A cistern that would normally be filled by an inlet in 9 hours takes 1 hour longer to be filled because of a leak at the bottom. If the cistern is full, in how many hours would the leak alone empty it?

Introduction / Context:
Let f be the inlet rate and l the leak (outlet) rate in tanks/hour. The observed slower fill time implies f − l equals the net fill rate. Solve for l and invert to get the emptying time for a full cistern.

Given Data / Assumptions:

• Inlet alone: f = 1/9 tank/hour.
• With leak: net = 1/10 tank/hour.

Concept / Approach:
f − l = 1/10 ⇒ l = f − 1/10 = 1/9 − 1/10.

Step-by-Step Solution:

Verification / Alternative check:
Net with leak: 1/9 − 1/90 = 10/90 − 1/90 = 9/90 = 1/10, matching the observed 10 hours.

Why Other Options Are Wrong:
80/85/95 hours give inconsistent net rates.

Common Pitfalls:
Interpreting “1 hour more” incorrectly; it changes the net rate, not the inlet rate.

Final Answer:
90 hours