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(a) (b)(i) \(C \cap A' = {15}\) (ii) \(A' \cap (B \cup C) = {2, 6, 12, 15}\)

(a) (i) First discount = 20% of N25,000 = \(\frac{20}{100} \times N25,000 = N5,000\) Further discount for prompt payment = \(2\frac{1}{2}%\) of N25,000. = \(\frac{5}{200} \times N25,000 = N625\) \(\therefore \text{Total amount saved by distributor} = N(5,000 + 625)\) = \(N5625\) (ii) Total %age discount for prompt payment = \((20 + 2\frac{1}{2})% = 22\frac{1}{2}%\) Therefore, the distributor paid \((100% - 22\frac{1}{2}%) \times Nx\) = \(77\frac{1}{2}% \times Nx\) \(\implies 77\frac{1}{2}% \times x = 11,700\) \(x = \frac{11,700 \times 200}{155}\) = \(N15,096.77\) (b) \(6y^{2} - 149y - 102 \) = \(6y^{2} - 153y + 4y - 102\) = \(3y(2y - 51) + 2(2y - 51)\) = \((3y + 2)(2y - 51)\) \(\therefore 6y^{2} - 149y - 102 = 0 \implies (3y + 2)(2y - 51) = 0\) \(3y + 2 = 0 \implies 3y = -2 \) \(y = -\frac{2}{3}\) \(2y - 51 = 0 \implies 2y = 51\) \(y = 25\frac{1}{2}\) \(\therefore y = -\frac{2}{3} ; y = 25\frac{1}{2}\)

(a) \(\frac{3}{\sqrt{3}}(\frac{2}{\sqrt{3}} - \frac{\sqrt{12}}{6})\) = \(\frac{6}{\sqrt{9}} - \frac{3\sqrt{12}}{6\sqrt{3}}\) = \(2 - \frac{2\sqrt{3}}{2\sqrt{3}}\) = \(2 - 1 = 1\) (b)(i) \(< ABC = 180^\circ - 68^\circ = 112^\circ\) \(< AOC = 2 \times 112^\circ = 224^\circ\) \(\hat{O} = < AOC (obtuse)\) = \(360^\circ - 224^\circ = 136^\circ\) (ii) In \(\Delta AOC\), OA = OC (radii) \(\therefore \Delta AOC = isosceles\) \(\hat{A} = < OAC ; \hat{C} = < ACO ; \hat{A} = \hat{C} = x\) \(\hat{O} + \hat{A} + \hat{C} = 180^\circ\) \(136^\circ + x + x = 180^\circ \implies 2x = 44^\circ\) \(x = 22^\circ\) In \(\Delta BOC\), OB = OC (radii) \(\therefore \Delta BOC = isosceles\) \(< OCB = 49^\circ + 22^\circ= 71^\circ\) \(< OBC = < OCB = 71^\circ\) \(< OBC + < OCB + < BOC = 180^\circ\) \(71^\circ + 71^\circ + < BOC = 180^\circ\) \(< BOC = 180^\circ - 142^\circ = 38^\circ\)

(a) \(P(Ade) = \frac{2}{3} ; P(Kujo) = \frac{5}{8} ; P(Fati) = \frac{3}{4}\) \(P(\text{all three pass the exam}) = \frac{2}{3} \times \frac{5}{8} \times \frac{3}{4}\) = \(\frac{5}{16}\) (b) P(Ade fails) = \(1 - \frac{2}{3} = \frac{1}{3}\) P(Kujo fails) = \(1 - \frac{5}{8} = \frac{3}{8}\) P(Fati fails) = \(1 - \frac{3}{4} = \frac{1}{4}\) P(none passes) = \(\frac{1}{3} \times \frac{3}{8} \times \frac{1}{4}\) = \(\frac{1}{32}\). (c) P(Ade and Kujo only pass) = \(\frac{2}{3} \times \frac{5}{8} \times \frac{1}{4}\) = \(\frac{5}{48}\)

