NCERT Exemplar Solutions Class 10 Maths Surface Areas and Volumes — Exercise 12.4

NCERT Exemplar Problems, Class 10 Mathematics — Surface Areas and Volumes, Exercise 12.4. All 20 questions solved.

Twenty questions, and the longest in the unit — though the geometry is rarely the hard part. What makes them long is that each has a unit conversion, a rate or a cost wrapped round a volume calculation.

Three patterns run through them.

Flow. Water moving through a pipe at speed $v$ delivers, in time $t$, a cylinder of liquid:

$$\text{volume} = \text{(cross-sectional area)} \times v \times t$$

Questions 5, 8 and 18 are this, and all three turn on converting the speed into the same length unit as the pipe’s radius.

Recasting and pouring. Volume is conserved, so set the two expressions equal. Questions 1, 2, 9, 13, 16 and 17 are variants.

Rise and displacement. An object submerged in a tank raises the level by its own volume spread over the tank’s base:

$$\text{rise} = \frac{\text{volume added}}{\text{area of base}}$$

Questions 10 and 19 are this, the second with rain falling on a roof.

A note on $\pi$ and on rounding. The chapter’s convention is $\pi = \frac{22}{7}$ unless a question says otherwise; Question 14 says to use $3.14$. Several answers here are costs or weights quoted to two decimals, and rounding an intermediate value can shift the last digit — Question 12’s printed answer differs from the exact one by two paise for exactly that reason, and Question 14’s printed volume uses a different $\pi$ from its printed surface area. Both are worked out in full below.

A note on the current syllabus. “Frustum of a Cone” and “Conversion of Solid from One Shape to Another” have both been removed from the Class 10 textbook, which affects Questions 1, 9 and 12 here.

Key insight. Fix your unit before you start, and convert everything into it on the first line. More answers are lost in this exercise to a centimetre meeting a metre than to any geometric error — and the wrong answer always looks reasonable, because it differs by a clean power of ten.

Question 1

A solid metallic hemisphere of radius $8$ cm is melted and recasted into a right circular cone of base radius $6$ cm. Determine the height of the cone.

Solution. Melting preserves volume.

The hemisphere:

$$\frac23\pi r^3 = \frac23\pi (512) = \frac{1024\pi}{3}$$

The cone:

$$\frac13\pi R^2 h = \frac13\pi(36)h = 12\pi h$$

Equate:

$$12\pi h = \frac{1024\pi}{3} \implies h = \frac{1024}{36} = \frac{256}{9} = 28.\overline{4} \text{ cm}$$

So $h \approx 28.44$ cm.

The $\pi$ cancels, so no value is needed — a feature of every pure recasting question. The answer is not a whole number, which is unusual for this book and worth double-checking: $\frac{256}{9} = 28.444\ldots$, and the key’s $28.44$ cm is that rounded to two places.

A plausibility check: the cone is narrower than the hemisphere ($6$ cm against $8$ cm) and a cone is only a third of its enclosing cylinder, so it has to be much taller than the hemisphere’s $8$ cm to hold the same material. Nearly $28.5$ cm is about right.

This question is a conversion of solids, which is not in the current Class 10 textbook.

Question 2

A rectangular water tank of base $11$ m $\times$ $6$ m contains water upto a height of $5$ m. If the water in the tank is transferred to a cylindrical tank of radius $3.5$ m, find the height of the water level in the tank.

Solution. Pouring preserves volume.

The water:

$$11 \times 6 \times 5 = 330 \text{ m}^3$$

In the cylinder, of radius $3.5$ m:

$$\pi r^2 h = \frac{22}{7} \times 12.25 \times h = 38.5h$$

using $\frac{22}{7} \times \frac{49}{4} = \frac{22 \times 7}{4} = 38.5$.

Equate:

$$38.5h = 330 \implies h = \frac{330}{38.5} = \frac{60}{7} = 8.57\ldots \text{ m}$$

So $h \approx 8.6$ m.

Exactly: $\frac{330 \times 7}{22 \times 12.25} = \frac{2310}{269.5} = \frac{60}{7}$.

A check on the shape of the answer: the cylinder’s base area is $38.5$ m² against the rectangle’s $66$ m², so it is narrower and the water must stand higher. It does — $8.6$ m against $5$ m, in the ratio $\frac{66}{38.5} = \frac{12}{7}$, and $5 \times \frac{12}{7} = \frac{60}{7}$ ✓.

