GMAT Prep - Absolute Values

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GMAT Prep - Absolute Values

by fleurdelisse » Wed Apr 22, 2009 4:27 am
Interesting problem - does anyone have an easier way to do it?

If zy < xy < 0, is |x-z| + |x| = |z|?
(1) z<x
(2) y>0

(1) Since z < x, then |x-z| = x-z
And since zy < xy < 0 and z < x, then x and z are negative (they must be of opposite sign of y since their product is less than zero, and knowing that z < x, then y must be positive)
so |x| = -x and |z| = -z
and the equality is therefore not correct x-z-x EQUAL to -z
So we can answer the question
SUFFICIENT

(2) y>0, then z and x must be negative for their product to be negative
And since zy < xy, than z < x (both negative) so their absolute values is their opposite.

In this case, we have: |x-z| = x-z, |x| = -x and |z| = -z
so, |x-z| + |x| = x-z-x EQUAL to -z
SUFFICIENT

So answer is D
Last edited by fleurdelisse on Wed Apr 22, 2009 8:58 am, edited 1 time in total.
Source: — Data Sufficiency |

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by rossmj » Wed Apr 22, 2009 8:18 am
Where is this question from? It seems that the stems are contradicting each other which is not possible in a GMAT question. I agree with your analysis of the question, but it is not a possible question.

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by fleurdelisse » Wed Apr 22, 2009 8:33 am
It's from GMAT Prep. What is contradictory in the question?

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by rossmj » Wed Apr 22, 2009 8:39 am
z>x and y>0 is not possible given the stem. z and x can't take on different values (positive in 1, negative in 2) within the same question. Are you sure you didn't write the question down wrong?

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by fleurdelisse » Wed Apr 22, 2009 8:54 am
You're right - it's z <x! I will edit it. Thanks for noting this.

But the idea is the same. Is there any other more straightforward way to solve this.

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by scoobydooby » Wed Apr 22, 2009 9:07 am

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by rossmj » Wed Apr 22, 2009 9:23 am
With that correction yes there is a quick way to solve this. As you can see x and z must be negative and so if we rearrange the equation to get

|x-z| = |z| - |x|

so with both x and z negative we get a positve z minus x on both sides.

-x--z=|z|- |x|

Note that this invalidates your explanation as the answer is yes |x-z| + |x| = |z|, and also that even without the correction your explanation to statement 2 was incorrect.

Hope this helps. For more help on ABS look for posts by Ian Stewart, he does a great job explaining ABS using a number line approach. It is usually very helpful, especially when you have integers and variables mixed with ABS, but on a problem like this the key is to break it down to something simple that will make the question easy.

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by schumi_gmat » Wed Apr 22, 2009 9:48 am
Hi,


I solved the question and found that 2 was sufficient.

We dont know the sign of y.
We know that -

i) y>0, z<0 and x<0; z<x
ii) y<0, z>0 and x>0;z>x


For (i) the ans is Yes
For (i) ans is No.

From stmt 1:
z<x is true only if y>0

Hence disregard (ii) and stmt 1 is Suff

Ans D