1. **Problem b)** Given the inequality $|c - 10| + 12 > 18$, find the interval of $c$. 2. Start by isolating the absolute value term: $$|c - 10| + 12 > 18$$ $$|c - 10| > 18 - 12$$ $$|c - 10| > 6$$ 3. Recall that $|x| > a$ means $x > a$ or $x < -a$. So, $$c - 10 > 6 \quad \text{or} \quad c - 10 < -6$$ 4. Solve each inequality: $$c > 16 \quad \text{or} \quad c < 4$$ 5. Therefore, the solution interval is: $$(-\infty, 4) \cup (16, \infty)$$ --- 6. **Problem 6a)** Write the first 4 terms of the sequence with nth term: $$a_n = \frac{(-1)^{n-1}}{\sqrt{2n}}$$ 7. Calculate each term: - $a_1 = \frac{(-1)^{0}}{\sqrt{2 \times 1}} = \frac{1}{\sqrt{2}}$ - $a_2 = \frac{(-1)^{1}}{\sqrt{4}} = \frac{-1}{2}$ - $a_3 = \frac{(-1)^{2}}{\sqrt{6}} = \frac{1}{\sqrt{6}}$ - $a_4 = \frac{(-1)^{3}}{\sqrt{8}} = \frac{-1}{2\sqrt{2}}$ 8. So, the first 4 terms are: $$\frac{1}{\sqrt{2}}, -\frac{1}{2}, \frac{1}{\sqrt{6}}, -\frac{1}{2\sqrt{2}}$$ --- 9. **Problem 6b)** Prize money distribution: - Total prize money = 10500 - Number of teams = 15 - First place prize = 1000 - Prize decreases by fixed amount $d$ each place 10. The prize money forms an arithmetic sequence: $$a_1 = 1000, \quad n = 15, \quad S_n = 10500$$ 11. Sum of arithmetic sequence: $$S_n = \frac{n}{2} (2a_1 + (n-1)d)$$ 12. Substitute known values: $$10500 = \frac{15}{2} (2 \times 1000 + 14d)$$ $$10500 = 7.5 (2000 + 14d)$$ $$\frac{10500}{7.5} = 2000 + 14d$$ $$1400 = 2000 + 14d$$ 13. Solve for $d$: $$14d = 1400 - 2000 = -600$$ $$d = -\frac{600}{14} = -\frac{300}{7} \approx -42.86$$ 14. The fixed decrease amount is approximately 42.86. 15. Find prize for 15th place: $$a_{15} = a_1 + (15-1)d = 1000 + 14 \times (-42.86) = 1000 - 600 = 400$$ 16. Find prize for 5th and 10th place: $$a_5 = 1000 + 4 \times (-42.86) = 1000 - 171.44 = 828.56$$ $$a_{10} = 1000 + 9 \times (-42.86) = 1000 - 385.74 = 614.26$$ --- 17. **Problem 7a)** House loan problem: - House cost = 345000 - Down payment = 10% of 345000 = 34500 - Loan amount = 345000 - 34500 = 310500 - Interest rate = 5% per year compounded monthly - Loan term = 15 years - Number of payments $n = 15 \times 12 = 180$ - Monthly interest rate $i = \frac{5\%}{12} = 0.0041667$ 18. Monthly payment formula: $$P = \frac{L i (1+i)^n}{(1+i)^n - 1}$$ 19. Substitute values: $$P = \frac{310500 \times 0.0041667 \times (1.0041667)^{180}}{(1.0041667)^{180} - 1}$$ 20. Calculate: $$(1.0041667)^{180} \approx 2.11383$$ $$P = \frac{310500 \times 0.0041667 \times 2.11383}{2.11383 - 1} = \frac{310500 \times 0.008807}{1.11383} \approx \frac{2734.5}{1.11383} = 2455.5$$ 21. Monthly payment is approximately 2455.5. 22. **Problem 7a ii)** After missing first 17 payments, find amount to settle on 18th payment. 23. Calculate remaining loan balance after 17 payments: $$B = L (1+i)^{17} - P \frac{(1+i)^{17} - 1}{i}$$ 24. Calculate: $$(1.0041667)^{17} \approx 1.0729$$ 25. Substitute: $$B = 310500 \times 1.0729 - 2455.5 \times \frac{1.0729 - 1}{0.0041667}$$ $$B = 333000 - 2455.5 \times 17.5 = 333000 - 42971 = 290029$$ 26. The 18th payment to settle the loan is approximately 290029. --- 27. **Problem 7b i)** Govind saves 300 every 3 months, interest 7.5% compounded quarterly for 9 years. 28. Number of quarters: $$n = 9 \times 4 = 36$$ 29. Interest rate per quarter: $$i = \frac{7.5\%}{4} = 0.01875$$ 30. Use future value of annuity formula: $$A = P \frac{(1+i)^n - 1}{i}$$ 31. Substitute: $$A = 300 \times \frac{(1.01875)^{36} - 1}{0.01875}$$ 32. Calculate: $$(1.01875)^{36} \approx 1.938$$ 33. So, $$A = 300 \times \frac{1.938 - 1}{0.01875} = 300 \times 49.97 = 14991$$ 34. Accumulated amount is approximately 14991. 35. **Problem 7b ii)** Find equivalent effective annual rate: 36. Effective annual rate formula: $$r = (1 + i)^m - 1$$ where $i=0.01875$ quarterly rate, $m=4$ quarters per year. 37. Calculate: $$r = (1.01875)^4 - 1 = 1.0776 - 1 = 0.0776$$ 38. Equivalent effective annual rate is approximately 7.76%.