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(a) \(\frac{1\frac{1}{4} + \frac{7}{9}}{1\frac{4}{9} - 2\frac{2}{3} \times \frac{9}{64}}\) \(1\frac{1}{4} + \frac{7}{9} = \frac{5}{4} + \frac{7}{9}\) = \(\frac{45 + 28}{36}\) = \(\frac{73}{36}\) \(1\frac{4}{9} - 2\frac{2}{3} \times \frac{9}{64} = \frac{13}{9} - (\frac{8}{3} \times \frac{9}{64})\) = \(\frac{13}{9} - \frac{3}{8}\) = \(\frac{104 - 27}{72}\) = \(\frac{77}{72}\) \(\therefore \frac{1\frac{1}{4} + \frac{7}{9}}{1\frac{4}{9} - 2\frac{2}{3} \times \frac{9}{64}} = \frac{73}{36} \div \frac{77}{72}\) = \(\frac{73}{36} \times \frac{72}{77}\) = \(\frac{146}{77}\) = \(1\frac{69}{77}\) (b) \(\sin x = \frac{2}{3}\) In \(\Delta ABC, 3^{2} = 2^{2} + |BC|^{2}\) \(\therefore |BC|^{2} = 9 - 4 = 5\) \(|BC| = \sqrt{5}\) \(\tan x = \frac{2}{\sqrt{5}} ; \cos x = \frac{\sqrt{5}}{3}\) \(\therefore \tan x - \cos x = \frac{2}{\sqrt{5}} - \frac{\sqrt{5}}{3}\) = \(\frac{6 - 5}{3\sqrt{5}}\) = \(\frac{1}{3\sqrt{5}}\) Rationalizing, we have \(\frac{1}{3\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5}}{15}\)

Let Wale's present age = x. Then Sonny's age = 2x Four years ago, Wale = x - 4 Sonny = 4(x - 4) = 2x - 4 \(4x - 16 = 2x - 4 \implies 4x - 2x = -4 + 16\) \(2x = 12 \implies x = 6\) Sonny = 2x = 2(6) = 12 years In z years' time, the sum of their ages = 66 Wale = 6 + z Sonny = 12 + z \(6 + z + 12 + z = 66 \implies 18 + 2z = 66\) \(2z = 66 - 18 = 48 \implies z = 24\) Therefore, in 24 years time, the sum of their ages will be 66.

(a) < UST = 100^\circ (exterior angle of a cyclic quad) < SUT = 36^\circ (angle in alternate segment) \(\therefore < UST + < SUT + < STU = 180^\circ\) (sum of angles in a triangle) \(\therefore 100^\circ + 36^\circ + < STU = 180^\circ\) \(< STU = 180^\circ - 136^\circ = 44^\circ\) (b) \(\frac{y}{6} = \frac{2}{5}\) \(\therefore 5y = 2(6) = 12\) \(\therefore y = \frac{12}{5} = 2.4 cm\) \(\therefore XQ + QZ = XZ\) \(QZ = XZ - XQ\) = \(6 cm - 2.4 cm = 3.6 cm\) \(\therefore QZ = 3.6 cm\).

(a) (i) No of red discs = \(\frac{1}{4} \times 40 = 10\) Total no of discs = 40 \(\therefore\) No of white balls = 40 - 10 = 30 (ii) Let the number of red balls to be added = z \(\therefore \frac{10 + z}{40 + z} = \frac{1}{3}\) \(3(10 + z) = 40 + z\) \(30 + 3z = 40 + z \implies 3z - z = 40 - 30\) \(2z = 10 \implies z = 5\) Therefore, adding 5 red discs, the probability of picking a red disc = \(\frac{1}{3}\). (b) CP of a plate = N50.00 Let the number of plates be x \(\therefore \text{CP of x plates} = N50x\) \(\text{SP of all plates} = N600\) \(\frac{100 + 20}{100} \times N50x = N600 \implies N60x = N600\) \(x = \frac{600}{60} = 10 plates\)

(a) Let the angle subtended at the centre by XY = \(\theta\) Length of chord = \(2r \sin \frac{\theta}{2}\) \(\implies 6 cm = 2 \times 5 \times \frac{\theta}{2}\) \(\sin \frac{\theta}{2} = \frac{6}{10} = 0.6\) \(\frac{\theta}{2} = \sin^{-1} (0.6) = 36.87^\circ\) \(\theta = 2 \times 36.87^\circ = 73.74^\circ\) (b) Area of sector = \(\frac{\theta}{360^\circ} \times \pi r^{2}\) = \(\frac{73.74}{360} \times \frac{22}{7} \times 5^{2}\) = \(16.094 cm^{2}\) Area of triangle = \(\sqrt{s(s - a)(s - b)(s - c)}\) where \(s = \frac{a + b + c}{2}\) \( a = 5 cm ; b = 5 cm ; c = 6 cm \therefore s = \frac{5 + 5 + 6}{2} = \frac{16}{2} = 8\) \(\therefore Area = \sqrt{8(8 - 5)(8 - 5)(8 - 6)}\) = \(\sqrt{8(3)(3)(2)}\) = \(\sqrt{144}\) = \(12 cm^{2}\) Area of shaded portion = \(16.094 cm^{2} - 12 cm^{2} = 4.094 cm^{2}\)

