Solved Questions

How to Use These Questions

The 36 questions below mix every solid in this chapter, in no fixed order, the way a paper does. A room can follow a capsule, and a bucket can follow a heap of wheat, so before any working, name the solid and write down what is asked: the volume, the curved surface, the total surface, or a count of pieces.

Work each one on paper before you read the answer. Mark the radius on a sketch, because a diameter used as a radius is behind a large share of wrong answers here, and change every length to one unit before you multiply. In a combined solid, list only the surfaces you could touch from outside; the circle where two parts meet is not one of them. In a melting question, write volume before = volume after and cancel $\pi$ straight away.

About a third of the questions are on cuboids, cubes, prisms and pyramids, a third on cylinders, cones and spheres, and the rest on combined solids, recasting, water levels and percentage change. The three marked [Tier 2] need a regular tetrahedron, a pyramid on a triangular base or an embankment round a well. A Tier 1 question of this kind should take you under a minute; if one takes much longer, find its fast route in Shortcuts & Speed Methods and try it again the next day.

Solved Examples

Question 1

The total surface area of a cube is 1350 cm². Find its volume and the length of its diagonal.

Answer. $6a^2 = 1350$, so $a^2 = 225$ and the edge is 15 cm. Volume $= 15^3$ = 3375 cm³. The diagonal of a cube is $a\sqrt{3}$, so here it is $15\sqrt{3}$ cm, about 25.98 cm.

Question 2

The radius and height of a cylinder are in the ratio 2 : 3, and its volume is 12936 cm³. Find its total surface area.

Answer. I take the radius as $2k$ and the height as $3k$. Then $\frac{22}{7} \times 4k^2 \times 3k = 12936$, which gives $k^3 = 12936 \times \frac{7}{264} = 343$, so $k = 7$. The radius is 14 cm and the height 21 cm. Total surface $= 2 \times \frac{22}{7} \times 14 \times (14 + 21) = 88 \times 35$ = 3080 cm².

Question 3

A toy is a cone of radius 4.2 cm standing on a hemisphere of the same radius. The whole toy is 9.8 cm tall. Find its total surface area.

Answer. The hemisphere takes 4.2 cm of the height, so the cone is 5.6 cm tall. Its slant is $\sqrt{5.6^2 + 4.2^2} = 7$ cm (3-4-5 times 1.4). Cone's curved surface $= \frac{22}{7} \times 4.2 \times 7 = 92.4$ cm². Hemisphere's curved surface $= 2 \times \frac{22}{7} \times 4.2^2 = 110.88$ cm². The flat circle where they meet is hidden, so I add only these two: 203.28 cm².

Question 4

The length, breadth and height of a cuboid are in the ratio 6 : 3 : 2, and its diagonal is 35 cm. Find its total surface area and its volume.

Answer. Take the sides as $6k$, $3k$ and $2k$. The diagonal is $\sqrt{36k^2 + 9k^2 + 4k^2} = 7k$, so $7k = 35$ and $k = 5$. The cuboid is 30 cm by 15 cm by 10 cm. Total surface $= 2(450 + 150 + 300)$ = 1800 cm², and volume $= 30 \times 15 \times 10$ = 4500 cm³.

Question 5

A conical tent is 9.6 m high and the radius of its base is 2.8 m. How many metres of canvas 1.1 m wide are needed to make it, and what is the cost at ₹125 per metre?

Answer. The slant height is $\sqrt{2.8^2 + 9.6^2} = \sqrt{100} = 10$ m (it is 7-24-25 times 0.4). Canvas needed $= \pi rl = \frac{22}{7} \times 2.8 \times 10 = 88$ m². With a width of 1.1 m, the length is $\frac{88}{1.1}$ = 80 m, which costs $80 \times 125$ = ₹10,000. A tent has no floor, so the base is not counted.

Question 6

A spherical balloon is blown up so that its radius grows from 10.5 cm to 21 cm. Find the ratio of its surface areas before and after, and the volume of air pumped in.

Answer. The radius doubles, so the surface becomes 4 times: the ratio is 1 : 4 (1386 cm² to 5544 cm²). The volume goes from 4851 cm³ to 38808 cm³, so the air pumped in is $38808 - 4851$ = 33957 cm³.

