OCR GCSE Mathematics: Algebraic formulae, rearranging and kinematics (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 OCR GCSE (9-1) Mathematics J560, Version 2.1 (August 2026), section 6.02, Algebraic formulae (6.02a to 6.02e). Questions 1 to 8 and 11 are for both tiers. Questions 9 and 10 use content that is Higher tier only and are labelled. The specification says no formula sheet is provided, so recall the standard formulae before you start. Show your working on every question.

Learn the method first in the study guide or the revision notes. See also the OCR GCSE Mathematics hub, the printable checklist and the free 10-minute diagnostics.

Questions

1. A hall costs £90 to hire plus £6.50 for each guest.

(a) Write a formula for the total cost £C for n guests. [2] (b) Find the cost for 48 guests. [1] (c) A club has a budget of £500. Find the greatest number of guests it can afford. [2]

2. Work out the value of each expression.

(a) P = 3a − 2b when a = 4 and b = −3. [2] (b) y = x³ − 3x² + 2 when x = −2. [2] (c) T = (p − q²)/(2q) when p = 7 and q = −3. [2]

3. Rearrange each formula.

(a) Make x the subject of y = 4x − 9. [2] (b) Make w the subject of V = lwh. [1] (c) Make f the subject of c = 5(f − 32)/9. [3]

4. Rearrange each formula.

(a) Make r the subject of A = πr². [2] (b) Make a the subject of v² = u² + 2as. [2] (c) Make x the subject of y = 5√(x − 1). [3]

5. Rearrange each formula.

(a) Make x the subject of y = (x + 4)/(x − 2). [4] (b) Make n the subject of 2/k = 1/m + 3/n. [3]

6. A circular lawn has diameter 9.6 m.

(a) Find its circumference to 3 significant figures. [2] (b) Find its area to 3 significant figures. [2] (c) A different circle has area 64π cm². Find its circumference in terms of π. [3]

7. A ramp rises 0.45 m over a horizontal distance of 5.4 m.

(a) Find the length of the ramp to 3 significant figures. [3] (b) Find the angle the ramp makes with the horizontal to 3 significant figures. [3]

8. Use the constant-acceleration formulae.

(a) A car travelling at 8 m/s accelerates at 1.2 m/s² for 10 s. Find its final velocity and the distance it travels in this time. [4] (b) A cyclist brakes from 20 m/s and stops after 80 m. Find the acceleration. [3]

9. (Higher tier only) Use the quadratic formula.

(a) Solve 3x² − 8x − 2 = 0. Give both answers to 2 decimal places. [4] (b) An object has initial velocity 4 m/s and constant acceleration 2 m/s². Find the time it takes to travel 30 m, to 2 decimal places. [4]

10. (Higher tier only) In triangle PQR, PQ = 8.4 cm, PR = 6.8 cm and angle QPR = 63°.

(a) Find QR to 3 significant figures. [3] (b) Find angle PQR to 1 decimal place. [3] (c) Find the area of the triangle to 3 significant figures. [2]

11. A stone is dropped from rest from a bridge. Take downwards as positive and a = 9.8 m/s².

(a) Show that the distance fallen is s = 4.9t² after t seconds. [1] (b) Make t the subject of s = 4.9t². [2] (c) Find the time taken to fall 44.1 m. [2] (d) Find the speed of the stone at that moment. [2] (e) Use v² = u² + 2as to check your answer to part (d). [2]

Answers

1. (a) Fixed charge 90 [1]; 6.5n for the guests, so C = 90 + 6.5n [1] (b) 90 + 6.5 × 48 = 90 + 312 = £402 [1] (c) 90 + 6.5n ≤ 500 gives n ≤ 410/6.5 = 63.08 [1]; so 63 guests (the next guest would take the cost over £500) [1] [5] Examiner insight: In part (c), a decimal such as 63.08 needs interpreting. Write the whole-number conclusion in words so the context is answered.

2. (a) 3(4) − 2(−3) = 12 + 6 [1]; 18 [1] (b) (−2)³ − 3(−2)² + 2 = −8 − 12 + 2 [1]; −18 [1] (c) Top 7 − 9 = −2, bottom 2(−3) = −6 [1]; T = 1/3 [1] [6] Examiner insight: Write the substituted line with brackets around each negative value. If a sign slips later, a clear substitution line still shows the method.

3. (a) y + 9 = 4x [1]; x = (y + 9)/4 [1] (b) w = V/(lh) [1] (c) 9c = 5(f − 32) [1]; 9c/5 = f − 32 [1]; f = 9c/5 + 32 [1] [6] Examiner insight: Make one change per line and keep the equals signs aligned, so each step can be followed. Two operations at once is where errors creep in.

