1. Problem 9: In a right triangle \(\triangle ABC\) with \(\angle B = 90^\circ\), given \(BC = 5\) cm and \(AC - AB = 1\) cm, find the value of \(\frac{1 + \sin C}{1 + \cos C}\). 2. Use Pythagoras theorem: \(AC^2 = AB^2 + BC^2\). 3. Let \(AB = x\), then \(AC = x + 1\). 4. Substitute: \((x + 1)^2 = x^2 + 5^2\) \(\Rightarrow x^2 + 2x + 1 = x^2 + 25\). 5. Simplify: \(2x + 1 = 25 \Rightarrow 2x = 24 \Rightarrow x = 12\). 6. So, \(AB = 12\) cm, \(AC = 13\) cm. 7. \(\sin C = \frac{BC}{AC} = \frac{5}{13}\), \(\cos C = \frac{AB}{AC} = \frac{12}{13}\). 8. Calculate \(\frac{1 + \sin C}{1 + \cos C} = \frac{1 + \frac{5}{13}}{1 + \frac{12}{13}} = \frac{\frac{18}{13}}{\frac{25}{13}} = \frac{18}{25}\). --- 9. Problem 10: LCM of two primes \(p > q\) is 221. Find \(3p - q\). 10. Since \(p, q\) are primes, LCM = product \(p \times q = 221\). 11. Factor 221: \(221 = 13 \times 17\). 12. So, \(p = 17, q = 13\). 13. Calculate \(3p - q = 3 \times 17 - 13 = 51 - 13 = 38\). --- 14. Problem 11: 12 spheres made from melting a cylinder of diameter 16 cm and height 2 cm. Find diameter of each sphere. 15. Volume of cylinder \(= \pi r^2 h = \pi (8)^2 (2) = 128\pi\). 16. Volume of 12 spheres \(= 12 \times \frac{4}{3} \pi r^3 = 16\pi r^3\). 17. Equate volumes: \(128\pi = 16\pi r^3 \Rightarrow r^3 = 8 \Rightarrow r = 2\). 18. Diameter of each sphere \(= 2r = 4\) cm. --- 19. Problem 12: For polynomial \(2x^2 + 5x - 4\), zeroes \(p, q\), find \((1 - p)(1 - q)\). 20. Sum of zeroes \(p + q = -\frac{5}{2}\), product \(pq = -\frac{4}{2} = -2\). 21. Expand \((1 - p)(1 - q) = 1 - (p + q) + pq = 1 - \left(-\frac{5}{2}\right) - 2 = 1 + \frac{5}{2} - 2 = \frac{1}{2} + \frac{5}{2} = \frac{4}{3}\) (recalculate carefully). 22. Correct calculation: \(1 - (p+q) + pq = 1 - (-\frac{5}{2}) + (-2) = 1 + \frac{5}{2} - 2 = 1 + 2.5 - 2 = 1.5 = \frac{3}{2}\). 23. So, value is \(\frac{3}{2}\). --- 24. Problem 13: Area of sector \(= 54\pi\), radius \(= 36\) cm, find arc length. 25. Area of sector \(= \frac{\theta}{360} \pi r^2\). 26. \(54\pi = \frac{\theta}{360} \pi (36)^2 \Rightarrow 54 = \frac{\theta}{360} \times 1296 \Rightarrow \theta = \frac{54 \times 360}{1296} = 15^\circ\). 27. Arc length \(= \frac{\theta}{360} \times 2\pi r = \frac{15}{360} \times 2\pi \times 36 = 3\pi\). --- 28. Problem 14: In \(\triangle ABC\), \(DE \parallel BC\), \(AD = 3\) cm, \(BD = 4\) cm, \(BC = 14\) cm, find \(DE\). 29. By similarity, \(\frac{DE}{BC} = \frac{AD}{AB}\). 30. \(AB = AD + BD = 3 + 4 = 7\) cm. 31. \(DE = BC \times \frac{AD}{AB} = 14 \times \frac{3}{7} = 6\) cm. --- 32. Problem 15: Two dice thrown, probability difference of numbers is 2. 