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(a) (b) A = {0, 1, 2, 3, 4, 5, 6, 7} B = {4, 6, 8, 10, 12} C = {2, 3, 5, 7} U = {0,1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} (i) \(B \cup C = {2, 3, 4, 5, 6, 7, 8, 10, 12}\) \((B \cup C)' = {0, 1, 9, 11}\) (ii) \(A' = {8, 9, 10, 11, 12}\) \(A' \cap B \cap C = { }\)

(a)(i) \(\log_{10} 35 = \log_{10} (7 \times 5) = \log_{10} 7 + \log_{10} 5\) = \(0.8451 + 0.6990 = 1.5441\) (ii) \(\log_{10} 2.8 = \log_{10} (\frac{4 \times 7}{10} = \log_{10} 4 + \log_{10} 7 - \log_{10} 10\) = \(\log_{10} 2^{2} + \log_{10} 7 - 1\) = \(2\log_{10} 2 + \log_{10} 7 - 1 = 2(0.3010) + 0.8451 - 1\) = \(0.4471\) (b) \(N^{0.8942} = 2.8\) \(0.8942 \log N = \log 2.8\) \(\log N = \frac{\log 2.8}{0.8942} = \frac{0.4471}{0.8942}\) \(\log N = 0.5\)

(a) \(2Ax - Kx^{2}\) When x = 1, \(2A(1) - K(1^{2}) = 7\) \(2A - K = 7 ... (1)\) When x = 2, \(2A(2) - K(2^{2}) = 4\) \(4A - 4K = 4 \implies A - K = 1 ... (2)\) From (2), \(A = 1 + K\) (1) becomes : \(2(1 + K) - K = 7\) \(2 + 2K - K = 7 \implies 2 + K = 7\) \(\therefore K = 5\) \(A = 1 + K = 1 + 5 = 6\) \(A, K = 6, 5\) (b) \(x^{2} - 3x - 1 = 0\) \(a = 1, b = -3, c = -1\) Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) \(x = \frac{-(-3) \pm \sqrt{(-3)^{2} - 4(1)(-3)}}{2(1)}\) \(x = \frac{3 \pm \sqrt{9 + 12}}{2}\) \(x = \frac{3 \pm \sqrt{21}}{2} = \frac{3 \pm 4.583}{2}\) \(x = \frac{3 + 4.583}{2} ; x = \frac{3 - 4.583}{2}\) \(x = \frac{7.583}{2} = 3.7915 \approxeq 3.8\) (to 1 d.p) \(x = \frac{-1.583}{2} = -0.7915 \approxeq -0.8\) (to 1 d.p)

Let one of the sides be x metres. The other is (x + 1.5) metres. (a) \(Area = Length \times breadth\) \(x(x + 1.5) = 13.5\) \(x^{2} + 1.5x = 13.5 \implies x^{2} + 1.5x - 13.5 = 0\) \(x^{2} + 4.5x - 3x - 13.5 = 0 \implies x(x + 4.5) - 3(x + 4.5) = 0\) \((x - 3)(x + 4.5) = 0 \implies x = 3 ; x = -4.5\) Since x cannot be negative, therefore x = 3m. The longer side = (3 + 1.5) = 4.5m (b) \(N250 = 1m^{2}\) \(N2000 = \frac{2000}{250} = 8m^{2}\) \(\therefore \text{The area of the floor that will be covered by the carpet} = 8m^{2}\)

(a) + 1 3 5 2 3 5 7 3 4 6 8 4 5 7 9 Sample space = 9 Sum greater than 3 but less than 7 = {5, 4, 6, 5} = 4 P(sum greater than 3 but less than 7) = \(\frac{4}{9}\). (b) Given that y + m = 90^\circ < ABE = x^\circ (vertically opp. angles) < BAD = (x + m)^\circ (exterior angles) < DCB = (x + y)^\circ (exterior angles) (x + m)^\circ + (x + y)^\circ = 180^\circ (opp. angles of a cyclic quadrilateral) (2x + y + m)^\circ = 180^\circ (2x + 90) = 180 2x = 180 - 90 = 90^\circ x = 45^\circ

