Two pipes A and B can fill a tank in 20 and 30 hours, respectively. Both pipes are opened to fill the tank, but when the tank is one-third full, a leak develops through which one-fourth of the water supplied by both pipes goes out. Find the total time (in hours) taken to fill the tank.

Correct — B. The leak only starts once the tank is one-third full, so the fill runs in two phases at two different rates.

Combined rate of A and B = 1⁄20 + 1⁄30 = 3⁄60 + 2⁄60 = 1⁄12 tank per hour.

Phase 1, no leak, filling the first 1⁄3 of the tank:
time = (1⁄3) ÷ (1⁄12) = 4 hours

Phase 2, the leak removes one-fourth of the supply, so 3⁄4 of it survives:
effective rate = (3⁄4) × (1⁄12) = 1⁄16 tank per hour
remaining 2⁄3 takes (2⁄3) ÷ (1⁄16) = 32⁄3 = 10 2⁄3 hours

Total = 4 + 10 2⁄3 = 14 2⁄3 hours → option (b), the image showing 14 2⁄3.

Why the others are wrong

• (a)
  Option (a) shows 11 2⁄5, which is shorter than 12 hours — the time A and B need with no leak at all. A leak can only slow a fill, so nothing below 12 can be right.
• (c)
  Option (c) shows 12 1⁄3. Since 12 hours is the leak-free time, this credits the leak with just 20 extra minutes across two-thirds of the tank, far too little when a quarter of the supply is being lost.
• (d)
  Option (d) shows 14, the figure you get if the leak wastes one-fifth of the supply instead of one-fourth: rate 4⁄5 × 1⁄12 = 1⁄15, and (2⁄3) ÷ (1⁄15) = 10 hours, plus the first 4.

A pipes-and-cisterns item becomes a two-phase problem the moment a leak, a closure or an extra pipe enters part-way through.

Work in rates, not times: a pipe that fills in 20 hours contributes 1⁄20 of the tank each hour, and rates add.

A leak stated as a fraction of the supply is a multiplier on the rate, not a fixed outflow. Losing one-fourth of what is poured in leaves 3⁄4 of the combined rate.

Split the tank at the boundary the stem names, time each part at its own rate, and add the two times.

Key facts

• A and B together fill at 1⁄20 + 1⁄30 = 1⁄12 of the tank per hour, so 12 hours with no leak.
• A leak taking one-fourth of the supply leaves an effective rate of 3⁄4 × 1⁄12 = 1⁄16.
• The first one-third of the tank takes 4 hours and the remaining two-thirds take 10 2⁄3 hours.

Common traps

• Applying the leak to the whole tank instead of only the last two-thirds.
• Treating one-fourth of the supply as one-fourth of the time.
• Answering 12 hours, the combined rate's time, and forgetting the leak.

No directly related past PYQ was found.