GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 04 Aug 2020, 03:49

### 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

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

# For a party, three solid cheese balls with diameters of 2 inches, 4 in

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 65785
For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

20 Oct 2015, 01:57
10
87
00:00

Difficulty:

45% (medium)

Question Stats:

75% (02:36) correct 25% (02:48) wrong based on 2271 sessions

### HideShow timer Statistics

For a party, three solid cheese balls with diameters of 2 inches, 4 inches, and 6 inches, respectively, were combined to form a single cheese ball. What was the approximate diameter, in inches, of the new cheese ball? (The volume of a sphere is $$\frac{4}{3}\pi r^3$$, where r is the radius.)

(A) 12

(B) 16

(C) $$\sqrt[3]{16}$$

(D) $$3\sqrt[3]{8}$$

(E) $$2\sqrt[3]{36}$$

Kudos for a correct solution.

_________________
GMAT Club Legend
Status: GMAT/GRE Tutor l Admission Consultant l On-Demand Course creator l A learner forever :)
Joined: 08 Jul 2010
Posts: 4512
Location: India
GMAT: QUANT EXPERT
Schools: IIM (A)
GMAT 1: 750 Q51 V41
WE: Education (Education)
Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

20 Oct 2015, 04:27
15
20
Bunuel wrote:
For a party, three solid cheese balls with diameters of 2 inches, 4 inches, and 6 inches, respectively, were combined to form a single cheese ball. What was the approximate diameter, in inches, of the new cheese ball? (The volume of a sphere is $$\frac{4}{3}\pi r^3$$, where r is the radius.)

(A) 12

(B) 16

(C) $$\sqrt[3]{16}$$

(D) $$3\sqrt[3]{8}$$

(E) $$2\sqrt[3]{36}$$

Kudos for a correct solution.

diameters of 2 inches, 4 inches, and 6 inches
i.e. Radie of 1 inches, 2 inches, and 3 inches

Sum of their volumes = $$\frac{4}{3}\pi (1^3+2^3+3^3)$$ = $$\frac{4}{3}\pi (36)$$

Volume of New Ball = $$\frac{4}{3}\pi (R^3)$$ = $$\frac{4}{3}\pi (36)$$

i.e. $$R^3 = 36$$
i.e. $$R = \sqrt[3]{36}$$
i.e. $$Diameter = 2\sqrt[3]{36}$$

_________________
Prepare with PERFECTION to claim Q≥50 and V≥40 !!!
GMATinsight .............(Bhoopendra Singh and Dr.Sushma Jha)
e-mail: info@GMATinsight.com l Call : +91-9999687183 / 9891333772
One-on-One Skype classes l Classroom Coaching l On-demand Quant course l Admissions Consulting

Most affordable l Comprehensive l 2000+ Qn ALL with Video explanations l LINK: Courses and Pricing
Our SUCCESS STORIES: From 620 to 760 l Q-42 to Q-49 in 40 days l 590 to 710 + Wharton l
FREE GMAT Resource: 22 FREE (FULL LENGTH) GMAT CATs LINKS l NEW OG QUANT 50 Qn+VIDEO Sol.
##### General Discussion
Intern
Joined: 29 Aug 2015
Posts: 8
For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

Updated on: 21 Oct 2015, 08:22
On combining the 3 cheese balls, we get a combined volume of
(4/3) π (1+8+27)= 48π.

Equating 48π with the formula for vol.of sphere
48π=4/3πr3
r3=36
So diameter= 236−−√3

Option E

Originally posted by HarrishGowtham on 20 Oct 2015, 07:06.
Last edited by HarrishGowtham on 21 Oct 2015, 08:22, edited 1 time in total.
GMAT Club Legend
Status: GMAT/GRE Tutor l Admission Consultant l On-Demand Course creator l A learner forever :)
Joined: 08 Jul 2010
Posts: 4512
Location: India
GMAT: QUANT EXPERT
Schools: IIM (A)
GMAT 1: 750 Q51 V41
WE: Education (Education)
Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

20 Oct 2015, 08:34
HarrishGowtham wrote:
On combining the 3 cheese balls, we get a combined volume of
(4/3) π (1+8+27)= 36π.

Equating 36π with the formula for vol.of sphere
36π=4/3πr3

r3=36
So diameter= 236−−√3

Option E

The answer that you have calculated is correct however the Highlighted calculation seems to be flawed.

