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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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peachfuzz
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

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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Are there any links anyone can provide that can help me with a strategy to solve these type of questions?
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Bunuel
If xy < 0, is x/y > z ?

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

Kudos for a correct solution.

Asked: If xy < 0, is x/y > z ?
Q. x/y > z ?


(1) xyz < 0
Since xy<0
z>0
x/y <0 & z>0
x/y <z
SUFFICIENT

(2) x > yz
(x-yz) >0
But y may be +ve or -ve
NOT SUFFICIENT


IMO A
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