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The diagram given simple illustrates that a pentagon contains three \(\Delta\)s. Two angles which are in the ratio 2:3 will have actual values 2xo, 3xo respectively. Thus 2xo + 3xo + 3 x 60 = 3x sum of angles of \(\Delta\)a i.e. 5xo + 180o = 3 x 180o 5xo + 180o = 540o 5xo = 540o - 180o 5xo = 360o xo = \(\frac{360^o}{5}\) = 72o Hence, the smaller of the two angles is 2 x 72o = 144o

In \(\Delta\) XYZ, 2m + 2n + 68o = 180o 2(m + n) + 68o = 180o...(1) in \(\Delta\) XOZ, m + n + q = 180o ...(2) (m + n) = 180o - q...(3) substituting 180o - q for (m + n) in (1) gives 2(180o - q) + 68o = 180o 360o - 2q = 180o - 68o 360o - 2q = 112o 360o - 112o = 2q 248o = 2q q = \(\frac{248^o}{2}\) = 124o hence, o

(a) \(3\frac{4}{9} \div (5\frac{1}{3} - 2\frac{3}{4}) + 5\frac{9}{10}\) \(\frac{31}{9} \div (\frac{16}{3} - \frac{11}{4}) + \frac{59}{10}\) \(\frac{31}{9} \div (\frac{64 - 33}{12}) + \frac{59}{10}\) \((\frac{31}{9} \div \frac{31}{12}) + \frac{59}{10}\) \((\frac{31}{9} \times \frac{12}{31}) + \frac{59}{10}\) \(\frac{12}{9} + \frac{59}{10} = \frac{120 + 531}{90}\) \(\frac{651}{90} = \frac{217}{30}\). (b) + 1 3 5 2 3 5 7 3 4 6 8 4 5 7 9 Let E be the event of the sum being greater than 3 and less than 7 and S be the total sample space. n(E) = 4; and n(S) = 9. P(E) = \(\frac{n(E)}{n(S)} = \frac{4}{9}\)

(a) \(4 + \frac{3}{4}(x + 2) \leq \frac{3}{8}x + 1\) Multiply through by the LCM of 4 and 8, to clear fractions : \(8 \times 4 + 8 \times \frac{3}{4}(x + 2) \leq 8 \times \frac{3}{8}x + 8 \) \(32 + 6(x + 2) \leq 3x + 8 \) \(32 + 6x + 12 \leq 3x + 8 \) \(44 + 6x \leq 3x + 8\) \(6x - 3x \leq 8 - 44 \implies 3x \leq -36\) \(x \leq - 12\). (b) Area \((A_{1})\), of rectangle PQRS = \(20(20 + x) cm^{2}\) Area \((A_{2})\), of square = \(x \times x = x^{2} cm^{2}\) \(\implies 20(20 + x) - x^{2} = 484\) \(400 + 20x - x^{2} = 484\) \(x^{2} - 20x + 484 - 400 = x^{2} - 20x + 84 = 0\) \(x^{2} - 14x - 6x + 84 = 0\) \(x( x - 14) - 6(x - 14) = 0 \) \((x - 14)(x - 6) = 0\) \(\therefore \text{x = 6 cm or 14 cm}\)

(a) Let the diagram below represent a section of the polygon SEE DIAGRAM ABOVE Also, let : \(i\) represent the size of an interior angle; \(e\) represent the size of an exterior angle. Then \(\frac{i}{e} = \frac{5}{2}\) \(\implies i = \frac{5}{2}e\) But \(i + e = 180^\circ\) (sum of angles on a straight line) Substitute \(\frac{5}{2}e\) for \(i\) in the equation \(\frac{5}{2}e + e = 180^\circ\) \(\frac{7}{2}e = 180^\circ\) \(e = \frac{180^\circ \times 2}{7} = \frac{360^\circ}{7}\) Number of sides of the polygon = \(\frac{360^\circ}{\text{size of one exterior angle}}\) = \(360^\circ \div \frac{360^\circ}{7}\) \(360^\circ \times \frac{7}{360^\circ} = 7 sides\). (b) In the diagram above, < PSR = 68^\circ (interior angle of a cyclic quad = opp exterior angle) < PSQ = 68^\circ - 40^\circ = 28^\circ < PRQ = 28^\circ (angles in same segment) < SRQ = 74^\circ (interior angle of a cyclic quad = opp exterior angle) hence, < PRS = 74^\circ - 28^\circ = 46^\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) \(T_{1} = a = -8\) \(\frac{T_{7}}{T_{9}} = \frac{5}{8}\) \(\frac{a + 6d}{a + 8d} = \frac{5}{8} \implies \frac{-8 + 6d}{-8 + 8d} = \frac{5}{8}\) \(-8 + 6d = \frac{5}{8}(-8 + 8d) \implies -8 + 6d = -5 + 5d\) \(6d - 5d = -5 + 8\) \(d = 3\) (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).

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

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

If log 7.5 = 0.8751 Then 2log75 + log750 = 2(1.8751) + 2.8751 = 3.7502 + 2.8751 = 6.6253

8x-2 = 2/25 x-2 = 2/25 x 1/8 x-2 = 2/200 x-2 = 1/100 1/x2 = 1/100 x2 = 100 x = 10

\(= \frac{2\sqrt{2} - \sqrt{3}}{\sqrt{2} + \sqrt{3}} \times \frac{\sqrt{2} - \sqrt{3}}{\sqrt{2} - \sqrt{3}}\) \(= \frac{2\sqrt{2}(\sqrt{2}) + (2\sqrt{2})(-\sqrt{3})-\sqrt{3}(\sqrt{2})-\sqrt{3}(-\sqrt{3})}{(\sqrt{2})^2-(\sqrt{3})^2}\) \(= \frac{2 \times 2 - 2\sqrt{6} - \sqrt{6} + 3}{2 - 3}\) \(= \frac{4 - 3\sqrt{6} + 3}{-1}\) \(= \frac{7 - 3\sqrt{6}}{-1}\) \(= \frac{7}{-1} - \frac{3\sqrt{6}}{-1}\) \(= -7 + 3\sqrt{6}\) \(= 3\sqrt{6}-7\)