NCERT Exemplar Solutions Class 10 Maths Surface Areas and Volumes — Exercises 12.2 and 12.3

NCERT Exemplar Problems, Class 10 Mathematics — Surface Areas and Volumes, Exercises 12.2 and 12.3. All 22 questions solved.

Exercise 12.2 is eight statements to judge, and Exercise 12.3 is fourteen calculations. Between them they cover the two things this chapter is really about.

Combining solids. When two solids are stuck together, the faces in contact stop being surface. So the total surface area of the combination is

$$S_1 + S_2 – 2 \times (\text{area of the face in contact})$$

— you lose it twice, once from each solid. Five of Exercise 12.2’s eight questions are that one rule, and getting it right is mostly a matter of drawing the join and asking what is now hidden.

Recasting. Melting changes the shape and nothing else, so

$$\text{volume before} = \text{volume after}$$

Set the two expressions equal and solve. Exercise 12.3’s Questions 1, 2, 11, 12 and 14 are all this.

A note on the current syllabus. The rationalised Class 10 chapter has lost two of its four sections: “Frustum of a Cone” and “Conversion of Solid from One Shape to Another” are both gone, and the words “frustum” and “conversion” occur nowhere in it. Ten of these twenty-two questions depend on one or the other. They are flagged where they appear.

Key insight. For a surface-area question, ask what you could still touch. For a volume question, ask what is still there. Those two questions decide every problem in Exercise 12.2 — a face in contact is no longer touchable, but a hemisphere scooped out of a cylinder removes volume while adding surface.

Exercise 12.2 — Write ‘True’ or ‘False’ and justify your answer in the following:

Question 1

Two identical solid hemispheres of equal base radius $r$ cm are stuck together along their bases. The total surface area of the combination is $6\pi r^2$.

Solution. Two hemispheres joined along their flat faces make a sphere, whose surface area is

$$4\pi r^2$$

The statement claims $6\pi r^2$, so it is false.

Where $6\pi r^2$ comes from. Each hemisphere on its own has total surface area

$$\underbrace{2\pi r^2}_{\text{curved}} + \underbrace{\pi r^2}_{\text{flat circle}} = 3\pi r^2$$

and two of those add to $6\pi r^2$. But when they are stuck together the two flat circles meet each other and become internal — you can no longer touch either. Subtracting them both:

$$6\pi r^2 – 2\pi r^2 = 4\pi r^2 \ \checkmark$$

That subtraction is the rule for every joined solid in this exercise, and the sphere is the check that it has been applied correctly.

Question 2

A solid cylinder of radius $r$ and height $h$ is placed over other cylinder of same height and radius. The total surface area of the shape so formed is $4\pi rh + 4\pi r^2$.

Solution. Two identical cylinders stacked make one cylinder of radius $r$ and height $2h$. Its total surface area is

$$2\pi r (2h) + 2\pi r^2 = 4\pi r h + 2\pi r^2$$

The statement claims $4\pi rh + 4\pi r^2$, which has twice the correct circular-face term. False.

The same accounting as Question 1. Separately the two cylinders have $2(2\pi rh + 2\pi r^2) = 4\pi rh + 4\pi r^2$ — exactly the claimed value. Joining them buries one circular face from each:

$$4\pi rh + 4\pi r^2 – 2\pi r^2 = 4\pi rh + 2\pi r^2 \ \checkmark$$

So the printed expression is what you get by adding the two totals and forgetting to remove the join. It is wrong by $2\pi r^2$, the area of the two hidden circles.

Question 3

A solid cone of radius $r$ and height $h$ is placed over a solid cylinder having same base radius and height as that of a cone. The total surface area of the combined solid is $\pi r\left[\sqrt{r^2 + h^2} + 3r + 2h\right]$.

Solution. Build the surface up piece by piece, asking each time whether the piece is still touchable.

• Curved surface of the cone: $\pi r l = \pi r\sqrt{r^2 + h^2}$ — visible.
• Curved surface of the cylinder: $2\pi rh$ — visible.
• Base of the cylinder: $\pi r^2$ — visible, it is the bottom of the whole solid.
• Base of the cone and top of the cylinder: these are the two faces in contact — hidden.

