Question

Four pipes A, B, C and D can fill a tank with water in 15, 20, 30 and 60 hours respectively. Pipe A is opened at 4 a.m, B at 5 a.m, C at 6 a.m and D at 7 a.m. When is the tank filled up completely?

• A. 9:30 a.m
• B. 10:00 a.m
• C. 10:30 a.m
• D. 11:00 a.m

Hint:

To solve such problems, calculate the individual rates of each pipe, find the water filled before all pipes are opened, and then calculate the time required to fill the remaining water with all pipes working together.

Answer 1

The Correct Option is D

Let the capacity of the tank be \( C \) (in liters).
- Pipe A can fill the tank in 15 hours, so the rate of A is \( \frac{C}{15} \).
- Pipe B can fill the tank in 20 hours, so the rate of B is \( \frac{C}{20} \).
- Pipe C can fill the tank in 30 hours, so the rate of C is \( \frac{C}{30} \).
- Pipe D can fill the tank in 60 hours, so the rate of D is \( \frac{C}{60} \).
Now, let's consider the times when the pipes are opened:
- Pipe A is opened at 4:00 a.m., so it works from 4:00 a.m. to 7:00 a.m. (3 hours). The amount of water filled by A is: \[ \text{Water filled by A} = 3 \times \frac{C}{15} = \frac{C}{5}. \] - Pipe B is opened at 5:00 a.m., so it works from 5:00 a.m. to 7:00 a.m. (2 hours). The amount of water filled by B is: \[ \text{Water filled by B} = 2 \times \frac{C}{20} = \frac{C}{10}. \] - Pipe C is opened at 6:00 a.m., so it works from 6:00 a.m. to 7:00 a.m. (1 hour). The amount of water filled by C is: \[ \text{Water filled by C} = 1 \times \frac{C}{30} = \frac{C}{30}. \] - Pipe D is opened at 7:00 a.m., and it will continue working from 7:00 a.m. onward. The combined rate of all pipes working together is: \[ \text{Rate of all pipes} = \frac{C}{15} + \frac{C}{20} + \frac{C}{30} + \frac{C}{60} = \frac{4C}{60} + \frac{3C}{60} + \frac{2C}{60} + \frac{C}{60} = \frac{10C}{60} = \frac{C}{6}. \] Now, the total amount of water filled by all pipes from 7:00 a.m. onward is: \[ \text{Total water filled by all pipes} = \frac{C}{5} + \frac{C}{10} + \frac{C}{30} = \frac{6C}{30} + \frac{3C}{30} + \frac{C}{30} = \frac{10C}{30} = \frac{C}{3}. \] Since the rate of all pipes working together is \( \frac{C}{6} \), the time taken to fill the remaining \( \frac{2C}{3} \) of the tank is: \[ \text{Time to fill remaining water} = \frac{\frac{2C}{3}}{\frac{C}{6}} = 4 \text{ hours}. \] So, the tank will be completely filled at 7:00 a.m. + 4 hours = 11:00 a.m.
Thus, the tank will be filled completely at 11:00 a.m.