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# RS Aggarwal MCQs: Volumes and Surface Area of Solids Notes | EduRev

## Class 9 : RS Aggarwal MCQs: Volumes and Surface Area of Solids Notes | EduRev

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Question:111
The length, breadth and height of a cuboid are 15 cm, 12 cm and 4.5 cm respectively. Its volume is
a 243 cm
3
b 405 cm
3
c 810 cm
3
d 603 cm
3
Solution:
c 810 cm
3
Volume of the cuboid = l ×b ×h                                          = 15 ×12 ×4. 5  cm
3
= 810 cm
3
Question:112
A cuboid is 12 cm long, 9 cm broad and 8 cm high. Its total surface area is
a 864 cm
2
b 552 cm
2
c 432 cm
2
d 276 cm
2
Solution:
b 552 cm
2
Total surface area of the cuboid = 2(lb +bh +lh) cm
2
= 2 12 ×9 +8 ×9 +12 ×8 cm
2
= 2(108 +72 +96) cm
2

Question:113
The length, breadth and height of a cuboid are 15 m, 6 m and 5 dm respectively. The lateral surface area of the cuboid is
a 45 m
2
b 21 m
2
c 201 m
2
d 90 m
2
Solution:
Length of the cuboid, l = 15 m
Breadth of the cuboid, b = 6 m
Height of the cuboid, h = 5 dm = 0.5 m           1m = 10dm
? Lateral surface area of the cuboid = 2h(l + b) = 2 × 0.5 × 15 +6 = 2 × 0.5 × 21 = 21 cm
2
Hence, the correct answer is option b.
Question:114
A beam 9 m long, 40 cm wide and 20 cm high is made up of iron which weighs 50 kg per cubic metre. The weight of the beam is
a 27 kg
b 48 kg
c 36 kg
d 56 kg
Solution:
c 36 kg
Length = 9 m
Breadth = 40 cm = 0.4 m
Height = 20 cm = 0.2 m
? Weight of the beam = volume of the beam ×weight of iron per cubic metre
= 9 ×0. 4 ×0. 2 ×50 = 36 kg
Question:115
The length of the longest rod that can be placed in a room of dimensions (10 m × 10 m × 5 m) is
a 15 m
b 16 m
c 10v 5 m
d 12 m
Solution:
a 15 m
Length of longest rod = diagonal of the room
= diagonal of a cuboid
=
v
l
2
+b
2
+h
2
= v 100 +100 +25 m = v 225 m = 15 m
Question:116
What is the maximum length of a pencil that can be placed in a rectangular box of dimensions (8 cm × 6 cm × 5 cm)?
a 8 cm
b 9.5 cm
c 19 cm
d 11.2 cm
( )
Page 2

Question:111
The length, breadth and height of a cuboid are 15 cm, 12 cm and 4.5 cm respectively. Its volume is
a 243 cm
3
b 405 cm
3
c 810 cm
3
d 603 cm
3
Solution:
c 810 cm
3
Volume of the cuboid = l ×b ×h                                          = 15 ×12 ×4. 5  cm
3
= 810 cm
3
Question:112
A cuboid is 12 cm long, 9 cm broad and 8 cm high. Its total surface area is
a 864 cm
2
b 552 cm
2
c 432 cm
2
d 276 cm
2
Solution:
b 552 cm
2
Total surface area of the cuboid = 2(lb +bh +lh) cm
2
= 2 12 ×9 +8 ×9 +12 ×8 cm
2
= 2(108 +72 +96) cm
2

Question:113
The length, breadth and height of a cuboid are 15 m, 6 m and 5 dm respectively. The lateral surface area of the cuboid is
a 45 m
2
b 21 m
2
c 201 m
2
d 90 m
2
Solution:
Length of the cuboid, l = 15 m
Breadth of the cuboid, b = 6 m
Height of the cuboid, h = 5 dm = 0.5 m           1m = 10dm
? Lateral surface area of the cuboid = 2h(l + b) = 2 × 0.5 × 15 +6 = 2 × 0.5 × 21 = 21 cm
2
Hence, the correct answer is option b.
Question:114
A beam 9 m long, 40 cm wide and 20 cm high is made up of iron which weighs 50 kg per cubic metre. The weight of the beam is
a 27 kg
b 48 kg
c 36 kg
d 56 kg
Solution:
c 36 kg
Length = 9 m
Breadth = 40 cm = 0.4 m
Height = 20 cm = 0.2 m
? Weight of the beam = volume of the beam ×weight of iron per cubic metre
= 9 ×0. 4 ×0. 2 ×50 = 36 kg
Question:115
The length of the longest rod that can be placed in a room of dimensions (10 m × 10 m × 5 m) is
a 15 m
b 16 m
c 10v 5 m
d 12 m
Solution:
a 15 m
Length of longest rod = diagonal of the room
= diagonal of a cuboid
=
v
l
2
+b
2
+h
2
= v 100 +100 +25 m = v 225 m = 15 m
Question:116
What is the maximum length of a pencil that can be placed in a rectangular box of dimensions (8 cm × 6 cm × 5 cm)?
a 8 cm
b 9.5 cm
c 19 cm
d 11.2 cm
( )
Solution:
d 11.2 cm
Maximum length of the pencil =  diagonal of the box
=
v
l
2
+b
2
+h
2
=
v
8
2
+6
2
+5
2
cm = v 64 +36 +25 cm = v 125 cm = 11. 2 cm
Question:117
The number of planks of dimensions (4 m × 5 m × 2 m) that can be stored in a pit which is 40 m long, 12 m wide and 16 m deep, is
a 190
b 192
c 184
d 180
Solution:
b  192
Number of planks =
volume of the pit
volume of 1 plank
=
40×12×16
4×5×2
=
7680
40
= 192

