InterviewSolution
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A takes 7 minutes more than water tap B for filling up a tank with water.A takes 16 minutes more than the time taken by both the taps together to filltank. Find the time each tap alone would take to fill the tankthe |
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Answer» Let Tap B takes x mins Tap B takes x + 7 mins Tap B takes x mins ⇒ 1 min = 1/x of the tank filled Tap A takes (x + 7) mins ⇒ 1 min = 1/(x + 7) of the tank filled Together: 1 min = 1/x + 1/(x + 7) 1 min = [ x + 7 + x] /x(x + 7) 1 min = (2x + 7) /x(x + 7) of the tank Total time needed to fill the tank together: mins needed = x(x + 7)/(2x + 7) The tap A takes 16 minutes more than the time taken by both the taps together to fill the tank: Tap A = Tap A and Tap B + 16 mins (x + 7) = x(x + 7)/(2x + 7) + 16 x + 7 - 16 = x(x + 7)/(2x + 7) x - 9 = x(x + 7)/(2x + 7) x(x + 7) = (x - 9)(2x + 7) x² + 7x = 2x² + 7x - 18x - 63 x² - 18x - 63 = 0 (x - 21)(x + 3) = 0 x = 21 or x = - 3 (rejected, time cannot be negative) Find the time needed: Tap B = x = 21 mins Tap A = x + 7 = 21 + 7 = 28 mins Answer: Tap A takes 28 mins and Tap B takes 21 mins |
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