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(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches: \(2x-2y=1\) --> \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.

Re: Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 [#permalink]

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11 May 2010, 03:40

I'm still not clear why X and Y has to be positive when X/Y > 1. Can you please explain the way you combined taking both X and Y to be positive and also X and Y as negative. Since in either case X/Y will be > 1.

I'm still not clear why X and Y has to be positive when X/Y > 1. Can you please explain the way you combined taking both X and Y to be positive and also X and Y as negative. Since in either case X/Y will be > 1.

Thanks -H

From (2) \(\frac{x}{y}>1\), we can only deduce that x and y have the same sigh (either both positive or both negative).

When we consider two statement together:

From (1): \(2x-2y=1\) --> \(x=y+\frac{1}{2}\)

From (2): \(\frac{x}{y}>1\) --> \(\frac{x}{y}-1>0\) --> \(\frac{x-y}{y}>0\) --> substitute \(x\) from (1) --> \(\frac{y+\frac{1}{2}-y}{y}>0\)--> \(\frac{1}{2y}>0\) (we can drop 2 as it won't affect anything here and write as I wrote \(\frac{1}{y}>0\), but basically it's the same) --> \(\frac{1}{2y}>0\) means \(y\) is positive, and from (2) we know that if y is positive x must also be positive.

OR: as \(y\) is positive and as from (1) \(x=y+\frac{1}{2}\), \(x=positive+\frac{1}{2}=positive\), hence \(x\) is positive too.

I am still struggling to understand how come both together are sufficient? What is common in both the answer choices that makes c a correct choice?

Here is the logic for C:

When we consider two statement together:

From (1): \(2x-2y=1\) --> \(x=y+\frac{1}{2}\)

From (2): \(\frac{x}{y}>1\) --> \(\frac{x}{y}-1>0\) --> \(\frac{x-y}{y}>0\) --> substitute \(x\) from (1) --> \(\frac{y+\frac{1}{2}-y}{y}>0\)--> \(\frac{1}{2y}>0\) (we can drop 2 as it won't affect anything here and write as I wrote \(\frac{1}{y}>0\), but basically it's the same) --> \(\frac{1}{2y}>0\) means \(y\) is positive, and from (2) we know that if y is positive x must also be positive.

OR: as \(y\) is positive and as from (1) \(x=y+\frac{1}{2}\), \(x=positive+\frac{1}{2}=positive\), hence \(x\) is positive too.

Re: Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 [#permalink]

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13 Jun 2010, 10:41

1

This post received KUDOS

gsaxena26 wrote:

x/y>1

But x can be both +ve and negative. But if x>y then answer choice will be true as it can only be positive when x >y. Is my understanding correct?

Your reasoning is almost correct. Statement II doesn't say x>y, it only says x/y>1... Be careful there is a difference, this is the mistake you made initially

Taking this statement alone, it means x and y are both same sign (either both + or both -) and |x|>|y|. Thus, it is insufficient and you need to combine it with Statement I.

From statement I, you know x=y+1/2, hence x>y. If you know x>y and also |x|>|y| this means both are positive.

Re: Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 [#permalink]

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07 Sep 2010, 13:15

Bunuel,

From the 1/y>0 case what happens when y=0?, how goes GMAT treat that case? I notice that the question is asking if they are positive, not nonnegative, this case would have accounted for 0? Thanks

Bunuel wrote:

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

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches: \(2x-2y=1\) --> \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.

From the 1/y>0 case what happens when y=0?, how goes GMAT treat that case? I notice that the question is asking if they are positive, not nonnegative, this case would have accounted for 0? Thanks

Bunuel wrote:

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

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches: \(2x-2y=1\) --> \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.

Answer: C.

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\(\frac{x}{y}>1\) means that \(y\neq{0}\) (\(\frac{1}{y}\) means \(y>0\)). Also as division by zero is undefined then in cases when denominator could be zero GMAT would most likely state that denominator does not not equal to zero.
_________________

Re: Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 [#permalink]

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18 Jun 2011, 10:17

LM wrote:

Please explain....

(1) x-y = 1/2 so 1/4 - (-1/4) = 1/2 [y could be -ve] again, 1 - 1/2 = 1/2 [y could be +ve] Insufficient. (2) This option refers both x and y are -ves or +ves. insufficient.

For 1+2: x/y>1 and in option 1 x is positive so y is also positive.

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