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555-605 (Medium)|   Geometry|      
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A sphere with a radius of 5 is hollowed out at the center. The part re 29 Apr 2019, 23:35
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79% (01:54) correct 21% (01:57) wrong based on 151 sessions

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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 \(V = \frac{4}{3}*πr^3\).)

A. 2/5
B. 16/25
C. 27/125
D. 98/125
E. 3/5
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D
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A sphere with a radius of 5 is hollowed out at the center. The part re 30 Apr 2019, 03:09
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Bunuel
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 \(V = \frac{4}{3}*π^3\).)

A. 2/5
B. 16/25
C. 27/125
D. 98/125
E. 3/5
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part left ;
4/3 * pi * (125-27 ) ; 4/3 * pi * 98
so part ratio left 98/125
IMO D
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A sphere with a radius of 5 is hollowed out at the center. The part re 09 Aug 2020, 10:30
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Bunuel
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 \(V = \frac{4}{3}*πr^3\).)

A. 2/5
B. 16/25
C. 27/125
D. 98/125
E. 3/5
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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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A sphere with a radius of 5 is hollowed out at the center. The part re 24 Aug 2020, 21:22
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Volume of the Original Sphere = \(\frac{4}{3}\pi *5^3 \) = \(125 * \frac{4}{3}\pi \)

Since the portion removed from the center is also a sphere, the remaining Volume is given by the equation

Remaining Volume = Volume of original Sphere of Radius 5 - Volume of inner Sphere of Radius 3

= \(\frac{4}{3}\pi *5^3 - \frac{4}{3}\pi * 3^3\)

= \(\frac{4}{3}\pi * [125 - 27]\)

= \(98 * \frac{4}{3}\pi \)

Therefore the Fraction = \(\frac{98 * \frac{4}{3}\pi}{125 * \frac{4}{3}\pi} = \frac{98}{125}\)

Option D


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A sphere with a radius of 5 is hollowed out at the center. The part re 25 Aug 2020, 06:26
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1- 4/3Pi 27 divided by 4/3Pi 125

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A sphere with a radius of 5 is hollowed out at the center. The part re 25 Aug 2020, 21:46
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ScottTargetTestPrep - what is wrong with calculating the initial volume first and then setting that in proportion to what is left from the initial sphere (4/3*Pi*2^3 - as new radius = 2)?
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A sphere with a radius of 5 is hollowed out at the center. The part re 18 Sep 2020, 07:19
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chrtpmdr
ScottTargetTestPrep - what is wrong with calculating the initial volume first and then setting that in proportion to what is left from the initial sphere (4/3*Pi*2^3 - as new radius = 2)?
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The problem in your reasoning is that if you take out a sphere of radius 3 from a sphere of radius 5, what's left will not be a sphere of radius 3; but instead it will be a sphere of radius 5 with a spherical hole in the center.
We can calculate volumes to verify this: the volume of the sphare with radius 5 is (4/3) * π * 5^3 = (4/3) * π * 125 = 500π/3 and the volume of the sphere with radius 3 is (4/3) * π * 3^3 = (4/3) * π * 27 = 108π/3. If we subtract the latter from the former, we see that the volume of the hollowed out shpere is 500π/3 - 108π/3 = 392π/3. On the other hand, the volume of a sphere with radius 2 is (4/3) * π * 2^3 = (4/3) * π * 8 = 32π/3, which is much smaller compared to the volume of the hollowed out sphere.
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A sphere with a radius of 5 is hollowed out at the center. The part re 25 Jan 2023, 05:50
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