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Is |x + y| > |x - y| ?

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Is |x + y| > |x - y| ?  [#permalink]

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New post Updated on: 28 Sep 2014, 05:32
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Is |x + y| > |x - y| ?

(1) |x| > |y|

(2) |x - y| < |x|

Originally posted by sonaml on 28 Sep 2014, 05:22.
Last edited by Bunuel on 28 Sep 2014, 05:32, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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Is |x + y| > |x - y| ?  [#permalink]

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New post 28 Sep 2014, 05:32
Is |x + y| > |x - y| ?

Is \(|x+y| > |x-y|\)? --> square it: is \(x^2+2xy+y^2>x^2-2xy+y^2\) --> is \(xy>0\)?

(1) |x| > |y|. We cannot get from this whether x and y have the same sign. Not sufficient.

(2) |x - y| < |x| --> square again: \(x^2-2xy+y^2<x^2\) --> \(xy>\frac{y^2}{2}\) --> since \(y^2\geq{0}\) (the square of any number is more than or equal to 0), then we have that \(xy>\frac{y^2}{2}\geq{0}\). Sufficient.

Answer: B.

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Re: Is |x + y| > |x - y| ?  [#permalink]

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New post 12 Oct 2014, 04:35
Hey bunuel,
please help me out with this..
when we see such equation questions,should we use the squaring method that u used above ? I mean,will it be always useful to use squaring method in such cases ?



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Re: Is |x + y| > |x - y| ?  [#permalink]

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New post 12 Oct 2014, 05:06
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vards wrote:
Hey bunuel,
please help me out with this..
when we see such equation questions,should we use the squaring method that u used above ? I mean,will it be always useful to use squaring method in such cases ?



-Vardhaman


I don't like the word "always". There are cases when you can square an inequality and cases when you cannot square, check here: inequalities-tips-and-hints-175001.html In many cases when you can square, and when you have modulus on both sides squaring might be a good approach.
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Re: Is |x + y| > |x - y| ?  [#permalink]

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New post 31 Dec 2015, 21:12
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 - y| ?

(1) |x| > |y|

(2) |x - y| < |x|

First we need to modify the original condition and the question. Since an absolute value is always a positive number, we can square each side. Then, the original question, which asks if |x + y| > |x - y| is true, can be changed, and we can see how the question is asking if |x + y|^2 > |x - y|^2 is true. Squaring an absolute value and squaring just the value yield the same result. Therefore, we can further modify the question to (x + y)^2 > (x - y)^2? Then, we get x^2+2xy+y^2>x^2-2xy+y^2? We can simplify the question further to 4xy>0? So, we can see, essentially, the question is asking if xy>0 is true.
In the case of the condition 1), since it states |x|>|y|, it does not prove if xy>0.
In the case of the condition 2), we can see that |x-y|<|x|=|x-0|. This means that the absolute difference between x and y is less than that of between x and 0. In order to satisfy this condition, we need xy>0. The answer is ‘yes’ and the condition is sufficient. Therefore, the correct answer is B.

Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
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Re: Is |x + y| > |x - y| ?  [#permalink]

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Re: Is |x + y| > |x - y| ? &nbs [#permalink] 22 Aug 2017, 06:09
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