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(a) see figure above a \(< QTR = < TRP = 70^\circ\) (alternate angle) \(< MNR = < TRP = 70^\circ\) (alternate angle; MN || ST) \(< MNR + < PNM = 180^\circ\) \(70^\circ + < PNM = 180^\circ \implies < PNM = 110^\circ\) \(\therefore x = 110^\circ\) (b)see figure b above \(< BDC = < BCD = 50^\circ\) (BD = BC) \(\therefore \Delta BDC = isosceles\) \(< ABD = 2(50^\circ) = 100^\circ\) (exterior angle) \(55^\circ + < BDA + 100^\circ = 180^\circ\) \(< BDA = 180^\circ - 155^\circ = 25^\circ\)

(a) x -3 -2 -1 0 1 2 3 4 \(y = x^{2}\) 9 4 1 0 1 4 9 16 (b) \(y = 2x + 3\) x -3 -2 -1 0 1 2 3 4 2x -6 -4 -2 0 2 4 6 8 3 3 3 3 3 3 3 3 3 y -3 -1 1 3 5 7 9 11 see graph above (c)(i) The roots of the equation are -1 and 3, from the graph. (ii) The gradient of \(y = x^{2}\) at x = -2 : \(\frac{8}{-2.5} = -3.2\)

(a) Frequency Distribution Table Birth 45 - 54 55 - 64 65 - 74 75 - 84 85 - 94 95 - 104 105 - 114 Freq(f) 1 2 15 21 7 3 1 (b) see histogram above. (c) Modal class is 75 - 84 Mode = \(\frac{75 + 84}{2} = 79.5\) (d) Birth freq (f) class midpoint (x) \(fx\) 45 - 54 1 49.5 49.5 55 - 64 2 59.5 119 65 - 74 15 69.5 1042.5 75 - 84 21 79.5 1669.5 85 - 94 7 89.5 626.5 95 - 104 3 99.5 298.5 105 - 114 1 109.5 109.5 Total 50 3915 Mean \(\bar{x} = \frac{\sum fx}{\sum f}\) = \(\frac{3915}{50}\) = \(78.3\).

(a)(i) (ii) \(21 + 4 + 5 + 10 + y + 35 - y + 32 - y = 100\) \(107 - y = 100\) \(y = 107 - 100 = 7\) Therefore, 7 traders sell oranges and mangoes alone. (b) \(312_{four} + 52_{x} = 96_{ten}\) \(312_{4} = 3 \times 4^{2} + 1 \times 4^{1} + 2 \times 4^{0}\) = \(48 + 4 + 2 = 54_{10}\) \(52_{x} = 5 \times x^{1} + 2 \times x^{0}\) = \((5x + 2)_{10}\) \(\therefore 54 + (5x + 2) = 96\) \(56 + 5x = 96 \implies 5x = 96 - 56 = 40\) \(x = 8\) \(\therefore 312_{4} + 52_{8} = 96_{10}\)

(a)(i) P(A) = \(\frac{2}{3}\) ; P(B) = \(\frac{5}{8}\) ; P(C) = \(\frac{3}{4}\). \(P(A \cap B \cap C) = P(A) \times P(B) \times P(C)\) = \(\frac{2}{3} \times \frac{5}{8} \times \frac{3}{4} = \frac{5}{16}\) (ii) P(A') = \(1 - \frac{2}{3} = \frac{1}{3}\) P(B') = \(1 - \frac{5}{8} = \frac{3}{8}\) P(C') = \(1 - \frac{3}{4} = \frac{1}{4}\) \(P(A' \cap B' \cap C') = P(A') \times P(B') \times P(C')\) = \(\frac{1}{3} \times \frac{3}{8} \times \frac{1}{4} \) = \(\frac{1}{32}\) (iii) P(A and B pass, C fails) = \(\frac{2}{3} \times \frac{5}{8} \times \frac{1}{4} = \frac{5}{48}\) P( A and C pass, B fails) = \(\frac{2}{3} \times \frac{3}{4} \times \frac{3}{8} = \frac{3}{16}\) P(B and C pass, A fails) = \(\frac{5}{8} \times \frac{3}{4} \times \frac{1}{3} = \frac{5}{32}\) P(only two of the boys pass) = \(\frac{5}{48} + \frac{3}{16} + \frac{5}{32} = \frac{43}{96}\) (b) Let the cost price be x % gain = \(\frac{\text{actual gain}}{\text{cost price}} \times 100%\) \(20 = \frac{36,000 - x}{x} \times 100%\) \(\frac{36,000 - x}{x} = \frac{1}{5}\) \(5(36,000 - x) = x \implies 180,000 - 5x = x\) \(180,000 = x + 5x = 6x\) \(x = N30,000\) Discount (5%) : \(\frac{5}{100} \times N36,000 = N1,800\) Selling price : \(N(36,000 - 1,800) = N34,200\) Actual gain : \(\frac{34,200 - 30,000}{30,000} \times 100% \) = 14%