(a) \(\frac{1}{2}\log_{10} 25 - 2\log_{10} 3 + \log_{10} 18\) = \(\log_{10} (25^{\frac{1}{2}}) - \log_{10} (3^{2}) + \log_{10} 18\) = \(\log_{10} (\frac{5 \times 18}{9})\) = \(\log_{10} 10\) = 1 (b) \(123_{y} = (1 \times y^{2}) + (2 \times y^{1}) + (3 \times y^{0})\) \(y^{2} + 2y + 3 = 83\) \(y^{2} + 2y + 3 - 83 = 0 \implies y^{2} + 2y - 80 = 0\) Equation : \(y^{2} + 2y - 80 = 0\) \(y^{2} - 8y + 10y - 80 = 0\) \(y(y - 8) + 10(y - 8) = 0\) \((y + 10)(y - 8) = 0 \implies y = -10 ; 8\) y = 8 since y cannot be negative. (c) \(\frac{9^{2x - 3}}{3^{x + 3}} = 1\) \(\implies 9^{2x - 3} = 3^{x + 3}\) \((3^{2})^{2x - 3} = 3^{x + 3}\) \(3^{4x - 6} = 3^{x + 3}\) \(\implies 4x - 6 = x + 3\) \(4x - x = 3 + 6 \implies 3x = 9\) \(x = 3\)

(a) \(a = 3; d = 4 ; n = 28\) \(S_{n} = \frac{n}{2}(2a + (n - 1)d)\) \(S_{28} = \frac{28}{2}(2(3) + (28 - 1) 4)\) = \(14(6 + 108)\) = \(14(114)\) = \(1596\) (b) \(x = \frac{2m}{1 - m^{2}} ; y = \frac{2m}{1 + m}\) \(2x - y = 2(\frac{2m}{1 - m^{2}}) - (\frac{2m}{1 + m})\) = \(\frac{4m}{1 - m^{2}} - \frac{2m}{1 + m}\) = \(\frac{4m - 2m(1 - m)}{1 - m^{2}}\) = \(\frac{4m - 2m + 2m^{2}}{1 - m^{2}}\) = \(\frac{2m^{2} + 2m}{1 - m^{2}}\) = \(\frac{2m(m + 1)}{(1 + m)(1 - m)}\) = \(\frac{2m}{1 - m}\) (c) Sum of interior angles = (2n - 4)90^\circ When n = 5 (pentagon), = \((2(5) - 4) \times 90^\circ\) = \(540^\circ\) \(x + 2x + 3x + 2x + (3x - 10) = 11x - 10\) = \(11x - 10 = 540^\circ\) = \(11x = 550^\circ\) = \(x = 50^\circ\)

(a) Width = x m; Length = 2x m; Height = (x - 4) m. Total surface area = 2Lh + WL + 2Wh = \(2(2x)(x - 4) + x(2x) + 2(x)(x - 4) = 1440 m^{2}\) = \(4x^{2} - 16x + 2x^{2} + 2x^{2} - 8x = 1440\) = \(8x^{2} - 24x = 1440\) = \(x^{2} - 3x - 180 = 0\) = \((x - 15)(x + 12) = 0\) \(x = 15m\) Length = 2(15m) = 30m. (b) \(I = \frac{PRT}{100}\) \(I_{P} + I_{Q} = I_{T}\) \(\frac{3000 \times x \times 1}{100} + \frac{2000 \times y \times 1}{100} = 640\) \(30x + 20y = 640 \implies 3x + 2y = 64 .... (1)\) \(\frac{2000 \times x \times 1}{100} + \frac{3000 \times y \times 1}{100} = 660\) \(20x + 30y = 660 \implies 2x + 3y = 66 .... (2)\) To eliminate x , multiply (1) by 2 and (2) by 3. \(6x + 4y = 128 ... (1)\) \(6x + 9y = 198 .... (2)\) (2) - (1) : \(9y - 4y = 198 - 128 \implies 5y = 70\) \(y = 14\) \(2x + 3y = 66 \implies 2x + 3(14) = 66\) \(2x + 42 = 66 \implies 2x = 24\) \(x = 12\) \(\therefore x = 12 ; y = 14\).