Question:118
How many planks of dimensions (5 m × 25 cm × 10 cm) can be stored in a pit which is 20 m long, 6 m wide and 50 cm deep?
a 480
b 450
c 320
d 360
Solution:
a  480
Length of the pit = 20 m
Breadth of the pit = 6 m
Height of the pit = 50 cm = 0.5 m
Length of the plank = 5m
Breadth of the plank = 25 cm = 0.25 m
Height of the plank = 10 cm = 0.1 m
? Number of planks =
volume of the pit
volume of 1 plank
=
20×6×0.5
5×0.25×0.1
=
60×1000
125
= 480
Question:119
How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick if each brick measures (25 cm × 11.25 cm × 6 cm)?
a 4800
b 5600
c 6400
d 5200
Solution:
c  6400
Length of the wall = 8 m = 800 cm
Breadth of the wall = 6 m = 600 cm
Height of the wall = 22.5 cm
Length of the brick = 25 cm
Breadth of the brick = 11.25 cm
Height of the brick = 6 cm
? Number of bricks required =
volume of the wall
volume of 1 brick
=
800×600×22.5
25×11.25×6
=
10800000
1687.5
= 6400
Question:120
How many persons can be accommodated in a dining hall of dimensions (20 m × 15 m × 4.5 m), assuming that each person requires 5 m
3
of air?
a 250
b 270
c 320
d 300
Solution:
b 270
Number of persons  =
volume of the hall
volume of air required by 1 person
=
20×15×4.5
5
= 20 ×3 ×4. 5 = 270
? 270 persons can be accommodated.
Question:121
A river 1.5 m deep and 30 m wide is flowing at the rate of 3 km per hour. The volume of water that runs into the sea per minute is
a 2000 m
3
b 2250 m
3
c 2500 m
3
d 2750 m
3
Solution:
Page 3

Question:111
The length, breadth and height of a cuboid are 15 cm, 12 cm and 4.5 cm respectively. Its volume is
a 243 cm
3
b 405 cm
3
c 810 cm
3
d 603 cm
3
Solution:
c 810 cm
3
Volume of the cuboid = l ×b ×h                                          = 15 ×12 ×4. 5  cm
3
= 810 cm
3
Question:112
A cuboid is 12 cm long, 9 cm broad and 8 cm high. Its total surface area is
a 864 cm
2
b 552 cm
2
c 432 cm
2
d 276 cm
2
Solution:
b 552 cm
2
Total surface area of the cuboid = 2(lb +bh +lh) cm
2
= 2 12 ×9 +8 ×9 +12 ×8 cm
2
= 2(108 +72 +96) cm
2