(4/3) π (1+8+27) is NOT equal to 36π.

and 36π is NOT equal to 4/3πr3

I am sure the idea that you used is correct and it's just some typo.
_________________
Prepare with PERFECTION to claim Q≥50 and V≥40 !!!
GMATinsight .............(Bhoopendra Singh and Dr.Sushma Jha)
e-mail: info@GMATinsight.com l Call : +91-9999687183 / 9891333772
One-on-One Skype classes l Classroom Coaching l On-demand Quant course l Admissions Consulting

Most affordable l Comprehensive l 2000+ Qn ALL with Video explanations l LINK: Courses and Pricing
Our SUCCESS STORIES: From 620 to 760 l Q-42 to Q-49 in 40 days l 590 to 710 + Wharton l
FREE GMAT Resource: 22 FREE (FULL LENGTH) GMAT CATs LINKS l NEW OG QUANT 50 Qn+VIDEO Sol.
Retired Moderator
Joined: 29 Apr 2015
Posts: 748
Location: Switzerland
Concentration: Economics, Finance
Schools: LBS MIF '19
WE: Asset Management (Investment Banking)
Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

20 Oct 2015, 08:38
1
Bunuel wrote:
For a party, three solid cheese balls with diameters of 2 inches, 4 inches, and 6 inches, respectively, were combined to form a single cheese ball. What was the approximate diameter, in inches, of the new cheese ball? (The volume of a sphere is $$\frac{4}{3}\pi r^3$$, where r is the radius.)

(A) 12

(B) 16

(C) $$\sqrt[3]{16}$$

(D) $$3\sqrt[3]{8}$$

(E) $$2\sqrt[3]{36}$$

Kudos for a correct solution.

The volume of the combined cheese ball is $$\frac{4}{3}\pi*36$$. This is based on the three single $$r^3$$ from each cheese ball with radius 1, 2 and 3.

Now the new diameter will be 2*r = $$2\sqrt[3]{36}$$, where $$r^3 = 36$$

Senior Manager
Joined: 01 Mar 2015
Posts: 294
For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

21 Oct 2015, 04:09
Bunuel wrote:
For a party, three solid cheese balls with diameters of 2 inches, 4 inches, and 6 inches, respectively, were combined to form a single cheese ball. What was the approximate diameter, in inches, of the new cheese ball? (The volume of a sphere is $$\frac{4}{3}\pi r^3$$, where r is the radius.)

(A) 12

(B) 16

(C) $$\sqrt[3]{16}$$

(D) $$3\sqrt[3]{8}$$

(E) $$2\sqrt[3]{36}$$

Kudos for a correct solution.

$$\frac{4}{3}\pi r^3 = \frac{1}{6}\pi d^3$$

equating the volumes of cheese balls, we get

$$\frac{1}{6}\pi d^3 = \frac{1}{6}\pi 2^3 + \frac{1}{6}\pi 4^3 + \frac{1}{6}\pi 6^3$$

=> $$d^3 = 2^3 + 4^3 + 6^3$$

=> $$d^3 = 2^3 ( 1 + 8 + 27)$$

=> $$d= 2\sqrt[3]{36}$$

Kudos, if you like the explanation
Manager
Joined: 12 Sep 2015
Posts: 69
For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

28 Oct 2015, 05:31
v12345 wrote:
Bunuel wrote:
For a party, three solid cheese balls with diameters of 2 inches, 4 inches, and 6 inches, respectively, were combined to form a single cheese ball. What was the approximate diameter, in inches, of the new cheese ball? (The volume of a sphere is $$\frac{4}{3}\pi r^3$$, where r is the radius.)

(A) 12

(B) 16

(C) $$\sqrt[3]{16}$$

(D) $$3\sqrt[3]{8}$$

(E) $$2\sqrt[3]{36}$$

Kudos for a correct solution.

$$\frac{4}{3}\pi r^3 = \frac{1}{6}\pi d^3$$

equating the volumes of cheese balls, we get

$$\frac{1}{6}\pi d^3 = \frac{1}{6}\pi 2^3 + \frac{1}{6}\pi 4^3 + \frac{1}{6}\pi 6^3$$

=> $$d^3 = 2^3 + 4^3 + 6^3$$

=> $$d^3 = 2^3 ( 1 + 8 + 27)$$

=> $$d= 2\sqrt[3]{36}$$

Kudos, if you like the explanation

How do you get $$\frac{4}{3}\pi r^3 = \frac{1}{6}\pi d^3$$ ?
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 9256
GMAT 1: 760 Q51 V42
GPA: 3.82
For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

29 Oct 2015, 01:51
1
1
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

For a party, three solid cheese balls with diameters of 2 inches, 4 inches, and 6 inches, respectively, were combined to form a single cheese ball. What was the approximate diameter, in inches, of the new cheese ball? (The volume of a sphere is 43πr3, where r is the radius.)

