Pipes and Cisterns – A leak slows one inlet; deduce leak strength and B’s new time: Pipe A fills the cistern in 30 minutes and pipe B in 40 minutes. Because of a crack (leak) at the bottom, pipe A now needs 40 minutes to fill the cistern alone. With the same leak present, how long will B now take to fill it, and how long would the leak alone take to empty a full cistern?

Introduction / Context:
The leak reduces each inlet’s effective rate by the same leak rate. Using the observed slowdown for A (from 30 to 40 minutes), we can compute the leak rate and then apply it to B to find B’s new filling time. The leak’s emptying time is the reciprocal of its rate.

Given Data / Assumptions:

• A normal = 1/30 per min; with leak = 1/40 per min.
• B normal = 1/40 per min; leak rate is the same for both.

Concept / Approach:
Leak rate l = (A normal) − (A with leak) = 1/30 − 1/40. Then B with leak = (B normal) − l.

Step-by-Step Solution:
l = 1/30 − 1/40 = (4 − 3)/120 = 1/120 per min.B with leak = 1/40 − 1/120 = (3 − 1)/120 = 1/60 per min ⇒ B now takes 60 min (1 hour).Leak emptying time = 1 / (1/120) = 120 min = 2 hours.

Verification / Alternative check:
For A, 1/30 − 1/120 = 1/40, which matches the stated slowdown, confirming leak computation.

Why Other Options Are Wrong:
Options implying a 1-hour leak or a 2-hour B time contradict the rate arithmetic derived from A’s slowdown.

Common Pitfalls:
Subtracting times rather than rates; mixing minutes and hours.

Final Answer:
B fills in an hour and the leak empties in 2 hours