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(a) Let the sum of \(3\frac{2}{3}\) and \(2\frac{1}{5}\) be less than 7 by z. Then z = \(7 - (3\frac{2}{3} + 2\frac{1}{5})\) = \(7 - (\frac{11}{3} + \frac{11}{5})\) = \(7 - \frac{88}{15}\) = \(\frac{17}{15} = 1\frac{2}{15}\) Hence, the sum of \(3\frac{2}{3}\) and \(2\frac{1}{5}\) is less than 7 by \(1\frac{2}{15}\). (b) \(h = 6 + 4\cos (15p)^\circ\) (Given) (i) When p = 4, \(h = 6 + 4\cos (15 \times 4)^\circ\) \(h = 6 + 4 \cos 60^\circ\) = \(6 + 4 \times 0.5\) = \(6 + 2 = 8 m\) (ii) When h = 9 m, the given equation becomes \(9 = 6 + 4\cos (15p)^\circ\) \(9 - 6 = 4 \cos (15p)^\circ\) \(\cos (15p)^\circ = \frac{3}{4}\) \((15p)^\circ = \cos^{-1} (0.75)\) \((15p)^\circ = 41.4^\circ\) \(p^\circ = \frac{41.4}{15}\) \(p^\circ = 2.76^\circ\) \(p \approxeq 2.8^\circ\) (2 significant figures).

(a) Area of trapezium PQRS = \(\frac{1}{2}(PQ + RS) \times 40\) = \(\frac{1}{2}(20 + 60) \times 40\) = \(40 \times 40\) = \(1600 cm^{2}\) (b) In \(\Delta\) SPT, \(|SP|^{2} = |ST|^{2} + |TP|^{2}\) (Pythagoras theorem) \(|ST|^{2} = |SP|^{2} - |TP|^{2}\) = \(50^{2} - 40^{2}\) = \(2500 - 1600\) \(|ST|^{2} = 900\) \(\therefore |ST| = 30 cm\) \(|UR| = 60 cm - (30 + 20) cm\) = 10 cm Let < QRS = \(\alpha\), \(\tan \alpha = \frac{40}{10} = 4\) \(\alpha = \tan^{-1} (4)\) = \(75.96^\circ \approxeq 76^\circ\)

(a) (i) Let : L = { all good literature students in the school} A = {all students in the General Arts class} Then the Venn diagram below represents the statement : 'All good Literature students in a school are in the General Arts class'. where \(\in\) = {all students in the school}. (ii) (1) Invalid conclusion (2) Invalid conclusion (3) Valid conclusion (b) (i) c = h + y where \(y \propto n \implies y = kn\) \(\implies c = h + kn\) When n = 600 bricks, c = GH¢ 950.00 \(950.00 = h + 600k ....... (1)\) When n = 1000 bricks, c = GH¢ 1,030.00 \(1,030.00 = h + 1000k ....... (2)\) Subtracting (1) from (2), we have \(1,030 - 950 = 1000k - 600k\) \(80 = 400k\) \(k = \frac{80}{400} = \frac{1}{5}\) \(\therefore c = h + \frac{1}{5}n\) From (1), \(950 = h + 600k\) \(950 = h + (600 \times \frac{1}{5})\) \(950 = h + 120\) \(h = 950 - 120 = 830\) \(c = 830 + \frac{1}{5}n\) (ii) Using \(c = 830 + \frac{1}{5}n\), we have \(c = 830 + \frac{1}{5} \times 500\) = \(830 + 100\) = \(GH¢ 930.00\) Hence, the cost of producing 500 bricks = GH¢ 930.00.

(a) (i) \(y = px^{2} - 5x + q\) x -3 -2 -1 0 1 2 3 4 5 y 21 6 -12 0 13 From the table, y = -12 when x = 0, i.e. \(-12 = p(0^{2}) - 5(0) + q\) \(q = -12\). y = 0 when x = 4, i.e. \(0 = p(4^{2}) - 5(4) + q\) \(0 = 16p - 20 + q \implies 20 = 16p + q .... (1) \) Substitute q = -12 in (1), \(20 = 16p - 12\) \(20 + 12 = 16p \implies p = \frac{32}{16} = 2\) Equation : \(y = 2x^{2} - 5x - 12 \). (ii) Given the equation \(y = 2x^{2} - 5x - 12\). The table becomes x -3 -2 -1 0 1 2 3 4 5 y 21 6 -5 -12 -15 -14 -9 0 13 (b) (c) From the graph, (i) When x = 1.8, y = -14.5 ; (ii) When y = -8, x = -0.65 or 3.15 or 3.2.

