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# If the radius of the base of a right circular cone is decreased by 20%

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Math Expert
Joined: 02 Sep 2009
Posts: 65807
If the radius of the base of a right circular cone is decreased by 20%  [#permalink]

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01 Jun 2020, 08:59
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35% (medium)

Question Stats:

84% (01:56) correct 16% (03:17) wrong based on 32 sessions

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If the radius of the base of a right circular cone is decreased by 20% and its height is increased by 50%, what is the percentage change in the volume of the cone? (The volume of a right circular cone = $$\frac{1}{3}πr^2h$$.)

A. 4% decrease
B. 6% decrease
C. No change
D. 4% increase
E. 6% increase

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Re: If the radius of the base of a right circular cone is decreased by 20%  [#permalink]

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01 Jun 2020, 09:15
1
Let Volume of Cone be 50 = 1/3πr^2h
Radius is decreased by 20% so R will be 0.8r
Height is increased by 50% so h will be 1.5h
New volume of cone is 1/3π(0.8r)^2*1.5h
= 1/3π(0.96)r^2h => 0.96*50 = 48
The change in volume =
(New-old/old)*100
(48-50/50)*100
(-2/50)*100
-4
Therefore the volume of cone decreased by 4%

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Re: If the radius of the base of a right circular cone is decreased by 20%  [#permalink]

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01 Jun 2020, 09:21
Bunuel wrote:
If the radius of the base of a right circular cone is decreased by 20% and its height is increased by 50%, what is the percentage change in the volume of the cone? (The volume of a right circular cone = $$\frac{1}{3}πr^2h$$.)

A. 4% decrease
B. 6% decrease
C. No change
D. 4% increase
E. 6% increase

V(new)=V*(8/10)^2 *3/2
Vn=(V*48)/50
(V-Vn)/V= 1-48/50
% change= 4%(Decrease)
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If the radius of the base of a right circular cone is decreased by 20%  [#permalink]

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01 Jun 2020, 12:11
Bunuel wrote:
If the radius of the base of a right circular cone is decreased by 20% and its height is increased by 50%, what is the percentage change in the volume of the cone? (The volume of a right circular cone = $$\frac{1}{3}πr^2h$$.)

A. 4% decrease
B. 6% decrease
C. No change
D. 4% increase
E. 6% increase

Solution

• Let us assume that r and h are radius of the base and height of the original right circular cone, respectively.
o So, the volume of the original right circular cylinder $$= \frac{1}{3}*\pi*r^2*h$$
• Radius of the base is decreased by 20% and height is increased by 50%.
o The radius of the base of the new cone $$= (1 – \frac{20}{100})*r = \frac{4}{5}*r$$
o The height of the new cone $$= (1 + \frac{50}{100})*h = \frac{3}{2}*h$$
o So, the volume of the new cone $$= \frac{1}{3}*\pi*(\frac{4}{5}*r)^2*\frac{3}{2}*h$$
• change in the volume of the cone = The volume of new cone – the volume of original cone = $$\frac{1}{3}*\pi*r^2*h ({\frac{16}{25}*\frac{3}{2} -1)$$
• Thus, required % change $$= \frac{change \space in\space the \space volume}{ the \space original \space volume \space of \space the \space cone}*100 = \frac{\frac{1}{3}*\pi*r^2*h (\frac{16}{25} *\frac{3}{2} -1)}{\frac{1}{3}*\pi*r^2*h}*100 = \frac{-1}{25}*100 = - 4$$%
Thus, the correct answer is Option A.
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If the radius of the base of a right circular cone is decreased by 20%   [#permalink] 01 Jun 2020, 12:11