SOLUTION: If the height of a right circular cone is decreased by 8 percent by what percent must the radius of the base be decreased so that the volume of the cone is decreased by 15 percent?

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Question 1183328: If the height of a right circular cone is decreased by 8 percent by what percent must the radius of the base be decreased so that the volume of the cone is decreased by 15 percent?
Found 2 solutions by josgarithmetic, ankor@dixie-net.com:
Answer by josgarithmetic(39620) About Me  (Show Source):
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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If the height of a right circular cone is decreased by 8 percent by what percent must the radius of the base be decreased so that the volume of the cone is decreased by 15 percent?
:
8% reduction = .92h
15% reduction = .85v
let a = reduced amt multiplier time the radius
1%2F3pi%2A%28ar%29%5E2%2A.92h = .85(1%2F3pi%2Ar%5E2%2Ah)
multiply both sides by 3
pi%2A%28a%5E2r%5E2%29%2A.92h = .85(pi%2Ar%5E2%2Ah)
divide both side by pi*r^2*h
a%5E2%2A.92+=+.85
a%5E2+=+.85%2F.92
a = sqrt%28.924%29
a = .9612 times the radius, therefore
Radius has to be reduced by 3.88%