Pearson Edexcel GCSE Mathematics Paper 2Calculator (Higher Tier) — every question, answered
Pearson Edexcel GCSE Mathematics (1MA1) Higher Tier Paper 2 is the first calculator paper. We analysed every sitting we could obtain the real question paper and mark scheme for: June 2019, June 2022 and June 2023 (June 2018 could not be located in Pearson's public archive, and June 2020 and June 2021 do not exist as normal exam sittings, since GCSE exams were cancelled in both years because of the pandemic). Like Paper 1, this is around 20 to 24 questions worth 1 to 6 marks each, but a calculator changes what gets tested, with more compound percentage change, trigonometry, and statistics from real data. Below is what each recurring skill has actually asked across the three sittings we have, with a complete worked answer written to the mark scheme for each one, every paragraph explained.
Questions © Pearson Education Ltd, 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 Pearson or Edexcel.
The paper is a template
Same 15 question types, every sitting“Two of the three sittings we have full papers for open with a linear inequality question, combining an algebraic solve with a number line diagram, and June 2023 adds a second algebraic solve worth 3 marks on its own.”
Every version rewards treating an inequality exactly like an equation, except for remembering to keep the direction of the inequality sign consistent throughout.
1×asked“(a) Solve 14n greater than 11n plus 6. (b) On a number line, show the set of values of x for which negative 2 is less than x plus 3, and x plus 3 is less than or equal to 4.”
“(a) Solve 14n greater than 11n plus 6. (b) On a number line, show the set of values of x for which negative 2 is less than x plus 3, and x plus 3 is less than or equal to 4.”
What it’s really asking
For part (a), collect the n terms and solve like an equation. For part (b), subtract 3 from every part of the three-part inequality before drawing it.
The full worked answer — June 2019
For part (a), subtracting 11n from both sides of 14n greater than 11n plus 6 gives 3n greater than 6, and dividing both sides by 3 gives n greater than 2.
For part (b), subtracting 3 from every part of negative 2 is less than x plus 3, and x plus 3 is less than or equal to 4, gives negative 5 is less than x, and x is less than or equal to 1.
This is drawn on the number line with an open circle at negative 5, since x cannot equal negative 5, a closed circle at 1, since x can equal 1, and a line joining them.
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 solving inequalities questionsWhat the mark scheme rewarded
- A correct method to isolate n in part (a), such as 3n greater than 6 (1 mark)
- The correct final answer for part (a), n greater than 2 (1 mark)
- A correct method to subtract 3 from every part of the inequality in part (b) (1 mark)
- A correctly drawn line from negative 5 to 1 on the number line (1 mark)
- The correct final diagram, with an open circle at negative 5 and a closed circle at 1 (1 mark)
- Solve an inequality exactly like an equation, remembering to reverse the inequality sign only when multiplying or dividing by a negative number
- Apply the same operation to all three parts of a three-part inequality at the same time, rather than solving two separate inequalities
Traps examiners saw
- Losing track of which end of the three-part inequality is strict and which is inclusive, drawing the wrong type of circle at each end
- Forgetting that subtracting 3 must be applied to the middle part as well as both outer parts of the inequality
Full-mark self-check 0 of 4
1×asked“Negative 2 is less than or equal to n, and n is less than 5, where n is an integer. (a) Write down the greatest possible value of n. (b) On a number line, show the inequality negative 4 is less than or equal to m, and m is less than 1. (c) Solve two fifths of g, minus 4, is less than 6.”
“Negative 2 is less than or equal to n, and n is less than 5, where n is an integer. (a) Write down the greatest possible value of n. (b) On a number line, show the inequality negative 4 is less than or equal to m, and m is less than 1. (c) Solve two fifths of g, minus 4, is less than 6.”
What it’s really asking
Read off the greatest integer satisfying the given inequality, draw the second inequality on a number line, then solve the third inequality using inverse operations.
The full worked answer — June 2023
Since n is an integer and n is strictly less than 5, the greatest possible value of n is 4.
For part (b), negative 4 is less than or equal to m, and m is less than 1, is drawn with a closed circle at negative 4, since m can equal negative 4, an open circle at 1, since m cannot equal 1, and a line joining them.
For part (c), adding 4 to both sides of two fifths of g, minus 4, is less than 6, gives two fifths of g is less than 10, and multiplying both sides by five halves gives g is less than 25.
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 solving inequalities questionsWhat the mark scheme rewarded
- The correct answer for part (a), 4 (1 mark)
- A correctly drawn line from negative 4 to 1 for part (b) (1 mark)
- The correct final diagram for part (b), with a closed circle at negative 4 and an open circle at 1 (1 mark)
- A correct first step to solve part (c), such as two fifths of g is less than 10 (1 mark)
- A correct second step, such as 2g is less than 50 (1 mark)
- The correct final answer for part (c), g is less than 25 (1 mark)
- Check carefully whether each end of a given inequality is strict or inclusive before choosing the greatest integer value, or before drawing the correct type of circle
- Clear a fraction in an inequality the same way as in an equation, by multiplying both sides by the same positive number
Traps examiners saw
- Giving 5 as the greatest value of n in part (a), forgetting that n is strictly less than 5, not less than or equal to 5
- Drawing the wrong type of circle at negative 4 or at 1 in part (b), by mixing up which end is inclusive
Full-mark self-check 0 of 3
The method for every Q1 (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.
- Solving a linear inequality using the same steps as solving an equation
- Applying the same operation to every part of a three-part inequality when isolating the variable
- Representing an inequality correctly on a number line, using an open circle for a strict inequality and a closed circle for an inclusive one
The steps
- Treat the inequality like an equation, using inverse operations to isolate the variable
- If there are three parts to the inequality, apply every operation to all three parts at once
- Draw an open circle for a strict inequality (less than or greater than) and a closed circle for an inclusive one (less than or equal to, or greater than or equal to)
- Draw the line between the two circles to show every value the inequality includes
Which of the following correctly describes how to represent x > 3 on a number line?
Solving a linear inequality and representing it on a number line comes up in two of the three sittings we have. Practise treating an inequality exactly like an equation, and matching each inequality sign to the correct type of circle.
Practise solving inequalities questions“All three sittings we have full papers for test trigonometry to find a missing length or angle, and June 2019 tests it twice: once in a simple right-angled triangle and once inside a 3D solid.”
Every version needs the correct trigonometric ratio, or the sine or cosine rule, identified and set up correctly before a calculator can finish the job.
1×asked“Triangle ABC has a right angle at B. The side AC, which is the hypotenuse, is 16 cm. Angle ACB is 38 degrees. Calculate the length of AB, correct to 2 decimal places.”
“Triangle ABC has a right angle at B. The side AC, which is the hypotenuse, is 16 cm. Angle ACB is 38 degrees. Calculate the length of AB, correct to 2 decimal places.”
What it’s really asking
Spot that AB is opposite the given angle and AC is the hypotenuse, then use the sine ratio directly.
What the sources actually showed — June 2019
A right-angled triangle with the right angle at B. The hypotenuse AC is labelled 16 cm, and angle ACB is labelled 38 degrees. AB is the side to be found.
The full worked answer — June 2019
Since the right angle is at B, the side AC is the hypotenuse and AB is the side opposite angle ACB, so the sine ratio connects them: sine of 38 degrees equals AB divided by 16.
Rearranging gives AB equals 16 times sine of 38 degrees, which is 9.85 to 2 decimal places.
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 trigonometry questionsWhat the mark scheme rewarded
- A correct trigonometric equation using the sine ratio, such as sine of 38 degrees equals AB over 16 (1 mark)
- The correct final answer, 9.85 cm, or any value in the range 9.76 to 9.92 (1 mark)
- Check which side is the hypotenuse before choosing sine, cosine or tangent, since AC (16 cm) is the longest side here, opposite the right angle
- Use the given angle, not the right angle, when deciding which sides are 'opposite' and 'adjacent'
Traps examiners saw
- Assuming AB is the hypotenuse just because it is asked for first, rather than checking which side is opposite the right angle
- Using cosine instead of sine, by mixing up which side is adjacent to the 38 degree angle
Full-mark self-check 0 of 3
1×asked“ABCD is the square base of a solid, with side length 15 cm. E is a point above B such that angle ABE and angle CBE are both right angles, meaning BE is perpendicular to the whole base. Angle EAB is 35 degrees. M is the point on DA such that DM to MA is in the ratio 2 to 3. Calculate the size of the angle between EM and the base, correct to 1 decimal place.”
“ABCD is the square base of a solid, with side length 15 cm. E is a point above B such that angle ABE and angle CBE are both right angles, meaning BE is perpendicular to the whole base. Angle EAB is 35 degrees. M is the point on DA such that DM to MA is in the ratio 2 to 3. Calculate the size of the angle between EM and the base, correct to 1 decimal place.”
What it’s really asking
Recognise that BE stands straight up from the base, so the angle between EM and the base is the angle EMB in the right-angled triangle formed by E, M and B.
What the sources actually showed — June 2019
A solid with square base ABCD of side 15 cm. Point E sits directly above B, since angle ABE and angle CBE are both right angles. Point M lies on side DA, splitting it in the ratio DM to MA of 2 to 3. Angle EAB is marked 35 degrees.
The full worked answer — June 2019
Since angle ABE and angle CBE are both right angles, BE stands perpendicular to the whole base, so E is directly above B. This means the angle between EM and the base is the angle EMB, in the right-angled triangle EBM.
In right-angled triangle ABE, with the right angle at B, angle EAB is 35 degrees and AB is 15 cm, so BE equals 15 times tangent of 35 degrees, which is 10.50 cm.
Since ABCD is a square, angle DAB is 90 degrees, so triangle ABM has a right angle at A, with AB equal to 15 cm and AM equal to three fifths of 15, which is 9 cm. Using Pythagoras, MB equals the square root of 15 squared plus 9 squared, which is the square root of 306, or 17.49 cm.
In right-angled triangle EBM, with the right angle at B, tangent of angle EMB equals BE over MB, which is 10.50 over 17.49, giving angle EMB equal to 31.0 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 trigonometry questionsWhat the mark scheme rewarded
- A correct process to find the vertical height BE, such as 15 times tangent of 35 degrees (1 mark)
- A correct process to find the base length MB, such as the square root of 15 squared plus 9 squared (1 mark)
- A correct trigonometric equation using BE and MB to set up the angle EMB (1 mark)
- The correct final answer, in the range 30.9 to 31 degrees (1 mark)
- Recognise that when a line stands perpendicular to a whole plane, the angle between any other line and that plane is measured at the point where the two lines meet on the plane
- Use the square base's right angle at A to build a second right-angled triangle for the base length MB, rather than guessing it
Traps examiners saw
- Measuring the wrong angle, such as angle AEM or angle EAM, instead of the angle EMB where the line actually meets the base
- Finding AM as two fifths of 15 instead of three fifths, by mixing up the order of the ratio DM to MA
Full-mark self-check 0 of 4
1×asked“Triangles PQR and QRS share the side QR. In triangle PQR, PQ is 11 cm, PR is 9.4 cm, and angle QPR is 27 degrees. In triangle QRS, angle QRS is 88 degrees and angle RSQ is 41 degrees. Calculate the length of QS, correct to 3 significant figures.”
“Triangles PQR and QRS share the side QR. In triangle PQR, PQ is 11 cm, PR is 9.4 cm, and angle QPR is 27 degrees. In triangle QRS, angle QRS is 88 degrees and angle RSQ is 41 degrees. Calculate the length of QS, correct to 3 significant figures.”
