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# Is xy > 0? 1) x-y > -2 2) x-2y < -6 Would you

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Is xy > 0? 1) x-y > -2 2) x-2y < -6 Would you [#permalink]

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05 Jun 2008, 09:24
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Is xy > 0?
1) x-y > -2
2) x-2y < -6
Would you explain this question please? I will post the OA shortly. Thank you!
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Re: GP #2 Is xy > 0? [#permalink]

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05 Jun 2008, 09:40
E for me

in either case x could be 0 thus we cannot find if xy>0
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Re: GP #2 Is xy > 0? [#permalink]

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05 Jun 2008, 09:51

#1 is insufficient.

Y < X+2 If Y is 3 greater than X, then if X is 10 and Y is 13, then 10 - 13 = -3, and that goes against #1. X could be -0.5 and Y could be 1. The answer doesn't require X and Y to be integers. Here, X*Y would be -0.5 or -1/2, but x*y could also be greater than 0, we just don't know.

#2

Let X = 10.

10 - 2y < -6
10 < -6 + 2y
4 < 2y
2<y
so X = 10 and y can be 3 because it's greater than 2...x*y = positive.

Look for somethign that would make either x or y negative, but not both.

let y = 1.

x - 2(1) < -6
x - 2 < -6
x < -4

Here y = 1, and x could be -5, because x must be less than -4. x * y would be negative (1 * -5 = -5).

#2 insufficient.

Together?

Take some values that work for #1 and plug them into #2 and see what happens.

So x = 10, y = 10. so x - y = 10 - 10 = 0 which is greater than -2.

Plug into #2
10 - 2(10) < -6
10 - 20 < -6 - True

X = $$-\frac{1}{2}$$, y = 1

Here $$-\frac{1}{2} - 1 = -1\frac{1}{2}$$ which is still greater than -2.

Plug into #2

$$-\frac{1}{2} - 2(1) < -6$$ = $$-2\frac{1}{2} < -6$$ NOT TRUE.

With both statements together, I believe we can answer and say that xy > 0. So answer C.

Y = 0 does not work for statement #2. So you cannot use it as a possibility when solving #2. Sets that work for both #1 & #2, as required to answer C, 0 is not an option and therefore x * y will never = 0.

utgirl826 wrote:
Is xy > 0?
1) x-y > -2
2) x-2y < -6
Would you explain this question please? I will post the OA shortly. Thank you!

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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Director Joined: 01 May 2007 Posts: 793 Followers: 1 Kudos [?]: 314 [0], given: 0 Re: GP #2 Is xy > 0? [#permalink] ### Show Tags 05 Jun 2008, 11:05 Hmm...I thought B, but I forgot to account for fractions...nevermind.. Director Joined: 01 May 2007 Posts: 793 Followers: 1 Kudos [?]: 314 [0], given: 0 Re: GP #2 Is xy > 0? [#permalink] ### Show Tags 05 Jun 2008, 11:11 I think C...If you combine the inequalities you should find x > 3 and y > 4...the question is asking is x and y the same sign...so in this case we know they are both positive. thoughts? Director Joined: 01 May 2007 Posts: 793 Followers: 1 Kudos [?]: 314 [0], given: 0 Re: GP #2 Is xy > 0? [#permalink] ### Show Tags 05 Jun 2008, 11:17 Calling Walker...Calling Walker... Intern Joined: 09 Jun 2007 Posts: 35 Followers: 0 Kudos [?]: 1 [0], given: 0 Re: GP #2 Is xy > 0? [#permalink] ### Show Tags 05 Jun 2008, 11:36 The OA is C but I am having trouble understanding jallenmorris's explanation. Can someone help me? or explain?? Thank you! SVP Joined: 30 Apr 2008 Posts: 1880 Location: Oklahoma City Schools: Hard Knocks Followers: 41 Kudos [?]: 581 [1] , given: 32 Re: GP #2 Is xy > 0? [#permalink] ### Show Tags 05 Jun 2008, 12:12 1 This post received KUDOS I'll try to make it more clear. Is xy > 0? 1) x-y > -2 2) x-2y < -6 When we have two variables, there is essentially an infinite number of possibilities. If X = 10, then Y must be less than 12 because 10 - 12 = -2, which is equal to, not less than 12. The key to answer the question Is xy > 0 is to find any values for x & y that have different signs. Since we are not told x and y are integers, fractions are possible. This is as simple as x = -1 and y = $$\frac{1}{2}$$. That would be $$-1\frac{1}{2}$$ which is greater than -2. -1 * 1/2 = - 1/2, and xy is NOT greater than 0, but at the same time, some values for X and Y that satisfy #1 have a product greater than zero. This means the information is insufficient because greater than 0 and less than 0 is possible. We cannot difinitively answer either way. Is xy > 0? 1) x-y > -2 2) x-2y < -6 Same idea for #2. Easiest way to start is X = 1. Then y < 3.5. If y is 3.5, then 1 - 2 * 3.5 = 1 - 7 < -6 = -6<-6 which isn't true. We need to find some way that either x or y is negative when the other is positive. If X is a negative, such as -1, then -1 - 2y < -6 -2y < -5 {switch the sign because we're dividing by a negative number} y > 5/2 This means we have x = -1 and y > $$\frac{5}{2}$$ then x * y = a number less than zero. If you take the equations together, we must find some values for x and y that satisfy both equations. Then we must determine if there is any possible way x or y can be negative, but not both. I take some of the numbers I've already come up with. Like #1, x = 10 and y =13. 10 - 13 = 3 (i.e., greater than -2). Works for #1, check for #2. x-2y < -6 10 - 2 (13) < -6 10 - 26 < -6 -16 < -6 TRUE We need to check answers for #2 that had 1 pos and 1 neg to see if they work in #1. So x = -1, and y = 5/2 -1 - 5/2 > -2 -3.5 > -2 FALSE! In statement #2, if x is negative and the value of y is such that the statement is true, then y must be between 2.5>y>0. I used the following possibilities: x= -1, y>2.5 x= -2, y>2 x= -3, y>1.5 x= -4, y>1 x= -5, y>0.5 x= -6, y>0 x = -7, y = -0.5 ( At this point we have 2 negative numbers, which would always make xy>0 true. we're looking to disprove this in some manner in order to answer the question. Not one of these possible combinations makes #1 true. -1 - 2.5 > -2 Nope -2 - 2 > -2 Nope -6 - 0 > -2 Nope etc... The only values that satisfy both equations will always lead to x * y being a positive number (i.e., greater than zero). I'm sorry this is so long. I tried to explain each step of my analysis in greater detail than I would actually use on the GMAT simple for the sake of time. utgirl826 wrote: The OA is C but I am having trouble understanding jallenmorris's explanation. Can someone help me? or explain?? Thank you! _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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Re: GP #2 Is xy > 0? [#permalink]

