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temps

by bacali » Tue Dec 02, 2008 11:56 am
When the wind speed is 9 miles per hour, the wind-chill factor w is given by w = -17.366+ 1.19t, where t is the temperature in degrees Fahrenheit. If at noon yesterday the wind speed was 9 miles per hour, was the wind-chill factor greater than 0 ?


(1) The temperature at noon yesterday was greater than 10 degrees Fahrenheit.

(2) The temperature at noon yesterday was less than 20 degrees Fahrenheit.


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Re: temps

by Stuart@KaplanGMAT » Tue Dec 02, 2008 2:34 pm
bacali wrote:When the wind speed is 9 miles per hour, the wind-chill factor w is given by w = -17.366+ 1.19t, where t is the temperature in degrees Fahrenheit. If at noon yesterday the wind speed was 9 miles per hour, was the wind-chill factor greater than 0 ?


(1) The temperature at noon yesterday was greater than 10 degrees Fahrenheit.

(2) The temperature at noon yesterday was less than 20 degrees Fahrenheit.

Well, to solve for w, we need info about t. This question asks us for a range of w, so let's plug in w = 0:

0 = -17.366+ 1.19t
17.366 = 1.19t
17.366/1.19 = t

Therefore, if t = 17.366/1.19, w = 0.

So, for w to be greater than 0, we need t > 17.366/1.19.

Let's estimate: 12*15 = 180, so 17.4/1.2 is between 14 and 15. So, the question can be reduced to:

Is t > 14.5ish?

(1) t > 10. Could be on either side of the range: insufficient.

(2) t < 20. Could be on either side of the range: insufficient.

Together: 10 < t < 20. Still could be on either side of the range: insufficient.

Choose (E), not enough information.
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