What it’s really asking
Use the cosine rule in the first triangle to find the shared side QR, then use the sine rule in the second triangle to find QS.
What the sources actually showed — June 2022
Triangle PQR and triangle QRS share the side QR. In triangle PQR, PQ is 11 cm, PR is 9.4 cm, and the angle at P is 27 degrees. In triangle QRS, the angle at R is 88 degrees and the angle at S is 41 degrees.
The full worked answer — June 2022
In triangle PQR, using the cosine rule, QR squared equals 11 squared plus 9.4 squared minus 2 times 11 times 9.4 times cosine of 27 degrees, which gives QR squared equal to 25.09, so QR equals 5.01 cm.
In triangle QRS, using the sine rule, QS divided by sine of 88 degrees equals QR divided by sine of 41 degrees, so QS equals 5.01 times sine of 88 degrees divided by sine of 41 degrees.
Evaluating this gives QS equal to 7.63 cm, correct to 3 significant figures.
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 trigonometry questionsWhat the mark scheme rewarded
- A correct cosine rule equation to find QR, such as QR squared equals 11 squared plus 9.4 squared minus 2 times 11 times 9.4 times cosine 27 (1 mark)
- The correct value for QR, such as 5.01 cm, found in the correct order of operations (1 mark)
- A correct sine rule equation linking QS, QR, and the two given angles (1 mark)
- The correct final answer, in the range 7.61 to 7.632 (1 mark)
- Spot that QR is the side shared by both triangles, and that it must be found first before the second triangle can be solved
- Use the cosine rule when all three sides are unknown but two sides and the angle between them are known, and the sine rule when an angle and its opposite side are both known
Traps examiners saw
- Trying to use the sine rule in triangle PQR first, where it cannot be set up because no opposite side and angle pair is known
- Rounding QR too early, before using it in the sine rule, which can shift the final answer outside the accepted range
Full-mark self-check 0 of 4
1×asked“Triangle ABC has angle ABC equal to 70 degrees, with AB equal to 15 cm and AC equal to 18 cm. Calculate the size of angle BAC, correct to 1 decimal place.”
“Triangle ABC has angle ABC equal to 70 degrees, with AB equal to 15 cm and AC equal to 18 cm. Calculate the size of angle BAC, correct to 1 decimal place.”
What it’s really asking
Use the sine rule to find angle BCA first, then use the angle sum of a triangle to find angle BAC.
What the sources actually showed — June 2023
A triangle with angle ABC marked 70 degrees. Side AB is 15 cm and side AC is 18 cm.
The full worked answer — June 2023
Using the sine rule, 18 divided by sine of 70 degrees equals 15 divided by sine of angle BCA, since AC is opposite angle ABC and AB is opposite angle BCA.
Rearranging gives sine of angle BCA equal to 15 times sine of 70 degrees divided by 18, which is 0.783, so angle BCA equals 51.5 degrees.
Since the angles in a triangle add up to 180 degrees, angle BAC equals 180 minus 70 minus 51.5, which is 58.5 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 trigonometry questionsWhat the mark scheme rewarded
- A correct sine rule equation linking the two given sides and their opposite angles (1 mark)
- A correct rearrangement to find angle BCA, such as 51.5 degrees (1 mark)
- A complete process using the angle sum of a triangle to find angle BAC (1 mark)
- The correct final answer, in the range 58.4 to 58.5 degrees (1 mark)
- Match each given side with the angle directly opposite it before setting up the sine rule, since AC is opposite the given angle ABC
- Use the angle sum of a triangle as the final step, once the missing angle from the sine rule has been found
Traps examiners saw
- Matching side AB with angle ABC by mistake, rather than with the angle opposite it, angle BCA
- Forgetting to subtract from 180 at the end, and giving 51.5 as the final answer instead of 58.5
Full-mark self-check 0 of 4
The method for every Q5, Q19 (Jun19) / Q18 (Jun22) / Q13 (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.
- Identifying which sides and angles are known, and choosing the right-angled ratio, or the sine or cosine rule, that connects them
- Setting up the trigonometric equation correctly before using a calculator to solve it
- In a 3D problem, spotting the right-angled triangle hidden inside the solid before applying trigonometry to it
The steps
- Label the triangle, or the hidden triangle inside a 3D solid, with what is known and what is being asked for
- Decide whether the triangle has a right angle (use SOHCAHTOA) or not (use the sine or cosine rule)
- Set up the correct equation and rearrange it to make the unknown the subject
- Use a calculator to evaluate, giving the answer to the accuracy the question asks for
Which trigonometric ratio connects the opposite side and the hypotenuse in a right-angled triangle?
Trigonometry to find a missing length or angle comes up in every sitting we have, including inside 3D solids and using the sine and cosine rules. Practise spotting which rule applies before reaching for a calculator.
Practise trigonometry questions“All three sittings we have full papers for test the error interval for a rounded value, and each sitting uses a slightly different rounding context, from a calculator display to the nearest whole number to 1 decimal place.”
Every version rewards the same underlying idea, that a rounded value could really be anything from half a unit below to half a unit above the last place it was rounded to.
1×asked“Sally used her calculator to work out the value of a number y. The answer on her calculator display began 8.3, with further digits following. Complete the error interval for y.”
“Sally used her calculator to work out the value of a number y. The answer on her calculator display began 8.3, with further digits following. Complete the error interval for y.”
What it’s really asking
Recognise that a display beginning 8.3 could be anywhere from exactly 8.3 up to, but not reaching, 8.4.
The full worked answer — June 2019
Since the display began 8.3, with more digits following, the smallest y could be is exactly 8.3, and the largest it could be is just under 8.4, since any value from 8.30 up to 8.399 recurring would still begin 8.3 on the display.
The completed error interval is 8.3 is less than or equal to y, and y is less than 8.4.
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 bounds and error interval questionsWhat the mark scheme rewarded
- 8.3 placed correctly as the lower bound (1 mark)
- 8.4 placed correctly as the upper bound (1 mark)
- Treat a truncated calculator display the same way as a rounded value: work out the range of numbers that would produce those exact leading digits
- Always use a strict less than sign for the upper bound, since a value of exactly 8.4 would show as 8.4, not 8.3
Traps examiners saw
- Writing 8.4 as included in the interval, with a less than or equal to sign, rather than a strict less than sign
- Guessing an interval based on rounding to 1 decimal place, rather than thinking about what digits would actually be displayed
Full-mark self-check 0 of 3
1×asked“The length of a football pitch is 90 metres, correct to the nearest metre. Complete the error interval for the length of the football pitch.”
“The length of a football pitch is 90 metres, correct to the nearest metre. Complete the error interval for the length of the football pitch.”
What it’s really asking
Find half of 1 metre either side of 90 metres to give the lower and upper bounds.
The full worked answer — June 2022
Since the length is rounded to the nearest metre, the true length could be up to half a metre smaller or larger than 90, so the lower bound is 90 minus 0.5, which is 89.5, and the upper bound is 90 plus 0.5, which is 90.5.
The completed error interval is 89.5 is less than or equal to the length, and the length is less than 90.5.
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 bounds and error interval questionsWhat the mark scheme rewarded
- 89.5 placed correctly as the lower bound (1 mark)
- 90.5 placed correctly as the upper bound (1 mark)
- Always halve the degree of accuracy given, here 1 metre, to find the gap either side of the rounded value
- Keep the lower bound as an included value and the upper bound as an excluded value, since 90.5 itself would round up to 91, not 90
Traps examiners saw
- Using a gap of 1 metre either side instead of half a metre
- Swapping which bound uses the strict less than sign
Full-mark self-check 0 of 3
1×asked“A number, d, is rounded to 1 decimal place. The result is 12.7. Complete the error interval for d.”
“A number, d, is rounded to 1 decimal place. The result is 12.7. Complete the error interval for d.”
What it’s really asking
Find half of 0.1 either side of 12.7 to give the lower and upper bounds.
The full worked answer — June 2023
Since d is rounded to 1 decimal place, the gap either side of 12.7 is half of 0.1, which is 0.05, so the lower bound is 12.7 minus 0.05, which is 12.65, and the upper bound is 12.7 plus 0.05, which is 12.75.
The completed error interval is 12.65 is less than or equal to d, and d is less than 12.75.
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 bounds and error interval questionsWhat the mark scheme rewarded
- 12.65 placed correctly as the lower bound (1 mark)
- 12.75 placed correctly as the upper bound (1 mark)
- Halve the degree of accuracy correctly when it is a decimal place rather than a whole unit, here half of 0.1 is 0.05, not 0.5
- Add and subtract the same gap from the rounded value to find both bounds
Traps examiners saw
- Using a gap of 0.5 instead of 0.05, by forgetting the value was rounded to 1 decimal place, not the nearest whole number
- Reversing which bound is included and which is excluded
Full-mark self-check 0 of 3
The method for every Q6 (Jun19) / Q3 (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.
- Identifying the degree of accuracy the value was rounded, or truncated, to
- Finding half of that degree of accuracy to use as the gap either side of the given value
- Writing the lower bound with a 'less than or equal to' sign and the upper bound with a strict 'less than' sign
The steps
- Work out exactly what place value the number was rounded, or truncated, to
- Halve that place value to find the gap either side of the given value
- Subtract the gap to find the lower bound and add it to find the upper bound
- Write the completed inequality with the lower bound included and the upper bound excluded
A length is 240 cm to the nearest 10 cm. What is the lower bound of the length?
An error interval question comes up in every sitting we have, worth 2 marks each time. Practise halving the degree of accuracy correctly, whether it's a whole unit, a decimal place, or a truncated calculator display.
Practise bounds and error interval questions“All three sittings we have full papers for build at least one ratio question around a real-life context, and June 2019 tests it twice, once combining a ratio with an algebraic relationship, and once combining two separate ratios.”
Every version needs the given ratio combined with an extra piece of information, a fact, a percentage, an algebraic relationship, or a second ratio, before the specific quantity asked for can be found.
1×asked“There are some small cubes and some large cubes in a bag. Each cube is either red or yellow. The ratio of the number of small cubes to the number of large cubes is 4 to 7. The ratio of the number of red cubes to the number of yellow cubes is 3 to 5. All the small cubes are yellow. Work out the least possible number of large yellow cubes in the bag.”
“There are some small cubes and some large cubes in a bag. Each cube is either red or yellow. The ratio of the number of small cubes to the number of large cubes is 4 to 7. The ratio of the number of red cubes to the number of yellow cubes is 3 to 5. All the small cubes are yellow. Work out the least possible number of large yellow cubes in the bag.”
What it’s really asking
Find the smallest total number of cubes that satisfies both ratios at once, then use that total to work out how many large cubes are yellow.
The full worked answer — June 2019
The small to large ratio has 4 plus 7 equals 11 parts in total, and the red to yellow ratio has 3 plus 5 equals 8 parts in total, so the total number of cubes must be a common multiple of both 11 and 8. The lowest common multiple of 11 and 8 is 88, so the least possible number of cubes in the bag is 88.
With a total of 88 cubes, the number of small cubes is four elevenths of 88, which is 32, and the number of yellow cubes is five eighths of 88, which is 55.
Since all 32 small cubes are yellow, the number of large yellow cubes is the total yellow cubes minus the small yellow cubes, which is 55 minus 32, equalling 23.
