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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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14 Jun 2011, 07:43
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Question Stats:

72% (01:03) correct 28% (00:57) wrong based on 78 sessions

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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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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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14 Jun 2011, 09:18
1
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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

$$2x<0$$
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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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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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.

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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