A train $M$ leaves station $X$ at 5 a.m and reaches station $Y$ at 9 a.m. Another train $N$ leaves station $Y$ at 7 a.m. and reaches station $X$ at 10.30 a.m. At what time do the two trains cross each other?

### Concept & Proportional Speed When exact distances are unknown but time intervals are given, we can assume the total distance to be a convenient common multiple (LCM) of the given times to find proportional speeds, or use variables. The crossing point depends on relative speed from a common starting time. ### Step-by-Step Solution 1. **Determine Travel Times:** Train M travel time = 5 a.m. to 9 a.m. = $4 \text{ hours}$. Train N travel time = 7 a.m. to 10:30 a.m. = $3.5 \text{ hours} = \frac{7}{2} \text{ hours}$. 2. **Assume Distance & Find Speeds:** Let the total distance between X and Y be $D$. Speed of Train M ($S_M$) = $\frac{D}{4}$ km/hr. Speed of Train N ($S_N$) = $\frac{D}{3.5} = \frac{2D}{7}$ km/hr. 3. **Synchronize Start Times:** Train N starts at 7 a.m. By 7 a.m., Train M has already travelled for 2 hours (from 5 a.m.). Distance covered by Train M in 2 hours = $2 \times \frac{D}{4} = \frac{D}{2}$. Remaining distance between them at 7 a.m. = $D - \frac{D}{2} = \frac{D}{2}$. 4. **Calculate Crossing Time:** They travel towards each other, so relative speed is the sum. Relative Speed = $S_M + S_N = \frac{D}{4} + \frac{2D}{7} = \frac{7D + 8D}{28} = \frac{15D}{28}$ km/hr. Time taken to cross = $\frac{\text{Remaining Distance}}{\text{Relative Speed}} = \frac{\frac{D}{2}}{\frac{15D}{28}} = \frac{D}{2} \times \frac{28}{15D} = \frac{14}{15} \text{ hours}$. 5. **Convert to Minutes:** $\frac{14}{15} \times 60 \text{ minutes} = 14 \times 4 = 56 \text{ minutes}$. Time of crossing = 7:00 a.m. + 56 minutes = 7:56 a.m. ### Exam Strategy & Shortcut Let Distance = LCM of 4 and 3.5. Let's use 28 km. Speed M = $\frac{28}{4} = 7$ km/h. Speed N = $\frac{28}{3.5} = 8$ km/h. At 7 a.m., M has travelled for 2 hrs = 14 km. Distance left = $28 - 14 = 14$ km. Relative speed = $7 + 8 = 15$ km/h. Time = $\frac{14}{15}$ hours = 56 mins. Meeting time = 7:56 a.m. This completely bypasses algebraic variables! ### Common Pitfall Calculating the relative speed from 5 a.m. without accounting for the fact that Train N hasn't started moving yet. Always align both moving bodies to a common time before applying relative speed. ### Final Answer Therefore, the correct answer is **7.56 a.m**.