Question

Two pipes can separately fill a tank in 20 hours and 30 hours respectively. Both of them are opened and when the tank is \(\frac13\) full, a leak was developed through which \(\frac13\) of the water supplied by the pipes drains out. The total time taken to fill the tank (in hours) is

• A. \(14\)
• B. \(15\)
• C. \(16\)
• D. \(18\)

Hint:

When leak starts after some work is done, always split the work into “before leak” and “after leak”.

Answer 1

The Correct Option is C

Concept: First find the combined filling rate. Then divide the work into two parts:
• Before leak
• After leak

Step 1: Find combined filling rate.
First pipe: \[ \frac1{20} \] Second pipe: \[ \frac1{30} \] Combined rate: \[ \frac1{20}+\frac1{30} \] LCM \(=60\): \[ =\frac3{60}+\frac2{60} \] \[ =\frac5{60} \] \[ =\frac1{12} \] So they fill the tank in: \[ 12 \text{ hours} \]

Step 2: Time to fill first \(\frac13\).
At rate: \[ \frac1{12} \] Time: \[ \frac{\frac13}{\frac1{12}} \] \[ =\frac13\times12 \] \[ =4 \text{ hours} \]

Step 3: Rate after leak develops.
Leak drains: \[ \frac13 \] of supplied water. So effective rate becomes: \[ \frac23\times \frac1{12} \] \[ =\frac1{18} \]

Step 4: Fill remaining \(\frac23\).
Remaining work: \[ \frac23 \] Time: \[ \frac{\frac23}{\frac1{18}} \] \[ =\frac23\times18 \] \[ =12 \text{ hours} \]

Step 5: Find total time.
\[ 4+12=16 \] Thus, the total time taken is: \[ \boxed{16 \text{ hours}} \]