A cistern is normally filled in 8 hours, but due to a leak at the bottom it takes 10 hours to fill. If the cistern is full, in how many hours will the leak alone empty it?

Introduction / Context:
When a leak is present, the net filling rate equals the inlet rate minus the leak rate. By comparing normal filling time to delayed filling time, we can deduce the leak’s emptying rate and hence its time to empty a full cistern.

Given Data / Assumptions:

• Normal fill time (no leak) = 8 hours ⇒ inlet rate = 1/8 tank/hour.
• With leak, fill time = 10 hours ⇒ net rate = 1/10 tank/hour.
• Leak rate is constant and acts continuously while the inlet runs.

Concept / Approach:
Net rate = inlet rate − leak rate. Rearranging gives leak rate = inlet rate − net rate. Time to empty (leak alone) is the reciprocal of the leak rate.

Step-by-Step Solution:
Inlet rate = 1/8.Net rate with leak = 1/10.Leak rate = 1/8 − 1/10 = (5 − 4)/40 = 1/40 tank/hour.Time for leak to empty a full cistern = 1 / (1/40) = 40 hours.

Verification / Alternative check:
Check: With both acting, 1/8 − 1/40 = (5 − 1)/40 = 4/40 = 1/10 tank/hour, matching the delayed 10-hour fill.

Why Other Options Are Wrong:
16, 20, and 25 hours correspond to incorrect differences or inverted fractions.

Common Pitfalls:
Subtracting times instead of rates, or forgetting to invert to get time from rate.

Final Answer:
40 hrs.