(A) 12

(B) 16

(C) 3√16

(D) 33√8

(E) 23√36

Since the diameters are 2, 4, 6 the radii are 1, 2, 3.
So the total volume is (4/3)*π*1^3 + (4/3)*π*2^3 + (4/3)*π*3^3 =(4/3)*π*(1^3 + 2^3 +3^3 )=(4/3)*π*36.
Let the radius of the new cheese ball be r. Then the volume of the new cheese ball is (4/3)*π*r^3.

So (4/3)*π*r^3 should be (4/3)*π*36. That means r=3√36. So diameter is 2 3√36.

_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only \$79 for 1 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"
Verbal Forum Moderator
Status: Greatness begins beyond your comfort zone
Joined: 08 Dec 2013
Posts: 2445
Location: India
Concentration: General Management, Strategy
Schools: Kelley '20, ISB '19
GPA: 3.2
WE: Information Technology (Consulting)
Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

01 Nov 2015, 04:47
Let R be the radius of the single combined cheese ball
Sum of volumes of the 3 solid cheese balls = Volume of the single combined cheese ball

=> 4/3 * pi *( 1^(3) + 2^(3) + 3^(3) ) = 4/3 * pi * R^(3)
=> R^(3) = 36
Diameter = 2R = 2(36^(1/3))
_________________
When everything seems to be going against you, remember that the airplane takes off against the wind, not with it. - Henry Ford
The Moment You Think About Giving Up, Think Of The Reason Why You Held On So Long
Intern
Joined: 05 Sep 2015
Posts: 2
Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

28 Feb 2016, 01:15
My mistake was to calculate (1+2+3)³ instead of (1³+2³+3³) - can somebody explicate on why exactly it is wrong?
I see the mistake but I just can't find a sound explanation..
Intern
Joined: 10 Aug 2015
Posts: 29
Location: India
GMAT 1: 700 Q48 V38
GPA: 3.5
WE: Consulting (Computer Software)
Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

04 May 2016, 23:24
momomo wrote:
My mistake was to calculate (1+2+3)³ instead of (1³+2³+3³) - can somebody explicate on why exactly it is wrong?
I see the mistake but I just can't find a sound explanation..

Hi, the problem with your approach is you combined their length of radius to create one large sphere. But in reality we are combining the volumes of the spheres to create a larger sphere.

So we have to take sum of their volumes and not take sum of their radius.
Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2800
Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

05 May 2016, 03:52
7
1
Bunuel wrote:
For a party, three solid cheese balls with diameters of 2 inches, 4 inches, and 6 inches, respectively, were combined to form a single cheese ball. What was the approximate diameter, in inches, of the new cheese ball? (The volume of a sphere is $$\frac{4}{3}\pi r^3$$, where r is the radius.)

(A) 12

(B) 16

(C) $$\sqrt[3]{16}$$

(D) $$3\sqrt[3]{8}$$

(E) $$2\sqrt[3]{36}$$

Kudos for a correct solution.

Solution:

We first need to determine the volume of each individual cheese ball. We have 3 cheese balls of diameters of 2, 4, and 6 inches, respectively. Therefore, their radii are 1, 2 and 3 inches, respectively. Now let’s calculate the volume for each cheese ball.

Volume for 2-inch diameter cheese ball

(4/3)π(1)^3 = (4/3)π

Volume for 4-inch diameter cheese ball

(4/3)π(2)^3 = (4/3)π(8) = (32/3)π

Volume for 6-inch diameter cheese ball

(4/3)π(3)^3 = (4/3)π(27) = (108/3)π

Thus, the total volume of the large cheese ball is:

(4/3)π + (32/3)π + (108/3)π = (144/3)π = 48π

We can now use the volume formula to first determine the radius, and then the diameter, of the combined cheese ball.

