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# If x and y are integers and x>0, is y>0? 1) 7x-2y>0

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If x and y are integers and x>0, is y>0? 1) 7x-2y>0 [#permalink]

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05 Feb 2007, 18:43
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If x and y are integers and x>0, is y>0?

1) 7x-2y>0

2) -y<x

Last edited by buckkitty on 06 Feb 2007, 09:38, edited 1 time in total.

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Re: DS - Is y>0? [#permalink]

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05 Feb 2007, 18:55
buckkitty wrote:
If x and y are integers and x>0, is y>0?

1) 7x-2y>0

2) -y>x

Question: x and y are integers. x>0, y>0 ?

Info(1): 7x-2y>0
-2y>-7x
y<7x>x

Last edited by devilmirror on 05 Feb 2007, 19:06, edited 1 time in total.

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Re: DS - Is y>0? [#permalink]

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05 Feb 2007, 19:05
devilmirror wrote:
Question: x and y are integers. x > 0, y > 0 ?

Info(1): 7x-2y>0
-2y>-7x
y<7x>x
y<x>0 or -x<0
we can conclude that y<-x<0
or y<0>0? The answer is NO. SUFF

Could anyone fix the bugs about inequality signs?
I do not want to type the answer all over again.

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06 Feb 2007, 00:05
(B) as well

I prefer the approach of XY plan uses.

Stat1
7x-2y>0
<=> y <7>0 and y>0), in cadran III (x<0 and y<0>0 and y<0> 0, we can be either in cadran I or IV. Thus, y <0> 0 (Fig1).

INSUFF.

Stat2
-y> x
<=> y < -x

Thus, we are below the line y = -x in the XY plan. We can be either in cadran II (x<0>0), in cadran III (x<0 and y<0>0 and y<0> 0, we are necesserary in the cadran IV with y < 0 (Fig2).

SUFF.
Attachments

Fig1_Below 3.5 x X and X positive.GIF [ 3.07 KiB | Viewed 1430 times ]

Fig2_Below X and X positive.GIF [ 2.53 KiB | Viewed 1426 times ]

Last edited by Fig on 06 Feb 2007, 00:13, edited 1 time in total.

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Re: DS - Is y>0? [#permalink]

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06 Feb 2007, 00:08
devilmirror wrote:
devilmirror wrote:
Question: x and y are integers. x > 0, y > 0 ?

Info(1): 7x-2y>0
-2y>-7x
y<7x>x
y<x>0 or -x<0
we can conclude that y<-x<0
or y<0>0? The answer is NO. SUFF

Could anyone fix the bugs about inequality signs?
I do not want to type the answer all over again.

In your post, click the option "Disable HTML in this post". U should not have any problem after that

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06 Feb 2007, 04:06
If x and y are integers and x>0, is y>0?

1) 7x-2y>0

2) -y>x

7x>2y..........insuff

from two

-y>x ...........suff ( x is +ve thus sure -y is too , and y is -ve

the answer is NO Y IS NOT +VE

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06 Feb 2007, 07:44
Sorry folks, typo in the question.

stmt (2) should be: -y<x>0, is y>0?

1) 7x-2y>0

2) -y<x

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06 Feb 2007, 09:36
B it is...

statment 1) easy...Insuff

statement 2) -y>x, we know x >0

therefore y has to be less than zero, has to be negative...suff

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13 Feb 2007, 19:57
buckkitty wrote:
Sorry folks, typo in the question.

stmt (2) should be: -y<x>0, is y>0?

1) 7x-2y>0

2) -y<x

then the answer i guess is E.

y (<) 3.5*x ; insuff

-y< x ; y can also be negative where |y|<x or simply be positive

and from both 1 & 2 does not constrain y hence E. Simply draw the lines y = 3.5 x and y= -x and the area between them is the solution and in both cases Y is not bounded to the first quadrant. hence E.

hence both are inSuff

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13 Feb 2007, 22:40
stat1: 7x-2y>0

=> y < 7*x/2 so even if x is 100, y can be -1 or 1 so insuff

stat2: -y<x> y > x so if x>0 y is certainly > 0 so suff

B

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13 Feb 2007, 22:40
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