The radius of a right circular cylinder is halved while its height is doubled. What happens to its volume?

Volume = π$r^2$h, and the radius enters SQUARED while the height enters only to the first power. Halving $r$ multiplies the volume by $(1/2{)}^2=1/4$ and doubling h multiplies it by 2, so overall the volume becomes $1/4\times 2=1/2$ of the original. 'Unchanged' is the trap for treating $r$ and h symmetrically; one-fourth applies the radius change and forgets the height.