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Worked Examples · Example 23

Q.Two pipes can fill a cistern in 8 and 12 hours respectively. The pipes are opened simultaneously and it takes 12 minutes more to fill the cistern due to leakage. If the cistern is full, what will be the time taken by the leakage to empty it?

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✓ Free question

The two pipes alone fill in 4.84.8 h (=288=288 min); with leakage it takes 300300 min (=5=5 h), so the leak rate is 524−15=1120\tfrac{5}{24}-\tfrac{1}{5}=\tfrac{1}{120} per hour, i.e. 120120 hours to empty.

Combined filling rate of pipes =18+112=\dfrac{1}{8}+\dfrac{1}{12} (cistern/hour). With a leak of rate 1L\dfrac{1}{L},

Effective rate=(18+112)−1L=1time with leak.\text{Effective rate}=\Big(\tfrac{1}{8}+\tfrac{1}{12}\Big)-\tfrac{1}{L}=\frac{1}{\text{time with leak}}.

  1. Rate of both pipes together: 18+112=3+224=524\dfrac{1}{8}+\dfrac{1}{12}=\dfrac{3+2}{24}=\dfrac{5}{24} cistern/hour.
  2. Time to fill without leak =245=\dfrac{24}{5} h =4.8=4.8 h =288=288 minutes.
  3. Time with leak =288+12=300=288+12=300 minutes =5=5 hours, so effective rate =15=\dfrac{1}{5} cistern/hour.
  4. Leak rate =524−15=25−24120=1120=\dfrac{5}{24}-\dfrac{1}{5}=\dfrac{25-24}{120}=\dfrac{1}{120} cistern/hour.
  5. So the leak empties a full cistern in L=120L=120 hours.
  6. Check: 524−1120=25−1120=24120=15\dfrac{5}{24}-\dfrac{1}{120}=\dfrac{25-1}{120}=\dfrac{24}{120}=\dfrac{1}{5}, i.e. 55 h to fill. ✓
✓Final answer

The leakage alone would empty the full cistern in 120 hours\mathbf{120\ \text{hours}}.

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