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sqrt96<x<sqrt6

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Carcass
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sqrt96<x<sqrt6 [#permalink] New post   09 Aug 2018, 03:54
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\(\sqrt{96} < x \sqrt{6}\) and \(\frac{x}{\sqrt{6}}\) \(< \sqrt{6}\). If x is an integer, which of the following is the value of x?

(A) 2

(B) 3

(C) 4

(D) 5

(E) 6
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D
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Re: sqrt96<x<sqrt6 [#permalink] New post   12 Aug 2018, 01:08
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Carcass wrote:
\(\sqrt96 < x \sqrt6\) and \(\frac{x}{\sqrt{6}}\) \(< \sqrt{6}\). If x is an integer, which of the following is the value of x?

(A) 2

(B) 3

(C) 4

(D) 5

(E) 6



\(\sqrt96 < x \sqrt6\)

or \(4\sqrt6 < x \sqrt6\)

or\(4 < x\)

From the 2nd statement,

\({\frac{x}{\sqrt{6}}} < \sqrt6\)

or \(x < \sqrt6 * \sqrt6\)

or\(x < 6\)

So the value of x lies between \(4 < x < 6\)

So the answer is D i.e. 5
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Re: sqrt96<x<sqrt6 [#permalink] New post   29 Dec 2021, 05:28
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Given that \(\sqrt{96} < x \sqrt{6}\) and \(\frac{x}{\sqrt{6}}\) \(< \sqrt{6}\) and we need to find the value of x

\(\sqrt{96} < x \sqrt{6}\) => \(\sqrt{16 * 6} < x \sqrt{6}\)
=> \(\sqrt{16} * \sqrt{6} < x \sqrt{6}\)
Cancelling \(\sqrt{6}\) from both the sides we get
\(\sqrt{16} < x \)
=> x > 4

\(\frac{x}{\sqrt{6}}\) \(< \sqrt{6}\)
Multiply with \(\sqrt{6}\) on both the sides we get
x < \(\sqrt{6} * \sqrt{6}\)
=> x < 6

=> 4 < x < 6

So, Answer will be D
Hope it helps!

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Re: sqrt96<x<sqrt6 [#permalink]
29 Dec 2021, 05:28
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