# Question #94789

Dec 14, 2016

y varies directly as x and inversely as the square of z.

So $y \propto \frac{x}{z} ^ 2$

$\implies y = k \frac{x}{z} ^ 2 , \text{ where k is variation constant}$

Given $y = 4 \text{ when } x = 50 \mathmr{and} z = 5$

So $4 = k \cdot \frac{50}{5} ^ 2 \implies k = 2$

Hence $y = 2 \cdot \frac{x}{z} ^ 2$

Now when $x = 56 \mathmr{and} z = 2$, y becomes
$y = 2 \cdot \frac{56}{2} ^ 2 = 28$