**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 ?lling 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 ?lled on Thursday if both pumps were used simultaneously all along?

4:36 pm | |

4:12 pm | |

4:24 pm | |

4:48 pm |

**Show Answer**

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Correct Answer: 3

Let 't' pm be the time when the tank is emptied everyday. Let 'a' and 'b' be the liters/hr filled by pump A and pump B respectively.

On Monday, A alone completed ?lling the tank at 8 pm. Therefore, we can say that pump A worked for (8 - t) hours. Hence, the volume of the tank = a*(8 - t) liters.

Similarly, on Tuesday, B alone completed ?lling the tank at 6 pm. Therefore, we can say that pump B worked for (6 - t) hours. Hence, the volume of the tank = b*(6 - t) liters.

On Wednesday, A alone worked till 5 pm, and then B worked alone from 5 pm to 7 pm, to fill the tank. Therefore, we can say that pump A worked for (5 - t) hours and pump B worked for 2 hours. Hence, the volume of the tank = a*(5 - t)+2b liters.

We can say that a*(8 - t) = b*(6 - t) = a*(5 - t) + 2b

a*(8 - t) = a*(5 - t) + 2b

\( \Rightarrow \) 3a = 2b ... (1)

a*(8 - t) = b*(6 - t)

Using equation (1), we can say that

\(a*(8 - t) = \frac{{3a}}{2}*(6 - t)\)

\(t = 2\)

Therefore, we can say that the tank gets emptied at 2 pm daily. We can see that A takes 6 hours and pump B takes 4 hours alone.

Hence, working together both can fill the tank in = \frac{6*4}{6+4} = 2.4 hours or 2 hours and 24 minutes.

The pumps started filling the tank at 2:00 pm. Hence, the tank will be filled by 4:24 pm.

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