimate how many students take another test.”
“Here is information about the test marks of 120 students, given in a frequency table with class widths of 10 marks. Complete the cumulative frequency table. Draw a cumulative frequency graph. Students who scored 15 marks or fewer take another test. Use your graph to estimate how many students take another test.”
What it’s really asking
Add each frequency to the running total to build the table, plot the graph, then read the curve at 15 marks.
What the sources actually showed — June 2019
A frequency table with class intervals 0 to 10, 10 to 20, 20 to 30, 30 to 40 and 40 to 50 marks, with frequencies 20, 28, 40, 20 and 12, alongside a blank grid for the cumulative frequency graph.
The full worked answer — June 2019
Adding each frequency to the running total gives cumulative frequencies of 20, 48, 88, 108 and 120 at marks 10, 20, 30, 40 and 50.
Plotting these five points at their upper class boundaries and joining them with a smooth curve gives the cumulative frequency graph.
Reading across from 15 marks up to the curve, which sits roughly halfway between the points at 10 and 20 marks, gives an estimate of around 34 students who scored 15 marks or fewer.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise cumulative frequency questionsWhat the mark scheme rewarded
- Correctly building the cumulative frequency table by adding each frequency to the running total (1 mark)
- A correct smooth curve drawn through the five plotted points, each at its upper class boundary (1 mark)
- The correct estimate, around 34 students, read consistently from the drawn curve at 15 marks (1 mark)
- Add each class frequency onto the running total, never copy a class frequency straight across as if it were already cumulative
- Plot cumulative frequency against the upper boundary of each class, not its midpoint, since cumulative frequency is a running total up to that value
Traps examiners saw
- Copying each frequency directly onto the graph instead of building a genuine running total first
- Plotting the five points against the midpoint of each class instead of its upper boundary, which shifts the whole curve
Full-mark self-check 0 of 3
1×asked“75 people attend a clinic. Their ages are recorded and a cumulative frequency diagram is drawn. A nurse says, 'More than twice as many people are in their 60s as in their 50s.' Is he correct? Show working to support your answer.”
“75 people attend a clinic. Their ages are recorded and a cumulative frequency diagram is drawn. A nurse says, 'More than twice as many people are in their 60s as in their 50s.' Is he correct? Show working to support your answer.”
What it’s really asking
Read the cumulative frequency at ages 50, 60 and 70, subtract to find the number in each decade, then compare.
What the sources actually showed — June 2022
A cumulative frequency diagram for the ages of 75 clinic patients, with readings confirmed against the mark scheme as approximately 13 people aged up to 50, 30 people aged up to 60, and 66 people aged up to 70.
The full worked answer — June 2022
Reading the cumulative frequency at ages 50, 60 and 70 gives approximately 13, 30 and 66, so the number of people in their 50s is 30 minus 13, which is 17, and the number in their 60s is 66 minus 30, which is 36.
Twice the number in their 50s is 2 times 17, which is 34, and since 36 is greater than 34, the nurse's statement is correct.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise cumulative frequency questionsWhat the mark scheme rewarded
- Correctly reading the cumulative frequency at ages 50, 60 and 70 from the curve (1 mark)
- Correctly subtracting to find the individual decade counts, 17 in their 50s and 36 in their 60s (1 mark)
- The correct final comparison and conclusion, that the nurse is correct (1 mark)
- To find the count within one decade, subtract two cumulative frequency readings, never read a single value as the count within a band directly
- Take all three readings from the same curve consistently before comparing anything, since the whole conclusion depends on accurate reading
Traps examiners saw
- Reading the cumulative frequency at age 60 and treating it directly as the number of people in their 60s, rather than subtracting the reading at age 50
- Comparing the wrong two values, for example comparing the raw cumulative frequencies instead of the individual decade counts the claim is actually about
Full-mark self-check 0 of 3
The method for every Q19 (Jun18) / Q15 (Jun19) / Q17 (Jun22) — same every sittingMark bands, steps, timing
What this question type rewards
The topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
- Building the cumulative frequency by adding each class frequency to the running total, not just copying the frequency itself
- Plotting cumulative frequency against the upper boundary of each class
- Reading estimates from the curve consistently, checking any statement against the same readings
The steps
- Add each class's frequency to the running total to build the cumulative frequency table
- Plot each cumulative frequency against the upper boundary of its class, then join with a smooth curve
- Read across from the value asked for, then down to the axis, for any estimate
- For a comparison, take two consistent readings and compare them directly
Cumulative frequency is:
Cumulative frequency diagrams come up in three of the four sittings we have. Practise building the running total correctly before reading any estimate.
Practise cumulative frequency questions“Two of the four sittings we have full papers for test a probability calculation beyond a single event, one using expected frequency and one using systematic counting.”
