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mean = \(\bar{x}\) = \(\frac{\sum x}{n}\) = \(\frac{5 + 3 + 0 + 7 + 2 + 1}{6}\) \(\frac{18}{6}\) = 3 \(\begin{array}{c|c} x & x - \bar{x} & |x - \bar{x}|\\ \hline 5 & -2 & 2 \\ \hline 3 & 0 & 0 \\ \hline 0 & -3 & 3 \\ \hline 7 & 4 & 4 \\ \hline 2 & -1 & 1 \\ \hline 1 & -2 & 2 \\ \hline & & & \sum|x - \bar{x}| \\ \hline & & & 12 \end{array}\) hence, MD = \(\frac{\sum|x - \bar{x}}{n}|\) \(\frac{12}{6}\) = 2

Area of floor = 3.5m x 2.2m = 7.7m2 Area of carpet = (3.5 - 0.5 - 0.5)m x (2.2 - 0.5 - 0.5)m 2.5m x 1.2m = 3m2 hence, area of margine = area of floor - area of carpet = (7.7 - 3)m2 = 4.7m2

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

Two word mathematics is first rewritten as mathematics, since every letter of mathematics appears once in the english alphabet. Hence the probability that letter selected is from the word mathematics is prob. (matheics) = \(\frac{8}{26}\) = \(\frac{4}{13}\)

In the diagram |AM| = |MB| - \(\frac{|AB|}{2}\) = \(\frac{16\sqrt{3}}{2}\)cm = 8\(\sqrt{3}\)cm in \(\Delta\) AMO, r2 = |AM|Z + |MO|2 r2 = (8\(\sqrt{3}\))2 + 102 = 64 x 3 + 100 = 192 + 100 = 292 r = \(\sqrt{292}\) 17.088cm 17cm

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) 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.