Could someone explain the solution (or provide an alternative one), seeing as I don't quite understand it?
Mrs. Fastdrive met her husband at the station with the family car promptly at 5.00 pm every day, averaging 70 km/h each way. One day Mr. Fastdrive, without telling his wife, caught an earlier train which arrived at 4.00 pm and started walking home. Mrs. Fastdrive picked him up part way and they got home fifteen minutes earlier than usual. How fast did Mr. Fastdrive walk.
The solution given in the book is:
Mrs. Fastdrive saved 15 minutes, i.e. 7.5 minutes each way. At 70 km/h this represents 17.5 km, i.e. 8.75 km each way.
Thus Mr. Fastdrive walked 8.75 km in 52.5 minutes, a rate of
2 Answers | Add Yours
on the particular day he saved 15min. That mean if we traveled it by car he saved 15/60*70 = 17.5km. This travelling of 17.5km accounts for both sides of the journey. so for one side he has saved 17.5/2 = 8.75km. Each way he has saved 15/2 = 7.5 minutes.
After arriving the station one hour(60min) earlier he has saved 7.5 minutes. This means he has walked for (60-7.5) = 52.5minues.
In this 52.5 minutes he has walked 8.75km.
So the speed of walk
= 8.75/52.5*60 km/h
here we have to assume that no time losses occurs such as getting out of the station, get in to the car.
I think the solution given in the book is an appropriate one. To explain the solution we can proceed as under:
Let the meeting point be A and it is x kilometers short of station.
Mrs. Fastdrives saves 15 minutes by not driving the car from point A to station and back because Mr. Fastdrive has reached point A on foot. The distance saved by Mrs. Fastdrive is 2x (not going from point A to station and back) which would have taken 15 minutes if she had to pick Mr. Fastdrive from station.
Time to travel 2x km = 15 minutes
Time for x km = 15/2 = 7.5 minutes.
At 70km/h this distance = 70/60*7.5 = 8.75km
Had Mr. Fastdrive not started walking towards home, Mrs. Fastdrive would have reached the station at 5pm but because of his walking she saved 7.5 minutes and reached point A at 4hours 52.5 minutes in the afternoon.
Mr. Fastdrive reaches at station at 4pm and walks till 4H52.5m in the evening thus the total walking time is 52.5 minutes in which it covers 8.75km which is distance of point A from station.
Hence the walking speed of Mr. Fastdrive was 8.75/52.8*60 = 10km/h
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