Two inlet pipes A and B can fill an empty cistern in 35 hours and 52.5 hours respectively while Pipe C is an outlet pipe that can drain the filled cistern in `17. 5` hours. When the cistern is full Pipe C is left open for an hour, then closed and Pipe A is opened for an hour, closed and Pipe B is opened for an hour. The process continues till the cistern is empty. How many hours will it take for the filled cistern to be emptied ?
(a)301
(b)313
(c)298
(d)315

Options

A. 301
B. 313
C. 298
D. 315

Answer

To solve the problem step by step, we will first determine the rates at which each pipe fills or empties the cistern, and then calculate how long it takes to empty the cistern based on the sequence of operations described.

### Step 1: Determine the rates of the pipes
1. **Pipe A** fills the cistern in 35 hours.
- Rate of Pipe A = \( \frac{1}{35} \) of the cistern per hour.

2. **Pipe B** fills the cistern in 52.5 hours.
- Rate of Pipe B = \( \frac{1}{52.5} \) of the cistern per hour.
- Converting 52.5 hours to a fraction: \( 52.5 = \frac{105}{2} \) hours.
- Rate of Pipe B = \( \frac{2}{105} \) of the cistern per hour.

3. **Pipe C** empties the cistern in 17.5 hours.
- Rate of Pipe C = \( \frac{1}{17.5} \) of the cistern per hour.
- Converting 17.5 hours to a fraction: \( 17.5 = \frac{35}{2} \) hours.
- Rate of Pipe C = \( \frac{2}{35} \) of the cistern per hour.

### Step 2: Calculate the net effect of the pipes over a cycle
The sequence of operations is:
- Pipe C is open for 1 hour.
- Pipe A is open for 1 hour.
- Pipe B is open for 1 hour.

**Net effect in one cycle (3 hours):**
1. **First hour (Pipe C)**:
- Empties \( \frac{2}{35} \) of the cistern.

2. **Second hour (Pipe A)**:
- Fills \( \frac{1}{35} \) of the cistern.

3. **Third hour (Pipe B)**:
- Fills \( \frac{2}{105} \) of the cistern.

### Step 3: Calculate the total amount emptied or filled in one cycle
- Total emptied by Pipe C in 1 hour: \( -\frac{2}{35} \)
- Total filled by Pipe A in 1 hour: \( +\frac{1}{35} \)
- Total filled by Pipe B in 1 hour: \( +\frac{2}{105} \)

**Finding a common denominator (105):**
- \( -\frac{2}{35} = -\frac{6}{105} \)
- \( +\frac{1}{35} = +\frac{3}{105} \)
- \( +\frac{2}{105} = +\frac{2}{105} \)

**Net effect in one cycle:**
\[ \text{Net effect} = -\frac{6}{105} + \frac{3}{105} + \frac{2}{105} = -\frac{1}{105} \]

This means that in one complete cycle of 3 hours, the cistern is emptied by \( \frac{1}{105} \) of its total capacity.

### Step 4: Calculate the total time to empty the cistern
Since the total capacity of the cistern is 1 (or 105 units), to empty the entire cistern:
\[ \text{Total cycles needed} = 105 \text{ cycles (since each cycle empties } \frac{1}{105} \text{ of the cistern)} \]

Each cycle takes 3 hours, so:
\[ \text{Total time} = 105 \text{ cycles} \times 3 \text{ hours/cycle} = 315 \text{ hours} \]

### Final Answer
The total time taken to empty the filled cistern is **315 hours**.