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(i) \(2\frac{2}{3} - (2\frac{1}{2} - 1\frac{4}{5})\) = \(2\frac{2}{3} - (\frac{5}{2} - \frac{9}{5})\) = \(\frac{8}{3} - (\frac{25 - 18}{10})\) = \(\frac{8}{3} - \frac{7}{10}\) = \(\frac{80 - 21}{30}\) = \(\frac{59}{30}\) (ii) \(\frac{3.25 - 1.64}{2.47 - 2.01}\) = \(\frac{1.61}{0.46}\) = \(\frac{7}{2}\) = \(3\frac{1}{2}\)

(a) No Log \(\sqrt[3]{1.376}\) \(0.1386 \div 3 \) = \(0.0462\) \(\sqrt[4]{0.007}\) \(\bar{3}.8451 \div 4\) = \(\bar{1}.4613\) \(\frac{\sqrt[3]{1.376}}{\sqrt[4]{0.007}}\) = \(0.5849\) Antilog = 3.845 \(\therefore \frac{\sqrt[3]{1.376}}{\sqrt[4]{0.007}} \approxeq 3.85\) (to 3 s.f) (b) \(\frac{\log 81}{\log \frac{1}{3}}\) = \(\frac{\log 3^{4}}{\log 3^{-1}}\) = \(\frac{4 \log 3}{-1 \log 3}\) = \(-4\)

(a) (i) Nobody takes Economics and Chemistry \(n(E \cap C) = \null \). (ii) 4 pupils take Economics and Government; \(n(E \cap G) = 4\) (b) (i) Suppose \(n(C \cap G) = x\) \(22 = 8 + 4 + 8 - 4 - x + x + 7 - x\) \(22 = 23 - x \implies x = 1\) Only 1 person does Chemistry and Government. (ii) Number taking only Government = \(8 - (4 + x)\) \(8 - (4 + 1) = 3\)

Total number of balls = 9, No of blue = 2, No of black = 3 \(\therefore\) No of red = 4. (a)(i) P(not blue) = \( 1 - P(blue)\) = \(1 - \frac{2}{9} = \frac{7}{9}\) (ii) P(not red) = \(1 - P(red)\) = \(1 - \frac{4}{9} = \frac{5}{9}\) (b) Picking two balls; (i) P(black) with no replacement = \(\frac{3}{9} \times \frac{2}{8} = \frac{1}{12}\) (ii) P(blue) with replacement = \(\frac{2}{9} \times \frac{2}{9} = \frac{4}{81}\)

(i) Sum of interior angles of a regular polygon = \((2n - 4) \times 90^\circ\) For a pentagon, \((2(5) - 4) \times 90^\circ = 6 \times 90^\circ = 540^\circ\) Interior angle = \(\frac{540^\circ}{5} = 108^\circ\) (ii) < BXA = 108^\circ - 90^\circ = 18^\circ < ABX = 108^\circ (as shown above) \(\therefore < BXA = 180^\circ - (108^\circ + 18^\circ) = 54^\circ\)

(a) \(2x^{2} + 7x - 11 = 0\) Using the quadratic formula, \(a = 2, b = 7, c = -11\) \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) \(x = \frac{-7 \pm \sqrt{7^{2} - 4(2)(-11)}}{2(2)}\) \(x = \frac{-7 \pm \sqrt{49 + 88}}{4}\) \(x = \frac{-7 \pm 11.705}{4}\) \(x = \frac{-7 + 11.705}{4} ; \frac{-7 - 11.705}{4}\) \(x = \frac{4.705}{4} = 1.17625\) or \(x = \frac{-18.705}{4} = -4.67625\) \(x \approxeq 1.18\) or \(x \approxeq -4.68\) (b) \(\frac{2}{x} + \frac{1}{y} = 3\) \(\frac{1}{x} - \frac{5}{y} = 7\) Substituting using \(P = \frac{1}{x}; Q = \frac{1}{y}\), we have \(2P + Q = 3 ... (i)\) \(P - 5Q = 7 ... (ii)\) Multiply (ii) by 2, we have \(2P - 10Q = 14 ... (iii)\) (iii) - (i) : \(-10Q - Q = 14 - 3 \implies -11Q = 11\) \(Q = -1\) Substitute Q = -1 in (i), \(2P - 1 = 3 \implies 2P = 3 + 1 = 4\) \(\implies P = 2\) P = 2; Q = -1 But \(P = \frac{1}{x} \implies 2 = \frac{1}{x}\) \(\therefore x = \frac{1}{2}\) Also, \(Q = \frac{1}{y} \implies -1 = \frac{1}{y}\) \(y = -1\) \(\therefore x = \frac{1}{2} ; y = -1\)

(a) Sheets expected from 5 packets = \(480 \times 5 = 2400\) Actual sheet contained in the packets = \(435 + 420 + 405 + 415 + 440 = 2115\) Error in supply = 2400 - 2115 = 285 sheets. \(\therefore\) Percentage error = \(\frac{285}{2400} \times 100% = 11.875%\) \(\approxeq 12%\) (b) If 1 packet = 480 sheets cost N35.00 \(\therefore \text{2400 sheets cost } \frac{2400}{480} \times N35.00 = N175.00\) Amount of buyer was cheated = \(\frac{Error \times N175}{\text{Expected no of sheets}}\) \(\frac{N285 \times 175}{2400} = N20.78\) The buyer was cheated by N21 to the nearest naira.

