alexBLR wrote:

If x is a positive integer, is

\sqrt{x} an integer?

(1)

\sqrt{4x} is an integer.

(2)

\sqrt{3x} is not an integer.

This is the question from GMAT Quant Review. 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.

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

Insuff

Hence A is our best choice here

Hope it helps

Let the kudos rain begin

Cheers!

J