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

Question

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.

(A) 12

(B) 16

(C)  \sqrt {16}

(D)  3\sqrt {8}

(E)  2\sqrt {36}

GMATinsight · Accepted Answer

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}

Answer: Option E

JeffTargetTestPrep · Answer

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

Answer: E

HarrishGowtham · Answer

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

GMATinsight · Answer

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.

MathRevolution · Answer

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 \(\frac{4}{3}\pi r^3\), 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.

(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.

So the answer is (E).

Shubhi09 · Answer

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

Sum of their volumes = =

Volume of New Ball = =

Posted from my mobile device

EMPOWERgmatRichC · Answer

Hi All,

We’re told that 3 solid cheese ball spheres with DIAMETERS of 2 inches, 4 inches and 6 inches are going to be combined into one larger sphere. We’re asked for the approximate DIAMETER of that sphere in inches. While a sphere is a really rare shape on the GMAT (you likely won’t have to deal with it on your Official GMAT), this prompt gives us the formula for Volume of a sphere: (4/3)(pi)(Radius^3), so we'll likely just be plugging numbers into that equation and working through the necessary Arithmetic steps.

Since we’re going to be dealing with a total volume, we need to first figure out the volume of the 3 individual spheres…

1st sphere: diameter = 2, radius = 1, V = (4/3)(pi)(1^3) = (4/3)(pi)

2nd sphere: diameter = 4, radius = 2, V = (4/3)(pi)(2^3) = (32/3)(pi)

3rd sphere: diameter = 6, radius = 3, V = (4/3)(pi)(3^3) = (108/3)(pi)

Total Volume = (4/3 + 32/3 + 108/3)(pi) = (144/3)(pi) = 48pi cubic inches

Using this total, we can now work ‘backwards’ to find the radius of the combined sphere…

V = (4/3)(pi)(R^3)
48pi = (4/3)(pi)(R^3)
48 = (4/3)(R^3)

We can eliminate the (4/3) by multiplying both sides by (3/4).

(3/4)(48) = R^3
144/4 = R^3
36 = R^3

Since 36 has no ‘perfect cubes’ (such as 8 or 64) among its factors, there’s no way to reduce the cube-root of 36, so the RADIUS of this larger sphere is the cube-root-of-36. The question asks us for the DIAMETER of this sphere though, so we have to multiply the radius by 2….

Final Answer:  E

GMAT Assassins aren’t born, they’re made,
Rich

sujoykrdatta · Answer

When you combine the three spheres to form a bigger sphere,  the volume remains unchanged ; i.e. sum of the volumes of the 3 spheres is equal to the volume of the big sphere. The radii of the smaller spheres are 1, 2 and 3 inches. Assuming the radius of the big sphere is r, we have:

Equating the volumes, we have:

4/3 pi * 1^3 + 4/3 pi * 2^3 + 4/3 pi * 3^3 = 4/3 pi * r^3

Dividing throughout by 4/3 pi, we have:

1 + 8 + 27 = r^3

=> r = 36^(1/3)

=> Diameter = 2r = 2 * 36^(1/3) =  2\sqrt {36}

Answer E

HansJK · Answer

One way to solve this question is by logic: the diameter cannot be less than the diameter of one of the smaller cheese balls, hence c and d are out. The diamater cannot be greater than the sum of the combined cheese balls, hence a and b are out. E is the only answer left.

GMATinsight · Answer

Answer: Option E

Video solution by  GMATinsight

https://www.youtube.com/watch?v=Q8ezI60yoz8

100mitra · Answer

Correct Option : E
Since the diameters of the cheese balls are given as 2 inches, 4 inches, and 6 inches, the radii of the cheese balls are 1 inch, 2 inches, and 3 inches, respectively. Using 
V=4/3πr^3) the combined volume of the 3 cheese balls is 
4/3π(13+23+33) or 4/3π(36) cubic inches.
Thus, if R represents the radius of the new cheese ball, then the volume of the new cheese ball is 
4/3πR^3=4/3π(36) and R ^3 = 36, from which it follows that R=3√36 inches. 
Therefore, the diameter of the new cheese ball is 2R=2(3√36) inches

avigutman · Answer

Video solution from Quant Reasoning: Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1 https://www.youtube.com/watch?v=ZeW83uzFqt0

achloes · Answer

JeffTargetTestPrep   EMPOWERgmatRichC   avigutman

Love all three approaches, however I can't understand why simply adding the radii of each of the 3 balls and plugging the result into the volume formula renders the wrong answer.

Appreciate your help!

avigutman · Answer

For the same reason that (a+b)^2 is not the same thing as a^2+b^2,  achloes .

EMPOWERgmatRichC · Answer

Hi achloes, The question asks for the approximate DIAMETER of the NEW sphere, so you should be suspicious of simply adding the three smaller radii (or diameters) together - as that would essentially ignore every other aspect of the question (including the formula for the sphere). How often have you seen a GMAT question ask you "what is 2 + 4 + 6?" From a calculation standpoint, the big factor is the exponent in the formula. For example, if you double a radius of a sphere, the volume actually becomes 8 TIMES greater (not 2 times) - and a calculation that 'adds' to the radius instead of multiplies it can be a bit more complicated (as shown by how you actually go about solving this question; the answer isn't an integer). GMAT assassins aren't born, they're made, Rich Contact Rich at: Rich.C@empowergmat.com

ScottTargetTestPrep · Answer

Suppose we have three spheres of radii a, b, and c. If we combine their radii, we get a + b + c; and if we substitute this in the formula, we get (4/3)π(a + b + c)³. The individual volumes of the spheres are (4/3)πa³, (4/3)πb³, and (4/3)πc³.

Adding the volumes, we obtain (4/3)πa³ + (4/3)πb³ + (4/3)πc³ = (4/3)π(a³ + b³ + c³). So basically, your question is why (4/3)π(a + b + c)³ is not equal to (4/3)π(a³ + b³ + c³), which can be simplified to asking why (a + b + c)³ is not equal to a³ + b³ + c³. It is because we cannot distribute exponents over a sum like the way we can distribute multiplication over addition or distribute exponents over multiplication. So I'll actually reaffirm one of the previous answers to your question: it is precisely because (a + b)² is not necessarily equal to a² + b².

bumpbot · Answer

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

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