Question

Time required by two pipes a and b working separately to fill a tank is 36 seconds and 45 seconds respectively. another pipe c can empty the tank in 30 seconds. initially, a and b are opened and after 7 seconds, c is also opened. in how much more time the tank would be completely filled ?

Answer 1

please mark it as a brainliest answer

Answer 2

The tank would fill in 46 seconds.

Given: Pipe A can fill the tank in 36 sec, Pipe B can fill the tank in 45 sec.

Pipe C can empty the tank in 30 sec.

To Find: Time required to fill the tank

Solution: Pipe A's 1-second filling = 1/36

Pipe B's 1-second filling = 1/45

Pipe A and B's filling in 1 second = $\frac{1}{36} + \frac{1}{45}$

$\frac{5 + 4}{180} = \frac{9}{180} = \frac{1}{20}$

C empties the tank by 1/30 units per second

Now, Pipe A and B's filling in 7 seconds = 7/20

Now, tank left for filling = 20-7 = 13/20

Total units filled together by A, B, and C in 1 second = $\frac{1}{20} - \frac{1}{30}$

$\frac{3 - 2}{60} = \frac{1}{60}$

Now, the time taken to fill the tank = $\frac{\frac{13}{20} }{\frac{1}{60} } = \frac{13 * 60}{20 * 1} = 39 seconds$

Therefore, time taken to fill the whole tank = 7 + 39 = 46 seconds

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