Correct Answer - Option 1 : 12 hours
Given:
Time taken by Pipe A is 10 hours
Time taken by Pipe B is \(13\frac{1}{3}\)= (40/3) hours
Time taken by all pipe (A, B, C) together is 8 hours
Concept Used:
When two pipes A and B can fill the tank in x, y hours respectively.
There is also an outlet pipe C. If all three pipes opened simultaneously tank will fill in z hours,
then the time taken by Pipe C to empty the full tank is given by
Time taken by Pipe C = (X × Y × Z)/(XZ + YZ - XY)
Calculation:
Let time taken by Pipe A, B and Pipe (A, B, C) together be denoted by X, Y and and Z respectively.
⇒ Time taken by Pipe C = [10 × (40/3) × 8]/[(10 × 8) + (40/3 × 8) - (10 × 40/3)]
⇒ Time taken by pipe C = (3200/3)/(160/3)
⇒ Time taken by Pipe C = 20 hours
Time taken by Pipe C to completely empty the tank is 20 hours.
Then,
Time taken by Pipe C to empty 60% of the tank will be 60% of 20 hours
⇒ 60% of 20 hours
⇒ (60/100) × 20
⇒ 60/5
⇒ 12 hours
∴ The time taken by Pipe C to empty 60% of tank is 12 hours.