RS Aggarwal Solutions for the Exercise 20C of Class 8 Maths Chapter 20, Volume and Surface Area of Solids are available here. BYJUâ€™S expert has uniquely solved the exercise. As this is the last exercise, it includes multiple-choice questions which enhances time management among the students. Practising as many times as possible helps in developing time management skills and also boosts the confidence level to achieve high marks. By practising the RS Aggarwal Solutions for Class 8, students will be able to grasp the concepts correctly.

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Exercise 20C Page No: 228

**Select the correct answer in each of the following:**

**1. The maximum length of a pencil that can be kept in rectangular box of dimensions 12cm Ã— 9cm Ã— 8cm, is**

**(a) 13cm (b) 17cm (c) 18cm (d) 19cm**

**Solution:**

(b) 17cm

**Explanation:**

We know that length of diagonal of the cuboid= âˆšÂ (l^{2}+b^{2}+h^{2})

=âˆšÂ (12^{2}+9^{2}+8^{2})

=âˆšÂ (144+81+64)

=17

**2. The total surface area of a cube is 150 cm ^{2}.Its volume is**

** (a)216 cm ^{3} (b) 125 cm^{3} (c) 64 cm^{3} (d) 1000 cm^{3}**

** Solution: **

- 125 cm
^{3}

**Explanation: **

Total surface area of the cube = 6 a^{2}

150 = 6 a^{2}

a^{2} = 150/6 = 25

a = 5

Volume of the cube = a^{3}

V= 5^{3}=125

**3. The volume of a cube is 343 cm ^{3}. Its total surface area is **

** (a)196 cm ^{2} (b) 49 cm^{2} (c) 294 cm^{2} (d) 147 cm^{2}**

** Solution: **

Â©294 cm^{2}

**Explanation: **

Volume of the cube = a^{3}

343= a^{3}

a = 7cm

Total surface area of the cube = 6 a^{2}

= 6 (7)^{2}

=294 cm^{2}

**4. The cost of painting the whole surface area of a cube at the rate of 10 paise per cm ^{2} is â‚¹ 264.40. then, the volume of the cube is**

**(a)6859 cm ^{2} (b) 9261 cm^{2} (c) 8000 cm^{2} (d) 10648 cm^{2}**

**Solution:**

(b)9261 cm^{2}

** Explanation**:

Total surface area of the cube = 6 a^{2}

^{ }Cost of the painting the cube = 6 a^{2}Ã— 10

246.40 = 6 a^{2}Ã— 10

a^{2}= 4.41

a= 2.1

Volume of the cube = a^{3}

V= (2.1)^{3}

**5. How many bricks, each measuring 25cm Ã— 11.25cm Ã— 6cm, will be needed to build a wall 8m long, 6m high and 22.5 cm thick?**

**(a)5600 (b)6000 (c)6400 (d)7200**

**Solution:**

Â©6400

**Explanation:**

We know that volume of cuboid= length Ã— breadth Ã— height

Volume of brick = 25cm Ã— 11.25cm Ã— 6cm

= 1687.5 cm^{3}

Volume of the wall = 800 Ã— 600 Ã— 22.5

= 10800000 cm^{3}

Number of bricks = 10800000/ 1687.5 = 6400

**6. How many cubes 0f 10cm edge can be put in a cubical box of 1m edge?**

**(a)10 (b)100 (c)1000 (d)10000**

**Solution:**

Â©1000

**Explanation:**

Volume of the cube = a^{3}

Volume of the smaller cube = 10^{3}

=1000

Volume of the box = a^{3}

= (100)^{3}

=1000000

Total number of cubes = 1000000/1000

=1000

**7. The edges of a cuboid have their volumes in the ratio 1:2:3 and its surface area is 88cm ^{2}. The volume of the cuboid is**

**(a)48 cm ^{3} (b) 64 cm^{3} (c) 96 cm^{3} (d) 120 cm^{3}**

** Solution: **

(a)48 cm^{3}

**Explanation:**

Let a be the length of smallest edge.