Adding the three visible parts:

$$\pi r\sqrt{r^2+h^2} + 2\pi rh + \pi r^2 = \pi r\left[\sqrt{r^2+h^2} + 2h + r\right]$$

The statement has $3r$ where this has $r$. False.

The discrepancy is exactly

$$\pi r(3r) – \pi r(r) = 2\pi r^2$$

— again the two faces at the join, counted rather than removed. All three of Questions 1, 2 and 3 fail by the same $2\pi r^2$, which is the most reliable sign that the rule is the one being tested.

Question 4

A solid ball is exactly fitted inside the cubical box of side $a$. The volume of the ball is $\frac43\pi a^3$.

Solution. “Exactly fitted” means the ball touches all six faces, so its diameter equals the side of the cube:

$$2R = a \implies R = \frac{a}{2}$$

Therefore

$$V = \frac43\pi R^3 = \frac43 \pi \left(\frac a2\right)^3 = \frac43 \pi \cdot \frac{a^3}{8} = \frac{\pi a^3}{6}$$

The claim of $\frac43\pi a^3$ is eight times too big. False.

A check that settles it without algebra: the cube itself has volume $a^3$, and the ball is inside it, so the ball’s volume must be less than $a^3$. But $\frac43\pi a^3 \approx 4.19a^3$ — more than four times the box that contains it. Whenever a claimed volume exceeds its container, a radius has been used where a diameter belongs.

The correct fraction is $\frac{\pi}{6} \approx 52.4\%$: an inscribed sphere fills a little over half its cube.

Question 5

The volume of the frustum of a cone is $\frac13 \pi h\left[r_1^2 + r_2^2 – r_1r_2\right]$, where $h$ is vertical height of the frustum and $r_1$, $r_2$ are the radii of the ends.

Solution. The correct formula has a plus before the cross term:

$$V = \frac13\pi h\left[r_1^2 + r_2^2 + r_1 r_2\right]$$

The statement prints a minus, so it is false.

A test that needs no memory. Set $r_1 = r_2 = r$ — a frustum with equal ends is just a cylinder, whose volume is $\pi r^2 h$. Check both versions:

$$\text{correct: } \ \frac13\pi h\left[r^2 + r^2 + r^2\right] = \pi r^2 h \ \checkmark$$

$$\text{printed: } \ \frac13\pi h\left[r^2 + r^2 – r^2\right] = \frac{\pi r^2 h}{3} \ \times$$

The printed version gives a third of the right answer. A second check: set $r_2 = 0$, so the frustum becomes a cone. The correct formula gives $\frac13\pi r_1^2 h$ ✓, and so does the printed one — which is why the error survives casual checking and why the equal-radii test is the better one.

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

Question 6

The capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom as shown in the Fig. 12.7 is $\frac{\pi r^2}{3}\left[3h – 2r\right]$.

Solution. Capacity means what the vessel can hold, not how much material it is made of. The base has been pushed up into a dome, so the space available is the cylinder minus the hemisphere bulging into it.

$$\text{capacity} = \pi r^2 h – \frac23\pi r^3$$

Factor out $\frac{\pi r^2}{3}$:

$$= \frac{\pi r^2}{3}\left(3h – 2r\right)$$

which is exactly the printed expression. True.

Why a maker would do this. The raised base is the punt in a wine bottle. It costs some capacity but makes the vessel stand more steadily and strengthens the base — and the formula quantifies the cost as $\frac23\pi r^3$.

Note that the condition $3h > 2r$ is implicit: the hemisphere must not be taller than the cylinder, or there would be no capacity at all. Since a hemisphere of radius $r$ is $r$ tall, this needs $h > r$, which is comfortably stronger than $3h > 2r$.

Question 7

The curved surface area of a frustum of a cone is $\pi l\left(r_1 + r_2\right)$, where $l = \sqrt{h^2 + \left(r_1 + r_2\right)^2}$, $r_1$ and $r_2$ are the radii of the two ends of the frustum and $h$ is the vertical height.

