RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Exercise 20A

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TextbookNCERT
ClassClass 8
SubjectMaths
ChapterChapter 20
Chapter NameVolume and Surface Area of Solids
Exercise20 A
CategoryRS Aggarwal Solutions

RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Exercise 20A Download PDF

Question 1.
Solution:
(i)
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 1.1
Length of cuboid (l) = 22 cm.
Breadth (b) = 12 cm
and height (h) = 7.5 cm.
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 1.2
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 1.3
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 1.4
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 1.5

Question 2.
Solution:
Length of water tank (l) = 2 m
75cm = 2.75 m
breadth (b) = 1 m 80cm = 1.80 m
and height (h) = 1 m 40 cm = 1.40 m
Volume of water filled in it = l.b.h = 2.75 x 1.80 x 1.40 m³
= 6.93 m³
Water in litres = 6.93 x 1000
= 6930 litres (1 m³ = 1000 litres) Ans.

Question 3.
Solution:
Length of iron (l) = 1.05 m
= 105 cm
breadth (b) = 70 cm and height (h) = 1.5 cm
volume of iron = l x b x h = 105 x 70 x 1.5 cm³
= 11025 cm³
weight of 1cm³ iron = 8 gram
Total weight = 11025 x g = 88200 g
= 882001000 kg
= 88.2 kg Ans.

Question 4.
Solution:
Area of courtyard = 3750 m²
Height of gravel = 1 cm.
Volume of gravel = 3750 x 1100 m³
= 37.50 m³
Cost of 1 m³ gravel = Rs. 6.40
Total cost = Rs. 6.40 x 37.50
= Rs. 240 Ans.

Question 5.
Solution:
Length of hall (l) = 16 m
Breadth (b) = 12.5 m
height (h) = 4.5 m
Volume of air in it = l x b x h
= 16 x 12.5 x 4.5 m3
= 900 m³
Air for one person is required = 3.6 m³
Number of person which can be accommodated in the hall = 900 ÷ 3.6
= 900×1036
= 250

Question 6.
Solution:
Length of cardboard box (l)
= 1.2 m = 120 cm.
breadth (b) = 72 cm.
Height (h) = 54 cm.
Volume of box = l x b x h
= 120 x 72 x 54 cm³
= 466560 cm³
Volume of one soap bar = 6 x 4.5 x 4 cm³
= 108 cm³
No. of bars to be kept in it = 466560108
= 4320 Ans.

Question 7.
Solution:
Volume of one match box = 4 x 2.5 x 1.5 cm³ = 15 cm³
Volume of 144 matchboxes = 15 x 144 cm³
or volume of one packet = 2160 cm³
Length of carton (l) = 1.5 m = 150 cm
Breadth (b) = 84 cm
and height (h) = 60 cm.
Volume of one carton = l x b x h
= 150 x 84 x 60 cm³
= 756000 cm³
No. of packets = 756000 ÷ 2160
= 350 Ans.

Question 8.
Solution:
Length of one plank = 2m
= 200 cm
Breadth (b) = 25 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 8.1

Question 9.
Solution:
Length of wall (l) = 8m = 800 cm
Height (h) = 5.4 m = 540 cm
Width (b) = 33 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 9.1

Question 10.
Solution:
Length of wall (l) = 15 m
Width (b) = 30 cm = 0.3 m
Height (h) = 4 m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 10.1

Question 11.
Solution:
Length of rectangular cistern (l) = 11.2 m
Breadth (b) = 6 m
Height (h) = 5.8 m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 11.1

Question 12.
Solution:
Volume of block of gold = 0.5 m³
= 0.5 x 1000000 cm³
= 500000 cm³
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 12.1

Question 13.
Solution:
Area of field = 2 hectare
= 20000 m²
Rainfall = 5 cm. = 0.05 m
Volume of water of rainfall
= Area of field x height of rainfall water
= 20000 x 0.05 m³
= 1000 m³ Ans.

Question 14.
Solution:
Speed of water = 3 km/h
Length of water flow in 1 minute
= 3km60m
= 300060
= 50 m
Width of river = 45 m
Depth of river = 2 m
Volume of water in 1 minute
= 45 x 2 x 50 m³
= 4500 m³ Ans.

Question 15.
Solution:
Length of pit (l) = 5m
Width (b) = 3.5 m
Let depth of pit = h
then volume of earth dug out
= l.b.h = 5 x 3.5 x h = 17.5 h m³
But volume of earth = 14 m³
17.5 h = 14
h = 1417.5 = 140175
=> h = 45 m
= 45 x 100
= 80 cm Ans.

