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CTK Exchange
Jack
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Jan-20-03, 12:53 PM (EST) |
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"simple puzzle-why?"
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This seems simple but....
A car racer needs to complete two one mile courses at an average speed of 60mph. He can only manage 30mph for the first mile. How fast must he go for the second?
It SEEMS the answer should be 90. BUT, to average 60mph, he must complete the two courses in 2 minutes and it takes him 2 minutes for the first, to it can't be done? What is the mathmatical reason? |
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alexb
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1811 posts |
Jan-20-03, 01:34 PM (EST) |
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1. "RE: simple puzzle-why?"
In response to message #0
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>This seems simple but....
>A car racer needs to complete two one mile courses at an
>average speed of 60mph. He can only manage 30mph for the
>first mile. How fast must he go for the second?
>It SEEMS the answer should be 90. Not at all. See https://www.cut-the-knot.com/arithmetic/Means.shtml >
>BUT, to average 60mph, he must complete the two courses in 2
>minutes and it takes him 2 minutes for the first, to it
>can't be done? What is the mathmatical reason? The mathematical reason for the fallacy is in uncritical usage of the word "SEEMS". Can't built a lot on what only "seems". |
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G Cleverley
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Jan-21-03, 10:00 AM (EST) |
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2. "RE: simple puzzle-why?"
In response to message #1
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No quarrel with your answer, but there is an ambiguity in the original definition of the racer's target. What has to be 60 mph - the average of the two speeds in the two courses, or the average speed over the two miles taken together? In the first case, if he does 30 mph for the first course and 90 for the second, then the average of the two speeds (not his average speed) is 60, and he has met his target. It works like this in ski-ing. He will have travelled the total two miles in 2 minutes 40 seconds, so his average speed over the two miles is 45 mph. In the second case then if he only averages 30 for the first mile then of course he can't average 60 over the two miles. |
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Sesshoumaru
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Feb-14-03, 07:10 AM (EST) |
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3. "RE: simple puzzle-why?"
In response to message #0
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Here's the "simple" reason: Since you go 30 mph the first time around, it takes you a minute, right? Since you want to avg. 60 MPH, you use the time of 60MPH, which is for two miles, 2 minutes. Since he has already spent 2 minutes... the end result would be something like if you use r*t=d (rate*time=distance) Total Distance
-------------- = Avg. Speed
Total Time 1
--- = r/60
2-2 You cannot divide by zero, so there is no answer. |
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