(a) \(2x^{2} + (p + 1)x + q = 0\) The roots of the equation are 1 and 3, therefore (x - 1)(x - 3) = 0. Equation : \(x^{2} - 3x - x + 3 = 0\) \(x^{2} - 4x + 3 = 0\) \(\therefore 2x^{2} + (p + 1)x + q \equiv x^{2} - 4x + 3\) \(\frac{2x^{2} + (p + 1)x + q}{2} = x^{2} + \frac{(p + 1)}{2} x + \frac{q}{2}\) \(\implies \frac{q}{2} = 3\) \(q = 6\) \(\frac{p + 1}{2} = -4 \implies p + 1 = -8\) \(p = -9\) (p, q) = (-9, 6). (b) \(W \propto \frac{1}{d^{2}} \) \(W = \frac{k}{d^{2}}\) \(k = Wd^{2}\) \(W = \frac{80 \times (6400)^{2}}{d^{2}}\) (equation of variation). \(W = \frac{80 \times (6400)^{2}}{(6400 + 800)^{2}}\) \(W = \frac{80 \times (6400)^{2}}{(7200)^{2}}\) = 63.21 kg

< PAB = < ABE = 53^\circ (alternate angles) < CBE = 180^\circ - < DBC = 180^\circ - 161^\circ = 19^\circ < ABC = < ABE + < CBE = 53^\circ + 19^\circ = 72^\circ (a) In \(\Delta ABC\), \(AC^{2} = AB^{2} + BC^{2} - 2(AB)(BC) \cos < ABC\) \(AC^{2} = 15^{2} + 18^{2} - 2(15)(18) \cos 72\) = \(225 + 324 - 540 \cos 72\) = \(549 - 166.869\) \(AC^{2} = 382.131\) \(AC = \sqrt{382.131} = 19.548m\) \(\approxeq 19.5m\) (b) \(\frac{\sin A}{18} = \frac{\sin 72}{19.548}\) \(\sin A = \frac{18 \times \sin 72}{19.548}\) \(\sin A = 0.8757\) \(A = \sin^{-1} (0.8757) = 61.13^\circ\) The bearing of C from A = 61.13^\circ + 53^\circ = 114.13^\circ \(\approxeq\) 114^\circ. (c) In \(\Delta ATB\), \(\frac{15}{BT} = \tan 58\) \(BT = \frac{15}{\tan 58}\) \(9.373 m \approxeq 9.37m\)

(a) \(d = 15 cm \implies r = \frac{15}{2} cm\) \(V = \frac{4\pi r^{3}}{3}\) \(V = \frac{4}{3} \times \frac{22}{7} \times (\frac{15}{2})^{3}\) \(V = \frac{1125 \times 22}{7 \times 2}\) \(V = 1,767.857 cm^{3}\) (b)(i) Longitude difference : 105^\circ + 35^\circ = 140^\circ Distance along the parallel of latitude = \(\frac{140}{360} \times 2 \times \frac{22}{7} \times 6400 \cos 28\) = \(\frac{140 \times 44 \times 6400 \times 0.8829}{360 \times 7}\) = \(13812.48 km\) \(\approxeq 13800 km\) (ii) Distance from the equator : \(\frac{28}{360} \times 2 \times \frac{22}{7} \times 6400\) = \(3128.89 km\) \(\approxeq 3130 km\)

From the diagram provided, x lies between -7 and or equal to 2, i.e. -7 < x \(\leq\) -1

"Moribund" means something in a state of decline or near death, and "crumbling" best fits this meaning.

"Transients" refers to people who stay temporarily, so "temporary guest" is the nearest in meaning.

"Coalescence" is the coming together or union of different groups, so the nearest meaning is "union."

"A serious reverse" refers to a setback or defeat after a period of success. Therefore, the rebels met with a serious defeat.

"Sumptuous" means rich, luxurious and expensive, so the clothes the village girl wore would be described as expensive.

"Radical changes" refers to new or innovative ideas that can be difficult to implement, so "innovations" is the most fitting choice.