Mitchell poured the contents of a completely filled cone into an empty cylinder, and the cylinder became two-thirds full. The cylinder had a radius of 5 cm and a height of 15 cm.

What were possible dimensions of the cone? Use 3.14 to approximate pi.

A.

h = 5 cm; r = 15 cm

B.

h = 7.5 cm; r = 10 cm

C.

h = 10 cm; r = 7.5 cm

D.

h = 15 cm; r = 5 cm

First, we define the volume of a cone as V = 1/3x Pi x r²xH and the volume of a cylinder as v = Pi x R² x h. So, according to the question we have V=2/3v = (2/3) xR²xPixh = (2/3) x 3.14x5²x15=785cm^3. V=785cm^3. So it is (1/3) x3.14xr²xh=785, this implies r=10, and h=7.5, because (1/3) x3.14x100x7.5=785. Finally, the dimensions of the cone are r=10, h=7.5 and V=785cm^3