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.
May 1912:00 PM PDT
-01:00 PM PDT
Scoring 329 on the GRE is not always about using more books, more courses, or a longer study plan. In this episode of GRE Success Talks, Ashutosh shares his GRE preparation strategy, study plan, and test-day experience, explaining how he kept his prep....
May 1501:00 PM IST
-11:00 AM IST
Start your journey with a fully customized action plan and work with a dedicated mentor to achieve a 735+ score.
May 2008:00 AM PDT
-08:30 AM PDT
What’s in it for you- Live Profile Evaluation Chat Session with Jenifer Turtschanow, CEO, ARINGO. Come with your details prepared and ARINGO will share insights! Pre-MBA Role/Industry, YOE, Exam Score, C/GPA, ECs Post-MBA Role/ Industry & School List.- Jun 10
06:00 AM PDT
-06:15 PM PDT
Register for the GMAT Club Virtual MBA Spotlight Fair – the world’s premier event for serious MBA candidates. This is your chance to hear directly from Admissions Directors at nearly every Top 30 MBA program..
26 Aug 2021, 22:08
Kudos
Bookmarks
Area of the shaded region is 3 times the area of the smaller circular region.
Let the smaller circle have r as radius and larger have R as radius.
pi(R)^2 - pi(r)^2 = 3 x pi r^2
=>4 pi r^2 = pi(R)^2
=> r/R = 1/2
=> R = 2r
Circumference of the larger circle C1= 2 * pi *R
= 2 * pi * 2r
= 4*pi*r
Circumference of the smaller circle C2 = 2 * pi * r
C1/C2 = 4*pi*r / 2*pi*r
= 2:1
(option c)
D.S
GMAT SME
Let the smaller circle have r as radius and larger have R as radius.
pi(R)^2 - pi(r)^2 = 3 x pi r^2
=>4 pi r^2 = pi(R)^2
=> r/R = 1/2
=> R = 2r
Circumference of the larger circle C1= 2 * pi *R
= 2 * pi * 2r
= 4*pi*r
Circumference of the smaller circle C2 = 2 * pi * r
C1/C2 = 4*pi*r / 2*pi*r
= 2:1
(option c)
D.S
GMAT SME
avigutman
Expert
Tutor
Joined: 17 Jul 2019
Last visit: 30 Sep 2025
Posts: 1,285
Own Kudos:
Given Kudos: 66
Location: Canada
Schools: NYU Stern (WA)
GMAT 1: 780 Q51 V45
GMAT 2: 780 Q50 V47
GMAT 3: 770 Q50 V45
Expert reply
04 Oct 2021, 04:37
Kudos
Bookmarks
Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
12 Oct 2021, 01:52
Kudos
Bookmarks
The area of the larger circle = The area of the smaller circle + The area of the shaded region
Given,
The area of the smaller circle = x
The area of the shaded region = 3x
So,
The area of the larger circle = x+3x = 4x
Let,
The radius of the smaller circle = 1
The radius of the larger circle = a
The area of the smaller circle= π (1)^2 = π
The area of the larger circle = 4π
so,
πa^2 = 4π
a^2 = 4
a= 2
Now,
Circumference of the smaller circle = 2π.1 = 2π
Circumference of the larger circle = 2π.2 = 4π
So, (4π/2π)= 2 times.
The Answer is C.
Is this the right approach Bunuel ?
Given,
The area of the smaller circle = x
The area of the shaded region = 3x
So,
The area of the larger circle = x+3x = 4x
Let,
The radius of the smaller circle = 1
The radius of the larger circle = a
The area of the smaller circle= π (1)^2 = π
The area of the larger circle = 4π
so,
πa^2 = 4π
a^2 = 4
a= 2
Now,
Circumference of the smaller circle = 2π.1 = 2π
Circumference of the larger circle = 2π.2 = 4π
So, (4π/2π)= 2 times.
The Answer is C.
Is this the right approach Bunuel ?
