Train A leaves station M at 6:20 AM and reaches station N at 3:20 PM on the same day. Train B leaves station N at 8:20 AM and reaches station M at 2:20 PM on the same day. Find the time when Trains A and B meet.

Correct option is C

Given:

Train A travels from M to N in 9:00 hours (from 6:20 AM to 3:20 PM)

Train B travels from N to M in 6:00 hours (from 8:20 AM to 2:20 PM)

Formula Used: 

Speed = $$\frac{\text{Distance}}{\text{Time}}$$

Solution:

Let the total distance between M and N be LCM(9, 6) = 18 units 

Speed of Train A = $$\frac{18}{9}$$ = 2 units/hour

Speed of Train B = $$\frac{18}{6}$$ = 3 units/hour

Since Train A starts at 6:20 AM and B at 8:20 AM, 

Distance covered by A by 8:20 AM = 2 × 2 = 4 units

Remaining distance = 18 − 4 = 14 units

Relative speed = 2 + 3 = 5 units/hour

Time to meet = $$\frac{14}{5} \approx 2.8 \text{ hours}$$ = 2 hours 48 minutes

Add 2 hours 48 minutes to 8:20 AM = 11:08 AM