Question 3

How many cubic centimetres of iron is required to construct an open box whose external dimensions are $36$ cm, $25$ cm and $16.5$ cm provided the thickness of the iron is $1.5$ cm. If one cubic cm of iron weighs $7.5$ g, find the weight of the box.

Solution. The iron is the shell: external volume minus the hollow inside. The subtlety is that the box is open, so one face has no iron.

The internal dimensions. Length and width each lose $1.5$ cm at both ends:

$$36 – 2(1.5) = 33 \text{ cm}, \qquad 25 – 2(1.5) = 22 \text{ cm}$$

The height loses $1.5$ cm at the bottom only, because there is no lid:

$$16.5 – 1.5 = 15 \text{ cm}$$

The two volumes:

$$\text{external} = 36 \times 25 \times 16.5 = 14\,850 \text{ cm}^3$$

$$\text{internal} = 33 \times 22 \times 15 = 10\,890 \text{ cm}^3$$

The iron:

$$14\,850 – 10\,890 = 3960 \text{ cm}^3$$

The weight:

$$3960 \times 7.5 = 29\,700 \text{ g} = 29.7 \text{ kg}$$

The word “open” is worth a mark. Subtracting $1.5$ twice from the height, as though there were a lid, gives an internal volume of $33 \times 22 \times 13.5 = 9801$ and an iron volume of $5049$ cm³ — about $27\%$ too high.

Question 4

The barrel of a fountain pen, cylindrical in shape, is $7$ cm long and $5$ mm in diameter. A full barrel of ink in the pen is used up on writing $3300$ words on an average. How many words can be written in a bottle of ink containing one fifth of a litre?

Solution. Convert to centimetres first. The barrel’s diameter is $5$ mm $= 0.5$ cm, so $r = 0.25$ cm.

One barrel:

$$\pi r^2 h = \frac{22}{7} \times 0.0625 \times 7 = 22 \times 0.0625 = 1.375 \text{ cm}^3$$

The $7$ cancels the $\frac{22}{7}$ exactly, which is why the length is $7$ cm.

The bottle, one fifth of a litre:

$$\frac{1000}{5} = 200 \text{ cm}^3$$

Number of barrels in the bottle:

$$\frac{200}{1.375} = \frac{1600}{11} = 145.4\overline{54}$$

Words:

$$3300 \times \frac{1600}{11} = 300 \times 1600 = 480\,000$$

The awkward fraction $\frac{1600}{11}$ cancels against the $3300$ — which is why the final answer is a round number and why it is worth not evaluating the division early. Rounding $145.45$ to $145$ would give $478\,500$, short by fifteen hundred words.

Question 5

Water flows at the rate of $10$ m/minute through a cylindrical pipe $5$ mm in diameter. How long would it take to fill a conical vessel whose diameter at the base is $40$ cm and depth $24$ cm?

Solution. Put everything in centimetres. The pipe’s diameter is $5$ mm $= 0.5$ cm so $r = 0.25$ cm; the flow is $10$ m/min $= 1000$ cm/min; the cone has base radius $20$ cm and height $24$ cm.

The vessel’s capacity:

$$\frac13\pi R^2 h = \frac13 \times \frac{22}{7} \times 400 \times 24 = \frac{22 \times 3200}{7} = \frac{70\,400}{7} \text{ cm}^3$$

The delivery rate. In one minute the pipe delivers a cylinder of water $1000$ cm long:

$$\pi r^2 \times 1000 = \frac{22}{7} \times 0.0625 \times 1000 = \frac{1375}{7} \text{ cm}^3\text{/min}$$

The time:

$$t = \frac{70\,400/7}{1375/7} = \frac{70\,400}{1375} = 51.2 \text{ minutes}$$

The sevens cancel, so again no decimal value of $\pi$ is needed.

Convert the decimal part to seconds:

$$0.2 \times 60 = 12 \text{ seconds}$$

So the time is $51$ minutes $12$ seconds.

The answer is expected in minutes and seconds, and $51.2$ minutes is not “$51$ minutes $20$ seconds” — a decimal fraction of a minute is not read off like a clock.

