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If w^2x^3y^4z^5<0, is xyz>0?

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Math Revolution GMAT Instructor
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If w^2x^3y^4z^5<0, is xyz>0?  [#permalink]

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New post 18 Jan 2018, 00:41
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Question Stats:

53% (01:34) correct 47% (01:23) wrong based on 47 sessions

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[GMAT math practice question]

If \(w^2x^3y^4z^5<0\), is \(xyz>0\)?

\(1) x<0\)
\(2) y<0\)

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Re: If w^2x^3y^4z^5<0, is xyz>0?  [#permalink]

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New post 18 Jan 2018, 04:08
MathRevolution wrote:
[GMAT math practice question]

If \(w^2x^3y^4z^5<0\), is \(xyz>0\)?

\(1) x<0\)
\(2) y<0\)


We're asked if a product is positive or negative so we'll look for an answer to do with number properties.
This is a Logical approach.

Since any even power is always positve, then w^2 and y^4 are positive.
Then x^3 * z^5 is negative meaning that one of x,z must be negative and the other must be positive.
So, x*z is negative. Therefore to know if xyz is positive or negative we need to know if y is positive or negative.

(1) adds no information about y.
Insufficient!

(2) tells us that y is negative.
Sufficient!

(B) is our answrr.
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Re: If w^2x^3y^4z^5<0, is xyz>0?  [#permalink]

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New post 21 Jan 2018, 17:35
=>

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, and then recheck the question.

Modifying the original condition and question:
The original condition \(w^2x^3y^4z^5<0\) is equivalent to \(xz<0\) since we can ignore terms with even exponents in this type of inequality (they are always positive).

Under the modified condition \(xz < 0\), the question, ‘is \(xyz > 0\)?’ is equivalent to ‘is \(y < 0\)?’, which is the same as condition 2).

Since condition 1) tells us nothing about the sign of y, the answer is B.

Therefore, the answer is B.
Answer: B
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Re: If w^2x^3y^4z^5<0, is xyz>0? &nbs [#permalink] 21 Jan 2018, 17:35
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