1. **Problem 8:** Given that $R$ varies directly as the square of $S$ and inversely as $T$, with $R=21$ when $S=6$ and $T=3$. Find $T$ when $R=18$ and $S=4$. 2. **Formula:** Since $R$ varies directly as $S^2$ and inversely as $T$, we write: $$R = k \frac{S^2}{T}$$ where $k$ is the constant of proportionality. 3. **Find $k$ using given values:** $$21 = k \frac{6^2}{3} = k \frac{36}{3} = 12k$$ So, $$k = \frac{21}{12} = \frac{7}{4}$$ 4. **Find $T$ when $R=18$ and $S=4$:** $$18 = \frac{7}{4} \frac{4^2}{T} = \frac{7}{4} \frac{16}{T} = \frac{28}{T}$$ Multiply both sides by $T$: $$18T = 28$$ Divide both sides by 18: $$T = \frac{28}{18} = \frac{14}{9}$$ --- 5. **Problem 9:** $m$ varies inversely as the cube root of $n$. Given $m=7$ when $n=27$. Find the relation between $m$ and $n$. 6. **Formula:** $$m = \frac{k}{\sqrt[3]{n}}$$ 7. **Find $k$:** $$7 = \frac{k}{\sqrt[3]{27}} = \frac{k}{3}$$ So, $$k = 7 \times 3 = 21$$ 8. **Relation:** $$m = \frac{21}{\sqrt[3]{n}}$$ --- 9. **Problem 10:** $p$ varies directly as $q$ and inversely as the square of $r$. Given $p=10$ when $q=5$ and $r=2$, and $p=15$ when $q=m$ and $r=4$. Find $m$. 10. **Formula:** $$p = k \frac{q}{r^2}$$ 11. **Find $k$:** $$10 = k \frac{5}{2^2} = k \frac{5}{4}$$ So, $$k = \frac{10 \times 4}{5} = 8$$ 12. **Find $m$ when $p=15$ and $r=4$:** $$15 = 8 \frac{m}{4^2} = 8 \frac{m}{16} = \frac{8m}{16} = \frac{m}{2}$$ Multiply both sides by 2: $$m = 30$$ --- 13. **Problem 11:** Distance $J$ varies directly as the square of time $t$. Given $J=84$ when $t=6$. Find $J$ when $t=9$. 14. **Formula:** $$J = k t^2$$ 15. **Find $k$:** $$84 = k \times 6^2 = 36k$$ So, $$k = \frac{84}{36} = \frac{7}{3}$$ 16. **Find $J$ when $t=9$:** $$J = \frac{7}{3} \times 9^2 = \frac{7}{3} \times 81 = 7 \times 27 = 189$$ --- 17. **Problem 12:** Resistance $R$ varies inversely as the square of radius $j$. Given $R=0.6$ when $j=0.05$. Find $j$ when $R=0.15$. 18. **Formula:** $$R = \frac{k}{j^2}$$ 19. **Find $k$:** $$0.6 = \frac{k}{(0.05)^2} = \frac{k}{0.0025}$$ So, $$k = 0.6 \times 0.0025 = 0.0015$$ 20. **Find $j$ when $R=0.15$:** $$0.15 = \frac{0.0015}{j^2}$$ Multiply both sides by $j^2$: $$0.15 j^2 = 0.0015$$ Divide both sides by 0.15: $$j^2 = \frac{0.0015}{0.15} = 0.01$$ Take square root: $$j = \sqrt{0.01} = 0.1$$ --- 21. **Problem 13:** Water level $h$ varies inversely as time $t$. Given $h=1260$ cm at $t=40$ minutes. Find $h$ at $t=80$ minutes (1 1/3 hours = 80 minutes). 22. **Formula:** $$h = \frac{k}{t}$$ 23. **Find $k$:** $$1260 = \frac{k}{40}$$ So, $$k = 1260 \times 40 = 50400$$ 24. **Find $h$ at $t=80$:** $$h = \frac{50400}{80} = 630$$ --- 25. **Problem 14:** Height $H$ varies directly as volume $V$ and inversely as square of radius $r$. Given $H=14$ cm when $V=2156$ cm³ and $r=7$ cm. Find $r$ when $H=35$ cm and $V=15840$ cm³. 26. **Formula:** $$H = k \frac{V}{r^2}$$ 27. **Find $k$:** $$14 = k \frac{2156}{7^2} = k \frac{2156}{49}$$ So, $$k = \frac{14 \times 49}{2156} = \frac{686}{2156} = \frac{7}{22}$$ 28. **Find $r$ when $H=35$ and $V=15840$:** $$35 = \frac{7}{22} \frac{15840}{r^2} = \frac{7 \times 15840}{22 r^2} = \frac{110880}{22 r^2} = \frac{5040}{r^2}$$ Multiply both sides by $r^2$: $$35 r^2 = 5040$$ Divide both sides by 35: $$r^2 = \frac{5040}{35} = 144$$ Take square root: $$r = 12$$ --- 29. **Problem 15:** Time $t$ varies inversely as number of workers $y$. Given $t=2.25$ hours when $y=4$. Find $t$ when $y=6$. 30. **Formula:** $$t = \frac{k}{y}$$ 31. **Find $k$:** $$2.25 = \frac{k}{4}$$ So, $$k = 2.25 \times 4 = 9$$ 32. **Find $t$ when $y=6$:** $$t = \frac{9}{6} = 1.5$$ --- **Final answers:** 8. $\frac{14}{9}$ (B) 9. $m = \frac{21}{\sqrt[3]{n}}$ (D) 10. $m = 30$ (C) 11. $189$ (A) 12. $0.1$ (B) 13. $630$ (D) 14. $12$ (C) 15. $1.5$ hours (C)