Hi there,
Not sure I understand the rephrasing of this DS question. I read the explanation but I simply don't get it. My answer was A. Statement 2 was easy to disregard but after reading the explanation, I clearly see that I disregarded it for the wrong reason.
Explanation:
We can rephrase the question by opening up the absolute value sign. In other words, we must solve all possible scenarios for the inequality, remembering that the absolute value is always a positive value. The two scenarios for the inequality are as follows:
If x > 0, the question becomes "Is x < 1?"
If x < 0, the question becomes: "Is x > -1?"
We can also combine the questions: "Is -1 < x < 1?"
Can you please clarify this explanation if possible - I understand that anything coming out of an absolute value sign will be positive but that's about it as it pertains to this question.
Thanks
A4Fever
GMAT Forum
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- RonPurewal
- Students
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- Joined: Tue Aug 14, 2007 8:23 am
the best way for you to handle these is simply to memorize the template on which their solutions are built.
if you have |x| < 1, that translates as -1 < x < 1. in general, if you have |x| < A, where A is a positive number, you can render that equation as -A < x < A.
the proof of this rephrase lies along almost exactly the same lines as the work you've shown here, but you should NOT perform that same work whenever you encounter one of these inequalities in the field; that's simply not reasonable with the sort of time constraints you're going to be dealing with.
instead, just memorize:
|x| < A --> -A < x < A
|x| > A --> x < -A or x > A
in the extremely unlikely event that A is negative, the first inequality has no solution at all, and the second will be solved by any value of x.
if you have |x| < 1, that translates as -1 < x < 1. in general, if you have |x| < A, where A is a positive number, you can render that equation as -A < x < A.
the proof of this rephrase lies along almost exactly the same lines as the work you've shown here, but you should NOT perform that same work whenever you encounter one of these inequalities in the field; that's simply not reasonable with the sort of time constraints you're going to be dealing with.
instead, just memorize:
|x| < A --> -A < x < A
|x| > A --> x < -A or x > A
in the extremely unlikely event that A is negative, the first inequality has no solution at all, and the second will be solved by any value of x.
4 posts Page 1 of 1