1. Problem 4a: Cost of 5 books and 10 copies are the same. Cost of 10 books and 20 copies is 3000. Find cost of 10 books. 2. Let cost of 1 book = $b$ and cost of 1 copy = $c$. 3. Given: Cost of 5 books = Cost of 10 copies $$5b = 10c \implies b = 2c$$ 4. Also, cost of 10 books and 20 copies = 3000 $$10b + 20c = 3000$$ 5. Substitute $b = 2c$ into the equation: $$10(2c) + 20c = 3000$$ $$20c + 20c = 3000$$ $$40c = 3000$$ $$c = \frac{3000}{40} = 75$$ 6. Find cost of 10 books: $$10b = 10 \times 2c = 20c = 20 \times 75 = 1500$$ 7. Final answer for 4a: Cost of 10 books = 1500. --- 8. Problem 4b: Cost of 3 pens = cost of 5 copies. Cost of 9 pens and 7 copies = 264. Find cost of 25 copies. 9. Let cost of 1 pen = $p$ and cost of 1 copy = $c$. 10. Given: $$3p = 5c \implies p = \frac{5}{3}c$$ 11. Cost of 9 pens and 7 copies: $$9p + 7c = 264$$ 12. Substitute $p = \frac{5}{3}c$: $$9 \times \frac{5}{3}c + 7c = 264$$ $$15c + 7c = 264$$ $$22c = 264$$ $$c = \frac{264}{22} = 12$$ 13. Cost of 25 copies: $$25c = 25 \times 12 = 300$$ 14. Final answer for 4b: Cost of 25 copies = 300. --- 15. Problem 5a: 500 people have provisions for 60 days. How many people should be added to finish provisions in 40 days? 16. Total provision = people $\times$ days = $500 \times 60 = 30000$ person-days. 17. Let $x$ be the number of people added. 18. New number of people = $500 + x$. 19. Provisions last 40 days: $$(500 + x) \times 40 = 30000$$ 20. Solve for $x$: $$500 + x = \frac{30000}{40} = 750$$ $$x = 750 - 500 = 250$$ 21. Final answer for 5a: 250 people should be added. --- 22. Problem 5b: 5 men finish work in 24 days. How many more men to finish in 15 days? 23. Total work = men $\times$ days = $5 \times 24 = 120$ man-days. 24. Let $y$ be men added. 25. New men = $5 + y$. 26. Work done in 15 days: $$(5 + y) \times 15 = 120$$ 27. Solve for $y$: $$5 + y = \frac{120}{15} = 8$$ $$y = 8 - 5 = 3$$ 28. Final answer for 5b: 3 more men should be added. --- 29. Problem 5c: 800 men have provisions for 120 days. How many men added to finish in 100 days? 30. Total provision = $800 \times 120 = 96000$ person-days. 31. Let $z$ be men added. 32. New men = $800 + z$. 33. Provisions last 100 days: $$(800 + z) \times 100 = 96000$$ 34. Solve for $z$: $$800 + z = \frac{96000}{100} = 960$$ $$z = 960 - 800 = 160$$ 35. Final answer for 5c: 160 men should be added. --- 36. Problem 5d: 200 students have provisions for 75 days. How many students should leave for provisions to last 125 days? 37. Total provision = $200 \times 75 = 15000$ person-days. 38. Let $w$ be students who leave. 39. New students = $200 - w$. 40. Provisions last 125 days: $$(200 - w) \times 125 = 15000$$ 41. Solve for $w$: $$200 - w = \frac{15000}{125} = 120$$ $$w = 200 - 120 = 80$$ 42. Final answer for 5d: 80 students should leave. --- 43. Problem 6a: If $\frac{3}{4}$ of income is 3000, find $\frac{1}{4}$ of income. 44. Let income = $I$. 45. Given: $$\frac{3}{4}I = 3000$$ 46. Find $\frac{1}{4}I$: $$\frac{1}{4}I = \frac{3000}{3} = 1000$$ 47. Final answer for 6a: $\frac{1}{4}$ of income = 1000. --- 48. Problem 6b: $\frac{1}{6}$ of ribbon = 42 cm. Find length of $\frac{1}{7}$ part. 49. Let length of ribbon = $L$. 50. Given: $$\frac{1}{6}L = 42$$ 51. Find $\frac{1}{7}L$: $$L = 42 \times 6 = 252$$ $$\frac{1}{7}L = \frac{252}{7} = 36$$ 52. Final answer for 6b: $\frac{1}{7}$ part = 36 cm. --- 53. Problem 6c: Supreme spends $\frac{1}{10}$ of income which is 10000. Find $\frac{3}{10}$ of income. 54. Let income = $I$. 55. Given: $$\frac{1}{10}I = 10000$$ 56. Find $\frac{3}{10}I$: $$I = 10000 \times 10 = 100000$$ $$\frac{3}{10}I = 3 \times 10000 = 30000$$ 57. Final answer for 6c: $\frac{3}{10}$ of income = 30000.