Is |x-y| > |x| - |y|?
1) y< x
2) xy< 0
Anyone know how to solve this algebraically?
Source: GMATPREP1
OA: B
1) y< x
2) xy< 0
Anyone know how to solve this algebraically?
Source: GMATPREP1
OA: B
1. is insufficient. In order to prove this, pick some numbers:
a. x = 2 and y = 1:
|x - y| = 1
|x| - |y| = 1
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No, |x - y| is not greater than |x| - |y| (in this example, they're equal).
b. x = 1 and y = -2
|x - y| = 3
|x| - |y| = 1 - 2 = -1
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Yes, |x - y| is greater than |x| - |y| (because 3 > -1).
So, even if you choose y < x, you will get two conflicting results for the two pairs.
2. What this tells us is that x and y have different signs. A product of two numbers is negative only if one has a positive sign and the other has a negative sign. Again, we have two cases:
a. x is positive and y is negative
Since y is negative, -y will be positive. So x - y will be a positive ("x") plus another positive ("-y").
Now, |x| = x (since x is positive). This means that the right hand side of the inequality reads something like: positive ("x") minus another positive ("|y|" = -y).
So, this is why, yes, the inequality holds true.
b. x is negative and y is positive
This time, it's in reverse. Since y is positive, -y will be a negative. So x - y will be a negative ("x") plus another negative ("-y"). However, since we have the absolute value, it will be equal to the value described in the previous point.
The right hand side will have the same issue: you subtract something from a positive value, instead of adding it.
Again, yes, the inequality holds true.
This is why the answer is indeed the one pointed out by you. If anything is unclear, let me know and I'll try to explain better.
