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(c) |XP| = 3.8 cm.

(a) Total surface area of a sphere = \(4\pi r^{2}\) Let the TSA of the smaller sphere be \(S_{1}\) with radius r and the bigger sphere be \(S_{2}\) with radius R. \(\frac{S_{1}}{S_{2}} = \frac{9}{49}\) \(\frac{9}{49} = \frac{4\pi r^{2}}{4\pi R^{2}}\) \(\implies \frac{9}{49} = \frac{12^{2}}{R^{2}}\) \(\frac{49 \times 144}{9} = R^{2}\) \(R = \sqrt{49 \times 16} = 28 cm\) Volume of bigger sphere = \(\frac{4}{3} \pi r^{3}\) = \(\frac{4}{3} \times \frac{22}{7} \times 28 \times 28 \times 28\) = \(\frac{275,968}{3}\) = \(91989.33 cm^{3}\) \(\approxeq 91989 cm^{3}\) (nearest whole number). (b) \(|ZX|^{2} = |YZ|^{2} + |YX|^{2} - 2|YZ||YX| \cos Y\) = \(5^{2} + 3^{2} - 2(5)(3) \cos 135\) = \(25 + 9 - 30(-0.7071)\) = \(34 + 21.213\) \(|ZX|^{2} = 55.213\) \(|ZX| = 7.43 km \approxeq 7 km\) (ii) Using sine rule, \(\frac{\sin \theta}{5} = \frac{\sin 135}{7.43}\) \(\sin \theta = \frac{5 \times \sin 135}{7.43}\) = \(\frac{3.5355}{7.43}\) \(\sin \theta = 0.4758\) \(\theta = \sin^{-1} (0.4758)\) = \(28.41^\circ \approxeq 28^\circ\) (to the nearest degree). Bearing of Z from X = 270^\circ + 28.41^\circ = 298.41^\circ \(\approxeq\) 298^\circ.

Score 1 2 3 4 5 6 Frequency 2 5 x 11 9 10 P(3) = 0.26 \(\frac{x}{37 + x} = 0.26\) \(x = 0.26(37 + x) \implies x = 9.62 + 0.26x\) \(x - 0.26x = 9.62 \implies 0.74x = 9.62\) \(x = \frac{9.62}{0.74} = 13\) Total toss frequency = 37 + x = 37 + 13 = 50. (a) Median = \(\frac{n + 1}{2}\) = \(\frac{50 + 1}{2}\) = 25.5 From the table, 25.5 occurs at 4 which is the median. (b) Standard deviation = \(\sqrt{\frac{\sum fd^{2}}{\sum f}}\) \(x\) \(f\) \(\fx\) \(d = |x - \bar{x}|\) \(d^{2}\) \(fd^{2}\) 1 2 2 -3 9 18 2 5 10 -2 4 20 3 13 39 -1 1 13 4 11 44 0 0 0 5 9 45 1 1 9 6 10 60 2 4 40 \(\sum\) 50 200 100 \(Mean(\bar{x}) = \frac{\sum fx}{\sum f}\) = \(\frac{200}{50} = 4 S.D = \(\sqrt{\frac{100}{50}}\) = \(\sqrt{2}\) = 1.4142

(a) \(\theta = \frac{60^\circ}{2} = 30^\circ\) Area of trapezium = \(\frac{a + b}{2} h\) \(60 = \frac{15 + 25}{2} h \implies 60 = 20h\) \(h = \frac{60}{20} = 3 cm\) (ii) Considering \(\Delta PSR\), \(\sin \theta = \frac{Opp}{Hyp}\) \(\sin 30 = \frac{3}{|PS|}\) \(|PS| = \frac{3}{\sin 30}\) = \(\frac{3}{0.5} = 6 cm\) (b) Let the total amount to be shared = k. Ade's share = \(\frac{3}{5} k \) Remainder = \(k - \frac{3}{5}k \) = \(\frac{2}{5}k\) Nelly's share : \(\frac{1}{3} \times \frac{2}{5}k\) = \(\frac{2}{15}k\) Austin's share : \(\frac{2}{5}k - \frac{2}{15}k\) = \(\frac{4}{15}k\) \(\frac{4}{15}k - \frac{2}{15}k = \frac{2}{15}k\) \(\frac{2}{15}k = N3,000\) \(\implies k = \frac{3,000 \times 15}{2}\) = \(N22,500\) Ade's share = \(\frac{3}{5}k\) = \(\frac{3}{5} \times N22,500 = N13,500\).

