Problem Solving

boomgoesthegmat 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 (4/3)(pi)r^3, where r is the radius.)

A) 12
B) 16
C) cuberoot16
D) 3(cuberoot8)
E) 2(cuberoot36)

OA: E

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)Ï€(1)^3 = (4/3)Ï€

Volume for 4-inch diameter cheese ball

(4/3)Ï€(2)^3 = (4/3)Ï€(8) = (32/3)Ï€

Volume for 6-inch diameter cheese ball

(4/3)Ï€(3)^3 = (4/3)Ï€(27) = (108/3)Ï€

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

(4/3)Ï€ + (32/3)Ï€ + (108/3)Ï€ = (144/3)Ï€ = 48Ï€

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

48Ï€ = 4/3Ï€(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

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 will likely just be plugging numbers into that formula 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