Two trains X and Y start from stations A and B towards B and A respectively. After passing each other, they take 4 hours 48 minutes and 3 hours 20 minutes to reach B and A respectively. If train X is moving at 40 km/hr., the speed of train Y is :

Options

A. 60km/hr.
B. 54km/hr.
C. 64.8 km/hr
D. 48 km/hr.

Answer

To solve the problem, we need to find the speed of train Y given the speed of train X and the time taken by each train to reach their respective destinations after they meet.

### Step-by-Step Solution:

1. **Convert Time to Minutes:**
- Train X takes 4 hours 48 minutes to reach station B after meeting.
- Convert this time to minutes:
\[ 4 \text{ hours} = 4 \times 60 = 240 \text{ minutes} \]
\[ 48 \text{ minutes} = 48 \text{ minutes} \]
\[ \text{Total time for Train X} = 240 + 48 = 288 \text{ minutes} \]

- Train Y takes 3 hours 20 minutes to reach station A after meeting.
- Convert this time to minutes:
\[ 3 \text{ hours} = 3 \times 60 = 180 \text{ minutes} \]
\[ 20 \text{ minutes} = 20 \text{ minutes} \]
\[ \text{Total time for Train Y} = 180 + 20 = 200 \text{ minutes} \]

2. **Determine the Speed Ratio:**
- The speed of train X (S1) is given as 40 km/hr.
- Let the speed of train Y be S2.
- The ratio of the speeds of the two trains is inversely proportional to the time taken to reach their respective destinations after they meet:
\[ \frac{S1}{S2} = \frac{T2}{T1} \]
- Here, T1 is the time taken by train X (288 minutes) and T2 is the time taken by train Y (200 minutes):
\[ \frac{40}{S2} = \frac{200}{288} \]

3. **Cross-Multiply to Solve for S2:**
- Cross-multiplying gives:
\[ 40 \times 288 = 200 \times S2 \]
- Calculate \(40 \times 288\):
\[ 40 \times 288 = 11520 \]
- Now, we have:
\[ 11520 = 200 \times S2 \]
- Divide both sides by 200 to find S2:
\[ S2 = \frac{11520}{200} = 57.6 \text{ km/hr} \]

4. **Final Answer:**
- The speed of train Y is 57.6 km/hr.