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

Question

\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

Bunuel · Accepted Answer

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.

Zarrolou · Answer

\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

Bunuel · Answer

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on roots problems: math-number-theory-88376.html

All DS roots problems to practice: search.php?search_id=tag&tag_id=49
All PS roots problems to practice: search.php?search_id=tag&tag_id=113

Tough and tricky exponents and roots questions (DS): tough-and-tricky-exponents-and-roots-questions-125967.html
Tough and tricky exponents and roots questions (PS): new-tough-and-tricky-exponents-and-roots-questions-125956.html

lchen · Answer

Here is how I solved this:
1) Follow what Bunuel did up to a point

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

ziko · Answer

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.

ankur1901 · Answer

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

unceldolan · Answer

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.

TeamGMATIFY · Answer

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

arnold100 · Answer

No need for any complex calculations.
( 10- + 2/(5+2*2+) )^1/2
= ( 92/9+ )^1/2
= 10-/3+
= between 3 and 4

AweG · Answer

The fastest way for my is this : we have \sqrt{96}=\sqrt{6*8*2}=4*\sqrt{6}  and  \frac{2}{(5+2*\sqrt{6})}  = \frac{2*(5-2*sqr(6))}{5^2-2^2*sqr(6)^2}  =  10-4*\sqrt{6}

so we have  \sqrt{10}  which must lie between the closest perfect squares  \sqrt{9}<\sqrt{10}<\sqrt{12}  therefore answer is C

1473 · Answer

root of 96 = 7 x root2 = 9. something

2 x root6 = root of 24 = ~ 5

using just these 2 approximations, you can carry out the requisite operations in all parts of the expression to arrive at the conclusion that we are looking at approximately root of 10 as the answer, which is an easily mapped approximation to answer choice C since root of 9 is 3.

Pls forgive the awful formatting, i will learn to get sq. root symbols etc soon. Wrote this one out while studying.

BrushMyQuant · Answer

↧↧↧ Weekly Video Solution to the Problem Series ↧↧↧

https://www.youtube.com/watch?v=yBsS67Itgqs

We need to find the range in which  \sqrt{\sqrt{96}+\frac{2}{5+2*\sqrt{6}}}  lies

In order to simplify this lets try and simplify  \sqrt{96}  by breaking 96 into product of integers and  \frac{2}{5+2*\sqrt{6}}  by rationalizing the expression

=>  \sqrt{\sqrt{4*4*6}+\frac{2}{5+2*\sqrt{6}} * \frac{5-2*\sqrt{6}}{5-2*\sqrt{6}} }

Now for the second part of the expression in the denominator we have terms in the form (a+b)*(a-b) and it will be equal to  a^2 - b^2

=>  \sqrt{4 * \sqrt{6}+\frac{10 - 4*\sqrt{6}}{5^2 - 2*2*6}}  =  \sqrt{4 *\sqrt{6}+\frac{10 - 4*\sqrt{6}}{25-24}}  =  \sqrt{4 *\sqrt{6}+ 10 - 4*\sqrt{6}}  =  \sqrt{10}

And we know that  \sqrt{10}  is between 3 and 4

So,  Answer will be C 
Hope it helps!

Watch the following video to learn How to Rationalize Roots

https://www.youtube.com/watch?v=LhR_gXt8CLw

ThatDudeKnows · Answer

The name of the section is Quantitative REASONING. If you write more than one or two radical signs on your scratch paper in order to get the answer to this question, you are focused too much on the Q word, not enough on REASONING, and you're wasting valuable time during your test. Business schools don't care whether you can manipulate equations with radicals...that's going to be the job of the quants that you hire. Business schools care whether you can spot what really matters to a bottom line and whether you can eyeball a sensitivity analysis. You should be ballparking on a question like this. Even the answer choices make this clear...if they wanted you to do the "real" math, they would have given precise numbers (and made them close enough to each other that ballparking wouldn't work).

\sqrt{6}  is between 2 and 3, so  5+2*\sqrt{6}  is between 9 and 11. Even if it's one of those extremes, it's not going to move the needle on the final answer.  \sqrt{96}  is between 9 and 10 and the addend won't change that.

Answer choice C.

AmazingANI · Answer

Begin by figuring out the approx. value of square root of 96. It will be about 9.8.

Then proceed to figure out the approx. value of root 6 which in this case we take as 2.5 for the sake of making our calculations easy. We will see that 2/5+square root 6 will be equal to 2/5+5 = 2/10 = 1/5 which equals 0.2. Now add 9.8 and 0.2, we will get 10 and now since 10 is under the square root, we can clearly see that it will be greater than 3 and less than 4.

The correct answer will be C.

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

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

Prep Toolkit

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