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Loading ...Received this in the mail not too long ago: Every day a man arrives at Central Station and a car is waiting to give him a ride to his office. One day, however, the man arrives 1 hour early, so he starts walking toward the office. At some point on his way, the car picks him up, and he arrives 20 minutes early to the office.
How much time did the man spend walking?
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I find the brain improvement articles interesting, but the puzzles have to be the main reason.
Either I’m missing something on the “How much time walking?” or something else is missing.
Let’s say he normally arrives at 12:00. Which means he arrived at 11:00 this time. If the car averages 1 hour at 30 mph (and always averages 30mph at everypoint of the trip), he arrives at 1:00, 1 hour and 30 miles later.
Let’s say he walks 3 mph. He gets to the office 20 minutes early, so say he got there at 12:40. Now we can solve for his walking time (assuming he got in the car instantaneously and used no time!)
3mph*t1 + 30 mph*t2 = 30 miles
t1 + t2 = 1.6666.. hours
2 equations 2 unkowns, yields
t1 = 20/27
which is how long he would have walked.
What if he was a fast walker and walked 5 mph? Then we would have…
5mph*t1 + 30mph*t2 = 30 miles
t1 + t2 = 1.6666…
t1 = 4/5
But, the car could be faster or slower, the man faster or slower, the distance could be shorter or farther, etc, etc… The variables are too many to calculate. However, I would guess the man walked something similar to this…
t1 = d - cs*(nt + 2/3) / (ms - cs)
t1 = man’s walking time
d = distance to office
cs = average car speed
nt = normal time it takes to get from station to office
ms = man’s speed
cs = car’s speed
in the first example
t1 = 30 - (30*1.6666..)/ (3-30)
t1 = -20/-27
t1 = 20/27
in the second…
t1 = 30 - (30*1.666)/(5-30)
t1 = -20/-25
t1 = 4/5
This is all guessing, the car’s speed and man speed always stays the same, and there is no time wasted getting in the car.
So that’s my answer.
Let…
t1 = man’s walking time
d = distance to office
cs = average car speed
nt = normal time it takes to get from station to office
ms = man’s speed
cs = car’s speed
t1 = d - cs*(nt + 2/3) / (ms - cs)
Again, I doubt that’s what the given answer turns out to be.
We can say that the man must have walked for more than 40 minutes.
x=# of minutes before work starts that the man usually arrives at the station
w=# of minutes spent walking
d=# of minutes spent driving
60=# of minutes the man is early today
Usually:
x=d
w=0
Today:
x-60-w-d=20
if d=x, then w=40
but because of the distance w walked by the man, d40
x+60-w-d=20
I get to the same conclusion than Shawn, my reasoning its a bit different, but I think that is compatible,
If the time the car take form the station to the office its T, the man arrives T-60 min to the station, now if the man arrives 20 min early to the office its like the car were at the station in T - 20 min, the difference its 40 min, but the man walk so the car did not get to the man in the station but some distance away from there, so the time of walking has to be more than 40 min, however Shawn I saw. this puzzle some time ago, and the answer was exact!!, did not agree with that answer and did not understand it also jajaja but cool.
There is a simpler form to my equation above…
t1 = (2/3)*cs / (cs - ms)
t1 = time man spends walking
cs = car speed
ms = man’s speed
I concur with Shawn and Suineg’s assessment.
As ms approaches cs, the time gets longer and longer.
As ms approaches 0 and cs approaches infinity, the time approaches 2/3 of an hour or 40 minutes.
I thought of trying a math equation but there is not enough information.
So, I think the man did not walk at all. It says, “he starts walking”. He arrived 20 minutes early because traffic was not as bad since he was an hour earlier than usual.
usual arrival: x (time d to drive there from train arrive-time)
assume car never waits, thus car departs at x-2d
today’s arrival @work: x-20
today’s arrival @train: x-60-d
(x-60-d) + w + h = x-20
(h is the drive time today, h<d)
reducing: w+h=40+d
also x-2d +2h = x-20 (car left @ normal time, got him then returned)
so simplifying h=d-10
thus substituting
w + d-10 = 40 +d
w = 50 min
Nice onfire, that´s the answer I saw some time ago, did not understand it back then, now I do, the problem was that I had to assume that its a two way road, so its the same road and that the car picks the man up, and go back inmediatly, I thought the man call tha car, and the car left before his usual time, but I also assume that has to get to the station and in the way back(other road) pick the man up, cool answer.
Wow a lot of assumptions..
From the wording of the problem all I could work out is that it took him 40 minutes longer to walk some of the distance and get driven the rest, than simply getting driven all the way.
(Because he sets off an hour earlier but arrives only 20 minutes before his usual time). But without the speed of his walk or the distance and not assuming anything else I would say impossible. Is the answer among these?
What if the car that picks him up is a taxi?