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B
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Difficulty:
95%
(hard)
Question Stats:
40% (02:15) correct
60%
(02:33)
wrong
based on 124
sessions
History
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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.
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
![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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Hoozan
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