(a) Amount shared = D(M - 15) Sister's share = \(\frac{D(M - 15)}{2}\) \(\implies \frac{D(M - 15)}{2} = \frac{1}{3}M\) \(2M = 3(M - 15) \implies 3M - 2M = 45\) \(M = 45.00\) (b) (i) (ii) (1) Total no interviewed = \(\frac{2}{5}x + \frac{4}{15}x + \frac{1}{6}x + 20 = x\) \(\therefore 12x + 8x + 5x + 600 = 30x\) (Multiplying through with their LCM which is 30) \(600 = 30x - 12x - 8x - 5x = 5x \implies 600 = 5x\) \(x = \frac{600}{5} = 120\) (2) Those that travelled by air only = \((\frac{13}{30} \times 120) - (\frac{4}{15} \times 120)\) = \(52 - 32 = 20\)

(a) \(\tan 50^\circ = \frac{|TF|}{20} \implies |TF| = 20 \tan 50\) = \(20 \times 1.1918\) = \(23.836 m\) \(\tan 30 = \frac{|PF|}{20} \implies |PF| = 20 \tan 30\) = \(20 \times 0.5774\) = \(11.548 m\) \(\therefore |PT| = 23.836 m - 11.548m\) = \(12.288 m \approxeq 12.3 m\) (b) \(|TH| = \frac{20}{\cos 50}\) = \(\frac{20}{0.6428}\) = \(31.11 m\) (c) \(|HF| = \frac{20 \tan 50^\circ}{\tan 40^\circ}\) = \(\frac{20 \times 1.1918}{0.8391}\) = \(28.41 m\)

(b) Two possible ways (c) The distances of the locations from town X = 27.6 km and 12.4 km. (d) 12.4 km

Class Interval Freq (f) Mid-point (x) \(fx\) \(x^{2}\) \(fx^{2}\) 60 - 64 2 62 124 3844 7688 65 - 69 3 67 201 4489 13487 70 - 74 6 72 432 5184 31104 75 - 79 11 77 847 5929 65219 80 - 84 8 82 656 6724 53792 85 - 89 7 87 609 7569 52983 90 - 94 2 92 184 8464 16928 95 - 99 1 97 97 9409 9409 \(\sum\) 40 3150 250590 (a) Mean \(\bar{x} = \frac{\sum fx}{\sum f}\) = \(\frac{3150}{40}\) = 78.75 (b) \(SD = \sqrt{\frac{\sum fx^{2}}{\sum f} - (\frac{\sum fx}{\sum f})^{2}}\) = \(\sqrt{\frac{250590}{40} - (\frac{3150}{40})^{2}}\) = \(\sqrt{6264.75 - (78.75)^{2}}\) = \(\sqrt{6264.75 - 6201.5625}\) = \(\sqrt{63.1875}\) = 7.95.

(a) \(3(3n - 1) + 5(2n + 1) = 135\) \(9n - 3 + 10n + 5 = 135\) \(19n + 2 = 135\) \(19n = 135 - 2 = 133\) \(n = \frac{133}{19} = 7\) (b) Radius of a cylinder = 14 cm side of a cube = 22 cm \(\therefore \frac{22}{7} \times 14 \times 14 \times h = 22 \times 22 \times 22\) \(44 \times 14 \times h = 22 \times 22 \times 22\) \(h = \frac{22 \times 22 \times 22}{44 \times 14}\) = \(\frac{121}{7} cm\) TSA of a cylinder = \(2\pi r (r + h) = 2 \times \frac{22}{7} \times 14 (14 + \frac{121}{7})\) = \(88(\frac{98 + 121}{7}) = \frac{88 \times 219}{7}\) = \(\frac{19272}{7} cm^{2}\) TSA of a cube = \(6s^{2}\) = \(6 \times 22 \times 22\) = \(2904 cm^{2}\) Ratio = \(\frac{19272}{7} : 2904\) = \(73 : 77\)

(a) x 0^\circ 30^\circ 60^\circ 90^\circ 120^\circ 150^\circ 180^\circ 210^\circ 240^\circ 270^\circ 300^\circ y -3.0 -2.5 -1.0 1.0 3.0 4.5 5.0 4.5 3.0 1.0 -1.0 (b) (c)(i) x = 75^\circ or x = 285^\circ (ii) When x = 105^\circ, y = 2. (iii) When y = 1.5, x = 96^\circ or x = 264^\circ.

(a) \(\cos R = \frac{5}{13} = 0.3846\) \(R = \cos^{-1} (0.3846) = 67.38^\circ\) \(/PS/^{2} = 5^{2} + 19^{2} - 2(5)(19)\cos 67.38\) = \(25 + 361 - 190(0.3872)\) = \(386 - 73.568\) \(/PS/^{2} = 312.432\) \(/PS/ = \sqrt{312.432}\) = \(17.68 \approxeq 17.7 cm\) (b) Slant height of a cone = \(l = \sqrt{7^{2} + 24^{2}}\) \(\sqrt{49 + 576} = \sqrt{625}\) = \(25 cm\) C.S.A of a cone = \(\pi rl\) = \(\frac{22}{7} \times 7 \times 25\) = \(550 cm^{2}\) Area of hemisphere = \(\frac{1}{2} \times 4\pi r^{2} = 2 \pi r^{2}\) \(2 \times \frac{22}{7} \times 7 \times 7 = 308 cm^{2}\) TSA of the structure = \(550 cm^{2} + 308 cm^{2} = 858 cm^{2}\)

Area of sector = \(\frac{\theta}{360^\circ} \times \pi r^{2}\) = \(\frac{150}{360} \times \pi \times 12^{2}\) = 60\(\pi\) cm\(^{2}\)

\(\tan 16 = \frac{x}{150}\) \(x = 150 \tan 16\) = \(43 cm\)

If \(y = \cos x\) \(\frac{\mathrm d y}{\mathrm d x}\) =—\(\sin x\)