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:

From 1, we can say have the following combinations. x=2, y=4. Ans: Not greater x=4, y=2. Ans: Not greater x=16, y=1. Ans: Not greater

All three possibilities of combination gives us a consistent answer as 'No'. Therefore A is sufficient.

What is the OA?

What if: x=-2, y=4 x=-4, y=2

I forgot to include signs. Well, since (1) says that the x^y = 16, this means that y has to be a positive, even number. So we can add two more combinations. x=-4, y=2 x=-2, y=4

Re: If x and y are integers, is x^y<y^x ? (1) x^y = 16 (2) x [#permalink]

Show Tags

24 Dec 2011, 10:45

guys, i was looking at this again.. and the OA seems debatable.

the question does not say if x<y or not.

so for 1, if x=16 and y = 1 then \(x^y\) = 16 and \(y^x\) = 1. then \(x^y\) > \(y^x\) but if y=16 and x =1 then \(x^y\) = 1 and \(y^x\) = 16. and \(x^y\) < \(y^x\).

so A is not sufficient.

combining the 2 statements, the only possibility will be (x,y) = (4,2) and (2,4). in both cases however, \(x^y\) = \(y^x\) and the ans to the question will be a No. So i think the ans to this question should be C.

Re: If x and y are integers, is x^y<y^x ? (1) x^y = 16 (2) x [#permalink]

Show Tags

24 Dec 2011, 12:44

\(\)

mmm..even i was confused, a bit.. X^Y can be -4^2, -2^4, 2^4 or 16^1 but in all cases above y^X is smaller than 16...so i guess its A In DS problem any concrete answer YES or NO can be an answer so i guess we do have an answer here.. _________________

Life is very similar to a boxing ring. Defeat is not final when you fall down… It is final when you refuse to get up and fight back!

Re: If x and y are integers, is x^y<y^x ? (1) x^y = 16 (2) x [#permalink]

Show Tags

01 Jan 2012, 12:32

1

This post received KUDOS

dreambeliever wrote:

guys, i was looking at this again.. and the OA seems debatable.

the question does not say if x<y or not.

so for 1, if x=16 and y = 1 then \(x^y\) = 16 and \(y^x\) = 1. then \(x^y\) > \(y^x\) but if y=16 and x =1 then \(x^y\) = 1 and \(y^x\) = 16. and \(x^y\) < \(y^x\).

so A is not sufficient.

combining the 2 statements, the only possibility will be (x,y) = (4,2) and (2,4). in both cases however, \(x^y\) = \(y^x\) and the ans to the question will be a No. So i think the ans to this question should be C.

please let me know if i'm missing something.

you dont need to know whether x>y.

there are 3 possibilities for statement 1. x=2 y =4 x=4 y=2 x=16 y=1 for all three the answer to the question is NO. so it is sufficient. _________________

Re: If x and y are integers, is x^y<y^x ? (1) x^y = 16 (2) x [#permalink]

Show Tags

07 Jan 2012, 16:21

1. x^y = 16. Only possible relationship is x=4,y=2. In this case x^y=y^x. Suff. 2. Say x=2, y=3, we get 2^3<3^2. Say x=3, y=4, we get 3^4>4^3. Insuff.

A _________________

I am the master of my fate. I am the captain of my soul. Please consider giving +1 Kudos if deserved!

DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

Re: If x and y are integers, is x^y<y^x? (1) x^y=16 [#permalink]

Show Tags

18 Jan 2012, 13:54

Expert's post

2

This post was BOOKMARKED

If x and y are integers, is x^y<y^x?

We have an YES/NO data sufficiency question. In a Yes/No Data Sufficiency question, each statement is sufficient if the answer is “always yes” or “always no” while a statement is insufficient if the answer is "sometimes yes" and "sometimes no".

(1) \(x^y=16\), since \(x\) and \(y\) are integers then following cases are possible: \(x=-4\) and \(y=2\) --> \(x^y=16>\frac{1}{16}=y^x\) --> the answer to the question is NO; \(x=-2\) and \(y=4\) --> \(x^y=16>\frac{1}{16}=y^x\) --> the answer to the question is NO; \(x=2\) and \(y=4\) --> \(x^y=16=y^x\) --> the answer to the question is NO; \(x=4\) and \(y=2\) --> \(x^y=16=y^x\) --> the answer to the question is NO; \(x=16\) and \(y=1\) --> \(x^y=16>1=y^x\) --> the answer to the question is NO.

As you can see in ALL 5 possible cases the answer to the question "is \(x^y<y^x\)?" is NO. Thus this statement is sufficient.

(2) x and y are consecutive even integers --> if \(x=2\) and \(y=4\) the answer will be NO but if \(x=0\) and \(y=2\) the answer will be YES. Not sufficient.

Since it is mentioned that x and y are integers. \(x^y = 16\) gives following solution, (x,y) = (-4,2), (-2,4), (2,4), (4,2) & (16,1)

to check \(x^y < y^x\) using above values of (x,y) \(16 = (-4)^2\) \(16 = (-2)^4\) \(16 = (2)^4\) \(16 = (4)^2\) \(16 = (16)^1\)

Although the answer is same (i.e, (A)), but I want to emphasis on the fact that all the cases should be considered. This would help in other DS questions.

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

Show Tags

16 Jan 2014, 09:57

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. _________________

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

Show Tags

09 Jul 2015, 10:29

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. _________________

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...