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sqrt96<x<sqrt6
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09 Aug 2018, 03:54
09 Aug 2018, 03:54
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Question Stats:
75% (01:19) correct
25% (02:04) wrong based on 100 sessions
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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
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
Re: sqrt96<x<sqrt6
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12 Aug 2018, 01:08
12 Aug 2018, 01:08
1
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
(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
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29 Dec 2021, 05:28
29 Dec 2021, 05:28
1
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!
Watch the following video to learn the Basics of Inequalities
\(\sqrt{96} < x \sqrt{6}\) => \(\sqrt{16 * 6} < x \sqrt{6}\)
=> \(\sqrt{16} * \sqrt{6} < x \sqrt{6}\)
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!
Watch the following video to learn the Basics of Inequalities
