Volume Calculation - Online Quiz


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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 - A rectangular marble stone 28 cm wide and 5 cm thick weighs 112 kg. In the event that 1 cubic cm of marble measures 25 gm, then the length of given stone is:

A - 36 cm

B - 37.5 cm

C - 32 cm

D - 26.5 cm

Answer : C

Explanation

Let the required length be x cm.
Then, its volume = (x*28*5*25/1000) kg = 7x/2 kg
∴ 7x/2= 112 ⇒x= 112*2/7= 32 cm
Required length = 32 cm

Q 2 - An open box is comprised of a metal. It is 50cm long, 40cm expansive and 23 m high. The thickness of the metal is 3cm. In the event that 1 cubic cm of the metal weighs 0.5 gm, then what is the heaviness of the void box?

A - 6.832 kg

B - 7.576 kg

C - 7.16 kg

D - 8.04 kg

Answer : D

Explanation

Volume of metal = [(50*40*23)-(44*34*20)]c m3
= 46000-29920=   16080 cm3
Weight of metal = (16080*5/10*1/1000) kg = 8.04 kg
Weight of the empty box = 8.04 kg

Q 3 - What a portion of a trench 48m long, 16.5 m expansive and 4m profound can be filled by the earth got by burrowing a round and hollow passage of distance across 4m what's more, length 56 m?

A - 1/9

B - 2/9

C - 7/9

D - 8/9

Answer : B

Explanation

Volume of the earth dug out as a tunnel = πr2h= (22/7*2*2*56) m3= 704 m3
Volume of the ditch = (48*33/2*4) m3=3168 m3
Required part = 704/3168 = 2/9

Q 4 - A reservoir of limit 8000 liters measures remotely 3.3m by 2.6m by 1.1m and its dividers are 5cm thick. The thickness of the base is:

A - 90 cm

B - 1 dm

C - 1 m

D - 1.1 m

Answer : B

Explanation

Volume of the cistern = 8000ltr. =8000dm3
External length = 33 dm, external breadth =26 dm and external depth = 11 dm
Internal length = (33-5/10*2) dm = 32 dm
Internal breadth = (26-5/10*2) dm = 25 dm
Internal depth = (11-x) dm
∴ 32*25*(11-x) = 8000 ⇒ (11-x) =8000/ (32*25) =10
⇒ x = (11-10) =1dm

Q 5 - The region of the card board expected to make a crate size 25 cm* 15cm *8 cm will be:

A - 390 cm2

B - 1390 cm2

C - 2780 cm 2

D - 1000 cm2

Answer : B

Explanation

Area needed = 2(25*15+15*8+25*8) cm2
=2(375+120+200) cm2= 2(695) cm2 = 1390 cm2

Q 6 - Three solid shapes of iron whose edge are 6 cm, 8cm and 10 cm are dissolved and framed into a solitary 3D shape. The Edge of the new solid shape framed is:

A - 12 cm

B - 14 cm

C - 16 cm

D - 18 cm

Answer : A

Explanation

Volume of new cube= [(6) 3+ (8) 3+ (10) 3] cm3
= (216+512+1000) cm3
= (1728) cm3= (23*63) cm3
Edge of this cube = (2*6) cm = 12 cm

Q 7 - If every side of a 3D square of volume v is multiplied, its volume gets to be ℏv where ℏ is equivalent to:

A - 2

B - 4

C - 8

D - 16

Answer : C

Explanation

Let each side be a, then, V= a3
New volume = (2a) 3= 8a3= 8V
∴ℏV= 8V⇒ℏ= 8

Q 8 - A well must be hole that is to be 22.5 m profound and of breadth 7m. The expense of putting the internal bended surface at rs. 30 for each sq. meter are:

A - Rs. 14650

B - Rs. 14850

C - Rs. 14750

D - Rs. 14950

Answer : B

Explanation

Here r= 7/2 m and h= 45/2 m
Curved surface area = 2πrh=(2*22/7*7/2*45/2)m2  = 495m2
Required cost = (495*30) = 14850 rs.

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 a right roundabout chamber, 14 cm in stature, is equivalent to that of a solid shape whose edge is 11 cm. The range of the base of the barrel is:

A - 5.2 cm

B - 5.5 cm

C - 11 cm

D - 22 cm

Answer : B

Explanation

Let the radius of the base of the cylinder be r cm. Then,
πr2*14 = (11)3
⇒22/7*r2*14= 11*11*11   ⇒r2 =121/4 = (11/2)2 ⇒r =11/2 =5.5 cm


aptitude_volume_calculation.htm

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