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GMATPrep2

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Re: GMATPrep2

by Jameschan168 » Mon Aug 27, 2007 7:41 pm
I think the answer is B.

From the question stem we can tell that x & z have to have the same sign. So the answer will depend on whether y > 0 or y <0.

Stmt 1 is insufficent since we don't know what y is.
For example, if y > 0 and x = -5 & z = -6, then abs (-5 + 6) + abs (-5) = abs (6), or simply 1 + 5 = 6, which is true.

However, if y < 0 and let's say x = 7 and z = 6, then abs (7-6) + abs (7) does not equal = abs (6).

Stmt 2 is sufficient based on the explanation above.

givemeanid wrote:If zy < xy < 0, is |x-z| + |x| = |z|?
1. z < x
2. y > 0

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Re: GMATPrep2

by gabriel » Tue Aug 28, 2007 12:10 am
givemeanid wrote:If zy < xy < 0, is |x-z| + |x| = |z|?
1. z < x
2. y > 0

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...Me going for D ... from the first statement we get z<x, this means that both z and x are negative and that the magnitude of z is greater than x .. why? read this https://www.beatthegmat.com/viewtopic.php?t=4586 ...

... now, let z=-5 and x = -3 ... so mod(x-z)= 2, mod(x)=3 and mod(z)=5 .. so the statement holds true .. so statement one is sufficient ..
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by abhik » Tue Aug 28, 2007 4:16 am
What is the OA
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by agps » Tue Aug 28, 2007 4:46 am
I would answer B.

from the question z<x if y>0
in 1) z<x then y<0 or y>0 depending on z and x being positive or negative (for z and x < 0 then y > 0 e.g. -2 and -1 y>0, for z and x > 0 then y < 0 e.g. 1 and 2 y>0), not sufficient because, if z = 1 and x = 2, |2-1|+2 = 3 different from 1 and if z=-2 and x=-1 |-1-(-2)|+|-1|= 2 equal to |-2|
in 2) if y> 0 then z and x < 0, for z=-1/2 and x=-1/4, |-1/4-(-1/2)|+|-1/4| =1/4+1/4 = 1/2 = |z|.
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by givemeanid » Tue Aug 28, 2007 5:58 am
If zy < xy < 0, is |x-z| + |x| = |z|?
1. z < x
2. y > 0


OA is D.

First realize that, both statements are giving you the same information. (I missed this critical part when I was taking the test since stmt 1 does not mention y and stmt 2 does not mention x and z, I rushed to a conclusion that either C or E should be the right choice. My bad.)

1. Since zy < xy and z < x, y > 0 (same as stmt 2)
Now, x > z, so |x-z| = x-z
Also zy < xy < 0 and y > 0
So z < x < 0
|x| = -x
|z| = -z

Now,
|x-z| + |x| ? |z|
x - z -x ? -z
0 = 0. So, true.
SUFFICIENT.

2. The reasoning above applies to stmt 2.


Hence, D
So It Goes
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by gabriel » Tue Aug 28, 2007 6:47 am
agps wrote:I would answer B.

from the question z<x if y>0
in 1) z<x then y<0 or y>0 depending on z and x being positive or negative (for z and x < 0 then y > 0 e.g. -2 and -1 y>0, for z and x > 0 then y < 0 e.g. 1 and 2 y>0), not sufficient because, if z = 1 and x = 2, |2-1|+2 = 3 different from 1 and if z=-2 and x=-1 |-1-(-2)|+|-1|= 2 equal to |-2|
in 2) if y> 0 then z and x < 0, for z=-1/2 and x=-1/4, |-1/4-(-1/2)|+|-1/4| =1/4+1/4 = 1/2 = |z|.
Thats where ur wrong .. if z<x then z and x both have to be negative and y > 0 ... as givemeaind has written in his post if u correctly analyze both the statements then you will realize that they both give then same information ( that is y > 0 ) .. tricky question btw ..
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by agps » Tue Aug 28, 2007 8:10 am
you're right.
what i forgot was that if y<0 both z and x > 0, but z > x.
(y = -1, z=2 and x=3 zy=-2 > xy=-3, contradicts the stem)
statement 1 says z < x, therefore y > 0
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