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555-605 (Medium)|   Geometry|      
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Bunuel
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If the radius of the base of a right circular cone is decreased by 20% 01 Jun 2020, 09:59
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A
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25% (medium)

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82% (01:53) correct 18% (03:00) wrong based on 49 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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A
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If the radius of the base of a right circular cone is decreased by 20% 01 Jun 2020, 10:15
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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%

Answer is A

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If the radius of the base of a right circular cone is decreased by 20% 01 Jun 2020, 10:21
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Bunuel
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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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% 01 Jun 2020, 13:11
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Bunuel
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
Show more

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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