(a) (i) , (ii) (b) (i) |QX| = 2.7 cm ; (ii) < XWZ = 75^\circ

(a) \(T_{1} = a = -8\) \(\frac{T_{7}}{T_{9}} = \frac{5}{8}\) \(\frac{a + 6d}{a + 8d} = \frac{5}{8} \implies \frac{-8 + 6d}{-8 + 8d} = \frac{5}{8}\) \(-8 + 6d = \frac{5}{8}(-8 + 8d) \implies -8 + 6d = -5 + 5d\) \(6d - 5d = -5 + 8\) \(d = 3\) (b) Let : the cost price \(CP_{1}\) of a basket of pawpaw be w; the cost price \(CP_{2}\) of a basket of mangoes be m. Then \(30w + 100m = N2,450 ..... (1)\) Also, let : The selling price of the baskets of pawpaw be \(SP_{1}\); the selling price of the baskets of mangoes be \(SP_{2}\). Then \(SP_{1} + SP_{2} = N(2,450 + 855) = N3,305 ...... (2)\) %age profit on the sale of pawpaw = \(\frac{SP_{1} - CP_{1}}{CP_{1}} \times 100%\) i.e. \(40% = \frac{SP_{1} - 30w}{30w} \times 100%\) \(\implies 40 \times 3w = (SP_{1} - 30w) \times 10\) \(12w = SP_{1} - 30w\) \(SP_{1} = 12w + 30w = 42w\) %age profit on the sale of mangoes = \(\frac{SP_{2} - CP_{2}}{CP_{2}} \times 100%\) i.e. \(30% = \frac{SP_{2} - 100m}{100m} \times 100%\) \(\implies 30m = SP_{2} - 100m\) \(SP_{2} = 30m + 100m = 130m\) But \(SP_{1} + SP_{2} = N3,305\) \(\implies 42w + 130m = 3305 .... (2)\) From (1), \(m = \frac{2450 - 30w}{100}\) \(m = \frac{245 - 3w}{10}\) Substitute \(m = \frac{245 - 3w}{10}\) in (3) \(42w + 130(\frac{245 - 3w}{10}) = 3305\) \(42w + 3,185 - 39w = 3305\) \(3w = 3305 - 3185 = 120\) \(w = N40.00\) (i) Hence, the cost price of a basket of pawpaw = N40.00. (ii) From (2), \(SP_{1} + SP_{2} = 3305\) \(42w + 130m = 3305 \implies 42 \times 40 + 130m = 3305\) \(1680 + 130m = 3305\) \(130m = SP_{2} = 3305 - 1680\) = \(N1,625.00\) Hence, the selling point of 100 baskets of mangoes = N1,625.00.

(a) From \(\Delta\) AMC, \(\sin 60^\circ = \frac{\sqrt{3}}{2}\) \(\cos 30^\circ = \frac{\sqrt{3}}{2}\) \(\tan 60^\circ = \frac{\sqrt{3}}{1} = \sqrt{3}\) Hence, \(\frac{2 \tan 60^\circ + \cos 30^\circ}{\sin 60} = \frac{2 \times \sqrt{3} + \frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}}\) = \(\frac{4\sqrt{3} + \sqrt{3}}{2} \div \frac{\sqrt{3}}{2}\) = \(\frac{5\sqrt{3}}{2} \times \frac{2}{\sqrt{3}} = 5\) (b) In \(\Delta\) POT, \(\tan 36^\circ = \frac{|TO|}{1050}\) \(|OT| = 1050 \tan 36^\circ = 1050 \times 0.7265\) = 762.87m In \(\Delta\) POB, \(\tan 41^\circ = \frac{|OB|}{1050}\) \(|OB| = 1050 \tan 41^\circ = 1050 \times 0.8693\) = 912.751m \(|TB| = 912.751m - 762.87m = 149.88m\) Hence, the height of the control tower = 150 m (to the nearest meter). (ii) In \(\Delta\) POB, \(\cos 41^\circ = \frac{1050}{|PB|}\) \(|PB| = \frac{1050}{\cos 41^\circ} = \frac{1050}{0.7547}\) = 1,391.28m The shortest distance between the aeroplane and the base of the control tower = 1,391m (to the nearest metre).

