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Is x(y/z) > 0?

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Is x(y/z) > 0?  [#permalink]

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New post Updated on: 26 Dec 2013, 03:53
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74% (01:15) correct 26% (01:19) wrong based on 182 sessions

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Is x(y/z) > 0?

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

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Originally posted by sananoor on 29 Dec 2012, 08:51.
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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New post 29 Dec 2012, 10:52
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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New post 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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New post 30 Dec 2012, 02:23
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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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New post 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.

Answer is A.
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Re: Is x(y/z) > 0?  [#permalink]

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New post 31 Dec 2012, 07:13
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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New post 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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New post 26 Dec 2013, 03:58
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.

Answer: A.

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

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New post 03 Apr 2019, 01:14
sananoor wrote:
Is x(y/z) > 0?

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


Question : Is xy/z > 0 ?

It is possible when :

xy > 0 and z > 0 OR xy < 0 and z < 0

Statement 1 :

xyz > 0

Case 1 : + + + > 0

Case 2 : - - + > 0

Case 3 : + - - > 0

Case 4 : - + - > 0

By matching these cases with the given requirement, this is SUFFICIENT

Statement 2 : yz > 0

Not sufficient as we don't know anything about x. INSUFFICIENT


Hence A.

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Re: Is x(y/z) > 0?   [#permalink] 03 Apr 2019, 01:14
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