Walking home from station

**LAF** · Jan23-09, 11:16 AM

Everyday a man comes from work and get out the train, at station, at 5 PM. At the same time, everyday, his wife, driving a car, takes him and they go home.
One day, the man leaves his work early and get on the sation at 4 PM. So, he decides to go home walking. On the way home, he meets his wife, get in the car and they go home. This day, they arrive at home 10 minutes early. How many time has he walked?

**DaveC426913** · Jan23-09, 12:52 PM

I think this diagram captures the essentials of what's given and what's being asked.

It shows that there are some assumptions that must be made in order to answer the question.

One assumption is that the wife's drive to the subway is as fast (slope of a) as the drive from the subway (slope of -a). Her speed to the subway affects how far home he gets before meeting her. To demonstrate this requirement, change the slope of the dotted line -the lengths of the blue lines will change.

We must know what her to-subway speed is; we can assume it's the same as to-home.

Another assumption: his wife did not know he was leaving early today, therefore she left at the same time as usual, right?

**jimmysnyder** · Jan23-09, 01:10 PM

Spoiler

Look at it from her point of view. If she gets home 10 minutes earlier than usual, then she must have met him 5 minutes earlier than usual. That would be 4:55, after he had walked for 55 minutes.

**LAF** · Jan23-09, 01:15 PM

I think it´s not important the different speed and distance anyway (to n from subway).
You right, his wife didn´t know!

**LAF** · Jan23-09, 01:18 PM

jimmysnyder got it. He was very fast!

**DaveC426913** · Jan23-09, 02:13 PM

Originally Posted by LAF

I think it´s not important the different speed and distance anyway (to n from subway).
You right, his wife didn´t know!

Jimmy's answer is quite elegant. However, it does make the two assumptions I stated.

(You can't conclude that the five minutes she saved are split evenly between coming and going unless you first assume that her speed is split evenly between coming and going.)

**ssilver0210** · Feb23-09, 01:29 AM

We must assume that the wife left at the time that she normally leaves, because she had no idea that the husband had arrived early. We know that the wife and husband arrived home 10 minutes earlier than usual, even though the wife left home at the regular time. For them to arrive home 10 minutes early, the wife would have had to have met up with her husband 5 minutes earlier than the time she usually meets him. (5pm). So, she met up with her husband at 4:55. 10 minutes would be saved because the wife would save the 5 minutes that would have been required to drive from the point she met her husband to the point of the station, as well as the five minutes back from the point of the station to the point she met her husband.

So, the husband usually arrives at 5pm. Today he arrives at 4pm. If the husband and wife arrive home 10 minutes earlier than usual, then the wife will meet up with the husband at 4:55, rather than 5pm. We know that the husband began his walk at 4pm, so the length of time the husband has been walking is 55 minutes.

It does not matter which numbers you plug in for the time, the result is always the same. If husband arrives usually at 4:30pm, then that means today he arrived at 3:30 pm. Wife needs to meet up with him at 4:25 pm to save the 10 minutes back and forth from point of husband to point of station. 3:30-4:25=50 minutes of walking for husband.