Imperfectly opened taps (reduced discharge) Two taps A and B nominally fill a tank in 25 min and 20 min. Due to partial opening, A allows only 5/6 of its normal flow and B allows only 2/3 of its normal flow. How long will they take together to fill the tank?

Introduction / Context:
When taps are partially opened, multiply each nominal rate by its effective fraction to get the reduced rate. Then add reduced rates as usual.

Given Data / Assumptions:

• Nominal A: 25 min ⇒ 1/25 per min; effective = (5/6)*(1/25).
• Nominal B: 20 min ⇒ 1/20 per min; effective = (2/3)*(1/20).
• Tank initially empty; both run together.

Concept / Approach:
Compute effective rates and add. Time is the inverse of the total effective rate.

Step-by-Step Solution:

Verification / Alternative check:
In 15 min, A contributes 15*(1/30)=1/2 and B contributes 1/2; sum 1 tank.

Why Other Options Are Wrong:
14, 13, 12 imply total effective rates greater than 1/15, which are not supported by the given fractions.

Common Pitfalls:
Reducing the times instead of the rates (that yields wrong arithmetic).

Final Answer:
15 min