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Is |x-y|<|x-z|? 1) |y|<|z| 2) x<0 * A solution w

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Is |x-y|<|x-z|? 1) |y|<|z| 2) x<0 * A solution w

by Max@Math Revolution » Wed Mar 23, 2016 5:54 pm
Is |x-y|<|x-z|?

1) |y|<|z|
2) x<0


* A solution will be posted in two days.
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Source: — Data Sufficiency |
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by GMATGuruNY » Fri Mar 25, 2016 4:28 am
Max@Math Revolution wrote:Is |x-y|<|x-z|?

1) |y|<|z|
2) x<0
|a-b| = the DISTANCE between a and b on the number line.

Question stem, rephrased:
On the number line, is the distance between x and y less than the distance between x and z?
In other words:
On the number line, is x closer to y than to z?

Both statements are satisfied by the following cases:
.....x=-3.....y=-2..............................z=3......
.....x=-3.....z=-2.....y=-1.............................

In the first case, x is closer to y than to z, so the answer to the question stem is YES.
In the second case, x is closer to z than to y, so the answer to the questions stem is NO.
Since the answer is YES in the first case but NO in the second case, the two statements combined are INSUFFICIENT.

The correct answer is E.
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by Max@Math Revolution » Sun Mar 27, 2016 12:35 am
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is |x-y|<|x-z|?

1) |y|<|z|
2) x<0


In the original condition, there are 3 variables(x,y,z), which should match with the number of equations. So you need 3 equations. For 1) 1 equation, for 2) 1 equation. So you need 1 equation, which is likely to make E the answer.
When 1) & 2), y=-2, x=-1, z=10 yes but y=2, z=-10, x=-9 no, which is not sufficient.
Thus, the answer is E.
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