The edges are in the proportion a: 2a: 3a

We know that total surface area of cuboid= 2(l b + b h + h l)

Surface area = 2 (2 a^{2} + 3 a^{2} + 6 a^{2})

88 = 22 a^{2}

a = 2

2a=4

3a = 6

We know that volume of cuboid= length Ã— breadth Ã— height

V = 2 Ã— 4 Ã— 6

V = 48

**8. 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**

**Solution:**

(b)1:9

**Explanation:**

Volume in the ratio 1:27

1/27 = 1/3= a^{3}/b^{3}

b/a = 3

Therefore on squaring 1:9

**9. The surface area of a 10cm Ã— 4cm Ã— 3cm brick is **

**(a)84cm ^{2 }(b) 124cm^{2 }(c)164cm^{2} (d) 180cm^{2}**

** **

** Solution:**

Â©164cm^{2}

^{ }Explanation:

We know that surface area of cuboid= 2(l b + b h + h l)

= 2 (10 x 4 + 4 x 3 + 3 x 10)

= 2 (40 + 12 + 30)

= 164

**10. The surface area is 9m long, 40cm wide and 20cm high. If 1 cubic meter of iron weighs 50 kg, what is the weight of the beam?**

**(a)56kg (b)48kg (c)36kg (d)27kg**

**Solution:**

Â©36kg

**Explanation:**

We know that volume of cuboid= length Ã— breadth Ã— height

V = 9 Ã— 0.2 Ã— 0.4

V = 0.72m^{3}

Weight = 0.72 Ã— 50 = 36kg

**11.** **A rectangular water reservoir contains 42000 litres of water. If the length of reservoir is 6m and its breadth is 3.5m, the depth of the reservoir is**

**(a)2m (b)5m (c)6m (d)8m**

**Solution:**

(a)2m

**Explanation:**

We know that volume of cuboid= length Ã— breadth Ã— height

h = volume / l b

h = 42/6 Ã— 3.5

h = 2

**12. The dimensions of a room are 10m Ã— 8m Ã— 3.3m. How many men can be accommodated in this room if each man requires 3m ^{3} of space?**

**(a)99 (b)88 (c)77 (d)75**

**Solution:**

(b)88

**Explanation:**

We know that volume of cuboid= length Ã— breadth Ã— height

V = 10 Ã— 8 Ã— 3.3

V = 264m^{3}

Total number of people can be accommodated = 264/3 = 88

**13.** **A rectangular water tank is 3m long, 2m wide and 5m high. How many litres of water can it hold?**

**(a)30000 (b)15000 (c)25000 (d)35000**

**Solution:**

(a)30000

**Explanation:**

We know that volume of cuboid= length Ã— breadth Ã— height

V = 3 Ã— 2 Ã— 5

V = 30m^{3}

V= 30000l

**14. The area of the cardboard needed to make a box of size 25cm Ã— 15cm Ã— 8cm will be **

**(a)390cm ^{2 }(b) 1390cm^{2 }(c)2780cm^{2} (d) 1000cm^{2}**

** Solution: **

- 1390cm
^{2}

**Explanation: **

We know that total surface area of cuboid= 2(l b + b h + h l)

= 2 (375 + 120 + 200)

= 1390

**15. The diagonal of a cube measures 4 âˆšÂ 3 cm. its volume is**

**(a)8cm ^{3 }(b) 16cm^{3 }(c)27cm^{3} (d) 64cm^{3}**

** Solution: **

(d) 64cm^{3}

Explanation:

Diagonal of cube = a âˆšÂ 3 = 4 âˆšÂ 3

a = 4

Volume of cube = a^{3}= 4^{3}=64 cm^{3}

### Access other exercises of RS Aggarwal Solutions for Class 8 Maths Chapter 20 – Volume and Surface Area of Solids

Exercise 20A Solutions 30 questions

Exercise 20B Solutions 21 questions

## RS Aggarwal Solutions for Class 8 Maths Chapter 20 – Volume and Surface Area of Solids Exercise 20C

Exercise 20C of RS Aggarwal Solutions for Class 8 Maths Chapter 20 Volume and Surface Area of Solids covers all the basic concepts related to the Surface Area of solids. We can say that this exercise mainly deals with the formulae to calculate the Volume and Surface Area of Solids that the students have learnt in this Chapter. Some of the topics focused prior to exercise 20C include the following.

- Definition of cube
- Definition of cuboid
- Volume of a solid
- A standard unit of volume
- Formulae for Volume and Surface Area of a cuboid
- Formulae for Volume and Surface Area of a cube
- Formulae for Volume and Surface Area of a cylinder

The RS Aggarwal Solutions can help the students in practising and learning each and every concept as it provides solutions to all questions asked in the RS Aggarwal textbook.