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A sphere with a radius of \(5\) is hollowed out at the center. The part removed from the sphere has the same center and

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by VJesus12 » Thu Jul 30, 2020 8:51 am

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A sphere with a radius of \(5\) is hollowed out at the center. The part removed from the sphere has the same center and a radius of \(3.\) What fractional part of the original sphere remained? (The formula for the volume of a sphere is \(V=\dfrac43 \pi r^3.\))

A. \(\dfrac25\)
B. \(\dfrac{16}{25}\)
C. \(\dfrac{27}{125}\)
D. \(\dfrac{98}{125}\)
E. \(\dfrac35\)

[spoiler]OA=D[/spoiler]

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Re: A sphere with a radius of \(5\) is hollowed out at the center. The part removed from the sphere has the same center

by Scott@TargetTestPrep » Wed Aug 05, 2020 10:13 am
VJesus12 wrote: ↑
Thu Jul 30, 2020 8:51 am
 A sphere with a radius of \(5\) is hollowed out at the center. The part removed from the sphere has the same center and a radius of \(3.\) What fractional part of the original sphere remained? (The formula for the volume of a sphere is \(V=\dfrac43 \pi r^3.\))

A. \(\dfrac25\)
B. \(\dfrac{16}{25}\)
C. \(\dfrac{27}{125}\)
D. \(\dfrac{98}{125}\)
E. \(\dfrac35\)

[spoiler]OA=D[/spoiler]

Solution:

The volume of the entire sphere is 4/3π(5)^3 = 4/3π(125), and the volume of the hollowed-out portion is 4/3π(3)^3 = 4/3π(27). Thus, we have hollowed out 4/3π(27) / 4/3π(125) = 27/125 of the sphere. Thus, the fractional part of the original sphere that remains is 1 - 27/125, or 98/125.

Answer: D

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