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Re: If x + y > 0, is xy^2 + x^2y > 0?
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12 Mar 2018, 02:12
MathRevolution wrote:
[GMAT math practice question]
If \(x + y > 0\), is \(xy^2 + x^2y > 0?\)
\(1) x > y\) \(2) xy > 1\)
xy^2 + x^2y can also be written as: xy*(x+y). So the question asks whether xy*(x+y) is > 0 or not? We are given that x+y is > 0 so all we need to know is whether xy is > 0 or not.
Statement 1: x > y But both could be positive, both could be negative or x positive and y negative. So we cannot say what the sign of xy will be. Not suffcient. Statement 2: xy > 1 If xy >1, then definitely xy> 0. Sufficient.
Re: If x + y > 0, is xy^2 + x^2y > 0?
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12 Mar 2018, 07:30
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Top Contributor
MathRevolution wrote:
[GMAT math practice question]
If \(x + y > 0\), is \(xy^2 + x^2y > 0?\)
\(1) x > y\) \(2) xy > 1\)
Given: x + y > 0
Target question:Is xy² + x²y > 0
This is a good candidate for rephrasing the target question.
Aside: Below is a video with tips on rephrasing the target question
Take the inequality: xy² + x²y > 0 Factor to get: xy(y + x)> 0 Since it is GIVEN than x + y > 0, we can write: xy(SOME POSITIVE NUMBER)> 0 For this inequality to be true, it must be the case that xy > 0 So, we can REPHRASE our target question.... REPHRASED target question:Is xy > 0?
Statement 1: x > y There are several values of x and y that satisfy statement 1. Here are two: Case a: x = 2 and y = 1. In this case, xy = 2. So, the answer to the REPHRASED target question is YES, xy IS greater than zero Case b: x = 2 and y = -1. In this case, xy = -2. So, the answer to the REPHRASED target question is NO, xy is NOT greater than zero Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: xy > 1 If xy>1, then it MUST be the case that xy > 0 So, the answer to the REPHRASED target question is YES, xy IS greater than zero Since we can answer the REPHRASED target question with certainty, statement 2 is SUFFICIENT
Re: If x + y > 0, is xy^2 + x^2y > 0?
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14 Mar 2018, 00:22
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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.
Question: \(xy^2 + x^2y > 0\) \(=> xy(x+y) > 0\) \(=> xy > 0\), since \(x + y > 0\) So, the question asks if \(xy > 0.\)
Since \(xy > 1 > 0\) can be derived from condition 2), it is sufficient. Condition 1) gives us no information about the sign of \(xy\).
Re: If x + y > 0, is xy^2 + x^2y > 0?
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29 Aug 2019, 10:11
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