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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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New post 18 Oct 2016, 07:10
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Question Stats:

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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

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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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New post 18 Oct 2016, 08:32
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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New post 08 Aug 2018, 21:53
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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New post 09 Aug 2018, 00:13
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][ &nbs [#permalink] 09 Aug 2018, 00:13
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