Question 7

From each corner of a rectangular sheet 32 cm by 24 cm, a square of side 4 cm is cut off, and the sides are folded up to make an open box. Find the volume of the box and the area of the sheet used in it.

Answer. Cutting 4 cm from both ends takes 8 cm off each side, so the base is 24 cm by 16 cm, and the folded-up flaps make the height 4 cm. Volume $= 24 \times 16 \times 4$ = 1536 cm³. The sheet used is the whole sheet minus the four squares: $768 - 64$ = 704 cm². As a check, base plus four walls is $384 + 2 \times 4 \times (24 + 16) = 384 + 320 = 704$.

Question 8

A pyramid has a square base of side 32 cm and a height of 12 cm. Find its volume and its total surface area.

Answer. Volume $= \frac{1}{3} \times 32^2 \times 12 = 1024 \times 4$ = 4096 cm³. For the side faces I need the slant height, from the height and half the base side: $\sqrt{12^2 + 16^2} = 20$ cm. Lateral surface $= \frac{1}{2} \times 128 \times 20 = 1280$ cm², and with the base, total surface $= 1280 + 1024$ = 2304 cm².

Question 9

A copper pipe 2.1 m long has an outer diameter of 10 cm and an inner diameter of 6 cm. If 1 cm³ of copper weighs 9 g, find the weight of the pipe.

Answer. In cm, $R = 5$, $r = 3$ and the length is 210. Metal $= \frac{22}{7} \times (5 + 3)(5 - 3) \times 210 = \frac{22}{7} \times 16 \times 210 = 10560$ cm³. Weight $= 10560 \times 9 = 95040$ g = 95.04 kg.

Question 10

How many coins, each 1.4 cm across and 0.25 cm thick, must be melted to make a cuboid 15.4 cm by 7 cm by 2.5 cm?

Answer. Each coin is a flat cylinder of radius 0.7 cm: $\frac{22}{7} \times 0.49 \times 0.25 = 0.385$ cm³. The cuboid is $15.4 \times 7 \times 2.5 = 269.5$ cm³. Number of coins $= \frac{269.5}{0.385}$ = 700.

Question 11

A room is 12 m long and 4.5 m high. Painting its four walls at ₹30 per m² costs ₹5,400. Find the breadth of the room, and the cost of carpeting its floor at ₹250 per m².

Answer. The walls have an area of $\frac{5400}{30} = 180$ m². Four walls $= 2(l + b)h$, so $2 \times (12 + b) \times 4.5 = 180$, which gives $12 + b = 20$ and a breadth of 8 m. The floor is $12 \times 8 = 96$ m², and carpeting it costs $96 \times 250$ = ₹24,000.

Question 12

A rectangular sheet 44 cm by 18 cm is rolled into a hollow cylinder, first with the 44 cm side going round, then with the 18 cm side going round. Find the volume in each case.

Answer. When the 44 cm side goes round, it is the circumference: $2\pi r = 44$ gives $r = 7$ and the height is 18, so the volume is $154 \times 18$ = 2772 cm³. When the 18 cm side goes round, $r = \frac{18 \times 7}{44} = \frac{63}{22}$ cm and the height is 44, so the volume is $\frac{22}{7} \times \frac{63^2}{22^2} \times 44 = \frac{63^2 \times 2}{7}$ = 1134 cm³. The two volumes are in the ratio $2772 : 1134 = 22 : 9$, the same as $44 : 18$. Rolling with the longer side going round gives the bigger cylinder.

Question 13 [Tier 2]

Find the total surface area, the height and the volume of a regular tetrahedron with every edge 6 cm.

Answer. All four faces are equilateral triangles of side 6, so the surface is $4 \times \frac{\sqrt{3}}{4} \times 36$ = $36\sqrt{3}$ cm². The apex sits right above the centre of the base, which is $\frac{6}{\sqrt{3}} = 2\sqrt{3}$ cm from each corner. So the height is $\sqrt{36 - 12} = \sqrt{24}$ = $2\sqrt{6}$ cm, about 4.9 cm. Volume $= \frac{1}{3} \times 9\sqrt{3} \times 2\sqrt{6} = 6\sqrt{18}$ = $18\sqrt{2}$ cm³, which matches the formula $\frac{a^3}{6\sqrt{2}} = \frac{216}{6\sqrt{2}}$.