Question:113
The length, breadth and height of a cuboid are 15 m, 6 m and 5 dm respectively. The lateral surface area of the cuboid is
a 45 m
2
b 21 m
2
c 201 m
2
d 90 m
2
Solution:
Length of the cuboid, l = 15 m
Breadth of the cuboid, b = 6 m
Height of the cuboid, h = 5 dm = 0.5 m           1m = 10dm
? Lateral surface area of the cuboid = 2h(l + b) = 2 × 0.5 × 15 +6 = 2 × 0.5 × 21 = 21 cm
2
Hence, the correct answer is option b.
Question:114
A beam 9 m long, 40 cm wide and 20 cm high is made up of iron which weighs 50 kg per cubic metre. The weight of the beam is
a 27 kg
b 48 kg
c 36 kg
d 56 kg
Solution:
c 36 kg
Length = 9 m
Breadth = 40 cm = 0.4 m
Height = 20 cm = 0.2 m
? Weight of the beam = volume of the beam ×weight of iron per cubic metre
= 9 ×0. 4 ×0. 2 ×50 = 36 kg
Question:115
The length of the longest rod that can be placed in a room of dimensions (10 m × 10 m × 5 m) is
a 15 m
b 16 m
c 10v 5 m
d 12 m
Solution:
a 15 m
Length of longest rod = diagonal of the room
= diagonal of a cuboid
=
v
l
2
+b
2
+h
2
= v 100 +100 +25 m = v 225 m = 15 m
Question:116
What is the maximum length of a pencil that can be placed in a rectangular box of dimensions (8 cm × 6 cm × 5 cm)?
a 8 cm
b 9.5 cm
c 19 cm
d 11.2 cm
( )
Solution:
d 11.2 cm
Maximum length of the pencil =  diagonal of the box
=
v
l
2
+b
2
+h
2
=
v
8
2
+6
2
+5
2
cm = v 64 +36 +25 cm = v 125 cm = 11. 2 cm
Question:117
The number of planks of dimensions (4 m × 5 m × 2 m) that can be stored in a pit which is 40 m long, 12 m wide and 16 m deep, is
a 190
b 192
c 184
d 180
Solution:
b  192
Number of planks =
volume of the pit
volume of 1 plank
=
40×12×16
4×5×2
=
7680
40
= 192

Question:118
How many planks of dimensions (5 m × 25 cm × 10 cm) can be stored in a pit which is 20 m long, 6 m wide and 50 cm deep?
a 480
b 450
c 320
d 360
Solution:
a  480
Length of the pit = 20 m
Breadth of the pit = 6 m
Height of the pit = 50 cm = 0.5 m
Length of the plank = 5m
Breadth of the plank = 25 cm = 0.25 m
Height of the plank = 10 cm = 0.1 m
? Number of planks =
volume of the pit
volume of 1 plank
=
20×6×0.5
5×0.25×0.1
=
60×1000
125
= 480
Question:119
How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick if each brick measures (25 cm × 11.25 cm × 6 cm)?
a 4800
b 5600
c 6400
d 5200
Solution:
c  6400
Length of the wall = 8 m = 800 cm
Breadth of the wall = 6 m = 600 cm
Height of the wall = 22.5 cm
Length of the brick = 25 cm
Breadth of the brick = 11.25 cm
Height of the brick = 6 cm
? Number of bricks required =
volume of the wall
volume of 1 brick
=
800×600×22.5
25×11.25×6
=
10800000
1687.5
= 6400
Question:120
How many persons can be accommodated in a dining hall of dimensions (20 m × 15 m × 4.5 m), assuming that each person requires 5 m
3
of air?
a 250
b 270
c 320
d 300
Solution:
b 270
Number of persons  =
volume of the hall
volume of air required by 1 person
=
20×15×4.5
5
= 20 ×3 ×4. 5 = 270
? 270 persons can be accommodated.
Question:121
A river 1.5 m deep and 30 m wide is flowing at the rate of 3 km per hour. The volume of water that runs into the sea per minute is
a 2000 m
3
b 2250 m
3
c 2500 m
3
d 2750 m
3
Solution:
b 2250 m
3
Length of the river = 1.5 m
Breadth of the river = 30 m
Depth of the river = 3 km = 3000 m
Now, volume of water that runs into the sea = 1. 5 ×30 ×3000 m
3
= 135000 m
3
? Volume of water that runs into the sea per minute =
135000
60
= 2250 m
3
Question:122
The lateral surface area of a cube is 256 m
2
. The volume of the cube is
a 64 m
3
b 216 m
3
c 256 m
3
d 512 m
3
Solution:
d 512  m
3
Suppose that a m be the edge of the cube.
We have:
4a
2
= 256 ? a
2
=
256
4
= 64 ? a = 8 m
? Volume of the cube = a
3
m
3
= 8
3
m
3
= 512  m
3

Question:123
The total surface area of a cube is 96 cm
2
. The volume of the cube is
a 8 cm
3
b 27 cm
3
c 64 cm
3
d 512 cm
3
Solution:
c  64 cm
3
Let a cm be the edge of the cube.
We have:
6a
2
= 96 ? a
2
= 16 ? a = 4 cm
? Volume of the cube = a
3
cm
3
= 4
3
cm
3
= 64 cm
3

