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

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

y\(^2\)+ 14y + k let a = 1, b = 14, c = k (where a and b represent the coefficient of y and y\(^2\) respectively) 4ac = b\(^2\) 4 x 1 x c = 14 x 14 = 196 4c = 196 c = \(\frac{196}{4}\) = 49

(a) \(\log_{10} \sqrt{30} - \log_{10} \sqrt{6} + \log_{10} \sqrt{2}\) = \(\log_{10} (\sqrt{\frac{30}{6}} \times \sqrt{2})\) = \(\log_{10} (\sqrt{5} \times \sqrt{2})\) = \(\log_{10} \sqrt{10}\) = \(\log_{10} (10^{\frac{1}{2}})\) = \(\frac{1}{2} \log_{10} 10\) = \(\frac{1}{2}\). (b) (ii) \(A \cap C = { } \) (iii) \(A \cap B' = {1, 4}\).

(a) \((\frac{64}{125})^{-\frac{2}{3}} = (\frac{125}{64})^{\frac{2}{3}}\) = \((\frac{5}{4})^{3})^{\frac{2}{3}} \) = \((\frac{5}{4})^{2}\) = \(\frac{25}{16}\) \(\therefore 2 \div (\frac{64}{125})^{-\frac{2}{3}} = 2 \div \frac{25}{16}\) = \(2 \times \frac{16}{25}\) = \(\frac{32}{25}\) (b) \(3x + 5 = - 4x - 1\) \(3x + 4x = - 1 - 5\) \(7x = - 6\) \(x = -\frac{6}{7}\) \(y = 3x + 5\) ( You can use any of the given equations to get y) \(y = 3(-\frac{6}{7}) + 5 = -\frac{18}{7} + \frac{35}{7}\) = \(\frac{17}{7}\) \(k = (x, y) = (-\frac{6}{7}, \frac{17}{7})\)

(a) \(\frac{5}{8} of 2\frac{1}{2} - \frac{3}{4} \div \frac{3}{5}\) = \((\frac{5}{8} \times \frac{5}{2}) - (\frac{3}{4} \div \frac{3}{5})\) = \((\frac{25}{16}) - (\frac{5}{4})\) = \(\frac{25 - 20}{16}\) = \(\frac{5}{16}\) (b) \(V = \frac{1}{3} \times A \times h = \frac{1}{3} \pi r^{2} h\) \(\therefore \frac{1}{3} \times 154 \times h = \frac{1}{3} \times \pi \times r^{2} \times h\) Comparing the two equations, \(154 = \frac{22}{7} \times r^{2}\) \(r^{2} = \frac{154 \times 7}{22}\) \(r^{2} = 49\) \(\therefore r = 7 cm\).

(a) Sample - space 1 2 3 4 5 6 1 1, 1 1, 2 1, 3 1, 4 1, 5 1, 6 2 2, 1 2, 2 2, 3 2, 4 2, 5 2, 6 3 3, 1 3, 2 3, 3 3, 4 3, 5 3, 6 4 4, 1 4, 2 4, 3 4, 4 4, 5 4, 6 5 5, 1 5, 2 5, 3 5, 4 5, 5 5, 6 6 6, 1 6, 2 6, 3 6, 4 6, 5 6, 6 (i) Same digit = {(1,1), (2,2),(3,3), (4,4),(5,5), (6,6)} P(same digit) = \(\frac{6}{36}\) (ii) P(total score greater than 5) = \(\frac{26}{36}\) = \(\frac{13}{18}\) (b) \(\frac{x^{2} + y^{2}}{y^{2} - x^{2}} = \frac{(\cos 30)^{2} + (\sin 30)^{2}}{(\sin 30)^{2} - (\cos 30)^{2}}\) = \(\frac{(\frac{\sqrt{3}}{2})^{2} + (\frac{1}{2})^{2}}{(\frac{1}{2})^{2} - (\frac{\sqrt{3}}{2})^{2}}\) = \(\frac{\frac{3}{4} + \frac{1}{4}}{\frac{1}{4} - \frac{3}{4}}\) = \(\frac{1}{-\frac{1}{2}}\) = \(-2\).

