Last visit was: 17 Jun 2024, 00:57 It is currently 17 Jun 2024, 00:57
Toolkit
GMAT Club Daily Prep
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

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.

# 3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}=

SORT BY:
Tags:
Show Tags
Hide Tags
Intern
Joined: 26 Aug 2010
Posts: 44
Own Kudos [?]: 710 [52]
Given Kudos: 18
Location: India
Concentration: Finance
Math Expert
Joined: 02 Sep 2009
Posts: 93696
Own Kudos [?]: 632381 [21]
Given Kudos: 82322
General Discussion
Manager
Joined: 22 Mar 2012
Posts: 83
Own Kudos [?]: 198 [0]
Given Kudos: 27
Location: United States (NY)
Concentration: General Management, Marketing
GPA: 3.1
WE:Accounting (Other)
Director
Joined: 02 Sep 2012
Status:Far, far away!
Posts: 857
Own Kudos [?]: 4910 [1]
Given Kudos: 219
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
1
Kudos
Lets analyze the first part $$3\sqrt{80} = 3\sqrt{5*16} = 3*4\sqrt{5}$$
The second term: Denominator $$(9+4\sqrt{5})*(9-4\sqrt{5})=9^2-4^2*5=1$$ Rule: (x+y)(x-y)=x^2-y^2
The second term: Numerator $$3*(9-4\sqrt{5})=27-12\sqrt{5}$$
Now putting all in one:
$$\sqrt{(}12\sqrt{5}+27-12\sqrt{5})=$$
$$\sqrt{27}=\sqrt{3*3^2}=3\sqrt{3}$$

Hope it's clear now
Intern
Joined: 20 Mar 2015
Posts: 1
Own Kudos [?]: 3 [3]
Given Kudos: 7
2
Kudos
1
Bookmarks
LM wrote:
$$\sqrt{3*\sqrt{80}+\frac{3}{9+4*\sqrt{5}}}=?$$

A) $$\sqrt{3*\sqrt{5}}$$

B) 3

C) $$3*\sqrt{3}$$

D) $$3+2*\sqrt{5}$$

E) $$9+4*\sqrt{5}$$

Though, I am quite comfortable with the method mentioned by Bunuel, I found an alternative way by The Economist.

$$9 + 4*\sqrt{5}$$ will be approx. equal to 9 + 4 x 2 = 17. Hence 3/ 17 will be quite less to contribute towards the value of expression.

$$3\sqrt{80}$$ is approx. 3 x 9 = 27. now $$\sqrt{27}$$ will be something more than 5.

Now coming to options:

(A)$$\sqrt{3*\sqrt{5}}$$ is approx $$\sqrt{6}$$ which is quite less than 5. Rejected.
(B) Rejected.
(C) 3 x 1.732 = 5.1 , which is in our desired range.
(D) 3 + 4 =7. Rejected
(E) 9 + 4 x 2 = 17. Rejected.

Hence the correct ans is (C).

If you like this Ballparking method, please press "Kudos".
Intern
Joined: 07 Mar 2016
Posts: 14
Own Kudos [?]: 12 [4]
Given Kudos: 3
Location: Indonesia
GPA: 3.06
1
Kudos
3
Bookmarks
samark wrote:
$$\sqrt{3\sqrt{80}+\frac{3}{9+4\sqrt{5}}} =$$

A) $$2\sqrt{3\sqrt{5}}$$
B) $$3$$
C) $$3\sqrt{3}$$
D) $$9+4\sqrt{5}$$
E) $$3+2\sqrt{5}$$

You can simplify this question by changing $$\sqrt{80}$$ to $$\sqrt{81}$$

$$4\sqrt{5}$$ is actually $$\sqrt{80}$$, change it also...

$$\sqrt{81}$$ is 9

so approximately the equation produce $$\sqrt{27}$$

= $$3\sqrt{3}$$
Target Test Prep Representative
Joined: 04 Mar 2011
Affiliations: Target Test Prep
Posts: 3042
Own Kudos [?]: 6458 [0]
Given Kudos: 1646
LM wrote:
$$\sqrt{3*\sqrt{80}+\frac{3}{9+4*\sqrt{5}}}=?$$

