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

 It is currently 24 May 2019, 10:11

### 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: 55271
For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

20 Oct 2015, 02:57
8
59
00:00

Difficulty:

45% (medium)

Question Stats:

74% (02:36) correct 26% (02:51) wrong based on 1859 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.

_________________
CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2933
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

20 Oct 2015, 05:27
8
14
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. $$Diameter = 2\sqrt[3]{36}$$

_________________
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
##### General Discussion
Intern
Joined: 29 Aug 2015
Posts: 12
For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

Updated on: 21 Oct 2015, 09:22
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, 08:06.
Last edited by HarrishGowtham on 21 Oct 2015, 09:22, edited 1 time in total.
CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2933
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

20 Oct 2015, 09:34
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.
_________________
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
Retired Moderator
Joined: 29 Apr 2015
Posts: 837
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, 09:38
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 $$r^3 = 36$$

_________________
Saving was yesterday, heat up the gmatclub.forum's sentiment by spending KUDOS!

PS Please send me PM if I do not respond to your question within 24 hours.
Manager
Joined: 01 Mar 2015
Posts: 74
For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

21 Oct 2015, 05:09
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: 75
For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

28 Oct 2015, 06:31
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: 7372
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, 02:51
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 \$149 for 3 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: 2291
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, 05:47
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
+1 Kudos if you find this post helpful
Intern
Joined: 06 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, 02:15
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: 32
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

05 May 2016, 00:24
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: 2823
Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

05 May 2016, 04:52
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

Jeff@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

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

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Manager
Joined: 21 Sep 2015
Posts: 75
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, 09:11
1
( 1^3 + 2^3 + 3^3) = r ^3

r = cube root (36)

d= 2r
_________________
Appreciate any KUDOS given !
Director
Joined: 23 Feb 2015
Posts: 927
GMAT 1: 720 Q49 V40
Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

17 Nov 2016, 09:57
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.
_________________
“The heights by great men reached and kept were not attained in sudden flight but, they while their companions slept, they were toiling upwards in the night.”

Do you need official questions for Quant?
3700 Unique Official GMAT Quant Questions
------
SEARCH FOR ALL TAGS
GMAT Club Tests
Manager
Joined: 03 Oct 2013
Posts: 84
Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

17 Nov 2016, 10:17
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.
_________________
P.S. Don't forget to give Kudos on the left if you like the solution
Manager
Joined: 27 Dec 2016
Posts: 232
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, 16:45
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: 55271
Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

### Show Tags

18 Jun 2017, 21:40
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: 190
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, 10:53
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
Senior Manager
Joined: 14 Feb 2017
Posts: 277
Location: Australia
Concentration: Technology, Strategy
GMAT 1: 560 Q41 V26
GMAT 2: 550 Q43 V23
GMAT 3: 650 Q47 V33
GPA: 2.61
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, 17:14

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
Re: For a party, three solid cheese balls with diameters of 2 inches, 4 in   [#permalink] 28 Nov 2018, 17:14
Display posts from previous: Sort by