Question

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

• A. 4 times the draining rate
• B. 3 times the draining rate
• C. 2.5 times the draining rate
• D. 2 times the draining rate

Answer 1

The Correct Option is A

Given a half-full tank is emptied in 30 minutes at a constant rate, let's assume the rate of draining is R (in tank volume per minute). Thus, to drain half the tank, it takes:

$$0.5 \text{ (tank volume) } = 30R \Rightarrow R = \frac{0.5}{30}$$

Now, we need the tank full in 10 minutes by adding a pumping rate of P. In 10 minutes, both pumping and draining occur:

$$\text{Net water increase in 10 minutes} = \text{Tank volume}$$

Hence,

$$(10P - 10R) = 1 \text{ (tank volume) }$$

Solving for pumping rate, P:

$$10P - 10R = 1$$

$$P - R = \frac{1}{10}$$

From the earlier calculation, $$R = \frac{1}{60}$$. Substitute to find P:

$$P - \frac{1}{60} = \frac{1}{10}$$

$$P = \frac{1}{10} + \frac{1}{60}$$

Finding a common denominator, $$P = \frac{6}{60} + \frac{1}{60} = \frac{7}{60}$$

To compare with R:

$$\frac{7}{60R} = 7$$

Thus, P, the pumping rate, is 4 times the draining rate. Therefore, the rate is 4R.