Edexcel International GCSE Mathematics A 4MA1: Proportion – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 2.5 Proportion of Topic 2 (Equations, formulae and identities) in the Pearson Edexcel International GCSE Mathematics A (4MA1) specification, Issue 2 (November 2017), for the January and June series examined on it. Section 2.5 is Higher tier only, so every question here is labelled (Higher). A calculator may be used on every 4MA1 paper, but show your working: most marks are for method.

Learn the content first in the study guide and the revision notes. The course hub is Edexcel IGCSE Mathematics and the printable checklist lists every statement. For rearranging practice, use the algebraic manipulation practice.

Questions

1. (Higher) y is directly proportional to x. When x = 8, y = 18.

(a) Find a formula for y in terms of x. [2] (b) Calculate the value of y when x = 12. [1]

2. (Higher) The time, T hours, to paint a hall is inversely proportional to the number of painters, n. Four painters take 15 hours.

(a) Find a formula for T in terms of n. [2] (b) Calculate how long six painters take. [1] (c) Calculate how many painters are needed to finish in 2.4 hours. [1]

3. (Higher) In a model of a falling object, the distance fallen, d metres, is directly proportional to the square of the time, t seconds. When t = 2, d = 19.6.

(a) Find a formula for d in terms of t. [2] (b) Calculate d when t = 5. [1] (c) Calculate t when d = 78.4. [2]

4. (Higher) y is inversely proportional to the square of x, where x > 0. When x = 5, y = 12.

(a) Find a formula for y in terms of x. [2] (b) Calculate y when x = 10. [1] (c) Calculate x when y = 48. [2]

5. (Higher) The brightness, I units, of a lamp seen from a distance d metres is inversely proportional to d². At 2 metres the brightness is 75 units. Calculate the brightness at 5 metres. [3]

6. (Higher) y is directly proportional to √x. When x = 36, y = 18.

(a) Calculate y when x = 100. [3] (b) Calculate x when y = 7.5. [2]

7. (Higher) Four graphs of y against x are described for x ≥ 0 and positive k.

• Graph A: a straight line through the origin.
• Graph B: a curve through the origin that gets steeper as x increases.
• Graph C: a curve through the origin that gets less steep as x increases.
• Graph D: a curve that falls as x increases and gets close to both axes without touching them.

(a) Match each graph to one of these relationships: y ∝ x², y ∝ 1/x, y ∝ x, y ∝ √x. [4] (b) Graph A passes through (5, 8). Find the equation of graph A. [2]

8. (Higher)

(a) y is directly proportional to x³. x is increased by 50%. Calculate the percentage increase in y. [2] (b) p is inversely proportional to q². q is decreased by 20%. Calculate the percentage change in p. [3]

9. (Higher) The table shows values of p and q.

q	2	3	6	a
p	27	12	3	0.75

(a) Show that p is inversely proportional to q². [2] (b) Write a formula for p in terms of q. [1] (c) Find the value of a, given a > 0. [2]

10. (Higher) y is directly proportional to x². When x = 5, y = 50. x is inversely proportional to √z. When z = 4, x = 6. All values are positive.

(a) Find a formula for y in terms of x. [2] (b) Find a formula for x in terms of z. [2] (c) Show that y = 288/z. [2] (d) Calculate z when y = 4.5. [2]

11. (Higher) In a model, the time, t seconds, for one swing of a pendulum is directly proportional to the square root of its length, l cm. A pendulum of length 36 cm takes 1.2 seconds.

(a) Find a formula for t in terms of l. [2] (b) Calculate t when l = 81. [1] (c) Calculate the length that gives a swing time of 2.5 seconds. [2] (d) The length of a pendulum is increased by 44%. Calculate the percentage increase in its swing time. [2] (e) Describe the shape of the graph of t against l. [1]

12. (Higher) A fixed mass of steel is made into identical ball bearings. The number made, N, is inversely proportional to the cube of the radius, r mm. When r = 3, N = 4000.

(a) Find a formula for N in terms of r. [2] (b) Calculate N when r = 6. [1] (c) Calculate r when N = 32 000. [2] (d) State the effect on N of doubling the radius. [1]

Answers

1. (a) y = kx and 18 = 8k [1]; k = 2.25, so y = 2.25x [1] (b) 2.25 × 12 = 27 [1] Examiner insight: The second mark in (a) needs the formula with the number in it; “k = 2.25” alone gains only the method mark.

2. (a) T = k/n and 15 = k/4 [1]; k = 60, so T = 60/n [1] (b) 60/6 = 10 hours [1] (c) 2.4 = 60/n, n = 60/2.4 = 25 painters [1] Examiner insight: Using T = kn here gives k = 3.75 and scores no marks; the words “inversely proportional” are where the method mark is won or lost.

3. (a) d = kt² and 19.6 = 4k [1]; k = 4.9, so d = 4.9t² [1] (b) 4.9 × 5² = 122.5 m [1] (c) 78.4 = 4.9t², so t² = 16 [1]; t = 4 s [1] Examiner insight: Stopping at t² = 16 earns the method mark only; the accuracy mark needs the square root taken.

