Direction: In the following question, two quantities I and II are given. Compare quantity I and quantity II on its basis. (Only quantity is to be considered) Two trains A and B are running in the same direction and they cross a 75 m long platform in 25 seconds and 15 seconds respectively. Speed of train B is 25.2 km/h more than that of train A and the length of train A is 45 m more than that of train B. Quantity I: If both the trains were running in opposite directions, then in what time would they cross each other? Quantity II: If the speed of train B were 3 m/s more than its original speed, then at what time would train B cross a 95 m long tunnel?

### Concept & Relative Speed When dealing with trains crossing platforms or other trains, use the fundamental formula linking distance, speed, and time. $$Time = \frac{Total Distance}{Relative Speed}$$ For opposite directions, relative speed is the sum of their speeds; for the same direction, it's the difference. Convert all units to meters and seconds. ### Step-by-Step Solution * **Convert Units:** The speed difference is $25.2 \text{ km/h}$. Convert to m/s: $25.2 \times \frac{5}{18} = 7 \text{ m/s}$. * Let $v_A$ be the speed of Train A. Then $v_B = v_A + 7$. * Let $L_B$ be the length of Train B. Then $L_A = L_B + 45$. * **Set up Equations:** * Train A crosses a 75m platform in 25s: $\frac{L_A + 75}{v_A} = 25 \Rightarrow L_A + 75 = 25v_A$. * Train B crosses a 75m platform in 15s: $\frac{L_B + 75}{v_B} = 15 \Rightarrow L_B + 75 = 15v_B$. * **Solve for Variables:** Substitute $L_A$ and $v_B$ into the equations: * $(L_B + 45) + 75 = 25v_A \Rightarrow L_B + 120 = 25v_A \Rightarrow L_B = 25v_A - 120$. * $L_B + 75 = 15(v_A + 7) \Rightarrow L_B = 15v_A + 105 - 75 \Rightarrow L_B = 15v_A + 30$. * Equate both expressions for $L_B$: $25v_A - 120 = 15v_A + 30$. * $10v_A = 150 \Rightarrow v_A = 15 \text{ m/s}$. * Then, $v_B = 15 + 7 = 22 \text{ m/s}$. * $L_B = 15(15) + 30 = 255 \text{ m}$. * $L_A = 255 + 45 = 300 \text{ m}$. * **Calculate Quantity I:** Trains crossing each other in opposite directions. * $\text{Time} = \frac{L_A + L_B}{v_A + v_B} = \frac{300 + 255}{15 + 22} = \frac{555}{37} = 15 \text{ seconds}$. * **Calculate Quantity II:** Train B with speed increased by $3 \text{ m/s}$ ($22 + 3 = 25 \text{ m/s}$) crossing a $95 \text{ m}$ tunnel. * $\text{Time} = \frac{L_B + 95}{25} = \frac{255 + 95}{25} = \frac{350}{25} = 14 \text{ seconds}$. * **Compare:** Quantity I ($15$) > Quantity II ($14$). ### Exam Strategy & Shortcut Whenever a problem involves "speed difference in km/h", immediately convert it to m/s before reading further. This prevents unit mismatch errors deep into the calculation. Also, set up a linear system of equations quickly by isolating the shared variable ($L_B$ in this case). ### Common Pitfall A classic mistake is forgetting to add the platform's length to the train's length when calculating the total distance covered. ### Final Answer Therefore, the correct answer is **Quantity I > Quantity II**.