Between two railway staions the fare of `1^(st), 2^(nd) and 3^(rd)` class are in the ratio 8:6:3. But later on the fare of `1^(st) and 2^(nd)` class is decreased by `(1)/(6) and (1)/(12)` spectively. In one year the no. of passengers of 1, 2nd and 3rd class are in the ratio for 4:9:24. If total tickets amounting Rs 355600 are sold then find the fare of all 3rd class.

Options

A. 6400
B. 170800
C. 18800
D. 172800

Answer

To solve the problem step by step, we will follow the given information and perform calculations accordingly.

### Step 1: Determine the original fares based on the ratio
The fares of the 1st, 2nd, and 3rd class are in the ratio 8:6:3. Let's assume the common multiple is \( x \).

- Fare of 1st class = \( 8x \)
- Fare of 2nd class = \( 6x \)
- Fare of 3rd class = \( 3x \)

### Step 2: Calculate the new fares after the decrease
The fare of the 1st class is decreased by \( \frac{1}{6} \) and the fare of the 2nd class is decreased by \( \frac{1}{12} \).

- New fare of 1st class:
\[ \text{New fare of 1st class} = 8x - \frac{1}{6} \times 8x = 8x \left(1 - \frac{1}{6}\right) = 8x \times \frac{5}{6} = \frac{40x}{6} = \frac{20x}{3} \]

- New fare of 2nd class:
\[ \text{New fare of 2nd class} = 6x - \frac{1}{12} \times 6x = 6x \left(1 - \frac{1}{12}\right) = 6x \times \frac{11}{12} = \frac{66x}{12} = \frac{11x}{2} \]

- The fare of 3rd class remains the same:
\[ \text{Fare of 3rd class} = 3x \]

### Step 3: Write the new fare ratio
Now we have the new fares:
- New fare of 1st class = \( \frac{20x}{3} \)
- New fare of 2nd class = \( \frac{11x}{2} \)
- Fare of 3rd class = \( 3x \)

To find a common ratio, we can express all fares with a common denominator. The least common multiple of 3 and 2 is 6.

- New fare of 1st class:
\[ \frac{20x}{3} = \frac{40x}{6} \]

- New fare of 2nd class:
\[ \frac{11x}{2} = \frac{33x}{6} \]

- Fare of 3rd class:
\[ 3x = \frac{18x}{6} \]

Thus, the new fare ratio becomes:
\[ 40 : 33 : 18 \]

### Step 4: Determine the number of passengers
The number of passengers for 1st, 2nd, and 3rd class are in the ratio 4:9:24. Let's assume the number of passengers for each class is \( 4y, 9y, 24y \).

### Step 5: Calculate total revenue
The total revenue from ticket sales is given as Rs 355600. The revenue from each class can be calculated as follows:

- Revenue from 1st class:
\[ \text{Revenue from 1st class} = \text{Fare of 1st class} \times \text{Number of passengers} = \frac{40x}{6} \times 4y = \frac{160xy}{6} \]

- Revenue from 2nd class:
\[ \text{Revenue from 2nd class} = \frac{33x}{6} \times 9y = \frac{297xy}{6} \]

- Revenue from 3rd class:
\[ \text{Revenue from 3rd class} = \frac{18x}{6} \times 24y = \frac{432xy}{6} \]

### Step 6: Total revenue equation
Now, we can write the total revenue equation:
\[ \frac{160xy}{6} + \frac{297xy}{6} + \frac{432xy}{6} = 355600 \]

Combining the terms:
\[ \frac{(160 + 297 + 432)xy}{6} = 355600 \]
\[ \frac{889xy}{6} = 355600 \]

### Step 7: Solve for \( xy \)
Multiply both sides by 6:
\[ 889xy = 2133600 \]
Now, divide by 889:
\[ xy = \frac{2133600}{889} \approx 2400 \]

### Step 8: Find the fare of 3rd class
Now we can find the fare of the 3rd class:
\[ \text{Fare of 3rd class} = 3x \]
To find \( x \), we can use the value of \( y \):
\[ xy = 2400 \implies x = \frac{2400}{y} \]

Using the total revenue:
\[ \text{Revenue from 3rd class} = \frac{432xy}{6} = \frac{432 \times 2400}{6} = 432 \times 400 = 172800 \]

### Conclusion
The fare of the 3rd class can be calculated as:
\[ \text{Fare of 3rd class} = 3x = 3 \times \frac{2400}{y} \]
To find the exact fare, we can calculate \( 3x \) based on the total revenue.

### Final Answer
The fare of the 3rd class is \( Rs 432 \).