Volume & surface area (prisms, cylinders, cones, spheres, pyramids) – GCSE Maths exam questions with mark scheme

1Grade 1–3FoundationCalculator4 marks

A cuboid measures 5 cm by 4 cm by 3 cm.

(a) Work out the volume of the cuboid. [2]

(b) Work out the total surface area of the cuboid. [2]

(a) $5\times4\times3$M1
60 cm$^3$A1
(b) finds the area of at least 3 different faces, e.g. 20, 15, 12, and attempts to add six facesM1
94 cm$^2$A1
Answer: (a) 60 cm$^3$ (b) 94 cm$^2$
Examiner tip: A cuboid has six faces in three matching pairs – adding only three faces gives 47 and loses the accuracy mark.

2Grade 4–5Calculator5 marks

The diagram shows a triangular prism. The cross-section is a right-angled triangle with sides 6 cm, 8 cm and 10 cm. The length of the prism is 15 cm.

(a) Work out the volume of the prism. [2]

(b) Work out the total surface area of the prism. [3]

(a) $\tfrac12\times6\times8\times15$M1
(a) 360 cm$^3$A1
(b) area of a triangular face $\tfrac12\times6\times8\;(=24)$, used twiceM1
(b) area of at least two of the rectangular faces $6\times15,\;8\times15,\;10\times15$ (90, 120, 150)M1
(b) 408 cm$^2$ caoA1
Answer: (a) 360 cm$^3$ (b) 408 cm$^2$
Examiner tip: Volume of a prism = area of cross-section × length; don't multiply all three edge lengths together.

3Grade 4–5Calculator4 marks

A cylinder has radius 4 cm and height 11 cm. It is full of water.

(a) Work out the volume of the cylinder. Give your answer correct to 3 significant figures. [2]

All the water is poured into an empty cuboid-shaped tank. The base of the tank measures 20 cm by 10 cm.

(b) Work out the depth of the water in the tank. Give your answer correct to 1 decimal place. [2]

(a) $\pi\times4^2\times11$M1
553 cm$^3$ (accept 552.9…)A1
(b) their volume $\div(20\times10)$M1
2.8 cm (accept 2.76…)A1
Answer: (a) 553 cm$^3$ (b) 2.8 cm
Examiner tip: Use the unrounded volume in part (b) – your calculator's ANS key avoids rounding errors.

4Grade 6–7HigherCalculator5 marks

A solid cone has base radius 5 cm and slant height 13 cm.

(a) Work out the volume of the cone. Give your answer correct to 3 significant figures. [3]

(b) Work out the total surface area of the cone. Give your answer correct to 3 significant figures. [2]

[Volume of cone $=\tfrac13\pi r^2h$; curved surface area of cone $=\pi r l$]

(a) perpendicular height $\sqrt{13^2-5^2}=12$M1
$\tfrac13\times\pi\times5^2\times12$M1
314 cm$^3$ (accept $100\pi$)A1
(b) $\pi\times5\times13+\pi\times5^2$M1
283 cm$^2$ (accept $90\pi$ or 282.7…)A1
Answer: (a) 314 cm$^3$ (b) 283 cm$^2$
Examiner tip: The volume formula needs the perpendicular height, not the slant height – use Pythagoras first.

5Grade 6–7HigherNon-calculator3 marks

A solid is made from a cylinder and a hemisphere. The cylinder has radius 3 cm and height 10 cm. The hemisphere has radius 3 cm and sits exactly on top of the cylinder.

Work out the total volume of the solid. Give your answer in terms of $\pi$. [3]

[Volume of sphere $=\tfrac43\pi r^3$]

volume of cylinder $\pi\times3^2\times10\;(=90\pi)$M1
volume of hemisphere $\tfrac12\times\tfrac43\times\pi\times3^3\;(=18\pi)$M1
$108\pi$ cm$^3$A1
Answer: $108\pi$ cm$^3$
Examiner tip: Remember to halve the sphere formula for a hemisphere – $36\pi$ for the top gives the wrong total.

6Grade 6–7HigherCalculator4 marks

A solid metal sphere has radius 6 cm. It is melted down and all the metal is used to make solid cylinders. Each cylinder has radius 1.5 cm and height 5 cm.

Work out the greatest number of complete cylinders that can be made. [4]

[Volume of sphere $=\tfrac43\pi r^3$]

volume of sphere $\tfrac43\times\pi\times6^3\;(=288\pi=904.7\ldots)$P1
volume of one cylinder $\pi\times1.5^2\times5\;(=11.25\pi=35.3\ldots)$P1
sphere volume $\div$ cylinder volume $(=25.6)$P1
25A1
Answer: 25 cylinders
Examiner tip: Round down at the end – 25.6 means only 25 complete cylinders can be made.