48π = 4/3π(r)^3

48 x 3 = 4(r)^3

144 = 4(r)^3

36 = r^3

r = (cube root)√36

Thus, the diameter = 2*(cube root)√36

_________________

# Jeffrey Miller | Head of GMAT Instruction | Jeff@TargetTestPrep.com

250 REVIEWS

5-STAR RATED ONLINE GMAT QUANT SELF STUDY COURSE

NOW WITH GMAT VERBAL (BETA)

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

Manager
Joined: 21 Sep 2015
Posts: 84
Location: India
GMAT 1: 730 Q48 V42
GMAT 2: 750 Q50 V41
Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

10 Jun 2016, 08:11
1
( 1^3 + 2^3 + 3^3) = r ^3

r = cube root (36)

d= 2r
SVP
Joined: 23 Feb 2015
Posts: 2005
Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

17 Nov 2016, 08:57
Bunuel wrote:
For a party, three solid cheese balls with diameters of 2 inches, 4 inches, and 6 inches, respectively, were combined to form a single cheese ball. What was the approximate diameter, in inches, of the new cheese ball? (The volume of a sphere is $$\frac{4}{3}\pi r^3$$, where r is the radius.)

(A) 12

(B) 16

(C) $$\sqrt[3]{16}$$

(D) $$3\sqrt[3]{8}$$

(E) $$2\sqrt[3]{36}$$

Kudos for a correct solution.

combined volume of 3 solid cheese balls=(4/3)*πr^3
---->(4/3)*π(1^3+2^3+3^3)
----> (4/3)*π(36)
Let, the radius of new ball=R,
So, the volume of new balls=(4/3)*πR^3
So, we can write,
(4/3)*πR^3=(4/3)*π(36)
---> R^3=36
---> R=∛36
So, diameter (2*R)=2∛36
So, the correct answer is E.
Manager
Joined: 03 Oct 2013
Posts: 76
Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

17 Nov 2016, 09:17
Key equation: Sum of individual volumes = Total volume

4/3*pi*(r1^3+r2^3+r3^3) = 4/3*pi*(R^3)

r1, r2 and r3 are known -> Solve for R.

Be careful, that the question is diameter do answer = 2*R.
Manager
Joined: 27 Dec 2016
Posts: 218
Concentration: Marketing, Social Entrepreneurship
GPA: 3.65
WE: Marketing (Education)
Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

18 Jun 2017, 15:45
1
GMATinsight wrote:
Bunuel wrote:
For a party, three solid cheese balls with diameters of 2 inches, 4 inches, and 6 inches, respectively, were combined to form a single cheese ball. What was the approximate diameter, in inches, of the new cheese ball? (The volume of a sphere is $$\frac{4}{3}\pi r^3$$, where r is the radius.)

(A) 12

(B) 16

(C) $$\sqrt[3]{16}$$

(D) $$3\sqrt[3]{8}$$

(E) $$2\sqrt[3]{36}$$

Kudos for a correct solution.

diameters of 2 inches, 4 inches, and 6 inches
i.e. Radie of 1 inches, 2 inches, and 3 inches

Sum of their volumes = $$\frac{4}{3}\pi (1^3+2^3+3^3)$$ = $$\frac{4}{3}\pi (36)$$

Volume of New Ball = $$\frac{4}{3}\pi (R^3)$$ = $$\frac{4}{3}\pi (36)$$

i.e. $$R^3 = 36$$
i.e. $$R = \sqrt[3]{36}$$
i.e. $$Diameter = 2\sqrt[3]{36}$$

Simple question : why we calculate the sum of their volumes? In my opinion, total volume of NEW cheese ball must be greater than total combined volume from three small cheese balls.
_________________
There's an app for that - Steve Jobs.
Math Expert
Joined: 02 Sep 2009
Posts: 65785
Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

18 Jun 2017, 20:40
septwibowo wrote:
GMATinsight wrote:
Bunuel wrote:
For a party, three solid cheese balls with diameters of 2 inches, 4 inches, and 6 inches, respectively, were combined to form a single cheese ball. What was the approximate diameter, in inches, of the new cheese ball? (The volume of a sphere is $$\frac{4}{3}\pi r^3$$, where r is the radius.)