Both questions need the probability, or the count of favourable outcomes, built up correctly before any final multiplication.
1×asked“A biased dice is thrown. The probabilities of each score are 0.25, 0.05, 0.15, 0.05, 0.3 and 0.2 for scores 1 to 6. The dice is thrown 200 times. Work out the expected number of times the score will be odd.”
“A biased dice is thrown. The probabilities of each score are 0.25, 0.05, 0.15, 0.05, 0.3 and 0.2 for scores 1 to 6. The dice is thrown 200 times. Work out the expected number of times the score will be odd.”
What it’s really asking
Add the probabilities of every odd score, then multiply by the number of trials.
The full worked answer — June 2018
The odd scores are 1, 3 and 5, with probabilities 0.25, 0.15 and 0.3, so P(odd) equals 0.25 plus 0.15 plus 0.3, which is 0.7.
The expected number of odd scores in 200 throws is 0.7 times 200, which equals 140.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise probability problemsWhat the mark scheme rewarded
- Correctly identifying every odd score, 1, 3 and 5, and adding their individual probabilities to reach 0.7 (1 mark)
- Correct method to multiply the combined probability by the number of trials, 200 (1 mark)
- The correct final expected frequency, 140 (1 mark)
- Add the probabilities of every individual outcome that satisfies the condition before multiplying by the number of trials
- Double check every odd score has been included, since missing just one of the three odd outcomes changes the whole answer
Traps examiners saw
- Missing one of the three odd scores when adding the probabilities, for example forgetting the probability of rolling a 5
- Multiplying by the wrong number of trials, or forgetting to multiply the combined probability by 200 at all
Full-mark self-check 0 of 3
1×asked“Work out how many 5-digit odd numbers can be made using the digits 2, 4, 6, 7 and 9 once each. Do not list them.”
“Work out how many 5-digit odd numbers can be made using the digits 2, 4, 6, 7 and 9 once each. Do not list them.”
What it’s really asking
Fix the restricted last digit first (it must be odd), then count the number of ways to arrange the remaining digits.
The full worked answer — June 2023
For the number to be odd, the last digit must be one of the two odd digits available, 7 or 9, giving 2 choices for the last position.
The remaining four digits can be arranged in the first four positions in 4 times 3 times 2 times 1, which is 24 ways, so the total number of 5-digit odd numbers is 2 times 24, which equals 48.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise probability problemsWhat the mark scheme rewarded
- Correctly fixing the restricted last digit first, identifying 2 valid choices, 7 or 9 (1 mark)
- The correct final count, 48, from correctly multiplying by the arrangements of the remaining four digits (1 mark)
- Deal with the most restricted position first, here the last digit which must be odd, before counting the unrestricted positions
- Once the restricted position is fixed, the remaining positions can be filled in any order, giving a standard factorial count
Traps examiners saw
- Counting all 5-digit arrangements first and trying to filter for odd ones afterwards, which is far slower and easy to miscount
- Forgetting that there are 2 possible choices, not 1, for the restricted last digit, since both 7 and 9 are odd
Full-mark self-check 0 of 3
The method for every Q15 (Jun18) / Q19 (Jun23) — same every sittingMark bands, steps, timing
What this question type rewards
The topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
- Adding the probabilities of every outcome that satisfies the condition asked about
- Multiplying a probability by the number of trials for an expected frequency
- Counting systematically, position by position, when arranging items under a restriction
The steps
- Identify every individual outcome that satisfies the condition in the question
- Add their probabilities together if more than one outcome qualifies
- For an expected frequency, multiply this probability by the number of trials
- For counting arrangements, fix the restricted position first, then count the remaining positions
In an experiment, a coin is flipped 50 times and lands heads 32 times. What is the relative frequency of heads?
Expected frequency and counting problems come up in two of the four sittings we have. Practise adding qualifying probabilities and counting restricted arrangements systematically.
Practise probability problems“Two of the four sittings we have full papers for give a diagram with several vectors labelled in terms of a and b, and ask for an unknown vector, constant, or straight-line proof, built from those given vectors.”
Each version gives 2 or 3 vectors on a labelled diagram and asks you to combine them along a valid path between two points to reach the answer, either a straight-line proof with an unknown constant (June 2019) or an unknown vector reached through several intermediate points (June 2022).
1×asked“ABC and ACD are triangles. Vector BA equals k times a, vector AC equals 3b, and vector AD equals 6a plus 7.5b, where k is a constant. Show that vector CD equals 6a plus 4.5b. BCD is a straight line. Work out the value of k. You must show your working.”
