GMATinsight wrote:
If x, y and z are Integers and z is not equal to 0, Find range of \(\frac{(x+y)}{z}\)
-2
< z
<2
-5
< x
< 10
-11
< y
< 4
(A) -8
< \(\frac{(x+y)}{z}\)
< 7
(B) 8
< \(\frac{(x+y)}{z}\)
< -7
(C) -16
< \(\frac{(x+y)}{z}\)
< 14
(D) -14
< \(\frac{(x+y)}{z}\)
< 14
(E) -16
< \(\frac{(x+y)}{z}\)
< 16
Source:
http://www.GMATinsight.comWe are asked to find the range of \(\frac{(x+y)}{z}\), i.e., \(Max(\frac{(x+y)}{z})\) & \(Min(\frac{(x+y)}{z})\)
Finding out max value:-
Numerator(x+y) has to be maximum and denominator(z) has to be minimum.
The ratio would be maximum if both the numerator and denominator have same polarity.
So x+y could be -5+(-11)=-16 with z=-1, which yields the ratio to be \(\frac{-16}{-1}\)=16
Finding out min value:-
Numerator(x+y) has to be minimum and denominator(z) has to be maximum with opposite polarity.
So, min(x+y)=min(x)+min(y)=-5+(-11)=-16
\(\frac{(x+y)}{z}\) would be minimum if we can allocate denominator with the lowest positive value. (z=1)
So, the minimum value of ratio=\(\frac{-16}{1}\)=-16
Therefore, -16
< \(\frac{(x+y)}{z}\)
< 16
Ans. (E)
_________________
Regards,
PKN
Rise above the storm, you will find the sunshine