A right circular cone and a right circular cylinder hold exactly the same amount, and the cone's base radius is twice the cylinder's. If the cylinder stands 16 cm tall, how tall is the cone, in cm?

Let the cylinder's radius be $r$, so the cone's is $2r$. Cylinder = π$r^{2}x$ 16 = 16π$r^2$. Cone = (1/3)π$(2r{)}^{2}H=(4/3)$π$r^2$$H$. Setting them equal gives $(4/3)H=16$, so $H=12$ cm. The trap $48=3\times 16$ applies the cone's 1/3 but ignores that its base is four times as wide; 4 = 16/4 allows for the wider base but forgets the 1/3; 8 simply halves 16 because the radius doubled, mixing up a linear change with a squared one.