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 3001:30 AM EDT
-02:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
May 0111:00 AM PDT
-12:00 PM PDT
Free- full-length test + 15 concept videos + 150+ short videos + study plan!
May 0306:30 AM PDT
-08:30 AM PDT
Verbal trouble on GMAT? Fix it NOW! Join Sunita Singhvi for a focused webinar on actionable strategies to boost your Verbal score and take your performance to the next level.
29 Oct 2014, 08:18
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
66% (03:04) correct
34%
(03:01)
wrong
based on 342
sessions
History
Date
Time
Result
Not Attempted Yet
Tough and Tricky questions: Geometry.
A cylindrical tank, with radius and height both of 10 feet, is to be redesigned as a cone, capable of holding twice the volume of the cylindrical tank. There are two proposed scenarios for the new cone: in scenario (1) the radius will remain the same as that of the original cylindrical tank, in scenario (2) the height will remain the same as that of the original cylindrical tank. What is the approximate difference in feet between the new height of the cone in scenario (1) and the new radius of the cone in scenario (2)?
(A) 13
(B) 25
(C) 30
(D) 35
(E) 40
29 Oct 2014, 19:35
Kudos
Bookmarks
Answer = D = 35
Area of Cylindrical tank \(= \pi 10^2 10 = 1000\pi\)
Required area of cone \(= 2 * 1000\pi = 2000\pi\)
Scenario (1) the radius will remain the same
Area of cone \(= \frac{1}{3} \pi 10^2 h = 2000\pi\)
h = 60
Scenario (2) the height will remain the same
Area of cone\(= \frac{1}{3} \pi r^2 10 = 2000\pi\)
\(r^2 = 600\)
\(r = 24.5 \approx 25\) (\(\sqrt{600}\) lies between \(\sqrt{576} = 24\) & \(\sqrt{625} = 25\))
Answer = 60 - 25 = 35
Area of Cylindrical tank \(= \pi 10^2 10 = 1000\pi\)
Required area of cone \(= 2 * 1000\pi = 2000\pi\)
Scenario (1) the radius will remain the same
Area of cone \(= \frac{1}{3} \pi 10^2 h = 2000\pi\)
h = 60
Scenario (2) the height will remain the same
Area of cone\(= \frac{1}{3} \pi r^2 10 = 2000\pi\)
\(r^2 = 600\)
\(r = 24.5 \approx 25\) (\(\sqrt{600}\) lies between \(\sqrt{576} = 24\) & \(\sqrt{625} = 25\))
Answer = 60 - 25 = 35
mvictor
Board of Directors
Joined: 17 Jul 2014
Last visit: 14 Jul 2021
Posts: 2,118
Own Kudos:
Given Kudos: 236
Location: United States (IL)
Concentration: Finance, Economics
Schools: Stanford '20 (WD)
GMAT 1: 650 Q49 V30
GPA: 3.92
WE:General Management (Transportation)
Products:
09 May 2016, 19:20
Kudos
Bookmarks
Bunuel
to be honest, I have never seen any official problem to be asking the volume of a cone..
volume of a cylinder = pi*r^2 *h
r=10
h=10
volume = 1000pi
volume of the cone: 2000pi
volume of the cone = pi*r^2 * h/3
2000=r^2 * h/3
1st case: r=10 -> h/3=20 => h=60
2nd case: h=10 -> r^2 = sqrt(600)
sqrt(600) = 10*sqrt(6)
sqrt(6)>sqrt(4)
sqrt(6)<sqrt(9)
so radius must be between 20 and 30. this approximation is enough.
60-30=30
60-20=40
so we are looking for smth between 30 and 40.
only D fits in here.
