GMATPrep DS - Difficult for me

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GMATPrep DS - Difficult for me

by chatekar » Sun Jul 15, 2007 10:23 am
This is from the GMATPrep

Is |x| = y - z?

(1) x + y = z
(2) x < 0

Please advice.

Thanks in advance

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by givemeanid » Sun Jul 15, 2007 1:23 pm
Is |x| = y - z

(1) x + y = z
Doesn't tell anything about whether y > z or y < z.
If x = -1, y = 3, z = 2. |-1| = 3 - 2. True.
If x = 1, y = 2, z = 3. |1| not equal to 2 - 3. False
NOT SUFFICIENT.

(2) x < 0
Doesn't say anything about y and z.
NOT SUFFICIENT.

Combining:
x = z - y and x < 0. This could mean z is -ve and y is -ve or z is +ve, y is +ve.
x = -3, y = -2, z = -5. |x| = 3 = -2 - (-5). True
x = -3, y = 4, z = 7. |x| = 3 not equal to 4 - 7.
NOT SUFFICIENT.

Answer is (E). What is OA?
So It Goes

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by jayhawk2001 » Sun Jul 15, 2007 1:41 pm
givemeanid wrote:
Combining:
x = -3, y = 4, z = 7. |x| = 3 not equal to 4 - 7.
givemeanid, we don't get x+y = z for the above case.

Close. give it another shot.

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by givemeanid » Sun Jul 15, 2007 1:58 pm
Damn carelessness. I need to be careful! Thanks Jay.
So It Goes

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by chatekar » Sun Jul 15, 2007 9:54 pm
givemeanid,

Thanks for the explanation.
I'm in office right now, once I get back to home, I will post the OA.

Thanks

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by highfive287 » Wed Jul 25, 2007 7:45 pm
Hi there,

the question is asking whether /x/ = y - z

(1) the equation above can be written as -x = y - z (I) or x = y-z (II). We

know from (II) that /x/ = y -z. We need to examine (I).

For (I), if you times (-1) to both sides of equation, (I) will become

x = -y + z. Hence, it can also be written as x + y = z, which will be as

same as the first assumption (x + y = z). Sufficient

(2) x < 0 . It is not sufficient to determine if /x/ = y - z

The answer is A

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by marvelxx35 » Mon Jul 30, 2007 6:05 am
1) x + y = z or x = z - y

multiply by -1 to get: -x = y - z

-x = y - z, therefore the statement can only be true when x is negative.
-(-x) = |x|, so when x is negative, it's true
-x != |x|, so when x is positive, it's false

2) x<0

Not enough information.

1 and 2 together tells us that since x<0, the above equation will satisfy the given problem b/c when x is negative, |x| = y - z.

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by bingojohn » Tue Aug 07, 2007 8:33 am
[quote="chatekar"]givemeanid,

Thanks for the explanation.
I'm in office right now, once I get back to home, I will post the OA.

Thanks[/quote]

The answer is [C], sufficient together but neither alone

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by lalitgmat » Tue Aug 07, 2007 5:35 pm
If you start from (B), x < 0 implies |x| = (-x), y & z does not come into picture. So we have to take stmt A.
from A alone we have x = -(y - z ).

Combining (A) and (B) x = (y -z )
and principle of modulus states, |x| = a implies -a <= x < a
in this case 'a' being ( y - z ).

Hence Both stmts are sufficient. However, first stmt deduction, x = (z -y) or -(y -z) holds true only when x < 0.

Hence option (C)...Nice question.
"We always take |x| as -x and +x , remeber here x can be -ve but |x| is always is +ve."

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by bingojohn » Wed Aug 08, 2007 4:54 am
Also, just wanted to point out to chatekar, correct usage of the word advice (vs advise).

Advice is an abstract noun, representing someone's advice to you.
Advise is a verb, the action of providing advice to someone.

Ignore this, in case that was just a mis-spell... (mis-spells like this one will kill you in AWA though).

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by chatekar » Wed Aug 08, 2007 5:05 am
Thanks bingojohn
Thats was pretty silly mistake