OCR GCSE Mathematics: Surface area and volume of prisms, cylinders, cones, spheres and pyramids (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 10.04 (10.04a to 10.04c). All of them suit both the Foundation tier and the Higher tier. Where a question says “in terms of π”, leave π in your answer.

Read the study guide or the revision notes first. The hub is OCR GCSE Mathematics and the list of skills is the checklist. Not sure where you stand? Try the free 10-minute diagnostics.

Questions

1. A cuboid measures 14 cm by 9 cm by 6.5 cm. Find (a) its volume [1] and (b) its total surface area [2].

2. A prism has a right-angled triangle as its cross-section. The shorter sides of the triangle are 8 cm and 15 cm. The prism is 25 cm long. Find (a) the volume [2] and (b) the total surface area [3].

3. A closed cylindrical tin has radius 3.5 cm and height 12 cm. Find (a) its volume in terms of π [2], (b) the area of the paper label that wraps once round the curved surface, in terms of π [1], and (c) the total surface area to 1 decimal place [2].

4. A cylinder has radius 25 cm. It stands upright and 150 litres of water are poured in. Find the depth of the water to 1 decimal place. [4]

5. (a) A ball is a sphere of radius 4.5 cm. Find its volume and its surface area, in terms of π. [3] (b) A solid hemisphere has volume 18π cm³. Find its radius. [3]

6. A cone has base radius 7.5 cm and perpendicular height 10 cm. Find (a) the slant height [2], (b) the volume in terms of π [2] and (c) the total surface area in terms of π [2].

7. A square-based pyramid has base side 18 cm and perpendicular height 12 cm. Find (a) its volume [2] and (b) its total surface area [4].

8. A toy is a cone placed on a solid hemisphere. They have the same radius, 6 cm, and the cone has perpendicular height 8 cm. Find (a) the volume of the toy in terms of π [3] and (b) the surface area of the toy in terms of π [3].

9. A solid metal sphere of radius 6 cm is melted and recast into small solid cylinders, each of radius 2 cm and height 9 cm. No metal is lost. How many cylinders can be made? [4]

10. A square-based pyramid has volume 192 cm³ and base side 6 cm. (a) Find its perpendicular height. [2] (b) A cuboid has the same square base and the same height. How many times larger is its volume than the pyramid’s? [2]

11. A storage tank is a cylinder with a cone-shaped roof on top. The cylinder has radius 2.5 m and wall height 4 m. The cone has the same radius and perpendicular height 6 m. The floor is not painted. (a) Find the volume of the whole tank in terms of π. [3] (b) Find the slant height of the roof. [1] (c) Find the area of wall and roof together, in terms of π. [2] (d) One litre of paint covers 8 m². Two coats are applied. Tins hold 5 litres. How many tins are needed? [3]

Answers

1. (a) 14 × 9 × 6.5 = 819 [1] 819 cm³ [1] (b) Faces: 14 × 9 = 126, 14 × 6.5 = 91, 9 × 6.5 = 58.5 [1]. 2 × (126 + 91 + 58.5) = 551 cm² [1] [2] Examiner insight: Writing the three face areas before doubling makes it easier to spot a slip in one product.

2. (a) Area of triangle = ½ × 8 × 15 = 60 [1]. Volume = 60 × 25 = 1500 cm³ [1] [2] (b) Hypotenuse = √(64 + 225) = √289 = 17 cm [1]. Rectangles: (8 + 15 + 17) × 25 = 1000 [1]. Total = 2 × 60 + 1000 = 1120 cm² [1] [3] Examiner insight: Part (b) needs the hypotenuse; an answer that includes only two rectangles will lose most of the credit for the area.

3. (a) V = π × 3.5² × 12 [1] = 147π cm³ [1] [2] (b) Label = 2π × 3.5 × 12 = 84π cm² [1] [1] (c) Ends: 2 × π × 3.5² = 24.5π [1]. Total = 24.5π + 84π = 108.5π = 340.9 cm² [1] [2] Examiner insight: Part (b) asks for the curved surface only, so adding the circles there is a common way to throw away a mark.

4. 150 litres = 150 000 cm³ [1]. 150 000 = π × 25² × h [1]. h = 150 000 ÷ (625π) = 240/π [1]. h = 76.4 cm [1] [4] Examiner insight: Show the conversion from litres to cm³ as its own line; a depth that is a factor of 1000 out is hard to credit without it.

