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

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When the wind speed is 9 miles per hour, the wind-chill factor w is  [#permalink]

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New post 23 May 2016, 03:35
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A
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E

Difficulty:

  25% (medium)

Question Stats:

76% (01:50) correct 24% (01:36) wrong based on 635 sessions

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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: When the wind speed is 9 miles per hour, the wind-chill factor w is  [#permalink]

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New post 23 May 2016, 05:27
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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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Re: When the wind speed is 9 miles per hour, the wind-chill factor w is  [#permalink]

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New post 20 Dec 2017, 08:52
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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


Given
=> w=-17.366+1.19t Is w>0

Now for 'w>0'
=> (-17.366+1.19t) should be >0
=> (-17.366+1.19t)>0
=> 1.19t>17.366
=> t>17.366/1.19 or
=> t>173.66/11.9 or
=> t>173.66/12
=> t>14

So if t>14 then w>0

Statement 1 t>10 therefore 2 scenario
a)if 10<t<14 then w<0
b)if t>14 then w>0
Since NO unique sol. Therefore NOT sufficient

Statement 2 t<20 therefore 2 scenario
a) if 14<t<20 then w>0
b) if t<14 then w<0
Since NO unique sol. Therefore NOT sufficient

BOTH 1 & 2
from Stat 1 t>10
from stat 2 t<20
Therefore 10<t<20
AGAIN 2 scenario.Therefore NOT sufficient

Therefore "E"

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Re: When the wind speed is 9 miles per hour, the wind-chill factor w is  [#permalink]

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New post 23 Apr 2019, 17:21
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
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Re: When the wind speed is 9 miles per hour, the wind-chill factor w is   [#permalink] 23 Apr 2019, 17:21
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