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!
Value of |x|
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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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Rich
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.
GMAT assassins aren't born, they're made,
Rich
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All is correct. Just another way to look at Statement 2 istanvis1120 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!
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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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.
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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Hey Guys ,
Can anyone pls solve this question.
I want to see the proper solution because i am bit confused in this.
Can anyone pls solve this question.
I want to see the proper solution because i am bit confused in this.
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Statement 1: |x + 1| = 2|x - 1|Is |x| < 1 ?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| > 0
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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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.
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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Changing the sign on both sides of an equation is the equivalent of multiplying both sides by -1.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.
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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