Problems on Volumes and Surface Area
Q11. The area of the base of the rectangular tank is 6500 cm2 and the volume of the water contained in it is 32.5 cubic meters. The depth of the water in the tank is
(a) 42m
(b) 50m
(c) 57m
(d) 62m
Answer: (b) 50m
Here l X b = 6500 cm2
v = 32.5 m3 = 32.5 X 100 X 100 X 100 cm3
We have v = l X b X h
⇒ 32.5 X 100 X 100 X 100 = 6500 X h
⇒ h = (32.5 X 100 X 100 X 100)/6500 cm = 5000 cm = 50m
Q12. A covered wooden box has the inner measures as 110cm, 70cm and 30cm and the thickness of wood is 2.5cm. The volume of the wood is
(a) 54982 cm3
(b) 62547 cm3
(c) 69832 cm3
(d) 70875 cm3
Answer: (d) 70875 cm3
Inner volume of box = 110 X 70 X 30 cm3 = 231000 cm3
Outer volume of box = (110 + 5) X (70 + 5) X (30 + 5) cm3
= 115 X 75 X 35 cm3
= 301875 cm3
∴ Volume of the wood = (301875 - 231000) cm3 = 70875 cm3
Q13. A hall is 15m long and 10m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of the 4 walls, the volume of the hall is
(a) 815 m3
(b) 875 m3
(c) 900 m3
(d) 950 m3
Answer: (c) 900 m3
here l = 15m b = 10m
Area of floor + Area of ceiling = lXb + lXb = 15X10 + 15X10 = 300m2
Sum of the areas of 4 walls = 2 lXh + 2 bXh = 2 X 15 X h + 2 X 10 X h = 30h + 20h = 50h
A/Q 50h = 300
⇒ h = 300/50 = 6m
∴ Volume of the ball = l X b X h = 15 X 10 X 6 m3 = 900 m3
Q14. The sum of the length, breadth and depth of a cuboid is 20cm and its diagonal is 5√ 5 cm. Its surface area is
(a) 230 cm2
(b) 275 cm2
(c) 333 cm2
(d) 350 cm2
Answer: (b) 275 cm2
Here l+b+h = 20
√ l2+b2+h2 = 5√ 5
Now (l+b+h)2 = 202 = 400
(√l2+b2+h2)2 = (5√ 5)2
⇒ l2+b2+h2 = 125
We have (l+b+h)2 = l2+b2+h2+2 (lb+bh+hl)
⇒ 400 = 125 + 2 (lb+bh+hl)
⇒ 2 (lb+bh+hl) = 400 - 125
⇒ 2 (lb+bh+hl) = 275 cm2
Q15. The size of a wooden block is (12cm X 10cm X 8cm). How many such blocks will be required to construct a solid wooden cube of minimum size ?
(a) 1300
(b) 1500
(c) 1600
(d) 1800
Answer: (d) 1800
L.C.M of 12, 10 and 8 is 120
Volume of a smallest cube = 120 X 120 X 120 cm3
Volume of one block = 12 X 10 X 8 cm3 = 960 cm3
∴ Required number of block = (120X120X120)/960 = 1800
Q16. The height of a cylinder is 12cm and its diameter is 10cm. The volume of the cylinder is
(a) 733.50 cm3
(b) 856.21 cm3
(c) 942.85 cm3
(d) 993.88 cm3
Answer: (c) 942.85 cm3
Volume = π r2h
= (22/7) X (10/2)2 X 12
= (22/7) X 5 X 5 X 12 = 942.85 cm3
Q17. The height of a cylinder is 14cm and its curved surface area is 264cm2. The volume of the cylinder is
(a) 396 cm3
(b) 451 cm3
(c) 482 cm3
(d) 527 cm3
Answer: (a) 396 cm3
Here h = 14cm
2π rh = 264
⇒ 2 X (22/7) X r X 14 = 264
⇒ r = (264 X 7)/(2 X 22 X 14) = 3
∴ Volume = π r2h
=(22/7) X 32 X 14 = (22 X 9 X 14)/7 = 396 cm3
Q18. A cylinder has a radius of 7 cm and the area of its curved surface is 176 cm2. The volume of the cylinder is
(a) 577 cm3
(b) 616 cm3
(c) 659 cm3
(d) 721 cm3
Answer: (b) 616 cm3
Here r = 7 cm
2πrh = 176
⇒ 2 X (22/7) X 7 X h = 176
⇒ h = 176/(2X22) = 4 cm
∴ Volume = πr2h = (22/7) X 72 X 4 = 616 cm3
Q19. The ratio of the radius of two cylinders is 2:3 and the ratio of their heights is 3:5. The ratio of their volumes will be
(a) 3:7
(b) 3:17
(c) 4:15
(d) 4:17
Answer: (c) 4:15
Let, the radius be 2r and 3r
And the heights be 3h and 5h
| ∴ Ratio of their volume = | π X (2r)2 X 3h |
| π X (3r)2 X 5h |
| = | 4r2 X 3h |
| 9r2 X 5h |
| = | 4 X 3 |
| 9 X 5 |
= 4/15 = 4:15
Q20. The number of coins, 1.5 cm in diameter and 0.2 cm thick to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm is
(a) 375
(b) 390
(c) 420
(d) 450
Answer: (d) 450
Here for coins r = 1.5/2 cm
h = 0.2 cm
∴ Volume of one coin = πr2h
= (π X (1.5/2) X (1.5/2) X 0.2)cm3 = 9π/80 cm3
For cylinder R = 4.5/2 cm H = 10 cm
Volume = πR2H
= (π X (4.5/2) X (4.5/2) X 10)cm3
= (45 X 9π)/8 cm3
| ∴ Required number of coins = | 45 X 9π | X | 80 |
| 8 | 9π |
= 450
Practice Test Exam