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A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is?

Correct answer: C. 15 hrs

  • A. 6 hrs
  • B. 10 hrs
  • C. 15 hrs
  • D. 30 hrs

Explanation

Let the first pipe take x hours, so the second takes x - 5 hours and the third takes x - 9 hours. Since the first two together take the same time as the third alone, 1/x + 1/(x - 5) = 1/(x - 9); solving gives x² - 18x + 45 = 0, whose valid root is x = 15 hours. The likely mistake is choosing 10 hours by using only one of the time differences and not applying the combined-rate condition.

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About Time and Work

Problems measure work through rates, treating each person's or machine's contribution as work completed per unit time. Questions cover combined work, efficiency, wages, alternate working schedules, pipes and cisterns, and men-days relationships, with care needed to distinguish time taken by one worker from the time required when workers act together.

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