Question:124
The volume of a cube is 512 cm
3
. Its total surface area is
a 256 cm
2
b 384 cm
2
c 512 cm
2
d 64 cm
2
Solution:
b 384 cm
2
Suppose that a cm is the edge of the cube.
We have:
a
3
= 512 ? a =
3
v 512 = 8 cm
? Total surface area of cube = 6a
2
cm
2
= 6 ×8 ×8 cm
2
= 384 cm
2
Question:125
The length of the longest rod that can fit in a cubical vessel of side 10 cm, is
a 10 cm
b 20 cm
c 10v 2 cm
d 10v 3 cm
Solution:
d  10v 3 cm
Length of the longest rod = body diagonal of the vessel
= v 3a = v 3 ×10 = 10v 3 cm
Question:126
If the length of diagonal of a cube is 8v 3 cm, then its surface area is
a 192 cm
2
b 384 cm
2
c 512 cm
2
d 768 cm
2
Solution:
b  384 cm
2
We have:
v 3a = 8v 3 ? a = 8 cm
? Surface area of the cube = 6a
2
= 6 ×8 ×8 = 384 cm
2
Question:127
If each edge of a cube is increased by 50%, then the percentage increase in its surface area is
10 cm
Page 4

Question:111
The length, breadth and height of a cuboid are 15 cm, 12 cm and 4.5 cm respectively. Its volume is
a 243 cm
3
b 405 cm
3
c 810 cm
3
d 603 cm
3
Solution:
c 810 cm
3
Volume of the cuboid = l ×b ×h                                          = 15 ×12 ×4. 5  cm
3
= 810 cm
3
Question:112
A cuboid is 12 cm long, 9 cm broad and 8 cm high. Its total surface area is
a 864 cm
2
b 552 cm
2
c 432 cm
2
d 276 cm
2
Solution:
b 552 cm
2
Total surface area of the cuboid = 2(lb +bh +lh) cm
2
= 2 12 ×9 +8 ×9 +12 ×8 cm
2
= 2(108 +72 +96) cm
2

Question:113
The length, breadth and height of a cuboid are 15 m, 6 m and 5 dm respectively. The lateral surface area of the cuboid is
a 45 m
2
b 21 m
2
c 201 m
2
d 90 m
2
Solution:
Length of the cuboid, l = 15 m
Breadth of the cuboid, b = 6 m
Height of the cuboid, h = 5 dm = 0.5 m           1m = 10dm
? Lateral surface area of the cuboid = 2h(l + b) = 2 × 0.5 × 15 +6 = 2 × 0.5 × 21 = 21 cm
2
Hence, the correct answer is option b.
Question:114
A beam 9 m long, 40 cm wide and 20 cm high is made up of iron which weighs 50 kg per cubic metre. The weight of the beam is
a 27 kg
b 48 kg
c 36 kg
d 56 kg
Solution:
c 36 kg
Length = 9 m
Breadth = 40 cm = 0.4 m
Height = 20 cm = 0.2 m
? Weight of the beam = volume of the beam ×weight of iron per cubic metre
= 9 ×0. 4 ×0. 2 ×50 = 36 kg
Question:115
The length of the longest rod that can be placed in a room of dimensions (10 m × 10 m × 5 m) is
a 15 m
b 16 m
c 10v 5 m
d 12 m
Solution:
a 15 m
Length of longest rod = diagonal of the room
= diagonal of a cuboid
=
v
l
2
+b
2
+h
2
= v 100 +100 +25 m = v 225 m = 15 m
Question:116
What is the maximum length of a pencil that can be placed in a rectangular box of dimensions (8 cm × 6 cm × 5 cm)?
a 8 cm
b 9.5 cm
c 19 cm
d 11.2 cm
( )
Solution:
d 11.2 cm
Maximum length of the pencil =  diagonal of the box
=
v
l
2
+b
2
+h
2
=
v
8
2
+6
2
+5
2
cm = v 64 +36 +25 cm = v 125 cm = 11. 2 cm
Question:117
The number of planks of dimensions (4 m × 5 m × 2 m) that can be stored in a pit which is 40 m long, 12 m wide and 16 m deep, is
a 190
b 192
c 184
d 180
Solution:
b  192
Number of planks =
volume of the pit
volume of 1 plank
=
40×12×16
4×5×2
=
7680
40
= 192