\(y = 3 + 2x - x^{2}\) and \(y = 2x - 3\) Table of values for the equation for \(-3 \leq x \leq 4\) x -3 -2 -1 0 1 2 3 4 \(y = 3 + 2x - x^{2}\) -12 -5 0 3 4 3 0 5 \(y = 2x - 3\) -9 -7 -5 -3 -1 1 3 5 (i) \(6 - x^{2} = 0\) \(6 - 3 + 2x - x^{2} = 2x - 3\) \(3 + 2x - x^{2} = 2x - 3\) \(\therefore y = 2x - 3\) Read the point where the two equations intersect on the graph. x = -2.6 and x = 2.5. (ii) Maximum value of \(3 + 2x - x^{2}\) is at y = 4. (iii) Range for which \(3 + 2x - x^{2} \leq 1\) is represented by the shaded portion in the graph.

(a) Given: Circle ABC with centre O To prove : < AOB = 2 < ACB Construction: Join CO produced to P. Proof: With lettering as in the figure, OA = OB (radii) ; \(x_{1} = x_{2}\) (base angles of an isosceles triangle) \(\therefore < AOP = x_{1} + x_{2}\) (exterior angle of triangle AOC) \(\therefore < AOP = 2x_{2} (x_{1} = x_{2})\) Also, \(< BOP = 2y_{2}\) (similar proof as the earlier done ones) \(\therefore < AOB = 2x_{2} + 2y_{2} = 2(x_{2} + y_{2})\) \(< AOB = 2 \times < ABC\) (proven) (b) From the diagram, < OQR = 32^\circ and < MPQ = 15^\circ. (i) \(\therefore < ORQ = 32^\circ\) (base angles of an isosceles triangle) \(< OQR = 180^\circ - (32^\circ + 32^\circ) = 180^\circ - 64^\circ = 116^\circ\) \(\therefore < QPR = \frac{116^\circ}{2} = 58^\circ\) (angle subtended at the centre) (ii) Join MO and MQ, Since < MPQ = 15^\circ then < MOQ = 2 < MPQ (angle at the centre) \(\therefore < MOQ = 2 \times 15^\circ = 30^\circ\) \(\therefore < OMQ = < MQO \) (base angles of an isosceles triangle) \(\therefore < MQO = \frac{180^\circ - 30^\circ}{2} = 75^\circ\)

Waiting time (in mins) Class boundaries No of customers Cum Freq 1.5 - 1.9 1.45 - 1.95 3 3 2.0 - 2.4 1.95 - 2.45 10 13 2.5 - 2.9 2.45 - 2.95 18 31 3.0 - 3.4 2.95 - 3.45 10 41 3.5 - 3.9 3.45 - 3.95 7 48 4.0 - 4.4 3.95 - 4.45 2 50 (b) (c) Interquartile range of the distribution = \(Q_{3} - Q_{1}\) \(Q_{3} = \frac{3 \times 51}{4} = 38.25\) \(Q_{1} = \frac{1 \times 51}{4} = 12.75\) From the graph, \(Q_{3} = 3.25 ; Q_{1} = 2.45\) \(\therefore \text{The interquartile range of the distribution } = 3.25 - 2.45 = 0.80\) (ii) Proportion of customers who must have waited for more than 3 minutes = \(\frac{16}{50} = \frac{8}{25}\)

(a) & (b)

Let T and B be the top and base of the control tower respectively; P, the point of pilot's observation and PS, the line of observation. (a) Shortest distance is |PB|. \(\therefore \cos 33^\circ = \frac{8.5}{PB}\) \(PB = \frac{8.5}{\cos 33} = \frac{8.5}{0.8387}\) \(|PB| = 10.135km \approxeq 10.1km\) (b) Height of the control tower = |SB| - |ST| \(\therefore \tan 33^\circ = \frac{|SB|}{8.5 km} \implies |SB| = 8.5 \tan 33^\circ\) \(|SB| = 5.52 km\) \(\tan 30^\circ = \frac{|ST|}{8.5 km} \implies |ST| = 8.5 \tan 30^\circ\) \(|ST| = 4.907 km\) \(\therefore\) The height of the tower = 5.52 km - 4.907 km = 0.613 km.

first work out the expression and then correct the answer to 4 s.f = 241.34..............(A) (3 x 10-\(^3\))\(^2\)............(B) = 3\(^2\)x\(^2\) = \(\frac{1}{10^3}\) x \(\frac{1}{10^3}\) (Note that x\(^2\) = \(\frac{1}{x^3}\)) = 24.34 x 3\(^2\) x \(\frac{1}{10^6}\) = \(\frac{2172.06}{10^6}\) = 0.00217206 = 0.002172(4 s.f)

\(\sqrt{27}\) + \(\frac{3}{\sqrt{3}}\) = \(\sqrt{9 \times 3}\) + \(\frac{3 \times {\sqrt{3}}}{{\sqrt{3}} \times {\sqrt{3}}}\) = 3\(\sqrt{3}\) + \(\sqrt{3}\) = 4\(\sqrt{3}\)

\(\frac{4\frac{3}{4} - 6\frac{1}{4}}{4\frac{1}{5} \text{ of } 1\frac{1}{4}}\) \(\frac{19}{4}\) - \(\frac{25}{4}\)............(A) \(\frac{21}{5}\) x \(\frac{5}{4}\).............(B) Now work out the value of A and the value of B and then find the value \(\frac{A}{B}\) A = \(\frac{19}{4}\) - \(\frac{25}{4}\) = \(\frac{-6}{4}\) B = \(\frac{21}{5}\) x \(\frac{5}{4}\) = \(\frac{105}{20}\) = \(\frac{21}{4}\) But then \(\frac{A}{B}\) = \(\frac{-6}{4}\) \(\div\) \(\frac{21}{4}\) = \(\frac{-6}{4}\) x \(\frac{4}{21}\) = \(\frac{-24}{84}\) = \(\frac{-2}{7}\)