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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...

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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...

by BTGmoderatorLU » Wed Feb 19, 2020 4:42 pm
Source: Princeton Review

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 \(V=\frac{4}{3}\cdot πr^3. \left. \right)\)

A. 2/5
B. 16/25
C. 27/125
D. 98/125
E. 3/5

The OA is D
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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 Jay@ManhattanReview » Thu Feb 20, 2020 10:38 pm
BTGmoderatorLU wrote: ↑
Wed Feb 19, 2020 4:42 pm
 Source: Princeton Review

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 \(V=\frac{4}{3}\cdot πr^3. \left. \right)\)

A. 2/5
B. 16/25
C. 27/125
D. 98/125
E. 3/5

The OA is D
Volume of the solid sphere (unhollowed sphere) = \(V=\frac{4}{3}\cdot πr^3=\frac{4}{3}\cdot π\cdot5^3\)

Volume of the part removed from the sphere = \(V=\frac{4}{3}\cdot πr^3=\frac{4}{3}\cdot π\cdot3^3\)

Volume of the hollowed sphere (part remained) =\(\frac{4}{3}\cdot π\cdot5^3 - \frac{4}{3}\cdot π\cdot3^3=\frac{4}{3}\cdot π(5^3-3^2)\)

Fractional part of the original sphere remained = \(\frac{\frac{4}{3}\cdot π(5^3-3^2)}{\frac{4}{3}\cdot π\cdot5^3}=\frac{98}{125}\)

The correct answer: D

Hope this helps!

-Jay
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