### 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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## Comments

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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