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GMATPrep DS - Difficult for me


 
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chatekar
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PostPosted: Sun Jul 15, 2007 11:23 am    Post subject: GMATPrep DS - Difficult for me Reply with quote

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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givemeanid
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PostPosted: Sun Jul 15, 2007 2:23 pm    Post subject: Reply with quote

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?

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jayhawk2001
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PostPosted: Sun Jul 15, 2007 2:41 pm    Post subject: Reply with quote

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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givemeanid
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PostPosted: Sun Jul 15, 2007 2:58 pm    Post subject: Reply with quote

Damn carelessness. I need to be careful! Thanks Jay.
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chatekar
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PostPosted: Sun Jul 15, 2007 10:54 pm    Post subject: Reply with quote

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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highfive287
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PostPosted: Wed Jul 25, 2007 8:45 pm    Post subject: Reply with quote

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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marvelxx35
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PostPosted: Mon Jul 30, 2007 7:05 am    Post subject: Reply with quote

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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bingojohn
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PostPosted: Tue Aug 07, 2007 9:33 am    Post subject: Reply with quote

[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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lalitgmat
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PostPosted: Tue Aug 07, 2007 6:35 pm    Post subject: Reply with quote

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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bingojohn
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PostPosted: Wed Aug 08, 2007 5:54 am    Post subject: Reply with quote

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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chatekar
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PostPosted: Wed Aug 08, 2007 6:05 am    Post subject: Reply with quote

Thanks bingojohn
Thats was pretty silly mistake
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