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

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18 Oct 2016, 08:10
1
9
00:00

Difficulty:

85% (hard)

Question Stats:

43% (01:53) correct 57% (02:00) wrong based on 121 sessions

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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.com

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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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18 Oct 2016, 09:32
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.com

for finding minimum range within given ranges of x,y,& z we must have Ix+yI to be maximum and IzI to be minimum , having both numerator and denominator opposite signs
thus Ix+yI= I-11-5I=16 and IzI=1
thus minimum we get -16/1=-16

For max. value Ix+yI to be maximum and IzI to be minimum , having both numerator and denominator same signs
thus Ix+yI= I-11-5I=16 and IzI=1----->16
So -16<= (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]

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08 Aug 2018, 22:53
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.com

Hi GMATinsight,

I'm having some trouble understanding how the maximum value for x+y is being set to 16? In Rohit's response above, I see he has used mods but there were no mods mentioned on the question. Could you please help?
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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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09 Aug 2018, 01:13
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.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] 09 Aug 2018, 01:13
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