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? (1) 2x-2y = 1 (2) x/y > 1 [#permalink]

Show Tags

20 Jul 2010, 12:41

14

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.

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

Show Tags

12 May 2012, 01:36

5

This post received KUDOS

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 [#permalink]

Show Tags

04 Jan 2012, 00:28

4

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.
_________________

\(2x-2y=1\) --> \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> 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 \(\frac{1}{y}>0\).

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? (1) 2x-2y = 1 (2) x/y > 1 [#permalink]

Show Tags

14 Sep 2013, 22:39

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.
_________________

Statement 1: 2x - 2y = 1 There are several pairs of numbers that satisfy this condition. Here are two: Case a: x = 1 and y = 0.5, in which case x and y are both positive Case b: x = -0.5 and y = -1, in which case x and y are not both positive Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: x/y > 1 This tells us that x/y is positive. This means that either x and y are both positive or x and y are both negative. Here are two possible cases: Case a: x = 4 and y = 2, in which case x and y are both positive Case b: x = -4 and y = -2, in which case x and y are not both positive Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 Statement 1 tells us that 2x - 2y = 1. Divide both sides by 2 to get: x - y = 1/2 Solve for x to get x = y + 1/2

Now take the statement 2 inequality (x/y > 1) and replace x with y + 1/2 to get: (y + 1/2)/y > 1 Rewrite as: y/y + (1/2)/y > 1 Simplify: 1 + 1/(2y) > 1 Subtract 1 from both sides: 1/(2y) > 0 If 1/(2y) is positive, then y must be positive.

Statement 2 tells us that either x and y are both positive or x and y are both negative. Now that we know that y is positive, it must be the case that x and y are both positive Since we can now answer the target question with certainty, the combined statements are SUFFICIENT

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

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
_________________

Jeffery Miller Head of GMAT Instruction

GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions

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

Show Tags

30 Sep 2010, 07:15

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.

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 .

Check Number Theory chapter of Math Book for more on number properties (link in my signature).
_________________

We’ve given one of our favorite features a boost! You can now manage your profile photo, or avatar , right on WordPress.com. This avatar, powered by a service...

Sometimes it’s the extra touches that make all the difference; on your website, that’s the photos and video that give your content life. You asked for streamlined access...

A lot has been written recently about the big five technology giants (Microsoft, Google, Amazon, Apple, and Facebook) that dominate the technology sector. There are fears about the...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...