33. Total outcomes = 36. 34. Pairs with difference 2: (1,3),(3,1),(2,4),(4,2),(3,5),(5,3),(4,6),(6,4) = 8. 35. Probability = \(\frac{8}{36} = \frac{2}{9}\). --- 36. Problem 16: Point \(P(x,y)\) with \(x = 2y\), equidistant from \(Q(2,-5)\) and \(R(-3,6)\). 37. Distance \(PQ = PR\). 38. \(\sqrt{(x-2)^2 + (y+5)^2} = \sqrt{(x+3)^2 + (y-6)^2}\). 39. Square both sides and substitute \(x=2y\): \((2y - 2)^2 + (y + 5)^2 = (2y + 3)^2 + (y - 6)^2\). 40. Simplify and solve for \(y\): \(4y^2 - 8y + 4 + y^2 + 10y + 25 = 4y^2 + 12y + 9 + y^2 - 12y + 36\). 41. \(5y^2 + 2y + 29 = 5y^2 + 0y + 45\). 42. \(2y + 29 = 45 \Rightarrow 2y = 16 \Rightarrow y = 8\). 43. \(x = 2y = 16\). 44. Coordinates of \(P = (16, 8)\). --- 45. Problem 17: Difference of mode and median is 24, find difference of median and mean. 46. Using empirical relation: \(\text{Mode} - \text{Median} = 3(\text{Median} - \text{Mean})\). 47. Given \(\text{Mode} - \text{Median} = 24\). 48. So, \(24 = 3(\text{Median} - \text{Mean}) \Rightarrow \text{Median} - \text{Mean} = 8\). --- 49. Problem 18: Maximum common tangents to two circles intersecting at two points is 2. --- 50. Problem 19: Assertion: For two primes \(p, q\), HCF = 1 and LCM = \(p + q\). 51. Reason: For any two natural numbers, HCF \(\times\) LCM = product. 52. Assertion is false because LCM of two primes is product \(p \times q\), not sum. 53. Reason is true. 54. Correct option: (D). --- 55. Problem 20: Assertion: If \(\sin(A-B) = 0\) and \(2\cos(A+B) - 1 = 0\), then \(A = 30^\circ\). 56. Reason: \(1 + \cot^2 \beta = \csc^2 \beta\). 57. Both Assertion and Reason are true but Reason is not explanation of Assertion. 58. Correct option: (B). --- 59. Problem 21(a): Sum of 5th and 9th terms of AP is 72, sum of 7th and 12th terms is 97. Find AP. 60. Let first term \(a\), common difference \(d\). 61. \(T_n = a + (n-1)d\). 62. \(T_5 + T_9 = (a + 4d) + (a + 8d) = 2a + 12d = 72\). 63. \(T_7 + T_{12} = (a + 6d) + (a + 11d) = 2a + 17d = 97\). 64. Subtract equations: \((2a + 17d) - (2a + 12d) = 97 - 72 \Rightarrow 5d = 25 \Rightarrow d = 5\). 65. Substitute \(d=5\) in first: \(2a + 12 \times 5 = 72 \Rightarrow 2a + 60 = 72 \Rightarrow 2a = 12 \Rightarrow a = 6\). 66. AP is \(6, 11, 16, 21, ...\). --- 67. Problem 21(b): Third term is 16, 7th term exceeds 5th term by 12. 68. \(T_3 = a + 2d = 16\). 69. \(T_7 - T_5 = (a + 6d) - (a + 4d) = 2d = 12 \Rightarrow d = 6\). 70. Substitute \(d=6\) in \(a + 2d = 16\): \(a + 12 = 16 \Rightarrow a = 4\). 71. AP is \(4, 10, 16, 22, ...\).