Train A leaves station M at 8:25 AM and reaches station N at 3:25 PM on the same day. Train B leaves station N at 10:25 AM and reaches station M at 3:25 PM on the same day. Find the time when trains A and B meet.

Correct option is B

Given:

Train A:

Starts from M at 8:25 AM

Reaches N at 3:25 PM → Total time = 7 hours

Train B:

Starts from N at 10:25 AM

Reaches M at 3:25 PM → Total time = 5 hours

Find the meeting time.

Concept Used:

Relative speed and proportional distance concept.

Ratio of speeds = inverse ratio of travel times for the same distance.

Meeting occurs when the sum of distances covered equals the total distance.

Solution:

Let distance = D

A’s speed = D/7

B’s speed = D/5

If meeting occurs x hours after A starts:

​$$\frac{x}{7}D + \frac{x-2}{5}D = D \\ \ \\\frac{x}{7} + \frac{x-2}{5} = 1 \\ \ \\5x + 7(x - 2) = 35 \\ \ \\5x + 7x - 14 = 35 \\ \ \\12x = 49$$​​

x = $$\frac{49}{12}$$​ hours = 4 hours 5 minutes

Meeting time:

8 : 25 AM + 4 h 5m = 12 : 30 PM 

Alternate Solution: ​

Let Distance = LCM of 7, 5  = 35 

A's Speed = 5 , B's speed = 7 

Distance cover by A in 2 hours = 5 × 2 = 10 

Remaining distance = 35 - 10 = 25 

Relative speed(opposite direction) = 7 + 5 = 12 

Time taken to cover remaining distance; 

= $$\frac{25}{12} = 2 h \, 5 min.$$ 

Time at which both train meet = 10 : 25 AM + 2 h 5 min = 12 : 30 PM