Question

Every day, Jack arrives at the train station from work at 5pm. His wife leaves home in her car to meet him there exactly at 5pm, and drives him home. One day, Jack gets to the station an hour earlier than usual. How long was he walking. please write the solution also.

Answer 1

He total had 60+15 minutes to move but he reached before 30 minutes, so total time he moved was 60=15-30=45.
so he must have been walking foe 45 minutes.

Answer 2

Answer:

Jack was walking for a time of 45 minutes.

Step-by-step explanation:

The journey of Jack’s wife is always the same each day. She will always reach home after a specific time, say, t hours from the time she leaves her house.  

Suppose if Jack wife left the house at some ‘u’ hours and she takes t hours to reach the station to pick Jack.

So,  

                                  U:00 + t = 5:00 PM ---(1)

As Jack left one hour earlier from station, they will meet after Jack has traveled for ‘v’ hrs

So, Jack will meet his wife at 4:00 +v hrs

Since Jack’s wife has saved ½ hour in total journey (from and to), she must have saved ¼ hr in each way of the journey. So, she should have reached the point of meeting at,

                                  u + t - ¼   --(2)

Subtracting (2) from (1)

                                 U + t - ¼ = 4:00 + v  

                                          ¼ = 1 + v

                                          V = 3/4  hours = 45 mintues

Taking absolute value, we get 45 minutes.