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**