4. (a) y = k/x² and 12 = k/25 [1]; k = 300, so y = 300/x² [1] (b) 300/100 = 3 [1] (c) 48 = 300/x², so x² = 300/48 = 6.25 [1]; x = 2.5 [1] Examiner insight: A common slip in (c) is x² = 48/300 = 0.16, which inverts the rearrangement and loses both marks; write x² = 300/48 before using the calculator.

5. I = k/d² and 75 = k/4 [1]; k = 300 [1]; I = 300/25 = 12 units [1] Examiner insight: The scale factor method also earns full marks: d is multiplied by 2.5, so I is divided by 2.5² = 6.25, giving 75 ÷ 6.25 = 12; both methods need the square to score.

6. (a) y = k√x and 18 = 6k [1]; k = 3 [1]; y = 3√100 = 30 [1] (b) 7.5 = 3√x, so √x = 2.5 [1]; x = 6.25 [1] Examiner insight: Taking k = 18/36 = 0.5 (forgetting the square root) loses the method mark and everything that follows from it.

7. (a) A: y ∝ x [1]; B: y ∝ x² [1]; C: y ∝ √x [1]; D: y ∝ 1/x [1] (b) y = kx and 8 = 5k [1]; y = 1.6x [1] Examiner insight: Each match is marked separately, so one swap (B and C, say) costs two marks; check “steeper” against “less steep” before you write.

8. (a) 1.5³ = 3.375 [1]; increase = 237.5% [1] (b) q is multiplied by 0.8, so p is multiplied by 1/0.8² [1]; 1/0.64 = 1.5625 [1]; p increases by 56.25% [1] Examiner insight: In (a), 3.375 is the multiplier, not the answer; the accuracy mark needs it converted to a percentage increase, and “337.5%” is wrong.

9. (a) p × q² = 27 × 4 = 108, 12 × 9 = 108, 3 × 36 = 108 [1]; the product is constant, so p is inversely proportional to q² [1] (b) p = 108/q² [1] (c) 0.75 = 108/a², a² = 144 [1]; a = 12 [1] Examiner insight: A “show that” needs every product shown and a conclusion; testing only one pair of values does not show a constant and scores at most one mark.

10. (a) y = kx² and 50 = 25k [1]; y = 2x² [1] (b) x = c/√z and 6 = c/2 [1]; x = 12/√z [1] (c) y = 2 × (12/√z)² [1]; = 2 × 144/z = 288/z [1] (d) 4.5 = 288/z [1]; z = 288/4.5 = 64 [1] Examiner insight: In (c) the method mark is for substituting the whole expression for x and squaring it; writing y = 288/z without the line 2 × 144/z gains nothing, because the answer is given.

11. (a) t = k√l and 1.2 = 6k [1]; k = 0.2, so t = 0.2√l [1] (b) 0.2 × 9 = 1.8 s [1] (c) 2.5 = 0.2√l, √l = 12.5 [1]; l = 156.25 cm [1] (d) l multiplied by 1.44, so t multiplied by √1.44 = 1.2 [1]; 20% increase [1] (e) A curve through the origin that gets less steep as l increases [1] Examiner insight: In (d), answering 44% (as if t were directly proportional to l) gains no credit; the method mark is for applying the square root to the multiplier.

12. (a) N = k/r³ and 4000 = k/27 [1]; k = 108 000, so N = 108 000/r³ [1] (b) 108 000/216 = 500 [1] (c) r³ = 108 000/32 000 = 3.375 [1]; r = 1.5 mm [1] (d) N is divided by 8 (2³ = 8) [1] Examiner insight: In (c) the cube root is needed for the accuracy mark; giving r = 3.375 or using a square root (1.84) earns only the method mark for r³ = 3.375.

Where marks are usually lost

• Writing a constant that is added (y = x² + k) instead of multiplied (y = kx²).
• Missing “inversely” or “square” in the wording and choosing the wrong form.
• Giving the value of k when the question asks for a formula in terms of x.
• Leaving t² = 16 or √x = 2.5 without finishing the last step.
• Rearranging an inverse formula upside down, so x² = y/k instead of x² = k/y.
• Quoting a multiplier (3.375, 1.5625) instead of the percentage change asked for.
• Testing a single pair of values in a “show that” question about a table.
• Adding the percentage change instead of using a power of the multiplier.
• Describing a graph without saying whether it passes through the origin, or whether it gets steeper or less steep.

Next steps

• Refresh the formulae: revision notes
• Re-learn anything you missed: study guide
• More Topic 2 work: use of symbols and algebraic manipulation
• Course hub: Edexcel IGCSE Mathematics
• Track coverage: printable checklist
• Check other topics: all free 10-minute diagnostics
• Want help from a teacher? Book a free trial class

Official syllabus

Pearson Edexcel International GCSE in Mathematics (Specification A) (4MA1), Specification, Issue 2, November 2017, Pearson Education Limited (first assessment June 2018). Topic 2, Equations, formulae and identities, section 2.5 Proportion (Higher tier only).