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The bearing of x from y = 270o + \(\theta\) where \(\theta\) + 50o = y in \(\bigtriangleup\) OXY O + X + Y = 180o Where O = 40o + 30o = 70o 70o + 30o + y = 180o y + 100o = 180o y = 180o - 100o = 30o \(\theta\) + 50o = 80o 80o - 50o = 30o The bearing of x from y = 270o + 30o = 300o

\(U = {2, 3, 4, 5, 6, 7, 8, 9}\) \(A = {2, 4, 6, 8}\) \(B = {2, 3, 7, 9}\) \(C = {4, 5, 6, 7, 8}\) (a) \(A \cap (B' \cap C')\) = \({2, 4, 6, 8} \cap ({4, 5, 6, 8} \cap {2, 3, 9})\) B' = 4,5,6,8 C' = 2,3 B' n C' = Ø \(A \cap (B' \cap C')\) = \({2, 4, 6, 8} \cap { }\) Ø = Ø (b) \((A \cup B) \cap (B \cup C) = {2, 3, 4, 6, 7, 8, 9} \cap {2, 3, 4, 5, 6, 7, 8, 9}\) = \({2, 3, 4, 6, 7, 8, 9}\)

(a) \(\therefore r = 36^\circ + 55^\circ = 91^\circ\) (b) \(< BAS = 60^\circ\) (angles in alternate segments) \(< BAE = 180^\circ - 5x^\circ\) (adjacent angles on a transversal) \(\therefore < BAS + < BAE + < EAT = 180^\circ\) \(\therefore 60^\circ + 180^\circ - 5x + x = 180^\circ\) \(\implies 240^\circ - 180^\circ = 4x\) \(60^\circ = 4x \implies x = \frac{60}{4} = 15^\circ\)

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

(a) M = {(4, 6), (5, 5), (6,4)} N = {(1, 4), (2, 5), (3, 6), (4, 1), (5, 2), (6, 3)} (b) Pr(M or N) = Pr(M) + Pr(N) Total sample = 36. \(Pr(M) = \frac{3}{36} = \frac{1}{12}\) \(Pr(N) = \frac{6}{36} = \frac{1}{6}\) \(Pr(\text{M or N}) = \frac{1}{12} + \frac{1}{6}\) = \(\frac{2 + 1}{12}\) = \(\frac{3}{12} = \frac{1}{4}\) (c) Yes, they are mutually exclusive since \(M \cap N = \emptyset\).

(a) 1km = 100,000 cm. Area = \(\frac{85 \times 100,000 \times 100,000}{20,000 \times 20,000}\) = \(85 \times 5 \times 5\) = \(2125 cm^{2}\) (b) Perimeter of path = \(2(length + width)\) = \(2(30 + (12 + a)) = 2(42 + a)\) = \(84 + 2a\) Area of field = \(length \times width\) = \(18 \times a\) = 18a \(\therefore 18a = 84 + 2a\) \(18a - 2a = 84 \implies 16a = 84\) \(a = \frac{84}{16}\) = \(5.25 cm\) (c) Area of major sector = \(\frac{360 - x}{360} \times \pi r^{2}\) \(27.5 = \frac{360 - x}{360} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\) \(27.5 = \frac{360 - x}{360} \times \frac{77}{2}\) \(360 - x = \frac{27.5 \times 360 \times 2}{77}\) \(360 - x = 257.14\) \(x = 360^\circ - 257.14^\circ = 102.86^\circ\) \(x \approxeq 103^\circ\) (to the nearest degree).

(a) Import duty : \(\frac{15}{100} \times 400 = GH¢ 60.00\) New cost price : GH¢400.00 + GH¢ 60 = GH¢ 460.00 Sales tax : \(\frac{10}{100} \times 460 = GH¢ 46.00\) Total cost price : \(GH¢(460 + 46) = GH¢ 506.00\) Profit = GH¢(660 - 506) = GH¢ 154.00 \(% profit = \frac{154}{506} \times 100%\) = \(30.43%\). (b) Let the number of girls be t. Therefore the total number of students = (1000 + t) students. The number that passed the exam = \(\frac{48}{100} \times (1000 + t)\) = \(0.48(1000 + t)\) \(\implies 0.48(1000 + t) = (\frac{50}{100} \times 1000 + \frac{40}{100} \times t)\) \(0.48(1000 + t) = 500 + 0.4t\) \(480 + 0.48t = 500 + 0.4t \implies 0.48t - 0.4t = 500 - 480\) \(0.08t = 20\) \(t = \frac{20}{0.08} = 250\) There are 250 girls in the school.

(a) (b) /XT/ = 3.4 cm.