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 problem questionsWhat the mark scheme rewarded
- A correct explanation showing that 88 is the lowest common multiple of 11 and 8 (1 mark, for the first part)
- The number of small cubes, or yellow cubes, correctly found from a total of 88 (1 mark)
- A complete process to find the large cubes and yellow cubes needed (1 mark)
- The correct final answer, 23 (1 mark)
- Add the parts of each ratio separately first, 11 for small to large and 8 for red to yellow, then find their lowest common multiple
- Subtract the small yellow cubes from the total yellow cubes, since 'all the small cubes are yellow' does not mean all the yellow cubes are small
Traps examiners saw
- Adding 11 and 8 together, or multiplying them incorrectly, instead of finding their lowest common multiple
- Forgetting that some large cubes could also be yellow, and subtracting the wrong count from the total yellow cubes
Full-mark self-check 0 of 3
1×asked“The ratio of Marta's hourly pay to Khalid's hourly pay is 6 to 5. Both Marta and Khalid get an increase of £1.50 in their hourly pay. The ratio of Marta's hourly pay to Khalid's hourly pay after this increase is 13 to 11. Work out the hourly pay before the increase for Marta and for Khalid.”
“The ratio of Marta's hourly pay to Khalid's hourly pay is 6 to 5. Both Marta and Khalid get an increase of £1.50 in their hourly pay. The ratio of Marta's hourly pay to Khalid's hourly pay after this increase is 13 to 11. Work out the hourly pay before the increase for Marta and for Khalid.”
What it’s really asking
Write both original hourly pays in terms of one unknown using the first ratio, then use the second ratio, after adding £1.50 to each, to solve for that unknown.
The full worked answer — June 2022
Let Marta's original hourly pay be 6x and Khalid's original hourly pay be 5x, matching the ratio 6 to 5.
After the increase, the new ratio gives (6x plus 1.5) over (5x plus 1.5) equals 13 over 11, which rearranges to 66x plus 16.5 equals 65x plus 19.5.
Subtracting 65x and 16.5 from both sides gives x equals 3, so Marta's original hourly pay is 6 times £3, which is £18, and Khalid's original hourly pay is 5 times £3, which is £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 ratio problem questionsWhat the mark scheme rewarded
- Both original hourly pays correctly expressed in terms of one unknown, using the first ratio (1 mark)
- A correct equation set up using the second ratio and the £1.50 increase (1 mark)
- A correct process to isolate the unknown, such as x equals 3 (1 mark)
- Both correct final hourly pays, £18 for Marta and £15 for Khalid (1 mark)
- Use a single unknown to represent one part of the original ratio, so the £1.50 increase can be added to each expression before the second ratio is applied
- Add £1.50 to both expressions, not just one, since both Marta and Khalid receive the same increase
Traps examiners saw
- Adding £1.50 to the ratio itself, rather than to each person's actual hourly pay expression
- Solving the equation for x correctly but then forgetting to multiply back to find both final hourly pays
Full-mark self-check 0 of 3
1×asked“Last year a family recycled 800 kg of household waste. 57% of this waste was paper and glass. The weight of paper recycled to the weight of glass recycled is in the ratio 12 to 7. Calculate the weight of glass the family recycled.”
“Last year a family recycled 800 kg of household waste. 57% of this waste was paper and glass. The weight of paper recycled to the weight of glass recycled is in the ratio 12 to 7. Calculate the weight of glass the family recycled.”
What it’s really asking
Find 57% of 800 kg first, then share that amount in the ratio 12 to 7 to find the glass portion.
The full worked answer — June 2023
57% of 800 kg is 456 kg, which is the combined weight of paper and glass.
Sharing 456 kg in the ratio 12 to 7 gives 12 plus 7 equals 19 parts in total, so one part is 456 divided by 19, which is 24 kg.
The glass makes up 7 parts, so the weight of glass recycled is 7 times 24 kg, which equals 168 kg.
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 problem questionsWhat the mark scheme rewarded
- 57% of 800 correctly found as 456 (1 mark)
- A complete process to share 456 in the ratio 12 to 7 (1 mark)
- The correct final answer, 168 kg (1 mark)
- Find the percentage of the total first, since the ratio applies only to the paper and glass, not to the full 800 kg of waste
- Add both parts of the ratio together to find the total number of parts before dividing
Traps examiners saw
- Sharing the full 800 kg in the ratio 12 to 7, instead of first finding the 456 kg that the ratio actually applies to
- Finding the value of one part correctly but then multiplying by 12 instead of 7, giving the paper weight instead of the glass weight
Full-mark self-check 0 of 3
The method for every Q7, Q17 (Jun19) / Q15 (Jun22) / 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.
- Combining the given ratio correctly with the extra information in the question, rather than using either fact alone
- Finding the value of one share, or the total number of shares, before scaling up to the specific quantity asked for
- Carrying out every stage of a multi-step problem in the right order
The steps
- Identify every piece of information given: the ratio, and any total, percentage, algebraic relationship or second ratio alongside it
- Combine them to find the value of one share, or the total number of shares
- Scale up to find the specific quantity the question actually asks for
- Check the final answer against the original context, such as the total given
A recipe uses flour and sugar in the ratio 5:2. What fraction of the mixture is flour?
Multi-stage ratio problems set in a real-life context come up in all three sittings we have, and twice in June 2019. Practise combining a ratio with a total, a percentage, an algebraic relationship, or a second ratio until the extra step feels automatic.
Practise ratio problem questions“Two of the three sittings we have full papers for ask for an algebraic fraction to be factorised and simplified down to a stated general form.”
Every version needs a quadratic factorised first, so that a common factor can cancel, before the fraction reaches the simplest form the question demands.
1×asked“Show that 6, plus the result of dividing (x plus 5) by the fraction with numerator (x squared plus 3x minus 10) and denominator (x minus 1), simplifies to (ax minus b) over (cx minus d), where a, b, c and d are integers.”
“Show that 6, plus the result of dividing (x plus 5) by the fraction with numerator (x squared plus 3x minus 10) and denominator (x minus 1), simplifies to (ax minus b) over (cx minus d), where a, b, c and d are integers.”
What it’s really asking
Factorise the quadratic, divide by flipping and multiplying, then combine the result with 6 using a common denominator.
The full worked answer — June 2019
Factorising the quadratic, x squared plus 3x minus 10 equals (x plus 5)(x minus 2).
Dividing (x plus 5) by this fraction means multiplying by its reciprocal, so (x plus 5) times (x minus 1) over (x plus 5)(x minus 2) equals (x minus 1) over (x minus 2), since (x plus 5) cancels from top and bottom.
Adding 6 to this fraction over a common denominator gives 6(x minus 2) plus (x minus 1), all over (x minus 2), which is 6x minus 12 plus x minus 1, all over (x minus 2).
Simplifying the numerator, 6x minus 12 plus x minus 1 equals 7x minus 13, so the final expression is (7x minus 13) over (x minus 2), matching the required form with a equal to 7, b equal to 13, c equal to 1 and d equal to 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 algebraic fraction questionsWhat the mark scheme rewarded
- Correct factorisation of the quadratic, (x plus 5)(x minus 2) (1 mark)
- A correct method to divide by the algebraic fraction, using its reciprocal (1 mark)
- A correct method to combine two fractions, or a fraction and a whole number, over a common denominator (1 mark)
- The correct final simplified form, (7x minus 13) over (x minus 2) (1 mark)
- Factorise the quadratic before doing anything else, since the common factor (x plus 5) only becomes visible once it is factorised
- Turn a division by a fraction into a multiplication by its reciprocal, rather than trying to divide the fractions directly
Traps examiners saw
- Forgetting to factorise the quadratic first, making the common factor of (x plus 5) impossible to spot
- Combining 6 with the fraction incorrectly, for example adding 6 only to the numerator without multiplying it by the denominator first
Full-mark self-check 0 of 4
1×asked“Show that the fraction with numerator (x squared minus x minus 6) and denominator (2x squared minus 5x minus 3) can be written in the form (ax plus b) over (cx plus d), where a, b, c and d are integers.”
“Show that the fraction with numerator (x squared minus x minus 6) and denominator (2x squared minus 5x minus 3) can be written in the form (ax plus b) over (cx plus d), where a, b, c and d are integers.”
What it’s really asking
Factorise both the numerator and the denominator, then cancel the factor they have in common.
The full worked answer — June 2023
Factorising the numerator, x squared minus x minus 6 equals (x minus 3)(x plus 2).
Factorising the denominator, 2x squared minus 5x minus 3 equals (x minus 3)(2x plus 1).
Since (x minus 3) is a common factor of both the numerator and the denominator, it cancels, leaving (x plus 2) over (2x plus 1), matching the required form with a equal to 1, b equal to 2, c equal to 2 and d equal to 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 fraction questionsWhat the mark scheme rewarded
- Correct factorisation of the numerator, (x minus 3)(x plus 2) (1 mark)
- Correct factorisation of the denominator, (x minus 3)(2x plus 1) (1 mark)
- The correct final simplified form, (x plus 2) over (2x plus 1), after cancelling the shared factor (1 mark)
- Factorise both the numerator and the denominator fully before looking for anything to cancel
- Check carefully for the common factor between the two quadratics, since it is what makes the whole simplification possible
Traps examiners saw
- Factorising only one of the two quadratics and guessing the rest
- Cancelling an x term from the numerator and denominator without factorising first, which is not a valid algebraic step
Full-mark self-check 0 of 3
The method for every Q13 (Jun19) / Q14 (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.
- Factorising each quadratic expression correctly before attempting to simplify
- Spotting the common factor that cancels between the numerator and denominator, or between two fractions
- Combining the resulting terms correctly into the exact general form the question asks for
The steps
- Factorise every quadratic expression in the fraction
- Cancel any common factor between the numerator and the denominator
- Combine or simplify what remains, showing every algebraic step clearly
- Check the final form exactly matches the general form the question specifies
Simplify 6x²/3x
A 'show that' question simplifying an algebraic fraction comes up in two of the three sittings we have. Practise factorising quadratics quickly, since spotting the common factor is the whole question.
Practise algebraic fraction questions“Two of the three sittings we have full papers for apply a percentage change repeatedly over several years, and June 2023 asks for two separate compound calculations to be compared.”
Every version needs the same percentage multiplier applied more than once, rather than simply multiplying the single-year change by the number of years.
1×asked“A new phone cost £679. The value of the phone decreases at a rate of 4% per year. Work out the value of the phone at the end of 3 years.”
“A new phone cost £679. The value of the phone decreases at a rate of 4% per year. Work out the value of the phone at the end of 3 years.”
What it’s really asking
Multiply £679 by 0.96, the multiplier for a 4% decrease, three times over.
The full worked answer — June 2022
A decrease of 4% per year means the value is multiplied by 0.96 each year, since 100% minus 4% is 96%, or 0.96 as a decimal.
Applying this multiplier three times, the value after 3 years is 679 times 0.96 times 0.96 times 0.96, which is 679 times 0.96 cubed.
Evaluating this gives a value of £600.74, correct to the nearest penny.
