RS Aggarwal solution class 8 chapter 20 Volume and Surface Area of Solid Exercise 20C

Exercise 20C

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Q1 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 1:

Tick (✓) the correct answer:
The maximum length of a pencil that can be kept in a rectangular box of dimensions 12 cm × 9 cm × 8 cm, is
(a) 13 cm
(b) 17 cm
(c) 18 cm
(d) 19 cm

Answer 1:

(b) 17
 
Length of the diagonal of a cuboid $=\sqrt{l^2+b^2+h^2}$

∴ $\sqrt{l^2+b^2+h^2}=\sqrt{{12}^2+9^2+8^2}$

$=\sqrt{144+81+64}=\sqrt{289}=17 cm$

Q2 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 2:

Tick (✓) the correct answer:
The total surface area of a cube is 150 cm². Its volume is
(a) 216 cm³
(b) 125 cm³
(c) 64 cm³
(d) 1000 cm³

Answer 2:

(b) $125 c{m}^3$

Total surface area $=6a^2=150 c{m}^2$, where a is the length of the edge of the cube.
$\Rightarrow 6a^2=150$
$\Rightarrow a=\sqrt{\frac{150}{6}}=\sqrt{25}=5 cm$
∴ Volume$=a^3=5^3=125 c{m}^3$

Q3 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 3:

Tick (✓) the correct answer:
The volume of a cube is 343 cm³. Its total surface area is
(a) 196 cm²
(b) 49 cm²
(c) 294 cm²
(d) 147 cm²

Answer 3:

(c) $294 c{m}^2$

Volume$=a^3=343 c{m}^3$
$\Rightarrow a=\sqrt[3]{343}=7 cm$
∴ Total surface area$=6a^2=6\times 7\times 7=294 c{m}^2$

Q4 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 4:

Tick (✓) the correct answer:
The cost of painting the whole surface area of a cube at the rate of 10 paise per cm² is Rs 264.60. Then, the volume of the cube is
(a) 6859 cm³
(b) 9261 cm³
(c) 8000 cm³
(d) 10648 cm³

Answer 4:

(b) $9261 c{m}^3$

Rate of painting = 10 paise per sq cm = Rs 0.1/cm²
Total cost = Rs 264.60
Now, total surface area  $=\frac{264.6}{0.1}=2646 c{m}^2$
Also, length of edge, a $=\sqrt{\frac{2646}{6}}=\sqrt{441}=21 cm$
$\therefore \mathrm{Volume}= {21}^3=9261 c{m}^3$

Q5 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 5:

Tick (✓) the correct answer:
How many bricks, each measuring 25 cm × 11.25 cm × 6 cm, will be needed to build a wall 8 m long, 6 m high and 22.5 cm thick?
(a) 5600
(b) 6000
(c) 6400
(d) 7200

Answer 5:

(c) 6400

Volume of each brick$=25\times 11.25\times 6=1687.5 c{m}^3$
Volume of the wall$=800\times 600\times 22.5=10800000 c{m}^3$
∴  No. of bricks =$\frac{10800000}{1687.5}=6400$

Q6 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 6:

Tick (✓) the correct answer:
How many cubes of 10 cm edge can be put in a cubical box of 1 m edge?
(a) 10
(b) 100
(c) 1000
(d) 10000

Answer 6:

(c) 1000

Volume of the smaller cube$=(10 cm{)}^3=1000 c{m}^3$
Volume of box$=(100 cm{)}^3=1000000 c{m}^3$     [1 m = 100 cm]
∴ Total no. of cubes $=\frac{100\times 100\times 100}{10\times 10\times 10}=1000$

Q7 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 7:

Tick (✓) the correct answer:
The edges of a cuboid are in the ratio 1 : 2 : 3 and its surface area is 88 cm². The volume of the cuboid is
(a) 48 cm³
(b) 64 cm³
(c) 96 cm³
(d) 120 cm³

Answer 7:

(a) $48 c{m}^3$

Let a be the length of the smallest edge.
Then the edges are in the proportion a : 2a : 3a.

