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(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\(^3\)).

a) Table \[\begin{array}{c|c} x & 0^o & 30^o & 60^o & 90^o & 120^o & 150^o & 180^o & 210^o & 240^o & 270^o& 300^o \\ \hline Y & 2.00 & 1.23 & 0.13 & -1.00 & -1.87 & -2.23 & -2.00 & -1.23 & -0.13 & 1.00 &1.87 \end{array}\] (b)(i) The value of x for which 2 cos x - sinx = 1 x = \((36 \pm 3)^o\) or \((270 \pm 3)^o\) (ii) The value of x for which tan x = 3 x = \((64 \pm 3)^o\) or \((244 \pm 3)^o\)

(a) Applying the laws of logarithm, log(y + 2) = log ax which amplies that y + 2 = ax and finding x to get x = \(\frac{y + 2}{a}\) (b)(i) The pie is shown (b)(i) The percentage of timber produced that year that was from Amenfi = \(\frac{900}{7200}\) x 100% = 12.5% (ii) Difference in revenue = $560.00(1800 - 600) = $672,000.00

(a) The quantity of shares bought \(\frac{8400}{2.5}\) = 3360 shares. Therefore, the dividend the man received is 3360 X 0.25 = N840.00. (b) Clearing fractions and simplifying they got the quadratic equations 9x\(^2\) - 9x + 2 = 0. Next, dividing by 9 and adding the square of the coefficient of x to both sides yields (x - \(\frac{1}{2}\))\(^2\) = \(\frac{1}{4} - \frac{2}{9}\). Then, taking square root and solving for x To obtain x = \(\frac{2}{3}\) or x = \(\frac{1}{3}\)

(a) it was expected that they substitute into the formula for sum of an Arithmetic Progression to have 130 = 1\(\frac{1}{2}\) [2a+(10 - 1)d] and simplifying will yield 2a + 9d = 26 -------(1). Also, U5 = a + (5 – 1)d = 3a and simplifying will yield a = 2d --------------------------------(2) Substituting for a in equation (1) will yield 2(2d) + 9d = 26 and solving for d, d = 2. (b), Substituting for d in equation (2) to get the first term:a = 2(2) = 4 (c) The number of terms required can be obtained by solving 28 = 4 + (n – 1)2 which resulted to n = 13.

(a) By the use of Pythagoras theorem, the adjacent side of the right – angled triangle is 4. Then, cos⁡(X = \(\frac{4}{5}\)) and substituting to get 5(cos X)\(^2\) – 3 = 5(\(\frac{4}{5}\))\(^2\) - 3 = \(\frac{16}{5}\) - 3 = \(\frac{(16 - 15)}{5}\) = \(\frac{1}{5}\) (b) find Volume of pyramid = \(\frac{1}{3}\) \times 42 \times 11 \times 12 = 1,848 (cm)\(^2\) Volume of conical container = \(\frac{1}{3}\) \times \(\frac{22}{7}\) x r\(^2\) \times 12 = 1,848 and simplifying to get r\(^2\) = 147 so that r =7\sqrt3 cm as required.

(a) The expected table with class boundaries was Price (N1,000,000.00) Class Boundaries Number of Vehicles 1.0 - 1.9 0.95 - 1.95 23 2.0 - 2.9 1.95 - 2.95 48 3.0 - 3.9 2.95 - 3.95 107 4.0 - 4.9 3.95 - 4.95 90 5.0 - 5.9 4.95 - 5.95 32 (b) From the first graph, the most preffered selling price for the brand of car is 2.95 + 0.75 = 3.70 = N1,000,000 x 3.70 = N3,700,000.00

(a) By summing 2y + x + 6y - 2x + 1 + 4y = 28 and simplified to arrive at 12y - x = 27 ...............(1). Similarly, IPRI = IRQI, which implied that 4y = 6y - 2x + 1. This simplified to 2y - 2x - 1 ............... (2) Solving equations (1) and (2) simultaneously gave 2y - 2(12y - 27) = -1. so that y = 2(\(\frac{1}{2}\)). Then, x = 12(\(\frac{5}{2}\)) - 27 = 30 - 27 = 3 (b) The lengths of the sides of the triangle were also obtained as |PO| = 2y + x = 2(\(\frac{5}{2}\)) + 3 = 8 cm. |QR| = 4y = 4(\(\frac{5}{2}\)) = 10 cm and IPRI = 6y - 2x + 1 = 6(\(\frac{5}{2}\)) - 2(3) + 1 = 10 cm

(a) x < 4 or -1 \(\geq\) x. The range of values of x is -1 \(\geq\) x < 4. (b) First, < PQR + 56^\circ= 180^\circ so that < PQR = 180^\circ- 56^\circ= 124^\circ From triangle PQR, < PQR + < QRP+ < QPR = 180^\circ. But IRQI = IQPI so that < QRP = < QPR. 124^\circ + 2 < QRP = 180^\circ and < QRP =28^\circ. But < PRS = 90^\circ (angle on a semi-circle), therefore, < QRS = 28^\circ+ 90^\circ = 118^\circ.

