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Is xy>0?

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Is xy>0?  [#permalink]

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New post 10 Sep 2018, 00:58
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[Math Revolution GMAT math practice question]

Is \(xy>0?\)

\(1) x^3y>0\)
\(2) x^2y>0\)

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Re: Is xy>0?  [#permalink]

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New post 10 Sep 2018, 01:25
From statement 1:

\(x^3\)y>0.

If x is positive then xy>0. If x is negative then y also has to be negative to satisfy \(x^3\)y>0.

In either the case xy will always be greater than 0.

1 is sufficient.

From statement 2:

\(x^2\)y>0.

If x is positive then y is also positive. xy>0.

If x is negative and y is positive then \(x^2\)y>0 is satisfied but xy will be less than 0.

2 is insufficient.

A is the answer.
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Re: Is xy>0?  [#permalink]

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New post 10 Sep 2018, 01:29
Is \(xy>0?\)
Both X and y should be of same sign...

\(1) x^3y>0\)
Even power will always result in positive value..
\(x^3y=x^2*xy>0\), so xy>0
Sufficient

\(2) x^2y>0\)
Just tells us that y>0
If x >0, yes but if x<0, no
Insufficient

A
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Is xy>0?  [#permalink]

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New post 10 Sep 2018, 01:38
MathRevolution wrote:
[Math Revolution GMAT math practice question]

Is \(xy>0?\)

\(1) x^3y>0\)
\(2) x^2y>0\)



we are asked whether xy>0. Two possibilities can make our dream true. Both x and y are positive or both negative.

Statement 1: \(x^3y>0\). If x is negative , y has to be negative. xy>0 in this case. If both x and y are positive still xy has to be positive. SUFFICIENT.

Statement 2: \(x^2\) is always positive but x is not the same as \(x^2\). x could be positive or negative. Thus , this one is NOT Sufficient as we don't know the exact sign of x.

The best answer is A.
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Re: Is xy>0?  [#permalink]

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New post 10 Sep 2018, 06:45
Top Contributor
MathRevolution wrote:
[Math Revolution GMAT math practice question]

Is \(xy>0?\)

\(1) x^3y>0\)
\(2) x^2y>0\)


Target question: Is xy > 0?

Statement 1: x³y > 0
Since x² is POSITIVE, we can safely divide both sides of the inequality by x²
When we do this, we get xy > 0
So, the answer to the target question is YES, xy IS greater than 0
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: x²y > 0
Once again, we can safely divide both sides of the inequality by x²
However, this time we get y > 0
So, we now know that y is POSITIVE, but we don't know whether x is positive or negative.
As such statement 2 is NOT SUFFICIENT

If you're not convinced, we can always test some values
There are several values of x and y that satisfy statement 2. Here are two:
Case a: x = 1 and y = 1. In this case, xy = (1)(1) = 1. So, the answer to the target question is YES, xy IS greater than 0
Case b: x = -1 and y = 1. In this case, xy = (-1)(1) = -1. So, the answer to the target question is NO, xy is NOT greater than 0
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

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Re: Is xy>0?  [#permalink]

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New post 12 Sep 2018, 01:38
=>

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.

Since even powers are always non-negative, we can ignore the even exponents.
Condition 1)
\(x^3y > 0\) is equivalent to \(xy > 0\), which is the question itself. Therefore, condition 1) is sufficient.
Condition 2)
\(x^2y>0\) is equivalent to \(y > 0\).
It is not sufficient

Therefore, A is the answer.
Answer: A
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Re: Is xy>0? &nbs [#permalink] 12 Sep 2018, 01:38
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