Train A leaves station M at 7:45 AM and reaches station N at 2:45 PM on the same day. Train B leaves station N at 9:45 AM and reaches station M at 2:45 PM on the same day. Find the time when Trains A and B meet.

Correct option is D

Given:

Train A: leaves M at 7:45 AM, reaches N at 2:45 PM
Time taken = 7 hours

Train B: leaves N at 9:45 AM, reaches M at 2:45 PM
Time taken = 5 hours

They travel on the same route, opposite directions.

Need: Time when they meet.

Concept Used:

Speeds are inversely proportional to times.

Solution :

Speed A : Speed B =$$\frac{1}{7} : \frac{1}{5} = 5 : 7$$​

Assume distance = LCM of 7 & 5 = 35 units

Speed of A = $$\frac{35}7 =$$​ 5 units/hr

Speed of B $$= \frac{35}5$$​= 7 units/hr

Time taken by A = t, by B = t − 2 (since B starts 2 hrs later)

Equation:

5t + 7(t - 2) = 35

5t + 7t - 14 = 35

12t = 49

t =$$\frac{49}{12} \approx 4 \, \text{hrs } 5 \, \text{min}$$​

A starts at 7:45 AM, so meeting time =

7 : 45 AM + 4h 5m = 11:50 AM