If x is not equal to 0, is |x| less than 1?
(1)x/|x|<x
(2) |x| > x
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|x| less than 1
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j_shreyans
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GMATGuruNY
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Question rephrased: Is x between -1 and 1?If x is not equal to 0, is |x| less than 1?
(1) x/|x|< x
(2) |x| > x
Statement 1: x/|x| < x
x < x|x|
0 < x|x| - x
0 < x (|x| - 1)
The CRITICAL POINTS are -1, 0 and 1.
These are the only values where x(|x|-1) = 0.
To determine the ranges where x(|x|-1) > 0, test one value to the left and right of each critical point.
Case 1: x<-1
Plug x = -2 into x/|x| < x:
-2/ |-2| < -2
-1 < -2.
Doesn't work.
Thus, x < -1 is not a valid range.
Case 2: -1<x<0
Plug x = -1/2 into x/|x| < x:
-1/2/ |-1/2| < -1/2
-1 < -1/2.
This works.
Thus, -1<x<0 is a valid range.
Case 3: 0<x<1
Plug x = 1/2 into x/|x| < x:
(1/2)/ |1/2| < 1/2
1 < 1/2
Doesn't work.
Thus, 0<x<1 is not a valid range.
Case 4: x>1
Plug x = 2 into x/|x| < x:
2/ |2| < 2
1 < 2.
This works.
Thus, x > 1 is a valid range.
Thus, -1<x<0 or x>1.
INSUFFICIENT.
Statement 2: |x| > x
Any negative value will satisfy this inequality.
If x=-1/2, then x is between -1 and 1.
If x=-2, then x is NOT between -1 and 1.
INSUFFICIENT.
Statements combined:
The only range that satisfies both statements is -1<x<0.
Thus, x is between -1 and 1.
SUFFICIENT.
The correct answer is C.
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Hi j_shreyans,
This DS question is built around some interesting Number Properties and patterns. If you can spot those patterns, then solving this problem should take considerably less time. This also looks like a question that can be beaten by TESTing VALUES.
We're told that X CANNOT = 0. We're asked if |X| < 1. This is a YES/NO question.
Fact 1: X/|X| < X
Before TESTing VALUES, I want to note a pattern in this inequality:
X/|X| will either equal 1 (if X is positive) OR -1 (if X is negative). This will save us some time when it comes to TESTing VALUES, since there are many values of X that will NOT fit this information.
If X = 2, then the answer to the question is NO.
X cannot be 1, any positive fraction, 0, or any negative integer.....
So what's left to TEST....?
If X = -1/2, then the answer to the question is YES.
Fact 1 is INSUFFICIENT
Fact 2: |X| > X
This tells us that X CANNOT be positive or 0.
If X = -1, then the answer to the question is NO.
If X = -1/2, then the answer to the question is YES.
Fact 2 is INSUFFICIENT
Combined, we have deal with the "overlapping restrictions" that we noted in the two Facts:
X cannot be....anything positive, 0, or any negative integer.
X can ONLY BE negative fractions between 0 and -1.
ALL of those answers (e.g. -1/2, -.4, etc.) lead to a YES answer.
Combined SUFFICIENT.
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
This DS question is built around some interesting Number Properties and patterns. If you can spot those patterns, then solving this problem should take considerably less time. This also looks like a question that can be beaten by TESTing VALUES.
We're told that X CANNOT = 0. We're asked if |X| < 1. This is a YES/NO question.
Fact 1: X/|X| < X
Before TESTing VALUES, I want to note a pattern in this inequality:
X/|X| will either equal 1 (if X is positive) OR -1 (if X is negative). This will save us some time when it comes to TESTing VALUES, since there are many values of X that will NOT fit this information.
If X = 2, then the answer to the question is NO.
X cannot be 1, any positive fraction, 0, or any negative integer.....
So what's left to TEST....?
If X = -1/2, then the answer to the question is YES.
Fact 1 is INSUFFICIENT
Fact 2: |X| > X
This tells us that X CANNOT be positive or 0.
If X = -1, then the answer to the question is NO.
If X = -1/2, then the answer to the question is YES.
Fact 2 is INSUFFICIENT
Combined, we have deal with the "overlapping restrictions" that we noted in the two Facts:
X cannot be....anything positive, 0, or any negative integer.
X can ONLY BE negative fractions between 0 and -1.
ALL of those answers (e.g. -1/2, -.4, etc.) lead to a YES answer.
Combined SUFFICIENT.
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
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Max@Math Revolution
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In the case of the inequality given in the question, squares and inclusive relationships are important. In other words, the core of solving this question comes down to 1) squares 2) if the question's range includes the condition's range, then the condition is sufficient and becomes the answer.
Changing the question, it becomes "-1<x<1?"
1) From "If x>0, x/|x|<x ==> x/x<x ==> 1<x", we get "1<x" and from "If x<0, x/|x|<x ==> x/-x<x ==> -1<x" we get -1<x<0
2) We get |x|>x ==> x<0 by substituting x=-1
question: -1<x<1?
1) From -1<x<0 or 1<x, The question's range does not include the range of condition 1, so this condition is not sufficient
2) For x<0 also, the question's range does not include the range of condition 2, so this condition is not sufficient
If we take both conditions together, 1) & 2) -1<x<0, then the question does include the range of both the conditions, so this is sufficient and the answer becomes C.
With inequality questions, don't forget that you can divide both sides by squares without affecting the direction of the inequality, since squares are always positive, and also that if the que's range includes the con's range, then the con becomes sufficient. These two facts will help you solve inequality DS problems.
Changing the question, it becomes "-1<x<1?"
1) From "If x>0, x/|x|<x ==> x/x<x ==> 1<x", we get "1<x" and from "If x<0, x/|x|<x ==> x/-x<x ==> -1<x" we get -1<x<0
2) We get |x|>x ==> x<0 by substituting x=-1
question: -1<x<1?
1) From -1<x<0 or 1<x, The question's range does not include the range of condition 1, so this condition is not sufficient
2) For x<0 also, the question's range does not include the range of condition 2, so this condition is not sufficient
If we take both conditions together, 1) & 2) -1<x<0, then the question does include the range of both the conditions, so this is sufficient and the answer becomes C.
With inequality questions, don't forget that you can divide both sides by squares without affecting the direction of the inequality, since squares are always positive, and also that if the que's range includes the con's range, then the con becomes sufficient. These two facts will help you solve inequality DS problems.
