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# are x and y both positive?

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are x and y both positive? [#permalink]  09 Jul 2007, 10:24
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0% (00:00) correct 0% (00:00) wrong based on 1 sessions
are x and y both positive?

a. 2x-2y = 1
b. x/y > 1

Senior Manager
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Re: gmatprep - DS question [#permalink]  09 Jul 2007, 10:48
dmittal wrote:
are x and y both positive?

a. 2x-2y = 1
b. x/y > 1

From stmt 1: 2*(x-y) = 1 => x-y = 1/2; x>y. x,y could be either +ve or -ve. Insufficient.

From stmt 2: x/y > 1 => |x| > |y| and x,y both have same sign (+ve or -ve). Insufficient.

From both stmts: Cannot tell if +ve or -ve. Insufficient. Hence, E.
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if x > y and either both x, and are +ve or -ve
then
based on equation x = 1/2 + y => both x and y are +ve. Right?
Intern
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dmittal wrote:
if x > y and either both x, and are +ve or -ve
then
based on equation x = 1/2 + y => both x and y are +ve. Right?

Hi all, I'm new here but figured I'd jump right in!!

I agree with you on this one. Taking the two statements together we know abs(x) must be greater than abs(y) and that x and y are either both positive or both negative.

We also know that 2x - 2y = 1.

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

There is no way abs(x) could be greater than abs(y) if x and y were both negative.

x = -2

-2 = 1/2 + y
-5/2 = y

abs(-2) < abs(-5/2)

x = -4
-4 = 1/2 + y
-9/2 = y

abs (-4) < abs(-9/2)

I'm going with C for my answer.

T
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Re: gmatprep - DS question [#permalink]  09 Jul 2007, 18:30
dmittal wrote:
are x and y both positive?

a. 2x-2y = 1
b. x/y > 1

(1) x - y = 1/2. Insuf.

(2) Insuf.

(1&2) From (2), x and y are both +ve or both -ve. There are infinite combinations of x and y so that (1) holds. What we must demonstrate is that there´s no combination of x, y -ve so that (1) holds.

Let x, y be both -ve and x/y >1, then x<y. Therefore, x-y<0, cannot be = 1/2 in any circumstance, then (1&2) are sufficient to state that both x and y are -ve, and the answer is C.
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st1:
x-y = 1/2. Can be x = -1/4, y = -3/4, or x = 1, y = 1/2. Insufficient.

st2:
x and y must be both positive or both negative. Insufficient.

using both, we still can't tell which set is applicable.

Ans E
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