1. **State the problem:** Given that $q$ varies inversely as the square of $p$, and the table: | p | 2 | 10 | 18 | | q | 15 | 120 | 240 | Find: (a) The missing value of $q$ when $p=10$. (b) The missing value of $p$ when $q=15$. --- 2. **Formula and explanation:** Since $q$ varies inversely as the square of $p$, we have: $$q = \frac{k}{p^2}$$ where $k$ is a constant. 3. **Find the constant $k$ using known values:** Using $p=2$ and $q=15$: $$15 = \frac{k}{2^2} = \frac{k}{4}$$ Multiply both sides by 4: $$k = 15 \times 4 = 60$$ 4. **(a) Find $q$ when $p=10$:** $$q = \frac{60}{10^2} = \frac{60}{100} = 0.6$$ 5. **(b) Find $p$ when $q=120$:** Start with: $$120 = \frac{60}{p^2}$$ Multiply both sides by $p^2$: $$120 p^2 = 60$$ Divide both sides by 120: $$\cancel{120} p^2 = \frac{60}{\cancel{120}}$$ $$p^2 = \frac{1}{2}$$ Take the square root: $$p = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ --- **Final answers:** (a) $q = 0.6$ (b) $p = \frac{\sqrt{2}}{2}$