(a) < ABC = 100^\circ \(\therefore b^{2} = 8^{2} + 13^{2} - 2(13)(8) \cos 100^\circ\) = \(64 + 169 - (208 \times - 0.1736)\) = \(233 + 36.12\) \(b^{2} = 269.12 \implies b = \sqrt{269.12} = 16.405 km\) \(\approxeq 16.4 km\) (3 significant figures) (b) \(\frac{\sin B}{b} = \frac{\sin A}{a}\) \(\frac{\sin 100}{16.4} = \frac{\sin A}{13}\) \(\sin A = \frac{13 \times \sin 100}{16.4}\) \(\sin A = \frac{12.803}{16.4}\) \(\sin A = 0.7806\) \(A = \sin^{-1} (0.7806) = 51.32^\circ\) \(\therefore\) Bearing of C from A = 180^\circ - (50^\circ + 51.32^\circ) = 180^\circ - 101.32^\circ = 78.68^\circ \(\approxeq\) 78.7^\circ (3 significant figure). (c) \(\cos 40 = \frac{BD}{13}\) \(BD = 13 \cos 40\) = \(13 \times 0.7660\) = 9.959 km \(\approxeq\) 9.96 km east of B.

(a) x -3 -2 -1 0 1 2 3 y -10 -4 0 2 2 0 -4 (b) (c) (ii) roots of the equation : \(x^{2} - x - 2 = 0\) are x = -1 or x = 2. (iii) gradient of the curve at x = -0.5 is 2.1.

(a) \((PT)^{2} + (TQ)^{2} = (PQ)^{2}\) \((PT)^{2} + 6^{2} = 10^{2}\) \((PT)^{2} = 100 - 36 = 64\) \(\therefore PT = 8 cm\) \(\therefore \frac{8}{6} = \frac{14.4}{SR}\) \(SR = \frac{6 \times 14.4}{8}\) \(SR = 10.8 cm\) Area of the quadrilateral QRST = \(\frac{1}{2} (6 + 10.8) \times 6.4\) = \(53.76 cm^{2}\) (b) Opp sides of a square = \((\frac{(100 - 10)x}{100} = 0.9x\) other two sides of the square = \(\frac{(100 + 15)x}{100} = 1.15x\) Area of the square : \(x \times x = x^{2}\) Area of the rectangle : \(0.9x \times 1.15x = 1.035x^{2}\) \(\therefore \text{The ratio of the area of the rectangle to the square} = 1.035x^{2} : x^{2}\) = \(1.035 : 1\)

(a) Class Intervals freq cum freq class boundaries 20 - 29 10 10 19.5 - 29.5 30 - 39 18 28 29.5 - 39.5 40 - 49 22 50 39.5 - 49.5 50 - 59 25 75 49.5 - 59.5 60 - 69 16 91 59.5 - 69.5 70 - 79 9 100 69.5 - 79.5 (b) (c) (i) Median (\(Q_{2}\)) = 49.5 (ii) Semi- Interquartile range = \(\frac{1}{2} (Q_{3} - Q_{1})\) \(Q_{1} = 38.5\); \(Q_{3} = 59.5\). Semi- Interquartile range = \(\frac{1}{2} (59.5 - 38.5)\) = \(\frac{1}{2} (21)\) = 10.5. (iii) Pr(\(\geq 60kg = \frac{25}{100} = \frac{1}{4}\)

\(T_{n} = ar^{n - 1}\) (Terms of a G.P) \(T_{3} = ar^{2} =24 .... (1)\) \(T_{7} = ar^{6} = 4\frac{20}{27} = \frac{128}{27} .......... (2)\) Divide (2) by (1), \(\frac{ar^{6}}{ar^{2}} = \frac{\frac{128}{27}}{24}\) \(r^{4} = \frac{16}{81}\) \(r = \sqrt[4]{\frac{16}{81}}\) = \(\frac{2}{3}\) Putting r in equation (1), \(ar^{2} = a(\frac{2}{3})^{2} = 24\) \(\frac{4}{9} a = 24 \implies a = \frac{24 \times 9}{4}\) a = 54. (b) \(y \propto x \propto \frac{1}{z^{2}}\) \(\implies y = \frac{kx}{z^{2}}\) \(4 = \frac{3k}{1^{2}} \implies k = \frac{4}{3}\) \(\therefore y = \frac{4x}{3z^{2}}\) When x = 3, z = 2, y = ? \(y = \frac{4(3)}{3(2^{2})}\) = \(\frac{12}{12} = 1\).

In \Delta TYX XY = TY ∴y =47^\circ base ∠s of ISC\Delta But y+x+34+47 = 180 interior opposite ∠s are supplementary 47 + x + 34 + 47 = 180 x + 128= 180 x = 180- 128 x = 52^\circ

X\(^2\) + x\(^2\) = 2\(^2\) 2x\(^2\) = 4 x\(^2\) = 4/2 x\(^2\) = 2 x = +/-\sqrt2 Area of \Delta = 1/2bh = 1/2 x \sqrt2 x \sqrt2 = 2/2 = 1 cm\(^2\)