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Aptitude - Volume and Surface Area

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You can easily solve all kind of Aptitude questions based on Volume and Surface Area by practicing the objective type exercises given below, also get shortcut methods to solve Aptitude Volume and Surface Area problems.

Home » Aptitude » Volume and Surface Area » General Questions

Exercise :: Volume and Surface Area - General Questions

1. 

A right triangle with sides 3 cm, 4 cm and 5 cm is rotated the side of 3 cm to form a cone. The volume of the cone so formed is:
A. 12 cm3
B. 15 cm3
C. 16 cm3
D. 20 cm3

Answer: Option A

Explanation:

Clearly, we have r = 3 cm and h = 4 cm.

Volume = 1 r2h = 1 x x 32 x 4 cm3 = 12 cm3.
3 3

Video Explanation: https://youtu.be/Njsid2mVwUY

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2. 

In a shower, 5 cm of rain falls. The volume of water that falls on 1.5 hectares of ground is:
A. 75 cu. m
B. 750 cu. m
C. 7500 cu. m
D. 75000 cu. m

Answer: Option B

Explanation:

1 hectare = 10,000 m2

So, Area = (1.5 x 10000) m2 = 15000 m2.

Depth = 5 m = 1 m.
100 20

Volume = (Area x Depth) = 15000 x 1 m3 = 750 m3.
20

Video Explanation: https://youtu.be/dDbVuBzYop8

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3. 

A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume of the hall is:
A. 720
B. 900
C. 1200
D. 1800

Answer: Option C

Explanation:

2(15 + 12) x h = 2(15 x 12)

h = 180 m = 20 m.
27 3

Volume = 15 x 12 x 20 m3 = 1200 m3.
3

Video Explanation: https://youtu.be/V8EQ1YIaH74

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4. 

66 cubic centimetres of silver is drawn into a wire 1 mm in diameter. The length of the wire in metres will be:

A. 84
B. 90
C. 168
D. 336

Answer: Option A

Explanation:

Let the length of the wire be h.

Radius = 1 mm = 1 cm. Then,
2 20

22 x 1 x 1 x h = 66.
7 20 20

h = 66 x 20 x 20 x 7 = 8400 cm = 84 m.
22

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5. 

A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weighs 8 g/cm3, then the weight of the pipe is:

A. 3.6 kg
B. 3.696 kg
C. 36 kg
D. 36.9 kg

Answer: Option B

Explanation:

External radius = 4 cm,

Internal radius = 3 cm.

Volume of iron
= 22 x [(4)2 - (3)2] x 21 cm3
7
= 22 x 7 x 1 x 21 cm3
7
= 462 cm3.

Weight of iron = (462 x 8) gm = 3696 gm = 3.696 kg.

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