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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.
Q 1 - A divider 24m long , 8 m high and 60cm thick is comprised of blocks, every measuring 24cm * 12cm *8cm , it being given that 10% of the divider comprises of mortar. What number of blocks will be required?
Answer : B
Explanation
Volume of wall = (24*8*60/100)m3 =576/5m3 Volume of bricks = (90% of 576/5) m3= (90/100*576/5) m3= (144*18/25) m3 Volume if 1 bricks = (24/100*12/100*8/100) m3 Number of bricks = [(144*18/25)*100/24*100/12*100/8) = 45000
Q 2 - A stream 2m profound and 45m wide is running at the rate of 3 km/hr. The measure of water that keeps running into the ocean every moment is:
Answer : A
Explanation
Speed per minute = (3*1000)/60 m= 50 m Volume of water running per minute= (45*2*50) m3=4500 m3
Q 3 - The whole of the length, broadness and profundity of a cuboid is 19 cm and its askew is 5√5cm. Its surface region is:
Answer : B
Explanation
(L+b+h) =19 and √ (L2+b2+h2) = 5√5 ⇒ (L+b+h) 2= (19) 2 and (L2+b2+h2) = 125 ⇒ (L2+b2+h2) +2(Lb+bh+Lh) = 361 and (L2+b2+h2) = 125 ⇒125+2(Lb+bh+Lh) =361 ⇒2 (Lb+bh+Lh) = (361-125) =236 ∴ Surface area = 236cm2
Q 4 - The result of the regions of three neighboring countenances of a rectangular box is equivalent to:
A - The volume of the container
B - twice of the volume of the crate
Answer : C
Explanation
Product of areas of 3 adjacent faces = (Lb*bh*Lh) =(L2*b2*h2) = (L*b*h) 2= V2
Q 5 - The volume of a solid shape is 512 cm3. Its aggregate surface zone is:
Answer : C
Explanation
a3=512 =83⇒a= 8 Surface area = 6a2= (6*8*8) cm2= 384cm2
Q 6 - If every edge of a solid shape is expanded by half, the rate increment in surface territory is:
Answer : D
Explanation
Let, Original edge = a, then surface area= 6a2 New edge = 150% of a = (150/100*a) = 3a/2 New surface area = 6*(3a/2) 2= 6*9a2/4= 27a2/2 Increase in surface area = (27a2/2-6a2) = 15a2/2 Increase %= (15a2/2*1/6a2*100)%= 125%
Q 7 - The measurement of the base of a tube shaped drum is 35dm and its tallness is 24 dm. It is brimming with lamp oil. What number of tins each of size 25cm *22cm* 35 cm can be loaded with lamp fuel from the drum?
Answer : D
Explanation
r= 35/2 dm=(35/2*10)cm= 175 cm , h=24 dm = 240cm Volume of drum = (22/7*175*175*240) cm3 =(22*25*175*240) cm3 Volume of a tin = (25*22*35) cm3 Number of tin = (22*25*175*240)/ (25*22*35) = 1200
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?
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 - The bended surface region and the aggregate surface zone of a barrel are in the proportion 1:2. In the event that the aggregate surface territory of the barrel is 616 cm2, then its volume:
Answer : A
Explanation
2πrh/2πr (h+r) =1/2⇒ h/(h+r)=1/2 ⇒2h= h+r⇒h=r 2πr (h+r) = 616 ⇒2πr*2r= 616 ⇒πr2= 154 ⇒22/7*r2=154 ⇒ r2 = (154*7/22) =49 ⇒ r = 7cm and h= 7cm ∴ Volume of cylinder = πr2h = (22/7*7*7*7) cm3= 1078 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:
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
