[GMAT math practice question]
$$If\ x+\ y>0,\ is\ xy^2+x^2y>0?$$
$$1)\ x>y$$
$$2)\ xy>1$$
If x + y > 0, is xy^2 + x^2y > 0?
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- Max@Math Revolution
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Since x+y>0, the inequality in the question stem can safely be divided by x+y, as follows:Max@Math Revolution wrote:[GMAT math practice question]
$$If\ x+\ y>0,\ is\ xy^2+x^2y>0?$$
$$1)\ x>y$$
$$2)\ xy>1$$
xy² + x²y > 0
xy(y + x) > 0
[xy(y+x)]/(x+y) > 0/(x+y)
xy > 0.
Question stem, rephrased:
Is xy > 0?
Statement 1: x>y
Case 1: x=2 and y=1, satisfying the constraints that x+y > 0 and x>y
In this case, xy > 0, so the answer to the rephrased question stem is YES.
Case 2: x=2 and y=-1, satisfying the constraints that x+y > 0 and x>y
In this case, xy < 0, so the answer to the rephrased question stem is NO.
Since the answer is YES in Case 1 but NO in Case 2, INSUFFICIENT.
Statement 2: xy>1
Here, xy>0, so the answer to the rephrased question stem is YES.
SUFFICIENT.
The correct answer is B.
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- Brent@GMATPrepNow
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Given: x + y > 0Max@Math Revolution wrote:[GMAT math practice question]
$$If\ x+\ y>0,\ is\ xy^2+x^2y>0?$$
$$1)\ x>y$$
$$2)\ xy>1$$
Target question: Is xy² + x²y > 0
This is a good candidate for rephrasing the target question.
Aside: Here's a video with tips on rephrasing the target question: https://www.gmatprepnow.com/module/gmat- ... cy?id=1100
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
Answer: B
Cheers,
Brent
- Max@Math Revolution
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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.
Therefore, B is the answer.
Answer: B
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.
Therefore, B is the answer.
Answer: B
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