If x≠0, is |x| < 1 ?
(1) x^2 < 1
(2) |x| < 1/x
GMAT Prep Absolute question
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Statement 1:Mo2men wrote:If x≠0, is |x| < 1 ?
(1) x^2 < 1
(2) |x| < 1/x
Since x is nonzero, only a positive or negative fraction between -1 and 1 will satisfy x²<1.
The absolute value of a positive or negative fraction must be less than 1.
SUFFICIENT.
Statement 2:
Only a positive fraction between 0 and 1 will satisfy |x| < 1/x.
The absolute value of a positive fraction must be less than 1.
SUFFICIENT.
The correct answer is D.
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Question: If x ≠0, is |x| < 1?
Rephrase: Is -1 < x < 1
Statement 1: x² < 1
If x² < 1, then -1 < x < 1.
Sufficient.
Statement 2: |x| < 1/x
Try some numbers. There are six categories that could be tried.
x < -1, x = -1, -1 < x < 0, 0 < x < 1, x = 1, 1 < x
-2: |-2| < 1/-2 This is impossible as the left side is positive and the right side is negative. A positive number is not less than a negative number.
No negative numbers work. Try the positive categories
1/2: |1/2| < 1/(1/2) This is true.
1: |1| < 1/1 This is not true.
2: |2| < 1/2 This is not true, and illustrates the fact that for any x such that x > 1, Statement 2 will not be true.
The only one that works is 1/2.
|1/2| < 1
Sufficient.
The correct answer is D.
There is a trick to Statement 2.
One may think that one can multiply both sides by x and get x|x| < 1.
However that idea is not entirely correct.
In working with the inequality have to consider the possibility that x could be negative, in which case when we multiply both sides by x, the sign of the inequality would have to be reversed.
In considering the case in which x is negative, when we multiply both sides by x, we get x|x| > 1.
Since a negative number multiplied by an absolute value must generate a negative product, if x is negative, x|x| > 1 is not true.
So once again we have showed that x cannot be negative.
We also have to consider the case in which x is positive. If x is positive when we multiply we get x|x| < 1.
That statement is true when x is a fraction.
So 0 < x < 1, and once again we have showed that Statement 2 is sufficient.
Rephrase: Is -1 < x < 1
Statement 1: x² < 1
If x² < 1, then -1 < x < 1.
Sufficient.
Statement 2: |x| < 1/x
Try some numbers. There are six categories that could be tried.
x < -1, x = -1, -1 < x < 0, 0 < x < 1, x = 1, 1 < x
-2: |-2| < 1/-2 This is impossible as the left side is positive and the right side is negative. A positive number is not less than a negative number.
No negative numbers work. Try the positive categories
1/2: |1/2| < 1/(1/2) This is true.
1: |1| < 1/1 This is not true.
2: |2| < 1/2 This is not true, and illustrates the fact that for any x such that x > 1, Statement 2 will not be true.
The only one that works is 1/2.
|1/2| < 1
Sufficient.
The correct answer is D.
There is a trick to Statement 2.
One may think that one can multiply both sides by x and get x|x| < 1.
However that idea is not entirely correct.
In working with the inequality have to consider the possibility that x could be negative, in which case when we multiply both sides by x, the sign of the inequality would have to be reversed.
In considering the case in which x is negative, when we multiply both sides by x, we get x|x| > 1.
Since a negative number multiplied by an absolute value must generate a negative product, if x is negative, x|x| > 1 is not true.
So once again we have showed that x cannot be negative.
We also have to consider the case in which x is positive. If x is positive when we multiply we get x|x| < 1.
That statement is true when x is a fraction.
So 0 < x < 1, and once again we have showed that Statement 2 is sufficient.
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Dear GMATGuru,GMATGuruNY wrote:Statement 1:Mo2men wrote:If x≠0, is |x| < 1 ?
(1) x^2 < 1
(2) |x| < 1/x
Since x is nonzero, only a positive or negative fraction between -1 and 1 will satisfy x²<1.
The absolute value of a positive or negative fraction must be less than 1.
SUFFICIENT.
Statement 2:
Only a positive fraction between 0 and 1 will satisfy |x| < 1/x.
The absolute value of a positive fraction must be less than 1.
SUFFICIENT.
The correct answer is D.
In statement 2,
if I want to use critical point. I struggle when x<0
is it -x<1/x or -x<1/-x?
Should replace every 'x' with '-x' or only 'x' inside modulus??
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Since an absolute value cannot be less than a negative, |x| < 1/x is valid only if x is nonnegative.Mo2men wrote: Dear GMATGuru,
In statement 2,
if I want to use critical point. I struggle when x<0
is it -x<1/x or -x<1/-x?
Should replace every 'x' with '-x' or only 'x' inside modulus??
Thus, we can simplify as follows:
x < 1/x
x² < 1
x < 1.
Since x≠0, 0 < x < 1.
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We are given that x is not equal to 0 and we need to determine whether the absolute value of x is less than 1.Mo2men wrote:If x≠0, is |x| < 1 ?
(1) x^2 < 1
(2) |x| < 1/x
Statement One Alone:
x^2 < 1
We can take the square root of both sides of the inequality x^2 < 1:
√(x^2) < √1.
Since √(x^2) = |x|, |x| < 1.
Statement one alone is sufficient. Eliminate answer choices B, C and E.
Statement Two Alone:
|x| < 1/x
Since |x| will always be positive, in order for 1/x to be greater than |x|, x must be positive. Furthermore, the only way |x| < 1/x holds true is if x is a proper fraction. Thus, x must be a positive proper fraction. Since a positive proper fraction is between zero and one, the absolute value of x is less than one. Statement two alone is also sufficient.
Answer: D
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We could start by rephrasing the prompt:
Is -1 < x < 1?
S1:
x*x < 1
It's probably easiest to see if we plug in numbers. If x ≥ 1, this won't work. If -1 ≥ x it won't work either. So we must have -1 < x < 1, and this answers our question! Sufficient.
S2:
|x| < 1/x
If x is positive, this is the same as statement one! We multiply both sides by x, and get |x| * x < 1.
If x is negative, this is impossible, since the left side is positive and the right side is negative.
So we must have x positive, and each statement is sufficient.
Is -1 < x < 1?
S1:
x*x < 1
It's probably easiest to see if we plug in numbers. If x ≥ 1, this won't work. If -1 ≥ x it won't work either. So we must have -1 < x < 1, and this answers our question! Sufficient.
S2:
|x| < 1/x
If x is positive, this is the same as statement one! We multiply both sides by x, and get |x| * x < 1.
If x is negative, this is impossible, since the left side is positive and the right side is negative.
So we must have x positive, and each statement is sufficient.