5. (a) V = (4/3) × π × 4.5³ = (4/3) × 91.125π [1] = 121.5π cm³ [1]. Surface area = 4 × π × 4.5² = 81π cm² [1] [3] (b) (2/3)πr³ = 18π [1]. r³ = 18 × 3/2 = 27 [1]. r = 3 cm [1] [3] Examiner insight: In a working-backwards question, cancel π from both sides first; it keeps the cube root simple.

6. (a) l = √(7.5² + 10²) = √(56.25 + 100) [1] = √156.25 = 12.5 cm [1] [2] (b) V = (1/3) × π × 7.5² × 10 [1] = 187.5π cm³ [1] [2] (c) Curved = π × 7.5 × 12.5 = 93.75π [1]. Base = 56.25π. Total = 150π cm² [1] [2] Examiner insight: Use the perpendicular height in part (b) and the slant height in part (c); swapping them is a frequent and avoidable error.

7. (a) V = (1/3) × 18² × 12 [1] = 1296 cm³ [1] [2] (b) Half the base side = 9 cm [1]. Slant height = √(12² + 9²) = 15 cm [1]. One triangle = ½ × 18 × 15 = 135, so four = 540 [1]. Total = 324 + 540 = 864 cm² [1] [4] Examiner insight: A sketch of the right-angled triangle inside the pyramid (height, half-side, slant) makes the slant height easy to find.

8. (a) Hemisphere = (2/3) × π × 6³ = 144π [1]. Cone = (1/3) × π × 6² × 8 = 96π [1]. Total = 240π cm³ [1] [3] (b) Slant height = √(36 + 64) = 10 cm [1]. Curved hemisphere = 2π × 36 = 72π, curved cone = π × 6 × 10 = 60π [1]. Total = 132π cm² [1] [3] Examiner insight: The circle where the two solids join is hidden, so including it in part (b) gives an answer that is too large.

9. Sphere = (4/3) × π × 6³ = 288π [1]. Cylinder = π × 2² × 9 = 36π [1]. 288π ÷ 36π [1] = 8 cylinders [1] [4] Examiner insight: Dividing the volumes shows π cancelling, so no decimal is needed; this is the sort of step a paper without a calculator rewards.

10. (a) 192 = (1/3) × 6² × h [1]. 192 = 12h, so h = 16 cm [1] [2] (b) Cuboid = 6 × 6 × 16 = 576 cm³ [1]. 576 ÷ 192 = 3 times [1] [2] Examiner insight: Explaining that the pyramid is one third of the matching cuboid is a good check on any pyramid volume.

11. (a) Cylinder = π × 2.5² × 4 = 25π [1]. Cone = (1/3) × π × 2.5² × 6 = 12.5π [1]. Total = 37.5π m³ [1] [3] (b) l = √(2.5² + 6²) = √42.25 = 6.5 m [1] [1] (c) Wall = 2π × 2.5 × 4 = 20π; roof = π × 2.5 × 6.5 = 16.25π [1]. Total = 36.25π m² (≈ 113.9 m²) [1] [2] (d) Area to cover = 2 × 113.88 = 227.77 m² [1]. Paint = 227.77 ÷ 8 = 28.47 litres [1]. 28.47 ÷ 5 = 5.69, so 6 tins [1] [3] Examiner insight: The last step is a rounding-up decision; writing “5.69, so 6 tins” shows why you did not give 5.7.

Where marks are usually lost

• Using the slant height where the perpendicular height is needed, and the reverse.
• Forgetting the hypotenuse when finding the perimeter of a triangular cross-section.
• Including a circle that is not on the outside of a composite solid, or leaving out a base that is.
• Using the diameter in place of the radius.
• Not converting litres to cm³, or converting with the wrong power of ten.
• Rounding in the middle of a calculation and ending with a final digit that is out.
• Giving a decimal when the question asks for the answer in terms of π.
• Rounding down on a “how many tins” question when you need to round up.

Next steps

• Re-read the revision notes and the study guide.
• Tick off the skills on the OCR GCSE Mathematics checklist and see the course hub.
• Take one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR, GCSE (9-1) Mathematics J560, Version 2.1 (August 2026), for first assessment in 2017, section 10.04 Volume and surface area calculations.