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(a) In triangle SPQ, \(|SQ|^{2} = 5^{2} + 12^{2}\) (Pythagoras theorem) = \(25 + 144 = 169\) \(|SQ| = \sqrt{169} = 13 cm\) Angle b is common to triangles SPQ and PRS are similar. Using \(\sin b = \frac{12}{|SQ|} = \frac{|PR|}{5}\) \(\sin b = \frac{12}{13} = \frac{|PR|}{5}\) \(|PR| = \frac{12 \times 5}{13} \approxeq 4.62 cm\) (to 3 s.f) (b)(i) \(h^{2} = 10^{2} - 8^{2} = 36\) \(h = \sqrt{36} = 6 cm\) (ii) In the smaller triangle, \(\cos x = \frac{8}{10} = 0.8\) \(\cos^{-1} (0.8) = 36.87^\circ\) Since these are corresponding angles, x = x in the bigger triangle. \(\sin x = \frac{y}{12}\) \(y = 12 \sin x = 12 \sin 36.87\) = \(12 \times 0.6\) = 7.20 m The ladder M touches the wall at a height 7.2 m above the horizontal ground.

(a) x -5 -4 -3 -2 -1 0 1 2 3 4 y 35 18 5 -4 -9 -10 -7 0 11 26 (b) (c)(i) 2x\(^{2}\) + x = 10 \(\equiv\) 2x\(^{2}\) + x - 10 = 0 Thus, the solution of 2x\(^{2}\) + x - 10 = 0 are the values of x at which the curve cuts the x- axis. x = -2.5 or x = 2. (ii) 2x\(^{2}\) + x - 10 = 2x To solve the given equation, we first draw the graph of y = 2x: when x = -2, y = 2(-2) = -4; when x = 3, y = 2(3) = 6. The values of x at which y = 2x\(^{2}\) + x - 10 and y = 2x intersect give the solution of 2x\(^{2}\) + x - 10 = 2x. These are x = -2 or x = 2.5.

(a) px + qy = z \(p \begin{pmatrix} 2 \\ 3 \end{pmatrix} + q \begin{pmatrix} 5 \\ -2 \end{pmatrix} = \begin{pmatrix} -4 \\ 13 \end{pmatrix}\) \(\begin{pmatrix} 2p + 5q \\ 3p - 2q \end{pmatrix} = \begin{pmatrix} -4 \\ 13 \end{pmatrix}\) Thus, \(2p + 5q = -4 .....(1)\) \(3p - 2q = 13 ......(2)\) Solving the equations, we have p = 3 and q = -2. (b)(i) (ii) See attached graph for the quadrilateral W(2, 3), X(4, -1), Y(-3, -4) and Z(-3,2). (iii) See attached for the image for the quadrilateral WXYZ under an anticlockwise rotation of 90^\circ is \(W_{1}(-3, 2), X_{1}(1, 4), Y_{1}(4, -3)\) and \(Z_{1}(-2, -3)\).

(a) Marks(x) Frequency (f) fx 10 1 10 20 1 20 30 x 30x 40 5 200 50 y 50y 60 1 60 70 4 280 80 3 240 90 1 90 Total 16 + x + y 900 + 30x + 50y \(\sum f = 16 + x + y = 30\) \(\implies x + y = 14 ... (1)\) \(\bar{x} = \frac{\sum fx}{\sum f}\) \(52 = \frac{900 + 30x + 50y}{30}\) \(1560 = 900 + 30x + 50y \implies 30x + 50y = 660\) \(3x + 5y = 66 .... (2)\) Solving equation (1) and (2), From (1), x = 14 - y \(3(14 - y) + 5y = 42 - 3y + 5y = 66\) \(2y = 24 \implies y = 12\) \(x = 14 - y = 14 - 12 = 2\) (x, y) = (2, 12). (b) Class interval Frequency Upper class boundary 1 - 10 1 10.5 11 - 20 1 20.5 21 - 30 2 30.5 31 - 40 5 40.5 41 - 50 12 50.5 51 - 60 1 60.5 61 - 70 4 70.5 71 - 80 3 80.5 81 - 90 1 90.5 (c) (d) From the histogram, Mode = 44.

x 1 2 3 4 1 1* 2* 3* 4* 2 2* 4* 6* 8 3 3* 6* 9 12 4 4* 8 12 16 The numbers that are not greater than 6 are either less than 6 or equal to 6. P(picking a number not greater than 6) = \(\frac{10}{16}\) = \(\frac{5}{8}\)

\(\begin{vmatrix} 3 & 5 & -4 \\ 6 & -3 & -5 \\ -2 & 2 & 1 \end{vmatrix}\) = \(3(-3 - (-10)) - 5(6 - 10) + (-4)(12 - 6)\) = \(21 + 20 - 24\) = 17