Now, surface area$=2(a\times 2a+a\times 3a+2a\times 3a)$

$=2(2a^2+3a^2+6a^2)$

$=22a^2=88 c{m}^2$
$\Rightarrow a=\sqrt{\frac{88}{22}}=\sqrt{4}=2$

Also, 2a = 4 and 3a = 6
∴ Volume$=a\times 2a\times 3a=2\times 4\times 6=48 c{m}^3$

Q8 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 8:

Tick (✓) the correct answer:
Two cubes have their volumes in the ratio 1 : 27. The ratio of their surface areas is
(a) 1 : 3
(b) 1 : 9
(c) 1 : 27
(d) none of these

Answer 8:

(b) 1: 9

$$\begin{gathered} \frac{\mathrm{Volume} 1}{\mathrm{Volume} 2}=\frac{1}{27}=\frac{a^3}{b^3} \\ \Rightarrow a=\frac{b}{\sqrt[3]{27}}=\frac{b}{3}\mathrm{or} b = 3a\mathrm{or} \frac{b}{a}=3 \end{gathered}$$
$\mathrm{Now}, \frac{\mathrm{surface} \mathrm{area} 1}{\mathrm{surface} \mathrm{area} 2}=\frac{6{\mathrm{a}}^2}{6{\mathrm{b}}^2}$

$=\frac{{\mathrm{a}}^2}{{\mathrm{b}}^2}=\frac{(\mathrm{b}/3{)}^2}{{\mathrm{b}}^2}= \frac{1}{9}$

$\therefore \mathrm{Ratio} \mathrm{of} \mathrm{the} \mathrm{surface} \mathrm{areas} = 1 : 9$

Q9 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 9:

Tick (✓) the correct answer:
The surface area of a (10 cm × 4 cm × 3 cm) brick is
(a) 84 cm²
(b) 124 cm²
(c) 164 cm²
(d) 180 cm²

Answer 9:

(c) 164 sq cm 

Surface area $=2(10\times 4+10\times 3+4\times 3)$

$=2(40+30+12)=164 c{m}^2$

Q10 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 10:

Tick (✓) the correct answer:
An iron beam is 9 m long, 40 cm wide and 20 cm high. If 1 cubic metre of iron weighs 50 kg, what is the weight of the beam?
(a) 56 kg
(b) 48 kg
(c) 36 kg
(d) 27 kg

Answer 10:

(c) 36 kg

Volume of the iron beam $=9\times 0.4\times 0.2=0.72 m^3$
∴ Weight$=0.72\times 50=36 kg$

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Q11 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 11:

Tick (✓) the correct answer:
A rectangular water reservoir contains 42000 litres of water. If the length of reservoir is 6 m and its breadth is 3.5 m, the depth of the reservoir is
(a) 2 m
(b) 5 m
(c) 6 m
(d) 8 m

Answer 11:

(a) 2 m

42000 L = 42 m³
$\mathrm{Volume}=lbh$
$\therefore \mathrm{Height} \left(h\right)=\frac{\mathrm{volume}}{lb}=\frac{42}{6\times 3.5}=\frac{6}{6\times 0.5}=2 m$

Q12 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 12:

Tick (✓) the correct answer:
The dimensions of a room are (10 m × 8 m × 3.3 m). How many men can be accommodated in this room if each man requires 3 m³ of space?
(a) 99
(b) 88
(c) 77
(d) 75

Answer 12:

(b) 88  

Volume of the room$=10\times 8\times 3.3=264 m^3$
One person requires 3 m³. 
∴ Total no. of people that can be accommodated$=\frac{264}{3}=88$

Q13 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 13:

Tick (✓) the correct answer:
A rectangular water tank is 3 m long, 2 m wide and 5 m high. How many litres of water can it hold?
(a) 30000
(b) 15000
(c) 25000
(d) 35000

Answer 13:

(a) 30000

$\mathrm{Volume}=3\times 2\times 5=30 m^3=30000 \mathrm{L}$

Q14 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 14:

Tick (✓) the correct answer:
The area of the cardboard needed to make a box of size 25 cm × 15 cm × 8 cm will be
(a) 390  cm²
(b) 1390 cm²
(c) 2780 cm²
(d) 1000 cm²

Answer 14:

(b) $1390 c{m}^2$

Surface area$=2(25\times 15+15\times 8+25\times 8)$

$=2(375+120+200)=1390 c{m}^2$

Q15 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 15:

Tick (✓) the correct answer:
The diagonal of a cube measures $4\sqrt{3}$ cm. Its volume is
(a) 8 cm³
(b) 16 cm³
(c) 27 cm³
(d) 64 cm³

Answer 15:

(d) $64 c{m}^2$

Diagonal of the cube$=a\sqrt{3}=4\sqrt{3} cm$
i.e., a = 4 cm
∴ Volume$=a^3=4^3=64 c{m}^3$

Q16 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 16:

Tick (✓) the correct answer:
The diagonal of a cube is $9\sqrt{3}$ cm long. Its total surface area is
(a) 243 cm²
(b) 486 cm²
(c) 324 cm²
(d) 648 cm²

Answer 16:

(b) 486 sq cm

Diagonal $=\sqrt{3}a cm = 9\sqrt{3}cm$
i.e., a = 9
∴ Total surface area $=6a^2=6\times 81=486 c{m}^2$

Q17 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 17:

Tick (✓) the correct answer:
If each side of a cube is doubled then its volume
(a) is doubled
(b) becomes 4 times
(c) becomes 6 times
(d) becomes 8 times

Answer 17:

(d) If each side of the cube is doubled, its volume becomes 8 times the original volume.