Solution. The formula for the area is right; the formula for the slant height is wrong. It should be

$$l = \sqrt{h^2 + \left(r_1 – r_2\right)^2}$$

with a minus inside the bracket. The statement prints a plus, so it is false.

Where the minus comes from. Stand the frustum up and drop a perpendicular from the rim of the smaller end to the plane of the larger. That makes a right-angled triangle whose vertical leg is the height $h$ and whose horizontal leg is the overhang — how much further out the bigger circle reaches:

$$r_1 – r_2$$

The slant side is the hypotenuse, so Pythagoras gives $l^2 = h^2 + (r_1-r_2)^2$.

A test of the printed version. Set $r_1 = r_2 = r$, making the frustum a cylinder, whose slant height is just its vertical height $h$. The correct formula gives $\sqrt{h^2 + 0} = h$ ✓. The printed one gives $\sqrt{h^2 + 4r^2}$, which is too big — and would be so for every cylinder.

This question uses the frustum formulae, which are not in the current Class 10 textbook.

Question 8

An open metallic bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The surface area of the metallic sheet used is equal to curved surface area of frustum of a cone $+$ area of circular base $+$ curved surface area of cylinder.

Solution. Go round the object and ask which surfaces are made of sheet.

• The sloping wall of the bucket — the curved surface of the frustum. Sheet ✓
• The bottom of the bucket — a flat circle, the smaller end of the frustum. Sheet ✓
• The cylindrical base ring — its curved surface. Sheet ✓
• The top of the bucket — the larger circular end. It is open, so there is no sheet there ✗
• The two ends of the hollow cylinder — it is a hollow ring, open at both ends, so neither is a face ✗

Adding what remains gives exactly the three terms listed. True.

The two exclusions are the question. Adding the top would be wrong because the bucket is open — that is why the word is in the sentence. Adding ends to the cylinder would be wrong because it is hollow — a ring, not a tin.

Notice also that “area of circular base” means the frustum’s smaller end, of radius $r_2$, which is the bucket’s floor. The cylindrical ring sits underneath it as a stand and hides nothing, because a hollow ring has no face to hide.

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

Exercise 12.3

Question 9

Three metallic solid cubes whose edges are $3$ cm, $4$ cm and $5$ cm are melted and formed into a single cube. Find the edge of the cube so formed.

(Exercise 12.3, Question 1.)

Solution. Melting preserves volume, so add the three and take the cube root.

$$3^3 + 4^3 + 5^3 = 27 + 64 + 125 = 216$$

$$\text{edge} = \sqrt[3]{216} = 6 \text{ cm}$$

The identity $3^3 + 4^3 + 5^3 = 6^3$ is a small curiosity in its own right — the only such relation among four consecutive-ish small cubes — and it is why the question uses these numbers.

Note that the edges do not simply add: $3 + 4 + 5 = 12$, not $6$. Volumes add, lengths do not. The new cube is only half as long as that sum, because doubling an edge multiplies volume by eight.

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

Question 10

How many shots each having diameter $3$ cm can be made from a cuboidal lead solid of dimensions $9$ cm $\times$ $11$ cm $\times$ $12$ cm?

(Exercise 12.3, Question 2.)

Solution. The cuboid:

$$9 \times 11 \times 12 = 1188 \text{ cm}^3$$

One shot, of diameter $3$ cm so radius $\frac32$ cm:

$$\frac43\pi r^3 = \frac43 \times \frac{22}{7} \times \frac{27}{8} = \frac{22}{7} \times \frac92 = \frac{99}{7} \text{ cm}^3$$

The count:

$$\frac{1188}{\frac{99}{7}} = 1188 \times \frac{7}{99} = 12 \times 7 = 84$$

using $1188 \div 99 = 12$.

The numbers are chosen so that the division is exact, which is the signal to keep $\frac{22}{7}$ as a fraction. In decimals the shot volume is $14.142857\ldots$ cm³ and a rounded value gives $83.9$ or $84.1$, leaving you to guess.

In practice one could not really get $84$ spheres from the block — spheres do not pack without gaps — but the question is about volume, not packing, and says nothing about waste.

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

Question 11

A bucket is in the form of a frustum of a cone and holds $28.490$ litres of water. The radii of the top and bottom are $28$ cm and $21$ cm, respectively. Find the height of the bucket.

