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?
Explanation:
The time taken by pump A alone to fill the tank completely = t hours.
The time taken by pump B alone to fill the tank completely = ‘t - 2’ hours.
On Wednesday, pump A was used for 3 hour less and pump B was used for 2 hours to complete the remaining work. ⇒ Work done by A in 3 hours = Work done by B in 2 hours. ⇒ efficiency of A : efficiency of B = 2 : 3
Let efficiency of A be '2x' and that of B be '3x'.
Now, total work is done by B in (t - 2) hours and by A in t hours. ⇒ 3x × (t - 2) = 2x × t ⇒ 3(t - 2) = 2t ⇒ t = 6 hours
∴ The tank starts filling at 2 pm.
If both pumps start filling, their combined efficiency = 5x Total work to be done = 3x(t - 2) = 12x
⇒ Time taken to fill the tank = 12x/5x = 2.4 hours = 2 hours 24 minutes.
Therefore on Thursday, the tank will be completely full at 4.24 pm.
Hence, option (d).
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