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 compound percentage questionsWhat the mark scheme rewarded
- A correct method to find one year's decrease, such as 679 times 0.96 (1 mark)
- A correct compound method applying the multiplier three times, such as 679 times 0.96 cubed (1 mark)
- The correct final answer, in the range £600.71 to £600.74 (1 mark)
- Use a single decimal multiplier, 0.96, and apply it once per year, rather than working out each year's decrease from a shrinking value manually
- Remember the multiplier is applied three times for three years, using powers rather than simple multiplication by 3
Traps examiners saw
- Multiplying the single year's decrease, 4% of £679, by 3 and subtracting that from £679, which ignores that the value shrinks each year
- Rounding the value after each year before continuing the calculation, which can shift the final answer outside the accepted range
Full-mark self-check 0 of 3
1×asked“Tamsin buys a house with a value of £150,000. The value of Tamsin's house increases by 4% each year. Rachel buys a house with a value of £160,000. The value of Rachel's house increases by 1.5% each year. At the end of 2 years, whose house has the greater value?”
“Tamsin buys a house with a value of £150,000. The value of Tamsin's house increases by 4% each year. Rachel buys a house with a value of £160,000. The value of Rachel's house increases by 1.5% each year. At the end of 2 years, whose house has the greater value?”
What it’s really asking
Apply each person's own percentage increase over 2 years to their own starting value, then compare the two final values.
The full worked answer — June 2023
Tamsin's house increases by 4% each year, so after 2 years its value is 150,000 times 1.04 squared, which is 150,000 times 1.0816, giving £162,240.
Rachel's house increases by 1.5% each year, so after 2 years its value is 160,000 times 1.015 squared, which is 160,000 times 1.030225, giving £164,836.
Since £164,836 is greater than £162,240, Rachel's house has the greater value after 2 years.
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 compound percentage questionsWhat the mark scheme rewarded
- A correct process to begin working with the percentage for either Tamsin or Rachel (1 mark)
- A correct compound process applied to one of the two houses (1 mark)
- A full process finding comparable figures for both houses after 2 years (1 mark)
- The correct conclusion, Rachel, with supporting figures such as £162,240 and £164,836 (1 mark)
- Apply each person's own percentage increase to their own starting value using the compound multiplier squared, since both are compared after the same 2 years
- State the final comparison clearly with both figures, since a bare answer of 'Rachel' without supporting values does not show the reasoning
Traps examiners saw
- Comparing the two percentage rates directly, 4% against 1.5%, without accounting for Rachel's higher starting value
- Working out only 1 year's growth for one house and 2 years for the other, rather than comparing both after the same length of time
Full-mark self-check 0 of 4
The method for every Q6 (Jun22) / Q8 (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 the percentage change into a single decimal multiplier
- Applying that multiplier repeatedly, once for each year, rather than adding the same amount every year
- Comparing final values correctly when two different situations are involved
The steps
- Write the annual percentage change as a decimal multiplier, such as 1.04 for a 4% increase
- Raise that multiplier to the power of the number of years, or multiply it out year by year
- Multiply the original value by this compound multiplier
- Compare final values if the question involves more than one situation
Which formula correctly calculates the amount A after compound interest at rate r% per year for n years on principal P?
Compound percentage change over several years comes up in two of the three sittings we have. Practise using a single decimal multiplier raised to a power, rather than working out each year one at a time.
Practise compound percentage questions“Two of the three sittings we have full papers for build a cumulative frequency graph from a table of data, and June 2019 also asks for an estimated percentage to be read from it.”
Every version needs the running total found correctly at each class boundary before the graph can be plotted and a value read from it.
1×asked“The grouped frequency table gives the times, in minutes, that 80 office workers take to get to work, in class intervals of 0 up to 20, 20 up to 40, 40 up to 60, 60 up to 80, 80 up to 100 and 100 up to 120 minutes, with frequencies 5, 30, 20, 15, 8 and 2. (a) Complete the cumulative frequency table. (b) Draw the cumulative frequency graph. (c) Use your graph to find an estimate for the percentage of these office workers who take more than 90 minutes to get to work.”
“The grouped frequency table gives the times, in minutes, that 80 office workers take to get to work, in class intervals of 0 up to 20, 20 up to 40, 40 up to 60, 60 up to 80, 80 up to 100 and 100 up to 120 minutes, with frequencies 5, 30, 20, 15, 8 and 2. (a) Complete the cumulative frequency table. (b) Draw the cumulative frequency graph. (c) Use your graph to find an estimate for the percentage of these office workers who take more than 90 minutes to get to work.”
What it’s really asking
Build the running total table, plot it, then read off the cumulative frequency at 90 minutes and use it to find the percentage taking longer than that.
What the sources actually showed — June 2019
A frequency table for the journey times, in minutes, of 80 office workers, with class intervals 0 up to 20, 20 up to 40, 40 up to 60, 60 up to 80, 80 up to 100 and 100 up to 120, with frequencies 5, 30, 20, 15, 8 and 2.
| Time (t minutes) | Frequency |
|---|---|
| 0 ≤ t < 20 | 5 |
| 20 ≤ t < 40 | 30 |
| 40 ≤ t < 60 | 20 |
| 60 ≤ t < 80 | 15 |
| 80 ≤ t < 100 | 8 |
| 100 ≤ t < 120 | 2 |
A blank grid with time in minutes, from 0 to 140, along the horizontal axis, and cumulative frequency, from 0 to 80, up the vertical axis, ready for the cumulative frequency curve to be drawn.
The full worked answer — June 2019
Adding a running total, the cumulative frequencies are 5, 35, 55, 70, 78 and 80, found by adding each class's frequency to the total so far.
Plotting these values at the upper boundary of each class, at (20, 5), (40, 35), (60, 55), (80, 70), (100, 78) and (120, 80), and joining them with a smooth curve, gives the cumulative frequency graph.
Reading across from 90 minutes on the graph gives a cumulative frequency of about 74, meaning 74 workers took 90 minutes or less.
The number taking more than 90 minutes is 80 minus 74, which is 6, so the percentage is 6 divided by 80 times 100, which is 7.5%.
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 cumulative frequency questionsWhat the mark scheme rewarded
- The cumulative frequency table correctly completed as 5, 35, 55, 70, 78, 80 (1 mark)
- 5 or 6 of the 6 points correctly plotted from the table (1 mark)
- A fully correct graph, drawn as a smooth curve or line segments (1 mark)
- A clear method to read off the cumulative frequency graph at 90 minutes (1 mark)
- A full method to find the percentage above 90 minutes, using 80 minus the reading, divided by 80 (1 mark)
- The correct final answer, 7.5%, or a value consistent with the candidate's own graph (1 mark)
- Plot each cumulative frequency at the upper boundary of its class, at 20, 40, 60 and so on, never at the midpoint
- Read up to the curve at exactly 90 minutes, then across to the cumulative frequency axis, before subtracting from the total of 80
Traps examiners saw
- Plotting the cumulative frequencies at the midpoint of each class instead of the upper boundary
- Reading off the cumulative frequency at 90 minutes but forgetting to subtract it from 80 to find the number above that time
Full-mark self-check 0 of 5
1×asked“The cumulative frequency table gives the ages of 80 people working for a company, with cumulative frequencies of 20, 48, 64, 75 and 80 at ages up to 30, 40, 50, 60 and 70 years. (a) Draw a cumulative frequency graph for this information. (b) Use your graph to find an estimate for the median age.”
“The cumulative frequency table gives the ages of 80 people working for a company, with cumulative frequencies of 20, 48, 64, 75 and 80 at ages up to 30, 40, 50, 60 and 70 years. (a) Draw a cumulative frequency graph for this information. (b) Use your graph to find an estimate for the median age.”
What it’s really asking
Plot the given cumulative frequencies at their class boundaries, then read the age at the middle of the cumulative frequency scale.
What the sources actually showed — June 2023
A cumulative frequency table for the ages of 80 people working for a company, showing cumulative frequency 20 for ages 20 up to 30, 48 for ages 20 up to 40, 64 for ages 20 up to 50, 75 for ages 20 up to 60, and 80 for ages 20 up to 70.
| Age (a years) | Cumulative frequency |
|---|---|
| 20 < a ≤ 30 | 20 |
| 20 < a ≤ 40 | 48 |
| 20 < a ≤ 50 | 64 |
| 20 < a ≤ 60 | 75 |
| 20 < a ≤ 70 | 80 |
A blank grid with age in years, from 20 to 70, along the horizontal axis, and cumulative frequency, from 0 to 80, up the vertical axis.
The full worked answer — June 2023
Plotting the given cumulative frequencies at the upper boundary of each age group, at (30, 20), (40, 48), (50, 64), (60, 75) and (70, 80), and joining them with a smooth curve, gives the cumulative frequency graph.
Since there are 80 people, the median is found at a cumulative frequency of 40, halfway up the scale.
Reading across from a cumulative frequency of 40 to the curve, and down to the age axis, gives an estimate for the median age of about 37 years.
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
- 4 or 5 of the 5 points correctly plotted from the cumulative frequency table (1 mark)
- A fully correct graph, drawn as a smooth curve or line segments (1 mark)
- An estimate for the median in the range 36 to 38 years, or a value consistent with the candidate's own graph (1 mark)
- Plot each given cumulative frequency at the upper boundary of its age group, at 30, 40, 50, 60 and 70
- Find the median by reading across from half of the total frequency, 40 out of 80, not from half of the age range
Traps examiners saw
- Reading the median from halfway along the age axis rather than halfway up the cumulative frequency axis
- Plotting a point at the wrong age boundary, shifting the whole curve out of shape
Full-mark self-check 0 of 3
The method for every Q11 (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.
- Building the cumulative frequency table correctly, adding each class's frequency to the running total
- Plotting the cumulative frequency at the upper boundary of each class, not the middle or lower boundary
- Drawing a smooth curve, or joined line segments, and reading a value from it accurately
The steps
- Add up a running total of the frequencies to build the cumulative frequency table
- Plot each cumulative frequency against the upper boundary of its class
- Join the points with a smooth curve or straight line segments
- Read the value asked for from the graph, drawing a line across or up to help read it accurately
Cumulative frequency is:
Building and reading a cumulative frequency graph comes up in two of the three sittings we have. Practise plotting at the upper class boundary every time, since that single habit decides most of the marks.
Practise cumulative frequency questions“Two of the three sittings we have full papers for calculate a volume using a given or standard formula, and June 2022 joins two solids, a cone and a hemisphere, into one composite shape.”
Every version needs the correct volume formula applied accurately, then the result carried through an extra real-life step, such as filling cups or finding a missing height.
1×asked“A container in the shape of a cuboid measures 30 cm by 6 cm by 19 cm. The container is two thirds full of water. A cup holds 275 ml of water. What is the greatest number of cups that can be completely filled with water from the container?”
“A container in the shape of a cuboid measures 30 cm by 6 cm by 19 cm. The container is two thirds full of water. A cup holds 275 ml of water. What is the greatest number of cups that can be completely filled with water from the container?”
What it’s really asking
Find the full volume of the cuboid, take two thirds of it for the water actually present, then see how many whole 275 ml cups that water fills.
What the sources actually showed — June 2019
A cuboid-shaped container measuring 30 cm by 6 cm by 19 cm, shown two thirds full of water.
The full worked answer — June 2019
The full volume of the cuboid container is 30 times 6 times 19, which equals 3420 cubic centimetres.
Since the container is two thirds full, the actual volume of water is two thirds of 3420, which equals 2280 millilitres, since 1 cubic centimetre holds 1 millilitre.
Dividing 2280 by 275 gives 8.29 recurring, so the greatest number of cups that can be completely filled is 8.
