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Is x/yz>0?

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Math Revolution GMAT Instructor
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Is x/yz>0?  [#permalink]

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New post 27 Mar 2018, 01:17
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Difficulty:

  35% (medium)

Question Stats:

74% (01:58) correct 26% (01:51) wrong based on 46 sessions

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[GMAT math practice question]

Is \(\frac{x}{yz}>0\)?

\(1) yz>x^2\)
\(2) x<y+z\)

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Re: Is x/yz>0?  [#permalink]

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New post 27 Mar 2018, 05:41
MathRevolution wrote:
[GMAT math practice question]

Is \(\frac{x}{yz}>0\)?

\(1) yz>x^2\)
\(2) x<y+z\)


x/yz will be > 0 if signs of both 'x' and 'yz' are same, i.e., either both are positive or both are negative.

Statement 1:
If yz > x^2 then yz has to be positive (because square of x can not be negative).
But sign of x is not given. So we cannot conclude. Not sufficient.

Statement 2:
x < y+z. This doesnt tell anything about signs of y, z or x. Not sufficient.

Combining the statements:
yz is positive so either both y/z are positive, or both y/z are negative.
Lets take the case of y=3, z=2. Here yz =6 and y+z = 5. Now, as per the two statements:

x^2 < 6 and x < 5. Lets take x=2, it satisfies both. And we can also take x=-1, it also satisfies both. But in first case, x/yz will be positive while in second case x/yz will be negative. So not sufficient, since we cannot determine the sign of x/yz.

Hence E answer
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Re: Is x/yz>0?  [#permalink]

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New post 27 Mar 2018, 17:58
MathRevolution wrote:
[GMAT math practice question]

Is \(\frac{x}{yz}>0\)?

\(1) yz>x^2\)
\(2) x<y+z\)


I have question for solution proposed here, from \(yx> x^2\), we know yz is positive since value of \(x^2\) can never be negative. We know at this point that y & z have same sign.
This also follows

\(xyz > 0\) we can multiply yz on both sides since multiplying +ve number on both side of in-equality will not chnage the sign. This would mean xyz >0 can only hold trye if and only if x is +ve, as only +ve * +ve results in >0.

Please let me know if I am over extending the problem
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Re: Is x/yz>0?  [#permalink]

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New post 29 Mar 2018, 00:08
=>

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.

Since we have 3 variables (x, y and z) and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2) together:
If \(x = 1, y = 2\) and \(z = 3\), then the answer is ‘yes’.
If \(x = -1, y = 2\) and \(z = 3\), then the answer is ‘no’.

Since we don’t have a unique answer, both conditions together are not sufficient.

Therefore, E is the answer.

Answer: E
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Re: Is x/yz>0?  [#permalink]

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New post 03 Apr 2018, 21:49
sameerspice wrote:
MathRevolution wrote:
[GMAT math practice question]

Is \(\frac{x}{yz}>0\)?

\(1) yz>x^2\)
\(2) x<y+z\)


I have question for solution proposed here, from \(yx> x^2\), we know yz is positive since value of \(x^2\) can never be negative. We know at this point that y & z have same sign.
This also follows

\(xyz > 0\) we can multiply yz on both sides since multiplying +ve number on both side of in-equality will not chnage the sign. This would mean xyz >0 can only hold trye if and only if x is +ve, as only +ve * +ve results in >0.

Please let me know if I am over extending the problem


You are talking about the condition 1) only.
It is valid for the condition 1). It not over-extending.
Depending on x, x/yz could be positive or negative.
Thus the condition 1) only is not sufficient.

If you consider the condition 2), the above description is not valid.
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Re: Is x/yz>0? &nbs [#permalink] 03 Apr 2018, 21:49
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