(a) \(< ADB = < ACB = 20^\circ\) (Angle in the same segment). In \(\Delta ACY\), \(\hat{C} = < ACB\) (i) \(\hat{C} + \hat{A} + \hat{Y} = 180^\circ\) \(20^\circ + \hat{A} + 60^\circ = 180^\circ\) \(\hat{A} = 180^\circ - 80^\circ = 100^\circ\) \(< CAY = \hat{A} = 100^\circ\) \(< CAX = 180^\circ - 100^\circ = 80^\circ\) (ii) In \(\Delta ACB\), \(\hat{B} = \hat{C} = \frac{180^\circ - 20^\circ}{2} \) = \(\frac{160^\circ}{2}\) = \(80^\circ\) In \(\Delta ABY\), \(< ABY = \hat{B} = 180^\circ - 80^\circ\) = \(100^\circ\) (b) \((m + 1)(m - 3) \equiv m^{2} - km + c\) \(m^{2} - 3m + m - 3 = m^{2} - 2m - 3\) \(m^{2} - 2m - 3 \equiv m^{2} - km + c\) \(\implies k = 2 ; c = -3\)

(a) \(\sqrt{\frac{0.18 \times 12.5}{0.05 \times 0.2}}\) \(\frac{0.18 \times 12.5}{0.05 \times 0.2} = \frac{18 \times 10^{-2} \times 125 \times 10^{-1}}{5 \times 10^{-2} \times 2 \times 10^{-1}}\) = \(9 \times 25 \times 10^{-3 - (-3)}\) = \(9 \times 25\) = \(225\) \(\therefore \sqrt{\frac{0.18 \times 12.5}{0.05 \times 0.2}} = \sqrt{225}\) = \(15\). (b) \(\frac{8 - 4\sqrt{18}}{\sqrt{50}}\) \(\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}\) \(\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}\) \(\frac{8 - 4(3\sqrt{2})}{5\sqrt{2}} = \frac{8 - 12\sqrt{2}}{5\sqrt{2}}\) Rationalising, we have = \(\frac{8 - 12\sqrt{2}}{5\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\) = \(\frac{8\sqrt{2} - 24}{10}\) = \(0.8\sqrt{2} - 2.4\) = \(-2.4 + 0.8\sqrt{2}\) (c) \(x \propto \frac{y^{3}}{z^{2}}\) \(\implies x = \frac{ky^{3}}{z^{2}}\) \(108 = \frac{k \times 3^{3}}{4^{2}}\) \(k = \frac{108 \times 16}{27}\) \(k = 64\) \(\therefore x = \frac{64y^{3}}{z^{2}}\) \(\therefore 4000 = \frac{64 \times 10^{3}}{z^{2}}\) \(4 = \frac{64}{z^{2}}\) \(z^{2} = \frac{64}{4} = 16\) \(z = \sqrt{16} = \pm 4\)

(a) (i) Let the exterior angle be x Let the interior angle be x + 120^\circ. Exterior angle + Interior angle = 180^\circ \(\therefore x + x + 120 = 180 \implies 2x + 120 = 180\) \(2x = 60 \implies x = 30^\circ\) Number of sides = \(\frac{360}{\text{exterior angle}}\) = \(\frac{360}{30} = 12 sides\) (ii) Sum of interior angle = \((2n - 4) \times 90^\circ\) = \((2(12) - 4) \times 90^\circ\) = \(20 \times 90\) = \(1800^\circ\) (b) Note : < PQR = 65^\circ + 34^\circ = 99^\circ\) \(|PR|^{2} = |PQ|^{2} + |QR|^{2} - 2(|QR|)(|PQ|) \cos 99\) \(|PR|^{2} = 6^{2} + 13^{2} - 2(6)(13) \cos 99\) \(|PR|^{2} = 205 - 156 \cos 99\) \(|PR|^{2} = 205 + 24.40 = 229.40\) \(|PR| = \sqrt{229.40}\) = \(15.146 km\) \(\approxeq 15 km\)

