nalinnair 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

In such Qs, it is possible on many occasions, May not be here, that one of the statements gives you a value which will give you a range that will suffice.. May not be here

w = -17.366 + 1.19t....

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

w = -17.366 + 1.19t so here w > -17.366 + 1.19*10 = -17.366+11.9= -6.abc.. so A value belw 0 is also possible and also above0.... ....

Insuff..

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

w = -17.366 + 1.19t so here w < -17.366 + 1.19*20 = -17.366+23.8= 6.xyz.. so A value below 0 is also possible and also above 0.... ....

Insuff..

Combined

again the range is -6 to 6..

Insuff

E

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