A cistern is normally filled in 8 hours but takes 2 hours longer (10 hours total) due to a bottom leak. If the cistern is full, how long will the leak alone take to empty it completely?

Introduction / Context:Compare normal filling versus delayed filling to deduce the leak’s emptying rate. The leak subtracts from the inlet’s rate, increasing the total fill time.

Given Data / Assumptions:

• No leak fill time = 8 hours ⇒ inlet rate = 1/8 tank/hour.
• With leak fill time = 10 hours ⇒ net rate = 1/10 tank/hour.
• Rates remain constant; the leak acts uniformly.

Concept / Approach:leak rate = inlet rate − net rate. Emptying time of the leak = reciprocal of leak rate.

Step-by-Step Solution:Inlet rate = 1/8.Net rate = 1/10.Leak rate = 1/8 − 1/10 = 1/40 tank/hour.Time to empty = 1 / (1/40) = 40 hours.

Verification / Alternative check:With both acting: 1/8 − 1/40 = 4/40 = 1/10 ⇒ matches 10 hours to fill.

Why Other Options Are Wrong:16, 20, and 25 hours arise from wrong fraction differences or inversion mistakes.

Common Pitfalls:Subtracting times rather than rates; forgetting to invert rate to get time.

Final Answer:40 hrs.