(a) Longitude difference = 85^\circ - 25^\circ = 60^\circ. Distance AB along the parallel of latitude = \(\frac{\theta}{360^\circ} \times 2\pi R \cos \alpha\) \(AB = \frac{60}{360} \times 2 \times \frac{22}{7} \times 6400 \cos 53\) = \(\frac{1}{6} \times \frac{44}{7} \times 3851.62\) = \(4,035.03 km\) \(\approxeq 4035 km\) (b) \(Speed = \frac{Distance}{Time}\) \(\therefore Time = \frac{Distance}{Speed}\) = \(\frac{4035}{400}\) \(\approxeq 10 hours\). (c) Distance BC measured along the meridian \(BC = \frac{\theta}{360} \times 2 \pi R\) \(2000 = \frac{\theta}{360} \times 2 \times \frac{22}{7} \times 6400\) \(\theta = 17.898^\circ \approxeq 18^\circ\) \(\theta\) = Latitude difference Let the latitude of C = x. \(\theta = 53 - x\) \(18 = 53 - x\) \(x = 35^\circ\)

(a) \(x + 8 < 4x - 15 \implies 4x - 15 > x + 8\) \(4x - x > 8 + 15 \implies 3x > 23\) \(x > \frac{23}{3}\) \(x > 7\frac{2}{3}\). (b) Extra income = 5k of N200,000 = \(\frac{5}{100} \times N200,000\) = \(N10,000\) Total income = Monthly income + commission = N(2,500 + 10,000) = N12,500. (c) Area of semi- circle = \(\frac{\pi r^{2}}{2}\) = \(\frac{1}{2} \times \frac{22}{7} \times 35 \times 35\) = \(1925 cm^{2}\) Length of rectangular part = 50 cm Breadth = Diameter of semi- circle = 2(35 cm) = 70 cm Area of rectangular part = \(length \times \breadth\) = \(50 \times 70\) = \(3500 cm^{2}\) Total area of window = \((3500 + 1925) cm^{2}\) = \(5425 cm^{2}\)

(a) \(|AB| = 85m = \frac{85}{10}\) = \(8.5cm\) \(|TC| = 6cm \equiv 60 m\).

(a) Class Interval Freq (f) Classmark (x) \(fx\) 10 - 14 7 12 84 15 - 19 18 17 306 20 - 24 25 22 550 25 - 29 17 27 459 30 - 34 9 32 288 35 - 39 4 37 148 80 1835 \(\bar{x} = \frac{\sum fx}{\sum f}\) = \(\frac{1835}{80}\) \(\approxeq 22.9yrs\) (b)(i) Class Interval Freq (f) Class boundaries 10 - 14 7 9.5 - 14.5 15 - 19 18 14.5 - 19.5 20 - 24 25 19.5 - 24.5 25 - 29 17 24.5 - 29.5 30 - 34 9 29.5 - 34.5 35 - 39 4 34.5 - 39.5 (ii) Modal age = \(\frac{20 + 24}{2} = 22\) (c) P(picking from the modal class) = \(\frac{25}{80}\) = \(\frac{5}{16}\).

(a) In \(\Delta ACB, < CAB = 180^\circ - (90^\circ + 61^\circ)\) = \(180^\circ - 151^\circ\) = \(29^\circ\) In \(\Delta ADB, < DAB = (29 + x)^\circ\) \(\cos (29 + x) = \frac{3}{5} = 0.6\) \((29 + x)^\circ = \cos^{-1} (0.6)\) \((29 + x)^\circ = 53.13^\circ\) \(x = 53.13^\circ - 29^\circ\) = \(24.13^\circ\) \(\approxeq 24.1^\circ\) (1 decimal place). (b) (i) \(DB = 2DN = 2NB\) \(OA = OB = OC = OD = 4cm\) \(AB = BC = CD = DA = 2cm\) \(DB^{2} = BC^{2} + DC^{2}\) = \(2^{2} + 2^{2}\) = \(\sqrt{8}\) = \(2\sqrt{2} cm\) \(2DN = DB \implies DN = \frac{2\sqrt{2}}{2} = \sqrt{2} cm\) In \(\Delta DON\), \(OD^{2} = ON^{2} + DN^{2}\) \(4^{2} = h^{2} + (\sqrt{2})^{2} \implies h^{2} = 16 - 2\) \(h^{2} = 14 \implies h = \sqrt{14}\) (ii) \(V = \frac{1}{3} \times \base area \times \perp distance\) \(V = \frac{1}{3} \times 2^{2} \times \sqrt{14}\) = \(\frac{4\sqrt{14}}{3}\) = \(4.99 cm^{3}\)