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Is xy > 0? (1) x-y > -2 (2) x-2y < -6 [#permalink]
 05 Jun 2011, 05:06
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Question Stats:
55% (02:17) correct
44% (01:08) wrong based on 3 sessions
Last edited by Bunuel on 26 May 2012, 03:06, edited 1 time in total.
Edited the question.
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(1) x = 1, y = 0 xy = 0 although x-y = 1 > -2 x = -3, y = -2 xy = 6 > 0 and x-y = -1 > -2 Insufficient (2) x - 2y < -6 x = 1, y = 4 x - 2y = 1 - 8 = -7 < -6 and xy > 0 x = 0, y = 4 x - 2y = 0 - 8 = -8 < -6 and xy = 0 Insufficient (1) + (2) x - y > -2 2y - x > 6 y > 4 Now x has to be such that x > y - 2 So x is positive, because y is 4.1, 5 etc. and x is > 2.1, 3 etc. So xy = positive Answer - C
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a+b x-y > -2 and -x+2y >6 (changing the signs for addition) y>4 and x >2.1. Thus xy>0 always. C
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Re: Is xy > 0? 1) x-y > -2 2) x-2y < -6 [#permalink]
25 May 2012, 19:15
Hi all,
The process that I went through was
reverse the sign of the second equation
-x + 2y > 6 (to make the signs same for both equations)
and
solve for y to get y>4
Now for solving x
Do you solve the equations again for x
or
how does it work by substituting y > 4 ? Could someone show this method and also tell me if this is the best way to solve these questions.
Thanks
Ravi
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Re: Is xy > 0? 1) x-y > -2 2) x-2y < -6 [#permalink]
26 May 2012, 03:09
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raviram80 wrote: Hi all,
The process that I went through was
reverse the sign of the second equation
-x + 2y > 6 (to make the signs same for both equations)
and
solve for y to get y>4
Now for solving x
Do you solve the equations again for x
or
how does it work by substituting y > 4 ? Could someone show this method and also tell me if this is the best way to solve these questions.
Thanks
Ravi Is xy>0?Note that question basically asks whether x and y have the same sign. (1) x-y > -2 --> we can have an YES answer, if for example x and y are both positive ( x=10 and y=1) as well as a NO answer, if for example x is positive and y is negative ( x=10 and y=-10). Not sufficient. (2) x-2y <-6 --> again it' easy to get an YES answer, if for example x and y are both positive ( x=1 and y=10) as well as a NO answer, if for example x is negative and y is positive ( x=-1 and y=10). Not sufficient. You can get that the the two statement individually are not sufficient in another way too: we have (1) y<x+2 and (2) y>\frac{x}{2}+3. We are asked whether x and y have the same sign or whether the points (x,y) are in the I or III quadrant ONLY. But all (x,y) points below the line y=x+2 (for 1) and all (x, y) points above the line y=\frac{x}{2}+3 can not lie only I or III quadrant: points above or below some line (not parallel to axis) lie at least in 3 quadrants. (1)+(2) Now, remember that we can subtract inequalities with the signs in opposite direction --> subtract (2) from (1): x-y-(x-2y)>-2-(-6) --> y>4. As y>4 and (from 1) x>y-2 then x>2 (because we can add inequalities when their signs are in the same direction, so: y+x>4+(y-2) --> x>2) --> we have that y>4 and x>2: both x and y are positive. Sufficient. Answer: C.
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Re: Is xy > 0? (1) x-y > -2 (2) x-2y < -6 [#permalink]
26 May 2012, 19:20
Thanks, it is clear now.
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Re: Is xy > 0? (1) x-y > -2 (2) x-2y < -6 [#permalink]
26 May 2012, 22:09
Hello, I have a doubt. After we have got y >4, do we check both the equations for consistency? With the first equation, m getting x >2 but with the second, I am receiving x<2 upon solving. What am I doing wrong Posted from my mobile device 
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Re: Is xy > 0? (1) x-y > -2 (2) x-2y < -6 [#permalink]
28 May 2012, 04:53
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Re: Is xy > 0? (1) x-y > -2 (2) x-2y < -6 [#permalink]
18 Jun 2012, 09:00
Bunuel wrote: tech3 wrote: Hello, I have a doubt. After we have got y >4, do we check both the equations for consistency? With the first equation, m getting x >2 but with the second, I am receiving x<2 upon solving. What am I doing wrong Posted from my mobile device Can you please show how you get x<2? Any easier method to solve this question folks?
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Re: Is xy > 0? (1) x-y > -2 (2) x-2y < -6 [#permalink]
12 Mar 2013, 10:28
Bunuel wrote: tech3 wrote: Hello, I have a doubt.
After we have got y >4, do we check both the equations for consistency? With the first equation, m getting x >2 but with the second, I am receiving x<2 upon solving. What am I doing wrong
Can you please show how you get x<2? _____________________________________________________________________________________________________________________________ Bunuel, I'm struggling to understand how the second statement proves that x is definitely positive. I follow that y>4 and I follow that x>2 if I only test equation (1). But I would think that I should be able to test either and come away with the fact that x is positive. However when I test equation (2) x proves to be positive or negative! Testing equation (2) x - 2y < -6 since y>4 If y = 4.1, then x - 8.2 < -6 and x < 2.2 (this suggests x can be positive or negative) If y = 6, then x - 12 < -6 and x < 6 (this suggests x can be positive or negative) and so on... I don't understand how we prove x is definitely positive if I test equation (2)
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Re: Is xy > 0? (1) x-y > -2 (2) x-2y < -6 [#permalink]
17 Mar 2013, 13:19
dolumoks wrote: Bunuel wrote: tech3 wrote: Hello, I have a doubt.
After we have got y >4, do we check both the equations for consistency? With the first equation, m getting x >2 but with the second, I am receiving x<2 upon solving. What am I doing wrong
Can you please show how you get x<2? _____________________________________________________________________________________________________________________________ Bunuel, I'm struggling to understand how the second statement proves that x is definitely positive. I follow that y>4 and I follow that x>2 if I only test equation (1). But I would think that I should be able to test either and come away with the fact that x is positive. However when I test equation (2) x proves to be positive or negative! Testing equation (2) x - 2y < -6 since y>4 If y = 4.1, then x - 8.2 < -6 and x < 2.2 (this suggests x can be positive or negative) If y = 6, then x - 12 < -6 and x < 6 (this suggests x can be positive or negative) and so on... I don't understand how we prove x is definitely positive if I test equation (2) Agree! I cant understand the same )
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Re: Is xy > 0? (1) x-y > -2 (2) x-2y < -6 [#permalink]
18 Mar 2013, 13:04
adding both equations, you would get y >4. Substituting this to equation 1, you would get x > 2! Hence both x and y are positive, so xy > 0
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Re: Is xy > 0? (1) x-y > -2 (2) x-2y < -6
[#permalink]
18 Mar 2013, 13:04
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