Question

A cistern is fitted with many inlets and outlet taps. All inlet taps fill the cistern at the same rate and outlet taps empty the cistern at the same rate. The empty cistern gets filled in 10 hours when 9 inlet and 5 outlet taps are opened. It takes 30 hours when 6 inlets and 5 outlets are opened. How much time does it take for the empty cistern to completely get filled if 2 inlets and 2 outlet taps are opened?
1. 18.75 days
2. 9.375 days
3. 9 days
4. 10.5 days
5.

Answer 1

Correct Answer - Option 2 : 9.375 days

Calculation:

Let x be the efficiency of the inlet pipe and -y be the efficiency of the outlet pipe.

Case 1: 

When 9 inlet taps and 5 outlet taps are opened then it takes 10 hours to fill the cistern

∴ The capacity of the tank is 10(9x - 5y)

Case 2: 

When 6 inlet taps and 5 outlet taps are opened then it takes 30 hours to fill the cistern

∴ The capacity of the tank is 30(6x - 5y)

The capacity of the tank remains the same.

∴ 10(9x - 5y) = 30(6x - 5y)

⇒ 90x - 50y = 180x - 150y

⇒ 100y = 90x

⇒ y = 0.9 x

To find the capacity of the cistern, substitute y in any one of the equations.

∴ The capacity of the tank is 10(9x - 5y) 

⇒ 90x - 45x

⇒ 45x

∴ The capacity of the cistern is 45x

2 inlets and 2 outlets are given by 2x - 2y

⇒ 2x - 1.8x = 0.2x

Time taken to fill the tank is 45x/0.2x = 225 hours which is 9.375 days.

If Pipe A can fill the tank in P hours and Pipe B can empty the tank in q hours then work done per hour will be

⇒ 1/p - 1/q