“ABC and ACD are triangles. Vector BA equals k times a, vector AC equals 3b, and vector AD equals 6a plus 7.5b, where k is a constant. Show that vector CD equals 6a plus 4.5b. BCD is a straight line. Work out the value of k. You must show your working.”
What it’s really asking
Find vector CD by combining the two given vectors along the path C to A to D, then use the fact that B, C and D lie on a straight line, so vector BC is a scalar multiple of vector CD, to find k.
What the sources actually showed — June 2019
Triangles ABC and ACD share side AC, with B, C and D lying on a straight line at the base. The diagram shows vector BA equals k times a (arrow pointing from B towards A), vector AC equals 3b (arrow pointing from A towards C), and vector AD equals 6a plus 7.5b (arrow pointing from A towards D).
The full worked answer — June 2019
Vector CD is found by going from C to A, then A to D. Going from C to A reverses the given vector AC, giving negative 3b, and adding vector AD gives negative 3b plus 6a plus 7.5b, which simplifies to 6a plus 4.5b, exactly matching the expression to show.
Since B, C and D lie on a straight line, vector BC must be a scalar multiple of vector CD. Vector BC is found by going from B to A, then A to C, giving k times a plus 3b. Comparing this with vector CD, which equals 6a plus 4.5b, the ratio of the a-coefficients must equal the ratio of the b-coefficients: k to 6 equals 3 to 4.5. Solving this ratio, k equals 3 times 6 divided by 4.5, which equals 4.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise vector geometry questionsWhat the mark scheme rewarded
- 1 mark for showing vector CD equals 6a plus 4.5b, and up to 3 marks for a valid method finding k equals 4, such as comparing BC and CD as a ratio or setting BC equal to a scalar multiple of CD
- Three points lie on a straight line exactly when the vector between any two of them is a scalar (number) multiple of the vector between any other two
- A vector's arrow shows its direction: travelling against the arrow reverses (negates) the vector
Traps examiners saw
- Using vector AC directly instead of reversing it to CA when finding the path from C to D
- Forgetting that k multiplies only vector a, not the whole expression ka plus 3b, when comparing coefficients
Full-mark self-check 0 of 4
1×asked“Five points, D, E, F, G and H, are connected by vectors. Vector ED equals 3a plus b, vector DH equals a plus 6b, and vector EG equals 2a plus 15b. Vector FG equals 2 times vector EH. Work out vector FE in terms of a and b.”
“Five points, D, E, F, G and H, are connected by vectors. Vector ED equals 3a plus b, vector DH equals a plus 6b, and vector EG equals 2a plus 15b. Vector FG equals 2 times vector EH. Work out vector FE in terms of a and b.”
What it’s really asking
Find vector EH by combining ED and DH, double it to find FG, then combine FG and GE to reach vector FE.
What the sources actually showed — June 2022
Five points D, E, F, G and H are connected by line segments forming two triangles sharing point E. The diagram shows vector ED equals 3a plus b (arrow pointing from E towards D), vector DH equals a plus 6b (arrow pointing from D towards H), and vector EG equals 2a plus 15b (arrow pointing from E towards G).
The full worked answer — June 2022
Vector EH is found by going from E to D, then D to H: EH equals ED plus DH, which is 3a plus b plus a plus 6b, giving 4a plus 7b.
Since vector FG equals 2 times vector EH, FG equals 2 times 4a plus 7b, which is 8a plus 14b.
Vector FE is found by going from F to G, then G to E: FE equals FG plus GE. Vector GE is the reverse of the given vector EG, so GE equals negative 2a minus 15b. Adding this to FG gives FE equals 8a plus 14b plus negative 2a minus 15b, which simplifies to 6a minus b.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise vector geometry questionsWhat the mark scheme rewarded
- 1 mark for vector EH equals 4a plus 7b, 1 mark (dependent) for vector FG equals 8a plus 14b, 1 mark (dependent) for any correct path from F to E, and 1 mark for the fully simplified answer 6a minus b
- Vectors can be combined along any valid path between two points, since the overall displacement between them is the same however you get there
- A vector labelled with an arrow pointing from one point to another represents the vector in that direction; reversing the arrow negates the vector
Traps examiners saw
- Reading a vector's direction backwards, for example using DE instead of ED when the diagram's arrow points from E to D, which flips the sign of that term
- Forgetting to double EH before continuing to find FG, since the question states FG equals 2 times EH
Full-mark self-check 0 of 4
The method for every Q22 (Jun19) / Q21 (Jun22) — same every sittingMark bands, steps, timing
What this question type rewards
The topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
- Combining vectors along any valid path between two points, adding vectors head to tail exactly as they are drawn
- Reading each labelled vector's direction correctly from its arrow, since reversing a vector negates it
- For a straight-line proof, showing one vector is a scalar multiple of another vector on the same line
The steps
- Identify every vector given on the diagram and the exact direction its arrow points
- Plan a path between the two points the question asks about, using only vectors you know or can find
- Add the vectors along that path head to tail, reversing (negating) any vector you travel against its arrow
- For a straight-line proof, compare coefficients of a and b between the two vectors to solve for the unknown constant
Vector AB goes from point A to point B. Which of the following describes vector BA?
