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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

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:

(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? GMAT PREP CAT [#permalink]

Show Tags

20 Jul 2010, 11:41

12

This post received KUDOS

I found this one easiest to solve by drawing a graph. Clearly 1) and 2) alone are not sufficient as discussed, so what remains to be seen is if 2) adds enough information to 1) to determine if both x and y are positive.

Drawing a quick graph of the line y=x-1/2 we find that the x-intercept of the line is (0.5,0) and the y-intercept is (0,-0.5). From this graph we can clearly see that we don't need to worry about anything in the 4th quadrant (+x/-y is not >1) or the 3rd quadrant (|x|<|y|, therefore x/y is not >1). All that is left is the 1st quadrant, in which x and y are both positive.

Sufficient.
_________________

If you find my posts useful, please award me some Kudos!

(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.

Statement (1): x-y = 1/2. We can have x=1,y=1/2. Can also have x=0,y=-1/2. Insufficient. Statement (2): x/y>1. We can have x=3,y=2. Can also have x=-3,y=-2. Insufficient.

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

Show Tags

03 Jan 2012, 23:28

2

This post received KUDOS

C is the answer. Question: Is x > 0 AND y > 0?

Statement 1: 2x - 2y = 1 => 2(x - y) = 1 => x - y = 1/2 This just tells us that the difference is positive. But this can be true for cases when both x and y are positive, and when both x and y are negative. For instance, x = 1.5, y = 1 => x - y = 0.5; also, x = -1, y = -1.5 => x - y = 0.5. Thus, INSUFFICIENT.

Statement 2: x/y > 1 This just tells us that x and y have the same sign. That is, both are positive or both are negative. INSUFFICIENT.

Combining these statements, we can use the same numbers used in Statement 1 to find out that both the cases together do not work for negative numbers. For instance, x = -1, y = -1.5 => x - y = 0.5. However, x/y < 1. This violates statement 2.

Thus, the combination of the given statements tells us that x and y both have to be positive. => x > 0 AND y > 0. SUFFICIENT.
_________________

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

Show Tags

14 Sep 2013, 21:39

2

This post received KUDOS

imhimanshu wrote:

Hello Bunuel, Request you to please provide your comments on the doubt posted here-

Usually, whenever I see combining an inequality and equation, I substitute the value of one of the variable in the inequality and then analyze the effect. So, going by that approach;

x-y=1/2 ---(1) x/y>1 --(2) Substituting the value of x in equation(2)

(y+1/2)/y>1

Lets assume that y is positive-

(y+1/2) > y

1/2>0 --This means that our assumption is true since 1/2 is greater than Zero. Hence, y > 0

Now, Lets assume that y is negative-

Now, here I'm stuck, I know that multiplying by a negative number changes the sign of the inequality. I'm sure that the sign will be changed but what would be the resulting equation. I mean, do we need to replace y with "-y" in the whole equation. Please clarify. Which of the following would be correct then

a) y+1/2 <y b) y+1/2 < -y c) -y+1/2 < -y

Please help. Thanks

Refer to the highlighted portion : Actually you don't have to take 2 cases at this point: The expression you have is : \(\frac{y+0.5}{y}>1 \to 1+\frac{0.5}{y}>1 \to \frac{1}{y}>0\)--> Hence, y>0.

As for your doubt, if y is negative, we cross-multiply it and get : \(y+0.5<y \to 0>0.5\), which is absurd.

If y is negative, then -y would be positive, and for multiplying a positive quantity, you don't need to flip signs. So , yes expression a is correct.
_________________

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

Hope it helps.

Sorry for the bump but could you elaborate on the last part where you go from x/y>1 to (x-y)/y>0 to 1/y>0 ..?

I don't quite follow this algebra

\(\frac{x}{y}>1\) --> \(\frac{x}{y}-1>\) --> \(\frac{x-y}{y}>0\). Now, substitute \(x=y+\frac{1}{2}\) there to get \(\frac{1}{2y}>0\), which further simplifies to \(\frac{1}{y}>0\).

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

Show Tags

24 Sep 2015, 15:21

2

This post received KUDOS

jitendra31 wrote:

What happens when y = o? X = 0.5 and y = 0, satisfies i and x/y is indeed > 0 since anything divided by 0 is infinity. But, despite this, y is not positive as it '0'. '

It may seem that 0.5/0 = infinity, but this is not the case. If we approach 0 from the positive side, then it looks like 0.5/0 is a REALLY BIG POSITIVE NUMBER 0.5/0.1 = 5 0.5/0.01 = 50 0.5/0.001 = 500 0.5/0.0001 = 5000 0.5/0.00001 = 50000 etc.

But what if we approach 0 from the NEGATIVE side: 0.5/(-0.1) = -5 0.5/(-0.01) = -50 0.5/(-0.001) = -500 0.5/(-0.0001) = -5000 0.5/(-0.00001) = -50000 Here it looks like 0.5/0 will be a REALLY BIG NEGATIVE NUMBER

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

Show Tags

27 Oct 2016, 16:16

2

This post received KUDOS

Manbehindthecurtain wrote:

Are x and y both positive?