Question 14

A hemispherical vessel has an inner radius of 21 cm. Find how many litres it holds, and the cost of tin-plating its inner surface at ₹4 per 100 cm².

Answer. Volume $= \frac{2}{3} \times \frac{22}{7} \times 21^3 = 44 \times 441 = 19404$ cm³, which is 19.404 litres. The inner surface is only the curved part, $2\pi r^2 = 2 \times 1386 = 2772$ cm². Tin-plating costs $\frac{2772}{100} \times 4$ = ₹110.88.

Question 15

A tent is a cylinder of base radius 2.1 m and height 4 m with a cone of the same base and height 2.8 m on top. Find the canvas needed, and its cost at ₹500 per m².

Answer. The cone's slant is $\sqrt{2.1^2 + 2.8^2} = 3.5$ m (3-4-5 times 0.7). Canvas = curved surface of the cylinder + curved surface of the cone $= 2 \times \frac{22}{7} \times 2.1 \times 4 + \frac{22}{7} \times 2.1 \times 3.5 = 52.8 + 23.1$ = 75.9 m². Cost $= 75.9 \times 500$ = ₹37,950. The floor and the circle where the cone sits on the cylinder get no canvas.

Question 16

A wall 18 m long, 30 cm thick and 2.5 m high is built with bricks 25 cm by 12.5 cm by 7.5 cm. If the mortar takes up $\frac{1}{10}$ of the wall, how many bricks are used?

Answer. In cm, the wall is $1800 \times 30 \times 250 = 13500000$ cm³, and the bricks fill $\frac{9}{10}$ of it, 12150000 cm³. One brick is $25 \times 12.5 \times 7.5 = 2343.75$ cm³. Number of bricks $= \frac{12150000}{2343.75}$ = 5184.

Question 17

The curved surface of a cone of radius 8 cm and height 6 cm is cut along a slant line and opened out flat. Find the radius and the angle of the sector formed.

Answer. The slant height is $\sqrt{64 + 36} = 10$ cm, and that becomes the radius of the sector: 10 cm. The arc of the sector is the rim of the cone's base, so $\frac{\theta}{360} = \frac{r}{l} = \frac{8}{10}$, and the angle is $0.8 \times 360$ = $288^\circ$.

Question 18

A hollow metal sphere has an outer radius of 9 cm and an inner radius of 6 cm. It is melted and made into a solid cylinder of radius 6 cm. Find the height of the cylinder.

Answer. Only the metal counts: $\frac{4}{3}\pi(9^3 - 6^3) = \frac{4}{3}\pi \times 513 = 684\pi$. The cylinder has radius 6, so $\pi \times 36 \times h = 684\pi$, and the height is $\frac{684}{36}$ = 19 cm.

Question 19

The base of a right prism is a triangle with sides 11 cm, 60 cm and 61 cm, and the prism is 8 cm tall. Find its volume and its total surface area.

Answer. $11^2 + 60^2 = 121 + 3600 = 3721 = 61^2$, so the right angle is between the 11 cm and 60 cm sides, and the base has area $\frac{1}{2} \times 11 \times 60 = 330$ cm². Volume $= 330 \times 8$ = 2640 cm³. Lateral surface $= \text{perimeter} \times \text{height} = 132 \times 8 = 1056$ cm², and the two triangular ends add 660 cm², so the total surface is 1716 cm².

Question 20

A bucket is a frustum of a cone, 45 cm deep, with a radius of 28 cm at the top and 7 cm at the bottom. How many litres of water can it hold?

Answer. Frustum volume $= \frac{\pi h}{3}(R^2 + Rr + r^2) = \frac{22}{7} \times 15 \times (784 + 196 + 49)$. The bracket is 1029, which is $7 \times 147$, so the 7 cancels: $22 \times 15 \times 147 = 48510$ cm³. That is 48.51 litres.

Question 21

A solid cone of radius 6 cm and height 16 cm is dropped into a cylindrical vessel of radius 8 cm that is partly full of water, and the cone goes fully under. By how much does the water rise?