(A) 12

(B) 16

(C) $$\sqrt[3]{16}$$

(D) $$3\sqrt[3]{8}$$

(E) $$2\sqrt[3]{36}$$

Kudos for a correct solution.

diameters of 2 inches, 4 inches, and 6 inches
i.e. Radie of 1 inches, 2 inches, and 3 inches

Sum of their volumes = $$\frac{4}{3}\pi (1^3+2^3+3^3)$$ = $$\frac{4}{3}\pi (36)$$

Volume of New Ball = $$\frac{4}{3}\pi (R^3)$$ = $$\frac{4}{3}\pi (36)$$

i.e. $$R^3 = 36$$
i.e. $$R = \sqrt[3]{36}$$
i.e. $$Diameter = 2\sqrt[3]{36}$$

Simple question : why we calculate the sum of their volumes? In my opinion, total volume of NEW cheese ball must be greater than total combined volume from three small cheese balls.

This does not make sense.

How can you make a ball larger in volume than the combined volume of three balls you are making it from?
_________________
Manager
Joined: 21 Jul 2017
Posts: 180
Location: India
GMAT 1: 660 Q47 V34
GPA: 4
WE: Project Management (Education)
Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

18 Aug 2017, 09:53
Tip: Volume remains same when a solid is transformed from one shape to another!

Sum of their volumes = 43π(13+23+33)43π(13+23+33) = 43π(36)43π(36)

Volume of New Ball = 43π(R3)43π(R3) = 43π(36)43π(36)

i.e. R3=36R3=36
i.e. R=36‾‾‾√3R=363
i.e. Diameter=236‾‾‾√3
VP
Joined: 14 Feb 2017
Posts: 1386
Location: Australia
Concentration: Technology, Strategy
GMAT 1: 560 Q41 V26
GMAT 2: 550 Q43 V23
GMAT 3: 650 Q47 V33
GMAT 4: 650 Q44 V36
GMAT 5: 650 Q48 V31
GMAT 6: 600 Q38 V35
GMAT 7: 710 Q47 V41
GPA: 3
WE: Management Consulting (Consulting)
Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

28 Nov 2018, 16:14

Easiest method to understand for me was:
1. Solve individual volumes by plugging in radii
3. Step 2 = 4/3 pi*r^3 - solve for r
4. Recall we are asked to find the diameter. Diameter = 2pi*r =2*(step3)*pi
_________________
Here's how I went from 430 to 710, and how you can do it yourself:
SVP
Joined: 23 Feb 2015
Posts: 2005
Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

25 Apr 2020, 00:07
Bunuel wrote:
septwibowo wrote:
GMATinsight wrote:
Bunuel wrote:
For a party, three solid cheese balls with diameters of 2 inches, 4 inches, and 6 inches, respectively, were combined to form a single cheese ball. What was the approximate diameter, in inches, of the new cheese ball? (The volume of a sphere is $$\frac{4}{3}\pi r^3$$, where r is the radius.)

(A) 12

(B) 16

(C) $$\sqrt[3]{16}$$

(D) $$3\sqrt[3]{8}$$

(E) $$2\sqrt[3]{36}$$

Kudos for a correct solution.

diameters of 2 inches, 4 inches, and 6 inches
i.e. Radie of 1 inches, 2 inches, and 3 inches

Sum of their volumes = $$\frac{4}{3}\pi (1^3+2^3+3^3)$$ = $$\frac{4}{3}\pi (36)$$

Volume of New Ball = $$\frac{4}{3}\pi (R^3)$$ = $$\frac{4}{3}\pi (36)$$

i.e. $$R^3 = 36$$
i.e. $$R = \sqrt[3]{36}$$
i.e. $$Diameter = 2\sqrt[3]{36}$$

Simple question : why we calculate the sum of their volumes? In my opinion, total volume of NEW cheese ball must be greater than total combined volume from three small cheese balls.

This does not make sense.

How can you make a ball larger in volume than the combined volume of three balls you are making it from?

Bunuel
Then what's the range of diameter of the single cheese ball could be, actually?

Also, WHY did the author use the word ''approximate'' in the question prompt though the correct answer (E) says that there is no ''approximation'' in the correct choice? It seems that the correct choice is the EXACT figure of the 'diameter'.
Thanks__
Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in   [#permalink] 25 Apr 2020, 00:07

Go to page    1   2    Next  [ 21 posts ]