Two cars X and Y start from places $A$ and $B$ respectively which are $700$ km apart at $9$ a.m. Both the cars run at an average speed of $60$ km/hr. Car X stops at $10$ a.m. and again starts at $11$ a.m. while the other car continues to run without stopping. The two cars cross each other at

### Concept & Managing Stoppage Time For moving bodies that stop at different intervals, establish a common synchronization point in time. Calculate the positions of both entities at that common time, find the remaining distance, and then apply standard relative speed formulas. ### Step-by-Step Solution * **Given:** Distance = $700$ km. Speed of X = $60$ km/hr. Speed of Y = $60$ km/hr. X stops from 10 a.m. to 11 a.m. * **Calculation:** Let's analyze the situation at 11 a.m., when both cars are finally moving simultaneously again. * Distance traveled by Car X from 9 a.m. to 11 a.m. (travels for only $1$ hour) = $60 \times 1 = 60$ km. * Distance traveled by Car Y from 9 a.m. to 11 a.m. (travels non-stop for $2$ hours) = $60 \times 2 = 120$ km. * Total distance covered by both cars by 11 a.m. = $60 + 120 = 180$ km. * Remaining distance = $700 - 180 = 520$ km. * From 11 a.m. onwards, both cars are moving towards each other. Relative speed = $60 + 60 = 120$ km/hr. * Time taken to cover the remaining $520$ km = $\frac{520}{120} = \frac{13}{3}$ hours = $4\frac{1}{3}$ hours. * $4\frac{1}{3}$ hours is exactly $4$ hours and $20$ minutes. * Adding this time to the synchronized 11 a.m. start: 11:00 a.m. + $4$ hours $20$ minutes = 3:20 p.m. ### Exam Strategy & Shortcut Calculate distance gap at 11 a.m.: $700 - (60 \times 1) - (60 \times 2) = 520$ km. Time to meet = $520 / 120 = 4$ hr $20$ min. Add to 11 a.m. to get 3:20 p.m. ### Common Pitfall A major pitfall is attempting to calculate the meeting time without pausing to calculate the independent distances traveled during Car X's 1-hour rest period. ### Final Answer Therefore, the correct answer is **$3 : 20$ p.m**.