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

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