Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

\(\sqrt{\sqrt{96}+\frac{2}{5+2*\sqrt{6}}}\) 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: \(\sqrt{\sqrt{96}+\frac{2}{5+2*\sqrt{6}}}\approx{\sqrt{9.5+\frac{2}{5+2*2.5}}}=\sqrt{9.5+\frac{1}{5}}\approx{\sqrt{10}}\approx{3.something}\)

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

Show Tags

05 Jun 2013, 08:15

2

This post received KUDOS

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

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

Show Tags

06 Jun 2013, 06:48

1

This post received KUDOS

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. _________________

If you found my post useful and/or interesting - you are welcome to give kudos!

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

Show Tags

06 Jun 2013, 07:30

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 _________________

“Confidence comes not from always being right but from not fearing to be wrong.”

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

Show Tags

16 Dec 2013, 04:49

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.

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

Show Tags

30 Sep 2015, 22:35

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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

Show Tags

01 Oct 2015, 23:58

Expert's post

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 \(5 - 2*\sqrt{6}\)

So, my final tally is in. I applied to three b schools in total this season: INSEAD – admitted MIT Sloan – admitted Wharton – waitlisted and dinged No...

A few weeks ago, the following tweet popped up in my timeline. thanks @Uber_Mumbai for showing me what #daylightrobbery means!I know I have a choice not to use it...

“This elective will be most relevant to learn innovative methodologies in digital marketing in a place which is the origin for major marketing companies.” This was the crux...