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
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