Question 6

A heap of rice is in the form of a cone of diameter $9$ m and height $3.5$ m. Find the volume of the rice. How much canvas cloth is required to just cover the heap?

Solution. The radius is $\frac92 = 4.5$ m.

The volume:

$$\frac13\pi r^2 h = \frac13 \times \frac{22}{7} \times 20.25 \times 3.5 = \frac{22 \times 20.25}{6} \times \frac{3.5 \times 6}{7 \times 3.5}$$

More simply, $\frac{3.5}{7} = 0.5$, so

$$= \frac13 \times 22 \times 20.25 \times 0.5 = \frac{222.75}{3} = 74.25 \text{ m}^3$$

The canvas. Cloth laid over a conical heap covers its sloping surface only — the base is the ground. So the area needed is the curved surface area, and that needs the slant height:

$$l = \sqrt{r^2 + h^2} = \sqrt{20.25 + 12.25} = \sqrt{32.5} \approx 5.7009 \text{ m}$$

$$\text{CSA} = \pi r l = \frac{22}{7} \times 4.5 \times 5.7009 \approx 80.61 \text{ m}^2$$

Two things to get right. The canvas covers the curved surface, not the total surface — including the base would add $\pi r^2 = 63.6$ m² of cloth spread on the ground. And the slant height must be computed; using the vertical height $3.5$ m in place of $l$ would give $49.5$ m², well under what is needed.

Question 7

A factory manufactures $120\,000$ pencils daily. The pencils are cylindrical in shape each of length $25$ cm and circumference of base as $1.5$ cm. Determine the cost of colouring the curved surfaces of the pencils manufactured in one day at ₹$0.05$ per dm².

Solution. The circumference is given directly, which is a kindness — the curved surface of a cylinder is circumference times length, so no radius is needed at all:

$$\text{CSA of one pencil} = 2\pi r \times h = 1.5 \times 25 = 37.5 \text{ cm}^2$$

For the day’s production:

$$120\,000 \times 37.5 = 4\,500\,000 \text{ cm}^2$$

Convert to square decimetres. One decimetre is $10$ cm, so one dm² is $100$ cm²:

$$\frac{4\,500\,000}{100} = 45\,000 \text{ dm}^2$$

The cost:

$$45\,000 \times 0.05 = ₹2250$$

The dm² conversion is where this is lost. A decimetre is ten centimetres, so a square decimetre is a hundred square centimetres — not ten. Dividing by $10$ instead gives ₹$22\,500$, and the unit is unfamiliar enough that the error passes unnoticed.

Working out the radius from the circumference would also work, but it is a detour: $r = \frac{1.5 \times 7}{44}$ is ugly and cancels back out in the very next line.

Question 8

Water is flowing at the rate of $15$ km/h through a pipe of diameter $14$ cm into a cuboidal pond which is $50$ m long and $44$ m wide. In what time will the level of water in pond rise by $21$ cm?

Solution. Work in metres, since the pond is in metres. The pipe’s diameter is $14$ cm $= 0.14$ m, so $r = 0.07$ m; the rise is $21$ cm $= 0.21$ m.

Volume needed:

$$50 \times 44 \times 0.21 = 462 \text{ m}^3$$

The delivery rate. In one hour the pipe delivers a cylinder $15\,000$ m long:

$$\pi r^2 \times 15\,000 = \frac{22}{7} \times 0.0049 \times 15\,000 = 0.0154 \times 15\,000 = 231 \text{ m}^3\text{/hour}$$

using $\frac{22}{7} \times 0.0049 = 22 \times 0.0007 = 0.0154$.

The time:

$$\frac{462}{231} = 2 \text{ hours}$$

The radius of $0.07$ m is chosen so that $\frac{22}{7} \times (0.07)^2$ is exact, and the answer is a whole number of hours — both signs that the conversion has been done correctly. If your answer is not clean, check whether a centimetre has been left among the metres.

Question 9

A solid iron cuboidal block of dimensions $4.4$ m $\times$ $2.6$ m $\times$ $1$ m is recast into a hollow cylindrical pipe of internal radius $30$ cm and thickness $5$ cm. Find the length of the pipe.

