If x is not equal to 0, is |x| less than 1?
(1)
x/|x|< x
(2) |x| > x
I tried number punching method for this problem. I am thinking each statement alone is sufficient to answer the question stem.
Can anyone explain me why official answer is C to this question.
How many times I have tried, I come to the same conclusion that each statement alone is sufficient to answer.
Any help will be greatly appreciated.
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- chintuiisc
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Re: MGMAT CAT 1 DS question - Inequalities Chintu
Hi,
When you post problems on here, please be specific about what you tried. It's difficult for us to give any meaningful feedback/advice if all you say is "I tried xxxxxxx method", and you don't tell us what you actually did.
The left-hand side of this equation is 1 if x is any positive number, and -1 if x is any negative number. (If you don't see why, just throw a few values in there, and it should become clear pretty quickly what is going on.)
So, this statement will be true under exactly one of 2 possible conditions:
1/
x is between -1 and 0. (In this case, the left-hand side is -1, so x is bigger.)
or
2/
x is greater than 1. (In this case, the left-hand side is 1, so x is bigger.)
If you pick a number from case #1 (e.g., -0.5), the answer to the question is "yes". If you pick a number from case #2 (e.g., 2, 3, 4, etc.), then the answer is "no".
Not sufficient.
If x is zero or positive, then |x| and x are the same. So, this statement is just a roundabout way of saying that x is negative.
Note, x can be any negative number in this statement.
If you pick a number between -1 and 0, then the answer to the question is "yes".
If you pick a number that's -1 or less, then the answer is "no".
Not sufficient.
If you put the two statements together, then the only x's that work are between -1 and 0. (If you plug numbers, these will be the only "plugged" numbers that will work.)
All such x's will give an answer of "yes" to the question, so, the statements together are sufficient. It's C.
When you post problems on here, please be specific about what you tried. It's difficult for us to give any meaningful feedback/advice if all you say is "I tried xxxxxxx method", and you don't tell us what you actually did.
chintuiisc Wrote:If x is not equal to 0, is |x| less than 1?
(1)
x/|x|< x
The left-hand side of this equation is 1 if x is any positive number, and -1 if x is any negative number. (If you don't see why, just throw a few values in there, and it should become clear pretty quickly what is going on.)
So, this statement will be true under exactly one of 2 possible conditions:
1/
x is between -1 and 0. (In this case, the left-hand side is -1, so x is bigger.)
or
2/
x is greater than 1. (In this case, the left-hand side is 1, so x is bigger.)
If you pick a number from case #1 (e.g., -0.5), the answer to the question is "yes". If you pick a number from case #2 (e.g., 2, 3, 4, etc.), then the answer is "no".
Not sufficient.
(2) |x| > x
If x is zero or positive, then |x| and x are the same. So, this statement is just a roundabout way of saying that x is negative.
Note, x can be any negative number in this statement.
If you pick a number between -1 and 0, then the answer to the question is "yes".
If you pick a number that's -1 or less, then the answer is "no".
Not sufficient.
If you put the two statements together, then the only x's that work are between -1 and 0. (If you plug numbers, these will be the only "plugged" numbers that will work.)
All such x's will give an answer of "yes" to the question, so, the statements together are sufficient. It's C.
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