Question:118
How many planks of dimensions (5 m × 25 cm × 10 cm) can be stored in a pit which is 20 m long, 6 m wide and 50 cm deep?
a 480
b 450
c 320
d 360
Solution:
a  480
Length of the pit = 20 m
Breadth of the pit = 6 m
Height of the pit = 50 cm = 0.5 m
Length of the plank = 5m
Breadth of the plank = 25 cm = 0.25 m
Height of the plank = 10 cm = 0.1 m
? Number of planks =
volume of the pit
volume of 1 plank
=
20×6×0.5
5×0.25×0.1
=
60×1000
125
= 480
Question:119
How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick if each brick measures (25 cm × 11.25 cm × 6 cm)?
a 4800
b 5600
c 6400
d 5200
Solution:
c  6400
Length of the wall = 8 m = 800 cm
Breadth of the wall = 6 m = 600 cm
Height of the wall = 22.5 cm
Length of the brick = 25 cm
Breadth of the brick = 11.25 cm
Height of the brick = 6 cm
? Number of bricks required =
volume of the wall
volume of 1 brick
=
800×600×22.5
25×11.25×6
=
10800000
1687.5
= 6400
Question:120
How many persons can be accommodated in a dining hall of dimensions (20 m × 15 m × 4.5 m), assuming that each person requires 5 m
3
of air?
a 250
b 270
c 320
d 300
Solution:
b 270
Number of persons  =
volume of the hall
volume of air required by 1 person
=
20×15×4.5
5
= 20 ×3 ×4. 5 = 270
? 270 persons can be accommodated.
Question:121
A river 1.5 m deep and 30 m wide is flowing at the rate of 3 km per hour. The volume of water that runs into the sea per minute is
a 2000 m
3
b 2250 m
3
c 2500 m
3
d 2750 m
3
Solution:
b 2250 m
3
Length of the river = 1.5 m
Breadth of the river = 30 m
Depth of the river = 3 km = 3000 m
Now, volume of water that runs into the sea = 1. 5 ×30 ×3000 m
3
= 135000 m
3
? Volume of water that runs into the sea per minute =
135000
60
= 2250 m
3
Question:122
The lateral surface area of a cube is 256 m
2
. The volume of the cube is
a 64 m
3
b 216 m
3
c 256 m
3
d 512 m
3
Solution:
d 512  m
3
Suppose that a m be the edge of the cube.
We have:
4a
2
= 256 ? a
2
=
256
4
= 64 ? a = 8 m
? Volume of the cube = a
3
m
3
= 8
3
m
3
= 512  m
3

Question:123
The total surface area of a cube is 96 cm
2
. The volume of the cube is
a 8 cm
3
b 27 cm
3
c 64 cm
3
d 512 cm
3
Solution:
c  64 cm
3
Let a cm be the edge of the cube.
We have:
6a
2
= 96 ? a
2
= 16 ? a = 4 cm
? Volume of the cube = a
3
cm
3
= 4
3
cm
3
= 64 cm
3

Question:124
The volume of a cube is 512 cm
3
. Its total surface area is
a 256 cm
2
b 384 cm
2
c 512 cm
2
d 64 cm
2
Solution:
b 384 cm
2
Suppose that a cm is the edge of the cube.
We have:
a
3
= 512 ? a =
3
v 512 = 8 cm
? Total surface area of cube = 6a
2
cm
2
= 6 ×8 ×8 cm
2
= 384 cm
2
Question:125
The length of the longest rod that can fit in a cubical vessel of side 10 cm, is
a 10 cm
b 20 cm
c 10v 2 cm
d 10v 3 cm
Solution:
d  10v 3 cm
Length of the longest rod = body diagonal of the vessel
= v 3a = v 3 ×10 = 10v 3 cm
Question:126
If the length of diagonal of a cube is 8v 3 cm, then its surface area is
a 192 cm
2
b 384 cm
2
c 512 cm
2
d 768 cm
2
Solution:
b  384 cm
2
We have:
v 3a = 8v 3 ? a = 8 cm
? Surface area of the cube = 6a
2
= 6 ×8 ×8 = 384 cm
2
Question:127
If each edge of a cube is increased by 50%, then the percentage increase in its surface area is
10 cm
a 50%
b 75%
c 100%
d 12%
Solution:
Let a be the edge of the cube.
Then the surface area is 6a
2
= S say
Now, increased edge = a +
50
100
a =
150
100
a =
3
2
a
Then, new surface area = 6
3
2
a
2
= 6 ×
9
4
a
2
=
9
4
SIncrease in surface area =
9
4
S - S  =
5
4
S ? Percentage increase in surface area =
5
4
S
S
=
5
4
×100 % = 125 %
Question:128
Three cubes of metal with edges 3 cm, 4 cm and 5 cm respectively are melted to form a single cube. The lateral surface area of the new cube formed is
a 72 cm
2
b 144 cm
2
c 128 cm
2
d 256 cm
2
Solution:
b  144  cm
2
Volume of the new cube formed = total volume of the three cubes
Suppose that a cm is the edge of the new cube, then
a
3
= 3
3
+4
3
+5
3
= 27 +64 +125      = 216     = 6 cm
? Lateral surface area of the new cube = 4a
2
= 4 ×6 ×6 = 144 cm
2
Question:129
In a shower, 5 cm of rain falls. What is the volume of water that falls on 2 hectares of ground?
a 500 m
3
b 750 m
3
c 800 m
3
d 1000 m
3
Solution:
d 1000 m
3

