Earlier Than Expected

Algebra Level pending

A man plans on arriving by train at the station at 3:00 p.m. and to have his wife drive him home. Instead, he catches an earlier train and arrives at the station at 2:00 p.m. He decides to walk towards home and is picked up by his punctual wife en route. They arrive home 20 minutes earlier than planned. How long did he walk? Assume that there is no time lapse at the meeting place Also assume that wife starts on her drive to the station after 2:00 pm.

(This is from a problem solving book which didn't give any solutions to the listed problems but listed them as challenges For anyone who wants to checks it out, the source is this: Adventures In Problem Solving By Shailesh Shirali)

40 minutes 15 minutes 30 minutes 50 minutes

Sachetan Debray by Sachetan Debray , 16, India

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2 solutions

Matthew Burr
Jun 2, 2020

Let t 0 t_0 be the time at which the wife departs towards the train station,

let t 1 t_1 be the planned pickup time (3:00pm),

and let t p t_p be the actual pickup time.

The time it takes to drive from the train station to home is then t 1 − t 0 t_1-t_0 , and so the expected time of arrival is t 1 + ( t 1 − t 0 ) = 2 t 1 − t 0 t_1+(t_1-t_0)=2t_1-t_0 . The couple arrives home 20 minutes early, so the actual time of arrival, T T , is T = 2 t 1 − t 0 − 20 T=2t_1-t_0-20 .

Additionally, the time it takes to drive from the actual pickup location to home is t p − t 0 t_p-t_0 , and so the actual time of arrival is T = t p + ( t p − t 0 ) = 2 t p − t 0 T=t_p+(t_p-t_0)=2t_p-t_0 .

Equating these two equations for the actual arrival time, we get 2 t 1 − t 0 − 20 = 2 t p − t 0 ⟹ t p = t 1 − 10 2t_1-t_0-20=2t_p-t_0 \implies t_p=t_1-10 . That is, the husband was picked up 10 minutes before 3:00pm, at 2:50pm, and thus he walked for 50 minutes.

0 Helpful 0 Interesting 1 Brilliant 0 Confused

Nice! This is a neat solution. all the variables for time assumed are measured from 2:00 p.m, right?

Sachetan Debray - 1 year ago
Sachetan Debray
May 28, 2020

This problem has a lot of variables involved, but the answer can be found using the given information, by manipulating the conditions.

let's assume the distance between the train station and the home is 'x' m.

Now let's assume the wife started out on her pick-up journey 't' minutes before 3:00 pm(that means she started out (60-t) minutes after 2:00 p.m.-this bit of information will be crucial later.)

Now she had planned to bring the man back home by ′ 2 t ′ '2t' minutes after she had started.

But she met the man when she had only traveled a distance, say 'y' m, so she traveled to their unexpected meeting-place in y t x \frac{yt}{x} minutes, and brought him back by twice of that time period.

So lets write

expected time-real time =20 minutes

this means

2 t − 2 y t x = 20 2t - \frac{2yt}{x} = 20

or

t ( 1 − y x ) = 10 t(1-\frac{y}{x})=10 - (equation 1)

Now, this is the fun part. The man obviously walked for some time before reaching the real meeting-place. He started from 2:00 pm. Measuring from 2:00 p.m the woman would have taken the same time to reach the real meeting-place as well.

So the woman started out '60-t' minutes after 2 p.m. Then she took y t x \frac {yt}{x} minutes to walk to the meeting-place.

This means that the man walked for

( 60 − t ) + y t x (60 - t) + \frac {yt}{x} minutes in total

this can be written as

60 − ( t ( 1 − y x ) ) 60- ( t(1-\frac{y}{x}) )

from equation 1 this becomes

60 − 10 = 50 60-10=50

Thus the man walks for 50 minutes.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

So one needs to keep track of what one has to find without getting fazed by the variables, and manipulate them in a clever way.

Although this approach is lengthy, it pays off in the end.

Sachetan Debray - 1 year ago

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