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root[root{96}+2/(5+2*root{6})] lies between

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root[root{96}+2/(5+2*root{6})] lies between [#permalink]

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New post 26 Jan 2012, 05:26
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\(\sqrt{\sqrt{96}+\frac{2}{5+2*\sqrt{6}}}\) lies between

A) 1 & 2
B) 2 & 3
C) 3 & 4
D) 4 & 5
E) 5 & 6
[Reveal] Spoiler: OA
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Re: Square Root [#permalink]

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New post 26 Jan 2012, 05:39
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LM wrote:
\(\sqrt{\sqrt{96}+\frac{2}{5+2*\sqrt{6}}}\) lies between

A) 1 & 2
B) 2 & 3
C) 3 & 4
D) 4 & 5
E) 5 & 6


Multiply both nominator and denominator of \(\frac{2}{5+2*\sqrt{6}}\) by \({5-2*\sqrt{6}}\) and apply \((a+b)(a-b)=a^2-b^2\): \(\frac{2*(5-2*\sqrt{6})}{(5+2*\sqrt{6})(5-2*\sqrt{6})}=\frac{2*(5-2*\sqrt{6})}{25-24}=2*(5-2*\sqrt{6})\).

Now, \(\sqrt{6}\) is a little bit more than 2 (~2.5), hence \(2*(5-2*\sqrt{6})\approx{2(5-2*2.5)}=0\);

So we have: \(\sqrt{\sqrt{96}+\frac{2}{5+2*\sqrt{6}}}\approx{\sqrt{\sqrt{96}+0}=\sqrt{\sqrt{96}}\) --> \(\sqrt{96}\) is more than 9 but less than 10 (~9.5), hence \(\sqrt{\sqrt{96}}\approx{\sqrt{9.5}}\), which is more than 3 but less than 4.

Answer: C.

Or another way, using the same approximations as above:
\(\sqrt{\sqrt{96}+\frac{2}{5+2*\sqrt{6}}}\approx{\sqrt{9.5+\frac{2}{5+2*2.5}}}=\sqrt{9.5+\frac{1}{5}}\approx{\sqrt{10}}\approx{3.something}\)

Answer: C.

Hope it's clear.
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Re: root[root{96}+2/(5+2*root{6})] lies between [#permalink]

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New post 05 Jun 2013, 03:30
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Re: root[root{96}+2/(5+2*root{6})] lies between [#permalink]

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New post 05 Jun 2013, 06:27
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LM wrote:
\(\sqrt{\sqrt{96}+\frac{2}{5+2*\sqrt{6}}}\) lies between

A) 1 & 2
B) 2 & 3
C) 3 & 4
D) 4 & 5
E) 5 & 6


\(\sqrt{\sqrt{96}+\frac{2}{5+2*\sqrt{6}}}=\sqrt{4\sqrt{6}+\frac{2}{5+2*\sqrt{6}}}=\sqrt{2(2*\sqrt{6}+\frac{1}{5+2*\sqrt{6}})}=\sqrt{2(\frac{25+10\sqrt{6}}{5+2*\sqrt{6}})}=\sqrt{2*5(\frac{5+2\sqrt{6}}{5+2*\sqrt{6}})}=\sqrt{2*5}\)

a bit more than \(3\),C
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Re: root[root{96}+2/(5+2*root{6})] lies between [#permalink]

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New post 05 Jun 2013, 07:15
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Here is how I solved this:
1) Follow what Bunuel did up to a point
Bunuel wrote:
LM wrote:
\(\sqrt{\sqrt{96}+\frac{2}{5+2*\sqrt{6}}}\) lies between

A) 1 & 2
B) 2 & 3
C) 3 & 4
D) 4 & 5
E) 5 & 6


Multiply both nominator and denominator of \(\frac{2}{5+2*\sqrt{6}}\) by \({5-2*\sqrt{6}}\) and apply \((a+b)(a-b)=a^2-b^2\): \(\frac{2*(5-2*\sqrt{6})}{(5+2*\sqrt{6})(5-2*\sqrt{6})}=\frac{2*(5-2*\sqrt{6})}{25-24}=2*(5-2*\sqrt{6})\).


