Let the rate of each filling pipes be ‘x lts/hr’ similarly, the rate of each draining pipes be ‘y lts/hr’. As per the first condition,

Capacity of tank = (6x-5y)×6. (i)

Similarly, from the second condition, Capacity of tank = (5x – 6y)×60 (ii)

On equating (i) and (ii), we get (6x – 5y) × 6 = (5x – 6y)×60

or, 6x – 5y = 50x – 60y or, 44x = 55y

or, 4x = 5y or, x = 1.25y

Therefore, the capacity of the tank = (6x – 5y) × 6 = (7.5y – 5y) × 6 = 15y lts Effective rate of 2 filling pipes and 1 draining pipe = (2x – y) = (2.5y – y) = 1.5y Hence, the required time = 15y/1.5y=10 hours.