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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SET 28 QUESTION 5 ABS VALUE
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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
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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jessicamuniz
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cramya
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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!
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!
-
cramya
- Legendary Member
- Posts: 2467
- Joined: Thu Aug 28, 2008 6:14 pm
- Thanked: 331 times
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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!
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!
-
jessicamuniz
- Junior | Next Rank: 30 Posts
- Posts: 26
- Joined: Thu Jul 17, 2008 1:02 pm
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!
-
rohangupta83
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