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2 votes
2 votes

It takes $30$ min to empty a half full tank by draining it at constant rate. It is decided to simultaneously pump water into the half full tank while draining it. What is the rate at which water should be pumped so that it gets full in $10$ minutes?

  1. 4 times draining rate
  2. 3 times draining rate
  3. 2.5 times draining rate
  4. 2 times draining rate

2 Answers

Best answer
7 votes
7 votes

Let $x$ is the length of Tank.

Rate of doing tank empty, by draining, is, say, $V_d$

it takes $30$ min to do tank empty by draining when it is half filled , $\frac{x}{2}$ distance, with rate $V_d$

$V_d = \frac{\frac{x}{2}}{30}$   or $\frac{x}{2} =30 V_d$

Say $V_u$ is the rate of filling tank

so $V_u-V_d$ is rate of filling tank while its draining

It take $10$min to fill , half filled tank ($\frac{x}{2}$) while it is draining .

$(V_u-V_d) \times 10 = \frac{x}{2}$

$10V_u - 10V_d = 30V_d$

$V_u = 4 V_d $

Rate of filling tank = $4 \times$ Rate of doing it empty

selected by
4 votes
4 votes
Let $d$ be Draining Rate, $p$ be Pumping Rate and $L$ be the capacity of the tank.

$$\begin{align}
30 \times d &= \frac L 2\\[1em]
(10 \times p) - (10 \times d) &= \frac L 2\\[1em]
\implies\\[1em]
10 \cdot (p-d) &= 30\cdot d\\[1em]
p - d &= 3\cdot d \\[1em]
p &= 4\cdot d
\end{align}$$

Hence, option A is the correct answer.

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