GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 19 Oct 2018, 23:25

Close

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

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

The volume of a cone is S × H/3, where S is the area of the base and H

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 50002
The volume of a cone is S × H/3, where S is the area of the base and H  [#permalink]

Show Tags

New post 29 Jul 2015, 03:27
1
3
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

57% (02:29) correct 43% (02:34) wrong based on 118 sessions

HideShow timer Statistics

Image
The volume of a cone is S × H/3, where S is the area of the base and H is the height. According to the figure above, if AB = BC, what is the ratio of the volume of the right section of the cone (with AB as height) to the volume of the left section of the cone (with BC as height)? (Note: Figure not drawn to scale.)

A. 1 : 3
B. 1 : 4
C. 1 : 6
D. 1 : 7
E. 1 : 8

Kudos for a correct solution.

Attachment:
cone-1.gif
cone-1.gif [ 7 KiB | Viewed 1835 times ]

_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

CEO
CEO
User avatar
P
Joined: 08 Jul 2010
Posts: 2559
Location: India
GMAT: INSIGHT
WE: Education (Education)
Reviews Badge
The volume of a cone is S × H/3, where S is the area of the base and H  [#permalink]

Show Tags

New post Updated on: 29 Jul 2015, 06:09
2
1
Bunuel wrote:
Image
The volume of a cone is S × H/3, where S is the area of the base and H is the height. According to the figure above, if AB = BC, what is the ratio of the volume of the right section of the cone (with AB as height) to the volume of the left section of the cone (with BC as height)? (Note: Figure not drawn to scale.)

A. 1 : 3
B. 1 : 4
C. 1 : 6
D. 1 : 7
E. 1 : 8

Kudos for a correct solution.

Attachment:
cone-1.gif


Ratio of Height of smaller and bigger cone = 1/2

CONCEPT: Ratio of Volume of similar solids = (Ratio of Height)^3

So Ratio of Volume of similar solids = (Ratio of Height)^3 = (1/2)^3 = 1/8

i.e. Volume of Right Section (Smaller Cone) = 1
then, Volume of Bigger Cone (including Smaller Cone) = 8
i.e. Volume of Left Section (Frustum) = Volume of Bigger Cone - Volume of Smaller Cone = 8-1 = 7
i.e. volume of Left Section = 8-1 =7

i.e. Volume of Right Section / Volume of Right Section = 1 / 7

Answer: Option D
_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION


Originally posted by GMATinsight on 29 Jul 2015, 04:47.
Last edited by GMATinsight on 29 Jul 2015, 06:09, edited 1 time in total.
Manager
Manager
User avatar
Joined: 14 Mar 2014
Posts: 147
GMAT 1: 710 Q50 V34
Re: The volume of a cone is S × H/3, where S is the area of the base and H  [#permalink]

Show Tags

New post 29 Jul 2015, 04:51
4
Bunuel wrote:
Image
The volume of a cone is S × H/3, where S is the area of the base and H is the height. According to the figure above, if AB = BC, what is the ratio of the volume of the right section of the cone (with AB as height) to the volume of the left section of the cone (with BC as height)? (Note: Figure not drawn to scale.)

A. 1 : 3
B. 1 : 4
C. 1 : 6
D. 1 : 7
E. 1 : 8

Kudos for a correct solution.





Let AB = BC = h. Then AC = 2h
Let the line extended from B meet at D and from C meet at E as shown in fig below

Attachment:
cone-1.gif
cone-1.gif [ 6.52 KiB | Viewed 1661 times ]


Triangle ABD and Triangle ACE are similar triangles ( ∠BAD = ∠ CAE , ∠B = ∠C =90)
Thus \(\frac{AB}{AC}\) = \(\frac{1}{2}\)
And \(\frac{BD}{CE}\) = \(\frac{1}{2}\) ==> \(\frac{x}{R}\) = \(\frac{1}{2}\)
R = 2x

volume of the right section of the cone (with AB as height) = π\(r^2\) * \(\frac{AB}{3}\)
= π\(x^2\) * \(\frac{h}{3}\)

volume of the left section of the cone (with BC as height) = volume of the left section of the cone (with AC as height) - volume of the right section of the cone (with AB as height)

volume of the left section of the cone (with AC as height) = π\(r^2\) * \(\frac{AC}{3}\)
=π\(R^2\) * \(\frac{2h}{3}\)
=π\(2x^2\) * \(\frac{2h}{3}\)
=π\(x^2\) * \(\frac{8h}{3}\)

Thus volume of the left section of the cone (with BC as height) = π\(x^2\) * \(\frac{8h}{3}\) - π\(x^2\) * \(\frac{h}{3}\)
= π\(x^2\) * \(\frac{7h}{3}\)