Area of the land = 2 sq hec = 2000 sq mAmount of rainfall = 5 cm = 0. 05 m ? Volume of the water = area of the land ×amount of rainfall                              = 2000 ×0. 05
Question:130
Two cubes have their volumes in the ratio 1 : 27. The ratio of their surface areas is
a 1 : 3
b 1 : 8
c 1 : 9
d 1 : 18
Solution:
c 1 : 9
Suppose that the edges of the cubes are a and b.
We have:
a
3
b
3
=
1
27
?
a
b
3
=
1
27
?
a
b
=
1
3
? Ratio of the surface areas =
6a
2
6b
2
=
a
b
2
=
1
3
2
=
1
9
Question:131
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
Solution:
d becomes 8 times
Suppose that the side of the cube is a.
When it is doubled, it becomes 2a.
New volume of the cube = (2a)
3
= 8a
3

Hence, the volume becomes 8 times the original volume.
Question:132
The diameter of the base of a cylinder is 6 cm and its height is 14 cm. The volume of the cylinder is
a 198 cm
3
b 396 cm
3
c 495 cm
3
d 297 cm
3
Solution:
b  396 cm
3
Volume of the cylinder = pr
2
h                                      =
22
7
×3
2
×14                                      = 22 ×9 ×2                                       = 396 cm
3
( )
( )
( )
( )
( ) ( )
Page 5

Question:111
The length, breadth and height of a cuboid are 15 cm, 12 cm and 4.5 cm respectively. Its volume is
a 243 cm
3
b 405 cm
3
c 810 cm
3
d 603 cm
3
Solution:
c 810 cm
3
Volume of the cuboid = l ×b ×h                                          = 15 ×12 ×4. 5  cm
3
= 810 cm
3
Question:112
A cuboid is 12 cm long, 9 cm broad and 8 cm high. Its total surface area is
a 864 cm
2
b 552 cm
2
c 432 cm
2
d 276 cm
2
Solution:
b 552 cm
2
Total surface area of the cuboid = 2(lb +bh +lh) cm
2
= 2 12 ×9 +8 ×9 +12 ×8 cm
2
= 2(108 +72 +96) cm
2

Question:113
The length, breadth and height of a cuboid are 15 m, 6 m and 5 dm respectively. The lateral surface area of the cuboid is
a 45 m
2
b 21 m
2
c 201 m
2
d 90 m
2
Solution:
Length of the cuboid, l = 15 m
Breadth of the cuboid, b = 6 m
Height of the cuboid, h = 5 dm = 0.5 m           1m = 10dm
? Lateral surface area of the cuboid = 2h(l + b) = 2 × 0.5 × 15 +6 = 2 × 0.5 × 21 = 21 cm
2
Hence, the correct answer is option b.
Question:114
A beam 9 m long, 40 cm wide and 20 cm high is made up of iron which weighs 50 kg per cubic metre. The weight of the beam is
a 27 kg
b 48 kg
c 36 kg
d 56 kg
Solution:
c 36 kg
Length = 9 m
Breadth = 40 cm = 0.4 m
Height = 20 cm = 0.2 m
? Weight of the beam = volume of the beam ×weight of iron per cubic metre
= 9 ×0. 4 ×0. 2 ×50 = 36 kg
Question:115
The length of the longest rod that can be placed in a room of dimensions (10 m × 10 m × 5 m) is
a 15 m
b 16 m
c 10v 5 m
d 12 m
Solution:
a 15 m
Length of longest rod = diagonal of the room
= diagonal of a cuboid
=
v
l
2
+b
2
+h
2
= v 100 +100 +25 m = v 225 m = 15 m
Question:116
What is the maximum length of a pencil that can be placed in a rectangular box of dimensions (8 cm × 6 cm × 5 cm)?
a 8 cm
b 9.5 cm
c 19 cm
d 11.2 cm
( )
Solution:
d 11.2 cm
Maximum length of the pencil =  diagonal of the box
=
v
l
2
+b
2
+h
2
=
v
8
2
+6
2
+5
2
cm = v 64 +36 +25 cm = v 125 cm = 11. 2 cm
Question:117
The number of planks of dimensions (4 m × 5 m × 2 m) that can be stored in a pit which is 40 m long, 12 m wide and 16 m deep, is
a 190
b 192
c 184
d 180
Solution:
b  192
Number of planks =
volume of the pit
volume of 1 plank
=
40×12×16
4×5×2
=
7680
40
= 192

