It is currently 17 Dec 2018, 03:58
My Tests

Close

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
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.

One of GRE question

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Intern
Intern
Joined: 07 Jan 2017
Posts: 10
Followers: 0

Kudos [?]: 6 [0], given: 3

One of GRE question [#permalink] New post 07 Jan 2017, 09:50
00:00

Question Stats:

100% (00:05) correct 0% (00:00) wrong based on 1 sessions
Hi everyone, I couldn't confirm my answer whether it is correct or not, can anyone confirm for me?? Or correct me if I'm wrong??
Thanks.

BC = (10√3)/3
BC = 5.77

angle of C = (cos θ = AC/BC)
since radius of the circle with center C is 5
AC = 5
cos θ = 5/5.77
θ = (sin-1) 5/5.77
angle of C ≈ 30

sin θ = AB/BC
sin 30 = AB/5.77
0.5 = AB/5.77
AB = 2.88

Therefore Quantity B is greater.
Attachments

question.jpg
question.jpg [ 81.4 KiB | Viewed 1893 times ]

1 KUDOS received
GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 07 Jun 2014
Posts: 4750
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 93

Kudos [?]: 1664 [1] , given: 396

CAT Tests
Re: One of GRE question [#permalink] New post 07 Jan 2017, 17:45
1
This post received
KUDOS
Expert's post
Hey,

Your answer is correct. But use Pythagoras theorem.

\(BC^2 = AC^2 + AB^2\)

or \((\frac{10}{3}\sqrt{3})= 5^2 + AB^2\)

or \(AB^2= 100/3 -25 = 25/3 \approx 8.3333\)

or \(AB \approx 2.88\)

Regards
_________________

Sandy
If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test

Intern
Intern
Joined: 07 Jan 2017
Posts: 10
Followers: 0

Kudos [?]: 6 [0], given: 3

Re: One of GRE question [#permalink] New post 07 Jan 2017, 20:58
Thanks for reply
1 KUDOS received
Intern
Intern
Joined: 26 Feb 2018
Posts: 4
GRE 1: Q162 V148
Followers: 0

Kudos [?]: 5 [1] , given: 1

Re: One of GRE question [#permalink] New post 26 Feb 2018, 02:19
1
This post received
KUDOS
darkdevil8z wrote:
Hi everyone, I couldn't confirm my answer whether it is correct or not, can anyone confirm for me?? Or correct me if I'm wrong??
Thanks.

BC = (10√3)/3
BC = 5.77

angle of C = (cos θ = AC/BC)
since radius of the circle with center C is 5
AC = 5
cos θ = 5/5.77
θ = (sin-1) 5/5.77
angle of C ≈ 30

sin θ = AB/BC
sin 30 = AB/5.77
0.5 = AB/5.77
AB = 2.88

Therefore Quantity B is greater.



Both of you are going a very long way.



Method 1 : darkdevil8z method simplified : Only Mental Mathematic required. :idea:


Two sides are given and one angle is 90 degree so try diving both the length and see if you recognise the number.
I just divide (10√3)/3 with 5 and remainder is 2/√3.

I don't remember Cos/tan/cosec/cot/sec but I do remember Sin θ i.e (perpendicular / hypotenuse OR P/B)and you too have to.
This is the only requirement for GRE as far as trigonometry application is required.

I remember 1 - 1/2 - 1/√2 - √3/2 - 1 for Sin θ and recall that the above number is familiar. It is reciprocal of (10√3)/3 by 5

Now I know Sin 60 = √3/2 = 5 divided by (10√3)/3 = AC/CB = perpendicular/Hypotenuse
Therefore angle ABC is 60 degree and side in-front will definitely be bigger.

AC is greater

This was simpler method but for those who can't recite Sin θ at that moment, go for Pythagoras theorem.



Method 2 : Pythagoras theorem. method simplified : Only Mental Mathematic required. :idea:


ABˆ2 = (10/√3) - 5ˆ2 = 100/3 - 25 = 25/3 = Less than 9
AB = Less than 3

AC=5 is greater.



To get more simplified solutions.
Start Follow and Hit Kudos / Like

Regards
Shekhar :wink:
Re: One of GRE question   [#permalink] 26 Feb 2018, 02:19
Display posts from previous: Sort by

One of GRE question

  Question banks Downloads My Bookmarks Reviews Important topics  


GRE Prep Club Forum Home| About| Terms and Conditions and Privacy Policy| GRE Prep Club Rules| Contact

Powered by phpBB © phpBB Group

Kindly note that the GRE® test is a registered trademark of the Educational Testing Service®, and this site has neither been reviewed nor endorsed by ETS®.