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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VJesus12: Legendary Member; Posts: 2276; Joined: Sat Oct 14, 2017 6:10 amFollowed by: 3 members

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]

Source: Princeton Review

Source: Beat The GMAT — Problem Solving | Thu Jul 30, 2020 8:51 am

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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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