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Re: GMATprep Inequalities [#permalink]
28 Jan 2010, 02:51
6
This post received KUDOS
An easy way to approach this ineq will be to analyze that: |x-z| + |x| = |z| means |z-x| = |z| - |x|. |z-x| is the distance between z and x on number line. It can only be equal to |z| - |x| if both z and x have the same signs.
a) z < x - implies that y > 0 because zy < xy. If y > 0 then z < x < 0. Therefore both have same signs. SUFF
b) y>0 then z < x < 0. Therefore both have same signs. SUFF
Re: GMATprep Inequalities [#permalink]
28 Jan 2010, 05:14
5
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Expert's post
1
This post was BOOKMARKED
This question was discussed before here is my post from there:
This is not a good question, as neither of statement is needed to answer the question, stem is enough to do so.
If \(zy<xy<0\) is \(|x-z|+|x| = |z|\)
Look at the inequality \(zy<xy<0\):
We can have two cases:
A. If \(y<0\) --> when reducing we should flip signs and we'll get: \(z>x>0\). In this case: as \(z>x\) --> \(|x-z|=-x+z\); as \(x>0\) and \(z>0\) --> \(|x|=x\) and \(|z|=z\).
Hence in this case \(|x-z|+|x|=|z|\) will expand as follows: \(-x+z+x=z\) --> \(0=0\), which is true.
And:
B. If \(y>0\) --> when reducing we'll get: \(z<x<0\). In this case: as \(z<x\) --> \(|x-z|=x-z\); as \(x<0\) and \(z<0\) --> \(|x|=-x\) and \(|z|=-z\).
Hence in this case \(|x-z|+|x|=|z|\) will expand as follows: \(x-z-x=-z\) --> \(0=0\), which is true.
So knowing that \(zy<xy<0\) is true, we can conclude that \(|x-z|+|x| = |z|\) will also be true. Answer should be D even not considering the statements themselves.
As for the statements:
Statement (1) says that \(z<x\), hence we have case B.
Statement (2) says that \(y>0\), again we have case B.
Re: If zy < xy < 0, is |x-z| + |x| = |z|? [#permalink]
17 Dec 2013, 03:17
I solved this question on number line, we can first analyze this eq |x-z| + |x| = |z|? This equality is only possible, if x and z are on the same side of the number line.
(-) ----z----x----0--- or ---0-----x----z---- (+)
|distance between xandz| + |distance of x from origin| = |distance of z from origin|
given zy<xy<0 which can happen in two case.
z x y - - + zy negative xy negative < both less than 0 + + - zy negative xy negative < both less than 0
Therefore we don't even need option 1 and 2 to validate this |x-z| + |x| = |z|.
Answer : D _________________
Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use?| 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".
Re: If zy < xy < 0, is |x-z| + |x| = |z|? [#permalink]
10 May 2014, 11:04
using Statement 1 : from the question , zy < xy y(z-x)<0 ------- (A) Now statement 1 tells me that (z-x)< 0 . This implies Y>0 So, if zy < xy < 0 and Y>0 This implies Z & X < 0 mod (x-z) + mod (x) = X-Z (since Z-X<0) + (-X) = -Z = |Z|
Using Statement 2 : From (A), y(z-x)<0 Since from Statement 2 we know that y > 0 That implies (z-x)< 0 .
Re: If zy < xy < 0, is |x-z| + |x| = |z|? [#permalink]
16 May 2015, 06:49
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