power prep DS

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power prep DS

by orel » Sun Dec 14, 2008 8:56 pm
is xy>0 ?

1. x-y>-2
2. x-2y<-6

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Re: power prep DS

by tritrantran » Sun Dec 14, 2008 9:09 pm
Feruza Matyakubova wrote:is xy>0 ?

1. x-y>-2
2. x-2y<-6
For xy> 0 when:

1) x = positive, y = positive

2) x = negative, y = negative

Statement (1)
x-y>-2
x-y+2>0

we can make up multiple values of x and y and this statement is still true...INSUFFICIENT

Statement (2)
x-2y<-6

x-2y+6<0

same as above, we can have multiple values of x and y and statement is still true...INSUFFICIENT

St (1) and (2)
x-y+2>0
x-2y+6<0

-x+y-2<0
x-2y+6<0

-y+4<0
y>4

If y>4, then x>2 or x<2

INSUFFICIENT (E)

What's the OA?

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by cramya » Sun Dec 14, 2008 9:11 pm
One more vote for E)

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by vittalgmat » Sun Dec 14, 2008 9:48 pm
It is E.

While typing this reply I figured out B is wrong.
Here is what I did.

Stmt 2: x -2y < -6 is xy > 0 ?? Yes or no question.

let x = -11, y = -2

x -2y => -11 -2(-2) = -11 +4 => -7 which is < -6.
however x*y = -11 * -2 = +22 > 0 YES

Let x = 2, y = 11
x -2y => 2 -2(11) => 2 - 22 => -22 which is < -6.
However x*y = 2*11 = 22 > 0 YES.


Let x = -8, y = 1/2
x-2y = -9 which is < -6.
but xy = -8*1/2 = -4 > 0 ?? NO.

So N/S

combining does not give a unique answer.

so E

HT Helps

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by orel » Mon Dec 15, 2008 12:01 am
hey guys!
I hoped the OA would be E, but...

OA is C


we have to work a bit harder here

cheers

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by cramya » Mon Dec 15, 2008 12:09 am
Should have solved it dligently instead of voting for an E)

y> 4 only possible wiht stmts I and II

1. x-y>-2
2. x-2y<-6


x-y>-2

x+2 > y

y < x+2

x has to be positive since anything negative would make y>4 false

xy>0


Hence C)

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Re: power prep DS

by logitech » Mon Dec 15, 2008 12:21 am
Feruza Matyakubova wrote:is xy>0 ?

1. x-y>-2
2. x-2y<-6
x-y>-2
-x+2y>6
+
----------------------

y>4

x>2

xy>0

C
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by ronniecoleman » Mon Dec 15, 2008 12:30 am
IMO C

This is can be solved through coordinate representation.
Last edited by ronniecoleman on Mon Dec 15, 2008 12:34 am, edited 1 time in total.
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by logitech » Mon Dec 15, 2008 12:34 am
ronniecoleman wrote:IMO E
Dude if you have an IMO , keep it for yourself.

If you have a solution, share so we all can learn.

Sorry for the tough love, I hate IMOs!
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by ronniecoleman » Mon Dec 15, 2008 12:35 am
logitech wrote:
ronniecoleman wrote:IMO E
Dude if you have an IMO , keep it for yourself.

If you have a solution, share so we all can learn.

Sorry for the tough love, I hate IMOs!
love you logitech :lol:
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by vittalgmat » Mon Dec 15, 2008 2:07 am
OOps I too fell for the E trap. took a risky leap of guess and broke my nose :(.

Anyway here is how I solved it... similar to cramya..

from 1 and 2
x - y > -2
-x +2y > 6

Applying an important takeaway I learnt from this forum:
when two inequalities are in the same direction u can solve the inequalities as if they are simultanaeous equations.


Applying that we get y > 4.

Now plug in y > 4 in x -y > -2
we get x > 2.

So xy is greater than 8 and hence > 0
So C

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by cramya » Mon Dec 15, 2008 3:02 am
Applying an important takeaway I learnt from this forum:
when two inequalities are in the same direction u can solve the inequalities as if they are simultanaeous equation
When two inequalities face the same direction we can add the inequalities together.

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by tritrantran » Mon Dec 15, 2008 8:20 am
Ah,

I forgot to go back and plug in the value to see if the statement y > 4 is still true.

Moral of story: For inequalities, go back and plug in your answer choices to validate if the statement is still true

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by vittalgmat » Mon Dec 15, 2008 10:13 am
cramya wrote:
Applying an important takeaway I learnt from this forum:
when two inequalities are in the same direction u can solve the inequalities as if they are simultanaeous equation
When two inequalities face the same direction we can add the inequalities together.
OOps Oh yeah... thanks Cramya for the correction.
We cannot technically solve like simultaneous eqn coz if we multiply/divide by a -ve number the direction will change and then we cannot do anything with them!!!...

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by NicoFalcon » Mon Dec 15, 2008 12:53 pm
I solved it in a rather informal way. Let´s see if my reasoning is OK:

The problem is asking as if it is possible to say that x and y are both positive or negative ( so as xy<0)

(1) Only relative info, impossible to tell about the main issue.

INSUFF

(2) The same, only relative info.

INSUFF

Now considering together (1) and (2).

From (1) we know the "distance" between x and y is 2, but we don´t know if both of them are in the same "side" of the number line (if both of them are negative or positive).

From (2) we get that, by susbstracting an addiotional "y" to x, we get a "lowe" number, so y is deffinetly negative. And we also get that the result is less than -6, so to go from -2 to -6, that additional "y" has to be greater than 4. With that info, if we look again in (1) if you substract a number bigger than 4 (y) from x it gives you a number bigger than -2, so X must be positive too.

Then (1) + (2) SUFF. Answer C