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I got this right. But I want to know whether anyone has a

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rgajare14
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I got this right. But I want to know whether anyone has a [#permalink] New post   29 Apr 2008, 17:01
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I got this right. But I want to know whether anyone has a better way to tackle this.

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



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Re: I got this right. But I want to know whether anyone has a [#permalink] New post   30 Apr 2008, 13:39
rgajare14 wrote:
I got this right. But I want to know whether anyone has a better way to tackle this.

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




It's easy to see from both the conditions seperatly that x and/or y could very well be zero or any nonzero numbers , so rule out options A and D and B

taking both conditions together

we can write x-y=-1
x-2y=7

looking at both the conditions , x and y !=0 for sure

Now can we deduce from this that y is always -ve (why because subtracting multiple of y(2y here) yields number which is greater than subtracting y) and since from condition 1 x=y-1 ; x is also -ve

so xy>0 and hence C
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Re: I got this right. But I want to know whether anyone has a [#permalink] New post   30 Apr 2008, 13:46
I did this with lines and co-ordinate axis, which took me much less time. But its kinda hard to explain here.
Still if it helps, draw two lines corresponding to the inequalities and check for the regions they are covering.
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Re: I got this right. But I want to know whether anyone has a [#permalink] New post   30 Apr 2008, 13:56
itsme291 wrote:
I did this with lines and co-ordinate axis, which took me much less time. But its kinda hard to explain here.
Still if it helps, draw two lines corresponding to the inequalities and check for the regions they are covering.


which co-ordinate do you see these lines ? it would be interesting to see that approach but I can't put it on the paper correctly :-(
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Re: I got this right. But I want to know whether anyone has a [#permalink] New post   30 Apr 2008, 19:08
rpmodi, If as per question, x - 2y < -6 , wouldn't it be incorrect to say x - 2y = 7. ??
Did you mean x -2y = -7.

And following that Subtracting 2y from x has yielded a smaller number (-7) as compared to subtracting y, from x, which yielded -1.
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Re: I got this right. But I want to know whether anyone has a [#permalink] New post   30 Apr 2008, 19:10
itsme291 wrote:
I did this with lines and co-ordinate axis, which took me much less time. But its kinda hard to explain here.
Still if it helps, draw two lines corresponding to the inequalities and check for the regions they are covering.


Wow interesting...I tried to plot lines on a co ordinate system for the above equations..But those did not lead me anywhere. So I am clueless w.r.t your explanation right now..
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Re: I got this right. But I want to know whether anyone has a [#permalink] New post   30 Apr 2008, 19:57
rgajare14 wrote:
rpmodi, If as per question, x - 2y < -6 , wouldn't it be incorrect to say x - 2y = 7. ??
Did you mean x -2y = -7.

And following that Subtracting 2y from x has yielded a smaller number (-7) as compared to subtracting y, from x, which yielded -1.



ya I meant we can assume x-2y=-7 and x-y=-1 following that y would be +ve and x would be +ve
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Re: I got this right. But I want to know whether anyone has a [#permalink] New post   30 Apr 2008, 21:41
rgajare14 wrote:
itsme291 wrote:
I did this with lines and co-ordinate axis, which took me much less time. But its kinda hard to explain here.
Still if it helps, draw two lines corresponding to the inequalities and check for the regions they are covering.


Wow interesting...I tried to plot lines on a co ordinate system for the above equations..But those did not lead me anywhere. So I am clueless w.r.t your explanation right now..


Wil l try to make a diagram and explain....
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Re: I got this right. But I want to know whether anyone has a [#permalink] New post   30 Apr 2008, 22:12
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We know the sign of xy, if on a co-ordinate axis, we know which quadrant x and y lies in.
(i.e. whether they are both +ve, or both –ve or +ve and -ve).
See the diagram attached. Blue line is the line x-y=-2 and the blue region shows the inequality.
Same for the red line and red region, which represents the equation x-2y<-6.
Both the statements individually don’t give us anything as the sign of xy will vary.
But the common area to the both lies only in quadrant 1, which means that taking both the inequalities, x>0 and y>0

Hence we can say that xy>0

Hope this explains 
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rgajare14
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Re: I got this right. But I want to know whether anyone has a [#permalink] New post   01 May 2008, 10:48
itsme291 wrote:
We know the sign of xy, if on a co-ordinate axis, we know which quadrant x and y lies in.
(i.e. whether they are both +ve, or both –ve or +ve and -ve).
See the diagram attached. Blue line is the line x-y=-2 and the blue region shows the inequality.
Same for the red line and red region, which represents the equation x-2y<-6.
Both the statements individually don’t give us anything as the sign of xy will vary.
But the common area to the both lies only in quadrant 1, which means that taking both the inequalities, x>0 and y>0

Hence we can say that xy>0

Hope this explains 


Excellent ! Got you...What a novel way !
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Re: I got this right. But I want to know whether anyone has a [#permalink] New post   01 May 2008, 21:47
itsme291 wrote:
But the common area to the both lies only in quadrant 1, which means that taking both the inequalities, x>0 and y>0

Hence we can say that xy>0

Hope this explains 



This is simply brilliant! I feel like I've just added a major weapon with this type of DS questions.



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Re: I got this right. But I want to know whether anyone has a [#permalink]
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