A) $$\sqrt{3*\sqrt{5}}$$

B) 3

C) $$3*\sqrt{3}$$

D) $$3+2*\sqrt{5}$$

E) $$9+4*\sqrt{5}$$

Let’s first simplify 3√80:

3√80 = 3 x √16 x √5 = 12√5

Next, let’s simplify 3/(9 + 4√5) by rationalizing the denominator:

3/(9 + 4√5) x (9 - 4√5)/(9 - 4√5) = (27 - 12√5)/(81 - 80) = 27 - 12√5

Finally, we have:

√(12√5 + 27 - 12√5) = √27 = √9 x √3 = 3√3

Tutor
Joined: 12 Oct 2010
Status:GMATH founder
Posts: 892
Own Kudos [?]: 1392 [0]
Given Kudos: 56
samark wrote:
$$\sqrt{3\sqrt{80}+\frac{3}{9+4\sqrt{5}}} =$$

A) $$2\sqrt{3\sqrt{5}}$$
B) $$3$$
C) $$3\sqrt{3}$$
D) $$9+4\sqrt{5}$$
E) $$3+2\sqrt{5}$$

$$?\,\,\,:\,\,\,{\rm{expression}}$$

$$\sqrt {80} \,\, = \,\,\underleftrightarrow {\sqrt {8 \cdot 10} = \sqrt {{2^4} \cdot 5} } = 4\sqrt 5 \,$$

$$\frac{1}{{9 + 4\sqrt 5 }} = \frac{1}{{9 + 4\sqrt 5 }} \cdot \frac{{9 - 4\sqrt 5 }}{{9 - 4\sqrt 5 }} = \frac{{9 - 4\sqrt 5 }}{{81 - 16 \cdot 5}} = 9 - 4\sqrt 5$$

$$3\left( {\sqrt {80} + \frac{1}{{9 + 4\sqrt 5 }}} \right)\,\, = \,\,3\,\left( 9 \right)\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,?\,\, = \,\,\sqrt {3 \cdot 9} = 3\sqrt 3$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Manager
Joined: 12 Sep 2017
Posts: 239
Own Kudos [?]: 117 [0]
Given Kudos: 132
samark wrote:
$$\sqrt{3\sqrt{80}+\frac{3}{9+4\sqrt{5}}} =$$

A) $$2\sqrt{3\sqrt{5}}$$
B) $$3$$
C) $$3\sqrt{3}$$
D) $$9+4\sqrt{5}$$
E) $$3+2\sqrt{5}$$

Hello math experts!

I think this might be a silly question but Why can't we square the whole term in order to get fid of the square root?

I'm a bit confused.

Thank you in advance!
Math Expert
Joined: 02 Sep 2009
Posts: 93696
Own Kudos [?]: 632381 [1]
Given Kudos: 82322
1
Kudos
jfranciscocuencag wrote:
samark wrote:
$$\sqrt{3\sqrt{80}+\frac{3}{9+4\sqrt{5}}} =$$

A) $$2\sqrt{3\sqrt{5}}$$
B) $$3$$
C) $$3\sqrt{3}$$
D) $$9+4\sqrt{5}$$
E) $$3+2\sqrt{5}$$

Hello math experts!

I think this might be a silly question but Why can't we square the whole term in order to get fid of the square root?

I'm a bit confused.

Thank you in advance!

Squaring to get rid of the square root is useful technique in some cases but here it does not give anything super helpful. We'd get $$3\sqrt{80}+\frac{3}{9+4\sqrt{5}}$$. How is this any better to calculate then what we had before? Plus we should not forget un-squaring back to get the correct answer.

Check the following question to see when squaring might be a good call: https://gmatclub.com/forum/new-tough-an ... l#p1029216
Non-Human User
Joined: 09 Sep 2013
Posts: 33616
Own Kudos [?]: 838 [0]
Given Kudos: 0