Two right circular cones X and Y are made such that X has a radius three times the radius of Y and Y has a volume half the volume of X. What is the ratio of heights of X and Y?

2 : 9

$R_x = 3R_y$ ($R_x$ → Radius of X, $R_y$ → Radius of Y)

$V_y = \frac{1}{2}V_x$ ($Vy$→ Volume of Y, $V_x$ → Volume of X)

$\frac{1}{3}πR_y^2H_y=\frac{1}{2}\frac{1}{3}πR_x^2H_x$

so $H_y:H_x=9:2$

$H_x:H_y=2:9$

$H_x, H_y$ height of X, Y respectively