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The radius of a sphere was increased by p percent
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Updated on: 12 Jul 2020, 07:18
Updated on: 12 Jul 2020, 07:18
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Question Stats:
22% (00:50) correct
77% (00:37) wrong based on 9 sessions
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The radius of a sphere was increased by p percent. If the volume of the sphere was increased by 33.1%, what is the value of p?
(A) 5
(B) 10
(C) 11
(D) 12
(E) 15
My channel URL: https://www.youtube.com/channel/UCvdY0k ... EOsT5PMFRQ
(A) 5
(B) 10
(C) 11
(D) 12
(E) 15
My channel URL: https://www.youtube.com/channel/UCvdY0k ... EOsT5PMFRQ
Re: The radius of a sphere was increased by p percent
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11 Jul 2020, 12:38
11 Jul 2020, 12:38
Expert Reply
The meaning of the question above is the following
The radius of a sphere is increased by 10% . Prove that the volume will be increased by 33.1%
Moreover, the question above is not properly well worded.
Regards
The radius of a sphere is increased by 10% . Prove that the volume will be increased by 33.1%
Moreover, the question above is not properly well worded.
Regards
Re: The radius of a sphere was increased by p percent
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11 Jul 2020, 22:45
11 Jul 2020, 22:45
Quote:
The radius of a sphere was increased by p percent. If the volume of the sphere was increased by 33.1%, what is the value of p?
Step 1: Understanding the question
When the initial radius is r
New radius is (1+p/100)*r
When the initial volume is V, final volume is 1.331V
Step 2: Calculation
Old Volume -> V = 4/3 * pi * r^3 --------(i)
New Volume -> 1.331 V = 4/3 * pi*(1+p/100)^3*r^3 ----(ii)
Dividing eq (ii) by eq(i)
1.331 = (1+p/100)^3
Taking cube root on both the sides
1.1 = 1+p/100
p/100 = 0.1
p=10
(D) is correct
