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B
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Difficulty:
95%
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Question Stats:
39% (02:19) correct
61%
(02:32)
wrong
based on 117
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If x > 1, is \(\sqrt{{3x}}\) an integer?
(a) \(\sqrt{\frac{x}{3}}\)is not an integer
(b) \( x ÷\frac{1}{\sqrt{3}}\) is an integer
Bunuel please could you edit this post for better understanding. I was not able to showcase the sqrt and fraction together.
(a) \(\sqrt{\frac{x}{3}}\)is not an integer
(b) \( x ÷\frac{1}{\sqrt{3}}\) is an integer
Bunuel please could you edit this post for better understanding. I was not able to showcase the sqrt and fraction together.
![If x > 1, is [m][square_root]{3x}[/square_root][/m] an integer? (a) s](/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "If x > 1, is [m][square_root]{3x}[/square_root][/m] an integer? (a) s"){kind=icon}
23 Mar 2021, 05:22
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If x > 1, is \(\sqrt{{3x}}\) an integer?
(a) \(\sqrt{\frac{x}{3}}\)is not an integer
If \(x=\frac{4}{3}\), then yes.
If \(x=4\), then no.
Insufficient
(b) \( x ÷\frac{1}{\sqrt{3}}\) is an integer
\( x ÷\frac{1}{\sqrt{3}}=x\sqrt{3}\)
So x has to be some perfect square *\(\sqrt{3}\)
Let \(x=a^2\sqrt{3}\), where a is an integer.
\(\sqrt{3x}=\sqrt{3*a^2*\sqrt{3}}=a\sqrt{\sqrt{3}}\)
Answer is always no.
Sufficient
B
(a) \(\sqrt{\frac{x}{3}}\)is not an integer
If \(x=\frac{4}{3}\), then yes.
If \(x=4\), then no.
Insufficient
(b) \( x ÷\frac{1}{\sqrt{3}}\) is an integer
\( x ÷\frac{1}{\sqrt{3}}=x\sqrt{3}\)
So x has to be some perfect square *\(\sqrt{3}\)
Let \(x=a^2\sqrt{3}\), where a is an integer.
\(\sqrt{3x}=\sqrt{3*a^2*\sqrt{3}}=a\sqrt{\sqrt{3}}\)
Answer is always no.
Sufficient
B
30 Mar 2021, 13:54
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Bookmarks
Hoozan
If x > 1, is \(\sqrt{{3x}}\) an integer?
(a) \(\sqrt{\frac{x}{3}}\)is not an integer
(b) \( x ÷\frac{1}{\sqrt{3}}\) is an integer
Bunuel please could you edit this post for better understanding. I was not able to showcase the sqrt and fraction together.
(a) \(\sqrt{\frac{x}{3}}\)is not an integer
(b) \( x ÷\frac{1}{\sqrt{3}}\) is an integer
Bunuel please could you edit this post for better understanding. I was not able to showcase the sqrt and fraction together.
We may also ask "Is 3x a square?"
In order to have \(\sqrt{3x} = i\), we need x in the form of \(\frac{i^2}{3}\), where \(i\) is a positive integer.
Statement 1:
We may let \(x = \frac{2^2}{3}\). Then \(\sqrt{3x}\) would be an integer. Insufficient.
Statement 2:
\(\sqrt{3}x = j\), so we have \(x = \frac{\sqrt{3}j}{3}\). Compare \(\sqrt{3}j\) and \(i^2\), the former cannot be any integer squared so there is no case where \(\sqrt{3}j = i^2\). Then x cannot fit the criteria in any case, sufficient.
Ans: B
