Inequality 7

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Source: — Data Sufficiency |

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by DanaJ » Tue Feb 03, 2009 4:17 am
1. I would again break this down in two cases.
a. x < 0 which means that |x| = -x. This in turn means that x/|x| = x/(-x) = -1. We get that - 1 < x, so for this case we get that -1 < x < 0.
b. x > 0, with |x| = x. In this case x/|x| = 1, so x is greater than 1.
Anyhow, we get lots of possible values for x: -0.5, -0.25, 3, 45, ....
So 1 is not sufficient.

2. |x| > x only happens when x is negative. If x is positive, you get that |x| = x. If x is negative, you get that |x| (which is always positive or equal to zero) > x (which is negative). So we cannot say if |x| > 1.

Now put the two stmts together: from stmt 2 you get that x is negative. Therefore you choose the negative case from stmt 1, which means that - 1 < x < 0. This in turn gives us an idea about the whole thing: |x| will be smaller than 1.

Answer C.

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by sanju09 » Tue Feb 03, 2009 5:28 am
DanaJ wrote:1. I would again break this down in two cases.
a. x < 0 which means that |x| = -x. This in turn means that x/|x| = x/(-x) = -1. We get that - 1 < x, so for this case we get that -1 < x < 0.
b. x > 0, with |x| = x. In this case x/|x| = 1, so x is greater than 1.
Anyhow, we get lots of possible values for x: -0.5, -0.25, 3, 45, ....
So 1 is not sufficient.

2. |x| > x only happens when x is negative. If x is positive, you get that |x| = x. If x is negative, you get that |x| (which is always positive or equal to zero) > x (which is negative). So we cannot say if |x| > 1.

Now put the two stmts together: from stmt 2 you get that x is negative. Therefore you choose the negative case from stmt 1, which means that - 1 < x < 0. This in turn gives us an idea about the whole thing: |x| will be smaller than 1.

Answer C.
yupp... C
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