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 volume questionsWhat the mark scheme rewarded
- The full volume of the cuboid correctly found, 3420 (1 mark)
- Two thirds of the full volume correctly found, 2280 (1 mark)
- 2280 divided by 275 correctly carried out (1 mark)
- The correct final answer, 8, rounded down from the division (1 mark)
- Find the full volume of the container first, then apply the fraction that is actually full of water, rather than guessing at the water volume directly
- Round down at the end, since a container of water cannot completely fill a cup with only part of the 275 ml needed
Traps examiners saw
- Rounding the final division up to 9 cups, rather than down to 8, since a cup only counts if it is completely filled
- Forgetting to take two thirds of the volume, and dividing the full 3420 by 275 instead
Full-mark self-check 0 of 4
1×asked“A solid cone is joined to a solid hemisphere to make a solid, T. The diameter of the base of the cone is 7 cm, and the diameter of the hemisphere is also 7 cm. The total volume of T is 120 pi cubic centimetres. The total height of T is y cm. (a) Calculate the value of y, correct to 3 significant figures. (b) The diameter of the base of the cone and the diameter of the hemisphere are both increased by the same amount, and the total volume of T does not change. Explain the effect this would have on your answer to part (a).”
“A solid cone is joined to a solid hemisphere to make a solid, T. The diameter of the base of the cone is 7 cm, and the diameter of the hemisphere is also 7 cm. The total volume of T is 120 pi cubic centimetres. The total height of T is y cm. (a) Calculate the value of y, correct to 3 significant figures. (b) The diameter of the base of the cone and the diameter of the hemisphere are both increased by the same amount, and the total volume of T does not change. Explain the effect this would have on your answer to part (a).”
What it’s really asking
Find the hemisphere's volume first, since its radius is fixed, then use the remaining volume to find the cone's height, and add the hemisphere's radius to get the total height y.
What the sources actually showed — June 2022
A solid, T, formed from a cone joined to a hemisphere. Both the cone's base and the hemisphere share a diameter of 7 cm. The total height of the solid is labelled y cm. The volume formulae for a sphere and a cone are given alongside the diagram.
The full worked answer — June 2022
Both the cone and the hemisphere share a radius of 3.5 cm, since the diameter given for each is 7 cm.
The volume of the hemisphere is half of four thirds times pi times 3.5 cubed, which equals approximately 89.8 cubic centimetres.
Since the total volume of T is 120 pi, which is approximately 377.0, the volume of the cone is 377.0 minus 89.8, which is approximately 287.2 cubic centimetres.
Using the cone volume formula, one third times pi times 3.5 squared times the cone's height equals 287.2, so the cone's height is 287.2 divided by (one third times pi times 3.5 squared), which is approximately 22.4 cm.
The total height y is the cone's height plus the hemisphere's radius, which is 22.4 plus 3.5, giving y equal to 25.9 cm, correct to 3 significant figures.
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 volume questionsWhat the mark scheme rewarded
- A correct process to find the volume of the hemisphere, using radius 3.5 (1 mark)
- A correct equation linking the hemisphere volume, the cone volume and the total 120 pi (1 mark)
- A correct process to isolate the cone's own height from this equation (1 mark)
- The correct final answer, in the range 25.8 to 26.3 (1 mark)
- A correct explanation for part (b), that the height would decrease (1 mark)
- Halve the diameter to get the radius, 3.5 cm, before substituting into either volume formula
- Remember the total height y includes both the cone's own height and the hemisphere's radius, not just the cone's height alone
Traps examiners saw
- Using the diameter, 7, directly in the volume formulae instead of the radius, 3.5
- Forgetting to add the hemisphere's radius back on at the end, giving only the cone's height as the final answer for y
Full-mark self-check 0 of 5
The method for every Q4 (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.
- Substituting the correct dimensions into the volume formula for each shape involved
- Adding the volumes of two joined solids correctly when a composite shape is involved
- Carrying the volume through an extra real-life step, such as a fraction full, or a given total volume, to reach the final answer
The steps
- Identify every 3D shape involved and the formula needed for each
- Substitute the given dimensions carefully, using radius rather than diameter where needed
- Add volumes together if the solid is a composite of two shapes
- Use the volume in the extra real-life step the question asks about
What is the formula for the volume of a cone with base radius r and height h?
Calculating the volume of a 3D shape in a real-life context comes up in two of the three sittings we have, including a composite solid made of two joined shapes. Practise substituting radius rather than diameter into every volume formula.
Practise volume questions“Two of the three sittings we have full papers for use vector notation to express one side of a shape in terms of two given vectors, and June 2019 goes further, using a straight-line condition to find an unknown ratio.”
Every version needs a path of known vectors traced correctly through the shape before the unknown vector, or the unknown ratio, can be found.
1×asked“CDEF is a quadrilateral. CD is the vector a, DE is the vector b, and FC is the vector (a minus b). (a) Express FE in terms of a and or b, giving your answer in its simplest form. M is the midpoint of DE. X is the point on FM such that FX to XM is in the ratio n to 1. CXE is a straight line. (b) Work out the value of n.”
“CDEF is a quadrilateral. CD is the vector a, DE is the vector b, and FC is the vector (a minus b). (a) Express FE in terms of a and or b, giving your answer in its simplest form. M is the midpoint of DE. X is the point on FM such that FX to XM is in the ratio n to 1. CXE is a straight line. (b) Work out the value of n.”
What it’s really asking
First trace a path from F to E using the three given vectors, then set up X's position two different ways and compare coefficients to find n.
The full worked answer — June 2019
Tracing a path from F to E through C and D, FE equals FC plus CD plus DE, which is (a minus b) plus a plus b, and this simplifies to 2a.
Since M is the midpoint of DE, and taking C as the origin, D is at position a and E is at position a plus b, so M is at position a plus half of b.
Since FC is a minus b, F is at position b minus a, so the vector from F to M is (a plus half of b) minus (b minus a), which simplifies to 2a minus half of b.
Since FX to XM is in the ratio n to 1, X is at position F plus (n over n plus 1) times the vector FM, which is (b minus a) plus (n over n plus 1) times (2a minus half of b).
Since C, X and E lie on a straight line, the vector CX must be a multiple of CE, which is a plus b. Comparing the coefficients of a and of b separately in this condition gives two equations, which solve together to give n equal to 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
- FE correctly found and simplified to 2a (2 marks)
- A correct vector expression found for two of FM, FX, or CX, in terms of a and b (1 mark)
- A correct vector expression found for CE, or CX expressed a second way (1 mark)
- A complete process to equate the coefficients of a and b (1 mark)
- The correct final answer, n equal to 4 (1 mark)
- Trace a path using only the vectors actually given, checking the direction of each arrow matches the direction being travelled
- Compare the coefficients of a and b separately when two vector expressions for the same straight line are set equal, since a and b are independent directions
Traps examiners saw
- Travelling around the quadrilateral in the wrong order, which reverses the sign of one of the vectors used
- Setting up the ratio n to 1 the wrong way round, dividing FM into the wrong fraction for X's position
Full-mark self-check 0 of 5
1×asked“ORT is a triangle. OT is the vector a, and RT is the vector b. M is the point on OR such that OM to MR is in the ratio 2 to 3. Express MT in terms of a and b, giving your answer in its simplest form.”
“ORT is a triangle. OT is the vector a, and RT is the vector b. M is the point on OR such that OM to MR is in the ratio 2 to 3. Express MT in terms of a and b, giving your answer in its simplest form.”
What it’s really asking
Find the position of R using the given vectors, use the ratio to find M's position, then trace a path from M to T.
The full worked answer — June 2023
Taking O as the origin, T is at position a. Since RT is b, R is at position a minus b.
Since M is on OR with OM to MR in the ratio 2 to 3, M is at two fifths of the way from O to R, so M is at position two fifths of (a minus b).
The vector MT is T minus M, which is a minus two fifths of (a minus b), and this simplifies to three fifths of a plus two fifths of 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
- A correct expression for the position of R, such as a minus b (1 mark)
- A correct expression for the position of M, such as two fifths of (a minus b) (1 mark)
- A complete process to find MT by subtracting position vectors (1 mark)
- The correct final answer, three fifths of a plus two fifths of b (1 mark)
- Set a clear origin, such as O, before writing any position vectors, so every point in the triangle can be compared consistently
- Use the ratio 2 to 3 as two out of five total parts when finding how far M is along OR, not two out of three
Traps examiners saw
- Using three fifths instead of two fifths for M's position, by mixing up which part of the ratio belongs to OM
- Finding TM instead of MT, which reverses the sign of the final answer
Full-mark self-check 0 of 4
The method for every Q20 (Jun19) / 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.
- Tracing a path of known vectors from one point to another around the shape
- Using a point that divides a line in a given ratio to write a position in terms of the given vectors
- Setting up and solving vector equations by comparing the coefficients of each given vector separately
The steps
- Trace a path using only the given vectors from the start point to the end point needed
- Use any given ratio to write the position of a point that divides a line
- Set two vector expressions for the same point equal to each other
- Compare the coefficients of each vector separately to solve for the unknown
Vector AB goes from point A to point B. Which of the following describes vector BA?
Expressing a vector in terms of two given vectors comes up in two of the three sittings we have. Practise setting a clear origin and tracing paths using only the vectors given.
Practise vector geometry questions“Two of the three sittings we have full papers for ask for a sketched graph to be transformed, and both include reflecting a curve in the y-axis using y equals f of minus x.”
Every version rewards knowing exactly what each transformation does to the coordinates of key points on the original graph.
1×asked“The graph of y equals f(x) is shown on a grid. (a) On the grid, sketch the graph of y equals f(minus x). Here is a sketch of the graph of y equals tan x degrees, with a point Q marked on it. The graph of y equals tan x degrees is translated to give the graph of y equals g(x). Following the translation, the point Q moves to point R, with coordinates (90, negative 5). (b) Find an expression for g(x) in terms of x.”
“The graph of y equals f(x) is shown on a grid. (a) On the grid, sketch the graph of y equals f(minus x). Here is a sketch of the graph of y equals tan x degrees, with a point Q marked on it. The graph of y equals tan x degrees is translated to give the graph of y equals g(x). Following the translation, the point Q moves to point R, with coordinates (90, negative 5). (b) Find an expression for g(x) in terms of x.”
What it’s really asking
Reflect every point of the first graph across the y-axis for part (a), then work out the horizontal and vertical shift needed to move Q to R for part (b).
What the sources actually showed — June 2022
A sketched curve labelled y equals f(x) on a set of axes, with no scale or specific coordinates marked beyond the origin.
A sketch of the standard tan x degrees curve, repeating every 180 degrees, with vertical asymptotes shown as dashed lines and no scale marked on either axis beyond the origin. Point Q is marked on the curve where it crosses the x-axis, in the branch three complete periods on from the origin.
The full worked answer — June 2022
Reflecting the graph of y equals f(x) in the y-axis gives the graph of y equals f(minus x), since replacing x with minus x flips every point across that axis.
Since point R has coordinates (90, negative 5), and the mark scheme confirms the horizontal shift needed is 270 (using the tan graph's repeating period of 180 degrees), the translation moves the curve left by 270 and down by 5.
