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 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 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.
23 Dec 2021, 07:00
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
75% (02:14) correct
25%
(02:31)
wrong
based on 164
sessions
History
Date
Time
Result
Not Attempted Yet
12 Days of Christmas GMAT Competition with Lots of Fun
The solid shows a truncated cone. R represents the radius of the base and r represents the radius of the truncated top. If the height (h) of the part of the cone removed from the top is 4, \(r = 5\), \(R = 10\), and \(h = 6\), what is the ratio of the volumes of the truncated cone and of the top removed?
A. 1:3
B. 1:9
C. 3:1
D. 9:1
E. 9:10
EG 12 days.png [ 16.5 KiB | Viewed 3046 times ]
The solid shows a truncated cone. R represents the radius of the base and r represents the radius of the truncated top. If the height (h) of the part of the cone removed from the top is 4, \(r = 5\), \(R = 10\), and \(h = 6\), what is the ratio of the volumes of the truncated cone and of the top removed?
A. 1:3
B. 1:9
C. 3:1
D. 9:1
E. 9:10
|
Attachment:
EG 12 days.png [ 16.5 KiB | Viewed 3046 times ]
General Discussion
23 Dec 2021, 07:37
Kudos
Bookmarks
Bunuel
Volume of cone = \(\frac{1}{3} \pi r^2 h \)
If the height of part of the cone that is removed is 4 and the height of the truncated cone is 6, the total height of original cone is 10.
Let's call this Big Cone
Therefore Volume of the whole cone = \(\frac{1}{3} \pi 10^2 10 \)
Height of the part of cone that is removed is 4. Lets call this Small cone.
Therefore volume of the Small Cone = \(\frac{1}{3} \pi 5^2 4 \)
Volume of the Truncated cone = Volume of Big Cone - Volume of Small Cone
= \(\frac{1}{3} \pi 10^2 10 \) - \(\frac{1}{3} \pi 5^2 4 \)
Therefore ratio of Truncated Cone to Small Cone is
\(\frac {\frac {1} {3} \pi 10^2 10 - \frac{1}{3} \pi 5^2 4}{\frac{1}{3} \pi 5^2 4}\)
(\(\frac {1}{3} \pi \) gets cancelled.)
=\(\frac {100(10)-100}{100}\\
= \frac {10-1}{1}\\
=\frac {9}{1}\)
Therefore ratio = 9:1
Answer = D