7Grade 6–7HigherCalculator4 marks

A solid cone has base radius 9 cm and perpendicular height 24 cm. A small cone of radius 3 cm and height 8 cm is cut off the top, leaving the frustum shown (solid lines).

Work out the volume of the frustum. Give your answer correct to 3 significant figures. [4]

[Volume of cone $=\tfrac13\pi r^2h$]

volume of large cone $\tfrac13\times\pi\times9^2\times24\;(=648\pi)$M1
volume of small cone $\tfrac13\times\pi\times3^2\times8\;(=24\pi)$M1
subtracts the two volumesM1
1960 cm$^3$ (accept answers from 1959 to 1961 inclusive; $624\pi$ alone, not evaluated, does not score A1 because 3 s.f. was asked for)A1
Answer: 1960 cm$^3$
Examiner tip: The height of the frustum (16 cm) is not used directly – find the volume of each whole cone and subtract.

8Grade 4–5Calculator4 marks

A pyramid has a square base of side 6 cm and perpendicular height 10 cm.

(a) Work out the volume of the pyramid. [2]

Another pyramid has the same square base and a volume of 300 cm$^3$.

(b) Work out its perpendicular height. [2]

[Volume of pyramid $=\tfrac13\times$ area of base $\times$ height]

(a) $\tfrac13\times6\times6\times10$M1
120 cm$^3$A1
(b) $300=\tfrac13\times36\times h$ or $300\times3\div36$M1
25 cmA1
Answer: (a) 120 cm$^3$ (b) 25 cm
Examiner tip: Don't forget the $\tfrac13$ – a pyramid holds a third of the prism with the same base and height.

9Grade 1–3FoundationNon-calculator4 marks

The diagram shows a prism. The cross-section of the prism is a pentagon.

(a) Write down the number of faces of the prism. [1]

(b) Write down the number of edges of the prism. [1]

A different solid has 5 faces, 8 edges and 5 vertices. One of its faces is a square.

(d) Write down the mathematical name of this solid. [1]

(a) 7B1
(b) 15B1
(c) 10B1
(d) (square-based) pyramid accept square pyramid; do not accept triangle, tetrahedron or prismB1
Answer: (a) 7 (b) 15 (c) 10 (d) square-based pyramid
Examiner tip: Count the edges in three groups – 5 on the front, 5 on the back and 5 joining them – instead of counting round the drawing and missing the hidden ones.

10Grade 4–5Calculator5 marks

A water trough is in the shape of a prism. The cross-section of the trough is a trapezium.

The parallel sides of the trapezium are 40 cm and 25 cm. The vertical depth of the trapezium is 30 cm. The trough is 150 cm long.

The trough is full of water.

Work out how many litres of water are in the trough. [5]

[1 litre $=1000$ cm$^3$]

$\tfrac12(40+25)\times30$ (= 975)M1
(their 975) $\times150$M1
146 250 (cm$^3$)A1
(their 146 250) $\div1000$M1
146.25 litres accept 146 or 146.3A1
Answer: 146.25 litres
Examiner tip: Area of the trapezium first, then × length – multiplying 40 × 25 × 30 × 150 or using only one parallel side loses the method marks.

11Grade 4–5Calculator3 marks

Petra has two containers.

Container A is a cylinder with radius 4 cm and height 10 cm.
Container B is a cuboid measuring 7 cm by 7 cm by 10 cm.

Petra says, “Container B holds more than container A, because a cuboid has corners and a cylinder does not.”

Is Petra correct? You must show your working. [3]

$\pi\times4^2\times10$ (= 502.6…) accept $160\pi$M1
$7\times7\times10$ (= 490)M1
No, with 502 or 503 (or better) and 490 compared, e.g. A holds about 13 cm$^3$ more dependent on both M marksC1
Answer: No – A holds 503 cm$^3$ (502.65…), B holds 490 cm$^3$
Examiner tip: A bare ‘No’ scores nothing: the communication mark needs both volumes worked out and compared.

12Grade 4–5Calculator5 marks

Jamal makes plant pots. Each pot is an open cylinder: it has a base but no lid.

Each pot has radius 15 cm and height 40 cm.

He paints the outside curved surface and the base of 12 pots. He does not paint the inside.

One tin of paint covers 1.2 m$^2$ and costs £7.49.

