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:

the explanation says since sqrt (4x) is an integer ,it follows that 4x must be square of an integer and so x must be square of an integer and therefore sqrt (x) is an integer . i was trying to solve it this way sqrt(4x) =integer ,=> 2. sqrt (x) =integer => sqrt (x) =integer/2 =integer or non integer for example if 2 .sqrt (x) = 4 ,=> sqrt (x) =2 and so sqrt (x) is integer but if 2. sqrt (x) =3 ,=> sqrt (x) =3/2 and so sqrt (x) is non integer

friends please help me in pointing out whee i am going wrong.

Re: If x is a positive integer, is x^1/2 an integer [#permalink]

Show Tags

10 Jan 2010, 07:42

if x is a positive integer ,is sqrt (x) an integer

(1) sqrt(4x) is an integer .

(2) sqrt (3x) is not an integer .

the explanation says since sqrt (4x) is an integer ,it follows that 4x must be square of an integer and so x must be square of an integer and therefore sqrt (x) is an integer . i was trying to solve it this way sqrt(4x) =integer ,=> 2. sqrt (x) =integer => sqrt (x) =integer/2 =integer or non integer for example if 2 .sqrt (x) = 4 ,=> sqrt (x) =2 and so sqrt (x) is integer but if 2. sqrt (x) =3 ,=> sqrt (x) =3/2 and so sqrt (x) is non integer yes. sqrt(x) is an integer.

how i worked it out : sqrt(4x) is an integer ==> 2 * sqrt(x) is an integer

in order that the product of 2 and sqrt(x) be an integer, sqrt(x) must either be 1) an integer 2) exactly half of an integer. i.e a number like 0.5, 1.5, 2.5 etc etc you worked that out as well. sqrt(x) = integer/2

we also know x is an integer. is there any integer whose square root is a half of an integer ? no!

therefore the only other alternative is sqrt(x) is a whole integer.

\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 cannot 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|

If x is a positive integer, is \sqrt{x} an integer? (1) \(\sqrt{4x}\) is an integer4. (2) \(\sqrt{3x}\) is an integer.

I am not satisfied with the official explanation. Please give yours, thanks.

I think (2) should be \(\sqrt{3x}\) is NOT an integer.

If \(x=integer\), is \(\sqrt{x}=integer\)?

(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.

Re: If x is a positive integer, is x^1/2 an integer [#permalink]

Show Tags

29 Sep 2010, 07:10

ezinis wrote:

If x is a positive integer, is \sqrt{x} an integer? (1) \sqrt{4x} is an integer4. (2) \sqrt{3x} is an integer.

I am not satisfied with the official explanation. Please give yours, thanks.

(1) \(\sqrt{4x} = 2 * \sqrt{x}\) If this is an integer, then \(\sqrt{x}\) has to be an integer

(2) \(\sqrt{3x} = \sqrt{3} * \sqrt{x}\) For this to be an integer, \(\sqrt{x}\) must be of the form \(\sqrt{3} * Integer\) So \(\sqrt{x}\) is not an integer

I am not sure if the question is correct as (1) and (2) are contradicting. Is it supposed to say \(\sqrt{3x}\) is not an integer ?
_________________

Re: If x is a positive integer, is x^1/2 an integer [#permalink]

Show Tags

09 Jan 2011, 08:40

I didn't get the explanation:

if root(4x) is an integer fact. Then 2 * root(x) is an integer. So what!!!!!!!!!! This doesn't mean that root(x) should be an integer. Because root(x) can be 1.5 and yet won't distort the fact that 2 * 1.5 = 3 is an integer.

So, if x = 2.25(a non integer) root(x)=1.5 and 2*1.5=3 is an integer. if x=4(an integer) root(x)=2 and 2*2=4 is also an integer. So, statement one would be true for two values of x (2.25 and 4). root(2.25) is 1.5, not an integer. root(4) is 2, an integer. This statement is insufficient to conclude whether root(x) is an integer.

What's wrong with my explanation??
_________________

if root(4x) is an integer fact. Then 2 * root(x) is an integer. So what!!!!!!!!!! This doesn't mean that root(x) should be an integer. Because root(x) can be 1.5 and yet won't distort the fact that 2 * 1.5 = 3 is an integer.

So, if x = 2.25(a non integer) root(x)=1.5 and 2*1.5=3 is an integer. if x=4(an integer) root(x)=2 and 2*2=4 is also an integer. So, statement one would be true for two values of x (2.25 and 4). root(2.25) is 1.5, not an integer. root(4) is 2, an integer. This statement is insufficient to conclude whether root(x) is an integer.

What's wrong with my explanation??

You forgot that x is a positive integer, so \(\sqrt{x}\) cannot equal to \(\frac{integer}{2}\). Generally \(\sqrt{integer}\) is either an integer or an irrational number.

Complete solution:

If x is a positive integer, is sqrt(x) an integer

If \(x=integer\), is \(\sqrt{x}=integer\)?

(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.

Re: If x is a positive integer, is x^1/2 an integer [#permalink]

Show Tags

19 Jul 2011, 05:29

Quote:

then sqrt(x) can not be integer/2

I think sqrt(x) can be integer/2 as long as (integer/2) itself is an integer i.e. that integer is multiple of 2. I think that is what bunuel meant. And the answer remains same.
_________________

Conquer the Hell and make it Haven. Brain is your hell and Success is your haven!

"Kudos" is significant part of GMAT prep. If you like it, you just click it

Re: If x is a positive integer, is x^1/2 an integer [#permalink]

Show Tags

03 Jan 2012, 12:06

Oh nice problem. Took quite some time to answer it but got A. Explanation by Bunuel is more than sufficient to understand the solution.
_________________

Re: If x is a positive integer, is x^1/2 an integer [#permalink]

Show Tags

03 Jan 2012, 13:31

Both equations and number plugging helps here.

1. sqrt(4x) = integer, that is 2 x sqrt(x) = integer. For this equation to be true, sqrt(x) has to be an integer. SUFFICIENT. If number plugging, use 4x4 and 4x9 combinations 2. sqrt(3x) = frac. Here, sqrt(3) x sqrt(x) = frac, that is frac x sqrt(x) = frac. Difficult to determine if this relationship can infer sqrt(x) as integer. So, let's go with number plugging. sqrt(3x4) = 12 satisfies, and sqrt(4) = integer. sqrt(3x5) = 15 satisfies, but sqrt(5) = frac. So, insufficient.

+1 for 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 is a positive integer, is x^1/2 an integer [#permalink]

Show Tags

15 Dec 2012, 10:21

Bunuel - can you further explain your explanation for Statement 1? I am confused because if x is a positive integer, sqrt(x) can equal an integer/2 if for example x was equal to 4. 4 is a positive integer and the square root of 4 is equal to 4/2. I think I am missing the overall takeaway, can you help clarify? Thanks!

Bunuel - can you further explain your explanation for Statement 1? I am confused because if x is a positive integer, sqrt(x) can equal an integer/2 if for example x was equal to 4. 4 is a positive integer and the square root of 4 is equal to 4/2. I think I am missing the overall takeaway, can you help clarify? Thanks!

4/2=2 and is not a fraction (integer/2 means reduced fraction).
_________________