Two pipes A and B can fill a tank in 18 and 24 minutes respectively. After both are opened together for 6 minutes, the rate of pipe A becomes half, and pipe B becomes double its original. In how many more minutes will the tank be full?

Correct — A.
Rates: A fills 1⁄18 and B fills 1⁄24 of the tank per minute

Together for 6 min: 6 × (1⁄18 + 1⁄24) = 6 × 7⁄72 = 7⁄12
Left to fill: 1 − 7⁄12 = 5⁄12

Halve A's rate: 1⁄18 ÷ 2 = 1⁄36
Double B's rate: 1⁄24 × 2 = 1⁄12
New combined rate: 1⁄36 + 3⁄36 = 1⁄9 per minute

Time: 5⁄12 ÷ 1⁄9 = 45⁄12 = 3.75 min
Convert: 0.75 min = 45 sec, so 3 min 45 sec → option (a)

Why the others are wrong

• (b)
  2 min 8 sec — 2 min 8 sec is 32⁄15 min. At the new rate of 1⁄9 tank per minute that fills only 32⁄135 of the tank, well short of the 5⁄12 still empty.
• (c)
  5 min 42 sec — 5 min 42 sec is 5.7 min. At 1⁄9 tank per minute that adds about 0.63 of a tank, but only 5⁄12 ≈ 0.42 is left, so the tank would overflow.
• (d)
  6 min 50 sec — 6 min 50 sec is 41⁄6 min, which at 1⁄9 per minute adds about 0.76 of a tank. That is far more than the 5⁄12 remaining.

Pipe problems run on rates, not times. A pipe that fills a tank in t minutes adds 1⁄t of it each minute, and the rates of pipes working together add.

When a rate changes midway, split the job into phases. Work done = rate × time in each phase, and the phases must add up to one full tank.

Half the rate means double the time: A now needs 36 minutes alone, and B's doubled rate cuts its 24 minutes to 12.

Key facts

• A pipe that fills a tank in t minutes fills 1⁄t of it per minute
• Halving a pipe's rate doubles its filling time
• Here the first 6 minutes fill 7⁄12 of the tank
• 0.75 minute = 45 seconds

Common traps

• Halving A's time instead of its rate, which makes A faster when it has slowed down
• Reading 3.75 minutes as 3 min 75 sec instead of 3 min 45 sec

No directly related past PYQ was found.