Thus Ratio = 1:7
_________________

I'm happy, if I make math for you slightly clearer
And yes, I like kudos
¯\_(ツ)_/¯ :-)

Intern
Intern
User avatar
Joined: 17 Jun 2014
Posts: 16
GMAT 1: 760 Q51 V42
GPA: 3.3
WE: Engineering (Manufacturing)
  [#permalink]

Show Tags

New post 29 Jul 2015, 05:09
1
Here's my solution

Answer, D

Attachments

math solution.PNG
math solution.PNG [ 17.8 KiB | Viewed 1645 times ]

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 50002
Re: The volume of a cone is S × H/3, where S is the area of the base and H  [#permalink]

Show Tags

New post 17 Aug 2015, 09:44
Bunuel wrote:
Image
The volume of a cone is S × H/3, where S is the area of the base and H is the height. According to the figure above, if AB = BC, what is the ratio of the volume of the right section of the cone (with AB as height) to the volume of the left section of the cone (with BC as height)? (Note: Figure not drawn to scale.)

A. 1 : 3
B. 1 : 4
C. 1 : 6
D. 1 : 7
E. 1 : 8

Kudos for a correct solution.

Attachment:
cone-1.gif


800score Official Solution:

1) We can create two similar triangles using AC and AB as the heights, R and x as the bases, and the edge of the cone as the hypotenuse. These triangles are similar because the angles for both are the same. The sides of similar triangles are proportional to each other.

2) Because AB = BC, and AC = AB + BC, then 2(AB) = AC.

3) Since the triangles are similar, and because 2(AB) = AC, 2x = R.

4) The volume of the left half of the cone equals the volume of the entire cone minus the volume of the right half of the cone, so the ratio we are looking for is:
(volume of right half) : (volume of cone – volume of right half).

5) Volume of the entire cone = S × H/3
= (πR²) × (AC)/3
= (π(2x)²) × (2AB)/3
= 8πx²AB/3

6) Volume of the right half of the cone = S × H/3
= πx²AB/3

7) Plug these values into the ratio equation above and solve:
πx²AB/3 : (8πx²AB/3 – πx²AB/3)
πx²AB/3 : 7πx²AB/3
1 : 7.

The correct answer is D.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Board of Directors
User avatar
P
Joined: 17 Jul 2014
Posts: 2657
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
GMAT ToolKit User Premium Member Reviews Badge
Re: The volume of a cone is S × H/3, where S is the area of the base and H  [#permalink]

Show Tags

New post 07 Nov 2015, 12:51
if AB is equal to 1/2H then x must be R/2
let's find the volume of the small cone with the base on the B
we know that radius of the small circle is R/2 of the big, and height is 1/2 of the H.
volume must be (R/2)^2 * pi * H/3 * 1/2 = R^2/4 * pi * H * 1/6 or (R^2 * pi * H) / 24
this is the volume of the small cone.

the volume of the cone from B to C is volume of the big A-C cone minus the small cone A->B.
volume of the big cone is R^2*pi * H/3 or (8 * R^2 * pi * H) / 24
subtract from the whole, the small one
(8 * R^2 * pi * H) / 24 minus
(R^2 * pi * H) / 24

we get (7 * R^2 * pi * H) / 24 - volume of the cone B-C

volume of the A->B to the B-C is then:
(R^2 * pi * H) / 24
divided everything by
(7 * R^2 * pi * H) / 24

or (R^2 * pi * H) * 24 / 24 * (7 * R^2 * pi * H)
simplify by 24, R^2, pi, and H, and remain with 1/7.


tough and tricky one!!!
Senior Manager
Senior Manager
avatar
G
Joined: 02 Apr 2014
Posts: 471
GMAT 1: 700 Q50 V34
Re: The volume of a cone is S × H/3, where S is the area of the base and H  [#permalink]

Show Tags

New post 16 Apr 2018, 11:38
One General idea i have,
whenever, we have two similiar triangles with sides in ratio 1:x ,
then ratio of the areas of the triangles will be of ratio 1 : x^2

now extending the concept to 3-D dimension, if there are two similiar cones of side ratio 1:x
then ratio of the volumes of cones will be of ratio 1:x^3

now we have small right cone and parent cone, small cone height/parent cone height = 1:2
so volume of small right cone/volume of parent cone = 1:8

so volume of remaining part of cone/volume of parent cone = 7:8
so volume of small right cone/volume of remaining part of cone = 1:7
GMAT Club Bot
Re: The volume of a cone is S × H/3, where S is the area of the base and H &nbs [#permalink] 16 Apr 2018, 11:38
Display posts from previous: Sort by

The volume of a cone is S × H/3, where S is the area of the base and H

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.