Solution. Convert the block to cubic centimetres, since the pipe is in centimetres:

$$4.4 \text{ m} \times 2.6 \text{ m} \times 1 \text{ m} = 440 \times 260 \times 100 = 11\,440\,000 \text{ cm}^3$$

The pipe’s cross-section is an annulus. Internal radius $30$ cm and thickness $5$ cm give external radius $35$ cm:

$$\pi\left(R^2 – r^2\right) = \frac{22}{7}\left(1225 – 900\right) = \frac{22}{7} \times 325 = \frac{7150}{7} \text{ cm}^2$$

The length. Volume is cross-section times length:

$$L = \frac{11\,440\,000}{7150/7} = \frac{11\,440\,000 \times 7}{7150} = 1600 \times 7 = 11\,200 \text{ cm}$$

using $11\,440\,000 \div 7150 = 1600$.

$$11\,200 \text{ cm} = 112 \text{ m}$$

Two traps. The thickness is added to the internal radius to get the external one — not treated as the external radius itself. And the pipe is hollow, so only the annulus is iron; using the full circle $\pi R^2$ would give a much shorter pipe.

This question is a conversion of solids, which is not in the current Class 10 textbook.

Question 10

$500$ persons are taking a dip into a cuboidal pond which is $80$ m long and $50$ m broad. What is the rise of water level in the pond, if the average displacement of the water by a person is $0.04$ m³?

Solution. Each person displaces their own volume of water, so the total volume added is

$$500 \times 0.04 = 20 \text{ m}^3$$

That volume spreads over the pond’s surface, which has area

$$80 \times 50 = 4000 \text{ m}^2$$

The rise:

$$\frac{20}{4000} = 0.005 \text{ m} = 0.5 \text{ cm}$$

The depth of the pond is never mentioned and is not needed — only the base area matters, because the added water forms a thin layer of that area across the top.

The answer is half a centimetre, which is the kind of small number worth sanity-checking: five hundred people displacing $40$ litres each is $20\,000$ litres, spread over a pond the size of half a football pitch. A few millimetres is exactly right.

Question 11

$16$ glass spheres each of radius $2$ cm are packed into a cuboidal box of internal dimensions $16$ cm $\times$ $8$ cm $\times$ $8$ cm and then the box is filled with water. Find the volume of water filled in the box.

Solution. The water occupies whatever the spheres leave behind.

The box:

$$16 \times 8 \times 8 = 1024 \text{ cm}^3$$

The sixteen spheres:

$$16 \times \frac43\pi r^3 = 16 \times \frac43 \times \frac{22}{7} \times 8 = 16 \times \frac{704}{21} = \frac{11\,264}{21} \approx 536.38 \text{ cm}^3$$

The water:

$$1024 – \frac{11\,264}{21} = \frac{21\,504 – 11\,264}{21} = \frac{10\,240}{21} \approx 487.6 \text{ cm}^3$$

A check that the spheres actually fit. Each has diameter $4$ cm, so the box takes $4$ along the length, $2$ across the width and $2$ in height: $4 \times 2 \times 2 = 16$ ✓. They fill $52.4\%$ of the box — the same $\frac{\pi}{6}$ fraction as a single sphere in its own cube, which is what a regular packing of touching spheres always gives.

The water is therefore a little under half the box, which is the plausibility check worth making before writing the answer down.

Question 12

A milk container of height $16$ cm is made of metal sheet in the form of a frustum of a cone with radii of its lower and upper ends as $8$ cm and $20$ cm respectively. Find the cost of milk at the rate of ₹$22$ per litre which the container can hold.

Solution. The capacity, from the frustum volume formula with $h = 16$, $r_1 = 20$, $r_2 = 8$:

$$V = \frac{\pi h}{3}\left(r_1^2 + r_2^2 + r_1 r_2\right) = \frac{22}{7} \times \frac{16}{3} \times \left(400 + 64 + 160\right)$$

The bracket is $624$, so

$$V = \frac{22 \times 16 \times 624}{21} = \frac{219\,648}{21} = 10\,459.43 \text{ cm}^3$$

In litres:

$$\frac{10\,459.43}{1000} = 10.45943 \text{ litres}$$

The cost:

$$10.45943 \times 22 = ₹230.107 \approx ₹230.11$$

A note on the printed answer. The key gives ₹$230.12$, which is $10.46 \times 22$ — the volume rounded to two decimal places before multiplying. Rounding the capacity first and the cost second gives $230.12$; carrying the exact value through and rounding only at the end gives $230.11$.

