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Pipes A and B can fill a tank in 10 hours and \(13\frac{1}{3}\) hours, respectively. Pipe C is an emptying pipe. When all three pipes are opened together, the tanks is filled in 8 hours. Pipe C alone can empty 60% part of the tank in:
1. 12 hours
2. 15 hours
3. 9 hours
4. 10 hours

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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.   

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