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05 Jun 2008, 13:36
utgirl826 wrote:
Is xy > 0?
1) x-y > -2
2) x-2y < -6
Would you explain this question please? I will post the OA shortly. Thank you!

OK lets see..

XY>0?

1)x>y-2

assume worse case x=y-2
y(y-2)>0?
well if y>2...yes if y<2 NO..Insuff..

2)
x<2y-6
2(y-3)*y>0?
if y>3 yes..if y<3 no..

together..

x-y=-2
+
-(x-2y=-6)
y=4..

OK.so we needed to know if Y>2 if Y>3...

we know Y=4..

Sufficient..

C it is..
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Re: GP #2 Is xy > 0? [#permalink]

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05 Jun 2008, 13:38
Fresh, I had similar reasoning. I combined the inequalities, but shouldn't those be inequalities and not equal to...
Intern
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Re: GP #2 Is xy > 0? [#permalink]

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05 Jun 2008, 13:43
Thank you for such a thorough explanation!
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Re: GP #2 Is xy > 0? [#permalink]

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05 Jun 2008, 14:00
It can be an equal (=) sign for one, and then it would read "When x = 1, Y > [some value]."

I assume your question refers to the fact that my explanation uses equals signs in some equations throughout.

jimmyjamesdonkey wrote:
Fresh, I had similar reasoning. I combined the inequalities, but shouldn't those be inequalities and not equal to...

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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.

GMAT Club Premium Membership - big benefits and savings

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Re: GP #2 Is xy > 0? [#permalink]

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05 Jun 2008, 14:01
jimmyjamesdonkey wrote:
Fresh, I had similar reasoning. I combined the inequalities, but shouldn't those be inequalities and not equal to...

i know they are in-equalities.. but remember i am trying to get an idea of how big Y is..not the exact value..
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Re: GP #2 Is xy > 0? [#permalink]

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05 Jun 2008, 19:51
I realized my mistake , I was fooled and was testing the Stmt-1 and 2 separately with two diff. satisfying values rather than testing with one set of value for both the equation, Jallen good explanation. Thanks
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Re: GP #2 Is xy > 0? [#permalink]

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05 Jun 2008, 21:47
Here is a short simple explanation

we change both equations to x is greater than or equal to:

1) x-y > -2 = x > y-2
2) x-2y < -6 = x < 2y-6

now we know that y-2< 2y-6
solve for y
y> 4 y has to be positive. Now plug a number greater than 4 back into equation 1 to solve for x

x-5 > -2
x> 3 now we know x has to be positive as well.

there fore xy > 0
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Re: GP #2 Is xy > 0? [#permalink]

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05 Jun 2008, 22:36
We can use addition of inequalities here
taking (2)x-2y<-6
x-y-y<-6
putting value from (1)
-2-y<-6
-2+6<y
4<y or y>4
putting in (1)
x-y>-2
x>2
Thus both x and y are positive hence C should be the answer

utgirl826 wrote:
Is xy > 0?
1) x-y > -2
2) x-2y < -6
Would you explain this question please? I will post the OA shortly. Thank you!
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Schools: Chicago (Booth) - Class of 2011
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Re: GP #2 Is xy > 0? [#permalink]

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05 Jun 2008, 22:51
C

My solution:

1. or 2. It is obvious that each condition along is insufficient.

1.&2.
(x-y)>-2
x-2y<-6 --> (x-y)-y<-6 --> From two inequalities, we can see that subtraction y from (x-y) results in value less than -6 only if y>4.

(x-y)>-2 --> x-4>(x-y)>-2 --> x>2
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Re: GP #2 Is xy > 0?   [#permalink] 05 Jun 2008, 22:51
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