Multiple choice

Three pipes A, B and C are connected to a tank. These pipes can fill the tank separately in 5 hours, 10 hours and 15 hours, respectively. When all the three pipes are opened simultaneously, it is observed that pipes A and B are supplying water at 3/4 of their normal rates for the first hour, after which they supply water at the normal rate. Pipe C supplies water at 2/3 of its normal rate for the first 2 hours, after which it supplies water at its normal rate. In how much time would the tank be filled?

  1. 1.05 hours

  2. 2.05 hours

  3. 3.05 hours

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Rates: A=1/5, B=1/10, C=1/15. First hour: A+B = (3/4)(1/5 + 1/10) = (3/4)(3/10) = 9/40. C = (2/3)(1/15) = 2/45. Total = 9/40 + 2/45 = 81/360 + 16/360 = 97/360. Second hour: A+B = 3/10, C = 2/45. Total = 3/10 + 2/45 = 27/90 + 4/90 = 31/90 = 124/360. Total in 2 hours = 221/360. Remaining = 139/360. Normal rate = 1/5 + 1/10 + 1/15 = 6/30 + 3/30 + 2/30 = 11/30. Time = (139/360) / (11/30) = 139/132 = 1.053 hours. Total time = 2 + 1.053 = 3.053 hours.

AI explanation

Using the LCM method, let the total capacity be 30 units, making the normal rates of pipes A, B and C equal to 6, 3 and 2 units per hour respectively. In the first hour, pipes A and B work at 3/4 of their rates to fill 4.5 and 2.25 units, while pipe C works at 2/3 of its rate to fill 1.33 units, totaling 8.08 units. In the second hour, A and B work normally while C continues at the reduced rate, filling 6, 3 and 1.33 units for a combined 10.33 units, bringing the total filled to 18.41 units. The remaining capacity of 11.59 units is filled by all three pipes working at their normal combined rate of 11 units per hour, taking 1.05 hours, which results in a total time of 3.05 hours.

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