(a) Area of the major sector = \(\frac{\theta}{360^\circ} \times \pi r^{2}\) = \(\frac{360 - 120}{360} \times \frac{22}{7} \times 7 \times 7\) = \((\frac{2}{3} \times 154) cm^{2}\) Volume = \(\text{Area of the major sector} \times \text{height (cross section)}\) = \(\frac{2}{3} \times 154 \times 15 = 1540 cm^{3}\) (b) Curved surface area = \(\frac{240}{360} \times 2\pi rh\) = \(\frac{240}{360} \times 2 \times \frac{22}{7} \times 7 \times 15\) = \(440 cm^{2}\) Area of rectangular surface = \(2 \times l \times b\) = \(2 \times 7 \times 15 = 210 cm^{2}\) Area of top and bottom of cylinder (major sector) = \(2 \times \frac{240}{360} \times \frac{22}{7} \times 7 \times 7\) = \(205.33 cm^{2}\) \(\therefore \text{Total surface area} = 440 + 205.33 + 210 = 855.33 cm^{2}\)

No expanded working has been attached for this question yet. The verified answer is Option .

(a)(i) < PQR = (180^\circ - 130^\circ) + 50^\circ = 100^\circ \(PR^{2} = PQ^{2} + QR^{2} - 2(PQ)(QR) \cos < PQR\) = \(8^{2} + 13^{2} - 2(8)(13) \cos 100\) = \(233 + 36.118\) \(PR^{2} = 269.118\) \(PR = 16.405 m \approxeq 16.4 m\) (ii) \(\frac{QR}{\sin < QPR} = \frac{PR}{\sin < PQR}\) \(\frac{13}{\sin < QPR} = \frac{16.4}{\sin 100}\) \(\sin < QPR = \frac{13 \times \sin 100}{16.4}\) \(\sin < QPR = 0.7806\) \(< QPR = 51.32^\circ \approxeq 51.3^\circ\) The bearing of R from P = 50^\circ + 51.3^\circ = 101.3^\circ \(\approxeq\) 101^\circ. (b) In \(\Delta PQD\), \(\frac{h}{8} = \sin 51.3\) \(h = 8 \sin 51.3\) = \(6.243 m \) \(\approxeq 6.24 m\) Area of \(\Delta PQR = \frac{1}{2} \times PR \times h\) \(\frac{1}{2} \times 6.24 \times 16.4 = 51.166 m^{2}\) \(\approxeq 51.2 m^{2}\)

(a) Class Interval Freq (f) Tally Mid-point (x) \(fx\) 45 - 49 1 | 47 47 50 - 54 3 ||| 52 156 55 - 59 6 |||| | 57 342 60 - 64 7 |||| || 62 434 65 - 69 6 |||| | 67 402 70 - 74 4 |||| 72 288 75 - 79 2 || 77 154 80 - 84 1 | 82 82 \(\sum\) 30 1905 (b) (c) Mean \(\bar{x} = \frac{\sum fx}{\sum f}\) = \(\frac{1905}{30}\) = \(63.5\). (d) P(iron rod in modal class) = \(\frac{7}{30}\)

(a) The roots of the equation are -1.5 or 2. (b) C = 6 (intercept on the positive vertical axis) \(ax^{2} + bx + c = 0\) When x = 2, c = 6 \(4a + 2b + 6 = 0 \implies 4a + 2b = -6 ....(1)\) When x = -1.5, c = 6 \(a(-1.5)^{2} + b(-1.5) = -6\) \(\frac{9a}{4} - \frac{3b}{2} = -6 \implies 9a - 6b = -24 ... (2)\) Multiply (1) by 3 : \(12a + 6b = -18 ... (1a)\) (1a) + (2) : \(12a + 9a = -18 - 24 \implies 21a = -42\) \(a = -2\) \(4a + 2b = -6 \implies 4(-2) + 2b = -6\) \(2b = -6 + 8 = 2 \implies b = 1\) (a, b, c) = (-2, 1, 6) Hence the relation illustrated in the graph is \(y = -2x^{2} + x + 6\). (c) When \(x = 0.2, y_{max} = 6.2\) (d) When \(y = 2, x = \text{ -1.2 or 1.7}\)

No expanded working has been attached for this question yet. The verified answer is Option C.

No expanded working has been attached for this question yet. The verified answer is Option A.

No expanded working has been attached for this question yet. The verified answer is Option D.