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SET 28 QUESTION 5 ABS VALUE

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by cramya » Mon Oct 20, 2008 8:20 pm
I just posted a long explanation for absolute value and am stumped on this one

Q5:
Is |x|< 1?

(1) |x + 1| = 2|x - 1|
(2) |x - 3| = 0

[spoiler]OA:C[/spoiler]

I am getting D.

From (1)
x=3 x = 1/3 but only x=3 satisfies the equation Therefore |x|>1
SUFFICIENT

From (2) x= 3 |x|>1
SUFFICIENT

Thoughts?
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Source: — Data Sufficiency |
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by cramya » Mon Oct 20, 2008 9:14 pm
Sorry folks; posted the wrong statement in (2)


Correct question:

Is |x|< 1?

(1) |x + 1| = 2|x - 1|
(2) |x - 3| > 0


Then its C

I figured it out!!!
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by stop@800 » Mon Oct 20, 2008 11:14 pm
Yes the answer is C

B
B will not give any info other than x is not equal to 3
bcoz
|x-3| will always be greater than 0 or may be equal to 0


A
solving this you will get
x=3 and x=1/3

so insuff


A+B
we know x!=3 from b
so final x is 1/3
hence ans is C
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Confuse

by jessicamuniz » Tue Oct 21, 2008 9:31 am
Hey guys, I have been reading the posts but i do not get why in the first option you get x=3 o x=1/3?
I am sorry but somebody can do the math for me please...I must missing something.
Thanks,
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by cramya » Tue Oct 21, 2008 6:05 pm
Jessica please refer to my post

https://www.beatthegmat.com/general-ques ... 20342.html

(1)

I am doing Case 1 and Case 4 (atleast we must do these for absolute value problems wiht x or whatever on both sides)

Case 1: Considering the abs value expressions positive on both sides +/+

(1) |x + 1| = 2|x - 1|

x+1 = 2x-2
x=3

Case 4: Considering the abs value expressions -ve on lhs and +ve on rhs
i.e - / +

-(x+1) = 2x-2
-x-1 = 2x-2
-3x = -1
x= 1/3

2 possible values. In these abs value problems u need to substitute thes evalues in the original abs value expression to check if these indeed are solutions. If u
|x + 1| = 2|x - 1|
Put x = 3 and 1/3 seperately and lhs/rhs will be equal. INSUFFICIENT SINCE WE HAVE 2 SOLUTION FOR X BUT WE NEED EXACTLY 1 VALUE

(2)
WIHT X ON JUST ONE SIDE WE HAVE 2 CASES + AND -

CASE 1:
|x - 3| > 0

X-3>0
X>3

CASE 2
-(X-3)>0
-X+3>0
-X>-3
X<3 (WHEN U DIVIDE/MULTIPLY AN EQUALITY WIHT A NEGATIVE NUMBER THE INEQUALITY REVERSES I.E in this case we are dividing by -1 on both sides)

From x<3 and x>3 we know x can be anyhting but 3

From (1) and (2) we know the only other value for x is 1/3 and hence from both statements we can get a defnite value for X

Let me know if u still have questions!

Good luck!
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by cramya » Tue Oct 21, 2008 6:10 pm
Made minor corrections to my post above.

Jessica please refer to my post

https://www.beatthegmat.com/general-ques ... 20342.html

(1)

I am doing Case 1 and Case 4 (atleast we must do these 2 cases for absolute value problems with x on both sides)

Case 1: Considering the abs value expressions positive on both sides +/+

(1) |x + 1| = 2|x - 1|

x+1 = 2x-2
x=3

Case 4: Considering the abs value expressions -ve on lhs and +ve on rhs
i.e - / +

-(x+1) = 2x-2
-x-1 = 2x-2
-3x = -1
x= 1/3

2 possible values. In these abs value problems u need to substitute thes evalues in the original abs value expression to check if these indeed are solutions. If u
|x + 1| = 2|x - 1|
Put x = 3 and 1/3 seperately and lhs/rhs will be equal. INSUFFICIENT since we have 2 solutions FOR X where one > 1 and the other is less than 1
(2)
WITH X ON JUST ONE SIDE WE HAVE 2 CASES + AND -

CASE 1:
|x - 3| > 0

X-3>0
X>3

CASE 2
-(X-3)>0
-X+3>0
-X>-3
X<3 (WHEN U DIVIDE/MULTIPLY AN EQUALITY WIHT A NEGATIVE NUMBER THE INEQUALITY REVERSES I.E in this case we are dividing by -1 on both sides)

From x<3 and x>3 we know x can be anytHing but 3 and we cannot still say if |x|<1 INSUFFICIENT

From (1) and (2) combined, we know the only other value for x is 1/3 and hence from both statements we can get a defnite value for X

Let me know if u still have questions!

Good luck!
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Thank you Cramya, your reply was very helpful ;)

by jessicamuniz » Thu Oct 23, 2008 9:11 am
cramya wrote:Made minor corrections to my post above.

Jessica please refer to my post

https://www.beatthegmat.com/general-ques ... 20342.html

(1)

I am doing Case 1 and Case 4 (atleast we must do these 2 cases for absolute value problems with x on both sides)

Case 1: Considering the abs value expressions positive on both sides +/+

(1) |x + 1| = 2|x - 1|

x+1 = 2x-2
x=3

Case 4: Considering the abs value expressions -ve on lhs and +ve on rhs
i.e - / +

-(x+1) = 2x-2
-x-1 = 2x-2
-3x = -1
x= 1/3

2 possible values. In these abs value problems u need to substitute thes evalues in the original abs value expression to check if these indeed are solutions. If u
|x + 1| = 2|x - 1|
Put x = 3 and 1/3 seperately and lhs/rhs will be equal. INSUFFICIENT since we have 2 solutions FOR X where one > 1 and the other is less than 1
(2)
WITH X ON JUST ONE SIDE WE HAVE 2 CASES + AND -

CASE 1:
|x - 3| > 0

X-3>0
X>3

CASE 2
-(X-3)>0
-X+3>0
-X>-3
X<3 (WHEN U DIVIDE/MULTIPLY AN EQUALITY WIHT A NEGATIVE NUMBER THE INEQUALITY REVERSES I.E in this case we are dividing by -1 on both sides)

From x<3 and x>3 we know x can be anytHing but 3 and we cannot still say if |x|<1 INSUFFICIENT

From (1) and (2) combined, we know the only other value for x is 1/3 and hence from both statements we can get a defnite value for X

Let me know if u still have questions!

Good luck!
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by rohangupta83 » Thu Oct 23, 2008 3:08 pm
yup C
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