# RS Aggarwal Class 8 Math Twentieth Chapter Volume and Surface Area of Solids Exercise 20A Solution

## EXERCISE 20A

(1) Find the volume, lateral surface area and the total surface area of the cuboid whose dimensions are:

(i) Length = 22 cm, breadth = 12 cm and height = 7.5 cm

Solution: Volume of the cuboid = (l × b × h) cubic units

= (22 × 12 × 7.5) cm3 = 1980 cm3

Lateral surface area of the cuboid = {2 (l + b) × h} cm2

= {2 (22 + 12) × 7.5} m2 = 510 cm2

Total surface area of the cuboid = 2(lb + bh + lh) sq units

= 2(22 × 12 + 12 × 7.5 + 22 × 7.5) cm2

= 2 (264 + 90 + 165) cm2 = 1038 cm2

(ii) Length = 15 m, breadth = 6 m and height = 9 dm

Solution: Here, 9 dm = 0.9 m

Volume of the cuboid = {15 × 6 × 0.9} m3 = 81 m3

Lateral surface area of the cuboid = {2 (15 + 6) × 0.9} m2 = 37.8 m2

Total surface area of the cuboid  = [2 {(15 × 6) + (6 × 0.9) + (15 × 0.9)}] m2

= [2 × (90 + 5.4 + 13.5)] m2 = 217.8 m2

(iii) Length = 24 m, breadth = 25 cm and height = 6 m

Solution: Here, 25 cm = 0.25 m

Volume of the cuboid = {24 × 0.25 × 6} m3 = 36 m3

Lateral surface area of the cuboid = {2 (24 + 0.25) × 6} m2 = 291 m2

Total surface area of the cuboid = [2 {(24 × 0.25) + (0.25 × 6) + (24 × 6)}] m2

= [2 × (6 + 1.5 + 144)] m2 = 303 m2

(iv) Length = 48 cm, breadth = 6 dm and height = 1 m

Solution: Here, 48 cm = 0.48 m and 6 dm = 0.6 m

Volume of the cuboid = {0.48 × 0.6 × 1} m3 = 0.288 m3

Lateral surface area of the cuboid = {2 (0.48 + 0.6) × 1} m2 = 2.16 m2

Total surface area of the cuboid = [2 {(0.48 × 0.6) + (0.6 × 1) + (0.48 × 1)}] m2

= [2 × (0.288 + 0.6 + 0.48)] m2 = 2.736 m2

(2) The dimensions of a rectangular water tank are 2 m 75 cm by 1 m 80 cm by 1 m 40 cm. How many litres of water does it hold when filled to the brim?

Solution: Here 2m 75 cm = 275 cm; 1 m 80 cm = 180 cm and 1 m 40 cm = 140 cm

Volume of the water = (275 × 180 × 140) cm3 = 6930000 cm3

We know, 1L = 1000 cm3

∴ Volume = (6930000/1000) L = 6930 L.

(3) A solid rectangular piece of iron measures 1.05 m × 70 cm × 1.5 cm. Find the weight of this piece in kilograms if 1 cm3 of iron weighs 8 grams.

Solution: Here, 1.05

Volume of the piece of iron = (105 × 70 × 1.5) cm3 = 11025 cm3

If 1 cm3 of iron weighs 8 grams then, = (11025 × 8) grams = 88200 gram = 88.2 kg

(4) The area of a courtyard is 3750 m2. Find the cost of covering it with gravel to a height of 1 cm if the gravel costs Rs 6.40 per cubic metre.

Solution: Here, 1 cm = 0.01 m

Volume of the gravel = (3750 × 0.01) m3 = 37.5 m3

Therefore, the total cost = Rs (37.5 × 6.40) = Rs 240.

(5) How many persons can be accommodated in hall of length 16 m, breadth 12.5 m and height 4.5 m, assuming that 3.6 m3 of air is required for each person?

Solution: Volume of the hall = (16 × 12.5 × 4.5) m3 = 900 m3

∴ Number of person whose accommodated in hall = (900/3.6) = 250 person.

