Question

A tank is emptied everyday at a fixed time point. Immediately thereafter, either pump A or pump B or both start working until the tank is full. On Monday, A alone completed filling the tank at 8 pm. On Tuesday, B alone completed filling the tank at 6 pm. On Wednesday, A alone worked till 5 pm, and then B worked alone from 5 pm to 7 pm, to fill the tank. At what time was the tank filled on Thursday if both pumps were used simultaneously all along?

Accepted Answer

B. Let the fixed emptying time be $t$ pm. Time taken by A alone $= 8 - t$ hours. Time taken by B alone $= 6 - t$ hours. On Wednesday, A works for $(5 - t)$ hours and B works for $2$ hours (5 pm to 7 pm). Fraction of tank filled $= \frac{5-t}{8-t} + \frac{2}{6-t} = 1$. Let $x = 8 - t$. Then $6 - t = x - 2$, and $5 - t = x - 3$. $$\frac{x - 3}{x} + \frac{2}{x - 2} = 1$$ $$1 - \frac{3}{x} + \frac{2}{x - 2} = 1$$ $$\frac{3}{x} = \frac{2}{x - 2} \implies 3x - 6 = 2x \implies x = 6$$ So $t = 8 - 6 = 2$ pm. Time taken by A alone $= 6$ hours. Time taken by B alone $= 4$ hours. When both work together, time taken $= \frac{6 	imes 4}{6 + 4} = \frac{24}{10} = 2.4	ext{ hours} = 2	ext{ hours } 24	ext{ minutes}$. Starting at 2:00 pm, the tank will be full at 4:24 pm.

Time to fill tank using both pumps

Explanation

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