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Re: If x, y and z are Integers and z is not equal to 0, Find range of [m][ [#permalink]
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: https://www.GMATinsight.com


We 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)
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Re: If x, y and z are Integers and z is not equal to 0, Find range of [m][ [#permalink]
Always remember, if a numerator and a denominator are both negative, then the outcome will be positive.

for \(\frac{x+y}{z}\) (minimum), we'll probably have a negative number with a high magnitude. For that reason, \(z=1\). and choose the least extreme values of \(x\), and \(y\).
\(=> -5-11 = -16\)

for \(\frac{x+y}{z}\) (maximum), think of a possible high number through the both extremes.sure, \(10+4\) is high, but it isn't an extreme. Why? Take the two least possible values for\(x\) and \(y\)again and add them, you'll get \(-16\). Take \(z = -1\), calculate \(\frac{x+y}{z}\) and you'll get \(16\).

Thus, \(-16< \frac{x+y}{z}<16\)
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Re: If x, y and z are Integers and z is not equal to 0, Find range of [m][ [#permalink]
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