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

Correct option is C

Given:

1. Tap A can fill the tank in 12 hours.
2. Tap B can fill the tank in 15 hours.
3. Tap C can fill the tank in 20 hours.
4. Tap A is always open, while B and C are open alternately for one hour each.
5. We need to determine the total time required to fill the tank.

Formula Used:

1. Rate of filling by a tap:

$$\text{Rate of a tap} = \frac{1}{\text{Time taken to fill the tank}}$$​​

2. Combined work in one hour when multiple taps are working:
​$$\text{Total rate} = \text{Rate of A} + \text{Rate of (B or C)}$$​​

Solution:

1. Calculate individual rates:

$$\text{Rate of A} = \frac{1}{12}, \quad \text{Rate of B} = \frac{1}{15}, \quad \text{Rate of C} = \frac{1}{20}$$​​

2. Work done in 2 hours (1 cycle):
- In the 1st hour, taps A and B are open:

$$\text{Work done in 1st hour} = \frac{1}{12} + \frac{1}{15}$$​​

LCM of 12 and 15 is 60:

$$\frac{1}{12} + \frac{1}{15} = \frac{5}{60} + \frac{4}{60} = \frac{9}{60} = \frac{3}{20}$$​​

- In the 2nd hour, taps A and C are open:

$$\text{Work done in 2nd hour} = \frac{1}{12} + \frac{1}{20}$$​​

LCM of 12 and 20 is 60:

$$\frac{1}{12} + \frac{1}{20} = \frac{5}{60} + \frac{3}{60} = \frac{8}{60} = \frac{2}{15}$$​​

- Total work done in 2 hours:

$$\text{Total work in 2 hours} = \frac{3}{20} + \frac{2}{15}$$​​

LCM of 20 and 15 is 60:

$$\frac{3}{20} + \frac{2}{15} = \frac{9}{60} + \frac{8}{60} = \frac{17}{60}$$​​

3. Determine the total time to fill the tank:

$$\text{Number of cycles} = \frac{60}{17} \approx 3.53$$​​

- In 3 full cycles (6 hours), the tank is:

$$3 \times \frac{17}{60} = \frac{51}{60} \text{ full}$$​​

- Remaining portion to fill:
​$$1 - \frac{51}{60} = \frac{9}{60} = \frac{3}{20}$$​​

4. Work done in the final hour:
- In the final hour, A and B work together:

$$\text{Work done in 1 hour by A and B} = \frac{1}{12} + \frac{1}{15}$$​​

LCM of 12 and 15 is 60:

$$\frac{1}{12} + \frac{1}{15} = \frac{5}{60} + \frac{4}{60} = \frac{9}{60} = \frac{3}{20}$$​​

Thus, the remaining portion $$\frac{3}{20}$$​ is filled in exactly 1 hour.

5. Total time taken:

$$\text{Total time} = 6 \text{ hours (from 3 cycles)} + 1 \text{ hour (final)} = 7 \text{ hours}$$​​

Final Answer:

The tank will be full in 7 hours
**Option C: 7 hours**