Question

A cistern has two pipes. One can fill it with water in 8 hours and other can empty it in 5 hours. In how many hours will the cistern be emptied if both the pipes are opened together when ³⁄₄ of the cistern is already full of water?

• A. $2\frac{1}{2}$ hours
• B. $3\frac{1}{3}$ hours
• C. 8 hours
• D. 10 hours

Hint:

Always check the rates first. If the "Emptying" pipe is faster (takes less time) than the "Filling" pipe, the tank can never be filled if both are open; it can only be emptied.

Answer 1

The Correct Option is D

Concept: This problem involves rates of work. Filling is positive work, and emptying is negative work. If the emptying rate is faster than the filling rate, a full (or partially full) tank will eventually be emptied.

Step 1: Determine the net rate of work.
Let the capacity of the tank be the LCM of 8 and 5, which is 40 units. Filling rate (Pipe 1) $= 40 / 8 = +5$ units/hour. Emptying rate (Pipe 2) $= 40 / 5 = -8$ units/hour. Net rate when both are open $= +5 - 8 = -3$ units/hour (The negative sign indicates the tank is emptying).

Step 2: Calculate the amount of water to be emptied.
The cistern is $\frac{3}{4}$ full. Amount of water in the tank $= \frac{3}{4} \times 40 = 30$ units.

Step 3: Find the time required.
Time $= \frac{\text{Amount to empty}}{\text{Net rate of emptying}}$ \[ \text{Time} = \frac{30 \text{ units}}{3 \text{ units/hour}} = 10 \text{ hours.} \]