Let the original side be a units.
Then original volume = a³ cubic units
Now, new side  = 2a units
Then new volume = (2a)³ sq units = 8 a³cubic units
Thus, the volume becomes 8 times the original volume.

Q18 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 18:

Tick (✓) the correct answer:
If each side of a cube is doubled, its surface area
(a) is doubled
(b) becomes 4 times
(c) becomes 6 times
(d) becomes 8 times

Answer 18:

(b) becomes 4 times.

Let the side of the cube be a units.
Surface area = 6a² sq units
Now, new side = 2a units
New surface area = 6(2a² ) sq units = 24a² sq units.
Thus, the surface area becomes 4 times the original area.

Q19 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 19:

Tick (✓) the correct answer:
Three cubes of iron whose edges are 6 cm, 8 cm and 10 cm respectively are melted and formed into a single cube. The edge of the new cube formed is
(a) 12 cm
(b) 14 cm
(c) 16 cm
(d) 18 cm

Answer 19:

(a) 12 cm

Total volume $=6^3+8^3+{10}^3=216+512+1000=1728 c{m}^3$
∴ Edge of the new cube$=\sqrt[3]{1728}=12 cm$

Q20 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 20:

Tick (✓) the correct answer:
Five equal cubes, each of edge 5 cm, are placed adjacent to each other. The volume of the cuboid so formed, is
(a) 125 cm³
(b) 375 cm³
(c) 525 cm³
(d) 625 cm³

Answer 20:

(d) $625 c{m}^3$

Length of the cuboid so formed = 25 cm
Breadth of the cuboid = 5 cm
Height of the cuboid = 5 cm
∴ Volume of cuboid$=25\times 5\times 5=625 c{m}^3$

Q21 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 21:

Tick (✓) the correct answer:
A circular well with a diameter of 2 metres, is dug to a depth of 14 metres. What is the volume of the earth dug out?
(a) 32 m³
(b) 36 m³
(c) 40 m³
(d) 44 m³

Answer 21:

(d) 44 m³

Diameter = 2 m
Radius = 1 m
Height = 14 m
$\therefore \mathrm{Volume}={\mathrm{\pi r}}^2\mathrm{h}=\frac{22}{7}\times 1\times 1\times 14=44 {\mathrm{m}}^3$

Q22 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 22:

Tick (✓) the correct answer:
If the capacity of a cylindrical tank is 1848 m³ and the diameter of its base is 14 m, the depth of the tank is
(a) 8 m
(b) 12 m
(c) 16 m
(d) 18 m

Answer 22:

(b) 12 m

Diameter = 14 m
Radius = 7 m
Volume = 1848 m³

$$\begin{gathered} \mathrm{Now}, \mathrm{volume}={\mathrm{\pi r}}^2\mathrm{h}=\frac{22}{7}\times 7\times 7\times \mathrm{h}=1848 {\mathrm{m}}^3 \\ \therefore \mathrm{h} = \frac{1848}{22\times 7}=12 \mathrm{m} \end{gathered}$$

Q23 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 23:

Tick (✓) the correct answer:
The ratio of the total surface area to the lateral surface area of a cylinder whose radius is 20 cm and height 60 cm, is
(a) 2 : 1
(b) 3 : 2
(c) 4 : 3
(d) 5 : 3

Answer 23:

(c) 4 : 3

$$\begin{gathered} \mathrm{Here}, \\ \frac{\mathrm{Total} \mathrm{surface} \mathrm{area}}{} \end{gathered}$$

$$\begin{gathered} \frac{\mathrm{Lateral} \mathrm{surface} \mathrm{area}}{}=\frac{2\pi r(h+r)}{2\pi rh} \\ =\frac{h+r}{h} \\ =\frac{20+60}{60} \\ =\frac{4}{3} \\ = 4:3 \end{gathered}$$

Q24 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 24:

Tick (✓) the correct answer:
The number of coins, each of radius 0.75 cm and thickness 0.2 cm, to be melted to make a right circular cylinder of height 8 cm and base radius 3 cm is
(a) 460
(b) 500
(c) 600
(d) 640

