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:

\sqrt{4x}=2*\sqrt{x}, so \sqrt{x} can either be integer or not an integer (for example \sqrt{x}=2.5) and the 2*\sqrt{x} is still an integer. So Statement 1 is insufficient.

\sqrt{3x}= \sqrt{3}*\sqrt{x}. As \sqrt{3} is not an integer, the \sqrt{x} can be either integer or non integer and the \sqrt{3}*\sqrt{x} will still be not integer. So Statement 2 is insufficient.

S1 and S2 together is still insufficient as \sqrt{x}=2 and \sqrt{x}=2.5 both satisfy statements requirement.

So I choose E as an answer.

Is there a flaw in my reasoning? OG Quant review answer to this question is different from E. Please advice.

Re: Is OG Quant question answer wrong? [#permalink]

Show Tags

20 Feb 2010, 14:14

alexBLR wrote:

This is the question from GMAT Quant Review:

If x is a positive integer , is \sqrt{x} an integer?

1) \sqrt{4x} is an integer. 2) \sqrt{3x} is not an integer.

My logic to solve this question:

\sqrt{4x}=2*\sqrt{x}, so \sqrt{x} can either be integer or not an integer (for example \sqrt{x}=2.5) and the 2*\sqrt{x} is still an integer. So Statement 1 is insufficient.

\sqrt{3x}= \sqrt{3}*\sqrt{x}. As \sqrt{3} is not an integer, the \sqrt{x} can be either integer or non integer and the \sqrt{3}*\sqrt{x} will still be not integer. So Statement 2 is insufficient.

S1 and S2 together is still insufficient as \sqrt{x}=2 and \sqrt{x}=2.5 both satisfy statements requirement.

So I choose E as an answer.

Is there a flaw in my reasoning? OG Quant review answer to this question is different from E. Please advice.

IMO D...

Ques: if x is a positive integer, is \(\sqrt{x}\) an integer?

S1: \(\sqrt{4x}\) is an integer

--> \(2* \sqrt{x}\) is an integer --> \(\sqrt{x}\) has to be an integer.. as x is a positive integer and hence cannot be a fraction. Therefore SUFF

S2: \(\sqrt{3x}\) is an integer --> \(\sqrt{3}*\sqrt{x}\) --> \(\sqrt{x}\) is not an integer as same could be a of a form of \(a\sqrt{3}\) where 'a' is a positive integer. Therefore SUFF
_________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

If x is a positive integer , is \sqrt{x} an integer?

1) \sqrt{4x} is an integer. 2) \sqrt{3x} is not an integer.

My logic to solve this question:

\sqrt{4x}=2*\sqrt{x}, so \sqrt{x} can either be integer or not an integer (for example \sqrt{x}=2.5) and the 2*\sqrt{x} is still an integer. So Statement 1 is insufficient.

\sqrt{3x}= \sqrt{3}*\sqrt{x}. As \sqrt{3} is not an integer, the \sqrt{x} can be either integer or non integer and the \sqrt{3}*\sqrt{x} will still be not integer. So Statement 2 is insufficient.

S1 and S2 together is still insufficient as \sqrt{x}=2 and \sqrt{x}=2.5 both satisfy statements requirement.

So I choose E as an answer.

Is there a flaw in my reasoning? OG Quant review answer to this question is different from E. Please advice.

IMO D...

Ques: if x is a positive integer, is \(\sqrt{x}\) an integer?

S1: \(\sqrt{4x}\) is an integer

--> \(2* \sqrt{x}\) is an integer --> \(\sqrt{x}\) has to be an integer.. as x is a positive integer and hence cannot be a fraction. Therefore SUFF

S2: \(\sqrt{3x}\) is an integer --> \(\sqrt{3}*\sqrt{x}\) --> \(\sqrt{x}\) is not an integer as same could be a of a form of \(a\sqrt{3}\) where 'a' is a positive integer. Therefore SUFF

If x is a positive integer, is \(\sqrt{x}\) an integer?

As given that \(x\) is a positive integer then \(\sqrt{x}\) is either an integer itself or an irrational number.

(1) \(\sqrt{4x}\) is an integer --> \(2\sqrt{x}=integer\) --> \(2\sqrt{x}\) to be an integer \(\sqrt{x}\) must be an integer or integer/2, but as \(x\) is an integer, then \(\sqrt{x}\) can not be integer/2, hence \(\sqrt{x}\) is an integer. Sufficient.

