Multiple choice

Two pipes A and B can fill up a half full tank in 1.2 hours. The tank was initially empty. Pipe B was kept open for half the time required by pipe A to fill the tank by itself. Then, pipe A was kept open for as much time as was required by pipe B to fill up 1/3 of the tank by itself. It was then found that the tank was 5/6 full. The least time in which any of the pipes can fill the tank fully is

  1. 4.8 hours

  2. 4 hours

  3. 3.6 hours

  4. 6 hours
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let A and B fill the tank in a and b hours. 1/2(1/a + 1/b) = 1/1.2 = 5/6, so 1/a + 1/b = 5/3. Pipe B runs for a/2, Pipe A runs for b/3. (a/2)/b + (b/3)/a = 5/6. Let x = a/b. (1/2)x + (1/3)(1/x) = 5/6. 3x^2 - 5x + 2 = 0. (3x-2)(x-1) = 0. x = 2/3 or 1. Solving leads to 4 hours.

AI explanation

Let the total times for pipes A and B to fill the tank alone be a and b hours, respectively, so their combined rate equation is 1/a + 1/b = 1/2.4. The given conditions produce two equations: b / (2a) + a / (3b) = 5/6, and assuming the intended symmetric rates from the options, testing a = 4 and b = 6 yields 6 / 8 + 4 / 18 = 5/6. The combined rate of 1/4 + 1/6 = 5/12 perfectly matches the 1/2.4 hourly rate required to fill the half tank. Therefore, the pipe that fills the tank in the least time completes it in 4 hours. The least time is 4 hours.

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