Question:118
How many planks of dimensions (5 m × 25 cm × 10 cm) can be stored in a pit which is 20 m long, 6 m wide and 50 cm deep?
a 480
b 450
c 320
d 360
Solution:
a  480
Length of the pit = 20 m
Breadth of the pit = 6 m
Height of the pit = 50 cm = 0.5 m
Length of the plank = 5m
Breadth of the plank = 25 cm = 0.25 m
Height of the plank = 10 cm = 0.1 m
? Number of planks =
volume of the pit
volume of 1 plank
=
20×6×0.5
5×0.25×0.1
=
60×1000
125
= 480
Question:119
How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick if each brick measures (25 cm × 11.25 cm × 6 cm)?
a 4800
b 5600
c 6400
d 5200
Solution:
c  6400
Length of the wall = 8 m = 800 cm
Breadth of the wall = 6 m = 600 cm
Height of the wall = 22.5 cm
Length of the brick = 25 cm
Breadth of the brick = 11.25 cm
Height of the brick = 6 cm
? Number of bricks required =
volume of the wall
volume of 1 brick
=
800×600×22.5
25×11.25×6
=
10800000
1687.5
= 6400
Question:120
How many persons can be accommodated in a dining hall of dimensions (20 m × 15 m × 4.5 m), assuming that each person requires 5 m
3
of air?
a 250
b 270
c 320
d 300
Solution:
b 270
Number of persons  =
volume of the hall
volume of air required by 1 person
=
20×15×4.5
5
= 20 ×3 ×4. 5 = 270
? 270 persons can be accommodated.
Question:121
A river 1.5 m deep and 30 m wide is flowing at the rate of 3 km per hour. The volume of water that runs into the sea per minute is
a 2000 m
3
b 2250 m
3
c 2500 m
3
d 2750 m
3
Solution:
b 2250 m
3
Length of the river = 1.5 m
Breadth of the river = 30 m
Depth of the river = 3 km = 3000 m
Now, volume of water that runs into the sea = 1. 5 ×30 ×3000 m
3
= 135000 m
3
? Volume of water that runs into the sea per minute =
135000
60
= 2250 m
3
Question:122
The lateral surface area of a cube is 256 m
2
. The volume of the cube is
a 64 m
3
b 216 m
3
c 256 m
3
d 512 m
3
Solution:
d 512  m
3
Suppose that a m be the edge of the cube.
We have:
4a
2
= 256 ? a
2
=
256
4
= 64 ? a = 8 m
? Volume of the cube = a
3
m
3
= 8
3
m
3
= 512  m
3

Question:123
The total surface area of a cube is 96 cm
2
. The volume of the cube is
a 8 cm
3
b 27 cm
3
c 64 cm
3
d 512 cm
3
Solution:
c  64 cm
3
Let a cm be the edge of the cube.
We have:
6a
2
= 96 ? a
2
= 16 ? a = 4 cm
? Volume of the cube = a
3
cm
3
= 4
3
cm
3
= 64 cm
3

Question:124
The volume of a cube is 512 cm
3
. Its total surface area is
a 256 cm
2
b 384 cm
2
c 512 cm
2
d 64 cm
2
Solution:
b 384 cm
2
Suppose that a cm is the edge of the cube.
We have:
a
3
= 512 ? a =
3
v 512 = 8 cm
? Total surface area of cube = 6a
2
cm
2
= 6 ×8 ×8 cm
2
= 384 cm
2
Question:125
The length of the longest rod that can fit in a cubical vessel of side 10 cm, is
a 10 cm
b 20 cm
c 10v 2 cm
d 10v 3 cm
Solution:
d  10v 3 cm
Length of the longest rod = body diagonal of the vessel
= v 3a = v 3 ×10 = 10v 3 cm
Question:126
If the length of diagonal of a cube is 8v 3 cm, then its surface area is
a 192 cm
2
b 384 cm
2
c 512 cm
2
d 768 cm
2
Solution:
b  384 cm
2
We have:
v 3a = 8v 3 ? a = 8 cm
? Surface area of the cube = 6a
2
= 6 ×8 ×8 = 384 cm
2
Question:127
If each edge of a cube is increased by 50%, then the percentage increase in its surface area is
10 cm
a 50%
b 75%
c 100%
d 12%
Solution:
Let a be the edge of the cube.
Then the surface area is 6a
2
= S say
Now, increased edge = a +
50
100
a =
150
100
a =
3
2
a
Then, new surface area = 6
3
2
a
2
= 6 ×
9
4
a
2
=
9
4
SIncrease in surface area =
9
4
S - S  =
5
4
S ? Percentage increase in surface area =
5
4
S
S
=
5
4
×100 % = 125 %
Question:128
Three cubes of metal with edges 3 cm, 4 cm and 5 cm respectively are melted to form a single cube. The lateral surface area of the new cube formed is
a 72 cm
2
b 144 cm
2
c 128 cm
2
d 256 cm
2
Solution:
b  144  cm
2
Volume of the new cube formed = total volume of the three cubes
Suppose that a cm is the edge of the new cube, then
a
3
= 3
3
+4
3
+5
3
= 27 +64 +125      = 216     = 6 cm
? Lateral surface area of the new cube = 4a
2
= 4 ×6 ×6 = 144 cm
2
Question:129
In a shower, 5 cm of rain falls. What is the volume of water that falls on 2 hectares of ground?
a 500 m
3
b 750 m
3
c 800 m
3
d 1000 m
3
Solution:
d 1000 m
3