(2) \(\sqrt{3x}\) is not an integer --> if \(x=9\), condition \(\sqrt{3x}=\sqrt{27}\) is not an integer satisfied and \(\sqrt{x}=3\) IS an integer, BUT if \(x=2\), condition \(\sqrt{3x}=\sqrt{6}\) is not an integer satisfied and \(\sqrt{x}=\sqrt{2}\) IS NOT an integer. Two different answers. Not sufficient.

Answer: A.

jeeteshsingh, you should have spotted that there was something wrong with your solution as in DS two statements can not give you TWO DIFFERENT answers to the question (as you've got).

Re: Is OG Quant question answer wrong? [#permalink]

Show Tags

20 Feb 2010, 15:13

Bunuel wrote:

jeeteshsingh, you should have spotted that there was something wrong with your solution as in DS two statements can not give you TWO DIFFERENT answers to the question (as you've got).

Hope it helps.

My Bad.... overlooked it... Infact today I was telling this to someone over the forum that both the statements in DS would always be in sync..

Thanks Bunuel... for pointing this out.
_________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Re: If x is a positive integer, is root(x) an integer? [#permalink]

Show Tags

05 Jun 2013, 06:58

[quote="alexBLR"]If x is a positive integer, is \(\sqrt{x}\) an integer?

(1) \(\sqrt{4x}\) is an integer. (2) \(\sqrt{3x}\) is not an integer.

(1) since \(\sqrt{4x}\) is an integer, it is the same as 2*\(\sqrt{x}\) that means \(\sqrt{x}\) also must be ans integer - sufficient

(2) \(\sqrt{3x}\) is not an integer, in this case \(\sqrt{x}\) could be or couldn't be an integer. Case 1) x=4, where \(\sqrt{x}\) is an integer, then \(\sqrt{3*4}\)=2\(\sqrt{3}\) still not an integer Case 2) x=5, where \(\sqrt{x}\) is NOT an integer, then \(\sqrt{3*5}\)=\(\sqrt{15}\) still not an integer. This statement is not sufficient. Answer is A
_________________

If you found my post useful and/or interesting - you are welcome to give kudos!

\sqrt{4x}=2*\sqrt{x}, so \sqrt{x} can either be integer or not an integer (for example \sqrt{x}=2.5) and the 2*\sqrt{x} is still an integer. So Statement 1 is insufficient.

\sqrt{3x}= \sqrt{3}*\sqrt{x}. As \sqrt{3} is not an integer, the \sqrt{x} can be either integer or non integer and the \sqrt{3}*\sqrt{x} will still be not integer. So Statement 2 is insufficient.

S1 and S2 together is still insufficient as \sqrt{x}=2 and \sqrt{x}=2.5 both satisfy statements requirement.

So I choose E as an answer.

Is there a flaw in my reasoning? OG Quant review answer to this question is different from E. Please advice.

From First statement we get 2 sqrt (x) is an integer Therefore sqrt (x) must be an integer

Just to elaborate a bit further:

The thing is that sqrt (x) can't be an integer/2, cause the square root of any positive integer (as stated in the givens) is always >=1. Therefore, sqrt (x) will have to be an integer.

Suff

From second setatement we get the sqrt (3x) is not an integer Therefore, 3x is not a perfect square or x is not perfect square that is multiple of 3 But it could very well be a perfect square Matter is we still don't know

Just to clarify, sqrt (x) could or could not be an integer. If sqrt root (x) is an integer then sure 1.7 (integer) not an integer. But, sqrt (x) could also be say x=2, therefore sqrt (2)=1.4. Then 1.4*1.7 is clearly not an integer either. Answer A stands

Re: If x is a positive integer, is root(x) an integer? [#permalink]

Show Tags

22 Dec 2015, 23:59

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 is a positive integer, is root(x) an integer? [#permalink]

Show Tags

24 Dec 2015, 03:30

sqrt (4x) is an integer implies that 2 * sqrt(x) is also an integer. That implies x is an integer. Statement 1 is sufficient

sqrt (3x) is not an integer. For x = 2 it is possible. For a number such as 4 also it is. But sqrt () is not an integer while sqrt (4) is. Insufficient

Hence A
_________________

Fais de ta vie un rêve et d'un rêve une réalité

gmatclubot

Re: If x is a positive integer, is root(x) an integer?
[#permalink]
24 Dec 2015, 03:30

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...

The words of John O’Donohue ring in my head every time I reflect on the transformative, euphoric, life-changing, demanding, emotional, and great year that 2016 was! The fourth to...