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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # 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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Math Expert V
Joined: 02 Sep 2009
Posts: 56275
96^(1/2) < x*6^(1/2) and x/6^(1/2) < 6^(1/2). If x is an integer, whic  [#permalink]

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Difficulty:   15% (low)

Question Stats: 86% (01:25) correct 14% (02:06) wrong based on 33 sessions

### HideShow timer Statistics $$\sqrt{96} < x \sqrt{6}$$ 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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Math Expert V
Joined: 02 Aug 2009
Posts: 7764
Re: 96^(1/2) < x*6^(1/2) and x/6^(1/2) < 6^(1/2). If x is an integer, whic  [#permalink]

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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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Joined: 11 Mar 2018
Posts: 95
Re: 96^(1/2) < x*6^(1/2) and x/6^(1/2) < 6^(1/2). If x is an integer, whic  [#permalink]

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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

$$\sqrt{96} < x \sqrt{6}$$

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]

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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

$$\sqrt{96}$$ =$$\sqrt{16*6}$$ =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. 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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# 96^(1/2) < x*6^(1/2) and x/6^(1/2) < 6^(1/2). If x is an integer, whic  