Pipes and Cisterns – Inlet with a leak delaying completion: A tank normally fills in 8 hours through a single inlet. Because of a leak at the bottom, it actually takes 10 hours to fill completely. If the tank is full, in how many hours would the leak alone empty it?

Introduction / Context:
An inlet and a leak act against each other: the inlet fills while the leak empties. The observed increase in total time from 8 h to 10 h allows us to deduce the leak’s rate and thus its emptying time.

Given Data / Assumptions:

• Inlet alone time = 8 h ⇒ rate = 1/8 per h.
• With leak, net time = 10 h ⇒ net rate = 1/10 per h.
• Capacity normalized to 1 unit.

Concept / Approach:
Net rate = inlet rate − leak rate. Rearranging gives leak rate = inlet rate − net rate. The leak’s emptying time is the reciprocal of this leak rate.

Step-by-Step Solution:
Inlet rate = 1/8; net rate = 1/10.Leak rate = 1/8 − 1/10 = (5 − 4)/40 = 1/40 per h.Leak emptying time = 1 / (1/40) = 40 h.

Verification / Alternative check:
Check the net: 1/8 − 1/40 = 5/40 − 1/40 = 4/40 = 1/10, matching the observed 10 hours.

Why Other Options Are Wrong:
16, 20, 24, and 32 hours do not reproduce the 10-hour net time when combined with the 8-hour inlet.

Common Pitfalls:
Subtracting times instead of rates or mixing units.

Final Answer:
40 h