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 - 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 - A spread wooden box has the inward measures as 115 cm, 75 cm and 35 cm and the thickness of wood is 2.5 cm. The volume of the wood is:

A - 81000 cu. cm

B - 81775 cu.cm

C - 82125 cu.cm

D - none of these

Answer : C

Explanation

Inner volume= (115*75*35) cm3= 301875 cm3
Outer volume = (115+5)* (75+5)*(35+5) cm3
=(120*80*40) cm3= 384000 cm3
Volume of wood = (384000-301875) cm3= 82125 cm3

Q 3 - A corridor is 15m long and 12 m expansive. On the off chance that the aggregate of the zones of the floor and the roof is equivalent to the whole of the ranges of the 4 dividers, the volume of the corridor is:

A - 720 m3

B - 900 m3

C - 1200 m3

D - 1800 m3

Answer : C

Explanation

(Lb+Lb) = 2(L+b)*h= 2*Lb = 2(L+b)*h ⇒Lb= (L+b)*h
⇒ (15+12)*h = (15*12) ⇒h = (15*12)/27m = 20/3 m
Volume = (L*b*h) = (15*12*20/3) m3= 1200 m3

Q 4 - If the zone of the three adjoining appearances of a cuboidal box is 120 cm2, 72 cm2 and 60 cm2 individually, then the volume of the crate is:

A - 720 cm3

B - 864 cm3

C - 7200 cm3

D - 5184 cm3

Answer : A

Explanation

Lb= 120, bh= 72 and Lh = 60
⇒ (Lb*bh*Lh) = 120*72*60⇒ (Lbh) 2= (120*72*60)
⇒ Lbh =√ (12*10*12*6*10*6) = (12*10*6) = 720
∴ Volume = 720 cm3

Q 5 - The aggregate surface zone of a solid shape of side 27 cm is:

A - 2916 cm2

B - 729 cm2

C - 4374 cm2

D - 19683 cm2

Answer : C

Explanation

Surface area =6a2= (6*27*27) cm2= 4374cm2

Q 6 - The aggregate surface zone of 3D square (cube) is 1734 cm2 .Its volume is:

A - 2197 cm3

B - 2744 cm3

C - 4096 cm3

D - 4913 cm3

Answer : D

Explanation

6a2=1734 ⇒a2=289= (17) 2 ⇒a= 17
Volume = a3= (17*17*17) cm3= 4913cm3

Q 7 - If the span of a barrel is diminished by half and the stature is expanded by half to frame another chamber, then the volume will be diminished by:

A - 1:4

B - 4:1

C - 2:5

D - none of these

Answer : A

Explanation

Initial volume = πr2h
New radius = r/2, new volume = [π*(r/2)2*4] = 1/4 πr2h
Required ratio = 1/4 πr2h: πr2h = 1/4:1 = 1:4

Q 8 - If the tallness of a barrel is expanded by 15% and the span of the base is diminished by 10%, then by what percent will its bended surface region change?

A - 3.5 % diminished

B - 3.5% expansion

C - 5% diminish

D - 5% expansion

Answer : B

Explanation

Let the original radius =r and height = h
Then curved surface area = 2πrh
New height = 115% of h = (115/100*h) = 23h/20
New radius = 90% of r = (90/100*r) = 9r/10
New curved surface area = (2π*9r/10*23h/20) = 207πrh/100
Increase = (207πrh/100-2 πrh) = 7 πrh/100
Increase %= (7 πrh/100*1/2 πrh*100) %= 3.5%

Q 9 - The range and the base and stature of a barrel are in the proportion 2:3 and its volume is 12936 cm3. The entire surface territory of the barrel is:

A - 3080 cm2

B - 38808 cm2

C - 25872 cm2

D - 2587.2 cm2

Answer : A

Explanation

Let radius =2x cm and height= 3x cm
Then volume = πr2h [22/7(2x) 2*3x] cm3= (264/7) x3 cm3
(264/7) x3= 12936 ⇒x3= (12936*7/264) =343= (7)3⇒x= 7
∴ Radius =14cm, height = 21 cm
Total surface area = 2πr (h+r) = [2*22/7*14(21+14)] cm2= 3080 cm2

Q 10 - If the sweep of circle is 6cm, then its volume is:

A - (288π) cm3

B - (388π) cm3

C - (684π) cm3

D - (864π) cm3

Answer : A

Explanation

Volume = {4/3π*(6)3} cm3 = (4/3π*6*6*6) cm3= (288π) cm3


aptitude_volume_calculation.htm

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