\(T_{n} = ar^{n - 1}\) (terms of a geometric progression) \(T_{3} = ar^{3 - 1} = ar^{2} = 360 ... (1)\) \(T_{6} = ar^{6 - 1} = ar^{5} = 1215 ... (2)\) Divide (2) by (1). \(\frac{ar^{5}}{ar^{2}} = \frac{1215}{360}\) \(r^{3} = \frac{27}{8}\) \(\therefore r = \sqrt[3]{\frac{27}{8}} = \frac{3}{2}\) (b) Putting r in (1), we have \(ar^{2} = 360\) \(a \times (\frac{3}{2})^{2} = 360 \implies \frac{9a}{4} = 360\) \(a = \frac{360 \times 4}{9} = 160\) (c) \(S_{n} = \frac{a(r^{n} - 1)}{r - 1}\) (sum of terms of a G.P) \(S_{4} = \frac{160((\frac{3}{2})^{4} - 1)}{\frac{3}{2} - 1}\) = \(\frac{160(\frac{81}{16} - 1)}{\frac{1}{2}}\) = \(\frac{160 \times \frac{65}{16}}{\frac{1}{2}}\) = \(1300\)

(a) Cost price of a radio = N6,000.00 \(\therefore\) 30% profit = \(\N6000 \times \frac{30}{100} = N1800\) \(\therefore\) Marked price = N(6000 + 1800) = N7800 Final Selling price = \(\frac{100 - 10}{100} \times N7800 = N7020\) Profit = N(7020 - 6000) = N1020. \(\therefore\) % profit = \(\frac{1020}{6000} \times 100% = 17%\) (b) \(2^{(2x + 1)} - 9(2^{x}) + 4 = 0\) \((2^{x})^{2} \times 2 - 9(2^{x}) + 4 = 0\) Let \(2^{x} = y\) \(2y^{2} - 9y + 4 = 0\) \(2y^{2} - 8y - y + 4 = 0\) \(2y(y - 4) - 1(y - 4) = 0\) \((2y - 1)(y - 4) = 0\) \(2y = 1 \implies y = \frac{1}{2}; y = 4\) \(y = \frac{1}{2} \implies 2^{x} = \frac{1}{2} = 2^{-1} \) \(x = -1\) \(y = 4 \implies 2^{x} = 4 = 2^{2}\) \(x = 2\) \(\therefore x = -1, 2\)

x -3 -2 -1 -0.5 0 1 2 y -28 -5 6 7 5 -8 -33 (b)(i) (c)(i) \(2 - 7x - 6x^{2} = 0\) \(2 + 3 - 7x - 6x^{2} = 3\) \(5 - 7x - 6x^{2} = 3\) \(\therefore \text{the roots at y = 3 are } x = -1.4; 0.2\) (ii) Maximum value of \(y = 5 - 7x - 6x^{2}\) is y = 7.

(a) (b)(i) PS = 6.0cm (ii) Perpendicular disstance between RS and PQ i.e. RK = 5.9cm (iii) QRSP is a trapezium.

XWZ = 90^\circ - (45^\circ + 23^\circ) = 22^\circ WYX = 180^\circ - (45^\circ + 95^\circ) = 40^\circ < XZW = 180^\circ - (89^\circ + 22^\circ) = 69^\circ (a)(i) \(\frac{XY}{\sin 95} = \frac{180}{\sin 40^\circ}\) \(XY = \frac{180 \sin 95}{\sin 40} = 278.96m\) \(\approxeq 279m\) (ii) \(\frac{ZW}{\sin 89^\circ} = \frac{180}{\sin 69^\circ}\) \(ZW = \frac{180 \sin 89^\circ}{\sin 69^\circ} = 192.76m\) \(\approxeq 193m\) (b) < ZNX =

Source: Myschool Past Questions Archive (Record #54452)