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a tough PS problem

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a tough PS problem [#permalink] New post 10 Mar 2007, 07:21
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hi,

i've just finished working on this one. in my view it's challenging...
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 [#permalink] New post 10 Mar 2007, 07:31
I think it is A

1 ft = 1/5280 mile

so 20 ft = 20/5280 miles

.5 sec = .5/(60)^2 hour

So speed = dist/time = (20/5280)*(3600/.5)
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 [#permalink] New post 10 Mar 2007, 07:54
it's A) indeed. While solving it i made a small technical mistake and it took me around 5 minutes to find the flaw. So my advice: be very careful when you have to translate minutes in hours, miles in meters and so on...
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 [#permalink] New post 10 Mar 2007, 19:10
20/.5 feet/sec => (20/.5)*(60*60/5250) miles/hr
  [#permalink] 10 Mar 2007, 19:10
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