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29 Aug 2019, 12:22
Kudos
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A
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Difficulty:
45%
(medium)
Question Stats:
100% (01:44) correct
0%
(00:00)
wrong
based on 7
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Find the percentage change in the volume of a cone, if the radius of its base is decreased by 20% and its height is increased by 50%.
a) 4% decrease
b) 6% decrease
c) 4% increase
d) 6% increase
a) 4% decrease
b) 6% decrease
c) 4% increase
d) 6% increase
06 Sep 2019, 10:47
Kudos
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Let new volume- V2; old volume- V1
New radius- R2; old radius- R1
New height- H2; old height- H1
Now given, R2=0.8*R1; H2=1.5*H1
V1= (pi*R1^2*H1)/3
V2= (pi*R2^2*H2)/3
=(pi*0.64*R1^2*1.5*H1)/3
=0.96*((pi*R1^2*H1)/3)
=0.96*V1
So, change in volume =(V2-V1)*100/V1%
=96%
=4% decrease (A)
Posted from my mobile device
New radius- R2; old radius- R1
New height- H2; old height- H1
Now given, R2=0.8*R1; H2=1.5*H1
V1= (pi*R1^2*H1)/3
V2= (pi*R2^2*H2)/3
=(pi*0.64*R1^2*1.5*H1)/3
=0.96*((pi*R1^2*H1)/3)
=0.96*V1
So, change in volume =(V2-V1)*100/V1%
=96%
=4% decrease (A)
Posted from my mobile device
