Train A crosses a platform of length $230$ meters in $29$ seconds, and Train B crosses a platform of length $150$ meters in $24$ seconds. Train B, which is $450$ meters long, overtakes Train A in $160$ seconds while both are moving in the same direction. Given that the speed of Train B is greater than Train A, find the time taken by Train A to cross a bridge of length $50$ meters.

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**Solution:**

1. Let the length of Train A be $l$ meters and its speed be $S$ m/s.

2. Calculate speed of Train B:
   - Train B crosses a $150$ m platform in $24$ seconds.
   - Speed of Train B = $\frac{450 + 150}{24} = \frac{600}{24} = 25$ m/s.

3. Express speed of Train A using the platform crossing data:
   - Speed of Train A = $\frac{l + 230}{29}$ m/s  ... (i)

4. When Train B overtakes Train A running in the same direction, relative speed = speed of B - speed of A.
   - Distance to be covered = sum of lengths = $450 + l$ meters.
   - Time taken = $160$ seconds.
   - So, relative speed = $\frac{450 + l}{160}$ m/s.

5. Set up equation for relative speed:
   $$25 - S = \frac{450 + l}{160}$$
   Substitute $S$ from (i):
   $$25 - \frac{l + 230}{29} = \frac{450 + l}{160}$$

6. Solve for $l$:
   Multiply both sides by $29 	imes 160$ to clear denominators:
   $$ (25 	imes 29 	imes 160) - (l + 230) 	imes 160 = (450 + l) 	imes 29 $$

Calculate constants:
   $$25 	imes 29 	imes 160 = 116000$$

So,
   $$116000 - 160l - 36800 = 13050 + 29l$$

Simplify:
   $$116000 - 36800 - 13050 = 160l + 29l$$
   $$66250 = 189l$$

$$l = \frac{66250}{189} \approx 350.79 \approx 350 	ext{ meters}$$

7. Calculate speed of Train A:
   $$S = \frac{350 + 230}{29} = \frac{580}{29} = 20 	ext{ m/s}$$

8. Time taken by Train A to cross a $50$ m bridge:
   - Total distance = length of Train A + length of bridge = $350 + 50 = 400$ meters.
   - Time = $\frac{400}{20} = 20$ seconds.

**Answer:** $20$ seconds

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