GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 Sep 2019, 19:10

### 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

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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 26 Aug 2010
Posts: 51
Location: India

### Show Tags

Updated on: 13 Jan 2016, 13:59
1
15
00:00

Difficulty:

45% (medium)

Question Stats:

71% (02:14) correct 29% (02:51) wrong based on 577 sessions

### HideShow timer Statistics

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

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

_________________

Originally posted by samark on 31 Oct 2010, 21:23.
Last edited by ENGRTOMBA2018 on 13 Jan 2016, 13:59, edited 2 times in total.
Reformatted the question and added the OA.
Math Expert
Joined: 02 Sep 2009
Posts: 58117

### Show Tags

31 Oct 2010, 21:45
4
8
samark wrote:
Attachment:
solve.jpg

$$\sqrt{3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}=\sqrt{12\sqrt{5}+\frac{3}{9+4\sqrt{5}}*\frac{9-4\sqrt{5}}{9-4\sqrt{5}}}=\sqrt{12\sqrt{5}+\frac{3(9-4\sqrt{5})}{81-80}}=\sqrt{12\sqrt{5}+27-12\sqrt{5}}=\sqrt{27}=3\sqrt{3}$$.

_________________
##### General Discussion
Current Student
Joined: 07 Mar 2016
Posts: 14
Location: Indonesia
GPA: 3.06

### Show Tags

05 May 2016, 01:17
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 $$\sqrt{27}$$

= $$3\sqrt{3}$$
Director
Joined: 12 Nov 2016
Posts: 705
Location: United States
Schools: Yale '18
GMAT 1: 650 Q43 V37
GRE 1: Q157 V158
GPA: 2.66

### Show Tags

19 Jun 2017, 12:40
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) $$3+2\sqrt{5}$$

We can deconstruct this problem by knowing 2 basic things:

We can find the normal form of a radical expression like 3 root 80 by multiplying the square root of the coefficient in front of the radical by the number inside. So 3 root 80 is 9 times 80 720. We also need to rationalize the denominator.
Attachments

image2.JPG [ 605.4 KiB | Viewed 3231 times ]

GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 937

### Show Tags

22 Oct 2018, 11:27
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.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
Senior Manager
Joined: 12 Sep 2017
Posts: 301

### Show Tags

13 Jan 2019, 16:09
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.

Math Expert
Joined: 02 Sep 2009
Posts: 58117

### Show Tags

13 Jan 2019, 23:39
1
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.

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
_________________
Re: 3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}=   [#permalink] 13 Jan 2019, 23:39
Display posts from previous: Sort by