## IBM Sample Problems On Volume And Surface Area

Below are three problems based on Volume of the objects dealing with usual parameters.

Question 1

The half of the volume of water in a rectangular base tank is 2.52 cubic meters. Find the depth of the tank if the area of the base of the tank is 5600 square cm.

a)9m b)3m c)5m d)6.5m

Solution :

Given that the half volume of the tank is 2.52 cubic meters.
Let us convert the volume into cubic cm for easy calculation.
Full volume of the tank = 2 x 2.52 x 100 x 100 x 100 cubic cm.

Given that the area of the rectangular base(lb) = 5600sq cm
We know that the volume of the rectangular base tank = V = lbh
Then the depth(height) of the tank = h = V/lb
Here, h = 2 x 2.52 x 100 x 100 x 100 / 5600
h = 5040000/5600
h = 900 cm or 9 m.
Hence the depth of the tank is 9 m.

Question 2

There are two tanks A and B of length 142 cm and width 240cm. If the depth of A and B is 91cm and 46cm respectively then how much litres of water will be there in A more than B?

a)10 ltrs b)1.5 ltrs c)20 ltrs d)2.5 ltrs

Answer : b)1.5 ltrs

Solution :

Given that the length and width of A and B are same.
l = 142 cm and b = 240cm and h1 = 91cm, h2 = 46cm

Volume of the tank = lbh cubic unit
Volume of the tank A = 142 x 240 x 91 cubic cm
Since the required amount of water is given in litres, we should convert the unit cm into meter.
i.e., Volume of tank A = 142 x 240 x 91 / 1000000 cubic meters
And the volume of the tank b = 142 x 240 x 46 / 1000000 cubic meters.

The excess of water contained in tank A = (142 x 240 x 91 / 1000000 - 142 x 240 x 46 / 1000000)litres.

= 142 x 240 / 1000000 x(91 - 46) = 142 x 240 x 45 / 1000000 = 1.53 litres

= 1.5 litres (approximately)

Question 3

There are two tanks such that the length of the square base tank is 22m and the radius of the circular base tank is 14m.If the capacity of two tanks are equal then what will be the ratio of their respective heights?

a)10:9 b)11:12 c)13:14 d)14:11

Solution :

Length of the square base tank is 22m.
We know the length and width of the square are equal.
Volume of the square base tank = lbh1 = 22 x 22 x h1

The radius of the circular base tank is 14m.
Volume of the circular base tank = area of base x height
pi x r x r x h2 = (22/7) x 14 x 14 = 28 x 22 x h2
The capacity(volume) of two tanks are equal.
i.e., 22 x 22 x h1 = 28 x 22 x h2
h1/h2 = 28/22 = 14/11
Hence the required ratio = 14:11

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