ABS

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logitech: Legendary Member; Posts: 2134; Joined: Mon Oct 20, 2008 11:26 pmFollowed by: 25 members; Thanked: 237 times; GMAT Score:730

by logitech » Wed Nov 05, 2008 11:02 am

Question2 - Is x > 0?

(1) |x + 3| = 4x – 3
(2) |x – 3| = |2x – 3|

LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"

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Source: Beat The GMAT — Data Sufficiency | Wed Nov 05, 2008 11:02 am

logitech: Legendary Member; Posts: 2134; Joined: Mon Oct 20, 2008 11:26 pmFollowed by: 25 members; Thanked: 237 times; GMAT Score:730

Re: ABS

by logitech » Wed Nov 05, 2008 1:09 pm

logitech wrote:Question2 - Is x > 0?

(1) |x + 3| = 4x – 3
(2) |x – 3| = |2x – 3|

No taker for this question ?

Well (0,2) satisfy both equation.

So?

LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"

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Fab: Senior | Next Rank: 100 Posts; Posts: 87; Joined: Tue Jul 15, 2008 10:23 amFollowed by: 1 member; Location: Lima; Thanked: 4 times

by Fab » Wed Nov 05, 2008 1:19 pm

I'll go with E.
X could be 0 or 2, if it's 0 ----> X = 0, if it's 2 ----> X > 0

THANKS.

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EricLien9122: Master | Next Rank: 500 Posts; Posts: 106; Joined: Thu May 22, 2008 10:50 am; Thanked: 3 times

by EricLien9122 » Wed Nov 05, 2008 1:45 pm

I don't think 0 satisfy statement 1:

|x+3|=4x-3
|0+3|=4(0)-3
3=-3?

I think it should be A.

Please correct me if I make a mistake.

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nitin86: Master | Next Rank: 500 Posts; Posts: 111; Joined: Sat Sep 01, 2007 11:00 pm; Thanked: 4 times

by nitin86 » Wed Nov 05, 2008 7:22 pm

For (1),
Case a => |x+3| > 0 , when x > -3
Case b => |x+3| < 0 , when x < -3

Now, solve equation (1), for case a, we get
|x+3| = x+3 (case a) => x+3 = 4x-3 => x = 2

Now, solve equation (1), for case b, we get
|x+3| = -(x+3) (case b) => -(x+3) = 4x-3 => x = 0

x=0 is not possible, as per case b, x < -3

so, from (1), we get only one value of x, and that is 2

--------------------------------------------------
Solve, similarily for equation (2)

wherein, possible cases are
(a) x>3 ;
(b) 3/2<x<3;
(c) x<3/2

here, for case (a) x>3
both |x – 3| and |2x – 3| are > 0
On solving, we get x=0, and this is not valid as per case (a), x > 3

for case (b) 3/2<x<3
|x – 3| < 0 and |2x – 3| > 0
On solving, we get x=2

for case (c) x<3
both |x – 3| and |2x – 3| are < 0
On solving, we get x=0

Hence, we neglect case (a), and from case(b) and case (c), we get 2 values of x. i.e 2 and 0
which is insufficient

Hope it helps.

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logitech: Legendary Member; Posts: 2134; Joined: Mon Oct 20, 2008 11:26 pmFollowed by: 25 members; Thanked: 237 times; GMAT Score:730

by logitech » Wed Nov 05, 2008 7:24 pm

EricLien9122 wrote:I don't think 0 satisfy statement 1:

|x+3|=4x-3
|0+3|=4(0)-3
3=-3?

I think it should be A.

Please correct me if I make a mistake.

Nope you are absolutely right! Thank you.

LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"

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jimmiejaz: Master | Next Rank: 500 Posts; Posts: 207; Joined: Sun Mar 11, 2007 6:16 pm; Location: Mumbai; Thanked: 11 times

by jimmiejaz » Thu Nov 06, 2008 7:29 am

logitech,

You gave a new method of solving abs values in a previous question.
going by that...
we have |x-3| = |2x-3|
squaring both sides and rearranging we get
3x^2 - 6x = 0
3x(x-2) = 0
x=0 or x=2

hence 2 is insuff....
I must say this is the easiest approach for solving abs equations when both sides have single terms on each side.
Thanks a lot.
Keep rocking!!!!!!

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logitech: Legendary Member; Posts: 2134; Joined: Mon Oct 20, 2008 11:26 pmFollowed by: 25 members; Thanked: 237 times; GMAT Score:730

by logitech » Thu Nov 06, 2008 8:50 am

jimmiejaz wrote:logitech,

You gave a new method of solving abs values in a previous question.
going by that...
we have |x-3| = |2x-3|
squaring both sides and rearranging we get
3x^2 - 6x = 0
3x(x-2) = 0
x=0 or x=2

hence 2 is insuff....
I must say this is the easiest approach for solving abs equations when both sides have single terms on each side.
Thanks a lot.
Keep rocking!!!!!!

I am glad you found that method easy man. I know your test day is coming up soon.

BEST OF LUCK! BEAT THE SHIT OUT OF THAT GMAT THING!!

LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"

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Fab: Senior | Next Rank: 100 Posts; Posts: 87; Joined: Tue Jul 15, 2008 10:23 amFollowed by: 1 member; Location: Lima; Thanked: 4 times

by Fab » Thu Nov 06, 2008 2:06 pm

Got it A.

Thanks.

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jimmiejaz: Master | Next Rank: 500 Posts; Posts: 207; Joined: Sun Mar 11, 2007 6:16 pm; Location: Mumbai; Thanked: 11 times

by jimmiejaz » Fri Nov 07, 2008 5:09 am

logitech wrote:
jimmiejaz wrote:logitech,

You gave a new method of solving abs values in a previous question.
going by that...
we have |x-3| = |2x-3|
squaring both sides and rearranging we get
3x^2 - 6x = 0
3x(x-2) = 0
x=0 or x=2

hence 2 is insuff....
I must say this is the easiest approach for solving abs equations when both sides have single terms on each side.
Thanks a lot.
Keep rocking!!!!!!
I am glad you found that method easy man. I know your test day is coming up soon.

BEST OF LUCK! BEAT THE SHIT OUT OF THAT GMAT THING!!

Thanks a lot Logitech. I read your story on the I just Beat the GMAT section. I admire your never say die attitude and i have full confidence that this time you will definitely get that magical 700+ score which we all aim for.
Best of luck.
Cheers mate!!!!

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