OCR GCSE Mathematics: Direct and inverse proportion, and growth and decay (J560) – 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 follow the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026), sections 5.02 and 5.03. Questions 1 to 7 suit both tiers. Questions 8 to 11 are marked (Higher tier only) because they use content the specification lists only for Higher tier.

Support pages: the study guide, the revision notes, the OCR GCSE Mathematics hub and the checklist.

Questions

1. The exchange rate is £1 = US$1.30.

(a) Convert US$650 to pounds. [2] (b) Convert £240 to US dollars. [2]

2. y is directly proportional to x. When x = 6.5, y = 26.

(a) Find an equation connecting y and x. [2] (b) Find y when x = 9.25. [1] (c) Find x when y = 15. [1]

3. A coach driver covers a route in 2 hours 30 minutes at an average speed of 84 km/h.

(a) Find the time taken at an average speed of 105 km/h. [2] (b) Find the average speed needed to take 1 hour 45 minutes. [2]

4. y is inversely proportional to x. When x = 8, y = 7.5.

(a) Find an equation connecting y and x. [2] (b) Find y when x = 25. [1] (c) Find x when y = 0.6. [1]

5. Sana invests £4,200 at 3.2% per year.

(a) Find the simple interest earned in 5 years. [2] (b) Find how much more interest she would earn in 5 years if the 3.2% were compound interest instead. [3]

6. A boat costs £34,000 new. Its value falls by 18% in the first year, 12% in the second year and 9% in the third year, each as a percentage of the value at the start of that year.

(a) Find its value after three years. [3] (b) Find the overall percentage decrease, correct to 3 significant figures. [2]

7. Compound interest of 4% per year is added to a savings account. After 2 years the account holds £2,704. How much was originally paid in? [3]

8. (Higher tier only) The power P kW produced by a wind turbine is proportional to the cube of the wind speed v m/s. P = 27 when v = 6.

(a) Find a formula for P in terms of v. [2] (b) Find P when v = 10. [1] (c) Find v when P = 64. [2]

9. (Higher tier only) The brightness B of a lamp is inversely proportional to the square of the distance d metres from it. B = 50 when d = 6.

(a) Find a formula for B in terms of d. [2] (b) Find B when d = 15. [1] (c) Find d when B = 2. [2]

10. (Higher tier only) y is directly proportional to the square root of x. y = 10.5 when x = 49. The quantity z is inversely proportional to the square root of x. z = 2 when x = 36.

(a) Write y in terms of x. [2] (b) Find y when x = 64. [1] (c) Find the value of x for which y = z. [3]

11. (Higher tier only) The value V pounds of a machine n years after it was bought is V = 45,000 × 0.85ⁿ.

(a) State the starting value and the percentage the value falls by each year. [2] (b) Find the value after 4 years, to the nearest pound. [2] (c) Find the number of whole years after which the value is first less than half of its starting value. Show your trials. [3]

Answers

1. (a) Dividing to get pounds: 650 ÷ 1.30 [1] = £500 [1] (b) Multiplying to get dollars: 240 × 1.30 [1] = US$312 [1] [4] Examiner insight: Answers tend to gain credit when you show the operation and the rate. A quick size check helps: dollars should be the larger number here, so (b) must exceed 240.

2. (a) 26 = k × 6.5 [1], so k = 4 and y = 4x [1] (b) y = 4 × 9.25 = 37 [1] (c) 15 = 4x, so x = 3.75 [1] [4] Examiner insight: An equation with the letter k left in usually does not answer “find an equation”. Write k as a number in the final line.

3. (a) Distance = 84 × 2.5 = 210 km [1]. Time = 210 ÷ 105 = 2 hours [1] (b) 1 hour 45 minutes = 1.75 hours [1]. Speed = 210 ÷ 1.75 = 120 km/h [1] [4] Examiner insight: Converting minutes to a decimal is a frequent slip. 45 minutes is 0.75 hours, not 0.45. Say the units in each answer.