Two paise, and neither method is wrong in itself. But the habit worth having is to round once, at the end — intermediate rounding compounds, and on a larger bill it would matter more than this.

This question uses the frustum volume formula, which is not in the current Class 10 textbook.

Question 13

A cylindrical bucket of height $32$ cm and base radius $18$ cm is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is $24$ cm, find the radius and slant height of the heap.

Solution. Emptying preserves volume.

The bucket:

$$\pi R^2 H = \pi (324)(32) = 10\,368\pi \text{ cm}^3$$

The heap, a cone of height $24$ cm and unknown radius $r$:

$$\frac13\pi r^2 (24) = 8\pi r^2$$

Equate:

$$8\pi r^2 = 10\,368\pi \implies r^2 = 1296 \implies r = 36 \text{ cm}$$

The slant height:

$$l = \sqrt{r^2 + h^2} = \sqrt{1296 + 576} = \sqrt{1872} \approx 43.27 \text{ cm}$$

Exactly, $\sqrt{1872} = \sqrt{144 \times 13} = 12\sqrt{13}$.

The heap is wider than the bucket ($36$ cm against $18$ cm) and shorter ($24$ cm against $32$ cm), which is what you would expect from sand poured out and left to spread: a cone holds only a third of its enclosing cylinder, so it has to be much broader to hold the same amount.

Question 14

A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of the cylinder. The diameter and height of the cylinder are $6$ cm and $12$ cm, respectively. If the slant height of the conical portion is $5$ cm, find the total surface area and volume of the rocket. [Use $\pi = 3.14$]

Solution. The radius is $3$ cm throughout, and the cone’s height comes from its slant height by Pythagoras:

$$h_{\text{cone}} = \sqrt{l^2 – r^2} = \sqrt{25 – 9} = 4 \text{ cm}$$

the $3$–$4$–$5$ triangle.

The total surface area. Go round the rocket and ask what is exposed:

• curved surface of the cone: $\pi r l = \pi (3)(5) = 15\pi$
• curved surface of the cylinder: $2\pi r h = 2\pi(3)(12) = 72\pi$
• base of the cylinder: $\pi r^2 = 9\pi$ — the rocket is closed at the lower end
• the join: the cone’s base and the cylinder’s top are in contact, so neither counts

$$\text{TSA} = 15\pi + 72\pi + 9\pi = 96\pi = 96 \times 3.14 = 301.44 \text{ cm}^2$$

The volume:

$$\pi r^2 h_{\text{cyl}} + \frac13 \pi r^2 h_{\text{cone}} = \pi(9)(12) + \frac13\pi(9)(4) = 108\pi + 12\pi = 120\pi$$

$$= 120 \times 3.14 = 376.8 \text{ cm}^3$$

A note on the printed answer. The key gives $301.44$ cm² and $377.1$ cm³. The first is exactly $96 \times 3.14$, following the question’s instruction. The second is not $120 \times 3.14 = 376.8$ — it is $120 \times \frac{22}{7} = 377.14$. So the key uses $3.14$ for the surface area and $\frac{22}{7}$ for the volume, within a single question that specifies $3.14$.

Following the question as printed gives $376.8$ cm³. Either value will be accepted provided you say which $\pi$ you used; the point is to use one of them throughout.

Question 15

A building is in the form of a cylinder surmounted by a hemispherical vaulted dome and contains $41\frac{19}{21}$ m³ of air. If the internal diameter of dome is equal to its total height above the floor, find the height of the building?

Solution. Set up one variable. Let the dome’s radius be $r$, so its diameter is $2r$.

The condition says that diameter equals the building’s total height:

$$\text{total height} = 2r$$

The total height is the cylinder plus the dome, and a hemisphere of radius $r$ stands $r$ tall:

$$h_{\text{cyl}} + r = 2r \implies h_{\text{cyl}} = r$$

So the cylindrical part is exactly as tall as the dome’s radius. Everything is now in terms of $r$.