(a) \(P \propto Q\) and \(P \propto \frac{1}{R^{2}}\). \(P \propto \frac{Q}{R^{2}} \implies P = \frac{kQ}{R^{2}}\) When P = 1, Q = 8, R = 2 \(1 = \frac{8k}{2^{2}} \implies 8k = 4\) \(k = \frac{4}{8} = 0.5\) \(P = \frac{Q}{2R^{2}}\) When P = 3, R = 5, Q = ? \(3 = \frac{Q}{2(5^{2})}\) \(3 = \frac{Q}{50} \implies Q = 150\). (b) Angular difference : 60^\circ - 20^\circ = 40^\circ \(\frac{\theta}{360} \times 2\pi r \) \(r = R\cos \theta\) \(\frac{40}{360} \times 2 \times 3.142 \times 6400 \cos 20 = \frac{1511687.278}{360}\) = \(4199.13 km\) \(Speed = \frac{Distance}{Time}\) = \(\frac{4199.13}{6}\) = \(699.85 km/hr\) \(\approxeq 700 km/hr\) (to 3 significant figures) (ii) \(D = \frac{\theta}{360} \times 2\pi r\) \(420 = \frac{\theta}{360} \times 2 \times 3.142 \times 6400\) \(\theta = \frac{420 \times 360}{2 \times 3.142 \times 6400}\) \(\theta = \frac{151200}{40217.6} = 3.76^\circ\) Latitude of C : 20^\circ + 3.76^\circ = 23.76^\circ Note that due North implies same longitude but different latitude.

Median = \(\frac{\text{N}}{2}\) th item = \(\frac{150}{2}\) th ≈ 36.5 from the graph.

No expanded working has been attached for this question yet. The verified answer is Option C.

∠RTS = 90o (∠ in semi circle) ∠UTS = 25o (∠ in alternate segment) X + 90 + 25 = 180(sum of ∠s on a str line) X + 115 = 180 X = 180 - 115 X = 65o

TUVW is a cyclic quad 3χ + 20 + 88 = 180 (opp ∠ s of a cyclic quad are supplementary) 3χ + 108 = 180 3χ = 180 - 108 3χ = 72 χ = 72/3 χ = 24o

cot\(\theta\) = \(\frac{1}{\cos \theta}\) = \(\frac{8}{15}\)(given) tan\(\theta\) = \(\frac{15}{8}\) By Pythagoras' theorem, x\(^2\) = 15\(^2\) + 8\(^2\) x\(^2\) = 225 + 64 = 289 x = \(\sqrt{289}\) = 17 Hence sin\(\theta\) = \(\frac{15}{x}\) = \(\frac{15}{17}\)

\(\left(\frac{81}{16}\right)^{\frac{-1}{4}}\times 2^{-1}\) = \(\frac{1}{\left(\frac{81}{16}\right)^{\frac{1}{4}}}\times \frac{1}{2}\) = \(\left(\frac{16}{81}\right)^{\frac{1}{4}}\times \frac{1}{2}\) = \(\left(\frac{2}{3}\right)^{4\times \frac{1}{4}}\) =\(\frac{2}{3}\times \frac{1}{2}\) = \(\frac{1}{3}\)

\(\frac{2\sqrt{3} + \sqrt{5}}{\sqrt{5} - \sqrt{3}}\) = \((\frac{2\sqrt{3} + \sqrt{5}}{\sqrt{5} - \sqrt{3}})(\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}})\) = \(\frac{2\sqrt{15} + 6 + 5 + \sqrt{15}}{5 - 3}\) = \(\frac{3\sqrt{15} + 11}{2}\)

0.21 x 0.34 = 0.0714 = 7.14 x 10-2

Let C n G = X n C only = 40 - x n G only = 30 - x 40 - X + 30 - X + X = 50 70 - X = 50 X = 70 - 50 X = 20

\(\frac{x^{3} + 3x^{2} - 10x}{2x^{2} - 8} = \frac{x(x^{2} + 3x - 10)}{2(x^{2} - 4)}\) = \(\frac{x(x^{2} - 2x + 5x - 10)}{2(x - 2)(x + 2)}\) = \(\frac{x(x - 2)(x + 5)}{2(x - 2)(x + 2)}\) = \(\frac{x(x + 5)}{2(x + 2)}\)