IBPS Clerk Mains 29 Nov 2025 Paper

Concept:When a train crosses a platform, the total distance covered is the sum of the train's length and the platform's length.Time is calculated as Distance ÷ Speed.Use the given 1-hour difference to find this effective distance, then scale it by 10.Explanation:Let the length of each train be $L$ km.The platform length is $400$ m $= 0.4$ km.Speed of Train A $= Y$ km/hr.Speed of Train B $= Y + 10$ km/hr, since Train A is 10 km/hr slower.Time taken by Train A to cross the platform is $\dfrac{L + 0.4}{Y}$.Time taken by Train B to cross the platform is $\dfrac{L + 0.4}{Y + 10}$.Train A takes 1 hour longer than Train B, so:$\dfrac{L + 0.4}{Y} - \dfrac{L + 0.4}{Y + 10} = 1$$(L + 0.4)\left[\dfrac{1}{Y} - \dfrac{1}{Y + 10}\right] = 1$$(L + 0.4)\left[\dfrac{10}{Y(Y + 10)}\right] = 1$$L + 0.4 = \dfrac{Y(Y + 10)}{10}$Now, both the train length and platform length are multiplied by 10.So the new distance becomes $10(L + 0.4)$.New time taken by Train A $= \dfrac{10(L + 0.4)}{Y}$Substitute $L + 0.4 = \dfrac{Y(Y + 10)}{10}$:New time $= \dfrac{10 \times \frac{Y(Y + 10)}{10}}{Y} = \dfrac{Y(Y + 10)}{Y} = Y + 10$ hours.Answer:The new time taken by Train A is $(Y + 10)$ hours.Correct option: D. $(Y + 10)$ hours.