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Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1

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Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 [#permalink] New post 14 May 2008, 16:30
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C
D
E

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Are x and y both positive?
(1) 2x - 2y = 1
(2) x/y > 1
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Re: GMATPrep DS [#permalink] New post 15 May 2008, 04:17
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I got E as response.
1=> not sufficient because for {x=-1/2 and y=-1} and {x=1/2 and y=0}, can't answer
2=> not sufficient also by the same way
Combining the two we have (1)=> x-y=1/2 and (2)=> x-y>0
So the first implies the second, can't tell.
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Re: GMATPrep DS [#permalink] New post 15 May 2008, 06:14
I got answer C

What is OA?

1=> not sufficient because of multiple sets -
{x= -1/2 , y=-1}
{x=1/2 , y=0},
{x= 1/4 ,y = -1/4},
{x=1, y=1/2}

2=> not sufficient because of multiple sets -
{x= 4 , y= 2}
{x= -4 , y= -2},
etc

However applying 2 for 1, x/y>1

only {x=1, y=1/2} is satisfied in which case both are positives

Please note - x/y > 1 => x>y => x-y> 0 only if we are sure y is +ve otherwise it may lead to wrong results
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Re: GMATPrep DS [#permalink] New post 15 May 2008, 06:42
sjgmat wrote:
I got answer C

What is OA?

1=> not sufficient because of multiple sets -
{x= -1/2 , y=-1}
{x=1/2 , y=0},
{x= 1/4 ,y = -1/4},
{x=1, y=1/2}

2=> not sufficient because of multiple sets -
{x= 4 , y= 2}
{x= -4 , y= -2},
etc

However applying 2 for 1, x/y>1

only {x=1, y=1/2} is satisfied in which case both are positives

Please note - x/y > 1 => x>y => x-y> 0 only if we are sure y is +ve otherwise it may lead to wrong results


Hello,

I don't agree with the bolded sentence :
Examples : x=1, y=1/2 => ok x-y> 0
x=-1, y=-2 => x-y>0 still!!!
So to me x/y>1 => (always) x-y > 0
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Re: GMATPrep DS [#permalink] New post 16 May 2008, 03:50
Hi Poullo,

the example you gave satisfies the x/y > 1 and x-y > 0

however , if we take x = -4 and y = -2 in this case x/y > 1 but x-y is not greater than zero

so it is not good practice to simplify x/y> 1 as x-y > 0 .. unless you are sure that y is positive ... as it might give you wrong results

Hope it clearifies :)
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Re: GMATPrep DS [#permalink] New post 16 May 2008, 08:26
Is there a way to solve these type of problems algebrically?
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Re: GMATPrep DS [#permalink] New post 16 May 2008, 11:43
jimmyjamesdonkey wrote:
Is there a way to solve these type of problems algebrically?


we know each statement alone is insuff

To resolve for C and E :

1) x-y=1/2

2) x/y>1

easy way; divide 1) by x

1-(y/x) =1/(2*x) ; left hand side quantity will be positive given the condition 2) and so x will be +ve ; since x/y>1 if x is positive y is also positive and you can click on C for answer.
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Re: GMATPrep DS [#permalink] New post 16 May 2008, 14:12
sorry rpmod, still don't understand your explanation..
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Re: GMATPrep DS [#permalink] New post 16 May 2008, 14:28
I think I understand how C is the right answer!

A) 2x - 2y = 1
This could be simplified into 2(x-y) = 1
(x-y) = 1/2
We are not given any info about the values of x or y, so this by itself is NS

B) x/y > 1
This could be simplified in two ways.
If y is positive, then x > y, subtract y from both sides of the inequality and you get x - y > 0
If y is negative, then x < y, subtract y from both sides of the inequality and you get x - y < 0
We aren't given any info on x or y so this by itself is NS

From looking at information A), we could see that x - y is 1/2, which is a positive number. From b) we could see that the first simplified inequality, x - y > 0, is the one we choose. Since y is positive, x must be a larger number than y so we have proof that both x and y are positive. Correct Answer: C.
Re: GMATPrep DS   [#permalink] 16 May 2008, 14:28
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