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Is x > 0?; (1) |x + 3| = 4x – 3; (2) |x – 3| = |2x – 3|

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srikant
 Post subject: Is x > 0?; (1) |x + 3| = 4x – 3; (2) |x – 3| = |2x – 3|
 Post Posted: Sun Nov 04, 2007 11:23 pm 
Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x – 3| = |2x – 3|

The CAT answer was wrong and I would like to find what others have to say.
Can anyone please tell the write answer with explanation.
I am more concerned about the second statement!
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srikant
 Post subject: Ans is A
 Post Posted: Mon Nov 05, 2007 1:49 am 
From first statement we can see that L.H.S. is always +ve i.e. 4x-3>=0; x>=3/4 sufficient.

From statement 2 there are 3 possible cases
x<3/2
3/2<x<3
3<x

for each case we see the following result
1) for x<3/2
x=0. satisfies
2) for 3/2<x<3
x=2 satisfies
3)for 3<x
x=0. not right

Since we get 2 values of x, one equal to 0 and another greater than 0, it can not be said whether x>0. Not sufficient

Therefore answer is A.
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StaceyKoprince
 Post subject:
 Post Posted: Sat Nov 10, 2007 9:16 pm 
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ManhattanGMAT Staff


Posts: 6077
Location: San Francisco
Yes, the answer is A. You said in your first post that the CAT answer was wrong, but I just looked it up in the system and the answer is correctly listed as A. (Unless the second post is from some other person who also chose srikant as a username... ?)

So - I'm confused as to your question. Let me know!
:)

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Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT
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sathishm.123
 Post subject: Eniqualities
 Post Posted: Tue Aug 10, 2010 3:39 pm 
Offline
Students


Posts: 1
Hi Stacey,

Can you please solve the problem, I am quite confused with the obsolute to eniquality conversion.

My approach was
1) x-3=2x-3 therefore x=0
2) x-3 = -(2x-3) therefore x=2 so sufficient, please point my mistake.
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mschwrtz
 Post subject: Re: Is x > 0?; (1) |x + 3| = 4x – 3; (2) |x – 3| = |2x – 3|
 Post Posted: Fri Sep 03, 2010 2:16 am 
Offline
ManhattanGMAT Staff


Posts: 506
I suspect that this is the same as the CAT explanation, but I think that I can explain it about as quickly as I can look it up.

Is x > 0?

(1) |x + 3| = 4x – 3

This admits, or seems to admit, two possibilities:

if x+3>= 0, then x+3=4x-3-->3x=6-->x=2
plug back in to verify. Yup, |2+3|=4*2-3.

if x+3<=0, then x+3=3-4x-->3x=0-->x=0
This is consistent with the assumption that x<=0, but plug back in to verify. Nope, |0-3| does not equal 4*0+3.
THIS CAN HAPPEN WHEN THERE IS A VARIABLE OUTSIDE THE ABSOLUTE VALUE AS WELL AS INSIDE.

So S1 means that x=2 SUFFICIENT
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