Pipes and Cisterns – Two inlet taps with different efficiencies: Two taps A and B are used to fill the same water tank. When both taps are opened together, the tank fills in 40 minutes. If only tap A is opened, it fills the tank in 60 minutes. Determine the time required for tap B alone to fill the tank.

Introduction / Context:
This classic pipes-and-cisterns problem asks for the individual filling time of one tap (B) given the combined time and the time for the other tap (A). The key idea is to convert given times to per-minute rates and then use rate addition or subtraction appropriately.

Given Data / Assumptions:

• Together time (A + B) = 40 min ⇒ combined rate = 1/40 tank per min.
• Tap A alone time = 60 min ⇒ A’s rate = 1/60 tank per min.
• Tank capacity is taken as 1 unit for ease of calculation.

Concept / Approach:
When inlets run together, their rates add: rate(A + B) = rate(A) + rate(B). Rearranging gives rate(B) = rate(A + B) − rate(A). The solo time of B is the reciprocal of its rate.

Step-by-Step Solution:
Combined rate = 1/40 per min.A’s rate = 1/60 per min.B’s rate = 1/40 − 1/60 = (3 − 2)/120 = 1/120 per min.Therefore, B’s time alone = 1 / (1/120) = 120 min.

Verification / Alternative check:
Check by recombining: 1/60 + 1/120 = (2 + 1)/120 = 3/120 = 1/40 per min ⇒ 40 minutes together, which matches the given data exactly.

Why Other Options Are Wrong:
64, 80, 96, and 90 minutes produce combined rates that do not equal 1/40 when added to A’s 1/60 rate.

Common Pitfalls:
Adding times instead of rates, or taking an average of 40 and 60 minutes instead of using reciprocals.

Final Answer:
120 min