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x+y)/Z>0 is x<0? (1) x<y (2) z<0

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x+y)/Z>0 is x<0? (1) x<y (2) z<0 [#permalink]

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New post 14 Jun 2011, 07:43
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(x+y)/Z>0 is x<0?

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-y-z-0-is-x-135550.html
[Reveal] Spoiler: OA

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Re: x+y)/Z>0 is x<0? (1) x<y (2) z<0 [#permalink]

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New post 14 Jun 2011, 07:50
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Baten80 wrote:
(x+y)/Z>0 is x<0?
(1) x<y
(2) z<0


Clean C
st. 1 doesnt tell abt Z . insufficient
st2. nothing abt X and Y . insufficient

X<Y and Z<0

(X+y)/Z >0 and Z<0 means X+y <0
we know X<Y
hence X<0

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Re: x+y)/Z>0 is x<0? (1) x<y (2) z<0 [#permalink]

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New post 14 Jun 2011, 09:18
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Baten80 wrote:
(x+y)/Z>0 is x<0?
(1) x<y
(2) z<0


Sol:

\(\frac{x+y}{z} > 0\)
Means,

If \(z>0\), \(x+y>0\)
If \(z<0\), \(x+y<0\)

1. x<y
x=1, y=2, z=3;
\(\frac{x+y}{z}=\frac{1+2}{3}=1\) AND x>0

x=-2, y=-1, z=-3;
\(\frac{x+y}{z}=\frac{-1-2}{-3}=1\) BUT x<0

Not Sufficient.

2. z<0

Means;
\(x+y<0\)
x=-10; y=1; x+y= -9; x<0
x=1; y=-10; x+y=-9; x>0

Not Sufficient.

Combining both;
From stem we know,
If \(z<0\), \(x+y<0\)-----1

From St1, \(x<y\) OR \(x-y<0\)-------2

Adding 1 and 2;
\(2x<0\)
OR
\(x<0\)
Sufficient.

Ans: "C"
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Re: x+y)/Z>0 is x<0? (1) x<y (2) z<0 [#permalink]

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New post 13 Jan 2016, 07:38
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Re: x+y)/Z>0 is x<0? (1) x<y (2) z<0 [#permalink]

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New post 13 Jan 2016, 08:36
Baten80 wrote:
(x+y)/Z>0 is x<0?

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



If \(\frac{x+y}{z}>0\), is x<0?

Noticee that: \(\frac{x+y}{z}>0\) means that \(x+y\) and \(z\) have the same sign: either both are positive or both are negative.

(1) x < y. No info about \(z\). Not sufficient.
(2) z < 0. This statement implies that \(x+y\) must also be negative: \(x+y<0\). But we cannot say whether \(x<0\). Not sufficient.

(1)+(2) From (1) we have that \(x < y\) and from (2) we have that \(x+y<0\). Sum these two inequalities (remember we can add inequalities with the sign in the same direction): \(x+y+x<y\) --> \(2x<0\) --> \(x<0\). Sufficient.

Answer: C.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-y-z-0-is-x-135550.html
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Re: x+y)/Z>0 is x<0? (1) x<y (2) z<0   [#permalink] 13 Jan 2016, 08:36
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