Question

There are two pipes A & B which can fill a tank<br />in 30 hours and 15 hours respectively. There is a<br />leak as well. when these all 3 ( pipes + leak) are<br />active, they take 12 hours 30 minutes to fill the<br />tank. find the time in which the leak can empty<br />the whole tank?​

Answer 1

The time in which the leak can empty the whole tank is 50 hrs.

Step-by-step explanation:

The pipe A can fill a tank in 30 hours

So, In 1 hr pipe A can fill = $\frac{1}{30}$ of the tank

The pipe B can fill a tank in 15 hours

So, In 1 hr pipe B can fill = $\frac{1}{15}$ of the tank

∴ In 1 hour, the part of the tank that both A and B will fill is,

= $\frac{1}{30} + \frac{1}{15}$

= $\frac{1+2}{30}$

= $\frac{3}{30}$

= $\frac{1}{10}$

It is given that the two pipes A and B and along with a leak they can fill the tank in 12 hrs 30 minutes = $12\frac{30}{60}$ = $12\frac{1}{2}$ = $\frac{25}{2}$ hrs

So, in 1 hr pipe A, B and the leak will fill = $\frac{2}{25}$ of the tank

 Therefore,

In 1 hr the part of the tank that the leak can empty, is given by

= [part of tank that pipe A and B will fill in 1 hour] - [part of the tank that pipe A, B and leak will fill in 1 hour]

= $\frac{1}{10} - \frac{2}{25}$

$= \frac{5 - 4}{50} \\= \frac{1}{50}$

Thus, the leak in the tank will empty the whole tank in 50 hours.

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