(6) A cardboard box is 1.2 m long, 72 cm wide and 54 cm high. How many bars of soap can be put into it if each bar measures 6 cm × 4.5 cm × 4 cm?

Solution: Here, 1.2 m = 120 cm

Volume of the cardboard = (120 × 72 × 54) cm3 = 466560 cm3

Volume of each bar soap = (6 × 4.5 × 4) cm3 = 108 cm3

∴ Total number of bar soap = (466560/108) = 4320 bars.

(7) The size of a matchbox is 4 cm × 2.5 cm × 1.5 cm. What is the volume of a packet containing 144 matchboxes? How many such packets can be placed in a carton of size 1.5m × 84 cm × 60 cm?

Solution: Volume of a matchbox = (4 × 2.5 × 1.5) cm3 = 15 cm3

Volume of packet containing 144 matchboxes = (144 × 15) cm3 = 2160 cm3

Here, 1.5 m = 150 cm

Volume of carton = (150 × 84 × 60) cm3 = 756000 cm3

∴ Total number of packets could placed in a carton = (756000/2160) = 350 packets.

(8) How many planks of size 2 m × 25 cm × 8 cm can be prepared from a wooden block 5 m long, 70 cm broad and 32 cm thick, assuming that there is no wastage?

Solution: Here, 5 m = 500 cm

Volume of the wooden block = (500 × 70 × 32) cm3 = 1120000 cm3

Total volume of each plank = (200 × 25 × 8) cm.3 = 40000 cm3

∴ Total number of the planks = (1120000/40000) = 28 planks.

(9) How many bricks, each of size 25 cm × 13.5 cm × 6 cm, will be required to build a wall 8 m long, 5.4 m high and 33 cm thick?

Solution: Volume of each brick = (25 × 13.5 × 6) cm3 = 2025 cm3

Volume of the wall = (800 × 540 × 33) cm3 = 14256000 cm3

Total number of bricks = (14256000/2025) = 7040 bricks.

(10) A wall 15 m long, 30 cm wide and 4 m high is made of bricks, each measuring 22 cm × 12.5 cm × 7.5 cm. If 1/12 of the total volume of the wall consists of mortar, how many bricks are there in the wall?

Solution: Here, 15 m = 1500 cm

Volume of the wall = (1500 × 30 × 400) cm3 = 18000000 cm3

The quantity of the mortar = {(1/12) × 18000000} cm3 = 1500000 cm3

Volume of the bricks = (18000000 – 1500000) cm3 = 16500000 cm3

Volume of each brick = (22 × 12.5 × 7.5) cm3 = 2062.5cm3

∴ Total number of bricks = (16500000 ÷ 2062.5) = 8000 bricks.

(11) Find the capacity of a rectangular cistern in litres whose dimensions are 11.2 m × 6 m × 5.8 m. Find the area of the iron sheet required to make the cistern.

Solution: Volume of cistern = (11.2 × 6 × 5.8) m3 = 389.76 m3 = (389.76 × 1000) L = 389760 L

Area of the iron sheet = Total surface area of the cistern = 2(11.2 × 6 + 6 × 5.8 + 11.2 × 5.8) m2 = 2 (67.2 + 34.8 + 64.96) m2 = 333.92 m2

(12) The volume of a block of gold is 0.5 m3. If it is hammered into a sheet to cover an area of 1 hectare, find the thickness of the sheet.

Solution: We know, 1 hectare = 10000 m2

∴ Thickness of the sheet = (0.5 ÷ 10000) m = 0.00005 m = 0.005 cm.

(13) The rainfall recorded on a certain day was 5 cm. Find the volume of water that fell on a 2-hectare field.

Solution: Here, 2 hectare = 20000 m2

Rainfall recorded = 5 cm = 0.05 m

∴ Total rain over the field = (0.05 × 20000) m3 = 1000 m3

(14) A river 2 m deep and 45 m wide is flowing at the rate of 3 km/h. find the quantity of water that runs into the sea per minute.

Solution: Area of the cross – section of river = (45 × 2) m2 = 90 m2

Rate of flowing = 3 km/hr = [(3 × 1000) ÷ 60] m/min = 50 m/min.

