A train passes two bridges of lengths 500 m and 250 m in 100 seconds and 60 seconds respectively. The length of the train is [SSC—CHSL (10+2) Exam, 2015]

### Concept & Logic The speed of a train remains constant whether it crosses one bridge or another. The speed is equal to the total distance (train length + bridge length) divided by the time taken. $$ \text{Speed} = \frac{L + \text{Bridge}_1}{t_1} = \frac{L + \text{Bridge}_2}{t_2} $$ ### Step-by-Step Solution 1. **Set Up Equation:** - Let train length be $L$. - $\frac{L + 500}{100} = \frac{L + 250}{60}$ 2. **Simplify and Solve:** - Cross-multiply: $60(L + 500) = 100(L + 250)$ - Divide by 20 to simplify: $3(L + 500) = 5(L + 250)$ - $3L + 1500 = 5L + 1250$ - $2L = 250$ - $L = 125$ m. ### Exam Strategy & Shortcut Reduce the time ratio first ($100 : 60 = 5 : 3$). Since speeds are equal, the distance ratios must also balance, making quick work of the linear equation. ### Common Pitfall Forgetting to add the train's length ($L$) to both bridge lengths when setting up the speed equivalence, treating the bridges as standalone distances. ### Final Answer Therefore, the correct answer is **125 m**.