Two pipes A and B can fill a tank in 48 minutes and 66 minutes, respectively. If both the pipes are opened simultaneously, then after how many minutes should pipe B be closed so that the tank gets filled in 32 minutes?

Correct — A. Pipe A stays open for the whole 32 minutes, pipe B for only t minutes, and the two shares must add to one full tank.

A's share: 32 ⁄ 48 = 2⁄3 of the tank
B must supply the rest: 1 − 2⁄3 = 1⁄3
B fills 1⁄66 of the tank a minute, so t ⁄ 66 = 1⁄3
t = 66 ⁄ 3 = 22 minutes → option (a)

Why the others are wrong

• (b)
  20 — Twenty minutes of B is 20⁄66 = 10⁄33 of the tank. Added to A's 22⁄33 that is 32⁄33, leaving the tank 1⁄33 short when the 32 minutes are up.
• (c)
  16 — Sixteen minutes of B gives 8⁄33, and 22⁄33 + 8⁄33 = 30⁄33. The tank is still 3⁄33 short when the 32 minutes are up.
• (d)
  18 — Eighteen minutes of B gives 9⁄33 when 11⁄33 is required, so the tank finishes 2⁄33 short.

Work problems add up in rates, never in times. A pipe that fills a tank in 48 minutes contributes 1⁄48 of it each minute, and the fractions of the job done by each pipe must sum to exactly 1.

The deadline is fixed at 32 minutes here, so start with the pipe that runs throughout — its share is computable immediately. Whatever it leaves undone is the second pipe's quota, and that quota divided by its rate is the time it must run.

Key facts

• A pipe filling a tank in 48 minutes does 1⁄48 of the job each minute.
• A alone over 32 minutes fills 32⁄48 = 2⁄3 of the tank.
• B must supply the remaining 1⁄3 at 1⁄66 per minute, which takes 22 minutes.

Common traps

• Closing B at 32 − 22 = 10 minutes by reading the question as the time remaining.
• Adding or averaging the two filling times instead of adding the rates.
• Letting both pipes run the full 32 minutes, which overshoots a full tank.

No directly related past PYQ was found.