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C
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Difficulty:
95%
(hard)
Question Stats:
43% (02:29) correct
57%
(02:23)
wrong
based on 868
sessions
History
Date
Time
Result
Not Attempted Yet
28 Jul 2010, 04:22
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Is \(|x| < 1\)?
Is \(|x| < 1\), means is \(x\) in the range (-1,1) or is \(-1<x<1\) true?
(1) \(|x + 1| = 2|x - 1|\)
Two key points: \(x=-1\) and \(x=1\) (key points are the values of \(x\) when absolute values equal to zero), thus three ranges to check:
---------{-1}--------{1}---------
A. \(x<-1\) (blue range) --> \(|x + 1| = 2|x - 1|\) becomes: \(-x-1=2(-x+1)\) --> \(x=3\), not OK, as this value is not in the range we are checking (\(x<-1\));
B. \(-1\leq{x}\leq{1}\) (green range) --> \(|x + 1| = 2|x - 1|\) becomes: \(x+1=2(-x+1)\) --> \(x=\frac{1}{3}\). OK, as this value is in the range we are checking (\(-1\leq{x}\leq{1}\));
C. \(x>1\) (red range) --> \(|x + 1| = 2|x - 1|\) becomes: \(x+1=2(x-1)\) --> \(x=3\). OK, as this value is in the range we are checking (\(x>1\)).
So we got TWO values of \(x\) (two solutions): \(\frac{1}{3}\) and \(3\), first is in the range (-1,1) but second is out of the range. Not sufficient.
Alternately we could just expand absolute values of RHS and LHS with same sign (for example both LHS and RHS positive: \(x + 1 = 2(x - 1)\)) and then with different sign (for example LHS positive and RHS negative \(x + 1 = 2(-x + 1)\)), solve for \(x\) both equations, and finally check whether the solutions satisfy \(|x + 1| = 2|x - 1|\).
OR: we can square given equation to get rid of the modulus: \((x + 1)^2 = 4(x - 1)^2\) --> \(3x^2-10x+3=0\) --> \(x=3\) or \(x=\frac{1}{3}\).
(2) \(|x - 3|>{0}\). Absolute value is always non-negative, more than or equal to zero: \(|some \ expression|\geq{0}\). We are told that absolute value of \(x-3\) is MORE than zero, so just it says that \(|x-3|\neq{0}\), which simply means that \(x\neq{3}\). But we don't know whether \(x\) is in the range (-1,1) or not.
(1)+(2) \(x=\frac{1}{3}\) or \(x=3\) AND \(x\neq{3}\) --> means \(x\) can have only value \(\frac{1}{3}\), which is in the range (-1,1). Sufficient.
Answer: C.
Hope it helps.
Is \(|x| < 1\), means is \(x\) in the range (-1,1) or is \(-1<x<1\) true?
(1) \(|x + 1| = 2|x - 1|\)
Two key points: \(x=-1\) and \(x=1\) (key points are the values of \(x\) when absolute values equal to zero), thus three ranges to check:
---------{-1}--------{1}---------
A. \(x<-1\) (blue range) --> \(|x + 1| = 2|x - 1|\) becomes: \(-x-1=2(-x+1)\) --> \(x=3\), not OK, as this value is not in the range we are checking (\(x<-1\));
B. \(-1\leq{x}\leq{1}\) (green range) --> \(|x + 1| = 2|x - 1|\) becomes: \(x+1=2(-x+1)\) --> \(x=\frac{1}{3}\). OK, as this value is in the range we are checking (\(-1\leq{x}\leq{1}\));
C. \(x>1\) (red range) --> \(|x + 1| = 2|x - 1|\) becomes: \(x+1=2(x-1)\) --> \(x=3\). OK, as this value is in the range we are checking (\(x>1\)).
So we got TWO values of \(x\) (two solutions): \(\frac{1}{3}\) and \(3\), first is in the range (-1,1) but second is out of the range. Not sufficient.
Alternately we could just expand absolute values of RHS and LHS with same sign (for example both LHS and RHS positive: \(x + 1 = 2(x - 1)\)) and then with different sign (for example LHS positive and RHS negative \(x + 1 = 2(-x + 1)\)), solve for \(x\) both equations, and finally check whether the solutions satisfy \(|x + 1| = 2|x - 1|\).
OR: we can square given equation to get rid of the modulus: \((x + 1)^2 = 4(x - 1)^2\) --> \(3x^2-10x+3=0\) --> \(x=3\) or \(x=\frac{1}{3}\).
(2) \(|x - 3|>{0}\). Absolute value is always non-negative, more than or equal to zero: \(|some \ expression|\geq{0}\). We are told that absolute value of \(x-3\) is MORE than zero, so just it says that \(|x-3|\neq{0}\), which simply means that \(x\neq{3}\). But we don't know whether \(x\) is in the range (-1,1) or not.
(1)+(2) \(x=\frac{1}{3}\) or \(x=3\) AND \(x\neq{3}\) --> means \(x\) can have only value \(\frac{1}{3}\), which is in the range (-1,1). Sufficient.
Answer: C.
Hope it helps.
mustu
As I have said before, most mod questions are best tackled using a number line. You don't need to do many calculations then.
|x| means the distance from 0.
|x-3| means the distance from 3.
etc. For details of this approach, check out:
Let's go on to this question now.
Is |x| < 1 i.e. Is the distance of point x from 0 less than 1?
Statement 1: |x + 1| = 2|x – 1|
This means 'distance of x from -1 is twice the distance of x from 1'. Draw the number line now. There will be 2 points where the distance from -1 will be twice the distance from 1.
Attachment:
Ques6.jpg [ 5.12 KiB | Viewed 10519 times ]
For one of these points, distance from 0 is less than 1, for the other it is more than 1. So not sufficient.
Statement 2: |x – 3| > 0
This statement tells us that distance of x from 3 is more than 0 i.e. x does not lie at 3. It can lie anywhere else.
You can look at it in another way: Mods are always more than or equal to 0. All this statement tells us is that this mod is not equal to zero i.e. x is not equal to 3.
For some of these points, distance from 0 will be less than 1, for the others it will be more than 1. So not sufficient.
Using both statements together, statement 1 says that x is either 3 or a point between 0 and 1 (which I don't really need to calculate). Statement 2 tells us that x is not 3. So together, x must be a point between 0 and 1 and its distance from 0 must be less than 1. Sufficient.
Answer C.
I dare say, this problem is a little bit harder than what the GMAT will ask of you. 