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

Solution:

We need to determine whether x and y are both positive.

Statement One Alone:

2x – 2y = 1

Simplifying statement one we have:

2(x – y) = 1

x – y = ½

The information in statement one is not sufficient to determine whether x and y are both positive. For example, if x = 1 and y = ½, x and y are both positive; however, if x = -1/2 and y = -1, x and y are both negative. We can eliminate answer choices A and D.

Statement Two Alone:

x/y> 1

Using the information in statement two, we see that x and y can both be positive or both be negative. Statement two alone is not sufficient to answer the question. We can eliminate answer choice B.

Statements One and Two Together:

Using the information in statements one and two, we know that x – y = ½ and that x/y > 1. Isolating x in the equation we have: x = ½ + y. We can now substitute ½ + y for x in the inequality x/y > 1 and we have:

(1/2 + y)/y > 1

(1/2)/y + y/y > 1

1/(2y) + 1 > 1

1/(2y) > 0

Thus, y must be greater than zero. Since x = ½ + y, x also must be greater than zero.

Answer: C
_________________

Jeffrey Miller Jeffrey Miller Head of GMAT Instruction

I guessed and got it right with a 50/50 guess at the end.

What I have done here is this

1) 2x - 2y = 1 hence x - y = \frac{1}{2} {Dividing both side by 2} In sufficient

2) x > y Alone in sufficient

When (1) + (2) We can say that if X is greater than y than x-y will yield a positive result.

Please correct me if I am wrong

First of all: the question is "are x and Y both positive?" not whether "x-y will yield a positive result".

Next, the red part is not correct.

\(\frac{x}{y}>1\) does not mean that \(x>y\). If both \(x\) and \(y\) are positive, then \(x>y\), BUT if both are negative, then \(x<y\). What you are actually doing when writing \(x>y\) from \(\frac{x}{y}>1\) is multiplying both parts of inequality by \(y\): never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.

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

See the complete solution of this problem in my previous post.

Re: Are X and Y both positive? GMAT PREP CAT [#permalink]

Show Tags

30 Sep 2010, 06:15

1

This post received KUDOS

Bunuel wrote:

zerotoinfinite2006 wrote:

Manbehindthecurtain wrote:

Are x and Y both positive?

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

I guessed and got it right with a 50/50 guess at the end.

What I have done here is this

1) 2x - 2y = 1 hence x - y = \frac{1}{2} {Dividing both side by 2} In sufficient

2) x > y Alone in sufficient

When (1) + (2) We can say that if X is greater than y than x-y will yield a positive result.

Please correct me if I am wrong

First of all: the question is "are x and Y both positive?" not whether "x-y will yield a positive result".

Next, the red part is not correct.

\(\frac{x}{y}>1\) does not mean that \(x>y\). If both \(x\) and \(y\) are positive, then \(x>y\), BUT if both are negative, then \(x<y\). What you are actually doing when writing \(x>y\) from \(\frac{x}{y}>1\) is multiplying both parts of inequality by \(y\): never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.

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

See the complete solution of this problem in my previous post.

Hope it helps.

I can clearly see how much weak I am in DS . I have no idea how to improve it. I am extremely weak in number system , including these kind of question. And day by day I am getting demoralize that I can't solve these kind of questions.

Anyways, Thanks a lot for your explanation Bunuel. You are genius as always. +1 more .
_________________

I don't want kudos.. I want to see smile on your face if I am able to help you.. which is priceless.

Just to add we can multiply y to both numerator and denominator of x/y the advantage is that the denominator becomes a square i.e in this case \(y^2\) so now we can safely cross multiply in \(xy/y^2>1\) since square of a no. is always +ve \(xy>y^2\) or \(y(x-y)>0\) This is a general method.

More usual way of doing this would be: \(\frac{x}{y}>1\) --> \(\frac{x}{y}-1>\) --> \(\frac{x-y}{y}>0\).
_________________

(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 was trying to solve this question by plugging in numbers. I too agree that staements A and B both alone are insufficient. SO now by taking both the statements together x>y so let us take x=3/2 and y=1 and plugging this value we can satisfy the equation 2x-2y = 1. Now let us take x=-1 and y=-3/2 and again plugging this value we can satisfy the equation 2x-2y = 1.

So the answer must be E. Please correct me where I am going wrong.

x=-1 and y=-3/2 don't satisfy the second statement: x/y=(-1)/(-3/2)=2/3<1.
_________________

Campus visits play a crucial role in the MBA application process. It’s one thing to be passionate about one school but another to actually visit the campus, talk...

Its been long time coming. I have always been passionate about poetry. It’s my way of expressing my feelings and emotions. And i feel a person can convey...

Marty Cagan is founding partner of the Silicon Valley Product Group, a consulting firm that helps companies with their product strategy. Prior to that he held product roles at...

Written by Scottish historian Niall Ferguson , the book is subtitled “A Financial History of the World”. There is also a long documentary of the same name that the...