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\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]

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20 Feb 2010, 15:14

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|

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

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20 Feb 2010, 15:57

4

This post received KUDOS

Expert's post

jeeteshsingh wrote:

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

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]

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20 Feb 2010, 16:13

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]

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05 Jun 2013, 07:58

[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]

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23 Dec 2015, 00:59

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

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24 Dec 2015, 04:30

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 _________________

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Re: If x is a positive integer, is root(x) an integer?
[#permalink]
24 Dec 2015, 04:30

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