Vector geometry questions come up in two of the four sittings we have. Practise combining vectors along a path and reversing any vector you travel against its arrow.
Practise vector geometry questionsWhat is guaranteed to come up, and what genuinely varies
Across the four sittings we have full papers for, Paper 1's overall structure and total marks (80) never changed, and the same handful of skills recur every year, though the exact numbers and context are different every time.
Not seen as a standalone, cleanly repeating question on Paper 1 in the four sittings we have full papers for
Circle theorem proof using the alternate segment theorem, with a tangent and a chord, tested only in June 2019 · Rearranging a formula so that the subject appears on both sides of the original equation, tested only in June 2022 · Using rounded approximations to estimate the answer to a calculation involving a formula, tested only in June 2019 · Reading a gradient as a rate, or an area as a total distance, from a real-life graph such as a pay graph or a speed-time graph, tested only in June 2022 · Forming an equation from a worded description and then solving it in context, tested with a different structure in June 2019 and June 2023 · Using a given relationship between statistical measures, such as the median being twice the lower quartile, to complete a list of values, tested only in June 2023 · Comparing a function evaluated at different inputs, such as f(6) over f(2) against f(3), tested only in June 2023 · True or false reasoning about similar and congruent triangles formed by the diagonals of a trapezium, tested only in June 2023 · Enlarging a shape by a negative scale factor, or reflecting a sketched graph in the x-axis, tested with a different diagram in June 2018 and June 2022 · Probability of winning a board game within her next two turns, on a 10-space board where a player starts exactly one space from home, tested only in June 2022 · Single, standalone multiple choice concept checks that did not repeat in the same shape across sittings, such as identifying a reason for congruence, the equation of a straight line, or the number of faces on a prism
These topics genuinely appeared in at least one of the four sittings, but we could not find a single sub-question shape that repeated across sittings with a diagram or structure we could verify from the real paper, so there is no dedicated cluster on this page for them. Do not assume any of the specific facts below are safe to skip, only that we have not found clean repeat evidence, or a diagram we could confidently read, for a dedicated page section.
Before you revise
Does Paper 1 always have the same structure?
Yes, in all four sittings we have full papers for. Every sitting totalled 80 marks in 1 hour 30 minutes, with no calculator allowed, and every paper covered a similarly broad spread of number, algebra, ratio, geometry, probability and statistics questions. Always check your own paper's front cover to confirm, since AQA can make real changes in any future series.
Is a calculator allowed on Paper 1?
No, in any of the four sittings we have full papers for. Paper 1 is explicitly the non-calculator paper, which is why every question on this page, from standard form to exact trigonometric values, is designed to be worked out by hand.
Why is there no June 2020 or June 2021 paper on this page?
Because those sittings do not exist. GCSE exams were cancelled in both 2020 and 2021 due to the pandemic, so there are no real question papers or mark schemes to analyse from those years. Our four sittings, June 2018, June 2019, June 2022 and June 2023, are the most recent full sittings we could obtain both the real question paper and the real mark scheme for.
Was a formulae sheet always provided?
No. June 2022 and June 2023 both list a Formulae Sheet as an enclosed insert in the materials for this paper, but June 2018 and June 2019 do not mention one, meaning more formulae had to be memorised in those earlier years. Always check your own paper's materials list, since this has changed once already.
How is a maths paper actually marked, compared to a paper with long written answers?
Nearly every question on this paper is marked using method marks (M), which reward a correct approach even if the final answer is wrong, and accuracy marks (A), which reward the correct final value following a correct method. Showing your working matters a great deal on a non-calculator paper, since a correct method with an arithmetic slip can still earn most of the marks, while a correct final answer with no working shown at all can score zero on questions that specifically ask you to show your working.
What is the single biggest way marks are lost on this paper?
According to the real mark schemes for these four sittings, marks are very often lost by skipping the working on a question that specifically asks for it, since several questions state that method marks are not awarded to students who show no working. On multi-stage questions such as ratio and algebra problems, marks are also commonly lost by applying a fraction or a step to the wrong quantity, for example giving a fraction of the wrong person's share in a ratio question.
Practise the questions that are guaranteed to come up
Every skill on this page has practice questions waiting in the app, built the way AQA actually structures Paper 1.
Start revising free