4. (a) k = xy = 8 × 7.5 = 60 [1], so y = 60/x [1] (b) y = 60/25 = 2.4 [1] (c) 0.6 = 60/x, so x = 60 ÷ 0.6 = 100 [1] [4] Examiner insight: Check direction: y falls as x rises, so 2.4 is smaller than 7.5 and 100 is bigger than 8. Wrong-direction answers suggest the direct method was used.

5. (a) Interest = 4200 × 3.2 × 5 / 100 [1] = £672 [1] (b) Compound: 4200 × 1.032⁵ [1] = £4,916.41, so interest = £716.41 [1]. Difference = 716.41 − 672 = £44.41 [1] [5] Examiner insight: Keep unrounded values until the end. Rounding 1.032⁵ early can change the penny figure in a “how much more” question.

6. (a) Multipliers 0.82, 0.88 and 0.91 [1]. Value = 34000 × 0.82 × 0.88 × 0.91 [1] = £22,326.30 [1] (b) Overall multiplier = 0.82 × 0.88 × 0.91 = 0.656656 [1]. Decrease = 1 − 0.656656 = 0.343344, which is 34.3% (3 s.f.) [1] [5] Examiner insight: Adding 18 + 12 + 9 = 39% is the common wrong route. A single overall multiplier is the clearest way to show the method.

7. After 2 years the account is multiplied by 1.04² [1] = 1.0816. Original = 2704 ÷ 1.0816 [1] = £2,500 [1] [3] Examiner insight: Working backwards means dividing by the multiplier. Taking 4% off £2,704 twice is a different calculation and gives a different value.

8. (a) P = kv³ [1]. 27 = k × 216, so k = 0.125 and P = 0.125v³ [1] (b) P = 0.125 × 1000 = 125 [1] (c) 64 = 0.125v³ [1], so v³ = 512 and v = 8 [1] [5] Examiner insight: Start with P = kv³ so the cube is visible in your working. Writing P = kv gives a different k and a wrong later answer.

9. (a) B = k/d² [1]. 50 = k/36, so k = 1800 and B = 1800/d² [1] (b) B = 1800 ÷ 225 = 8 [1] (c) 2 = 1800/d², so d² = 900 [1] and d = 30 [1] [5] Examiner insight: After finding d² you still need the square root. Stopping at 900 is an easy way to lose credit. A distance is positive, so reject −30.

10. (a) y = k√x [1]. 10.5 = 7k, so y = 1.5√x [1] (b) y = 1.5 × 8 = 12 [1] (c) z = k/√x with 2 = k/6, so k = 12 [1]. Set 1.5√x = 12/√x, so 1.5x = 12 [1], giving x = 8 [1] [6] Examiner insight: Find both constants before comparing. Multiplying both sides by √x removes the fractions and keeps the equation simple.

11. (a) The starting value is £45,000 [1] and 0.85 = 1 − 0.15, so the value falls by 15% each year [1] (b) V = 45000 × 0.85⁴ [1] = 23,490.28, which is £23,490 to the nearest pound [1] (c) Half the starting value is 22,500 [1]. n = 4 gives £23,490, which is above 22,500. n = 5 gives £19,967, which is below [1]. So it is first less than half after 5 years [1] [7] Examiner insight: A table of trial values makes the stopping point clear. Check the value just before as well as the one that passes the target. State the answer in whole years.

Where marks are usually lost

• Using a direct method for an inverse problem. Check which way y moves when x increases.
• Leaving k unknown in the final equation, or finding k and not using it.
• Not converting minutes to decimal hours in speed and time questions.
• Multiplying by an exchange rate when the question needs division, and not checking the size of the answer.
• Adding percentages across years instead of multiplying multipliers.
• Rounding an interim multiplier or value, which shifts the last digits.
• Forgetting to take the final square root, or to reject a negative length.
• Giving a decay multiplier as the percentage itself, for example 0.15 instead of 0.85.
• Not interpreting a formula or a trial result in the words of the question.

Next steps

• Recap quickly with the revision notes
• Re-read the worked examples in the study guide
• Tick off topics on the OCR GCSE Mathematics checklist and visit the course hub
• Try all free 10-minute diagnostics at /diagnostics/
• Book a free trial class

Official syllabus

OCR, GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026), first assessment 2017, sections 5.02a, 5.02b and 5.03a.