Applying both parts of the translation to y equals tan(x degrees) gives the new function g(x) equal to tan(x plus 270) degrees, minus 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 graph transformation questionsWhat the mark scheme rewarded
- A correct sketch for part (a), a reflection of y equals f(x) in the y-axis (1 mark)
- A correct description of part of the translation, such as identifying a shift of 270 (1 mark)
- The correct final expression, g(x) equal to tan(x plus 270) degrees, minus 5 (1 mark)
- Learn the rule that y equals f(minus x) reflects in the y-axis, and y equals minus f(x) reflects in the x-axis, since these are easy to mix up
- Work out the horizontal shift using the periodic nature of a trig graph, since a shift of 270 degrees on a tan graph, with period 180 degrees, has the same effect as a shift of 90 degrees
Traps examiners saw
- Sketching a reflection in the x-axis instead of the y-axis for part (a), by confusing f(minus x) with minus f(x)
- Forgetting to apply the vertical shift of negative 5 after finding the horizontal shift, giving an incomplete expression for g(x)
Full-mark self-check 0 of 4
1×asked“The graph of y equals f(x) is shown on a grid. (a) On a grid, draw the graph of y equals f(x) minus 4. (b) On another grid, draw the graph of y equals f(minus x).”
“The graph of y equals f(x) is shown on a grid. (a) On a grid, draw the graph of y equals f(x) minus 4. (b) On another grid, draw the graph of y equals f(minus x).”
What it’s really asking
Move every point of the original graph down by 4 for part (a), then reflect every point across the y-axis for part (b).
What the sources actually showed — June 2023
A sketched curve labelled y equals f(x) on a set of axes, with the curve crossing and turning at a small number of visible grid points, repeated on two further blank grids for parts (a) and (b).
The full worked answer — June 2023
For part (a), y equals f(x) minus 4 moves every point on the original graph down by 4 units, since subtracting a number outside the function shifts the whole graph vertically.
For part (b), y equals f(minus x) reflects every point on the original graph in the y-axis, since replacing x with minus x swaps the left and right sides of the graph.
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 graph transformation questionsWhat the mark scheme rewarded
- A correct sketch for part (a), the original curve moved down by 4 units, keeping the same shape (1 mark)
- A correct sketch for part (b), the original curve reflected in the y-axis (1 mark)
- Move every key point on the curve down by exactly 4 units for part (a), keeping the shape of the curve identical, just shifted
- Reflect every key point across the y-axis for part (b), so a point to the right of the axis moves to the same distance on the left
Traps examiners saw
- Moving the curve up instead of down in part (a), by misreading the subtraction as an addition
- Reflecting in the x-axis instead of the y-axis in part (b), by confusing f(minus x) with minus f(x)
Full-mark self-check 0 of 2
The method for every Q21 (Jun22) / 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.
- Knowing that y equals f of minus x reflects the graph in the y-axis
- Knowing that y equals f of x plus a number translates the graph vertically
- Combining a horizontal and a vertical change correctly when finding a new function from a description
The steps
- Identify exactly which transformation is being applied, and to which axis or direction
- Apply the transformation to each key point on the original graph in turn
- Sketch the new curve through the transformed points, keeping its overall shape
- If a new function must be stated, describe the transformation algebraically instead of just sketching it
The graph of y = f(x) is transformed to y = f(x) + 3. Which of the following describes this transformation?
Sketching a transformed graph comes up in two of the three sittings we have, and both use a reflection in the y-axis. Learn the effect of each transformation rule so you can sketch confidently without a table of values.
Practise graph transformation questions“Two of the three sittings we have full papers for combine a circle theorem with other angle facts, and both require a correctly named reason at every stage of the working.”
Every version needs more than one angle fact chained together, a circle theorem plus an isosceles triangle or a straight line, before the final angle can be found.
1×asked“The points A, B, C and D lie on a circle. CDE is a straight line. BA equals BD, and CB equals CD. Angle ABD is 40 degrees. Work out the size of angle ADE, giving a reason for each stage of your working.”
“The points A, B, C and D lie on a circle. CDE is a straight line. BA equals BD, and CB equals CD. Angle ABD is 40 degrees. Work out the size of angle ADE, giving a reason for each stage of your working.”
What it’s really asking
Use the two isosceles triangles, ABD and BCD, together with the cyclic quadrilateral ABCD, to build up to angle ADE.
What the sources actually showed — June 2019
A circle with points A, B, C and D on its circumference. CDE is a straight line, with E beyond D. BA equals BD and CB equals CD. Angle ABD is marked 40 degrees.
The full worked answer — June 2019
Since BA equals BD, triangle ABD is isosceles, so angle BAD equals angle BDA, which is (180 minus 40) divided by 2, giving 70 degrees each, since the base angles of an isosceles triangle are equal.
Since ABCD is a cyclic quadrilateral, opposite angles add up to 180 degrees, so angle BCD equals 180 minus angle BAD, which is 180 minus 70, giving 110 degrees.
Since CB equals CD, triangle BCD is isosceles, so angle CBD equals angle CDB, which is (180 minus 110) divided by 2, giving 35 degrees each.
Angle ADC is angle ADB plus angle BDC, which is 70 plus 35, giving 105 degrees, and since CDE is a straight line, angle ADE is 180 minus 105, giving 75 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 circle theorem questionsWhat the mark scheme rewarded
- Angle BAD, or angle BDA, correctly found as 70 degrees, using the isosceles triangle ABD (1 mark)
- Angle BCD correctly found as 110 degrees, using the cyclic quadrilateral (1 mark)
- The correct final angle, angle ADE equal to 75 degrees (1 mark)
- Two correctly linked reasons given, such as the base angles of an isosceles triangle are equal and opposite angles of a cyclic quadrilateral add up to 180 (2 marks)
- Work through both isosceles triangles in turn, ABD first and then BCD, using the cyclic quadrilateral rule to bridge between them
- Name each reason precisely, since 'isosceles triangle' alone is not enough; the reason must state that base angles of an isosceles triangle are equal
Traps examiners saw
- Assuming angle ADC and angle ADE are the same angle, rather than recognising they are supplementary because CDE is a straight line
- Giving a reason that does not match the working shown, such as naming a circle theorem for a step that actually used an isosceles triangle
Full-mark self-check 0 of 5
1×asked“A, B, C and D are points on the circumference of a circle, centre O. ADE and BCE are straight lines. Angle BOD is 132 degrees, and angle CED is 16 degrees. Work out the size of angle CDE, giving a reason for each stage of your working.”
“A, B, C and D are points on the circumference of a circle, centre O. ADE and BCE are straight lines. Angle BOD is 132 degrees, and angle CED is 16 degrees. Work out the size of angle CDE, giving a reason for each stage of your working.”
What it’s really asking
Use the angle at the centre to find an angle at the circumference, use that to find an angle on a straight line, then use the angle sum of a triangle to reach angle CDE.
What the sources actually showed — June 2022
A circle, centre O, with points A, B, C and D on its circumference. ADE and BCE are straight lines, meeting at an external point E. Angle BOD, at the centre, is marked 132 degrees. Angle CED, at the external point, is marked 16 degrees.
The full worked answer — June 2022
The angle at the centre of a circle is twice the angle at the circumference subtended by the same arc, so angle BAD equals half of angle BOD, which is half of 132, giving 66 degrees.
Since ABCD is a cyclic quadrilateral, opposite angles add up to 180 degrees, so angle BCD equals 180 minus 66, giving 114 degrees.
Since BCE is a straight line, angle DCE is 180 minus angle BCD, which is 180 minus 114, giving 66 degrees.
In triangle CDE, the angles add up to 180 degrees, so angle CDE equals 180 minus angle DCE minus angle CED, which is 180 minus 66 minus 16, giving 98 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 circle theorem questionsWhat the mark scheme rewarded
- Angle BAD correctly found as 66 degrees, using the angle at the centre theorem (1 mark)
- Angle BCD correctly found as 114 degrees, using the cyclic quadrilateral (1 mark)
- The correct final angle, angle CDE equal to 98 degrees (1 mark)
- A correctly named and linked reason given, such as the angle at the centre of a circle is twice the angle at the circumference (1 mark)
- Use the angle at the centre theorem first, since angle BOD is the only angle directly given at the circle's centre
- Use angles on the straight line BCE to move from the cyclic quadrilateral's angle BCD into triangle CDE
Traps examiners saw
- Halving angle BOD incorrectly, or using the reflex angle at the centre without adjusting the circle theorem to match
- Missing the straight line step through C, and trying to use angle BCD directly inside triangle CDE without first finding angle DCE
Full-mark self-check 0 of 5
The method for every Q18 (Jun19) / Q20 (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.
- Correctly identifying which circle theorem applies to the given angle, such as the angle at the centre or a cyclic quadrilateral
- Chaining that circle theorem together with another angle fact, such as an isosceles triangle or an angle on a straight line
- Naming every reason correctly and linking it clearly to the specific angle it explains
The steps
- Mark every given angle on the diagram, and identify the circle theorem that connects to it
- Use that first angle to find a second angle, using an isosceles triangle, a cyclic quadrilateral, or a straight line
- Continue chaining angle facts until the specific angle asked for is reached
- Write out the correct name for each reason used, linked clearly to the working
AB is a diameter of a circle. Point C lies on the circle. What is the size of angle ACB?
Circle theorems combined with other angle facts, and full reasons, come up in two of the three sittings we have. Practise naming each reason precisely, since a correct angle with the wrong reason still loses marks.
Practise circle theorem questions“Two of the three sittings we have full papers for give a named person's specific, wrong mathematical statement and ask for an explanation of why it is incorrect.”
Every version rewards identifying the specific mathematical error the person made, not just restating that their final answer happens to be wrong.
1×asked“A box plot shows the sales, in thousands of pounds, of an online store each month. Andrew says, 'Three quarters of the given data lies between the lower quartile value and the upper quartile value, because these are the values of the lower quartile and the upper quartile.' Andrew is wrong. Explain why.”
“A box plot shows the sales, in thousands of pounds, of an online store each month. Andrew says, 'Three quarters of the given data lies between the lower quartile value and the upper quartile value, because these are the values of the lower quartile and the upper quartile.' Andrew is wrong. Explain why.”
What it’s really asking
Explain that the interquartile range, between the lower and upper quartile, always contains half of the data, not three quarters.
What the sources actually showed — June 2022
A box plot showing the sales, in thousands of pounds, of an online store each month, with a box drawn between the lower quartile and the upper quartile, and whiskers extending to the least and greatest values.
The full worked answer — June 2022
The lower quartile and the upper quartile always mark off the middle half of the data, since one quarter of the data lies below the lower quartile and one quarter lies above the upper quartile, leaving one half, not three quarters, in between.
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 reasoning and explanation questionsWhat the mark scheme rewarded
- A correct explanation stating that the interquartile range represents half of the data, not three quarters (1 mark)
- Remember that the lower quartile always cuts off the bottom 25% and the upper quartile always cuts off the top 25%, leaving exactly 50% in the middle, whatever the data set
- Avoid simply saying Andrew's answer is wrong without stating the correct fraction, since the mark is for the mathematical reason, not the disagreement
Traps examiners saw
- Describing the range or the actual sales values without addressing the specific error, that three quarters should be one half
- Giving a vague answer such as 'the values are wrong' without any correct mathematical reasoning
Full-mark self-check 0 of 2
1×asked“There are 48 counters in a bag. There are only red counters and blue counters in the bag. The number of red counters to the number of blue counters is in the ratio 1 to 2. Helen has to work out how many red counters are in the bag. She says, 'There are 24 red counters in the bag because 1 is half of 2 and 24 is half of 48.' Is Helen correct? You must give a reason for your answer.”
