If a train takes 1.75 sec to cross a telegraphic post and 1.5 sec to overtake a cyclist racing along a road parallel to the track @ 10 metres per second, then the length of the train is

### Concept & Relative Speed Anomaly Usually, "overtake" implies moving in the same direction. However, if the time taken to cross a moving object (1.5s) is *less* than the time taken to cross a stationary object (1.75s), the relative speed must be higher than the train's speed. This means the cyclist is actually moving in the *opposite* direction, despite the word "overtake". $$Length = Speed \times Time$$ ### Step-by-Step Solution 1. Let the length of the train be $L$ meters and its speed be $v$ m/s. 2. **Stationary Post:** - Time = 1.75 s. - $L = v \times 1.75 \implies L = 1.75v$ 3. **Cyclist:** - Cyclist speed = 10 m/s. Time = 1.5 s. - Because 1.5s < 1.75s, relative speed must be $v + 10$ (opposite direction). - $L = (v + 10) \times 1.5 \implies L = 1.5v + 15$ 4. **Equate and Solve:** - $1.75v = 1.5v + 15$ - $0.25v = 15$ - $v = \frac{15}{0.25} = 60$ m/s 5. **Find Length:** - $L = 1.75 \times 60 = 105$ meters. ### Exam Strategy & Shortcut Set up a quick ratio. $L/1.75 = v$ and $L/1.5 = v + 10$. Subtracting them gives $L(1/1.5 - 1/1.75) = 10$. $L(\frac{1.75 - 1.5}{1.5 \times 1.75}) = 10 \implies L(\frac{0.25}{2.625}) = 10 \implies L = 105$ m. ### Common Pitfall Taking "overtake" strictly as same-direction and formulating the equation $L = (v - 10) \times 1.5$. This results in a negative speed and impossible geometry. You must infer the physical reality from the time variables given. ### Final Answer Therefore, the correct answer is **105 m**.