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Absolute value inequality

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Absolute value inequality

by iLdern » Sat Sep 04, 2010 3:35 pm
I just picked this question up from another forum... I want to know if what i'm doing is the correct approach to these kinds of problems. Here goes:

1. Is |x+y|>|x-y|?
(1) |x| > |y|
(2) |x-y| < |x|

My way of approaching this problem:

x or y can either be positive or negative, x will always have the same sign (pos/neg), so it can be canceled out.

1. -x-y>-x+y <=> 0>2y <=> y < 0
2. -x+y > -x-y <=> 2y > 0 <=> y > 0
3. x+y > x-y <=> 2y > 0 <=> y > 0
4. x-y > x+y <=> 0 > 2y <=> y < 0

So, essentially they are asking:is y<0 or y>0 ?

Which would be the same as saying: Does y = 0 ?

Statement 1 doesn't give me much.

Statement 2 tells me (i just cancel out the x's because they will have the same sign anyway)

either y < 0, or y > 0.

Hence B

Am i right or is there another, more efficient, way to solve these kinds of problems?

Thanks for your help
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by Jim@Grockit » Sun Sep 05, 2010 8:02 am
iLdern wrote:I just picked this question up from another forum... I want to know if what i'm doing is the correct approach to these kinds of problems. Here goes:

1. Is |x+y|>|x-y|?
(1) |x| > |y|
(2) |x-y| < |x|

My way of approaching this problem:

x or y can either be positive or negative, x will always have the same sign (pos/neg), so it can be canceled out.

1. -x-y>-x+y <=> 0>2y <=> y < 0
2. -x+y > -x-y <=> 2y > 0 <=> y > 0
3. x+y > x-y <=> 2y > 0 <=> y > 0
4. x-y > x+y <=> 0 > 2y <=> y < 0

So, essentially they are asking:is y<0 or y>0 ?

Which would be the same as saying: Does y = 0 ?

Statement 1 doesn't give me much.

Statement 2 tells me (i just cancel out the x's because they will have the same sign anyway)

either y < 0, or y > 0.

Hence B

Am i right or is there another, more efficient, way to solve these kinds of problems?

Thanks for your help
Faster might be using number properties. Is |x+y|>|x-y|? If one or both variables equals 0, then no (the expressions are equal); likewise, if the signs of x and y are different, then no, |x+y|<|x-y|. If the signs of x and y are the same, then yes, |x+y|>|x-y|. We need to know first 1) whether either number is 0, and then 2) what the signs of the respective nonzero numbers are. If their signs are the same, then |x+y|>|x-y| every time, as adding two numbers of the same sign together takes them further from 0 on the number line. If their signs are different, adding them together brings the total closer to 0, and subtracting them takes them further from 0.

Statement 1 tells us that x is not 0 (there is no way that the absolute value of 0 could be greater than the absolute value of another number) but not much else. INSUFFICIENT
Statement 2 tells us that x is not 0, because no absolute value can be less than 0, and y is not 0 (subtracting y has an effect on x, so y is not 0). Also, if we accept |x-y| < |x| is true, then x and y cannot have opposite signs. As shown above, if they have the same signs and are not 0, |x+y|>|x-y|. SUFFICIENT
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by tanviet » Mon Sep 06, 2010 2:08 am
For yes-no question, alway pick numbers. if the expression is complicated, we simplify a little then pick number. This question have not comclicated expression. So, we pick number



a, pick x=3, y=2
pick x=3, y=-2

not sufficient

b, pick x=3, y=-2
pick x=3, y=2

sufficient.

for yes-no question, gmat test us the characteristics of number, no more. That is why picking number is fastest
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