Two trains A and B start from stations X and Y towards Y and X respectively. After passing each other, they take 4 hours 48 minutes and 3 hours 20 minutes to reach Y and X respectively. If train A is moving at 45 km/hr., then the speed of the train B is

Options

A. 60km/hr
B. 64.8 km/hr
C. 54km/hr
D. 37.5km/hr

Answer

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Speed of Train A (S_A) = 45 km/hr
- Time taken by Train A after passing Train B to reach station Y (T_A) = 4 hours 48 minutes
- Time taken by Train B after passing Train A to reach station X (T_B) = 3 hours 20 minutes

2. **Convert Time into Hours:**
- For Train A:
- 4 hours 48 minutes = 4 + (48/60) hours = 4 + 0.8 hours = 4.8 hours
- For Train B:
- 3 hours 20 minutes = 3 + (20/60) hours = 3 + (1/3) hours = 3.333 hours (or 10/3 hours)

3. **Use the Relation Between Speeds and Times:**
- The relationship between the speeds of the two trains and the times taken after they meet can be expressed as:
\[ \frac{S_A}{S_B} = \frac{T_B}{T_A} \]
- Substitute the known values:
\[ \frac{45}{S_B} = \frac{10/3}{4.8} \]

4. **Simplify the Right Side:**
- Convert 4.8 into a fraction:
\[ 4.8 = \frac{24}{5} \]
- Now substitute:
\[ \frac{10/3}{24/5} = \frac{10}{3} \times \frac{5}{24} = \frac{50}{72} = \frac{25}{36} \]

5. **Set Up the Equation:**
- Now we have:
\[ \frac{45}{S_B} = \frac{25}{36} \]

6. **Cross-Multiply to Solve for S_B:**
- Cross-multiplying gives:
\[ 45 \times 36 = 25 \times S_B \]
- Calculate \( 45 \times 36 \):
\[ 45 \times 36 = 1620 \]
- Thus, we have:
\[ 1620 = 25 \times S_B \]

7. **Solve for S_B:**
- Divide both sides by 25:
\[ S_B = \frac{1620}{25} = 64.8 \text{ km/hr} \]

### Final Answer:
The speed of Train B (S_B) is **64.8 km/hr**.