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:

Re: If x and y are integers, is x > y? [#permalink]

Show Tags

07 Dec 2012, 08:55

6

This post received KUDOS

Expert's post

12

This post was BOOKMARKED

If x and y are integers, is x > y?

(1) x + y > 0. Given that the sum of two numbers is greater than zero, but we cannot determine which one is greater. Not sufficient.

(2) y^x < 0. This statement implies that y is a negative number. Now, if y=-1 and x=1, then x>y BUT if y=-1 and x=-1, then x=y. Not sufficient.

(1)+(2) Since from (2) we have that y is a negative number, then -y is a positive number. Therefore from (1) we have that x>-y=positive, which means that x is a positive number. So, we have that x=positive>y=negative. Sufficient.

Re: If x and y are integers, is x > y? [#permalink]

Show Tags

08 Oct 2013, 09:29

Bunuel wrote:

If x and y are integers, is x > y?

(1) x + y > 0. Given that the sum of two numbers is greater than zero, but we cannot determine which one is greater. Not sufficient.

(2) y^x < 0. This statement implies that y is a negative number. Now, if y=-1 and x=1, then x>y BUT if y=-1 and x=-1, then x=y. Not sufficient.

(1)+(2) Since from (2) we have that y is a negative number, then -y is a positive number. Therefore from (1) we have that x>-y=positive, which means that x is a positive number. So, we have that x=positive>y=negative. Sufficient.

Re: If x and y are integers, is x > y? [#permalink]

Show Tags

08 Oct 2013, 17:12

shelrod007 wrote:

Bunuel wrote:

If x and y are integers, is x > y?

(1) x + y > 0. Given that the sum of two numbers is greater than zero, but we cannot determine which one is greater. Not sufficient.

(2) y^x < 0. This statement implies that y is a negative number. Now, if y=-1 and x=1, then x>y BUT if y=-1 and x=-1, then x=y. Not sufficient.

(1)+(2) Since from (2) we have that y is a negative number, then -y is a positive number. Therefore from (1) we have that x>-y=positive, which means that x is a positive number. So, we have that x=positive>y=negative. Sufficient.

Answer: C.

Should the answer not be E ?

My explanation as followa

Case 1 : -2>-5 Stmt 1 correct

(-5)^(-2) < 0

(1/25) < 0 Stmt 2 correct

-2 > 5 No ( Is x > y ? )

Case 2 :

5 > -2 Stmt 1 correct

(5) ^ (-2 )< 0

(1/25 ) < 0 Stmt 2 correct

5 > 2 Yes ( Is x > y ? )

(-5)^(-2)=\(\frac{1}{25}\) also 1/25>0

Remember when you square a negative number you get a positive number.

(5)^-2=\(\frac{1}{25}\), and again \(\frac{1}{25}\)>0 not less

when you have X^-Y, it's written out as \(\frac{1}{(X^Y)}\)

Re: If x and y are integers, is x > y? [#permalink]

Show Tags

26 Oct 2014, 13:04

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

(1) x+y>0 we can not determine wether x>y or not, but this statement says that one of the vaiables must be positive to satisfy it --> Not sufficient (2) Y^x<0 the only way it to be negative is when Y is negative, but X could be also negative (-2^-3 = -1/8) --> Not Sufficient, as we don't know wether x>y

(1)+(2) Statement 1 says that one of the variables must be positive + Statement 2 says that Y is in all cases negative --> if Y is negative X must be positive and is > Y

Answer (C) _________________

When you’re up, your friends know who you are. When you’re down, you know who your friends are.

Share some Kudos, if my posts help you. Thank you !

Re: If x and y are integers, is x > y? [#permalink]

Show Tags

04 May 2015, 09:47

Would anybody please explain why we ruled out statement 2. I thought the that the only was for y^x<0 is for y to be negative and X to be an odd power. I ruled out the negative values of X thinking that if X is negative then y would be a negative fraction and the question stem says that y is an integer. Why was I wrong here? Please help!

Re: If x and y are integers, is x > y? [#permalink]

Show Tags

04 May 2015, 21:21

1

This post received KUDOS

Expert's post

naeln wrote:

Would anybody please explain why we ruled out statement 2. I thought the that the only was for y^x<0 is for y to be negative and X to be an odd power. I ruled out the negative values of X thinking that if X is negative then y would be a negative fraction and the question stem says that y is an integer. Why was I wrong here? Please help!

From statement-II, we can deduce that \(y\) is a negative integer but we can't say if \(x\) is a negative or a positive odd integer. Let's evaluate both the cases:

Case-I: \(x\) is positive If \(x\) is a positive odd integer, then \(y^x < 0\). For example, assuming \(y = -2\) and \(x = 3\) would give \(y^x = -2^3 = -8 < 0\)

Case-II: \(x\) is negative When \(x\) is negative, \(y^x\) would become a negative fraction and not \(y\) itself.

For example: if \(y = -2\) and \(x = -3\), then \(x\) & \(y\) both are integers and \(y^x = -2^{-3} = \frac{-1}{8} < 0\). Here \(y^x = \frac{-1}{8}\) is a fraction and not \(y\) itself.

Hope its clear why statement-II does not give us a unique answer.

(1) x + y > 0. Given that the sum of two numbers is greater than zero, but we cannot determine which one is greater. Not sufficient.

(2) y^x < 0. This statement implies that y is a negative number. Now, if y=-1 and x=1, then x>y BUT if y=-1 and x=-1, then x=y. Not sufficient.

(1)+(2) Since from (2) we have that y is a negative number, then -y is a positive number. Therefore from (1) we have that x>-y=positive, which means that x is a positive number. So, we have that x=positive>y=negative. Sufficient.