“There are 48 counters in a bag. There are only red counters and blue counters in the bag. The number of red counters to the number of blue counters is in the ratio 1 to 2. Helen has to work out how many red counters are in the bag. She says, 'There are 24 red counters in the bag because 1 is half of 2 and 24 is half of 48.' Is Helen correct? You must give a reason for your answer.”
What it’s really asking
Explain that Helen has used the wrong relationship, since sharing in the ratio 1 to 2 means dividing the total by 3 parts, not simply halving it.
The full worked answer — June 2023
Helen is not correct, since sharing 48 counters in the ratio 1 to 2 means dividing them into 1 plus 2, which is 3 equal parts, not into 2 equal halves, so there are 16 red counters, not 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 reasoning and explanation questionsWhat the mark scheme rewarded
- A correct explanation stating that Helen is not correct, with a valid reason such as there being 16 red counters, not 24, or that the ratio has 3 parts in total, not 2 (1 mark)
- Add both parts of the ratio together, 1 plus 2, to find the true number of equal parts the total should be divided by
- State the correct number of red counters, 16, as supporting evidence for why Helen's reasoning does not work
Traps examiners saw
- Simply stating 'no' without identifying that the ratio should be divided into 3 parts, not 2
- Agreeing with Helen because 24 looks like a reasonable share of 48, without checking it against the actual ratio given
Full-mark self-check 0 of 3
The method for every Q10a (Jun22) / 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.
- Identifying precisely what mathematical rule or fact the person has misapplied
- Explaining the correct rule clearly, rather than simply disagreeing with the conclusion
- Using the numbers or context given in the question to support the explanation where helpful
The steps
- Read the person's statement carefully and work out the correct answer or method yourself first
- Compare the person's reasoning step by step against the correct method
- Identify exactly where their reasoning breaks down
- Write a clear explanation naming the specific error, rather than a vague disagreement
A recipe uses flour and sugar in the ratio 5:2. What fraction of the mixture is flour?
Explaining why a given statement or method is wrong comes up in two of the three sittings we have, always worth a single, precise mark. Practise identifying the exact mathematical error, not just disagreeing with the conclusion.
Practise reasoning and explanation questions“Two of the three sittings we have full papers for ask for a formula to be rearranged, and June 2019 goes further, requiring the subject to be collected together after it appears on both sides.”
Every version needs the formula cleared of any fractions first, before the required letter can be isolated on one side.
1×asked“Make m the subject of the formula f equals (3m plus 4) over (m minus 1).”
“Make m the subject of the formula f equals (3m plus 4) over (m minus 1).”
What it’s really asking
Clear the fraction first, then collect every m term onto one side and factorise, since m appears in both the numerator and the denominator.
The full worked answer — June 2019
Multiplying both sides by (m minus 1) clears the fraction, giving f times (m minus 1) equals 3m plus 4, which expands to fm minus f equals 3m plus 4.
Collecting every term containing m onto one side, fm minus 3m equals f plus 4.
Factorising m out of the left-hand side gives m times (f minus 3) equals f plus 4, so m equals (f plus 4) over (f minus 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 rearranging formulae questionsWhat the mark scheme rewarded
- A correct method to clear the fraction by multiplying both sides by (m minus 1) (1 mark)
- A correct method to collect terms containing m onto one side of the equation (1 mark)
- The correct final answer, m equals (f plus 4) over (f minus 3) (1 mark)
- Clear the fraction by multiplying both sides by the denominator before doing anything else, since m cannot be collected while it is still inside a fraction
- Factorise m out once every m term is on the same side, rather than trying to divide term by term
Traps examiners saw
- Attempting to collect the m terms before clearing the fraction, which leaves m trapped inside the denominator
- Factorising incorrectly, for example writing m(f minus 3) as m times f minus 3 instead of mf minus 3m
Full-mark self-check 0 of 3
1×asked“A box in the shape of a cuboid is placed on a horizontal floor. The box exerts a force of 180 newtons on the floor, and a pressure of 187.5 newtons per square metre, using the formula pressure equals force divided by area. The face in contact with the floor is a rectangle of length 1.2 metres and width x metres. Work out the value of x.”
“A box in the shape of a cuboid is placed on a horizontal floor. The box exerts a force of 180 newtons on the floor, and a pressure of 187.5 newtons per square metre, using the formula pressure equals force divided by area. The face in contact with the floor is a rectangle of length 1.2 metres and width x metres. Work out the value of x.”
What it’s really asking
Use the pressure formula to find the area of the rectangle first, then divide by the known length to find the unknown width.
The full worked answer — June 2022
Using pressure equals force divided by area, 187.5 equals 180 divided by the area, so the area equals 180 divided by 187.5, which is 0.96 square metres.
Since the area of the rectangle is length times width, 0.96 equals 1.2 times x.
Dividing both sides by 1.2 gives x equal to 0.96 divided by 1.2, which is 0.8 metres.
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 rearranging formulae questionsWhat the mark scheme rewarded
- A correct process to find the area from the pressure formula, such as 180 divided by 187.5 (1 mark)
- A complete process to find the width, using the area and the given length (1 mark)
- The correct final answer, 0.8 (1 mark)
- Rearrange the pressure formula to find the area first, since neither the length nor the width can be found directly from pressure and force alone
- Use the area of a rectangle, length times width, as the link between the pressure formula and the actual dimensions given
Traps examiners saw
- Substituting the given length, 1.2, directly into the pressure formula as if it were the whole area
- Dividing force by length instead of finding the area first, which skips a necessary step
Full-mark self-check 0 of 3
The method for every Q15 (Jun19) / Q9 (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.
- Clearing any fraction in the formula by multiplying both sides, before attempting to isolate the subject
- Collecting every term containing the required letter onto one side, when it appears more than once
- Factorising out the required letter, when necessary, to complete the rearrangement
The steps
- Multiply both sides by any denominator to clear fractions from the formula
- Expand any brackets that result from this multiplication
- Collect every term containing the required letter onto one side of the equation
- Factorise out the required letter if it appears in more than one term, then divide to isolate it
Make b the subject of the formula: a = b + c
Rearranging a formula, including one where the subject appears twice or a compound measure formula, comes up in two of the three sittings we have. Practise clearing any fraction first, before collecting the subject onto one side.
Practise rearranging formulae questions“Two of the three sittings we have full papers for use the squared relationship between the linear scale factor and the area scale factor of similar shapes, and June 2019 tests it twice in the same question.”
Every version needs the linear scale factor squared to reach the area scale factor, or the area scale factor square rooted to reach the linear scale factor, before the final ratio can be found.
1×asked“The circumference of circle B is 90% of the circumference of circle A. (a) Find the ratio of the area of circle A to the area of circle B. Square E has sides of length e cm, and square F has sides of length f cm. The area of square E is 44% greater than the area of square F. (b) Work out the ratio e to f.”
“The circumference of circle B is 90% of the circumference of circle A. (a) Find the ratio of the area of circle A to the area of circle B. Square E has sides of length e cm, and square F has sides of length f cm. The area of square E is 44% greater than the area of square F. (b) Work out the ratio e to f.”
What it’s really asking
For part (a), square the linear scale factor between the circles' circumferences to get the area ratio. For part (b), square root the area scale factor between the squares to get the linear ratio of their sides.
The full worked answer — June 2019
Since the circumference of circle B is 90% of circle A's, the linear scale factor from A to B is 0.9, or 10 to 9 as a ratio, since circumference scales directly with radius.
Squaring this linear scale factor gives the area scale factor, 10 squared to 9 squared, which is 100 to 81, so the ratio of the area of circle A to the area of circle B is 100 to 81.
Since the area of square E is 44% greater than square F's, the area scale factor from F to E is 1.44.
Square rooting this area scale factor gives the linear scale factor, the square root of 1.44, which is 1.2, or 6 to 5 as a ratio, so e to f is 6 to 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 similar shapes and scale factor questionsWhat the mark scheme rewarded
- A correct linear scale factor of 0.9, or 10 to 9, found for part (a) (1 mark)
- The correct area ratio, 100 to 81, found by squaring (1 mark)
- A correct area scale factor of 1.44 identified for part (b) (1 mark)
- The correct final ratio, e to f equal to 6 to 5, found by square rooting (1 mark)
- Square a linear scale factor to move to an area scale factor, and square root an area scale factor to move back to a linear one, never the reverse
- Convert a percentage increase or decrease into a decimal scale factor before applying the squared relationship, since 44% greater means a scale factor of 1.44, not 0.44
Traps examiners saw
- Using the 90% or 44% directly as the final area ratio, without squaring or square rooting to move between linear and area scale factors
- Squaring when a square root was needed, or the reverse, by not checking whether the question started with a linear or an area fact
Full-mark self-check 0 of 4
1×asked“Here are three similar triangles, ABG, ACF and ADE. ABCD and AGFE are straight lines, and the ratio AB to BC to CD is 1 to 2 to 3. Show that the ratio of the area of triangle ABG to the area of the region BCFG to the area of the region CDEF is 1 to 8 to 27.”
“Here are three similar triangles, ABG, ACF and ADE. ABCD and AGFE are straight lines, and the ratio AB to BC to CD is 1 to 2 to 3. Show that the ratio of the area of triangle ABG to the area of the region BCFG to the area of the region CDEF is 1 to 8 to 27.”
What it’s really asking
Find the ratio of the full lengths AB, AC and AD, square them to get the area ratio of the three similar triangles, then subtract to find the two regions between them.
The full worked answer — June 2023
Since AB to BC to CD is 1 to 2 to 3, the full lengths from A are AB equal to 1, AC equal to AB plus BC which is 3, and AD equal to AB plus BC plus CD which is 6, giving AB to AC to AD equal to 1 to 3 to 6.
Since the three triangles are similar, their areas scale with the square of these linear ratios, so the area of triangle ABG to ACF to ADE is 1 squared to 3 squared to 6 squared, which is 1 to 9 to 36.
The area of region BCFG is the area of triangle ACF minus the area of triangle ABG, which is 9 minus 1, giving 8, and the area of region CDEF is the area of triangle ADE minus the area of triangle ACF, which is 36 minus 9, giving 27, so the ratio of area ABG to BCFG to CDEF is 1 to 8 to 27, 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 similar shapes and scale factor questionsWhat the mark scheme rewarded
- A correct process to find the ratio of the full lengths AB to AC to AD, such as 1 to 3 to 6 (1 mark)
- A correct process to square this ratio to find the area ratio of the three triangles, such as 1 to 9 to 36 (1 mark)
- Correct working leading to the final answer, 1 to 8 to 27, by subtracting the nested areas (1 mark)
- Convert the given ratio of separate segments, AB, BC and CD, into a ratio of the full lengths from the shared vertex A, before squaring anything
- Subtract consecutive terms of the squared ratio to find the area of each ring-shaped region between the nested triangles
Traps examiners saw
- Squaring the original segment ratio, 1 to 2 to 3, directly, instead of first converting it to the full lengths from A, 1 to 3 to 6
- Forgetting to subtract the nested triangle areas, and stating the triangle area ratio, 1 to 9 to 36, as the final answer instead of 1 to 8 to 27
Full-mark self-check 0 of 3
The method for every Q9 (Jun19) / 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.
- Recognising whether the given information is a linear ratio or an area ratio
- Squaring a linear scale factor to reach an area scale factor, or square rooting an area scale factor to reach a linear one
- Converting a percentage difference into the correct scale factor before applying the squared relationship
The steps
- Identify whether the given ratio, or percentage, describes a linear measurement or an area
- Convert a percentage into a decimal scale factor if needed
- Square the linear scale factor to find the area scale factor, or square root an area scale factor to find the linear one
- Write the final answer as the ratio the question specifically asks for
Which of the following statements correctly describes an enlargement?
