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# If x, y and z are Integers and z is not equal to 0, Find range of [m][

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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
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

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