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# sqrt96<x<sqrt6

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sqrt96<x<sqrt6 [#permalink]  09 Aug 2018, 03:54
Expert's post
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Question Stats:

68% (01:16) correct 31% (01:39) wrong based on 19 sessions
$$\sqrt{96} < x \sqrt{6}$$ 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
[Reveal] Spoiler: OA

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Joined: 20 Apr 2016
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WE: Engineering (Energy and Utilities)
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Re: sqrt96<x<sqrt6 [#permalink]  12 Aug 2018, 01:08
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] 12 Aug 2018, 01:08
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