(Exercise 12.3, Question 3.)

Solution. Convert the capacity to cubic centimetres first:

$$28.490 \text{ litres} = 28\,490 \text{ cm}^3$$

Now use the frustum volume formula with $r_1 = 28$, $r_2 = 21$:

$$V = \frac{\pi h}{3}\left(r_1^2 + r_2^2 + r_1r_2\right) = \frac{22}{7} \times \frac h3 \times \left(784 + 441 + 588\right)$$

The bracket is $1813$, and $1813 \div 7 = 259$:

$$= 22 \times 259 \times \frac h3 = \frac{5698h}{3}$$

Setting this equal to $28\,490$:

$$h = \frac{28\,490 \times 3}{5698} = \frac{85\,470}{5698} = 15 \text{ cm}$$

All three terms of the bracket are needed — dropping the cross term $r_1r_2 = 588$ would give $1225$ and a height of about $22$ cm.

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

Question 12

A cone of radius $8$ cm and height $12$ cm is divided into two parts by a plane through the mid-point of its axis parallel to its base. Find the ratio of the volumes of two parts.

(Exercise 12.3, Question 4.)

Solution. The cut is halfway up the axis and parallel to the base, so the small cone above it is similar to the whole cone with scale factor $\frac12$. Its radius is therefore $4$ cm and its height $6$ cm.

The small cone:

$$\frac13\pi(4)^2(6) = 32\pi \text{ cm}^3$$

The whole cone:

$$\frac13\pi(8)^2(12) = 256\pi \text{ cm}^3$$

The frustum is what is left:

$$256\pi – 32\pi = 224\pi \text{ cm}^3$$

The ratio:

$$32\pi : 224\pi = 1 : 7$$

taking the small cone first. The printed answer gives $7 : 1$, which is the same result with the frustum named first — the question does not say which part to list first, so either is correct provided you say which is which.

The quick way, worth knowing. Volumes of similar solids go as the cube of the scale factor, so the small cone is $\left(\frac12\right)^3 = \frac18$ of the whole. That leaves $\frac78$ for the frustum, and

$$\frac18 : \frac78 = 1 : 7$$

with no volume ever computed. Halving the height does not halve the volume — it divides it by eight, and that is the whole content of the question.

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

Question 13

Two identical cubes each of volume $64$ cm³ are joined together end to end. What is the surface area of the resulting cuboid?

(Exercise 12.3, Question 5.)

Solution. Find the edge first:

$$\text{edge} = \sqrt[3]{64} = 4 \text{ cm}$$

Joining two such cubes end to end gives a cuboid of dimensions

$$8 \times 4 \times 4 \text{ cm}$$

Its surface area:

$$2(lb + bh + hl) = 2\left(8{\times}4 + 4{\times}4 + 4{\times}8\right) = 2\left(32 + 16 + 32\right) = 2 \times 80 = 160 \text{ cm}^2$$

A check using the joining rule. Each cube alone has surface area $6 \times 16 = 96$ cm², so two have $192$ cm². Joining buries one face from each, of area $16$ cm²:

$$192 – 2(16) = 192 – 32 = 160 \text{ cm}^2 \ \checkmark$$

The same $-2 \times (\text{face})$ correction as Questions 1, 2 and 3 of Exercise 12.2, this time with the arithmetic done both ways so the two agree.

Question 14

From a solid cube of side $7$ cm, a conical cavity of height $7$ cm and radius $3$ cm is hollowed out. Find the volume of the remaining solid.

(Exercise 12.3, Question 6.)

Solution. Hollowing out removes volume, so subtract.

The cube:

$$7^3 = 343 \text{ cm}^3$$

The conical cavity:

$$\frac13 \pi r^2 h = \frac13 \times \frac{22}{7} \times 9 \times 7 = \frac{22 \times 9}{3} = 66 \text{ cm}^3$$

The remaining solid:

$$343 – 66 = 277 \text{ cm}^3$$

The cone’s height equals the cube’s side, so it runs the full depth — the cavity opens on one face and its tip just reaches the opposite one. And its radius of $3$ cm is less than half the side, so the opening fits comfortably within the face.