(a) \(h = \frac{mt}{d(m + p)}\) \(dh(m + p) = mt\) \(dhm + dhp = mt \implies dhp = mt - dhm\) \(dhp = m(t - dh) \implies m = \frac{dhp}{t - dh}\) (b) In the diagram above, < WXM = 90^\circ (angle in a semicircle) < WMX = 180^\circ - (90^\circ + 48^\circ) = 42^\circ < XMZ = 180^\circ - 42^\circ (angles on a straight line) = 138^\circ < WYZ = 180^\circ - 138^\circ (opp. angles of a cyclic quadrilateral) = 42^\circ (c) \(\star\) 1 3 5 6 1 4 6 1 2 3 6 1 3 4 5 1 3 5 6 6 2 4 6 0 (ii) From the table, the truth set of : (I) \(3 \star n = 3; n = {5}\) (II) \(n \star n = 3; n = { }\)

(a) (i) Using volume of reservoir = volume of hemisphere + volume of cone. \(\implies 333\frac{1}{3}\pi = \frac{2}{3}\pi r^{3} + \frac{1}{3}\pi r^{2} h\) \(\implies \frac{1000}{3}\pi = \frac{1}{3}\pi (2r^{3} + r^{2}h)\) Let the radius of the hemisphere be x m; Then \(\frac{1000}{3} = \frac{1}{3}(2x^{3} + x^{2}(6x))\) \(1000 = 2x^{3} + 6x^{3}\) \(1000 = 8x^{3} \implies x^{3} = \frac{1000}{8} = 125\) \(x = \sqrt[3]{125} = 5 m\) Hence, the volume of the hemisphere = \(\frac{2}{3}\pi r^{3}\) (where r = x = 5 m) = \(\frac{2}{3} \times \frac{22}{7} \times 5^{3} m^{3}\) = \(\frac{5500}{21} m^{3} \approxeq 261.905 m^{3}\) = \(262 m^{3} \) (nearest whole number) (ii) Total surface area of reservoir = surface area of hemisphere + surface area of cone = \(2\pi r^{2} + \pi rl\) From the diagram above, \(l^{2} = 30^{2} + 5^{2}\) \(l^{2} = 900 + 25 = 925\) \(l = \sqrt{925} = 30.41 m\) Hence, the total surface area of the reservoir = \(2 \times \frac{22}{7} \times 5 \times 5 + \frac{22}{7} \times 5 \times 30.41\) = \(157.143 + 477.871\) = \(635.104 m^{2}\) \(\approxeq 635 m^{2}\) (to the nearest whole number).

Marks (%) frequency (f) Cumulative frequency Upper class boundaries 0 - 9 7 7 9.5 10 - 19 11 18 19.5 20 - 29 17 35 29.5 30 - 39 20 55 39.5 40 - 49 29 84 49.5 50 - 59 34 118 59.5 60 - 69 30 148 69.5 70 - 79 25 173 79.5 80 - 89 21 194 89.5 90 - 99 6 200 99.5 (b) (i) If 5% of the candidates passed the examination, then (100 - 5)% = 95% passed with a mark \(\leq\) the lowest mark for distinction. \(\text{95% of 200} = \frac{95}{100} \times 200 \) = 190 candidates. From the ogive, 190 corresponds to 79.5 + 8 = 87.5 marks (to one decimal place) (ii) From the ogive, the number of candidates who scored at most 45% is 69. Hence, the probability of selecting a candidate who scored at most 45% = \(\frac{69}{200} = 0.345 \approxeq 0.3\) (1 d.p)

< URT = < TRQ (angle alternate a tangent and a chord equal to angle in the alternate segment) < RUT = xo In \(\bigtriangleup\) URT < RUT + < RUT + < UTR = 180o (sum of int. < s of \(\bigtriangleup\)) < URT + x + x = 180o < URT = 180o - 2x

To find \( 110111_2 + 10100_2 \): Convert to decimal: \(110111_2 = 55_{10}, \quad 10100_2 = 20_{10}\) Add: \(55 + 20 = 75\) Convert back to binary: \(75_{10} = 1001011_2\) Final result: \(110111_2 + 10100_2 = 1001011_2\) OR \(\begin{array}{r} 110111_2 \\ + 010100_2 \\ \hline 1001011_2 \end{array}\) \(\therefore \; 110111_2 + 10100_2 = 1001011_2\)

SP = Nx CP = N60,000 Percentage loss = 15% Using percentage loss... \(\frac{CP - SP}{CP} = \text{100%}\) \(\text{15%} = \frac{60000 - x}{60000} \times \text{100%}\) \(\text{15%} \times 60000 = (60000 - x)\text{100%}\) \(60000 - x = \frac{\text{15%} \times 60000}{\text{100%}}\) 60000 - x = 3 x 3000 60000 -x = 9000 x = 60000 - 9000 x = N51,000

0.00043 x 2000 = 43 x 10-5 x 2 x 103 = 43 x 2 x 10-5+3 = 86 x 10-2 = 8.6 x 101 x 10-2 = 8.6 x 10-1

Let; M = Man monthly net earnings Then; \(\text{N4500} = \frac{\text{10%}}{\text{100%}} \times M \) \(M = \frac{\text{N4500} \times \text{100%}}{\text{10%}} \) M = N45,000