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The radius of a sphere was increased by p percent

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kapil1
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The radius of a sphere was increased by p percent [#permalink] New post   Updated on: 12 Jul 2020, 07:18
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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

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B

Originally posted by kapil1 on 11 Jul 2020, 10:34.
Last edited by kapil1 on 12 Jul 2020, 07:18, edited 2 times in total.
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Re: The radius of a sphere was increased by p percent [#permalink] New post   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.

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Re: The radius of a sphere was increased by p percent [#permalink] New post   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
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kapil1
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Re: The radius of a sphere was increased by p percent [#permalink] New post   11 Jul 2020, 22:46
Carcass wrote:
Moreover, the question above is not properly well worded.

Don't know why, but it seems okay to me.

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Re: The radius of a sphere was increased by p percent [#permalink] New post   12 Jul 2020, 01:34
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If the radius of a sphere was increased by p percent, then the volume of the sphere was increased by 33.1%. What is the value of p??

I hope it makes sense

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Re: The radius of a sphere was increased by p percent [#permalink] New post   12 Jul 2020, 03:56
Let the radius be 3. Volume of sphere is 4/3 *pie *r^3.
When radius is 3, the volume is 108pie.
If we increase the radius by 10%, then news radius is 3.3.
New volume is 143.7.
Thus, 143.7-108 / 108 * 100 = 33.1%
Hence, the answer is B.
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Re: The radius of a sphere was increased by p percent [#permalink] New post   12 Jul 2020, 04:00
kapil1 wrote:
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


If your answer is 10, then B is 10, and not D.
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Re: The radius of a sphere was increased by p percent [#permalink]
12 Jul 2020, 04:00
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