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dsineuality.JPG PFA

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vishalgc
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dsineuality.JPG PFA [#permalink] New post   22 Oct 2008, 08:26
00:00
A
B
C
D
E

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(N/A)

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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
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Attachment:
dsineuality.JPG
PFA



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Re: dsineuality.JPG PFA [#permalink] New post   22 Oct 2008, 08:55
while E seems very tempting..i think C is the answer..

7x-2y>0 well we know both x and y are integers so x>2/7 Y which means y=7 or -7 we dont know..insuff

2) -y<x well this just means y>-x or insuff

together
7x-2y>0
x+y>0
solve both for y you get -y>0

which means y has to be less than 0..

C it is
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Re: dsineuality.JPG PFA [#permalink] New post   22 Oct 2008, 10:10
fresinha12 wrote:
while E seems very tempting..i think C is the answer..

7x-2y>0 well we know both x and y are integers so x>2/7 Y which means y=7 or -7 we dont know..insuff

2) -y<x well this just means y>-x or insuff

together
7x-2y>0
x+y>0
solve both for y you get -y>0

which means y has to be less than 0..

C it is


IMO E

I agree with fresniha's assessment for separate a and b
When combined if we take an example, it makes a bit simple to understand

assume x = 5 and y =2
It would satisfy both cases ie 7(5) - 2(2) > 0 and -(2) < 5

now assume x=5 and y= (-2)
It will still satisfy both cases also ie 7(5) - 2(-2) > 0 and -(-2) < 5

Hence even together a and b are not enough to give an answer where we can conclude y >0 or otherwise

Hence E
OA ?
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Re: dsineuality.JPG PFA [#permalink] New post   22 Oct 2008, 11:24
vishalgc wrote:
Attachment:
dsineuality.JPG
PFA


IMO: A
For A, 7x -2y>0
7x>2y
x>2/7y
because x>0, y must be >0

For B, we cannot determine - insuff

So my choice is A.
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Re: dsineuality.JPG PFA [#permalink] New post   22 Oct 2008, 11:43
nganle08 wrote:
vishalgc wrote:
Attachment:
dsineuality.JPG
PFA


IMO: A
For A, 7x -2y>0
7x>2y
x>2/7y
because x>0, y must be >0

For B, we cannot determine - insuff

So my choice is A.


this logic won't work
if y < 0, a postive x will still be greater than 2/7y
what if y = -7 and x =1
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Re: dsineuality.JPG PFA [#permalink] New post   22 Oct 2008, 11:54
kandyhot27 wrote:
nganle08 wrote:
vishalgc wrote:
Attachment:
dsineuality.JPG
PFA


IMO: A
For A, 7x -2y>0
7x>2y
x>2/7y
because x>0, y must be >0

For B, we cannot determine - insuff

So my choice is A.


this logic won't work
if y < 0, a postive x will still be greater than 2/7y
what if y = -7 and x =1


Thanks for pointing my mistake out.
I would then vote for C.
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Re: dsineuality.JPG PFA [#permalink] New post   22 Oct 2008, 14:22
E

1. 7x-2y>0
7x>2y
y<=0 or y>0
NS

2. -y<x
since x>0, y>0 or y<=0
NS

Together
y<=0 or y>0 - NS
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Re: dsineuality.JPG PFA [#permalink] New post   23 Oct 2008, 00:21
E for me.

Fromt stmt1: x > 2/7y, y can be 0, +ve or -ve.
From stmt2: x > -y, y can be 0, +ve or -ve.

Combining two,
x > 2/7y and x > - y

I will take few examples here.

Case 1: x = 3, y = 7, hence x > 2/7y and x > -y
Case 2: x = 3, y = 0, hence x > 2/7y and x > -y
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Re: dsineuality.JPG PFA [#permalink] New post   23 Oct 2008, 09:32
C.

1) 7x > 2y
2) x > - y

1) + 2)

8x > y

If x > 0, then y must be > 0

Any takers?
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Re: dsineuality.JPG PFA [#permalink] New post   23 Oct 2008, 09:37
bigfernhead wrote:
C.

1) 7x > 2y
2) x > - y

1) + 2)

8x > y

If x > 0, then y must be > 0

Any takers?


even I got the same ...

vishal .. whats the OA ?
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Re: dsineuality.JPG PFA [#permalink] New post   23 Oct 2008, 09:41
fresinha12 wrote:
i agree ans should be E...


but you posted above that the answer should be C ??
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Re: dsineuality.JPG PFA [#permalink] New post   23 Oct 2008, 11:20
yes I made a mistake, since the stem said x and y are integers..i wrongly assumed that

x>2/7 y then y must be a multiple of 7..however, if you actually take into account any value of y would do, including negative values..

7x-2y>0

-y<x

any value of y works positive and negative..



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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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Re: dsineuality.JPG PFA [#permalink]
23 Oct 2008, 11:20
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