(15) A pit 5 m long and 3.5 m wide is dug to a certain depth. If the volume of earth taken out of it is 14 m3, what is the depth of the pit?

Solution: Let the depth of the pit be x m.

Volume of pit = (5 × 3.5 × x) m3 = 17.5x m3

∴ 17.5x = 14

⇒ x = (14 ÷ 17.5) = 0.8m = 80 cm

Therefore, depth of the pit is 80 cm.

(16) A rectangular water tank is 90 cm wide and 40 cm deep. If it can contain 576 litres of water, what is its length?

Solution: Here, 576 L = 0.576 m3

Width = 90 cm = 0.9 m and depth = 40 cm = 0.4 m

Length = {0.576 ÷ (0.9 × 0.4)} m = (0.576 ÷ 0.36) m = 1.6 m.

(17) A beam of wood is 5 m long and 36 cm thick. It is made of 1.35 m3 of wood. What is the width of the beam?

Solution:  Thickness of the beam = 36 cm = 0.36 m.

∴ Width = {1.35 ÷ (5 × 0.36)} m = (1.35 ÷ 1.8) m = 0.75 m = 75 cm.

(18) The volume of a room is 378 m3 and the area of its floor is 84 m2. Find the height of the room.

Solution: Height of the area = (378 ÷ 84)m = 4.5 m.

(19) A swimming pool is 260 m long and 140 m wide. If 54600 cubic metres of water is pumped into it, find the height of the water level in it.

Solution: Height of the water = {54600 ÷ (260 × 140)} m = (54600 ÷ 36400) m = 1.5 m

(20) Find the volume of wood used to make a closed box of outer dimensions are 60 cm × 45 cm × 32 cm, the thickness of wood being 2.5 cm all around.

Solution: External volume of the box = (60 × 45 × 32) cm3 = 86400 cm3

Internal length = {60 – (2.5 × 2)} cm = 55 cm

Internal breadth = {45 – (2.5 × 2)} cm = 40 cm

Internal height = {32 – (2.5 × 2)} cm = 27 cm

∴ Internal volume = (55 × 40 × 27) cm3 = 59400 cm3

Volume of the wood = (86400 – 59400) cm3 = 27000 cm3

(21) Find the volume of iron required to make an open box whose external dimensions are 36 cm × 25 cm × 16.5 cm, the box being 1.5 cm thick throughout. If 1 cm3of iron weighs 8.5 grams, find the weight of the empty box in kilograms.

Solution: External volume of the box = (36 × 25 × 16.5) cm3 = 14850 cm3

Therefore, Internal length = {36 – (1.5 × 2)} cm = 33 cm

Internal breadth = {25 – (1.5 × 2)} cm = 22 cm

Internal height of open box = {16.5 – 1.5} cm = 15 cm

∴ Internal volume of the box = (33 × 22 × 15) cm3 = 10890 cm3

Therefore, volume of the iron = (14850 – 10890) cm3 = 3960 cm3

If 1 cm3 iron = 8.5 grams

∴ The weight of the box = (3960 × 8.5) grams = 33660 grams.

(22) A box with a lid is made of wood which is 3 cm thick. Its external length, breadth and height are 56 cm, 39 cm and 30 cm respectively. Find the capacity of the box. Also find the volume of wood used to make the box.

Solution: External volume of the box = (56 × 39 × 30) cm3 = 65520 cm3

Therefore, internal length = {56 – (3 × 2)} cm = 50 cm

Internal width = {39 – (3 × 2)} cm = 33 cm

Internal height = {30 – (3 × 2)} cm = 24 cm

∴ Internal volume = (50 × 33 × 24) cm3 = 39600 cm3

Volume of the wood = (65520 – 39600) cm3 = 25920 cm3

(23) The external dimensions of closed wooden box are 62 cm, 30 cm and 18 cm. If the box is made of 2 cm thick wood, find the capacity of the box.

