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Length of arc, L = 21.4 - 2 x 4.2cm = 21.4 - 8.4 = 13cm But L = \(\frac{\theta}{360^o}\) x 2\(\pi r\) i.e 13 = \(\frac{\theta}{360^o}\) x 2 x \(\frac{22}{7}\) x 4.2 = 13 x 360\(^o\) x 7 = \(\theta\) x 2 x 22 x 4.2 \(\theta\) = \(\frac{13 \times 360^o \times 7}{44 \times 4.2}\) = \(\approx\) 177.27\(^o\) \(\approx\) 177\(^o\) (to the nearest degree)

From the diagram, h\(^2\) = 4\(^2\) + 3\(^2\) (pythagoras') h\(^2\) = 16 + 9 = 25 h = \(\sqrt{25}\) = 5 Hence, sin x - cos x = \(\frac{4}{5} - \frac{3}{5}\) = \(\frac{1}{5}\)

In \(\bigtriangleup\)YSC, sin 30\(^o\) = \(\frac{YS}{20}\) |YS| = 20 sin 30\(^o\) = 20 x 0.5 10m

Let the length of a side of the rhombus be n Then, n\(^2\) = 5\(^2\) + 12\(^2\) = 25 + 144 = 169 n = \(\sqrt{169}\) = 13cm Hence, perimeter of rhombus = 4n = 4 x 13 = 52cm

Let n(M \(\cup\) N \) = x Then 20 - x + x + 30 - x = n(M \(\cup\) N) 50 - x = 40 50 - 40 = x 10 = x x = 10 Hence, n(M \(\cup\N)' = 8 + (20 - 10) + (30 + 10) = 8 + 10 + 20 = 38

y = x\(^2\) ....(1) y = x ......(2) y = y x\(^2\) - x x\(^2\) - x = 0 x(x - 1) = 0 x = 0 or x - 1 = 0 x = 0 or x = 1 when x = 0, y = 0\(^2\) = 0 when x = 1, y = 1\(^2\) = 1 Hence; the two graphs interest at (0, 0) and (1, 1)

Beginning of 1st year (March 2011 - February 2012) - N900,000 Depreciation (30%) = \(\frac{30}{100} \times N900,000\) = N270,000 Price at the end of 1st year = N(900,000 - 270,000) = N630,000. Beginning of 2nd year (March 2012 - February 2013) - N630,000 Depreciation (22%) = \(\frac{22}{100} \times N630,000\) = N138,600 Price at the end of 2nd year = N(630,000 - 138,600) = N491,400. Beginning of 3rd year (March 2013 - February 2014) - N491,400 Depreciation (22%) = \(\frac{22}{100} \times N491,400\) = N108,108 Price at the end of the 3rd year = N(491,400 - 108,108) = N383,292. Beginning of 4th year (March 2014 - February 2015) - N383,292 Depreciation (22%) = \(\frac{22}{100} \times N383,292\) = N84,324.24 Price at the end of the 4th year = N(383,292 - 84,324.24) = N298,967.76 \(\approxeq\) N299,000 (to the nearest hundred naira).

(a) \(y = 2px^{2} - p^{2}x - 14\) At point (3, 10), y = 10 when x = 3. \(\implies 10 = 2p(3^{2}) - p^{2}(3) - 14\) \(10 = 18p - 3p^{2} - 14\) \(-10 = 3p^{2} - 18p + 14 \implies 3p^{2} - 18p + 24 = 0\) \(3p^{2} - 12p - 6p + 24 = 0\) \((p - 4)(3p - 6) = 0 \implies \text{p = 2 or 4}\). (b) \(3y - 2x = 21 ... (1)\) \(4y + 5x = 5 ....(2)\) Using elimination method, multiply (1) by 4 and (2) by 3. \((1) \times 4 : 12y - 8x = 84 ... (3)\) \((2) \times 3 : 12y + 15x = 15 ... (4)\) Subtracting (3) - (4), we have: \(-23x = 69 \implies x = -3\) Putting x = -3 in (1), we have \(3y - 2(-3) = 3y + 6 = 21\) \(3y = 15 \implies y = 5\) Hence, the coordinates of Q are (-3, 5).

(a) see diagram above Let a side of the rhombus be n. Then \(n^{2} = 4.65^{2} + 5.1^{2}\) = \(21.6225 + 26.01 \) \(n^{2} = 47.6325\) \(n = \sqrt{47.6325} = 6.902 cm\) Hence, the perimeter of the rhombus = 4n = \(4 \times 6.902\) = \(27.608 cm \approxeq 27.6 cm\) (to 1 d.p) (b) Right-angled triangle ABC see diagram above \(\sin x = \frac{3}{5}\), Thus, \(m^{2} = 5^{2} - 3^{2} = 25 - 9 = 16\) \(m = 4\). \(\cos x = \frac{4}{5} \implies 5\cos x = 5 \times \frac{4}{5} = 4\) \(\tan x = \frac{3}{4} \implies 4\tan x = 4 \times \frac{3}{4} = 3\) \(5\cos x - 4\tan x = 4 - 3 = 1\).

(a) (i) In the diagram, \(x^\circ + 90^\circ = (3x + 15)^\circ\) (Sum of opposite interior angles) Thus, \(90^\circ - 15^\circ = 3x - x \implies 75^\circ = 2x\) \(x = 37.5^\circ\) (ii) < RSQ = 90^\circ - 37.5^\circ = 52.5^\circ (b) \(2N4_{seven} = 15N_{nine}\) \(2N4_{seven} = (2 \times 7^{2}) + (N \times 7^{1}) + (4 \times 7^{0})\) = \(98 + 7N + 4\) = \(102 + 7N\) \(15N_{nine} = (1 \times 9^{2}) + (5 \times 9^{1}) + (N \times 9^{0})\) = \(81 + 45 + N\) = \(126 + N\) \(\implies 102 + 7N = 126 + N\) \(7N - N = 126 - 102 = 24\) \(6N = 24 \implies N = 4\) Therefore, \(244_{seven} = 154_{nine}\).

