y varies directly as x and inversely as the square of z. When $x = 5$ and $z = 2$, $y = 10$. What is y when $x = 12$ and $z = 4$?

Explanation

1. 01

   $y = \frac{kx}{z^2}$, so $10 = \frac{5k}{4}$ and $k = 8$.

2. 02

   When $x = 12$ and $z = 4$: $y = \frac{8 \times 12}{16} = 6$.