Question

Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If tap A is open all the time and tap B and tap C are open for one hour each alternately, the tank will be filled in:

• A. 7 hours
• B. 6 hours
• C. 5 hours
• D. None of these

Hint:

When working with problems involving multiple rates, ensure to calculate the total work done in each cycle and multiply accordingly.

Answer 1

The Correct Option is A

Let the capacity of the tank be \( C \). Step 1: Calculate the rate of filling for each tap:
- Tap A can fill the tank in 12 hours, so its rate is \( \frac{1}{12} \) of the tank per hour.
- Tap B can fill the tank in 15 hours, so its rate is \( \frac{1}{15} \) of the tank per hour.
- Tap C can fill the tank in 20 hours, so its rate is \( \frac{1}{20} \) of the tank per hour.

Step 2: Determine the rate of filling in one cycle (A, B, and C working alternately):
In each cycle, tap A works for one hour continuously, tap B works for one hour, and tap C works for one hour. The total work done in one cycle is: \text{Rate of A in 1st hour} = \( \frac{1}{12} \) \quad \text{(since A works all the time)}
\text{Rate of B in 2nd hour} = \( \frac{1}{15} \)
\text{Rate of A + C in 3rd hour} = \( \frac{1}{12} \) + \( \frac{1}{20} \).

Step 3: Work done in each hour: - In the first hour, \( A + B \) are open: The rate is \( \frac{1}{12} + \frac{1}{15} \), so work done = 9 units of work.
- In the second hour, \( A + C \) are open: The rate is \( \frac{1}{12} + \frac{1}{20} \), so work done = 8 units of work. Thus, in 2 hours, the total work done = 17 units of work.

Step 4: Calculate the total time taken to fill the tank: - In 6 hours, the total work done will be \( 51 \) units (3 cycles of 2 hours each).
- The remaining work is done in the next 1 hour (together, \( A + B \)) which fills the remaining part of the tank. Thus, the total time to fill the tank is 7 hours. Final Answer: The correct answer is (a) 7 hours.