Volume Calculation - Online Quiz



Following quiz provides Multiple Choice Questions (MCQs) related to Volume Calculation. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.

Questions and Answers

Q 1 - The region of the base of a rectangular tank is 6500 cm2 and the volume of the water contained in it in 2.6 cubic meter. The profundity of the water in the tank is:

A - 2.5 m

B - 3 m

C - 5.5 m

D - 4 m

Answer : D

Explanation

L*b= 6500cm2 , L*b*d=2.6m3=(2.6*100*100*100) cm3
∴ d = (2.6*100*100*100)/6500 cm = (2.6*100*100*100)/6500*100 = 4m
∴ Depth = 4m

Q 2 - Half cubic meter of gold sheet is stretched out by pounding in order to spread a region of 1 hectare. The thickness of the sheet is:

A - 0.5 cm

B - 0.05 cm

C - 0.005 cm

D - 0.0005 cm

Answer : C

Explanation

Area = 1 hectare= 10000m2, Volume = 0.5 m3
Thickness =Volume/Area   = (0.5/10000)m =(0.5*100/10000)cm=0.005cm

Q 3 - A rectangular tank measuring (5m*4.5 m*2.1 m) is dove in the focal point of field measuring 13.5m*2.5 m. The earth burrow is spread equitably over the remaining part of the field. What amount is the level of the field raised?

A - 4 m

B - 4.1 m

C - 4.2 m

D - 4.3 m

Answer : C

Explanation

Volume of the earth dugout = (5*9/2*21/10) m3= 189/4 m3
Area of the remaining field = [(27/2*5/2) - (5*9/2)] m2
= (135/4-45/2) m2= 45/4m2
Let the level of the field be raised h meter. Then,
45/4*h = 189/4 ⇒h= 189/45 m =21/5 m =4.2 m

Q 4 - A room is 10 m long, 8m wide and 3.3 m high. What number of men can be obliged in this room if every man requires 3m3 of space?

A - 99

B - 88

C - 77

D - 75

Answer : B

Explanation

Volume of the room = (10*8*3.3) m3= 264m3
Volume required by 1man = 3m3
Required number of men = 264/3 = 88

Q 5 - If every side of a solid shape is multiplied, then its volume:

A - is multiplied

B - get to be 4 times

C - gets to be 6 times

D - Gets to be 8 times

Answer : D

Explanation

Let the edge of the original cube be x. Then its volume = x3
New edge = 2x, new volume (2x) 3=8x3
So, it?s become 8 times.

Q 6 - The volume of a cuboid is twice that of a solid shape. On the off chance that the measurement of the cuboid is (9cm *8 cm* 6cm), the aggregate surface region of the block is:

A - 72 cm2

B - 216 cm2

C - 108 cm2

D - 432 cm2

Answer : B

Explanation

2*volume of cube= volume of cuboid= (9*8*6) cm3
Volume of cube = (1/2 *9*8*6) cm3= 216cm3
∴ a3= 216 = (6) 3 ⇒a =6 cm
Total surface area = 6a2 = (6*6*6) cm2= 216cm2

Q 7 - The tallness of the barrel is 14 cm and its bended surface range is 264 cm2. The volume of the barrel is:

A - 308 cm3

B - 396 cm3

C - 1232cm3

D - 1848 cm3

Answer : B

Explanation

Given h= 14cm and 2πrh= 264
∴ 2*22/7*r *14= 264 ⇒88r= 264 ⇒r =3
Volume = πr2h = (22/7*3*3*14) cm3= 396 cm3

Q 8 - A funnel of 2 inch measurement fills the water tank in 60 minutes. In the event that the measurement of the funnel is 4inch, in what the reality of the situation will become obvious eventually pipe fills the same tank?

A - 30 min.

B - 45 min.

C - 15 min.

D - 10 min.

Answer : C

Explanation

r = 1 inch,  volume = πr2h= π*1*1*h= πh
r =2 inch volume = π (2)2*h= 4πh
4 time water pass at the same time.
∴Rates are 1:4 so, time taken =4:1
∴ 2nd pipe will take = (1/4*60) = 15 min

Q 9 - A empty greenery enclosure roller 63cm wide with a circumference of 440 cm is made of iron 4 cm thick. The volume of the iron utilized is:

A - 56372 cm3

B - 58752 cm3

C - 54982 cm3

D - 57636 cm3

Answer : B

Explanation

2πr= 440 ⇒2*22/7*r= 440 ⇒r= (440*7/44) = 70 cm
Outer radius = 70cm, inner radius = (70-4) = 66cm
Volume of iron = π [(70)2-(66)2*63cm3= (22/7*136*4*63)
= 58752 cm3

Q 10 - The volume of the circle is 2145(11/21) cm3. Its sweep is:

A - 7 cm

B - 8 cm

C - 9 cm

D - none of these

Answer : B

Explanation

4/3πr3=45056/21 ⇒4/3*22/7* r3 = 45056/21
⇒r3= (45056/21 *21/88) = 512 = (8) 3 ⇒r =8
∴Radius of the sphere =8cm
aptitude_volume_calculation.htm
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