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96^(1/2) < x*6^(1/2) and x/6^(1/2) < 6^(1/2). If x is an integer, whic

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Bunuel
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96^(1/2) < x*6^(1/2) and x/6^(1/2) < 6^(1/2). If x is an integer, whic [#permalink] New post   14 Aug 2018, 02:18
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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: 96^(1/2) < x*6^(1/2) and x/6^(1/2) < 6^(1/2). If x is an integer, whic [#permalink] New post   14 Aug 2018, 02:23
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Bunuel wrote:
\(\sqrt{96} < x \sqrt{6}\) 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



m]\sqrt{96} < x \sqrt{6}........4\sqrt{6}<x\sqrt{6}..........x>4[/m] and
\(\frac{x}{\sqrt{6}} < \sqrt{6}.......x<6\).
So 4<x<6...
Since x is an integer, x must be 5..

D
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Re: 96^(1/2) < x*6^(1/2) and x/6^(1/2) < 6^(1/2). If x is an integer, whic [#permalink] New post   14 Aug 2018, 02:33
Bunuel wrote:
\(\sqrt{96} < x \sqrt{6}\) 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


\(\sqrt{96} < x \sqrt{6}\)

Dividing by \(\sqrt{6}\) on both sides.

Therefore,

x > \(\sqrt{16}\)

x > 4 ---------- (1)

Also,

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

Multiplying both sides with \(\sqrt{6}\)

Therefore,

x < 6 ------------- (2)

From (1) and (2), we get

4 < x < 6

As x is an integer, therefore,

x = 5 ---------- Option (D)
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Re: 96^(1/2) < x*6^(1/2) and x/6^(1/2) < 6^(1/2). If x is an integer, whic [#permalink] New post   14 Aug 2018, 02:38
Bunuel wrote:
\(\sqrt{96} < x \sqrt{6}\) 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



\(\sqrt{96}\) =\(\sqrt{16*6}\) =4\(\sqrt{6}\)

So , x > 4 .

again ,

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

x< \(\sqrt{6*6}\)

x < 6.

4<x<6

So, x is 5.

The best answer is D.
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Re: 96^(1/2) < x*6^(1/2) and x/6^(1/2) < 6^(1/2). If x is an integer, whic [#permalink]
14 Aug 2018, 02:38
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