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If xy < 0, is x/y > z ? (1) xyz < 0 (2) x > yz

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If xy < 0, is x/y > z ? (1) xyz < 0 (2) x > yz  [#permalink]

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New post 16 Sep 2015, 23:53
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Difficulty:

  55% (hard)

Question Stats:

61% (01:56) correct 39% (01:56) wrong based on 222 sessions

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Re: If xy < 0, is x/y > z ? (1) xyz < 0 (2) x > yz  [#permalink]

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New post 17 Sep 2015, 07:29
1
1
If xy < 0, is x/y > z ?
2 scenarios
x is neg and y is pos
x is pos and y is neg


(1) xyz < 0
x is neg, y pos, z pos
x is pos, y neg, z pos
Both scenarios make z larger than x/y
Sufficient

(2) x > yz
x is neg, y pos, z neg
X is pos, y negative, z pos
If z is neg the question is yes
If z is positive question is no
Insufficient

Answer: A
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Re: If xy < 0, is x/y > z ? (1) xyz < 0 (2) x > yz  [#permalink]

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New post 19 Sep 2015, 11:35
Bunuel wrote:
If xy < 0, is x/y > z ?

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

Kudos for a correct solution.


Since xy<0, x and y have opposite signs. And hence x/y is negative.

Statement 1:-
xyz<0
This is possible only when xy and z have opposite sign. We are already given that xy is negative.
So, z must be positive. and hence our question is answered.(x/y is not greater than z)
SUFFICIENT

Statement 2:-
Doesn't provide any information about sign of z.
INSUFFICIENT

Answer:- A
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Re: If xy < 0, is x/y > z ? (1) xyz < 0 (2) x > yz  [#permalink]

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New post 15 Jan 2017, 17:47
If xy < 0, is x/y > z ?

From the prompt,

x & y have different signs

(1) xyz < 0

xy is negative ....so z must be positive..... answer to question is NO (negative value can't be greater than positive one).

Sufficient

(2) x > yz

Let x =negative value then y is positive value. So z is negative.

Divide by y

x/y>z

Let x =positive value then y is negative value. So z is either positive or negative.

Divide by y then flip sign

x/y<z

Insufficient

Answer: A
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Re: If xy < 0, is x/y > z ? (1) xyz < 0 (2) x > yz  [#permalink]

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New post 03 Feb 2017, 01:03
peachfuzz wrote:
If xy < 0, is x/y > z ?
2 scenarios
x is neg and y is pos
x is pos and y is neg


(1) xyz < 0
x is neg, y pos, z pos
x is pos, y neg, z pos
Both scenarios make z larger than x/y
Sufficient

(2) x > yz
x is neg, y pos, z neg
X is pos, y negative, z pos
If z is neg the question is yes
If z is positive question is no
Insufficient

Answer: A

I'm just talking about the red part..
In the question stem: if xy<0, then
x-->(+)
y-->(-)

or,
x-->(-)
y-->(+)
If z is negative, the answer can't be YES always.
Here is test cases:
suppose,
x=-10
y=5
and z=-6
then, x/y>z?
-->-10/5>-6?
-->-2>-6?
--> YES
But, if
x=-10
y=5
and z=-1,
then, x/y>z?
-->-10/5>-1?
-->-2>-1?
-->NO....
Am I right Bunuel?
Thank you brother...
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If xy < 0, is x/y > z ? (1) xyz < 0 (2) x > yz  [#permalink]

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New post 12 Feb 2017, 01:54
If xy < 0, is x/y > z ?

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

From stmt (1) and xy < 0,we can conclude that z > 0.
Since x and y must have the opposite sign,x/y <0.Thus,x/y < z.SUFFICIENT

From stmt (2)
x > yz.It could be transform to x/y > z ONLY if y > 0.We don't know for sure that y must be > 0.INSUFFICIENT

Ans A
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Re: If xy < 0, is x/y > z ? (1) xyz < 0 (2) x > yz  [#permalink]

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New post 02 Sep 2018, 12:05
Are there any links anyone can provide that can help me with a strategy to solve these type of questions?
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Re: If xy < 0, is x/y > z ? (1) xyz < 0 (2) x > yz  [#permalink]

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New post 02 Sep 2018, 20:23
1
justgottamakeit1592 wrote:
Are there any links anyone can provide that can help me with a strategy to solve these type of questions?


9. Inequalities




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Re: If xy < 0, is x/y > z ? (1) xyz < 0 (2) x > yz &nbs [#permalink] 02 Sep 2018, 20:23
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