A train leaves station A at 8 am and reaches station B at 12 noon. A car leaves station B at 8:30 am and reaches station A at the same time when the train reaches station B. At what time do they meet?

10:08 am

Time taken by the first train = 4 hours

Time taken by the second train = 3.5 hours

Ratio of time taken by the two trains = 8 : 7

As we know, the ratio of speed will be the opposite of the ratio of time taken

So ratio of speed = 7 : 8

Let The distance traveled by the two trains = 4 × 7 = 28 km

Distance = (Speed × time)

In first 30 minutes . Only 1st traon travels

Thus, the distance travelled by the first train in 30 min = 7 × 0.5 = 3.5 km

Distance remaining = 28 - 3.5 = 24.5 km

Now both the trains are moving towards each other and the distance is 24.5 km

Relative speed = 7 + 8 = 15

(When two bodies move in the opposite directions, then the Relative Speed = Sum of Speeds)

Thus, the time of meeting = \(\frac{24.5}{15}\) 

= 1 hour 38 minutes 

So , meeting time is 10:08 am