Answer 24:

(d) 640 
$\mathrm{Total} \mathrm{no}. \mathrm{of} \mathrm{coins} =\frac{\mathrm{volume} \mathrm{of} \mathrm{cylinder}}{}$

$\frac{\mathrm{volume} \mathrm{of} \mathrm{each} \mathrm{coin}}{}=\frac{\mathrm{\pi }\times 3\times 3\times 8}{\mathrm{\pi }\times 0.75\times 0.75\times 0.2}$

$=640$

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Q25 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 25:

Tick (✓) the correct answer:
66 cm³ of silver is drawn into a wire 1 mm in diameter. The length of the wire will be
(a) 78 m
(b) 84 m
(c) 96 m
(d) 108 m

Answer 25:

(b) 84 m
$\mathrm{Length}=\frac{\mathrm{volume}}{\mathrm{\pi }{r}^{\mathit{2}}2}=\frac{66\times 7}{22\times 0.05\times 0.05}=8400 cm = 84 m$

Q26 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 26:

Tick (✓) the correct answer:
The height of a cylinder is 14 cm and its diameter is 10 cm. The volume of the cylinder is
(a) 1100 cm³
(b) 3300 cm³
(c) 3500 cm³
(d) 7700 cm³

Answer 26:

(a) 1100 cm³
Volume$={\mathrm{\pi r}}^2\mathrm{h}=\frac{22}{7}\times 5\times 5\times 14=1100 {\mathrm{cm}}^3$

Q27 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 27:

Tick (✓) the correct answer:
The height of a cylinder is 80 cm and the diameter of its base is 7 cm. The whole surface area of the cylinder is
(a) 1837 cm²
(b) 1760 cm²
(c) 1942 cm²
(d) 3080 cm²

Answer 27:

(a) 1837 cm²
Diameter = 7 cm
Radius  =3.5 cm
Height = 80 cm

∴ Total surface area$=2\mathrm{\pi r}(\mathrm{r}+\mathrm{h})=2\times \frac{22}{7}\times 3.5(3.5+80)$

$=22(83.5)=1837 {\mathrm{cm}}^2$

Q28 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 28:

Tick (✓) the correct answer:
The height of a cylinder is 14 cm and its curved surface area is 264 cm². The volume of the cylinder is
(a) 308 cm³
(b) 396 cm³
(c) 1232 cm³
(d) 1848 cm³

Answer 28:

(b) 396 cm³
Here, curved surface area$=2\mathrm{\pi rh}=264 {\mathrm{cm}}^3$
$\Rightarrow r=\frac{264\times 7}{2\times 22\times 14}=3 cm$
$\therefore \mathrm{Volume}={\mathrm{\pi r}}^2\mathrm{h}=\frac{22}{7}\times 3\times 3\times 14=396 {\mathrm{cm}}^3$

Q29 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 29:

Tick (✓) the correct answer:
The diameter of a cylinder is 14 cm and its curved surface area is 220 cm². The volume of the cylinder is
(a) 770 cm³
(b) 1000 cm³
(c) 1540 cm³
(d) 6622 cm³

Answer 29:

(a) 770 cm³
Diameter = 14 cm
Radius = 7 cm
Now, curved surface area$=2\mathrm{\pi rh}=220 {\mathrm{cm}}^2$
$\Rightarrow h=\frac{220\times 7}{2\times 22\times 7}=5 cm$
$\therefore \mathrm{Volume}={\mathrm{\pi r}}^2\mathrm{h}=\frac{22}{7}\times 7\times 7\times 5=770 {\mathrm{cm}}^3$

Q30 | Ex-20C | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid

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Question 30:

Tick (✓) the correct answer:
The ratio of the radii of two cylinders is 2 : 3 and the ratio of their heights is 5 : 3. The ratio of their volumes will be
(a) 4 : 9
(b) 9 : 4
(c) 20 : 27
(d) 27 : 20

Answer 30:

(c) 20:27

$$\begin{gathered} \mathrm{We} \mathrm{have} \mathrm{the} \mathrm{following}: \\ \frac{r_1}{r_2}=\frac{2}{3} \\ \frac{h_1}{h_2}=\frac{5}{3} \\ \therefore \frac{V_1}{V_2}=\frac{{{\mathrm{\pi r}}_1}^2{\mathrm{h}}_1}{{{\mathrm{\pi r}}_2}^2{\mathrm{h}}_2}=\frac{20}{27} \end{gathered}$$