Note that the surface area would go up, not down. Hollowing removes the circular opening ($\pi r^2$) from one face but adds the cone’s curved surface ($\pi r l$), and since $l > r$ the exchange is a net gain. Volume and surface area move in opposite directions here, which is worth holding on to.

Question 15

Two cones with same base radius $8$ cm and height $15$ cm are joined together along their bases. Find the surface area of the shape so formed.

(Exercise 12.3, Question 7.)

Solution. Joined base to base, the two cones form a double cone — like a spinning top with points at both ends. The two circular bases meet and become internal, so the surface is the two curved surfaces only.

The slant height, by Pythagoras:

$$l = \sqrt{r^2 + h^2} = \sqrt{64 + 225} = \sqrt{289} = 17 \text{ cm}$$

the $8$–$15$–$17$ triangle.

The surface area:

$$2 \times \pi r l = 2 \times \frac{22}{7} \times 8 \times 17 = \frac{5984}{7} = 854.857\ldots \approx 855 \text{ cm}^2$$

Adding the two bases would give a further $2\pi r^2 = 402$ cm² of surface that does not exist — the same error as Questions 1 to 3 of Exercise 12.2, and here it would inflate the answer by nearly half.

The key prints $855$ cm² (approx); the exact value is $\frac{5984}{7}$.

Question 16

Two solid cones $A$ and $B$ are placed in a cylindrical tube as shown in the Fig. 12.9. The ratio of their capacities are $2 : 1$. Find the heights and capacities of cones. Also, find the volume of the remaining portion of the cylinder.

(Exercise 12.3, Question 8.)

Solution. The tube is $21$ cm long and $6$ cm across, so the shared radius is $r = 3$ cm. The two cones sit at opposite ends with their tips meeting, so their heights fill the tube:

$$h_A + h_B = 21$$

The ratio fixes the heights. Both cones have the same radius, so

$$\frac{V_A}{V_B} = \frac{\frac13\pi r^2 h_A}{\frac13\pi r^2 h_B} = \frac{h_A}{h_B} = \frac21$$

Equal radii mean volume is proportional to height alone. With $h_A = 2h_B$ and $h_A + h_B = 21$:

$$3h_B = 21 \implies h_B = 7 \text{ cm}, \qquad h_A = 14 \text{ cm}$$

The capacities:

$$V_A = \frac13 \times \frac{22}{7} \times 9 \times 14 = 132 \text{ cm}^3, \qquad V_B = \frac13 \times \frac{22}{7} \times 9 \times 7 = 66 \text{ cm}^3$$

and $132 : 66 = 2 : 1$ ✓.

The remaining portion. The cylinder holds

$$\pi r^2 h = \frac{22}{7} \times 9 \times 21 = 594 \text{ cm}^3$$

so what is left after the cones is

$$594 – 132 – 66 = 396 \text{ cm}^3$$

Worth noticing: the two cones together occupy $198$ cm³, which is exactly one third of the cylinder — as they must be, since a cone of the same radius and height is a third of its cylinder, and here the two heights add to the full length.

Question 17

An ice cream cone full of ice cream having radius $5$ cm and height $10$ cm as shown in the Fig. 12.10. Calculate the volume of ice cream, provided that its $\frac16$ part is left unfilled with ice cream.

(Exercise 12.3, Question 9.)

Solution. The $10$ cm is the total height, not the cone’s, and that is the whole question. The dimension runs from the top of the hemispherical scoop down to the tip of the cone.

The hemisphere has radius $5$ cm, so it stands $5$ cm tall. The cone below it therefore has height

$$10 – 5 = 5 \text{ cm}$$

The cone:

$$\frac13\pi r^2 h = \frac13 \times \frac{22}{7} \times 25 \times 5 = \frac{2750}{21} \text{ cm}^3$$

The hemispherical top:

$$\frac23\pi r^3 = \frac23 \times \frac{22}{7} \times 125 = \frac{5500}{21} \text{ cm}^3$$

Total volume of the shape:

$$\frac{2750 + 5500}{21} = \frac{8250}{21} = \frac{2750}{7} \approx 392.86 \text{ cm}^3$$

The ice cream fills $\frac56$ of it, since $\frac16$ is left unfilled:

$$\frac56 \times \frac{2750}{7} = \frac{13750}{42} = \frac{6875}{21} \approx 327.4 \text{ cm}^3$$

What taking $10$ cm as the cone’s height would give. The total would be $\frac{250\pi}{3} + \frac{250\pi}{3} = \frac{500\pi}{3} \approx 523.8$ cm³, and five-sixths of that is about $436.5$ cm³ — a third too large. The figure settles it: its dimension arrow spans the entire solid.

