Answer: $\frac{1}{5}$

Set up the proportionalities.

• $y \propto x^2 \Rightarrow y = k x^2$
• $x \propto \frac{1}{z} \Rightarrow x = \frac{c}{z}$, i.e. $xz = c$

Find the constants using $z = \tfrac{7}{25},\ x = 5,\ y = 50$.

• From $y = kx^2$: $50 = k \cdot 25 \Rightarrow k = 2$
• From $xz = c$: $c = 5 \cdot \tfrac{7}{25} = \tfrac{7}{5}$

So $y = 2x^2$ and $xz = \tfrac{7}{5}$.

Use $y = 98$ to find $x$.

$2x^2 = 98 \Rightarrow x^2 = 49 \Rightarrow x = 7$

Find $z$.

$z = \frac{c}{x} = \frac{7/5}{7} = \frac{1}{5}$

Correct option: $\frac{1}{5}$