Answer. The cone pushes up its own volume: $\frac{1}{3}\pi \times 36 \times 16 = 192\pi$ cm³. The vessel's base is $\pi \times 64$ cm², so the water rises by $\frac{192\pi}{64\pi}$ = 3 cm. I never needed a value of $\pi$.

Question 22

A garden roller is 1.4 m long and 70 cm across. It makes 500 complete turns to level a lawn. Find the area of the lawn and the cost of levelling it at ₹2.50 per m².

Answer. In one turn the roller presses its own curved surface: $2 \times \frac{22}{7} \times 0.35 \times 1.4 = 3.08$ m². In 500 turns that is 1540 m², and the cost is $1540 \times 2.5$ = ₹3,850.

Question 23

A medicine capsule is a cylinder with a hemisphere at each end. It is 14 mm long from end to end and 5 mm across. Find its surface area.

Answer. The radius is 2.5 mm, so the two hemispheres take 5 mm of the length and the cylinder is 9 mm long. Surface $= 2\pi rh + 4\pi r^2 = 2\pi r(h + 2r) = 2\pi \times 2.5 \times 14$. That is $\pi \times 5 \times 14 = \frac{22}{7} \times 70$ = 220 mm². So a capsule's surface is just $\pi \times \text{width} \times \text{full length}$.

Question 24

The radius of a cylinder is decreased by 20%. By what percentage must its height be increased so that its volume stays the same?

Answer. The base area becomes $0.8^2 = 0.64$ times, so the height must become $\frac{1}{0.64} = \frac{25}{16}$ times. That is an increase of $\frac{9}{16} \times 100$ = 56.25%. Taking 40% (double the 20%) is not enough, because the radius is squared.

Question 25

The base of a right prism is a regular hexagon of side 8 cm, and its volume is $1440\sqrt{3}$ cm³. Find its height and its lateral surface area.

Answer. The hexagon is six equilateral triangles of side 8: area $= 6 \times \frac{\sqrt{3}}{4} \times 64 = 96\sqrt{3}$ cm². Height $= \frac{1440\sqrt{3}}{96\sqrt{3}}$ = 15 cm. Lateral surface $= \text{perimeter} \times \text{height} = 48 \times 15$ = 720 cm².

Question 26 [Tier 2]

A well 3 m across is dug 14 m deep. The earth taken out is spread evenly all round it to make an embankment 4 m wide. Find the height of the embankment.

Answer. Earth dug out $= \frac{22}{7} \times 1.5^2 \times 14 = 99$ m³. The embankment is a ring from radius 1.5 m out to $1.5 + 4 = 5.5$ m, so its area is $\frac{22}{7} \times (5.5^2 - 1.5^2) = \frac{22}{7} \times 28 = 88$ m². Height $= \frac{99}{88}$ = 1.125 m. The usual slip is to take the outer radius as 4 m: the 4 m is the width of the ring, measured from the edge of the well.

Question 27

A heap of wheat is a cone of base diameter 12 m and height 8 m. Find its volume, and the area of canvas needed to just cover it. (Take $\pi = 3.14$.)

Answer. $r = 6$ and $h = 8$, so the slant is 10 m. Volume $= \frac{1}{3} \times 3.14 \times 36 \times 8 = 3.14 \times 96$ = 301.44 m³. Canvas covers only the sloping surface: $3.14 \times 6 \times 10$ = 188.4 m².

Question 28

A closed wooden box measures 52 cm by 40 cm by 29 cm on the outside, and the wood is 2 cm thick. Find the volume of wood used.

Answer. The wood is on both sides of every measurement, so each inside length is 4 cm less: 48 cm by 36 cm by 25 cm. Outside volume $= 52 \times 40 \times 29 = 60320$ cm³ and inside volume $= 48 \times 36 \times 25 = 43200$ cm³. Wood used $= 60320 - 43200$ = 17120 cm³.

Question 29

An ice-cream cone is a cone of radius 4 cm and height 10 cm, filled with ice cream and topped with a hemisphere of the same radius. How many such cones can be filled from a cylindrical tub of radius 8 cm and height 18 cm?

Answer. One cone holds $\frac{1}{3}\pi \times 16 \times 10 + \frac{2}{3}\pi \times 64 = \frac{160\pi}{3} + \frac{128\pi}{3} = 96\pi$ cm³. The tub holds $\pi \times 64 \times 18 = 1152\pi$ cm³. Number of cones $= \frac{1152\pi}{96\pi}$ = 12. I kept $\pi$ as a letter, so it cancelled at the end.

