Ray who lives in the countryside, caught a train for home earlier than usual yesterday.His wife normally drives to the station to meet him. But yesterday he set out on foot from the station to meet his wife on the way.He reached home 12 mins earlier than he would have done had he waited at the station for his wife. The car travels at uniform speed which is 5 times Ray's speed on foot. Ray reached home exactly at 6' o clock. At what time would he have reached home if his wife, forewarned of his plan, had met him at the station.

A. 5:48
B. 5:42
C. 5:00
D. 5:36

I got 5.36

take: a=distance covered by ray's walk
x=ray walk's speed
5x=car's speed
12 mins is the time needed by the car to cover twice a (car met ray in the middle)
2a=5x.12
a=30x -->ray walk for 30 mins

If car meet ray 30mins earlier in station then for a, car only need t=30x/5x=6 mins
Time saved is 30 mins minus 6 mins which is 24 mins
so instead of 6, he could arrive at 5.36

OA please

Can someone please explain the solution? Bunuel?

why did you add 6min in the end to make the total to 36min? Bunuel pls explain

Ray who lives in the countryside, caught a train for home earlier than usual yesterday. His wife normally drives to the station to meet him. But yesterday he set out on foot from the station to meet his wife on the way. He reached home 12 mins earlier than he would have done had he waited at the station for his wife. The car travels at uniform speed which is 5 times Ray's speed on foot. Ray reached home exactly at 6' o clock. At what time would he have reached home if his wife, forewarned of his plan, had met him at the station.

As they "reached home 12 mins earlier than he would have done had he waited at the station for his wife" means that usually they reach home at 6:12 (6:00+12min). Now if we knew how many minutes earlier did Ray arrive at the station than 6:12 minus that time would be the time he have reached home if his wife, forewarned of his plan, had met him at the station.

As they arrived 12 minutes earlier than usual, wife saved 12 minutes on round trip from home to station (home-station-home) --> 6 minutes in each direction (home-station) --> so wife met husband on the road 6 minutes earlier than the usual time of their meeting on the station.

As wife met husband 6 minutes earlier than the usual time of their meeting --> so distance usually covered by wife in 6 minutes was covered by Roy --> as rate of wife is 5 times that of Roy's then Roy spent 5*6=30 minutes covering that distance. So Roy arrived 30 minutes + 6 minutes = 36 minutes earlier than usually.

6:12 - 36 minutes = 5:36.

Answer: D.


Hi Bunuel,
I still don't get why 6 mins are added to 30 mins (that Ray walked) ?

Let's assume the wife takes 1 hr to drive from home to station.
Normal Scenario:
Wife leaves home at 4:12 pm, reaches station at 5:12 pm (picks up Ray) and then reaches home at 6:12 pm.

Early arrival Scenario:
Wife leaves home at 4:12 pm, reaches mid-way point at 5:06 pm (Picks up Ray), and then reaches home at 6:00 pm. (6 mins saved on each trip. 12 mins total)

Ray has walked for 30 mins. So, Ray arrived at the station at 5:12 - 30 mins = 4:42 pm.

So,
Wife (now knowing that Ray is taking the earlier train)
leaves home at 3:42 pm, reaches station at 4:42 pm (Picks up Ray), reaches home at 5:42 pm.

Please explain.

Regards,
Nikhil.