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.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
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
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
10 Sep 2022, 09:45
Kudos
Bookmarks
C
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:
83% (02:44) correct
17%
(03:31)
wrong
based on 41
sessions
History
Date
Time
Result
Not Attempted Yet
If circle A has 3 times the area of circle B, and circle B has one-sixth of the area of square WXYZ, what is the ratio of the area of a circle inscribed within WXYZ to that of the area of circle A?
A. \(\frac{\pi}{12}\)
B. \(\frac{\pi}{6}\)
C. \(\frac{\pi}{2}\)
D. \(\frac{2}{\pi}\)
E. \(\sqrt {12\pi}\)
A. \(\frac{\pi}{12}\)
B. \(\frac{\pi}{6}\)
C. \(\frac{\pi}{2}\)
D. \(\frac{2}{\pi}\)
E. \(\sqrt {12\pi}\)
10 Sep 2022, 23:00
Kudos
Bookmarks
Let the side of the square = s
Area of circle inscribed inside a square =\(\pi\)*\((\frac{s}{2})^2\) ==\(\pi\)*\(\frac{s^2}{4}\)
Area of circle B = \(\frac{s^2}{6}\)
Area of circle A = \(\frac{s^2}{2}\)
Required Ratio = \(\frac{\pi}{2}\)
Option C
Area of circle inscribed inside a square =\(\pi\)*\((\frac{s}{2})^2\) ==\(\pi\)*\(\frac{s^2}{4}\)
Area of circle B = \(\frac{s^2}{6}\)
Area of circle A = \(\frac{s^2}{2}\)
Required Ratio = \(\frac{\pi}{2}\)
Option C
Attachments
Screenshot 2022-09-11 112623.png [ 96.47 KiB | Viewed 1234 times ]
11 Sep 2022, 03:55
Kudos
Bookmarks
Bunuel
Solution:
Let us take a few variables: \(r_1=\) radius of circle A, \(r_2=\) radius of circle B, \(a=\) side of square WXYZ
Attachment:
WXYZ.png [ 8.57 KiB | Viewed 1173 times ]
Area of circle inscribed inside square WXYZ \(= \pi (\frac{a}{2})^2\)
We are asked the ratio of area of a circle inscribed within WXYZ to that of the area of circle A i.e., the value of \(\frac{\pi (\frac{a}{2})^2}{\pi (r_1)^2}=\frac{a^2}{4(r_1)^2}\)
This means we need the relation between \(a^2\) and \((r_1)^2\) to get our answer
According to the question, circle A has 3 times the area of circle B which means \(\pi (r_1)^2=3\pi (r_2)^2\) OR \((r_2)^2=\frac{(r_1)^2}{3}\)
Secondly, we are given circle B has one-sixth of the area of square WXYZ which means \(\pi (r_2)^2=\frac{1}{6}a^2\)
\(⇒\pi \frac{(r_1)^2}{3}=\frac{1}{6}a^2\)
\(⇒\frac{a^2}{(r_1)^2}=2\pi\)
We need the value of \(\frac{a^2}{4(r_1)^2}=\frac{2\pi}{4}=\frac{\pi}{2}\)
Hence the right answer is Option C
