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DS Question Help?

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DS Question Help?

by tkherrmann » Sat Jun 13, 2009 4:16 pm
Hi everyone,

I've been checking stuff out for a while around here. I'll be taking the GMAT on June 27th. So far, I've taken two GMATPrep practice tests (same score both times: 760, Q50, V44). Obviously I'll be happy even if I get lower than this, but my aim is 750+. I came across a question on my practice test today that was really confusing. I'd post the question with my thoughts, but I think I'll just post the question and see what everyone thinks, with what I was stuck on later. Here the question is.

If zy < xy < 0, is |x-z| + |x| = |z|?

1) z < x
2) y < 0

Let me know what you think!

Tim
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Source: — Data Sufficiency |
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by ssmiles08 » Sat Jun 13, 2009 5:51 pm
IMO D.

I usually work backwards on DS questions so lets take choice 2:

2) y<0: since y is (-) and multiplied on both x and z, and if we divide both sides by y we would get: z>x>0

Here ultimately we know both z and x are positive so we can remove the absolute brackets b/c we know it will not make a difference.

Is x-z +x = z ? NO b/c 2z =/= 2x for the very reason that z>x>0

So since statement 2 is a definite no, it is a SUFFICIENT statement.

1) z<x: Since x and z are both the same signs depending on which sign y is, we can take the following two scenarios:

if y is positive, we can divide by y and get z<x<0. If y is negative z>x>0.

from this we can tell y is positive b/c choice 1 tells us that z<x.

From here you can choose any two negative values of x and y to put into the equation:

let z = -5; x= -3

|-3+5| + |-3| = |-5|
2 + 3 = 5: YES

This holds true for all negative values as long as z<x. So statement 1 is also SUFFICIENT.

Hope I did not miss anything. What is the OA?

good luck for your GMAT soon btw.
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by raleigh » Sat Jun 13, 2009 6:10 pm
Note that x, y, z are non-zero.

(1) z < x can be re-written as x - z > 0. So |x-z| = x - z.

Then |x - z| + |x| = x - z + x = 2x - z. We need to figure out of this equals |z|.

Case z > 0: Then |z| = z
Suppose |x - z| + |x| = |z| (proof by contradiction). Then:

|x - z| + |x| = |z|
2x - z = z
2x = 2z
x = z

But z < x. Contradiction.

Case z < 0: Then |z| = -z
Suppose |x - z| + |x| = |z| (proof by contradiction). Then:
|x - z| + |x| = |z|
x - z + x = -z
2x = 0
x = 0

But x is non-zero. Contradiction.

Therefore this equality never holds. SUFFICIENT.

(2) y < 0. Then zy < xz implies z > x or x - z < 0.

So |x - z| = -(x - z) = z - x

Then |x - z| + |x| = z - x + x = z = |z|

SUFFICIENT.

The answer is D. Can you tell me where this problem came from? Traditionally, (1) and (2) do not give conflicting answers, but (1) says the equality never holds, and (2) says it always holds.
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by Vemuri » Sat Jun 13, 2009 11:59 pm
IMO D as well. Is it the OA?
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by iamcste » Sun Jun 14, 2009 6:25 am
raleigh wrote: Can you tell me where this problem came from? Traditionally, (1) and (2) do not give conflicting answers, but (1) says the equality never holds, and (2) says it always holds.
Check out here..Key is you dont even need statements to answer the question. Read King Ian's comments here.

https://www.beatthegmat.com/gmat-prep-ma ... t5848.html
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by ghacker » Sun Jun 14, 2009 7:51 am
I don't think that You have to do a lot of things to solve this easy question

what we need to find out is whether Z and X are of the same sign and relative position

So statement I gives it and statement II also gives that they are of the same sign

Hence answer is D
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by tkherrmann » Thu Jun 18, 2009 5:49 pm
ghacker -- I'm sure you get every problem correct and never make stupid mistakes. :roll:

Thank you for all the helpful replies! And sorry I'm so late in giving the official answer, but yes it is D.

Cheers,
Tim
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