Question 18

Marbles of diameter $1.4$ cm are dropped into a cylindrical beaker of diameter $7$ cm containing some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by $5.6$ cm.

(Exercise 12.3, Question 10.)

Solution. Each marble pushes aside its own volume of water, so the rise in level corresponds to the total volume of marbles dropped in.

Water displaced, a cylinder of radius $3.5$ cm and height $5.6$ cm:

$$\pi r^2 h = \frac{22}{7} \times 12.25 \times 5.6 = 22 \times 1.75 \times 5.6 = 215.6 \text{ cm}^3$$

One marble, of radius $0.7$ cm:

$$\frac43\pi r^3 = \frac43 \times \frac{22}{7} \times 0.343 = \frac{88}{21} \times 0.343 = \frac{30.184}{21} = 1.437\overline{3} \text{ cm}^3$$

The count:

$$\frac{215.6}{1.4373\ldots} = 150$$

Exactly: $\dfrac{215.6 \times 21}{30.184} = \dfrac{4527.6}{30.184} = 150$.

The marbles must be fully submerged for this to work, and they will be — the water rises $5.6$ cm, which is four marble diameters, so there is ample depth.

Notice the question never mentions how much water was in the beaker to start with. It does not matter: only the rise depends on the marbles.

Question 19

How many spherical lead shots each of diameter $4.2$ cm can be obtained from a solid rectangular lead piece with dimensions $66$ cm, $42$ cm and $21$ cm?

(Exercise 12.3, Question 11.)

Solution. The block:

$$66 \times 42 \times 21 = 58\,212 \text{ cm}^3$$

One shot, of radius $2.1$ cm:

$$\frac43 \pi r^3 = \frac43 \times \frac{22}{7} \times 9.261 = \frac{88}{21} \times 9.261 = 38.808 \text{ cm}^3$$

The count:

$$\frac{58\,212}{38.808} = 1500$$

Every dimension here is a multiple of $21$ or $2.1$, which is what makes the $\frac{22}{7}$ cancel cleanly. Exactly: $r^3 = \left(\frac{21}{10}\right)^3 = \frac{9261}{1000}$, so one shot is $\frac{88 \times 9261}{21000} = \frac{814\,968}{21\,000}$, and dividing $58\,212$ by that gives $1500$.

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

Question 20

How many spherical lead shots of diameter $4$ cm can be made out of a solid cube of lead whose edge measures $44$ cm?

(Exercise 12.3, Question 12.)

Solution. The cube:

$$44^3 = 85\,184 \text{ cm}^3$$

One shot, of radius $2$ cm:

$$\frac43\pi r^3 = \frac43 \times \frac{22}{7} \times 8 = \frac{704}{21} \text{ cm}^3$$

The count:

$$\frac{85\,184}{\frac{704}{21}} = 85\,184 \times \frac{21}{704} = 121 \times 21 = 2541$$

using $85\,184 \div 704 = 121$.

The edge of $44$ cm is chosen so that $44^3$ divides by $704 = \frac{32 \times 22}{1}$ exactly — keeping $\frac{22}{7}$ as a fraction is what reveals it. In decimals the shot volume is $33.523809\ldots$ cm³ and the quotient looks like $2541.0000\ldots$, which is right but gives no confidence.

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

Question 21

A wall $24$ m long, $0.4$ m thick and $6$ m high is constructed with the bricks each of dimensions $25$ cm $\times$ $16$ cm $\times$ $10$ cm. If the mortar occupies $\frac1{10}$ th of the volume of the wall, then find the number of bricks used in constructing the wall.