Question 16.
Solution:
Width of tank = 90 cm = 90100 m
Depth = 40 cm = 40100 m
Water = 576 litre
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 16.1

Question 17.
Solution:
Volume of wood = 1.35 m³
Length of beam = 5m
Thickness = 36 cm = 36100 m.
Width = Volumelength×thickness
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 7.1

Question 18.
Solution:
Volume of a room = 378 m³
Area of its floor = 84 m²
Height = VolumeArea
= 37884 m
= 4.5 m Ans.

Question 19.
Solution:
Length of pool = 260 m
and width = 140 m.
Volume of water = 54600 m³
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 19.1

Question 20.
Solution:
Outer length of wooden box (L) = 60 cm
Width (B) = 45 cm
and height (H) = 32 cm.
Thickness of wood used = 2.5 cm.
Inner length (l) = 60 – 2 x 2.5
= 60 – 5
= 55 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 20.1

Question 21.
Solution:
Outer length of open box = 36 cm
breadth = 25 cm
and height = 16.5 cm
thickness of iron = 1.5 cm.
∴ Inner length = 36 – 2 x 1.5
= 36 – 3
= 33 cm
breadth = 25 – 2 x 1.5
= 25 – 3
= 22 cm
and height = 16.5 – 1.5
= 15 cm .
∴ Volume of iron used in it = Outer volume – Inner volume
= 36 x 25 x 16.5 cm3 – 33 x 22 x 15 cm³
= 14850 – 10890
= 3960 cm³
weight of 1 cm³ = 8.5 gram
∴ Total weight = 3960 x 8.5 g
= 33660 g
= 33.660 kg
= 33.66 kg Ans.

Question 22.
Solution:
Outer length of the box = 56 cm
Width = 39 cm
and height = 30 cm
Volume = 56 x 39 x 30
= 65520 cm³
Thickness of wood used = 3cm.
∴ Inner length = 56 – 2 x 3
= 56 – 6
= 50 cm
Width = 39 – 2 x 3
= 39 – 6
= 33 cm
and height = 30 – 2 x 3
= 30 – 6
= 24 cm
∴ Inner volume of the box = 50 x 33 x 24 cm³
= 39600 cm³
and volume of wood used = Outer volume – Inner volume
= (65520 – 39600)cm³
= 25920 cm³ Ans.

Question 23.
Solution:
Outer length of box = 62 cm.
Outer width = 30 cm.
Outer height = 18 cm.
Thickness of wood = 2 cm.
∴ Internal length = 62 – 2 x 2
= 58 cm.
Internal width = 30 – 2 x 2
= 26 cm.
Internal height =18 – 2 x 2
= 14 cm.
Capacity of the box = lbh
= 58 x 26 x 14 cm³
= 21112 cm³ Ans.

Question 24.
Solution:
Outer length = 80 cm.
Outer width = 65 cm.
Outer height = 45 cm.
Total volume = 80 x 65 x 45 cm³
= 234000 cm³
Thickness of wood = 2.5 cm.
∴ Inner length = 80 – 2 x 2.5 = 75 cm.
Inner width = 65 – 2 x 2.5 = 60 cm.
Inner height = 45 – 2 x 2.5 = 40 cm.
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 24.1

Question 25.
Solution:
(i) Edge of cube (a) = 7 m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 25.1
(a) Volume = a³ = (7)³
= 7 x 7 x 7 m³
= 343 m³ Ans.
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 25.2
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 25.3

Question 26.
Solution:
Surface area of a cube = 1176 cm²
Let edge of the cube = a
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 26.1

Question 27.
Solution:
Volume of a cube = 729 cm³
Let edge of cube = a
then a³ = 729 = (9)³
a = 9 cm.
Hence surface area = 6a² = 6 (9)² cm²
= 6 x 81
= 486 cm² Ans.

Question 28.
Solution:
Length of metal block (l) = 2.25 m = 225 cm
Width (b) = 1.5 m = 150 cm
and height (h) = 27 cm
Volume of block = l x b x h
= 225 x 150 x 27 cm³
= 911250 cm³
Side of each cube = 45 cm.
Volume of each cube = a³
= 45 x 45 x 45
= 94125 cm³
Number of cubes = 91125091125
= 10 Ans.

Question 29.
Solution:
Let edge of given cube = a
Volume = a³
and surface area = 6a²
By doubling the edge of cube3 the side of new cube = a x 2 = 2a
Volume (2a)³ = 8a³
and surface area = 6 (2a)² = 6 x 4a²
= 24a² = 4 x 6a²
It is clear from the above that
Volume is increased 8 times and surface area is 4 times. Ans.

Question 30.
Solution:
Total cost of wood = Rs. 256
Rate = Rs.,500 per m³
Volume of wood = 256500 = 0.512 m³
= 0.512 x 100 x 100 x 100 cm³
= 512000 cm³
Let length of each side = a
then a³ = 512000 = (80)³
a = 80
Hence length of each side = 80 cm. Ans.

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