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 2512:30 AM EDT
-01:30 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.
Apr 2601:30 AM EDT
-02:30 AM EDT
Learn how Keshav, a Chartered Accountant, scored an impressive 705 on GMAT in just 30 days with GMATWhiz's expert guidance. In this video, he shares preparation tips and strategies that worked for him, including the mock, time management, and more.
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.
May 0111:00 AM PDT
-12:00 PM PDT
Free- full-length test + 15 concept videos + 150+ short videos + study plan!
06 Dec 2021, 06:45
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
68% (02:18) correct
32%
(01:59)
wrong
based on 38
sessions
History
Date
Time
Result
Not Attempted Yet
A gold ingot in the shape of a cylinder is melted and the resulting molten metal is molded into few identical conical ingots. If the height of each conical ingot is half the height of the original cylinder and the area of the circular base of the cone is one fifth that of the cylinder, how many conical ingots can be made?
A. 60
B. 10
C. 30
D. 20
E. 40
Are You Up For the Challenge: 700 Level Questions
A. 60
B. 10
C. 30
D. 20
E. 40
Are You Up For the Challenge: 700 Level Questions
20 Dec 2021, 05:40
Kudos
Bookmarks
let
r1 = radius of cylinder
r2 = radius of cone
h1 = radius of cylinder
h2 = radius of cone
As per given information
h2 = h1\2
pi*(r2)^2 = 1/5 * pi * (r1)^2 -- (area of circular base of cone is 1/5th area of circular base of cylinder)
from above equation we have r1^2 = 5(r2)^2
cylinder is melted into conical ingots hence
vol(cylinder) = x (vol of cone) -- ( x is the number of conical ingots formed)
pi * r1^2 * h1 = x( 1/3 * pi * r2^2 *h2)
substituting above values , we will get value of x as 30
Answer is 30
Option C
r1 = radius of cylinder
r2 = radius of cone
h1 = radius of cylinder
h2 = radius of cone
As per given information
h2 = h1\2
pi*(r2)^2 = 1/5 * pi * (r1)^2 -- (area of circular base of cone is 1/5th area of circular base of cylinder)
from above equation we have r1^2 = 5(r2)^2
cylinder is melted into conical ingots hence
vol(cylinder) = x (vol of cone) -- ( x is the number of conical ingots formed)
pi * r1^2 * h1 = x( 1/3 * pi * r2^2 *h2)
substituting above values , we will get value of x as 30
Answer is 30
Option C
29 Dec 2021, 08:09
Kudos
Bookmarks
Let,
Height of Cylinder = H
Height of Cone = H/2
Radius of Cylinder = R
Area of circular base of the Cylinder = πR^2
Area of circular base of the Cone = (πR^2)/5
No. of conical ingots = N
Volume of cylinder = π*R^2*H
Volume of cone = 1/3*π*R^2*H
Volume of conical ingot = 1/3*(π*R^2/5)*(H/2) = (π*R^2*H)/30
Volume of cylindrical ingot = N * Volume of conical ingot
π*R^2*H = N * (π*R^2*H)/30
∴ N = 30 (C)
Height of Cylinder = H
Height of Cone = H/2
Radius of Cylinder = R
Area of circular base of the Cylinder = πR^2
Area of circular base of the Cone = (πR^2)/5
No. of conical ingots = N
Volume of cylinder = π*R^2*H
Volume of cone = 1/3*π*R^2*H
Volume of conical ingot = 1/3*(π*R^2/5)*(H/2) = (π*R^2*H)/30
Volume of cylindrical ingot = N * Volume of conical ingot
π*R^2*H = N * (π*R^2*H)/30
∴ N = 30 (C)
