Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

(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.

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.

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.

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.

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.

Posted from my mobile device

\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.
_________________

(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.