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# Is x(y/z) > 0?

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29 Dec 2012, 08:51
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Is x(y/z) > 0?

(1) xyz > 0
(2) yz > 0
[Reveal] Spoiler: OA

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Success isn't something that just happens - success is learned, success is practiced and then it is shared. -Sparky Anderson
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Last edited by Bunuel on 26 Dec 2013, 03:53, edited 1 time in total.
Edited the OA.
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Re: Is x(y/z) > 0? [#permalink]

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29 Dec 2012, 10:52
Expert's post
sananoor wrote:
Is x(y/z) > 0?

(1) xyz > 0
(2) yz > 0

Statement$$\frac{xy}{z}$$ $$>0$$ both are positive or negative (the numerator and the denominator)

1) $$xyz$$ $$> 0$$ this says that our variables are both positive or negative but we can't determine which of them are positive or negative

2) the product of $$y*z$$ are both positive or negative

1) + 2) we still do not know which of our variable are positive or negative

E is the best
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Re: Is x(y/z) > 0? [#permalink]

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29 Dec 2012, 11:17
statement 1) says xyz> 0 from this we conclude that either all three are positive or two are negative. statement 2 says that yz> 0 it means either both are negative or positive....by combining two statements we come to know that x is positive...x cant be negative in order to make the statement one right....xyz>0 and yz> 0....so x is positive...y and z are either positive or negative....x(y/z) >0 if y and z are negative then negative signs cancel each other and answer could be greater than 0...i am so much confused regarding this question...
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Re: Is x(y/z) > 0? [#permalink]

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30 Dec 2012, 02:23
guys, dont think E is the correct answer... As per me , its A

here is the explanation:

1) it says xyz>0 which means either two of these variables are negative or all three are positive...you can take out any combinations from these and you will find out that every combination gives you x (y/z) >0.. so, this seems sufficient

2) yz>0 doesn't give any information about x so insufficient
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Re: Is x(y/z) > 0? [#permalink]

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30 Dec 2012, 05:13
Is x(y/z) > 0?

(1) xyz > 0
(2) yz > 0

We need to know whether expression E= {x * y * (1/y)} positive or negative.
Let E = A*B*C where { A=x, B=y, C=1/y }
Expression will be -ve if, 1 OR 3 terms among A,B & C are –ve. That is..
1. Any 1 term out of of A,B and C is –ve and 2 are +ve. OR
2. All 3 A, B & C are –ve.

Option 1: xyz>0..
Again we can write this as ABC>0.. So, this option is sufficient. Multiplication and Division provide similar signs.

Option 2: yz>0.
This means, Both y & z are –ve Or Both are +ve. We can’t determine whether x(y/z)>0.

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Re: Is x(y/z) > 0? [#permalink]

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31 Dec 2012, 07:13
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sananoor wrote:
Is x(y/z) > 0?

(1) xyz > 0
(2) yz > 0

Multiplication or division doesn't change the sign of a term. So x(y/z) or xyz or any formation of xyz will be same as long as one of that is established.

In option 1. xyz>0 is sufficient to determine thus that x(y/z) > 0. Suff

Option 2, nithing is known about x so not suff.

Ans A
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Re: Is x(y/z) > 0? [#permalink]

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25 Dec 2013, 17:17
sananoor wrote:
Is x(y/z) > 0?

(1) xyz > 0
(2) yz > 0

So is it A or E finally?

Cheers!
J
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Re: Is x(y/z) > 0? [#permalink]

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26 Dec 2013, 03:58
Expert's post
1
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jlgdr wrote:
sananoor wrote:
Is x(y/z) > 0?

(1) xyz > 0
(2) yz > 0

So is it A or E finally?

Cheers!
J

Correct answer is A. Edited the OA. Thank you.

Is x(y/z) > 0?

We need to find whether x*y*1/z is positive.

(1) xyz > 0. Since z and 1/z have the same sign, then this statement implies that x*y*1/z > 0. Sufficient.

(2) yz > 0. If x > 0, then x*y*1/z is positive but if x <= 0, then x*y*1/z is not positive. Not sufficient.

Hope it's clear.
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Re: Is x(y/z) > 0? [#permalink]

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31 May 2015, 20:22
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Re: Is x(y/z) > 0?   [#permalink] 31 May 2015, 20:22
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