(a) They were expected to obtain 20,000 = \(\frac{x^2}{8} + 5x\) and thereafter multiply through by 8 and simplify to get \(x ^2\) + 40x - 16,000. Using the quadratic formula x = \(\frac {-b \pm \sqrt{b^2 - 4ac} }{2a}\) where a = 1, b = 40 and c = - 160,000 x = \(\frac{-40 \pm \sqrt{(40)^2 - 4(1) (-160000)}} {2(1)}\) = \(\frac{-40 \pm \sqrt{(40)^2 - 4(1) (-160000)}} {2}\) = \(\frac{-40 \pm \sqrt{1600 + 640000}}{2}\) = 38.4997 or -420.4997 The number of books required to make the profit approximately 380,500, correct to the nearest whole number. (b) The required diagram is From triangle PQR, tan x = \(\frac{y}{2}\)W y = 2w tan x..........(1) Also, from \(\bigtriangleup\) PTQ, tan 28\(^o\) = \(\frac{y}{w}\) y = w tan 28\(^o\).........(2) Equating (1) and (2) yielded 2w tan X = w tan 28\(^o\) so that tanX = \(\frac{\text{w} tan 28^o}{2w}\) = \(\frac{0.5317}{2}\) which implies that tan X = 0.2659 X = tan\(^{-1}\)(0.2659) = 15.42\(^o\). The angle of elevation of R from P = 15\(^o\), correct to the nearest degree.

(a). They integrated the given function correctly to obtain (\(\frac{x^4}{2} - 2x^2 + 6x\)). Substituting the limits 2 and 1 for x and simplifying yielded 12 - \(\frac{9}{2}\) = 7\(\frac{1}{2}\) (b). Find the determinant of which was P(\(^{-1}\))\(^{-1}\) = |P=\(^{-1}\)| (-3 x -1) - (4 x 1) = -1. Using the concept of P(\(^{-1}\))\(^{-1}\) = P, then, P = -1 \(\begin{pmatrix} -3 & -1 \\ 4 & -3 \end{pmatrix}\) = \(\begin{pmatrix} 3 & 1 \\ 4 & 1 \end{pmatrix}\)

(a). Completing the table as follows \(\oplus\) 2 3 4 2 4 0 1 3 0 1 2 4 1 2 3 \(\otimes\) 2 3 4 2 4 1 3 3 1 4 2 4 3 2 1 (a) (ii) (a), they obtained the value of e as 1. (a) (ii) (b), the value of n was 2. (b), they couldn't answer this question correctly. If p = q, then: (i) q \(\Rightarrow\) p, this is invalid (ii) \(sin\) q \(\Rightarrow\) \(sin\) p, this is valid (iii) \(sin\) p \(\Rightarrow\) \(sin\) q, this is invalid

(a) The following solution was expected from the candidates (b) (i) |QK| = 9.3cm (ii) < QRS = 55\(^o\)

(a) Marks Class Boundaries Frequency Cumulative Frequency 10 - 19 9.5 - 19.5 4 4 20 - 29 19.5 - 29.5 7 11 30 - 39 29.5 - 39.5 12 23 40 - 49 39.5 - 49.5 18 41 50 - 59 49.5 - 59.5 20 61 60 - 69 59.5 - 69.5 14 75 70 - 79 69.5 - 79.5 9 84 80 - 89 79.5 - 89.5 4 88 90 - 99 89.5 - 99.5 2 90 (b) The cumulative frequency curve (c)(i) using the curve, the median was 51.5. (ii) Using the curve, the estimate of tbe number of students who scored at least 75% was 90 - 80 = 10 Therefore, Prob(that a student selected at random obtained distinction) = \(\frac{10}{90} = \frac{1}{9}\)

(a) (i) The diagram can be represented as: (ii) first they were expected to convert from litres to cm\(^3 \) by using the fact that 1000 cm\(^3\) = 1 litre, then 4.158 litres = 4158 cm\(^3\). If the volume and height of the container were V\(_1\) and h\(_1\) respectively, then, V\(_1\) = \(\frac{1}{3}\) x \(\frac{22}{7}\) x 21 x 21 x h\(_1\) = 4,158 which simplified to 462h\(_1\) = 4,158. Therefore, h\(_1\) = 9cm. (b) let h\(_2\) be the height of the water remaining in the container. Then by similar triangles, \(\frac{h_2}{9}\) = \(\frac{14}{21}\), h\(_2\) = 6 cm. If we let V\(_2\) be the volume of water remaining in the container, V\(_2\) = \(\frac{1}{3} \times \frac{22}{7}\) x 14 x 14 x 6 = 1,232 cm\(^2\). Volume of water drawn out = (4,158 — 1,232) cm\(^3\) = 2,926 cm\(^3\)