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Q.

A pump can be used either to fill or to empty a tank. The capacity of the tank is (3600text{ m}^3). The emptying capacity of the pump is (10 text{ m}^3/text{min}) higher than its filling capacity. What is the emptying capacity of the pump if the pump needs 12 more minutes to fill the tank than to empty it?

 A.

\(10 \text{ m}^3/\text{min}\)

 B.

\(60 \text{ m}^3/\text{min}\)

 C.

\(45 \text{ m}^3/\text{min}\)

 D.

\(90 \text{ m}^3/\text{min}\)

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Solution:
Option(B) is correct

Let f  \(\text m^3/\text{min}\) be the filling capacity of the pump.

Therefore, the emptying capacity of the pump will be = $f+10$ \(\text m^3/\text{min}\)

The time taken to fill the tank will be = \(\dfrac{3600}{f}\) minutes

And the time taken to empty the tank will be = \(\dfrac{3600}{f+10}\)

We know that it takes 12 more minutes to fill the tank than to empty it

\(\dfrac{3600}{f}-\dfrac{3600}{f+10}=12\)

\(3600f+36000-3600f=12(f^2+10f)\)

\(f^2+10f-3000=0\)

Solving for positive value of $f$ we get, $f=50$

Therefore, the emptying capacity of the pump = $50+10=60$  \(\text m^3/\text{min}\).


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