The Chief Examiner reported that in part (a), some Candidates' carried out the requirement of BODMAS correctly tile few couldn't. However, in part (b), most Candidates' could not sketch the diagram correctly and hence failed to earn the marks allocated. In part (a), expressing as improper fraction we have (\(\frac{1}{4} \times 9\frac{1}{7} + \frac{2}{5} (\frac{2}{2} + \frac{3}{4})) \div (\frac{2}{5} - \frac{1}{4})\) Next, using BODMAS to get (\(\frac{16}{7} + \frac{2}{5} \times \frac{17}{12}) \div \frac{17}{12}\). Simplifying and inverting yields \(\frac{1198}{420} \times \frac{20}{3} = 19\frac{1}{63}\) In part (b), they were expected to sketch Then, using trigonometric ratio to have sin 48\(^o\) = \(\frac{x}{250}\) or cos 42\(^o\) = \(\frac{x}{250}\) and simplifying the vertical distance moved to be 186 m.

(a) If Musa's age = x and Manya's age = y, then x = y + 3.... (1) and (x — 7) = 2(y— 7) .......(2) Solving both equations will yield Musa's age = 13 years and Manya's age = 10 years. (b) Let n be the year when the sum of their ages will be 45, so that (10 + n) + (13 + n) = 45 and solving will give n = 11 years.

The Chief Examiner reported that quite a handful of the Candidates' got part (a) correctly but couldn't establish the relationship between km/h and m/s. (a) The distance covered = (2 x 60) + (2 x \(\frac{22}{7}\)) = 120 + 377.142 = 497.1429m so that if the athlete runs two times will covered = 2 x 497.1429 m = 994.2857m (at least one decimal place). (b) The speed m/s = \(\frac{994,2857}{200}\) and converting to km/h we will have \(\frac{994,2857}{1000} \times \frac{200}{3600}\) = 17.8973 km/h

(a) First joining XZ and WY, we have < XWY = < XZY = \(\frac{1}{2}\)(180^\circ — 80^\circ) = 50^\circ. This implies that < ZXY = 50^\circ(base angles of an isosceles triangle). So that < XWZ = 180^\circ - 80^\circ = 100^\circ. (b) The rate in every N1.00 = \(\frac{2}{100}\) so that the rate of interest charged per annum will be \(\frac{2 \times 12}{100}\) x 100 which gives 24%.

(a) The bar chart for the distribution is as shown (b) They were able to construct the frequency distribution table as Score (x) 1 2 3 4 5 6 Total Frequency (f) 2 5 13 11 9 10 50 fx 2 10 39 44 45 60 200 Mean = \(\frac{200}{50}\) = 4

(a) By the use of Pythagoras theorem, the hypotenuse side is 13 and also sin x = \(\frac{5}{13}\) and cos x = \(\frac{12}{13}\) Substituting and simplifying yields \(\frac{65}{161}\) (b) we have (11 + 2x)(14 + 2x) - (11 x 14) =186. Expanding and simplifying to get the quadratic equation 2x\(^2\) + 25x - 93 = 0. Solving the quadratic equation results in x = 3 m as the width of the path.

(a) let the original price of the shirt be x, then we have two percentages reduction leaving 60% and 70%. This implies that \(\frac{60}{100} \times \frac{70}{100} \times x\) = 35. Solving for x yields x = GH₵ 83.33. (b) (i), the first reduction = (\(\frac{40}{100}\)) x 180 = 72.00 and the second reduction = (\(\frac{30}{100}\) ) x 108 = 32.40. Then total reduction = (72.00 + 32. 40) = GH₵ 104.40. (b) (ii), the percentage reduction in the price of the article = (\(\frac{104.40}{180} \times 100\)) = 58%

(a) The trapezium PQRS is as shown (b) |QR| = (3.9 \pm 0.1) cm, < PSR = (60 \pm 1)\(^o\) and |QY| = (5.2 \pm 0.1) cm.

(a) first is to sketch the diagram Using Pythagoras theorem, x\(^2\) = 17\(^2\) - 15\(^2\) = 64 and x = 8 m. Then, multiplying by 2 we have 16m as the length of the fence within the reach of the donkey. (b)(i) PR = \(\sqrt{15^2 + 15^2}\) = 21.2132 cm and PO = OR = \(\frac{21.2132}{2}\) = 10.61 cm. Then height of the pyramid is 32\(^2\) = h\(^2\) + (10.6066)\(^2\) and h yields h = 30.2 cm. (b)(ii) Volume = \(\frac{1}{3}\) x 15 x 15 x 30.19106 = 2264 cm\(^2\) (correct to the nearest cm\(^3\)).