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

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Difficulty:   45% (medium)

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

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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) $$\sqrt{16}$$

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

(E) $$2\sqrt{36}$$

Kudos for a correct solution.

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

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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) $$\sqrt{16}$$

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

(E) $$2\sqrt{36}$$

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. $$R = \sqrt{36}$$
i.e. $$Diameter = 2\sqrt{36}$$

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

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

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

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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) $$\sqrt{16}$$

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

(E) $$2\sqrt{36}$$

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 = $$2\sqrt{36}$$, where $$r^3 = 36$$

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

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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) $$\sqrt{16}$$

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

(E) $$2\sqrt{36}$$

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

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

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

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

=> $$d= 2\sqrt{36}$$

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

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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) $$\sqrt{16}$$

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

(E) $$2\sqrt{36}$$

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

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

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

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

=> $$d= 2\sqrt{36}$$

Kudos, if you like the explanation

How do you get $$\frac{4}{3}\pi r^3 = \frac{1}{6}\pi d^3$$ ?
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For a party, three solid cheese balls with diameters of 2 inches, 4 in  [#permalink]

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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.

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

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

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

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

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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) $$\sqrt{16}$$

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

(E) $$2\sqrt{36}$$

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

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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]

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1
( 1^3 + 2^3 + 3^3) = r ^3

r = cube root (36)

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

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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) $$\sqrt{16}$$

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

(E) $$2\sqrt{36}$$

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

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

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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) $$\sqrt{16}$$

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

(E) $$2\sqrt{36}$$

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. $$R = \sqrt{36}$$
i.e. $$Diameter = 2\sqrt{36}$$

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

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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) $$\sqrt{16}$$

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

(E) $$2\sqrt{36}$$

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. $$R = \sqrt{36}$$
i.e. $$Diameter = 2\sqrt{36}$$

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

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

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

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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) $$\sqrt{16}$$

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

(E) $$2\sqrt{36}$$

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. $$R = \sqrt{36}$$
i.e. $$Diameter = 2\sqrt{36}$$

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

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