A train which is moving at an average speed of 40km/h reaches its destination on time. When its average speed reduces to 35 km/h, then it reaches its destination 15 minutes late. The distance travelled by the train, is [CLAT Exam, 2016]

### Concept & Formula When a fixed distance is covered at two different speeds resulting in a time difference $\Delta t$, the distance $D$ can be found using the formula: $$ D = \frac{S_1 \times S_2}{S_1 - S_2} \times \Delta t $$ ### Step-by-Step Solution 1. **Given Data:** - Speed 1 ($S_1$) = 40 km/h. - Speed 2 ($S_2$) = 35 km/h. - Time difference ($\Delta t$) = 15 minutes = $\frac{15}{60} = \frac{1}{4}$ hours. 2. **Apply Distance Formula:** - $D = \frac{40 \times 35}{40 - 35} \times \frac{1}{4}$ - $D = \frac{1400}{5} \times \frac{1}{4}$ - $D = 280 \times \frac{1}{4} = 70$ km. ### Exam Strategy & Shortcut Memorize the product-of-speeds formula for constant distance problems. It bypasses setting up complex algebraic fractions entirely and yields the answer in seconds. ### Common Pitfall Forgetting to convert the time difference from minutes to hours ($\frac{15}{60}$) before applying it to speeds given in km/h. ### Final Answer Therefore, the correct answer is **70km**.