2) multiply out \(2(5-2\sqrt{6}) = 10-4\sqrt{6}\)
3) \(\sqrt{96} = 4\sqrt{6}\)
4) \(\sqrt{4\sqrt{6}+10-4\sqrt{6}} = \sqrt{10}\)
5) 3² = 9 and 4² = 16, so \(\sqrt{10}\) is between 3 and 4
Answer is C
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Re: root[root{96}+2/(5+2*root{6})] lies between [#permalink]

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New post 06 Jun 2013, 05:48
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LM wrote:
\(\sqrt{\sqrt{96}+\frac{2}{5+2*\sqrt{6}}}\) lies between

A) 1 & 2
B) 2 & 3
C) 3 & 4
D) 4 & 5
E) 5 & 6


In approximation questions we can approximate, so:
\(\sqrt{96}\) is between 9 and 10, \(\sqrt{6}\) is between 2 and 3, and the expression \(\frac{2}{5+2*\sqrt{6}}\) is about 1/5 (0.20) which means it does not affect the final answer too much in our case. Since the first number is bigger than 9 and smaller than 10 its square root will be more than 3 but less than 4.

Answer is C.
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Re: root[root{96}+2/(5+2*root{6})] lies between [#permalink]

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New post 06 Jun 2013, 06:30
ziko wrote:
LM wrote:
\(\sqrt{\sqrt{96}+\frac{2}{5+2*\sqrt{6}}}\) lies between

A) 1 & 2
B) 2 & 3
C) 3 & 4
D) 4 & 5
E) 5 & 6


In approximation questions we can approximate, so:
\(\sqrt{96}\) is between 9 and 10, \(\sqrt{6}\) is between 2 and 3, and the expression \(\frac{2}{5+2*\sqrt{6}}\) is about 1/5 (0.20) which means it does not affect the final answer too much in our case. Since the first number is bigger than 9 and smaller than 10 its square root will be more than 3 but less than 4.

Answer is C.


No need to solve for the second part. moment we get to know that sqrt of 96 will be approx 9 and sqrt of 9 is 3. So ans should be between 3 and 4
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Re: root[root{96}+2/(5+2*root{6})] lies between [#permalink]

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New post 16 Dec 2013, 03:49
LM wrote:
\(\sqrt{\sqrt{96}+\frac{2}{5+2*\sqrt{6}}}\) lies between

A) 1 & 2
B) 2 & 3
C) 3 & 4
D) 4 & 5
E) 5 & 6


I used ballparking. root96 rounded up to root100 is 10. root6 rounded down to root4 is 2. so I have 10+ 2/9 which is a little bigger than root9 which is 3.

Thus answer C.
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Re: root[root{96}+2/(5+2*root{6})] lies between [#permalink]

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New post 30 Sep 2015, 21:35
Hello from the GMAT Club BumpBot!

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Re: root[root{96}+2/(√+25+2∗6√−−−−−−−−−−−−√ 5+2*root{6})] lies between [#permalink]

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New post 01 Oct 2015, 22:58
NOTE: When ever we have an expression which contains a square root in the denominator, always rationalize the denominator.
Example: \(a/b+\sqrt{c}\)
We should multiply both the numerator and the denominator by \(b - \sqrt{c}\)


Coming to the question, first step is to rationalize \(2/(5 + 2*\sqrt{6})\)
Multiplying both numerator and denominator by \(5 - 2*\sqrt{6}\)

The expression becomes:

\((4 \sqrt{6} + 2*( 5 - 2*\sqrt{6)}/(25 - 24))\)^(1/2)
\((4 *\sqrt{6} + 10 - 4* \sqrt{6})\)^(1/2)
\(\sqrt{10} = 3. xx\)

Option C
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Re: root[root{96}+2/(5+2*root{6})] lies between [#permalink]

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New post 15 Nov 2016, 08:24
Hello from the GMAT Club BumpBot!

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Re: root[root{96}+2/(5+2*root{6})] lies between   [#permalink] 15 Nov 2016, 08:24
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