The volume of air:

$$\pi r^2 h_{\text{cyl}} + \frac23 \pi r^3 = \pi r^3 + \frac23 \pi r^3 = \frac53 \pi r^3$$

Set equal to the given value. Convert the mixed number first:

$$41\frac{19}{21} = \frac{41 \times 21 + 19}{21} = \frac{861 + 19}{21} = \frac{880}{21}$$

So

$$\frac53 \times \frac{22}{7} \times r^3 = \frac{880}{21} \implies \frac{110}{21}r^3 = \frac{880}{21} \implies r^3 = 8 \implies r = 2$$

The height of the building:

$$2r = 4 \text{ m}$$

The awkward mixed number is there so that $r^3$ comes out as $8$ — if yours is not a perfect cube, the condition linking the diameter to the total height has probably been used wrongly.

Question 16

A hemispherical bowl of internal radius $9$ cm is full of liquid. The liquid is to be filled into cylindrical shaped bottles each of radius $1.5$ cm and height $4$ cm. How many bottles are needed to empty the bowl?

Solution. Pouring preserves volume.

The bowl:

$$\frac23\pi r^3 = \frac23\pi(729) = 486\pi \text{ cm}^3$$

One bottle:

$$\pi R^2 h = \pi (2.25)(4) = 9\pi \text{ cm}^3$$

The count:

$$\frac{486\pi}{9\pi} = 54$$

The $\pi$ cancels — as it does in every question where both solids are round — so no value is needed and the answer is exact.

$54$ bottles, and the answer is a whole number, which is what the question expects. Had it come out as, say, $54.3$, the honest answer would be $55$ bottles, since a fraction of a bottle still needs a whole bottle to hold it. Questions of this kind always round up, never to the nearest.

Question 17

A solid right circular cone of height $120$ cm and radius $60$ cm is placed in a right circular cylinder full of water of height $180$ cm such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is equal to the radius of the cone.

Solution. The cone displaces its own volume of water, which overflows. What is left is the cylinder minus the cone.

The cylinder, radius $60$ cm and height $180$ cm:

$$\pi r^2 h = \pi(3600)(180) = 648\,000\pi \text{ cm}^3$$

The cone, radius $60$ cm and height $120$ cm:

$$\frac13\pi r^2 h = \frac13 \pi (3600)(120) = 144\,000\pi \text{ cm}^3$$

Water left:

$$648\,000\pi – 144\,000\pi = 504\,000\pi = 504\,000 \times \frac{22}{7} = 1\,584\,000 \text{ cm}^3$$

Convert to cubic metres. One cubic metre is $100^3 = 1\,000\,000$ cm³:

$$\frac{1\,584\,000}{1\,000\,000} = 1.584 \text{ m}^3$$

The cone is shorter than the cylinder ($120$ cm against $180$ cm) and the same width, so it sits wholly inside — no part of it projects above the rim, and the whole of it displaces water.

Note the conversion factor: a million, not a hundred. Three lengths each multiplied by $100$ means the volume is multiplied by $10^6$.

Question 18

Water flows through a cylindrical pipe, whose inner radius is $1$ cm, at the rate of $80$ cm/sec in an empty cylindrical tank, the radius of whose base is $40$ cm. What is the rise of water level in tank in half an hour?

Solution. Convert the time. Half an hour is

$$30 \times 60 = 1800 \text{ seconds}$$

Volume delivered. In $1800$ seconds the pipe delivers a cylinder of water $80 \times 1800 = 144\,000$ cm long and $1$ cm in radius:

$$\pi (1)^2 (144\,000) = 144\,000\pi \text{ cm}^3$$

The rise. That volume spreads over the tank’s base, of radius $40$ cm:

$$\text{rise} = \frac{144\,000\pi}{\pi(40)^2} = \frac{144\,000}{1600} = 90 \text{ cm}$$

The $\pi$ cancels, so the answer is exact and needs no value for it.

A faster route. The rise is the pipe’s length of flow scaled by the ratio of the cross-sections:

$$144\,000 \times \left(\frac{1}{40}\right)^2 = \frac{144\,000}{1600} = 90 \text{ cm}$$

The tank is $40$ times as wide, so $1600$ times the area, so the water rises $1600$ times more slowly than it travels down the pipe. That ratio form is worth having, because it makes the size of the answer obvious before any arithmetic.

Question 19

The rain water from a roof of dimensions $22$ m $\times$ $20$ m drains into a cylindrical vessel having diameter of base $2$ m and height $3.5$ m. If the rain water collected from the roof just fill the cylindrical vessel, then find the rainfall in cm.

