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5 + 6 + 4 + 8 + 5 + 2 + x = 35 30 + x = 35 x = 35 - 30 x = 5

at least 2 kinds of food two or more foods i.e(R and p) + (p and Y) + (y and R) + R and y and p = 6 + 2 + 4 + 5 = 17

x : y = 3.5, y : z = 4 : 7 x = 3 x 4 = 12 y = 5 x 4 = 20 y = 4 x 5 = 20 z = 7 x 5 = 35 then x : y : z = 12 : 20 : 35

2y + Reflex angle = 360 Reflex angle = 360 - 2y 360 - 2y = 2(106)(angles at a centre = 2ce < of circum.) 360 - 212 = 2y 148 = 2y y \(\frac{148}{2}\) y = 74o

\(\frac{n}{4} = \frac{m}{4 + 12}\) \(\frac{n}{4} = \frac{m}{16}\) 4m = 16n \(\frac{4m}{4} = \frac{16n}{4}\) m = 4n

Area of shaded portion = Area of semicircle Area of \(\bigtriangleup\) RSO Area of semicircle = \(\frac {\pi r^2}{2} = \frac{22\times 4 \times 4}{7 \times 2}\) = 25.14cm2; Area of \(\bigtriangleup\)RSO =\(\sqrt{s(s - 1)(s - b)(s - c)}\); where s = \(\frac{a + b + c}{2}\) s = \(\frac{4 + 4 + 4}{2}\) = 6cm = \(\sqrt{6(6 - 4)(6 - 4) (6 - 4)}\) = \(\sqrt{6(2) (2) (2)}\) = \(\sqrt{18}\) = 6.93cm2 Area of shaded region = 25.14 - 6.93 = 18.21cm2

Convert \( (42_5) \) to base three: First, convert to decimal: \(42_5 = 4 \times 5 + 2 = 22_{10}\) Now convert \( (22_{10}) \) to base three: 22 \(\div 3 = 7 \quad ; \, r1 \) 7 \(\div 3 = 2 \quad ; \, r1 \) 2 \(\div 3 = 0 \quad ; \, r2\) Read the remainders from bottom to top: \(22_{10} = 211_3\).

(a) \((3.69 \times 10^{5}) \div (1.64 \times 10^{-3})\) = \(\frac{3.69}{1.64} \times 10^{5 - (-3)}\) = \(2.25 \times 10^{8}\) (b) Interest from bank A : \(\frac{y}{100} \times N20,000 = N200y\) Interest on bank B :\(\frac{1.5y}{100} \times N25,000 = N375y\) \(\therefore N(200y + 375y) = N4,600\) \(575y = 4,600 \implies y = 8\)

(a) \(\frac{x^{2} - 8x + 16}{x^{2} - 7x + 12}\) = \(\frac{(x - 4)(x - 4)}{(x - 4)(x - 3)}\) = \(\frac{(x - 4)}{(x - 3)}\) (b) Arithmetic Progression: \(\frac{1}{2}, \frac{1}{x}, \frac{1}{3}\) \(\frac{1}{x} - \frac{1}{2} = \frac{1}{3} - \frac{1}{x}\) \(\frac{2 - x}{2x} = \frac{x - 3}{3x}\) \(\frac{2 - x}{x - 3} = \frac{2x}{3x}\) \(\implies \frac{2 - x}{x - 3} = \frac{2}{3}\)

From \(\Delta\) BCD, \(\sin \theta = \frac{50}{56}\) \(\theta = \sin^{-1} (0.8929)\) \( < BCD = 63.23^\circ\) \(\therefore < ACB = 63.23^\circ \times 2 = 126.46^\circ\) \(< ACB (reflex) = 360^\circ - 126.46^\circ\) = \(233.54^\circ\) Perimeter of the arc = \(\frac{233.54}{360} \times 2 \times \frac{22}{7} \times 56\) = \(228.35 m\) Perimeter of the railway tunnel = Perimeter of the arc + |AB| = 228.35 m + 100 m = 328.35 m.

(a) \(P(Good) = \frac{18}{24} = \frac{3}{4}\) \(P(Bad) = \frac{6}{24} = \frac{1}{4}\) P(2 are good and 1 bad) = \(\frac{3}{4} \times \frac{3}{4} \times \frac{1}{4}\) = \(\frac{9}{64}\) (b) P(the 3 are bad) = \(\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4}\) = \(\frac{1}{64}\) (c) P(the 3 are good) = \(\frac{3}{4} \times \frac{3}{4} \times \frac{3}{4}\) = \(\frac{27}{64}\)

(a) \(\frac{3\frac{1}{12} + \frac{7}{8}}{2\frac{1}{4} - \frac{1}{6}}\) \(3\frac{1}{12} + \frac{7}{8} = \frac{37}{12} + \frac{7}{8}\) = \(\frac{74 + 21}{24}\) = \(\frac{95}{24}\) \(2\frac{1}{4} - \frac{1}{6} = \frac{9}{4} - \frac{1}{6}\) = \(\frac{27 - 2}{12}\) = \(\frac{25}{12}\) \(\therefore \frac{3\frac{1}{12} + \frac{7}{8}}{2\frac{1}{4} - \frac{1}{6}} = \frac{95}{24} \div \frac{25}{12}\) \(\frac{95}{24} \times \frac{12}{25} = \frac{19}{10}\) = \(1.9\) (b)(i) \(p = \frac{m}{2} - \frac{n^{2}}{5m}\) \(\frac{n^{2}}{5m} = \frac{m}{2} - p\) \(n^{2} = 5m(\frac{m}{2} - p)\) \(n = \pm \sqrt{5m(\frac{m}{2} - p)}\) (ii) When p = 14 and m = -8, \(n = \sqrt{5(-8)(\frac{-8}{2} - 14)}\) \(n = \sqrt{-40(- 4 - 14)}\) \(n = \sqrt{720}\) \(n = 12\sqrt{5}\) = \(\pm 26.83\)

(a) No Log 0.004592 \(\bar{3}.6620\) \((0.004592)^{\frac{1}{3}}\) \(\frac{\bar{3}.6620}{3} = \bar{1}.2207\) Antilog of \(\bar{1}.2207 = 0.1663\) (b) \(\log_{10} y + 3 \log_{10} x = 2\) \(\log_{10} y + \log_{10} x^{3} = 2\) \(\log_{10} (yx^{3}) = 2\) \(yx^{3} = 10^{2}\) \(yx^{3} = 100\) \(y = \frac{100}{x^{3}}\) (c) \(3x - 2y = 21 .... (1)\) \(4x + 5y = 5 ..... (2)\) Multiply (1) by 4 and (2) by 3, \(12x - 8y = 84 .... (1a)\) \(12x + 15y = 15 ... (2a)\) (1a) - (2a), \(- 8y - 15y = 84 - 15\) \(- 23y = 69 \implies y = -3\) \(3x - 2y = 21 \implies 3x - 2(-3) = 21\) \(3x + 6 = 21 \implies 3x = 15\) \(x = 5\)

The diagram above is the solution