When the wind speed is 9 miles per hour, the wind-chill factor w is

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

Hi,
Let's see when W=0

17.366=1.19t
t= \(\frac{17.366}{1.19}\)
rounding the numerator and denominator we have

t= \(\frac{17.4}{1.2}\)
t= 14.5 (approx)
So if t> 14.5 w will be >0

Now we need to know if 14.5<t

Statement 1 : NS
Statement 2 : NS

Combined No new information

Hence Choice E

what is the significance of the wind speed in this question?

Nothing, I believe it's there to confuse so that ppl will spare few seconds on it's usability and waste time.

\(w = -17.366 + 1.19t\)

(1) If the temperature was greater than 10 degrees Fahrenheit, if t = 11 then the wind chill will be less than 0. if t = 20 then the wind chill greater than 0. INSUFFICIENT.

(2) If temperature was less than 20 degrees Fahrenheit, if t = 1 then wind chill will be less than 0. if t = 19, then wind chill will be greater than 0. INSUFFICIENT.

(1&2) Combined, the temperature will be between 10 and 20 degrees Fahrenheit. If t = 11, then wind chill be be less than 0. If t = 19, then wind chill will be greater than 0. INSUFFICIENT.

Answer is E.

(1) If t=12, t=20 The answer will be Yes and No; Insufficient.

(2) t can be 1 t can be 19, The answer will be Yes and No; Insufficient.

Considering both:
If t=12, t=19 The answer will be Yes and No; Insufficient.

The answer is E

Given: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.

Asked: 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.
w = -17.366 + 1.19t ; where t>10 : Let t = 10 + x
w = - 17.366 + 11.9 + 1.19x
When t = 11; w < 0
But when t = 20; w>0
NOT SUFFICIENT

(2) The temperature at noon yesterday was less than 20 degrees Fahrenheit
w = -17.366 + 1.19t ; where t>10 : Let t = 20 - y
w = - 17.366 + 23.8 - 1.19y
When t = 20; w > 0
But when t = 0; w<0
NOT SUFFICIENT

(1) + (2)
(1) The temperature at noon yesterday was greater than 10 degrees Fahrenheit.
(2) The temperature at noon yesterday was less than 20 degrees Fahrenheit
w = -17.366 + 1.19t
When t = 11; w < 0
But when t = 20; w>0
NOT SUFFICIENT

IMO E

­I worked with the question stem first to determine that 1.19t > 17.366 if we have to have w>0 . 

This means that t> 8683/595. Since I did not want to get into calculations, I apprxoimated the values to be 8600/600 which is approx 14. 

True, I may have to get back to a more precise answer later, but seeing the statements given, i realised that it may not be necessary. 

Solving the statements have been explained well in this forum by other people. 

Hope this helps you...