4. (a) A/π = r² [1]; r = √(A/π) [1] (b) v² − u² = 2as [1]; a = (v² − u²)/(2s) [1] (c) y/5 = √(x − 1) [1]; y²/25 = x − 1 [1]; x = y²/25 + 1 [1] [7] Examiner insight: In part (c) isolate the root before squaring. Squaring 5√(x − 1) as it stands would also need the 5 squared, which is easy to forget.

5. (a) y(x − 2) = x + 4 [1]; xy − 2y = x + 4 [1]; xy − x = 4 + 2y [1]; x = (2y + 4)/(y − 1) [1] (b) 3/n = 2/k − 1/m = (2m − k)/(km) [1]; n/3 = km/(2m − k) [1]; n = 3km/(2m − k) [1] [7] Examiner insight: When the subject appears twice, the factorising line (x(y − 1) = 2y + 4) is the key step. Writing it out shows you have collected the subject terms correctly.

6. (a) C = π × 9.6 [1]; 30.2 m [1] (b) r = 4.8, so A = π × 4.8² [1]; 72.4 m² [1] (c) πr² = 64π, so r = 8 [1]; C = 2π × 8 [1]; 16π cm [1] [7] Examiner insight: Part (a) gives the diameter, so use C = πd. The radius is needed for the area. Units differ too: m for length, m² for area.

7. (a) l² = 0.45² + 5.4² [1]; l² = 29.3625 [1]; l = 5.42 m [1] (b) tan θ = 0.45/5.4 [1]; θ = tan⁻¹(1/12) [1]; θ = 4.76° [1] [6] Examiner insight: A sketch with the right angle and sides labelled hypotenuse, opposite and adjacent helps you choose between Pythagoras and a trigonometric ratio. Use the original values in part (b) and not the rounded length from part (a).

8. (a) v = 8 + 1.2 × 10 [1]; v = 20 m/s [1]; s = 8 × 10 + ½ × 1.2 × 10² [1]; s = 140 m [1] (b) 0 = 20² + 2a(80) [1]; 160a = −400 [1]; a = −2.5 m/s² [1] [7] Examiner insight: “Stops” tells you v = 0. List u, v, a, s and t before choosing a formula, and note the negative sign means the cyclist is slowing down.

9. (a) a = 3, b = −8, c = −2 [1]; b² − 4ac = 64 + 24 = 88 [1]; x = (8 ± √88)/6 [1]; x = 2.90 or x = −0.23 [1] (b) 30 = 4t + ½ × 2 × t², so t² + 4t − 30 = 0 [1]; t = (−4 ± √(16 + 120))/2 [1]; t = 3.83 or −7.83 [1]; time cannot be negative, so t = 3.83 s [1] [8] Examiner insight: Both ± values are worth stating in part (a). In part (b), explain why you reject the negative root, since time must be positive.

10. (a) QR² = 8.4² + 6.8² − 2(8.4)(6.8)cos 63° [1]; QR² = 64.94 [1]; QR = 8.06 cm [1] (b) sin Q/6.8 = sin 63°/QR [1]; sin Q = 0.7519 [1]; Q = 48.8° [1] (c) ½ × 8.4 × 6.8 × sin 63° [1]; 25.4 cm² [1] [8] Examiner insight: Keep the unrounded value of QR in part (b). Using the rounded 8.06 gives 48.7°, which would change the last digit. Round only the final answer.

11. (a) s = ut + ½at² with u = 0 and a = 9.8 gives s = 4.9t² [1] (b) t² = s/4.9 [1]; t = √(s/4.9) [1] (c) t = √(44.1/4.9) = √9 [1]; t = 3 s [1] (d) v = u + at = 0 + 9.8 × 3 [1]; v = 29.4 m/s [1] (e) v² = 29.4² = 864.36 [1]; 0 + 2 × 9.8 × 44.1 = 864.36, which agrees [1] [9] Examiner insight: In a “show that” part, write the formula, state u = 0, and substitute, so the link between the two is complete. A bare final line does not demonstrate the result.

Where marks are usually lost

• Dropping brackets around a negative number, so (−3)² becomes −9.
• Leaving the new subject on both sides of the formula.
• Squaring only part of an expression when removing a square root.
• Dividing only one term of a numerator by a number.
• Using the diameter as the radius in the area formula, or the other way round.
• Rounding intermediate values, which can shift the final digit of an answer.
• Missing hidden information such as “dropped” or “from rest” meaning u = 0, or “stops” meaning v = 0.
• Giving a decimal as the answer to a context question when a whole number is asked for.
• Reporting both roots of a quadratic when one is impossible in context.
• Leaving out units, or giving cm where cm² is needed.

Next steps

• Review with the revision notes.
• Re-read the explanations in the study guide.
• Go to the OCR GCSE Mathematics hub and tick off the printable checklist.
• Try the all free 10-minute diagnostics.
• Book a free trial class on the trial page.

Official syllabus

OCR GCSE (9-1) Mathematics J560, Version 2.1 (August 2026), for first assessment in 2017, published by OCR. Section 6.02, Algebraic formulae.