DS - GMATPrep - (are x and y both positive)

[nick_sun]

I think there is an error in GMATPrep. The answer must be E, but OA is C. Please see it attached.

[vnf]

Originally Posted by nick_sun

I think there is an error in GMATPrep. The answer must be E, but OA is C. Please see it attached.

Are x and y both positive?

(1) 2x - 2y = 1
(2) x/y > 1

(1) x - y = 1/2 => insuff
(2) x/y > 1

If y > 0 => x > y => x - y > 0
If y < 0 => x < y => x - y < 0

(1) & (2) => x - y = 1/2 > 0 => x > y > 0 => suff

Hence, (C)

[nick_sun]

Originally Posted by vnf

Are x and y both positive?

(1) 2x - 2y = 1
(2) x/y > 1

(1) x - y = 1/2 => insuff
(2) x/y > 1

If y > 0 => x > y => x - y > 0
If y < 0 => x < y => x - y < 0

(1) & (2) => x - y = 1/2 > 0 => x > y > 0 => suff

Hence, (C)

Well, it seems to be true but why -

Originally Posted by vnf

If y > 0 => x > y => x - y > 0
If y < 0 => x < y => x - y < 0

x > y can be 3>2 or 0.25>- 0.25 (=>0.25 - (- 0.25) = 1/2)
=> x and y can be both positive or positive and negative...

What am I missing?

[vnf]

Originally Posted by nick_sun

Well, it seems to be true but why -



x > y can be 3>2 or 0.25>- 0.25 (=>0.25 - (- 0.25) = 1/2)
=> x and y can be both positive or positive and negative...

What am I missing?

x > y and y > 0 => x > y > 0

[nick_sun]

Originally Posted by vnf

x > y and y > 0 => x > y > 0

I can't understand why "y > 0"... Could you please explain it

[kevinspain]

Originally Posted by nick_sun

I think there is an error in GMATPrep. The answer must be E, but OA is C. Please see it attached.

2 tells us that x and y are the same sign and that x is farther from 0 than is y

[nick_sun]

Originally Posted by kevinspain

2 tells us that x and y are the same sign and that x is farther from 0 than is y

Ok. I've got it

[jaybird]

nick,

remember when you get stuck the next best thing is to plug in numbers.

So check it:

Are x and y both positive?

(1) 2x - 2y = 1
(2) x/y > 1


Part I: 2x - 2y = 1 is the same as 2(x-y) = 1.

If we plug in 1/4 for "X" and (-1/4) for "Y" we get

2(1/4 - (-1/4) = 1. This is true. If we plug in: 2 for "X' and 3/2 for "Y" we get:
2(2 -3/2) = 1. Once again this is true. So here, X and Y can be both positive, or X can be positive and Y can be negative.

Insufficient:

Part II states: x/y > 1

Here X and Y can both be negative: -5/-4 = 5/4 which > 1. Or they can both be positive. 5/4 > 1.

Once again insufficient.

Combine both equations:

Equation 1 says: (X and Y are both positive) or (X is positive and Y is negative)
Equation 2 says: (X and Y are both positive) or (X and Y are negative)

(X are Y are positive) allows both equations to be true. Therefore C is sufficient.

[nick_sun]

Thank you Jaybird! Your explanation is very clear and comprehensive!

[tushar84]

Originally Posted by jaybird

nick,

remember when you get stuck the next best thing is to plug in numbers.

So check it:

Are x and y both positive?

(1) 2x - 2y = 1
(2) x/y > 1


Part I: 2x - 2y = 1 is the same as 2(x-y) = 1.

If we plug in 1/4 for "X" and (-1/4) for "Y" we get

2(1/4 - (-1/4) = 1. This is true. If we plug in: 2 for "X' and 3/2 for "Y" we get:
2(2 -3/2) = 1. Once again this is true. So here, X and Y can be both positive, or X can be positive and Y can be negative.

Insufficient:

Part II states: x/y > 1

Here X and Y can both be negative: -5/-4 = 5/4 which > 1. Or they can both be positive. 5/4 > 1.

Once again insufficient.

Combine both equations:

Equation 1 says: (X and Y are both positive) or (X is positive and Y is negative)
Equation 2 says: (X and Y are both positive) or (X and Y are negative)

(X are Y are positive) allows both equations to be true. Therefore C is sufficient.

Equation 1 doesn't rule out x negative and y negative.
x=-1/2; y = -1 -> -1/2 - (-1) = 1/2.
Am I missing something?