Solution. “Rainfall” is the depth of water that fell — the height of the layer the rain would have made on the roof had none drained away.

The vessel’s capacity, radius $1$ m and height $3.5$ m:

$$\pi r^2 h = \frac{22}{7} \times 1 \times 3.5 = 11 \text{ m}^3$$

The roof’s area:

$$22 \times 20 = 440 \text{ m}^2$$

The rainfall:

$$\text{depth} = \frac{\text{volume}}{\text{area}} = \frac{11}{440} = 0.025 \text{ m} = 2.5 \text{ cm}$$

The same relation as Question 10 — volume divided by area gives a depth — but read the other way round: there the volume was known and the rise was wanted; here the depth is the answer and the volume comes from the vessel.

The result is a plausible figure for a heavy shower. If yours comes out in metres, the conversion to centimetres has been missed: $0.025$ m is $2.5$ cm, not $25$ cm.

Question 20

A pen stand made of wood is in the shape of a cuboid with four conical depressions and a cubical depression to hold the pens and pins, respectively. The dimension of the cuboid are $10$ cm, $5$ cm and $4$ cm. The radius of each of the conical depressions is $0.5$ cm and the depth is $2.1$ cm. The edge of the cubical depression is $3$ cm. Find the volume of the wood in the entire stand.

Solution. The wood is the cuboid with five holes scooped out of it — four conical and one cubical.

The cuboid:

$$10 \times 5 \times 4 = 200 \text{ cm}^3$$

The four conical depressions. Each has radius $0.5$ cm and depth $2.1$ cm:

$$\frac13\pi r^2 h = \frac13 \times \frac{22}{7} \times 0.25 \times 2.1 = \frac{22 \times 0.25 \times 2.1}{21} = \frac{11.55}{21} = 0.55 \text{ cm}^3$$

The depth of $2.1$ cm cancels the $7$ in $\frac{22}{7}$, which is why it was chosen. Four of them:

$$4 \times 0.55 = 2.2 \text{ cm}^3$$

The cubical depression:

$$3^3 = 27 \text{ cm}^3$$

The wood remaining:

$$200 – 2.2 – 27 = 170.8 \text{ cm}^3$$

A check that the holes fit. The cuboid is $4$ cm deep, and the deepest hole is the $3$ cm cube, so nothing breaks through the bottom. The conical depressions are only $1$ cm across and $2.1$ cm deep — also comfortably inside.

Both kinds of depression are subtracted; neither is added. Scooping wood out always reduces the volume, whatever it does to the surface area.

Common mistakes

• Questions 5, 7, 8, 9, 17 and 19 — mixing units. A metre meeting a centimetre is the commonest way to lose a mark here, and the answer always comes out wrong by a clean power of ten, so it never looks absurd. Convert everything on the first line.
• Question 7 — converting cm² to dm². A decimetre is $10$ cm, so a square decimetre is $100$ cm². Dividing by $10$ gives a cost ten times too high.
• Question 17 — converting cm³ to m³. The factor is $10^6$, not $100$. Three lengths each scale by $100$, so the volume scales by $100^3$.
• Question 3 — forgetting that the box is open. With no lid the height loses only one thickness, not two. Subtracting $1.5$ twice overstates the iron by about $27\%$.
• Question 6 — using the vertical height as the slant height. Canvas covers the sloping surface, so $l = \sqrt{r^2 + h^2}$ has to be computed. And it covers the curved surface only; the base is the ground.
• Question 14 — mixing values of $\pi$ within a question. Pick the one the question specifies and use it for every part. The printed key here does not, which is worth knowing if your answer and the book’s differ in the last digit.
• Questions 12 and 14 — rounding intermediate values. Round once, at the end. Rounding the capacity to two decimals before multiplying by the rate is what puts the key’s answer two paise above the exact one.
• Question 16 — rounding a count of containers down. A fraction of a bottle still needs a whole bottle. These counts always round up.

Practise next

• Exercises 13.1 and 13.2 — Statistics and Probability, the final unit, and a complete change of subject after five units of geometry.
• Exercise 12.1 — worth a second pass, especially Questions 8, 11 and 13, where the same exact-arithmetic discipline decides between options that are one apart.
• Exercise 11.4 — its cost, rate and annulus questions are the two-dimensional version of the reasoning here.