Solution: External volume of the box = (62 × 30 × 18) cm3 = 33480 cm3

Therefore, Internal length = {62 – (2 × 2)} cm = 58 cm

Internal width = {30 – (2 × 2)} cm = 26 cm

Internal height = {18 – (2 × 2)} cm = 14 cm

∴ Internal volume = (58 × 26 × 14) cm3 = 21112 cm3

Therefore, the capacity of the box is 21112 cm3.

(24) A closed wooden box 80 cm long, 65 cm wide and 45 cm high, is made of 2.5 cm thick wood. Find the capacity of the box and its weight if 100 cm3 of wood weighs 8 g.

Solution: External volume of the box = (80 × 65 × 45) cm3 = 234000 cm3

Thickness = 2.5 cm

Therefore, internal length = {80 – (2.5 × 2)} cm = 75 cm

Internal wide = {65 – (2.5 × 2)} cm = 60 cm

Internal height = {45 – (2.5 × 2)} cm = 40 cm

∴ Internal volume = (75 × 60 × 40) cm = 180000 cm3

Volume of the wood = (234000 – 180000) cm3 = 54000 cm3

If 100 cm3 of wood weighs 8 g, then,

∴ Weight of the wood = {(54000 × 8) ÷ 100} g = 4320 g = 4.32 kg

(25) Find the volume, lateral surface area and the total surface area of a cube each of whose edges measures: (i) 7 m (ii) 5.6 cm (iii) 8 dm 5 cm

Solution: (i) Volume = 73 = 7×7×7 = 343 m3

Lateral surface = (4 × 72) = 4 × 7 × 7 = 196 m2

Total surface = (6 × 72) = 6 × 7 × 7 = 294 m2

(ii) Volume = (5.6)3 = 5.6 × 5.6 × 5.6 = 175.616 m3

Lateral surface = {4 × (5.6)2} m2 = 125.44 m2

Total surface = {6 × (5.6)2) m2 = 188.16 m2

(iii) Here, 8 dm 5 cm = 85 cm

Volume = (85)3 = 85 × 85 × 85 = 614125 cm3

Lateral surface = {4 × (85)2} m2 = 28900 m2

Total surface = {6 × (85)2} m2 = 43350 m2

(26) The surface area of a cube is 1176 cm2. Find its volume.

Solution: Let the length of the each edge of cube be a cm.

Total surface area = 6a2 cm2

∴ 6a2 = 1176

⇒ a2 = 196

⇒ a = √196 = 14 cm

Volume = (14)3 = 14 × 14 × 14 = 2744 cm3

(27) The volume of a cube is 729 cm3. Find its surface area.

Solution: Let the length of the each edge of cube be a cm.

Volume of cube = a3 cm3

∴ a3 = 729

⇒ a = ∛729 = 9 cm

∴ Total surface area = {6 × 92} cm2 = (6 × 9 × 9) cm2 = 486 cm2

(28) The dimensions of a metal block are 2.25 m by 1.5 m by 27 cm. It is melted and recast into cubes, each of side 45 cm. How many cubes are formed?

Solution: Here, 2.25 m = 225 cm and 1.5 m = 150 cm

Volume of the metal block = (225 × 150 × 27) cm3 = 911250 cm3

The volume of the cube = (45 × 45 × 45) cm3 = 91125 cm3

∴ Number of cubes = (911250 ÷ 91125) = 10 cubes.

(29) If the length of each edge of a cube is doubled, how many times does its volume become? How many times does its surface area become?

Solution: Let the length of each edge be a.

Volume of the cube = a3

Total surface area = 6a2

If the length of edge is double, then length becomes 2a.

∴ New volume = (2a)3 = 8a3

Total surface = 6 × (2a)2 = 24a2

Therefore while the surface area become increased by 4 then the volume increased by a factor of 8.

(30) A solid cubical block of fine wood casts Rs 256 at Rs 500 per m3. Find its volume and the length of each side.

Solution: Volume of the cubical block = (256 ÷ 500) m3 = 0.512 m3

∴ Length of the edge = (∛0.512) m = 0.8 m = 80 cm.

For more exercise solution, Click below –

Updated: May 30, 2022 — 2:37 pm

1. 2. 3. 