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Value of |x|

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Value of |x|

by tanvis1120 » Tue Aug 19, 2014 2:41 pm
Is |x| < 1 ?

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

(2) |x - 3| > 0

My approach:
Statement 1:
|x+1|=2|x-1|

Removing modulus gives,
+-(x+1) = +-2(x-1)

The above equation can be solved for 4 values (Using the combinations of x+1 and x-1 below:

+ + ; x = 3
+ - ; x = 1/3
- + ; x = 1/3
- - ; x = 3
Thus, x= 3 or 1/3


Statement 2:
|x - 3| > 0

Removing the Modulus:
+-(x-3) > 0
This gives, x>3 or x<3

Statements 1+2:
If x= 3
|x-3| = 0

If x = 1/3
|x-3| >0

So, if x = 1/3, it satisfies statement 2.
Thus,-1<x<1, |x|<1 is true.

So, answer is C.

Please let me know if it is correct, experts!

Thank You!
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by [email protected] » Tue Aug 19, 2014 5:49 pm
Hi tanvis1120,

Yes, your solution is correct.

Fact 1 has just 2 solutions (the "equals" sign and the absolute value symbols are a bit of a hint that there are probably 2 solutions). Fact 2 has an unlimited number of solutions (anything BUT the number 3). When combining statements, there is just 1 overlapping answer.

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by GMATinsight » Tue Aug 19, 2014 5:58 pm
tanvis1120 wrote:Is |x| < 1 ?

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

(2) |x - 3| > 0

My approach:
Statement 1:
|x+1|=2|x-1|
Thus, x= 3 or 1/3


Statement 2:
|x - 3| > 0

Removing the Modulus:
+-(x-3) > 0
This gives, x>3 or x<3

Statements 1+2:
If x= 3
|x-3| = 0

If x = 1/3
|x-3| >0

So, if x = 1/3, it satisfies statement 2.
Thus,-1<x<1, |x|<1 is true.

So, answer is C.

Please let me know if it is correct, experts!

Thank You!
All is correct. Just another way to look at Statement 2 is

Given : |x - 3| > 0
Ask yourself what's so special about it, Modulus of anything is always greater than or Equal to Zero
Here this only says it's Greater than which mean that |x - 3| is just not zero
i.e. x is just NOT 3

WHich when combining the two statements removes one of the solutions from the two solutions obtained from first statement and we get x = 1/3 only

Answer: Option C
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by Jim@StratusPrep » Sun Aug 24, 2014 10:52 am
Your approach is perfect. Well done with the math!

One thing to add is to check your solutions when doing the +/- calculation with absolute value. It is a bit of a work around and sometimes you can get solutions that in fact do not work when plugged back in. Not the case here, but worth remembering.
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by j_shreyans » Thu Oct 02, 2014 11:34 pm
Hey Guys ,

Can anyone pls solve this question.

I want to see the proper solution because i am bit confused in this.
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by GMATGuruNY » Fri Oct 03, 2014 2:12 am
Is |x| < 1 ?

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

(2) |x - 3| > 0
Statement 1: |x + 1| = 2|x - 1|
Case 1: No signs changed
x+1 = 2(x-1)
x+1 = 2x - 2
3 = x.

Case 2: Signs changed in ONE of the absolute values
-(x+1) = 2(x-1)
-x-1 = 2x - 2
-3x = -1
x = 1/3.

Since |x|>1 in Case 1 but |x|<1 in Case 2, INSUFFICIENT.

Statement 2: |x - 3| > 0
Case 3: No signs changed
x-3 > 0
x > 3

Case 4: Signs changed in the absolute value
-(x-3) > 0
-x + 3 > 0
-x > -3
x < 3.

The resulting inequalities -- x<3 or x>3 -- imply that x≠3.
If x=0, then |x|<1.
If x=10, then |x|>1.
INSUFFICIENT.

Statements combined:
Statement 1 requires that x=3 or x=1/3.
Statement 2 requires that x≠3.
Thus, x=1/3, with the result that |x| < 1.
SUFFICIENT.

The correct answer is C.
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by j_shreyans » Fri Oct 03, 2014 9:24 am
Hi Guru ,

Thanks for the explanation , it helps me a lot .

One more thing which i want to be cleared is that.

Case 2: Signs changed in ONE of the absolute values
-(x+1) = 2(x-1)
-x-1 = 2x - 2
-3x = -1
x = 1/3.

In the above case why cant we change the sign on both side like below.

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

Pls suggest me.
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by GMATGuruNY » Fri Oct 03, 2014 10:32 am
j_shreyans wrote:Hi Guru ,

Thanks for the explanation , it helps me a lot .

One more thing which i want to be cleared is that.

Case 2: Signs changed in ONE of the absolute values
-(x+1) = 2(x-1)
-x-1 = 2x - 2
-3x = -1
x = 1/3.

In the above case why cant we change the sign on both side like below.

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

Pls suggest me.
Changing the sign on both sides of an equation is the equivalent of multiplying both sides by -1.
When both sides of an equation are multiplied by the same value, the equation does not change.
Thus, the solution for the -(x+1) = -2(x-1) will the same as the solution for x+1 = 2(x-1):
-(x+1) = -2(x-1)
-x - 1 = -2x + 2
x = 3.
As shown in the my post above, x=3 is also the solution for x+1 = 2(x-1).

Since both cases will yield the same solution, there is no reason to change the sign on both sides of the equation.
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by j_shreyans » Fri Oct 03, 2014 10:45 am
Thank you so much Guru for your help , it really helps.
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