Answer: C.

Thanks for the explanation. I got the correct answer using the same approach, however at the beginning of my solution I tried a different way to understand statement 2: Since \(y^0=1\) and \(0<1\), so if \(y^x<0\), it should mean that \(y^x<y^0\) and hence \(x<0\). I know this is not correct but I am unable to understand what am I doing wrong here. Can you please help confirm that?

Re: If x and y are integers, is x > y? [#permalink]

Show Tags

27 Dec 2015, 04:47

1

This post received KUDOS

Expert's post

Dienekes wrote:

Bunuel wrote:

If x and y are integers, is x > y?

(1) x + y > 0. Given that the sum of two numbers is greater than zero, but we cannot determine which one is greater. Not sufficient.

(2) y^x < 0. This statement implies that y is a negative number. Now, if y=-1 and x=1, then x>y BUT if y=-1 and x=-1, then x=y. Not sufficient.

(1)+(2) Since from (2) we have that y is a negative number, then -y is a positive number. Therefore from (1) we have that x>-y=positive, which means that x is a positive number. So, we have that x=positive>y=negative. Sufficient.

Answer: C.

Thanks for the explanation. I got the correct answer using the same approach, however at the beginning of my solution I tried a different way to understand statement 2: Since \(y^0=1\) and \(0<1\), so if \(y^x<0\), it should mean that \(y^x<y^0\) and hence \(x<0\). I know this is not correct but I am unable to understand what am I doing wrong here. Can you please help confirm that?

Thanks again!

If \(y\neq{0}\), then y^0=1. y^x < 0 does not mean that y^x < (1 = y^0). _________________

Re: If x and y are integers, is x > y? [#permalink]

Show Tags

27 Dec 2015, 16:44

Bunuel wrote:

Dienekes wrote:

Bunuel wrote:

If x and y are integers, is x > y?

(1) x + y > 0. Given that the sum of two numbers is greater than zero, but we cannot determine which one is greater. Not sufficient.

(2) y^x < 0. This statement implies that y is a negative number. Now, if y=-1 and x=1, then x>y BUT if y=-1 and x=-1, then x=y. Not sufficient.

(1)+(2) Since from (2) we have that y is a negative number, then -y is a positive number. Therefore from (1) we have that x>-y=positive, which means that x is a positive number. So, we have that x=positive>y=negative. Sufficient.

Answer: C.

Thanks for the explanation. I got the correct answer using the same approach, however at the beginning of my solution I tried a different way to understand statement 2: Since \(y^0=1\) and \(0<1\), so if \(y^x<0\), it should mean that \(y^x<y^0\) and hence \(x<0\). I know this is not correct but I am unable to understand what am I doing wrong here. Can you please help confirm that?

Thanks again!

If \(y\neq{0}\), then y^0=1. y^x < 0 does not mean that y^x < (1 = y^0).

Thanks! One follow-up question - What if the question stem said that y is a non-zero integer, would the inference \(y^x<0<y^0\) still be incorrect?

(1) \(x+y>0\) : For this to be true, we will have the following 3 senarios: a) \(x > 0\) and \(y > 0\) b) \(x > 0\) and \(y <= 0\), which means \(x > y\) c) \(y > 0\) and \(x <= 0\), which means \(y > x\)

Therefore, this statement is Insufficient

(2) \(y^x < 0\): For this to be true, \(y\) must be less than 0 and \(x\) must be an odd integer. Therefore, this statement is insufficient.

(1) and (2) together: From statement (2), we know \(y < 0\), therefore the only viable scenario for this in statement (1) is option (b): \(x > 0\) and \(y < 0\), which means \(x > y\) => Sufficient.

Re: If x and y are integers, is x > y? [#permalink]

Show Tags

29 Dec 2015, 23:10

Expert's post

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution

If x and y are integers, is x > y? (1) x + y > 0 (2) y^x < 0

There are two variables (x and y) in the original condition. In order to match the number of variables and the number of equations, we need 2 equations. Since the condition 1) and 2) each has 1 equation, there is high chance that C is going to be the answer. Using both the condition 1) and 2), we know that the condition 2) states y^x<0, which means y<0 and x=odd. Also, the condition 1) states x+y>0, which means y<0. So, x>0 and it is always the case that x>y. The answer is ‘yes’ and the conditions are sufficient. Therefore, since the condition 1) and 2) are not sufficient alone, the correct answer is C.

For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E. _________________

Re: If x and y are integers, is x > y? [#permalink]

Show Tags

19 Jun 2016, 06:09

is x> y?

State1:- x+y>0 .. the sum of two numbers is greater than zero does not tell us anything about how they rank relative to each other state 2:- y^x<0 --> this implies that x must be odd and y<0 .. If y <0 then x can be -1, +1. let y=-1. x=1 => y^x = -1 this satisfies y=-1 . x=-1 => 1/-1^1 => -1 this also satisfies. so x < y and x> y can both betrue. So not sufficient can y>0 .

combine the two statements y<0 and x+y>0 implies x has to be positive. Also note that since y<0 x+y>0 => x+y-y>0 => x>0. When the two opposite inequality, we can subtract, smaller from larger and get positive result.

gmatclubot

Re: If x and y are integers, is x > y?
[#permalink]
19 Jun 2016, 06:09

Part 2 of the GMAT: How I tackled the GMAT and improved a disappointing score Apologies for the month gap. I went on vacation and had to finish up a...

Cal Newport is a computer science professor at GeorgeTown University, author, blogger and is obsessed with productivity. He writes on this topic in his popular Study Hacks blog. I was...

So the last couple of weeks have seen a flurry of discussion in our MBA class Whatsapp group around Brexit, the referendum and currency exchange. Most of us believed...