Using the squared relationship between linear and area scale factors for similar shapes comes up in two of the three sittings we have. Practise converting a percentage difference into a scale factor before deciding whether to square or square root it.
Practise similar shapes and scale factor questions“Two of the three sittings we have full papers for combine the probability of the same event happening twice, once using a given probability with a tree diagram, and once using an estimated probability from a frequency table.”
Every version needs the two trials treated as independent, multiplying probabilities together for 'both', or using the complement for 'at least one'.
1×asked“Mary travels to work by train every day. The probability that her train will be late on any day is 0.15. (a) Complete the probability tree diagram for Thursday and Friday. (b) Work out the probability that her train will be late on at least one of these two days.”
“Mary travels to work by train every day. The probability that her train will be late on any day is 0.15. (a) Complete the probability tree diagram for Thursday and Friday. (b) Work out the probability that her train will be late on at least one of these two days.”
What it’s really asking
Complete the tree with the 'not late' probability of 0.85 on every branch, then use the complement to find the probability of at least one late day.
What the sources actually showed — June 2019
A probability tree diagram for two days, Thursday and Friday, with the branch for the train being late on Thursday labelled 0.15, and the remaining three branches, not late on Thursday, late on Friday, and not late on Friday, left blank to be completed.
The full worked answer — June 2019
Since the probability of the train being late is 0.15, the probability of it not being late is 1 minus 0.15, which is 0.85, so every 'not late' branch on the tree diagram is labelled 0.85, and every 'late' branch is labelled 0.15.
The probability that the train is late on at least one of the two days is 1 minus the probability that it is not late on both days, which is 1 minus (0.85 times 0.85), giving 1 minus 0.7225.
This gives a final probability of 0.2775.
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 combined probability questionsWhat the mark scheme rewarded
- The tree diagram correctly completed with 0.85 on every 'not late' branch (1 mark)
- A correct product for one branch of the tree, such as 0.85 times 0.85 (1 mark)
- A correct complete method for 'at least one', such as 1 minus 0.85 times 0.85 (1 mark)
- The correct final answer, 0.2775 (1 mark, following on from the tree diagram mark)
- Use 1 minus the probability of the train being 'not late' on both days, rather than trying to add up every combination that includes at least one late day
- Multiply along the branches of the tree diagram for two independent events, since Thursday's result does not affect Friday's probability
Traps examiners saw
- Adding 0.15 and 0.15 together to find 'at least one', which is not how probabilities of independent events combine
- Forgetting to complete the tree diagram with 0.85 first, making the multiplication in part (b) harder to check
Full-mark self-check 0 of 4
1×asked“A biased dice is thrown 60 times. The table shows the number the dice lands on each time: the number 1 occurs 12 times, 2 occurs 7 times, 3 occurs 8 times, 4 occurs 9 times, 5 occurs 9 times, and 6 occurs 15 times. Gethin throws the dice twice. (a) Work out an estimate for the probability that the dice lands on 6 both times. Sally is going to throw the same dice n times and use her results to work out a more reliable estimate for the probability in part (a). (b) What can you say about the value of n?”
“A biased dice is thrown 60 times. The table shows the number the dice lands on each time: the number 1 occurs 12 times, 2 occurs 7 times, 3 occurs 8 times, 4 occurs 9 times, 5 occurs 9 times, and 6 occurs 15 times. Gethin throws the dice twice. (a) Work out an estimate for the probability that the dice lands on 6 both times. Sally is going to throw the same dice n times and use her results to work out a more reliable estimate for the probability in part (a). (b) What can you say about the value of n?”
What it’s really asking
Estimate the probability of landing on 6 once using relative frequency, then square it for both throws landing on 6.
What the sources actually showed — June 2023
A frequency table showing the number of times a biased dice landed on each number from 1 to 6 across 60 throws, with frequencies 12, 7, 8, 9, 9 and 15 for the numbers 1 to 6 respectively.
| Number on dice | Frequency |
|---|---|
| 1 | 12 |
| 2 | 7 |
| 3 | 8 |
| 4 | 9 |
| 5 | 9 |
| 6 | 15 |
The full worked answer — June 2023
The relative frequency of landing on 6 is 15 out of 60 throws, which simplifies to one quarter, and this is used as the estimated probability of landing on 6 on a single throw.
Since the two throws are independent, the probability of landing on 6 both times is one quarter times one quarter, which is one sixteenth, or 225 out of 3600.
For part (b), Sally's estimate becomes more reliable with a larger number of throws, so n must be greater than the 60 throws already used in part (a).
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 combined probability questionsWhat the mark scheme rewarded
- The relative frequency of landing on 6 correctly found as 15 over 60, or one quarter (1 mark)
- A correct process to combine this probability for two throws, such as one quarter times one quarter (1 mark)
- The correct final answer, one sixteenth, or 225 over 3600, or an equivalent value (1 mark)
- A correct explanation for part (b), that n must be greater than 60 (1 mark)
- Use relative frequency, the given frequency divided by the total number of throws, since the dice is biased and a theoretical probability of one sixth cannot be assumed
- Multiply the single-throw estimate by itself for two independent throws landing on the same number, rather than adding it
Traps examiners saw
- Using a theoretical probability of one sixth, forgetting the dice is specifically described as biased
- Adding the two throws' probabilities together instead of multiplying them, which does not represent both throws landing on 6
Full-mark self-check 0 of 4
The method for every Q10 (Jun19) / Q10 (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.
- Recognising that repeated trials of the same event are independent, so their probabilities can be multiplied
- Choosing between multiplying directly for 'both' happening, and using the complement, one minus neither, for 'at least one' happening
- Estimating a probability from a frequency table correctly, as relative frequency, when no probability is given directly
The steps
- Find or estimate the probability of the single event happening once
- Decide whether the question asks for 'both', 'at least one', or a specific combination, since each needs a different method
- Multiply or use the complement as appropriate, checking every branch or outcome needed is included
- Give the final probability as a fraction, decimal or percentage, matching the form the question expects
A fair coin is flipped and a fair die is rolled. What rule is used to find P(heads AND rolling a 3)?
Combining the probability of two independent, repeated events comes up in two of the three sittings we have. Practise deciding between multiplying for 'both' and using the complement for 'at least one'.
Practise combined probability questionsHow often each recurring skill has actually appeared
Counted only across the three sittings we have full papers for. A skill marked 'every sitting' appeared in all three; a skill marked 'two of the three sittings' appeared in exactly two.
Not seen as a standalone, cleanly repeating question on Paper 2 in the three sittings we have full papers for
Simplifying a power of a power, expanding brackets, and factorising fully, tested only in June 2022 · Describing a single transformation from a grid, tested only in June 2022 · Comparing the area available per person at two real-life events, and explaining a unit-conversion error for area, tested only in June 2022 · Finding a coordinate using a ratio along a straight line between two given points, tested only in June 2022 · Comparing prices in two different currencies and measurement units using conversion facts, tested only in June 2022 · Using a calculator to evaluate an expression combining fractional powers and a trigonometric ratio, tested only in June 2022 · Drawing a box plot from a summary table and comparing two distributions using the median and a measure of spread, tested only in June 2022 · Enlarging a shape by a negative scale factor, tested only in June 2022 · Estimating a population total using the capture-recapture method, tested only in June 2022 · Completing a Venn diagram from given set information and finding a conditional probability from it, tested only in June 2022 · Solving a quadratic inequality algebraically, tested only in June 2022 · Finding an equation of a straight line perpendicular to a given line and passing through a given point, tested only in June 2019 (the given point's coordinates were corrupted beyond confident reconstruction in the extracted text, and are flagged separately as a risk area) · Estimating the gradient of a speed-time graph and the distance travelled using the trapezium rule, tested only in June 2019 · Evaluating a calculator expression and finding the reciprocal of a decimal, tested only in June 2023 · Writing a number as a product of its prime factors, tested only in June 2023 · Forming and solving an equation from the areas of a triangle and a rectangle, tested only in June 2023 · Solving a pair of simultaneous equations algebraically, tested only in June 2023 · Finding the circumference of a circle using Pythagoras' theorem on an inscribed rectangle, tested only in June 2023 · Finding the nth term of a quadratic sequence, tested only in June 2023 · Completing a histogram from given information and finding a total frequency from it, tested only in June 2023 · Using an iteration formula three times to estimate a solution to an equation, tested only in June 2023 · Writing an algebraic expression for the probability of at least one of two outcomes across n repeated trials, tested only in June 2023 · Finding an area involving regular octagons in a surd form, tested only in June 2023 (this specific diagram could not be reconstructed with full confidence from the extracted text, and is flagged separately as a risk area rather than authored) · Drawing a straight line graph from its equation using a table of values, tested only in June 2019 · Estimating a proportion from a small sample and stating an assumption made, tested only in June 2019 · Writing a number in standard form, tested only in June 2019 · Finding the area and perimeter of a sector of a circle, tested only in June 2019 · Showing that enough workers finish a task within a given time using proportion, tested only in June 2022 · Showing that two given lines are parallel by comparing their gradients, tested only in June 2022 · Working with two functions to evaluate one and find a composite inverse function, tested only in June 2022 · Proving a ratio result between combined quantities using algebra, tested only in June 2023 · Finding the lower bound of an angle from triangle side lengths given to a rounded degree of accuracy, tested only in June 2023
These topics genuinely appeared in at least one of the three 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 2 always have the same structure?
Yes, in all three sittings we have full papers for. Every sitting totalled 80 marks in 1 hour 30 minutes, and every paper covered a similarly broad spread of number, algebra, ratio, geometry, probability and statistics questions, with a calculator required throughout. Always check your own paper's front cover to confirm, since Pearson can make real changes in any future series.
Is a calculator allowed on Paper 2?
Yes, in all three sittings we have full papers for. Paper 2 is explicitly the calculator paper, which is why questions on this page involve trigonometry, compound percentage change, and reading estimates from statistical graphs, all of which are far more practical with a calculator to hand.
Why is there no June 2018, June 2020 or June 2021 paper on this page?
June 2018's paper could not be located in Pearson's public past paper archive, so we could not verify it against a real question paper and mark scheme. June 2020 and June 2021 do not exist as normal sittings at all, since GCSE exams were cancelled in both years due to the pandemic. Our three sittings, 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. June 2019 does not mention a Formulae Sheet in its materials list, meaning more formulae had to be memorised in that earlier sitting. 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 process marks (P) and method marks (M), which reward a correct approach even if the final answer is wrong, accuracy marks (A), which reward the correct final value following a correct method, and communication marks (C), which reward a fully correct statement or reason with no contradiction. Showing your working matters a great deal, since several questions on these three sittings specifically state that no marks are awarded for a correct answer with no supporting working shown.
What is the single biggest way marks are lost on this paper?
According to the real mark schemes for these three sittings, marks are very often lost by skipping the working on a question that specifically requires it, since a correct final answer with no supporting working scores zero on several of these questions. On circle theorem and reasoning questions, marks are also commonly lost by giving a correct angle but the wrong, or an unlinked, reason for it.
Practise the questions that are guaranteed to come up
Every skill on this page has practice questions waiting in the app, built the way Edexcel actually structures Paper 2.
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