(Exercise 12.3, Question 13.)

Solution. Work in one unit. The wall is in metres and the bricks in centimetres, so convert the wall:

$$24 \text{ m} \times 0.4 \text{ m} \times 6 \text{ m} = 2400 \times 40 \times 600 = 57\,600\,000 \text{ cm}^3$$

Space for bricks. The mortar occupies $\frac1{10}$, so the bricks get $\frac9{10}$:

$$57\,600\,000 \times \frac9{10} = 51\,840\,000 \text{ cm}^3$$

One brick:

$$25 \times 16 \times 10 = 4000 \text{ cm}^3$$

The count:

$$\frac{51\,840\,000}{4000} = 12\,960$$

This is Question 11 of Exercise 12.1 with a different fraction and metres in the data. The two places it goes wrong are the same: converting the wall (a factor of $10^6$, since three lengths are each multiplied by $100$), and remembering that the mortar’s share is subtracted from the brick’s, not added.

Question 22

Find the number of metallic circular disc with $1.5$ cm base diameter and of height $0.2$ cm to be melted to form a right circular cylinder of height $10$ cm and diameter $4.5$ cm.

(Exercise 12.3, Question 14.)

Solution. Both solids are cylinders, so the $\pi$ cancels and no value for it is needed.

One disc, radius $0.75$ cm, height $0.2$ cm:

$$\pi (0.75)^2 (0.2) = 0.1125\pi \text{ cm}^3$$

The cylinder, radius $2.25$ cm, height $10$ cm:

$$\pi (2.25)^2 (10) = 50.625\pi \text{ cm}^3$$

The count:

$$\frac{50.625\pi}{0.1125\pi} = \frac{50.625}{0.1125} = 450$$

The quick way. The cylinder’s radius is three times the disc’s, so its cross-sectional area is $3^2 = 9$ times as large; and it is $\frac{10}{0.2} = 50$ times as tall. So it takes

$$9 \times 50 = 450$$

discs. Ratios of similar quantities are usually faster than the volumes themselves, and here they avoid the decimals entirely.

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

Common mistakes

• Exercise 12.2 Questions 1, 2 and 3 — adding total surface areas. Each fails by exactly $2\pi r^2$, the two faces at the join. When two solids are stuck together, subtract the contact face twice, once from each.
• Exercise 12.2 Question 4 — treating the side of the box as the ball’s radius. The ball’s diameter equals the side, so its volume is $\frac{\pi a^3}{6}$, about $52\%$ of the box. A claimed volume larger than its container is always a halving error.
• Exercise 12.2 Questions 5 and 7 — misremembering a frustum formula. Test any candidate by setting $r_1 = r_2 = r$: the frustum becomes a cylinder, so the volume must reduce to $\pi r^2h$ and the slant height to $h$. Both printed versions fail that test.
• Exercise 12.3 Question 17 — taking the stated height as the cone’s. The $10$ cm spans the whole solid, so the cone is $10 – 5 = 5$ cm tall. Reading it the other way inflates the answer by a third.
• Exercise 12.3 Question 12 — expecting half the height to mean half the volume. Similar solids scale as the cube: a half-height cut leaves $\frac18$ above and $\frac78$ below, hence $1 : 7$.
• Exercise 12.3 Questions 10, 19, 20 and 22 — rounding before dividing. Every one of these is built so that the count comes out whole. Keep $\frac{22}{7}$ as a fraction, cancel, and the exactness confirms the method.
• Exercise 12.3 Question 21 — converting the wall wrongly. Three lengths each multiplied by $100$ means the volume is multiplied by $10^6$, not by $100$. Convert every dimension before multiplying, not the product afterwards.
• Exercise 12.3 Question 15 — including the bases. Two cones joined base to base share both circles, which become internal. The surface is the two curved parts only.

Practise next

• Exercise 12.4 — twenty long-answer questions, where these same combinations appear with flow rates, costs and unit conversions attached.
• Exercise 12.1 — worth a second pass, especially Questions 8, 11 and 14, which are the exact-arithmetic habit this exercise depends on.
• Exercise 11.4 — its annulus and packing questions use the same decomposition thinking in two dimensions.