One pipe can fill a tank 6 times faster than another pipe. If both the pipes together can fill the tank in 40 minutes, then in how many minutes will the slower pipe alone be able to fill the tank?

Correct — D. Let the slower pipe alone take T minutes, and add rates — never times.

Slower pipe = 1/T of the tank per minute
Faster pipe fills 6 times faster = 6/T per minute

Together = 1/T + 6/T = 7/T, and that equals 1/40

T = 7 × 40 = 280 minutes → option (d)
Check: 1/280 + 6/280 = 7/280 = 1/40

Why the others are wrong

• (a)
  290 — If the slower pipe took 290 minutes the pair would finish in 290/7 ≈ 41.4 minutes, not the 40 the question fixes.
• (b)
  275 — 275 gives 275/7 ≈ 39.3 minutes together — quicker than the stated 40, so this slower pipe is being made too fast.
• (c)
  285 — 285 gives 285/7 ≈ 40.7 minutes together. It is close, but 7/T = 1/40 has the single solution T = 280.

Pipe-and-cistern items are work items in disguise, and the one rule that carries them is that rates add, times do not. A pipe that fills a tank in T minutes works at 1/T of the tank a minute.

Watch the comparison in the stem. One pipe is 6 times faster, so its rate is 6/T and the pair fill at 7/T.

Setting 7/T against the given combined time of 40 minutes returns T in a single step, with no need to find the faster pipe's own time at all.

Key facts

• A pipe filling a tank in T minutes works at 1/T of the tank per minute.
• Pipes opened together add their rates: 1/T + 6/T = 7/T.
• 7/T = 1/40 gives T = 280 minutes for the slower pipe and about 46.7 minutes for the faster one.

Common traps

• Averaging the two times instead of adding the two rates
• Answering with the faster pipe's time of about 46.7 minutes when the stem asked for the slower one

No directly related past PYQ was found.