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Stand By Me
John was trying to take a short cut through a very narrow tunnel when he heard the whistle of an approaching train behind him. Having reached three-eighths of the length of the tunnel, he could have turned back and cleared the entrance of the tunnel, running at 10 miles per hour, just as the train entered. Alternatively, if he kept running forward at that same rate of speed, the train would reach him the moment he would jump clear of the tracks. At what speed was the train moving?
Answer
At 40 miles per hour, the train would enter the tunnel when John was still two-eighths from the exit or a quarter of the tunnel's length. If the train was to reach him at the exit, it would have to travel at four times John's speed, i.e. 40 miles per hour.Hide Answer Show Answer
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Comments
That was hard 4 me. I am not good at math. Good quiz noetheless. thanxx.
Didn't really like this one.
clever!
A good follow-up question is : if after 24 minutes of continuous running at the same contant speed of 10 miles per hour, he began to tire and slow down, what is the maximum length of the tunnel that will allow him to escape from his predicament? Hehe..
(There is an answer, of course...)
(There is an answer, of course...)
Let l be the distance from train to the tunnel
l' be the length of the tunnel
v the speed of the train, then we have 2 equations( equating time of travel by train and a boy in 2 cases)
l 3/8l'
- = -----
v 10
l+l' 5/8l'
---- = -----
v 10
Dividing first equation by second we get:
l = 1.5l' and substitution into first equation yields:
v = 40
l' be the length of the tunnel
v the speed of the train, then we have 2 equations( equating time of travel by train and a boy in 2 cases)
l 3/8l'
- = -----
v 10
l+l' 5/8l'
---- = -----
v 10
Dividing first equation by second we get:
l = 1.5l' and substitution into first equation yields:
v = 40
I just did 10 / (5/8 - 3/5) = 40. Took maybe 5 seconds to see the solution.
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