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

 It is currently 19 Feb 2019, 07:01

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

Events & Promotions

Events & Promotions in February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar
• Free GMAT Prep Hour

February 20, 2019

February 20, 2019

08:00 PM EST

09:00 PM EST

Strategies and techniques for approaching featured GMAT topics. Wednesday, February 20th at 8 PM EST

February 21, 2019

February 21, 2019

10:00 PM PST

11:00 PM PST

Kick off your 2019 GMAT prep with a free 7-day boot camp that includes free online lessons, webinars, and a full GMAT course access. Limited for the first 99 registrants! Feb. 21st until the 27th.

Radius of circle center O is 3 times the radius of circle center C.

Author Message
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 52971

Show Tags

13 May 2016, 01:12
00:00

Difficulty:

25% (medium)

Question Stats:

79% (01:46) correct 21% (01:53) wrong based on 124 sessions

HideShow timer Statistics

Radius of circle center O is 3 times the radius of circle center C. Angle ACB = Angle POQ. If the shaded area of circle C is 2 then what is the area of the shaded part of circle O ?

A. 6
B. 12
C. 18
D. 36
E. 3/2

Attachment:

2016-05-13_1307.png [ 3.57 KiB | Viewed 4335 times ]

_________________
SC Moderator
Joined: 13 Apr 2015
Posts: 1687
Location: India
Concentration: Strategy, General Management
GMAT 1: 200 Q1 V1
GPA: 4
WE: Analyst (Retail)

Show Tags

13 May 2016, 01:43
2
1
Radius of circle with center O = 3*(Radius of circle with center C)
Area of circle with center O = 9*(Area of circle with center C)

Since the central angle is the same, Area of shaded region in Circle with center O = 9 * (Area of shaded region in circle with center C) = 9 * 2 = 18

Intern
Status: GMAT_BOOOOOOM.............. Failure is not an Option
Joined: 22 Jul 2013
Posts: 11
Location: India
Concentration: Strategy, General Management
GMAT 1: 510 Q38 V22
GPA: 3.5
WE: Information Technology (Consulting)

Show Tags

13 May 2016, 02:45
Area of the sector = (θ/360*) * π r2

So for C : (θ/360*) * π r2 = 2

r2 = (2/pi)*6 = 12/pi

Therefore r = 2 root 3 /Pi

Now for O : r = 3*2 root 3 /Pi = 6root3/pi

Area for O = 1/6 * pi(6root3/pi)2 = 1/6* 36 * 3 = 18

_________________

Kudos will be appreciated if it was helpful.

Cheers!!!!
Sit Tight and Enjoy !!!!!!!

Senior SC Moderator
Joined: 22 May 2016
Posts: 2479

Show Tags

31 May 2017, 14:50
1
Bunuel wrote:

Radius of circle center O is 3 times the radius of circle center C. Angle ACB = Angle POQ. If the shaded area of circle C is 2 then what is the area of the shaded part of circle O ?

A. 6
B. 12
C. 18
D. 36
E. 3/2

Attachment:
2016-05-13_1307.png

1. New area = old area * (scale factor)$$^2$$

2. Call the scale factor K.*

3. Radius of Circle O is 3 times the radius of Circle C. $$\frac{3r}{r}$$= 3. Scale factor K = 3

4. The scale factor for the sector areas is K$$^2$$, or 3$$^2$$= 9

5. Circle C's sector area is 2. Circle O sector area?

New area = old area * K$$^2$$

2 * 9 = 18

I tested values of sector angles = 60, Circle C radius = 6, Circle O radius therefore = 18 to be positive that 9 was the multiplier/scale factor.
With those values you get area of sector CAB = 6$$\pi$$ and area of sector OPQ = 54$$\pi$$. Correct.

*For a change from one length to another length, multiply by K.
For a change from one area to another area, multiply by K$$^2$$ (because area is length * length)
For a change from one volume to another volume, multiply by K$$^3$$ (because volume is length * length * length)
_________________

To live is the rarest thing in the world.
Most people just exist.

Oscar Wilde

Intern
Joined: 30 May 2017
Posts: 11

Show Tags

01 Jun 2017, 13:09
Another way to do this one:

Solve total area for both circles

Tiny circle = 4pi
Big circle = 36pi

Now lets just assume that the shaded region is an arbitrary amount of space of the overall circle (and since we know the interior angle is the same, we can assume that the % taken up by each region is the same % for both circles).

If the region is 10% of the circle, then it equals

Tiny circle shaded = .1 * 4pi = .4pi
Big circle shaded = .1 * 36pi = 3.6pi

This tells us two things, 1) the shaded region can't be bigger than 36pi since that is the overall area (so throw away answer D). 2) The regions are different by a factor of 9 (.4 * 9 = 3.6), so, the answer should also be able to divide by 9 evenly. THe only one that can do this is 18, or C.

Intern
Joined: 14 Jul 2016
Posts: 7

Show Tags

07 Oct 2018, 09:30
R=3.r
x= angle
Area of small circle(A1) = (x/360)*pi.r^2
Area of big circle (A2) = (x/360)*pi.R^2

A1/A2=1/9

A2 = 9.A1 = 18(C)
Re: Radius of circle center O is 3 times the radius of circle center C.   [#permalink] 07 Oct 2018, 09:30
Display posts from previous: Sort by