Pipes and Cisterns – Two inlets with a bottom leak causing delay: Two pipes can fill a cistern in 14 hours and 16 hours, respectively. Owing to a leak at the bottom, the cistern actually takes 92 minutes longer to fill than it would without the leak. When the cistern is full, in how many hours will the leak alone empty it?

Introduction / Context:
The leak reduces the net filling rate, increasing the total time by a known amount. By comparing the “no-leak” combined rate with the “with-leak” rate, we can isolate the leak’s emptying rate and invert it.

Given Data / Assumptions:

• Pipe rates: 1/14 and 1/16 per h.
• No-leak time = (14*16)/(14+16) = 224/30 = 112/15 h = 7 h 28 min.
• Observed time with leak = 7 h 28 min + 92 min = 9 h.

Concept / Approach:
Let r = 1/14 + 1/16. With leak, net = 1/9. Leak rate l = r − 1/9. The leak’s emptying time is 1/l.

Step-by-Step Solution:
r = 1/14 + 1/16 = (8 + 7)/112 = 15/112 per h.l = r − 1/9 = 15/112 − 1/9 = (135 − 112)/1008 = 23/1008 per h.Leak emptying time = 1 / (23/1008) = 1008/23 h ≈ 43.83 h.

Verification / Alternative check:
Check: 1/14 + 1/16 − 1/(1008/23) = 15/112 − 23/1008 = (135 − 23)/1008 = 112/1008 = 1/9 ⇒ 9 hours with leak.

Why Other Options Are Wrong:
Other choices do not reproduce the net 9-hour fill time when combined with the two inlet rates.

Common Pitfalls:
Adding times instead of rates, or converting 92 minutes incorrectly.

Final Answer:
1008/23 h