Aptitude Discussion

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

**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 = $50+10=60$ \(\text m^3/\text{min}\).