Brain Teasers
Street Cars
A man is walking down a street along which runs a streetcar line. He notices that, for every 40 streetcars
which pass him traveling in the same direction as he, 60 pass in the opposite direction. If the man`s walking speed is 3 miles per hour, what is the average speed of the streetcars?
which pass him traveling in the same direction as he, 60 pass in the opposite direction. If the man`s walking speed is 3 miles per hour, what is the average speed of the streetcars?
Answer
15 miles per hourHide Answer Show Answer
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Ok, that's great and all but you know what would make this site even better? If you had explanations to all the teasers.
This is how you answer it right?
let x represent the average speed of the cars:
[40 times the average speed of the cars, plus the speed of the man is the same as 60 times the average speed of the cars from the opposite direction]
40x + 3(1) = 60x
20x = 3
x = 3/20
x = 0.15
Therefore the average speed of the cars is 15 miles an hour.
let x represent the average speed of the cars:
[40 times the average speed of the cars, plus the speed of the man is the same as 60 times the average speed of the cars from the opposite direction]
40x + 3(1) = 60x
20x = 3
x = 3/20
x = 0.15
Therefore the average speed of the cars is 15 miles an hour.
Not quite!
Let D = the distance between streetcars.
Let d = the distance walked.
The 40th streetcar behind the walker travels 40D + d. The 60th streetcar ahead travels 60D - d. It's the same time period, so
40D + d = 60D - d.
2d = 20D.
d = 10D.
It is clear both of the remote cars moved 50D while the walker moved ony 10D making them 5X faster.
5 X 3mph = 15mph.
Reality non-factored, you can get the same results with 5% the streetcars. 2 chasing the walker and 3 confronting him.
2D + d = 3D - d.
2d = D.
d = D/2.
The cars travel five d while the walker travels one.
Let D = the distance between streetcars.
Let d = the distance walked.
The 40th streetcar behind the walker travels 40D + d. The 60th streetcar ahead travels 60D - d. It's the same time period, so
40D + d = 60D - d.
2d = 20D.
d = 10D.
It is clear both of the remote cars moved 50D while the walker moved ony 10D making them 5X faster.
5 X 3mph = 15mph.
Reality non-factored, you can get the same results with 5% the streetcars. 2 chasing the walker and 3 confronting him.
2D + d = 3D - d.
2d = D.
d = D/2.
The cars travel five d while the walker travels one.
I think this is easier than Stil's explanation:
If x is the speed of the street car then
(x - 3)/40 = (x + 3)/60
Taking the cross-product gives
40x + 120 = 60x - 180
20x = 300
x = 15
If x is the speed of the street car then
(x - 3)/40 = (x + 3)/60
Taking the cross-product gives
40x + 120 = 60x - 180
20x = 300
x = 15
Sane is insane. Otherwise good teaser and thanks for all the good solutions.
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