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

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

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.

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.

So the answer is (E).

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

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:

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

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) = \(2\sqrt[3]{36}\)

Answer E

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.

Answer: Option E

Video solution by GMATinsight



Get TOPICWISE: Concept Videos | Practice Qns 100+ | Official Qns 50+ | 100% Video solution CLICK.
Two MUST join YouTube channels : GMATinsight (1000+ FREE Videos) and GMATclub

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

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

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!

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

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

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.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.