Is the answer 1:4? Let me know if this is correct, so that I can post the solution.pepeprepa wrote:PS with a cone
Cone Geometry
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Source: Beat The GMAT — Problem Solving |
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parallel_chase
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sudhir3127
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sudhir3127
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this is how i approached the problem...
let r be the radius of the base of the orginal cone and h be the height...
thus the smaller cone will have a radius of r/2 and height h/2.
volume of the cone = 1/3 pi*r^2h
1/3*pi (r/2)^2*h/2= pi*r^2*h/24...................................1
the remaining part will
1/3pi*r^2h- pi*r^2*h/24
= 7*pi*r^2*h/24.....................................................2
ratio of 1 by 2
(pi*r^2*h/24)/(7*pi*r^2*h/24)
which is 1:7.
please let me know if u have any doubts..
let r be the radius of the base of the orginal cone and h be the height...
thus the smaller cone will have a radius of r/2 and height h/2.
volume of the cone = 1/3 pi*r^2h
1/3*pi (r/2)^2*h/2= pi*r^2*h/24...................................1
the remaining part will
1/3pi*r^2h- pi*r^2*h/24
= 7*pi*r^2*h/24.....................................................2
ratio of 1 by 2
(pi*r^2*h/24)/(7*pi*r^2*h/24)
which is 1:7.
please let me know if u have any doubts..
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pepeprepa
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The thing that annoyed me in this question is something which is obvious for Sudhir I think; the link between x and R (names on the original graph).
Indeed, you can see that the two triangles in the cone are similar so the ratio is the same between AB:AC and x:R that is 1:2
Indeed, you can see that the two triangles in the cone are similar so the ratio is the same between AB:AC and x:R that is 1:2
