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# Is xy > 0? 1. x-y>-2 2. x-2y<-6

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Manager
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Is xy > 0? 1. x-y>-2 2. x-2y<-6 [#permalink]  28 Oct 2006, 14:37
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Is xy > 0?

1. x-y>-2
2. x-2y<-6
Senior Manager
Joined: 31 Jul 2006
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Re: inequalities DS [#permalink]  28 Oct 2006, 17:05
Sam Kana wrote:
Is xy > 0?

1. x-y>-2
2. x-2y<-6

E also.

x or y can be zero and satisfy both 1 and 2 or they can be non-zero and satisfy
Senior Manager
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[#permalink]  28 Oct 2006, 19:51
I am getting C

Individually not sufficient.

Hint 2 can be written as
(x - y) â€“ y < -6
Here (x â€“y) part is > -2, so y has to be positive and greater than 4

So y > 4

(x â€“ y) > -2, here if y > 4, x has to be positive. Negative x will make the value of (x - y) lesser than -4.

As both x and y are positive, xy > 0

Last edited by anindyat on 28 Oct 2006, 20:33, edited 1 time in total.
VP
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[#permalink]  28 Oct 2006, 20:11
anindyat wrote:
I am getting C ..but not sure

Individually not sufficient.

Hint 2 can be written as
(x - y) â€“ y < -6
Here (x â€“y) part is > -2, so y has to be positive and greater than 4

So y > 4

(x â€“ y) > -2, here if y > 4, x has to be positive. Negative x will make the value of (x - y) lesser than -4.

As both x and y are positive, xy > 0

I think it is C the same explanation above
SVP
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[#permalink]  29 Oct 2006, 00:56
For me (C) as well.

I prefer to draw a XY plan and identify the points (X,Y) such that the equations alone and then together are verified. I believe it's very fast

Stat 1

x-y>-2
<=> y < x + 2 : it means that we consider all points (x,y) that are "under" the line x+2.

I represent the area in green in Fig-1. As we can see, the whole cadran (IV) is in the area, a cadran where x > 0 and y < 0. Thus, x*y < 0.
Meanwhile, part of cadran (I) is covered, in which x > 0 and y > 0. For this case, x*y > 0.

INSUFF.

Stat 2

x-2y<-6
<=> y > x/2 + 3 : it means that we consider all points (x,y) that are "above" the line x/2 + 3

I represent the area in green in Fig-2.
We can see here:
o Part of cadran (II) is covered. That means x < 0 and y > 0 and so x*y <0
o Part of cadran (I) is covered. That means x > 0 and y > 0 and so x*y >0

INSUFF.

Using (1) and (2):

We have to be at same time "under" the line x+2 and "above" the line x/2 + 3. I used a "flashy" green colour to reperesent the area.

Thus, the area is only a part of the cadran (I) where x > 0 and y > 0. Thus, x*y > 0

SUFF.
Attachments

Fig-1_y-inf-to-x+2.jpg [ 19.2 KiB | Viewed 365 times ]

Fig-2_y-sup-to-x-div-2+3.jpg [ 17.03 KiB | Viewed 366 times ]

Fig-3_both.jpg [ 22.67 KiB | Viewed 365 times ]

Manager
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[#permalink]  29 Oct 2006, 03:50
Yes 'C' it is...I like the graphics method, it is more generic and eleminates the thinking process...!
SVP
Joined: 01 May 2006
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[#permalink]  29 Oct 2006, 08:20
Sam Kana wrote:
Yes 'C' it is...I like the graphics method, it is more generic and eleminates the thinking process...!

I agree... We can conclude very fast with it
[#permalink] 29 Oct 2006, 08:20
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# Is xy > 0? 1. x-y>-2 2. x-2y<-6

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