Quantitative Aptitude
VOLUME AND SURFACE AREA MCQs
Total Questions : 820
| Page 10 of 82 pages
Answer: Option B. -> 60 kg
 - Volume of water displaced = (3 x 2 x 0.01) m3
= 0.06 m3.
Mass of man = Volume of water displaced x Density of water
= (0.06 x 1000) kg
= 60 kg.
 - Volume of water displaced = (3 x 2 x 0.01) m3
= 0.06 m3.
Mass of man = Volume of water displaced x Density of water
= (0.06 x 1000) kg
= 60 kg.
Answer: Option D. -> 25 cm
 - Total volume of water displaced = (4 x 50) m3 = 200 m3.
Rise in water level = 200 m 0.25 m = 25 cm. 40 x 20
 - Total volume of water displaced = (4 x 50) m3 = 200 m3.
Rise in water level = 200 m 0.25 m = 25 cm. 40 x 20
Answer: Option A. -> 49 sq mt
 - Area of the wet surface = [2(lb + bh + lh) - lb]
= 2(bh + lh) + lb
= [2 (4 x 1.25 + 6 x 1.25) + 6 x 4] m2
= 49 m2.
 - Area of the wet surface = [2(lb + bh + lh) - lb]
= 2(bh + lh) + lb
= [2 (4 x 1.25 + 6 x 1.25) + 6 x 4] m2
= 49 m2.
Answer: Option B. -> 5120
 - Clearly, l = (48 - 16)m = 32 m,
b = (36 -16)m = 20 m,
h = 8 m.
Volume of the box = (32 x 20 x 8) m3 = 5120 m3
 - Clearly, l = (48 - 16)m = 32 m,
b = (36 -16)m = 20 m,
h = 8 m.
Volume of the box = (32 x 20 x 8) m3 = 5120 m3
Answer: Option B. -> 1 dm
 - Let the thickness of the bottom be x cm.
Then, [(330 - 10) x (260 - 10) x (110 - x)] = 8000 x 1000
320 x 250 x (110 - x) = 8000 x 1000
(110 - x) = 8000 x 1000 = 100 320 x 250
 - Let the thickness of the bottom be x cm.
Then, [(330 - 10) x (260 - 10) x (110 - x)] = 8000 x 1000
320 x 250 x (110 - x) = 8000 x 1000
(110 - x) = 8000 x 1000 = 100 320 x 250
Answer: Option B. -> 498.35 sq m
 - h = 14 cm, r = 7 cm.
So, l = (7)2 + (14)2 = 245 = 75 cm.
Total surface area =22/7 (rl + r2) = 22 x 7 x 75 + 22 x 7 x 7 cm2 7 7
= [154(5 + 1)] cm2
= (154 x 3.236) cm2 = 498.35 cm2.
 - h = 14 cm, r = 7 cm.
So, l = (7)2 + (14)2 = 245 = 75 cm.
Total surface area =22/7 (rl + r2) = 22 x 7 x 75 + 22 x 7 x 7 cm2 7 7
= [154(5 + 1)] cm2
= (154 x 3.236) cm2 = 498.35 cm2.
Answer: Option C. -> 25 : 18
 - Volume of the large cube = (33 + 43 + 53) = 216 cm3.
Let the edge of the large cube be a.
So, a3 = 216 a = 6 cm.
Required ratio = 6 x (32 + 42 + 52) = 50 = 25 : 18. 6 x 62
 - Volume of the large cube = (33 + 43 + 53) = 216 cm3.
Let the edge of the large cube be a.
So, a3 = 216 a = 6 cm.
Required ratio = 6 x (32 + 42 + 52) = 50 = 25 : 18. 6 x 62
Answer: Option C. -> 6400
 - Number of bricks = Volume of the wall = 800 x 600 x 22.5 = 6400. Volume of 1 brick 25 x 11.25 x 6
 - Number of bricks = Volume of the wall = 800 x 600 x 22.5 = 6400. Volume of 1 brick 25 x 11.25 x 6
Answer: Option D. -> $$47\frac{2}{3}{\text{ c}}{{\text{m}}^3}$$
Answer: Option C. -> $$\frac{14}{3}$$ cm