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x and y both positive?

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Source: — Data Sufficiency |
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by shovan85 » Wed Sep 29, 2010 10:29 am
IMO C

from 2 and 1

x>y taking x and y both positive
say x=1, y=0.5 => 2x-2y = 1.
Or
x<y taking x an y both negative
and when x<y there will be no case 2x-2y=1 as 2x < 2y .

So we can say if both x and y are positive.

What is OA?
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by Fractal » Wed Sep 29, 2010 11:41 am
shovan85 wrote:IMO C

from 2 and 1

x>y taking x and y both positive
say x=1, y=0.5 => 2x-2y = 1.
Or
x<y taking x an y both negative
and when x<y there will be no case 2x-2y=1 as 2x < 2y .

So we can say if both x and y are positive.

What is OA?
thx for your explanation, but i don't understand it ;-)

answer: c
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by shovan85 » Wed Sep 29, 2010 11:49 am
from option 2
either both x and y will be positive or both will be negative.
if positive x>y
if negative x<y as sign will change.

now with this data if you want to validate option 1 then x>y will satisfy not x <y.
so u got a concrete answer that x and y are positive.

hope this helps
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by kvcpk » Wed Sep 29, 2010 12:02 pm
Fractal wrote:(1) 2x - 2y = 1
(2) x/y > 1

neither statement alone is sufficient. does anybody know a good way to check
whether the statements together are sufficient?

thx
Can you tell what the question is? I can only see 2 stmts. What is the question asking for?
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by Fractal » Wed Sep 29, 2010 12:06 pm
kvcpk wrote:
Fractal wrote:(1) 2x - 2y = 1
(2) x/y > 1

neither statement alone is sufficient. does anybody know a good way to check
whether the statements together are sufficient?

thx
Can you tell what the question is? I can only see 2 stmts. What is the question asking for?
are x and y both positive?
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by kvcpk » Wed Sep 29, 2010 12:19 pm
Fractal wrote:(1) 2x - 2y = 1
(2) x/y > 1

neither statement alone is sufficient. does anybody know a good way to check
whether the statements together are sufficient?

thx
x/y >1
x-y = 1/2

Let X and Y both be negative.
implies x<0, y<0

When an inequality is multiplied by negative value, its inequality sign changes.

x/y >1

Multiply by y [y is negative]
Hence x<y
which means x-y<0
but x-y =1/2 >0 [according to stmt1]
Hence X and Y both cant be negative

Let x be positive and y be negative.
Same situation as above arises.

Let x be negative and y be positive.
x/y>1
x>y
But a negative value can never be greater than a positive value.
hence this possibility is also lost.

only left out possibility is both are positive.

We dont have to check it. But let us see how it goes.

x>0,y>0
x/y>1
x>y
x-y>0
x-y = 1/2
All the stmts are satisfied.


Hope this helps!!
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by ankur.agrawal » Wed Sep 29, 2010 1:26 pm
Can we do it this way:

2x-2y=1 can be reframed as;

2(x-y) =1; So (x-y) = 1/2;-------(1)

& from 2;

(x/y)-1>0

(x-y)/y>0

Now since x-y=1/2 means x>y

& since (x-y)/y>0 y has to be + since x-y is positive.

Hence both are positive.

DOES THAT MAKES SENSE?
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by HPengineer » Wed Sep 29, 2010 8:37 pm
@ankur.agrawal

i solved it same as you..im also wondering if its dangerous to solve this way...
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by [email protected] » Wed Oct 20, 2010 9:53 pm
IMO E:
St:1: x-y=0.5
St 2: x>y
poss options.. x=0;y=-0.5
x can be either positive o negative right?? hence not sufficient
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by diebeatsthegmat » Thu Oct 21, 2010 10:24 am
[email protected] wrote:IMO E:
St:1: x-y=0.5
St 2: x>y
poss options.. x=0;y=-0.5
x can be either positive o negative right?? hence not sufficient
ohh this is confusing... some books say O is not negative or positive.. its just zero...
so i think 0 is not consider here....
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by shovan85 » Thu Oct 21, 2010 10:33 am
[email protected] wrote:IMO E:
St:1: x-y=0.5
St 2: x>y
poss options.. x=0;y=-0.5
x can be either positive o negative right?? hence not sufficient
After combining both if you consider x=0;y=-0.5 then how will you agree with the second option which says x/y > 1 as from ur choices x/y = 0 which is < 1.

Hope this makes sense
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by [email protected] » Thu Oct 21, 2010 7:14 pm
After combining both if you consider x=0;y=-0.5 then how will you agree with the second option which says x/y > 1 as from ur choices x/y = 0 which is < 1.
Ah shovan thanks man..
t prob dr i realised.. i multiplied the inequality not considering the sign o it..
this is a big no no.. thanks anyway
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