## RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20C.

**Other Exercises**

- RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20A
- RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20B
- RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20C

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

**Question 1.**

**Solution:**

Answer = (b)

Length (l) = 12 cm

Breadth (b) = 9cm

height (h) = 8 cm

**Question 2.**

**Solution:**

Total surface area of cube = 150 cm^{2}

Side = \( \sqrt { \frac { 150 }{ 6 } } \)

= √25

= 5 cm

Volume = (side)^{3}

= (5)^{3}

= 125 cm^{3} (b)

**Question 3.**

**Solution:**

Volume of cube = 343 cm^{2}

Side = \( \sqrt [ 3 ]{ 343 } =\sqrt [ 3 ]{ 7\times 7\times 7 } \)

= 7 cm

Total surface area = 6 (side)^{2}

= 6 x (7)^{2}

= 6 x 49 cm^{2}

= 294 cm^{2} (c)

**Question 4.**

**Solution:**

Rate of painting = 10 paise per cm^{2}

Total cost = Rs. 264.60

**Question 5.**

**Solution:**

Answer = (c)

Length of wall (l) = 8m = 800 cm

Breadth (b) = 22.5 cm

Height (h) = 6 m

= 600 cm

**Question 6.**

**Solution:**

Answer = (c)

Edge of cube = 10 cm

Volume = a^{3} = (10)^{3} = 1000 cm^{3}

Edge of box = 1 m = 100 cm

**Question 7.**

**Solution:**

Answer = (a)

Ratio in sides of a cuboid = 1 : 2 : 3

Surface area = 88 cm^{2}

**Question 8.**

**Solution:**

Ratio in the two volumes = 1 : 27

Let volume of first volume = x^{3}

and volume of second volume = 27x^{3}

Side of first cube = x

**Question 9.**

**Solution:**

Surface area of a brick of measure 10 cm x 4 cm x 3 cm

= 2 (l x b + b x h + h x l)

= 2 [10 x 4 + 4 x 3 + 3 x 10] cm^{2}

= 2 [40 + 12 + 30]

= 82 x 2

= 164 cm^{2} (c)

**Question 10.**

**Solution:**

Length of beam (l) = 9 m

**Question 11.**

**Solution:**

Water in rectangular reservoir = 42000

Volume = \(\\ \frac { 42000 }{ 1000 } \) = 42 m^{3}

Length (l) = 6 m

Breadth (b) = 3.5 m

Depth = \(\\ \frac { volume }{ l\times b } \)

= \(\\ \frac { 42 }{ 6\times 3.5 } \)

= 2 m (c)

**Question 12.**

**Solution:**

Dimensions of a room are 10 m, 8 m, 3.3 m

Volume of air in it = lbh

= 10 x 8 x 3.3 = 264 m^{3}

Air required for one man = 3 m^{3}

No. of men = \(\\ \frac { 264 }{ 3 } \)

= 88 (b)

**Question 13.**

**Solution:**

Length of water tank (l) = 3 m

Width (b) = 2 m

and height (h) = 5 m

Volume = lbh = 3 x 2 x 5 = 30 m^{3}

Water in it = 30 x 1000

= 30000 (a)

**Question 14.**

**Solution:**

Size of box = 25 cm, 15 cm, 8 cm

Surface area = (lb + bh + hl)

= 2 ( 25 x 15 + 15 x 8 + 8 x 25) cm^{2}

= 2 (375 + 120 + 200) cm^{2}

= 2(695)

= 1390 cm^{2 }(b)

**Question 15.**

**Solution:**

Diagonal of cube = 4√3

Side = \( \frac { 4\sqrt { 3 } }{ \sqrt { 3 } } \)

= 4 cm

Volume = a^{3} = (4)^{3}

= 64 cm^{3} (d)

**Question 16.**

**Solution:**

Diagonal of cube = 9√3 cm

Side = \( \frac { 9\sqrt { 3 } }{ \sqrt { 3 } } \)

= 9 cm

Surface area = 6a^{2}

= 6 (9)^{2} = 6 x 81 cm^{2}

= 486 cm^{2} (b)

**Question 17.**

**Solution:**

Let side of cube in first case = a

Then volume = a^{3}

If side of cube is doubled, then side = 2a

Volume (2a)^{3} = 8a^{3}

Becomes 8 times (d)

**Question 18.**

**Solution:**

Let side of cube in first case = a

Then surface area = 6a^{2}

and side of second cube = 2a

Surface area = 6 (2a)^{2} = 6 x 4a^{2} = 24a^{2}

Ratio = \(\frac { { 24a }^{ 2 } }{ { 6a }^{ 2 } } \) = 4

Becomes 4 times (b)

**Question 19.**

**Solution:**

Sides (edges) of 3 cubes are 6 cm, 8 cm, and 10 cm respectively

Volume of first cube = (6)^{3} = 216 cm^{3}

Volume of second cube = (8)^{3} = 512 cm^{3}

and volume of third cube

= (10)^{3} = 1000 cm^{3}

Sum of volumes of 3 cubes = 216 + 512 + 1000

= 1728 cm^{3}

Volume of new single cube = 1728 cm^{3}

Edge = \(\sqrt [ 3 ]{ 1728 } \)

\(\sqrt [ 3 ]{ { \left( 12 \right) }^{ 3 } } \)

= 12 cm (a)

**Question 20.**

**Solution:**

Each edge of 5 cubes = 5 cm

Placing than adjacent to each other

Length of new cuboid (l)

= 5 x 5 = 25 cm

Breadth (b) = 5 cm

and height (h) = 5 cm

Volume of new cuboid = lbh

= 25 x 5 x 5 cm^{3}

= 625 cm^{3} (d)

**Question 21.**

**Solution:**

Diameter of circular well = 2n

Radius = \(\\ \frac { 2 }{ 2 } \) = 1 m

Depth(h) = 14 m

Volume of earth dug out = πr^{2}h

= \(\\ \frac { 22 }{ 7 } \) x 1 x 1 x 14

= 44 m (d)

**Question 22.**

**Solution:**

Capacity of cylindrical tank = 1848 m^{3}

Diameter = 14 m

**Question 23.**

**Solution:**

Radius of a cylinder (r) = 20 cm

and height (h) = 60 cm

**Question 24.**

**Solution:**

Radius of each coin (r) = 0.75 cm

and thickness (h) = 0.2 cm

**Question 25.**

**Solution:**

Volume of silver = 66 cm^{3}

Diameter of wire = 1 mm = \(\\ \frac { 1 }{ 10 } \)

**Question 26.**

**Solution:**

Diameter of cylinder = 10 cm

Radius (r) = \(\\ \frac { 10 }{ 2 } \) = 5 cm

**Question 27.**

**Solution:**

Diameter of cylinder = 7 cm

Radius (r) = \(\\ \frac { 7 }{ 2 } \) cm

**Question 28.**

**Solution:**

Curved surface area of a cylinder = 264 cm^{3}

Height (h) = 14 cm

**Question 29.**

**Solution:**

Diameter of cylinder = 14 cm

Radius (r) = 7 cm

Curved surface area = 220 cm^{2}

**Question 30.**

**Solution:**

Ratio in radii of two cylinder = 2 : 3

and ratio in their height = 5 : 3

Let radii of two cylinder = 2x and 3x

and corresponding heights = 5y, 3y

Hope given RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20C are helpful to complete your math homework.

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