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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7599
GMAT 1: 760 Q51 V42 GPA: 3.82

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Difficulty:   15% (low)

Question Stats: 78% (00:50) correct 22% (01:19) wrong based on 58 sessions

### HideShow timer Statistics [GMAT math practice question]

Is $$\frac{x}{y^3}>0$$?

1) $$x-y>0$$
2) $$xy>0$$

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MathRevolution wrote:
[GMAT math practice question]

Is $$\frac{x}{y^3}>0$$?

1) $$x-y>0$$
2) $$xy>0$$

we are looking for whether x/y^3 >0 or not

in order to answer this question we have do know that quotient will be positive if both numerator and denominator are positive or both negetive.

1) x-y > 0
from this statement , we only know x> y but we know anything about their signs.
clearly insufficient.

2) xy> 0
it means that either x and y both are positive or both negative. thus , it's sufficient.

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MathRevolution wrote:
[GMAT math practice question]

Is $$\frac{x}{y^3}>0$$?

1) $$x-y>0$$
2) $$xy>0$$

Good question x/y^3 >0 so the question basically ask whether x and y have same sign as x/y^>0 if they have same sign .
From 1
x=3 y=2 then 3/8 >0
but when x=3 and y=-2 3/-8^3 <0 so insufficient

From 2
They have same sign therefore x/y^>0 hence sufficient .

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Math Revolution GMAT Instructor V
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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.

The question can be modified as follows:
$$\frac{x}{y^3}>0 ⇔ xy>0.$$
This can be seen by multiplying both sides by $$y^4$$.
It is same as condition 2).

Condition 1)
If $$x = 2$$ and $$y = 1, xy = 2 > 0$$, and the answer is ‘yes’.
If $$x = 2$$ and $$y = -1, xy = -2 < 0$$, and the answer is ‘no’.
Since we don’t have a unique answer, condition 1) is not sufficient.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
_________________ Re: Is x/y^3>0?   [#permalink] 08 Mar 2018, 02:22
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