Question Library

\(\frac{2}{3} - \frac{1}{5}\) = \(\frac{10 - 3}{15}\) = \(\frac{7}{15}\) \(\frac{1}{3}\) Of \(\frac{2}{5}\) = \(\frac{1}{3}\) x \(\frac{2}{5}\) = \(\frac{2}{15}\) (\(\frac{2}{3} - \frac{1}{5}\)) - \(\frac{1}{3}\) of \(\frac{2}{5}\) = \(\frac{7}{15} - \frac{2}{15}\) = \(\frac{1}{3}\) 3 - \(\frac{1}{1 \frac{1}{2}}\) = 3 - \(\frac{2}{3}\) = \(\frac{7}{3}\) \(\frac{\frac{2}{3} - \frac{1}{5} \text{of} \frac{2}{15}}{3 - \frac{1}{1 \frac{1}{2}}}\) = \(\frac{\frac{1}{3}}{\frac{7}{3}}\) = \(\frac{1}{3}\) x \(\frac{3}{7}\) = \(\frac{1}{7}\)

Let x represent the price of an orange and y represent the number of oranges that can be bought xy = 240k, y = \(\frac{240}{x}\).....(i) If the price of an oranges is raised by \(\frac{1}{2}\)k per orange, number that can be bought for N240 is reduced by 16 Hence, y - 16 = \(\frac{240}{x + \frac{1}{2}\) = \(\frac{480}{2x + 1}\) = \(\frac{480}{2x + 1}\).....(ii) subt. for y in eqn (ii) \(\frac{240}{x}\) - 16 = \(\frac{480}{2x + 1}\) = \(\frac{240 - 16x}{x}\) = \(\frac{480}{2x + 1}\) = (240 - 16x)(2x + 1) = 480x = 480x + 240 - 32x2 - 16 480x = 224 - 32x2 x2 = 7 x = \(\sqrt{7}\) = 2.5 = 2\(\frac{1}{2}\)k

Total yield = N3,700 Total amount invested = N 50,000 Let x be the amount invested at 6% interest and let y be the amount invested at 8% interest then yield on x = \(\frac{6}{100}\) x and yield on y = \(\frac{8}{100}\)y Hence, \(\frac{6}{100}\)x + \(\frac{8}{100}\)y = 3,700.........(i) x + y = 50,000........(ii) 6x + 8y = 370,000 x 1 x + y = 50,000 x 6 6x + 8y = 370,000.........(iii) 6x + 6y = 300,000........(iv) eqn (iii) - eqn(2) 2y = 70,000 y = \(\frac{70,000}{2}\) = 35,000 Money invested at 8% is N35,000

\(\frac{\log_{3} 27 - \log_{\frac{1}{4}} 64}{\log_{3} (\frac{1}{81})}\) \(\log_{3} 27 = \log_{3} 3^{3} = 3\log_{3} 3 = 3\) \(\log_{\frac{1}{4}} 64 = \log_{\frac{1}{4}} (\frac{1}{4})^{-3} = -3\) \(\log_{3} (\frac{1}{81}) = \log_{3} 3^{-4} = -4\) \(\therefore \frac{\log_{3} 27 - \log_{\frac{1}{4}} 64}{\log_{3} (\frac{1}{81})} = \frac{3 - (-3)}{-4}\) = \(\frac{6}{-4} = \frac{-3}{2}\)

P(x, y), P(0, 3) If x increases by 1 unit and y by 4 units, then ratio of x : y = 1 : 4 \(\frac{x}{1}\) = \(\frac{y}{4}\) y = 4x Hence the sign of the graph is y + 3 = 4x

Total cost of 1035 oranges at N145 each = 1035 x 145 = 20025 Total selling price at N245 each = (103)5 x 245 = 30325 Hence his gain = 30325 - 20025 = 10305

\(\frac{5\sqrt{7} - 7\sqrt{5}}{\sqrt{7} - \sqrt{5}}\) = \(\frac{5\sqrt{7} - 7\sqrt{5}}{\sqrt{7} - \sqrt{5}}\) x \(\frac{\sqrt{7} + \sqrt{5}}{\sqrt{7} + \sqrt{5}}\) = \(\frac{(5 \times 7) + (5 \sqrt{7} \times 5) - (7 \times \sqrt{5} \times 7) (-7 \times 5)}{(\sqrt{7})^2}\) = \(\frac{5 \sqrt{35} - 7\sqrt{35}}{2}\) = \(\frac{-2\sqrt{35}}{2}\) = - \(\sqrt{35}\)

\(\frac{3^n - 3^{n - 1}}{3^3 \times 3^n - 27 \times 3^{n - 1}}\) = \(\frac{3^n - 3^{n - 1}}{3^3(3^n - 3^{n - 1})}\) = \(\frac{3^n - 3^{n - 1}}{27(3^n - 3^{n - 1})}\) = \(\frac{1}{27}\)

P \(\alpha\) \(\frac{q^2}{r}\) P = \(\frac{kq^2}{r}\) k = \(\frac{pr}{q^2}\) = \(\frac{36 x 4}{(3)^2}\) p = \(\frac{16q^2}{r}\) = \(\frac{16 \times 25}{2}\) = 200

6x2 - 14x - 12 = 6x2 - 18x + 4x - 12 (3x + 2)(2x - 6) = 3x(2x - 6) + 2(2x - 6) = (3x + 2) 2(x - 3)

When y = 5, y = x2 - 12x + 40, becomes x2 - 12x + 40 = 5 x2 - 12x + 40 - 5 = 0 x2 + 12x + 35 = 0 x2 - 7x - 5x + 35 = 0 x(x - 7) - 5(x - 7) = 0 = (x - 5)(x - 7) x = 5 or 7

If the graph of y = x3 - 3x + 1 is plotted,the graph crosses the x-axis in the ranges -2 < x < -1 and 0 < x < 1

2x + 3y = 1 x 2.......(i) x - 2y = 11 x 3.......(ii) 4x + 6y = 2........(iii) 3x - 6Y = 33........(4) 7x = 35 x = 5 Subt. for x 10 + 3y = 1 3y = -9 y = -3 x + y = 5 + -3 5 - 3 = 2

f(x) = 2(x - 3)\(^2\) + 3(x - 3) + 4 = (2 + 3) (x - 3) + 4 = 5(x - 3) + 4 = 5x - 15 + 4 = 5x - 11 f(3) = 5 x 3 - 11 = 4 g(f(3)) = g(4) = \(\sqrt{5 + 4}\) = \(\sqrt{9}\) = 3 g(4) = 3 f(g(4)) = f(3) = 4 g[f(3)] and f[g(4)] = 3 and 4 respectively.

4a2 - 9b2, 8a3 + 27b3, (4a + 6b)2 = (2a + 3b)(2a - 3b) 8a3 + 27b3 = (2a)3 + (3b)3 = (2a + 3b)(4a - 6ab = 9a2) (4a + 6b)2 = 2(2a + 3b)2