Pipe A usually fills a tank in 6 hours. But due to a leak at the bottom of the tank, it takes extra 2 hours to fill the tank. If the tank is full, then how much time will it take to get emptied due to the leak?

Correct — C. Work in rates (tank per hour), never in times.

Pipe A alone fills 1⁄6 of the tank per hour.
With the leak it takes 6 + 2 = 8 hours, so the combined rate is 1⁄8 per hour.

The leak is the shortfall between the two rates:
1⁄6 − 1⁄8 = (4 − 3)⁄24 = 1⁄24 per hour.

The leak drains 1⁄24 of a full tank each hour, so a full tank empties in 24 hours → option (c).

Why the others are wrong

• (a)
  16 hours — 16 hours means a leak rate of 1⁄16, giving a net fill rate of 1⁄6 − 1⁄16 = 5⁄48. The tank would then fill in 9.6 hours, not the 8 the question states.
• (b)
  12 hours — 12 hours means a leak rate of 1⁄12, so the net fill rate is 1⁄6 − 1⁄12 = 1⁄12 and the tank takes 12 hours to fill. The question says 8.
• (d)
  20 hours — 20 hours gives 1⁄6 − 1⁄20 = 7⁄60, filling the tank in 60⁄7 ≈ 8.57 hours. Close to 8, but the question fixes the delayed time at exactly 8 hours.

Rates add and subtract. Times do not. A pipe that fills in t hours contributes 1⁄t of the tank per hour, and a leak that would empty it in T hours contributes −1⁄T.

The question gives you a delay, not a time: "extra 2 hours" means the combined operation takes 8 hours, so the net rate is 1⁄8. Everything else is one subtraction.

Reciprocate at the end and you have the leak's own emptying time.

Key facts

• A pipe filling in t hours works at 1⁄t per hour, and a leak emptying in T hours works at −1⁄T per hour.
• With a leak the balance reads 1⁄6 − 1⁄T = 1⁄8, which solves to T = 24.
• For fill time t and delay d, the leak alone takes t(t + d)⁄d hours — here 6 × 8 ⁄ 2 = 24.

Common traps

• Subtracting the times, 8 − 6 = 2, instead of the rates.
• Taking the extra 2 hours as the leak's own emptying time.
• Using 6 hours as the combined time and 8 as the pipe's own time.

No directly related past PYQ was found.