volume and algebra problems

• Jan 26th 2008, 11:02 AM
sarahh
volume and algebra problems
The radius of a right circular cone is doubled and its height is tripled to form a new right circular cone. What is the ratio of the volume of the original cone to the new one?
Seems to be a 1:12 ratio but I always come up with 1:6??

The variables a, b, and c are all integers and a + b + c = 50. If a < 0, and 0 < b < 30, then the minimum possible value for c is 22. Why?
• Jan 26th 2008, 11:12 AM
janvdl
Quote:

Originally Posted by sarahh
The variables a, b, and c are all integers and a + b + c = 50. If a < 0, and 0 < b < 30, then the minimum possible value for c is 22. Why?

The maximum value for a is -1 and for b is 29. When you add them you get 28, and you need another 22 to make 50. :)

(I'll look up the formula for a cone just now, i forgot it again :( )
• Jan 26th 2008, 11:16 AM
janvdl
Quote:

Originally Posted by sarahh
The radius of a right circular cone is doubled and its height is tripled to form a new right circular cone. What is the ratio of the volume of the original cone to the new one?
Seems to be a 1:12 ratio but I always come up with 1:6??

Right the first cone's volume is $\frac{1}{3} \pi r^2 h$

For the second cone we have: $\frac{1}{3} \pi (2r)^2 (3h)$

$\frac{1}{3} \pi r^2 h \ \ : \ \ \frac{1}{3} \pi (2r)^2 (3h)$

$r^2 h \ \ : \ \ 4r^2 (3h)$

$1 : 12$ (Whew) :D
• Jan 26th 2008, 11:19 AM
sarahh
Thanks for your help jandvl--those were driving me crazy!
• Jan 26th 2008, 11:20 AM
janvdl
Quote:

Originally Posted by sarahh
Thanks for your help jandvl--those were driving me crazy!

You're welcome :)