(a) \(\frac{m + n + s + p + q}{5} = 12 \) \(\implies m + n + s + p + q = 12 \times 5 = 60\) \(\frac{(m + 4) + (n - 3) + (s + 6) + (p - 2) + (q + 8)}{5} = \frac{m + n + s + p + q + 4 - 3 + 6 - 2 + 8}{5}\) = \(\frac{60 + 13}{5} = \frac{73}{5}\) = 14.6. (b) In the community, there are 500 people. The 75th percentile age = 65 years Number of people in the 75th percentile = \(\frac{75 \times 500}{100} = 375\) The 25th percentile = 15 years Number of people in the 25th percentile = \(\frac{25 \times 500}{100} = 125\) Number of people between 15 years and 65 years = 375 - 125 = 250 people.

(a) (b) From the Venn Diagram above, 60% passed implies that 40% passed : 40% of 240 = 96. 28 + 2x + x + 12 + 6 + 6 + 8 = 96 60 + 3x = 96 3x = 36 x = 12 (i) Faulty brakes = 12 + 12 + 8 + 6 =38 (ii) Cars with only one fault = 28 + 2x + x = 28 + 3x = 28 + 3(12) = 28 + 36 = 64

(a) Using the two- point form, \(\frac{y - y_{1}}{y_{2} - y_{1}} = \frac{x - x_{1}}{x_{2} - x_{1}}\) \(\frac{y - 5}{-7 - 5} = \frac{x - 2}{-4 - 2}\) \(\frac{y - 5}{- 12} = \frac{x - 2}{- 6}\) \(y - 5 = 2(x - 2)\) \(y - 5 = 2x - 4 \implies y - 2x = -4 + 5 = 1\) \(Equation : y = 2x + 1\) (b) (i)see figure above In \(\Delta{\text PQR,}\) (ii) (1) In \(\Delta PQR, |QR|^{2} = 5^{2} + 8^{2}\) \(25 + 64 = 89\) \(|QR| = \sqrt{89} = 9.434 km \approxeq 9.43 km\) (2) In the diagram above, the bearing of R from Q is the obtuse angle NQR. But \(\tan \alpha = \frac{8}{5} = 1.6\) \(\alpha = \tan^{-1} (1.6) = 58^\circ\) Hence, angle NQR = 360^\circ - (a + 60^\circ + 90^\circ) = 360^\circ - (58^\circ + 60^\circ + 90^\circ) = 360^\circ - 208^\circ = 152^\circ

(a) S.P = \(C.P + \frac{% profit}{100} \times C.P\) = \(300 + (\frac{x}{100} \times 300)\) S.P = N(300 + 3x) Therefore, Bola bought it at N(300 + 3x). James paid \(N(6x + \frac{3}{4})\) extra from what Lamin paid, therefore Bola's S.P = \(N(300 + 6x + \frac{3}{4})\) = N(300.75 + 6x). Profit for Bola = \(N(300.75 + 6x - (300 + 3x)) = N(0.75 + 3x)\) \(\frac{x}{100} \times N(300 + 3x) = N(0.75 + 3x)\) \(300x + 3x^{2} = 75 + 300x\) \(\implies 3x^{2} = 75\) \(x^{2} = 25 \therefore x = 5\) (b) \(3x - 2 < 10 + x < 2 + 5x\) \(3x - 2 < 10 + x \implies 3x - x < 10 + 2\) \(2x < 12 \implies x < 6\) \(10 + x < 2 + 5x\) \(x - 5x < 2 - 10\) \(-4x < -8 \implies x > 2\) The range = \(2 < x < 6\)

\(|TQ|^{2} = 4^{2} + 6^{2} - 2 \times 4 \times 6 \times \cos 30^\circ\) = \(16 + 36 - 48 \times 0.8660\) = \(52 - 41.568 = 10.432\) \(\therefore |TQ| = \sqrt{10.432} \approxeq 3.23 cm\) (a) By the rules of similar triangles, \(\frac{|PT|}{|TQ|} = \frac{|PS|}{|SR|}\) \(\frac{4}{3.23} = \frac{10}{|SR|}\) \(|SR| = \frac{3.23 \times 10}{4}\) = \(8.075 cm \approxeq 8 cm\) (b) Using sine rule, \(\frac{|PQ|}{\sin \alpha} = \frac{|TQ|}{\sin 30}\) \(\frac{6}{\sin \alpha} = \frac{3.23}{\sin 30}\) \(\sin \alpha = \frac{6 \sin 30}{3.23}\) sin \(\alpha\) = 0.9288 From the diagram, \(\alpha\) = \(\beta\) \(\sin \alpha = \sin \beta = 0.9288\) \(\sin \beta = \frac{h}{6}\) h = \(6 \sin \beta\) Hence, area of quadrilateral TQRS = \(\frac{1}{2} (TQ + SR) \times h\) = \(\frac{1}{2} (3.23 + 8.075) \times 6\sin \beta\) = \(\frac{1}{2} (11.305) (6 \times 0.9288)\) = 31.501 cm\(^2\) \(\approxeq\) 32 cm\(^2\) (to the nearest whole number)