**Puzzle:**

Every day, Jack arrives at the train station from work at 5 pm.

His wife leaves home in her car to meet him there at exactly 5 pm, and drives him home. One day, Jack gets to the station an hour early, and starts walking home, until his wife meets him on the road. They get home 30 minutes earlier than usual. How long was he walking?

Distances are unspecified. Speeds are unspecified, but constant.

Give a number which represents the answer in minutes.

**Puzzle Solution:**

The best way to think about this problem is to consider it from the perspective of the wife. Her round trip was decreased by 30 minutes, which means each leg of her trip was decreased by 15 minutes.

Jack must have been walking for 45 minutes.

Hi, sorry can you explain it mathematically? For some reason, I just can’t piece the answer together? Thanks

The Wives journey is constant each day. She will always get home after time T from leaving her house.

Where T = Drive to the station + Drive from the station or T = 2 x Each leg for his wife

We are assuming if Jack is walking that the time to drive to the station will be less, and the return drive will be less (by the same amount). So if T is 30 minutes shorter, she has driven for 15 minutes less on either leg of the journey.

If she has driven for 15 minutes less, she has picked him up at 4.45pm. If his train arrived at 4pm, he must have been walking for 45 minutes.

had he not started walking he would have waited for 1 hour (60 mins) to meet his wife. but he started walking that reduced his wife trip from home to station by (30/2=) 15 mins or in other worlds you can say that jack met his wife 15 mins before. so he waited 15 mins less that means he spent (60 -15 =) 45 mins while walking.

let the wife left at Xhrs ad it took T hrs to reach train station to receive Jack

T hr

5:00 pm ———————- X:00 pm, So X:00+T hrs = 5:00 PM ——–(1)

Now he left train station at 4 pm, let they meet on the way after Jack has traveled for Y hrs

met at 4:00+Yhrs

4:00 pm ——————x——— here x is the meeting point

since his wife has saved in total 30 min, so she saved 15min=0.25 hrs in one leg

so, she should reach point x at X+T-0.25 hrs

thus we have X+T-0.25hrs=4:00+Yhrs —————(2)

from (1) – (2)

0.25=1:00+Y hrs

Y hrs=0.75 hrs= 45 min

she caught him 15 mins earlier ( 30 mins

divide by 2 for round trip ) than expected . Expected time to meet him is 5.pm , so she met her husbadd at 4.45 pm. which means husband who reached station at 4pm ( 1 hour before ) walked for 45 minutes

Not good

let consider jack’s wife start from home at X p.m and she takes Y mints to reach to station.(at 5p.m according to the question).so jack returns back to home at (5+Y) p.m. so (5+Y=X+2Y). i.e

X+Y=5;on that day jack return back to home at (X+2Y-30) p.m and his wife took Z mints to meet him.so X+2Y-30=X+2Z ;i.e Y-Z=15;jack was walking for(( X-4)+Z) mints (as jack got to the station at 4p.m).so (X-4)+Z=(5-Y-4)+Z=1(hr)-(Y-Z)=(60-15)=45 mints.