Question 30

Three solid metal spheres of radii 4 cm, 17 cm and 22 cm are melted into one sphere. Find its radius, and the ratio of the total surface of the three spheres to the surface of the new one.

Answer. Volumes add: $R^3 = 64 + 4913 + 10648 = 15625$, so the new radius is 25 cm. Surfaces go with $r^2$: before, $16 + 289 + 484 = 789$; after, $625$. The ratio is 789 : 625, so some surface is lost when small spheres are melted into one big one.

Question 31 [Tier 2]

A right pyramid stands on an equilateral triangle of side $6\sqrt{3}$ cm. The slant height of each side face is 5 cm. Find the height and the volume of the pyramid, and its lateral surface area.

Answer. The foot of the height is the centre of the triangle, and its distance from each side is the inradius $\frac{a}{2\sqrt{3}} = \frac{6\sqrt{3}}{2\sqrt{3}} = 3$ cm. The slant height runs from the apex to the middle of a side, so height $= \sqrt{5^2 - 3^2}$ = 4 cm (the 3-4-5 triangle). The base has area $\frac{\sqrt{3}}{4} \times 108 = 27\sqrt{3}$ cm², so the volume is $\frac{1}{3} \times 27\sqrt{3} \times 4$ = $36\sqrt{3}$ cm³. Lateral surface $= \frac{1}{2} \times 18\sqrt{3} \times 5$ = $45\sqrt{3}$ cm².

Question 32

A hollow cylinder, open at both ends, is 15 cm long. Its outer radius is 10.5 cm and its inner radius is 7 cm. Find its total surface area: outside, inside and the two rims.

Answer. Outside $= 2 \times \frac{22}{7} \times 10.5 \times 15 = 990$ cm², inside $= 2 \times \frac{22}{7} \times 7 \times 15 = 660$ cm², and the two rims $= 2 \times \frac{22}{7} \times (110.25 - 49) = 385$ cm². Total = 2035 cm². In one line: $2\pi(R + r)(h + R - r) = 2 \times \frac{22}{7} \times 17.5 \times 18.5 = 110 \times 18.5$, the same 2035.

Question 33

The largest possible hemisphere is scooped out of one face of a solid cube of edge 14 cm. Find the surface area of what is left.

Answer. The hemisphere's radius is half the edge, 7 cm. Scooping removes a circle of $\pi r^2 = 154$ cm² from the top face but opens up a bowl of $2\pi r^2 = 308$ cm² inside. So the surface is $6 \times 196 - 154 + 308 = 1176 + 154$ = 1330 cm², more than the cube had.

Question 34

A prism has 30 edges. How many faces and vertices does it have? How many faces, edges and vertices does a pyramid on the same base have?

Answer. A prism on an $n$-sided base has $3n$ edges, so $n = 10$. It has $n + 2$ = 12 faces and $2n$ = 20 vertices ($12 + 20 - 30 = 2$, as Euler's rule says). The pyramid on a 10-sided base has $n + 1$ = 11 faces, $2n$ = 20 edges and $n + 1$ = 11 vertices.

Question 35

A frustum of a cone has end radii 20 cm and 12 cm and a height of 15 cm. Find its curved surface area and its total surface area, in terms of $\pi$.

Answer. The slant height comes from the height and the difference of the radii: $\sqrt{15^2 + 8^2} = 17$ cm. Curved surface $= \pi(R + r)l = \pi \times 32 \times 17$ = $544\pi$ cm². Adding the two circular ends, $400\pi + 144\pi$, the total surface is $1088\pi$ cm².

Question 36

A cone 32 cm high is cut by a plane parallel to its base so that the curved surface of the small cone on top is $\frac{1}{16}$ of the curved surface of the whole cone. At what height above the base is the cut made?

Answer. Surfaces go as squares of lengths, so the small cone's height is $\sqrt{\frac{1}{16}} = \frac{1}{4}$ of the whole: 8 cm, measured from the apex. The cut is $32 - 8$ = 24 cm above the base. Two slips to avoid: taking the cube root (that is for volumes), and answering 8 cm, which is measured down from the top.