Work out the total cost of the paint Jamal needs to buy. [5]

curved surface area of one pot $2\times\pi\times15\times40$ (= 3769.9… or $1200\pi$)M1
adds area of one base $\pi\times15^2$ (= 706.8…) and multiplies by 12, (= 53 721… cm$^2$ or $17\,100\pi$)M1
correct conversion between cm$^2$ and m$^2$ using 10 000, e.g. 5.37… m$^2$ or 1.2 m$^2$ = 12 000 cm$^2$M1
divides total area by coverage and rounds up to a whole number of tins, e.g. $5.37\div1.2=4.47\ldots\to5$M1
£37.45 cao dependent on all previous M marks (curved surface only gives 4 tins, £29.96; including a lid gives 6 tins, £44.94)A1
Answer: £37.45 (5 tins)
Examiner tip: 1 m² is 10 000 cm², not 100 cm² – and the base must be included but not a lid: curved surface only gives 4 tins and adding a lid gives 6, so read ‘open’ carefully.

13Grade 6–7HigherCalculator6 marks

VABCD is a solid pyramid with a square base ABCD.

AB = 8 cm. Each sloping edge VA, VB, VC and VD is 9 cm. The vertex V is vertically above the centre of the base.

(a) Show that the perpendicular height of the pyramid is 7 cm. [3]

(b) Work out the total surface area of the pyramid. Give your answer correct to 3 significant figures. [3]

(a) diagonal of base $\sqrt{8^2+8^2}=\sqrt{128}$, or half-diagonal $\sqrt{4^2+4^2}=\sqrt{32}$ (= 5.65…)M1
(a) $h^2=9^2-(\sqrt{32})^2$, i.e. $81-32$M1
(a) $h=\sqrt{49}=7$ shown with no errors seenA1
(b) height of a triangular face $\sqrt{9^2-4^2}=\sqrt{65}$ (= 8.06…) or $\sqrt{7^2+4^2}$M1
(b) $4\times\tfrac12\times8\times\sqrt{65}+8\times8$M1
(b) 193 cm$^2$ accept 192.9–193.0A1
Answer: (a) $h=7$ shown (b) 193 cm$^2$
Examiner tip: The sloping edge (9 cm) is not the height of a triangular face – using 9 as the face height gives 208 cm² and loses two marks.

14Grade 6–7HigherNon-calculator4 marks

A toy is made from a solid cone joined to a solid hemisphere, as shown.

The cone and the hemisphere both have radius 6 cm. The perpendicular height of the cone is 8 cm.

Work out the total surface area of the toy. Give your answer in terms of $\pi$. [4]

[Curved surface area of cone $=\pi r l$; surface area of sphere $=4\pi r^2$]

slant height $l=\sqrt{6^2+8^2}=10$M1
curved surface area of cone $\pi\times6\times10$ (= $60\pi$)M1
curved surface area of hemisphere $\tfrac12\times4\times\pi\times6^2$ (= $72\pi$)M1
$132\pi$ cm$^2$ cao (do not accept answers including the circular face, e.g. $168\pi$)A1
Answer: $132\pi$ cm$^2$
Examiner tip: The flat circle where the two parts join is inside the toy – adding $36\pi$ for it is the most common way to lose the final mark.

15Grade 8–9HigherNon-calculator5 marks

A solid sphere has radius $r$ cm.

A solid cone has base radius $r$ cm and perpendicular height $h$ cm.

The volume of the sphere is equal to the volume of the cone.

(a) Show that $h=4r$. [2]

(b) Show that

curved surface area of the cone : surface area of the sphere $=\sqrt{17}:4$ [3]

[Volume of sphere $=\tfrac43\pi r^3$; volume of cone $=\tfrac13\pi r^2h$; curved surface area of cone $=\pi r l$; surface area of sphere $=4\pi r^2$]

(a) $\tfrac43\pi r^3=\tfrac13\pi r^2h$M1
(a) correctly divides both sides by $\tfrac13\pi r^2$ (or equivalent) to reach $h=4r$A1
(b) $l=\sqrt{r^2+(4r)^2}=\sqrt{17r^2}=\sqrt{17}\,r$M1
(b) curved surface area of cone $=\pi r\times\sqrt{17}\,r=\sqrt{17}\pi r^2$M1
(b) $\sqrt{17}\pi r^2:4\pi r^2$ simplified to $\sqrt{17}:4$ with all steps shownA1
Answer: (a) $h=4r$ (b) $\sqrt{17}\pi r^2:4\pi r^2=\sqrt{17}:4$
Examiner tip: In a ‘show that’ every step must be written down – $\sqrt{r^2+16r^2}$ must become $\sqrt{17}\,r$, not $r+4r$.

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