Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Nov 1912:30 PM EST
-01:30 PM EST
Learn how Keshav, a Chartered Accountant, scored an impressive 705 on GMAT in just 30 days with GMATWhiz's expert guidance. In this video, he shares preparation tips and strategies that worked for him, including the mock, time management, and more
Nov 2001:30 PM EST
-02:30 PM IST
Learn how Kamakshi achieved a GMAT 675 with an impressive 96th %ile in Data Insights. Discover the unique methods and exam strategies that helped her excel in DI along with other sections for a balanced and high score.
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2407:00 PM PST
-08:00 PM PST
Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!
Nov 2510:00 AM EST
-11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
31 Mar 2012, 06:52
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
79% (01:49) correct
21%
(02:13)
wrong
based on 252
sessions
History
Date
Time
Result
Not Attempted Yet
The shaded region in the gure above represents a circular tire. If the distance from point O to point T is 18 inches and the area of the shaded region is equal to the area of the unshaded region, what is the radius of the unshaded region, in inches?
(A) \(\frac{9}{2}\sqrt{2}\)
(B) \(\frac{9}{2}\sqrt{2}\)
(C) 9
(D) \(9\sqrt{2}\)
(E) \(9\sqrt{3}\)
Attachment:
Circle.GIF [ 5.99 KiB | Viewed 8405 times ]
31 Mar 2012, 07:19
Kudos
Bookmarks
enigma123
The area of the big circle is \(\pi{R^2}=324\pi\);
The area of the small circle (the unshaded region) is \(\pi{r^2}\);
The area of the shaded region is \(324\pi-\pi{r^2}\);
We are told that the area of the shaded region is equal to the area of the unshaded region, so \(\pi{r^2}=324\pi-\pi{r^2}\) --> \(2\pi{r^2}=324\pi\) --> \(r^2=162\) --> \(r=9\sqrt{2}\).
Answer: D.
31 Mar 2012, 14:17
Kudos
Bookmarks
Outer ring's area = Bigger circle area - inner circle area = pi*R_out ^2 - pi*R_in ^2
it also equals to the inner circle area, by the question stem: pi*R_out ^2 - pi*R_in ^2 = pi*R_in ^2
So:
2*pi*R_in^2 = pi R_out^2
R_in^2 = (R_out^2)/2 => (18^2)/2
R_in = sqrt[ (18^2)/2 ] = 18/sqrt(2) = 9*sqrt(2) ===> answer D.
it also equals to the inner circle area, by the question stem: pi*R_out ^2 - pi*R_in ^2 = pi*R_in ^2
So:
2*pi*R_in^2 = pi R_out^2
R_in^2 = (R_out^2)/2 => (18^2)/2
R_in = sqrt[ (18^2)/2 ] = 18/sqrt(2) = 9*sqrt(2) ===> answer D.
