A train leaves station A at 6 AM and reaches station B at 10 AM on the same day. Another train leaves station B at 8 AM and reaches station A at 11:30 AM on the same day. The time at which the two trains meet is:
Answer & explanation
Correct answer: option 2
LCM of 4 and 3.5 = 28
Let distance = 28 km
SA = \(\frac{28}{4}\) = 7 km/hr
SB = \(\frac{28}{3.5}\) = 8 km/hr
From 6 AM to 8 AM, A runs alone, in first 2 hours it travels = 14 km
After 8 am: Relative speed = 7 + 8 = 15 km/hr
Time taken to meet after 8 am = \(\frac{Remaining\;Distance}{Relative\;speed}\) = \(\frac{14}{15}\) = 56 mins
So at ⇒ 8:56 AM both the trains meet.