Area of the land = 2 sq hec = 2000 sq mAmount of rainfall = 5 cm = 0. 05 m ? Volume of the water = area of the land ×amount of rainfall                              = 2000 ×0. 05
Question:130
Two cubes have their volumes in the ratio 1 : 27. The ratio of their surface areas is
a 1 : 3
b 1 : 8
c 1 : 9
d 1 : 18
Solution:
c 1 : 9
Suppose that the edges of the cubes are a and b.
We have:
a
3
b
3
=
1
27
?
a
b
3
=
1
27
?
a
b
=
1
3
? Ratio of the surface areas =
6a
2
6b
2
=
a
b
2
=
1
3
2
=
1
9
Question:131
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
Solution:
d becomes 8 times
Suppose that the side of the cube is a.
When it is doubled, it becomes 2a.
New volume of the cube = (2a)
3
= 8a
3

Hence, the volume becomes 8 times the original volume.
Question:132
The diameter of the base of a cylinder is 6 cm and its height is 14 cm. The volume of the cylinder is
a 198 cm
3
b 396 cm
3
c 495 cm
3
d 297 cm
3
Solution:
b  396 cm
3
Volume of the cylinder = pr
2
h                                      =
22
7
×3
2
×14                                      = 22 ×9 ×2                                       = 396 cm
3
( )
( )
( )
( )
( ) ( )
Question:133
If the diameter of a cylinder is 28 cm and its height is 20 cm, then its curved surface area is
a 880 cm
2
b 1760 cm
2
c 3520 cm
2
d 2640 cm
2
Solution:
b  1760 cm
2
Curved surface area of the cylinder = 2 prh                                                             = 2 ×
22
7
×14 ×20                                                            = 44 ×40                                                             = 1760
Question:134
If the curved surface area of a cylinder is 1760 cm
2
and its base radius is 14 cm, then its height is
a 10 cm
b 15 cm
c 20 cm
d 40 cm
Solution:
c 20 cm
Curved surface area = 1760 cm
2
Suppose that h cm is the height of the cylinder.
Then we have:
2 prh = 1760 ? 2 ×
22
7
×14 ×h = 1760 ? h =
1760×7
44×14
= 20 cm
Question:135
The height of a cylinder is 14 cm and its curved surface area is 264 cm
2
. The volume of the cylinder is
a 308 cm
3
b 396 cm
3
c 1232 cm
3
d 1848 cm
3
Solution:
b  396 cm
3
Curved surface area = 264 cm
2
.
Let r cm be the radius of the cylinder.
Then we have:
2 prh = 264 ? 2 ×
22
7
×r×14 = 264 ? r =
264×7
44×14
= 3 cm
? Volume of the cylinder =
22
7
×3
2
×14                                = 22 ×9 ×2                               = 396 cm
3
Question:136
The curved surface area of a cylindrical pillar is 264 m
2
and its volume is 924 m
3
. The height of the pillar is
a 4 m
b 5 m
c 6 m
d 7 m
Solution:
c  6 m
Curved surface area = 264 m
2

Volume = 924 m
3
Let r m be the radius and h m be the height of the cylinder.
Then we have:
2 prh = 264  and  pr
2
h = 924 ? rh =
264
2 p
? h =
264
2r× p
Now, pr
2
h = p ×r
2
×
264
2r× p
= 924 ? r =
924×2
264
? r = 7 m ? h =
264×7
2×7×22
= 6 m
Question:137
The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their curved surface area is
a 2 : 5
b 8 : 7
c 10 : 9
d 16 : 9
Solution:
c  10 : 9
Suppose that the radii of the cylinders are 2r and 3r and their respective heights are 5h and 3h.
Then, ratio of the curved surface areas =
2 p (2r)(5h)
2 p (3r)(3h)
=10 : 9
Question:138
The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their volumes is
a 27 : 20
b 20 : 27
c 4 : 9
d 9 : 4
Solution:
b  20 : 27
Suppose that the radii of the cylinders are 2r and 3r and their respective heights are 5h and 3h..
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## Mathematics (Maths) Class 9

46 videos|323 docs|108 tests

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