Question Library

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

(a) x 10 11 100 101 10 100 110 1000 1010 11 110 1001 1100 1111 100 1000 1100 10000 10100 (b) \(11.011_{two} = 1 \times 2^{1} + 1 \times 2^{0} + 0 \times 2^{-1} + 1 \times 2^{-2} + 1 \times 2^{-3}\) = \(2 + 1 + 0 + \frac{1}{4} + \frac{1}{8}\) = \(3\frac{3}{8}\) = \(3.375\) (c) \(\frac{9.6 \times 10^{18}}{0.24 \times 10^{5}} \equiv \frac{9.6 \times 10^{18}}{2.4 \times 10^{4}}\) = \(\frac{9.6}{2.4} \times 10^{18 - 4}\) = \(4 \times 10^{14}\)

(a) \(\frac{15.05 \times \sqrt{0.00695}}{6.95 \times 10^{2}}\) No Log 15.05 \(1.1775 = 1.1775\) \(\sqrt{0.00695}\) \(\bar{3}.8420 \div 2 = \bar{2}.9210\) = 0.0985 \(6.95 \times 10^{2}\) - 2.8420 Antilog = 0.001805 = \(\bar{3}.2565\) \(\therefore \frac{15.05 \times \sqrt{0.00695}}{6.95 \times 10^{2}} \approxeq 0.00181\) (3 sig. figs) (b) No of boys = 2, No of girls = 3 \(\therefore\) Total students = 5 (i) P(both are boys) = \(\frac{2}{5} \times \frac{1}{4} = \frac{1}{10}\) (ii) P(both are of different sexes) = P(first a boy, then a girl) or P(first a girl, then a boy) = \(\frac{2}{5} \times \frac{3}{4} + \frac{3}{5} \times \frac{2}{4} \) = \(\frac{6}{20} + \frac{6}{20}\) = \(\frac{12}{20} = \frac{3}{5}\)

(a) \(\log_{10} x + \log_{10} y = 4 ... (1)\) \(\log_{10} x + 2 \log_{10} y = 3 ...(2)\) \((1) \implies \log_{10} (xy) = 4 \therefore xy = 10^{4} ... (3)\) \((2) \implies \log_{10} (xy^{2}) = 3 \therefore xy^{2} = 10^{3} ... (4)\) Divide (4) by (3) : \(y = 10^{-1} = 0.1\) Put y in (3), we have \(\frac{x}{10} = 10^{4} \implies x = 10^{5}\) \((x, y) = (10^{5}, 10^{-1})\) (b) (i) \(t \propto \frac{V}{P}\) \(t = \frac{kV}{P}\) when t = 10, V = 20 and P = 5. \(10 = \frac{20k}{5} \implies 50 = 20k\) \(k = \frac{5}{2}\) \(\therefore t = \frac{5V}{2P}\) (ii) When V = 50, P = 2, we have \(t = \frac{5(50)}{2(2)} = \frac{250}{4}\) = \(62.5 minutes\) (iii) when V = 40, t = 20 minutes \(20 = \frac{5(40)}{2P} \implies 40P = 200\) \(P = \frac{200}{40} = 5 pumps\)

Log\(_{10}\)4 + Log\(_{10}\)25 = Log\(_{10}\)(4 + 25) = Log\(_{10}\)100 = Log\(_{10}\)10\(^2\) = 2Log\(_{10}\)10 = 2 x 1 = 2

From the diagram above, using Pythagoras' theorem y = \(\sqrt{10^2 - 8^2}\) = \(\sqrt{36}\) = 6 cm Area of Rectangle = length x breath = 6 x 8 = 48 cm\(^2\)

(a) \(6y + 5x = -12 ... (1)\) \(4y - 3x = 11 ... (2)\) Multiply (1) by 2 and (2) by 3, \(12y + 10x = -24 ... (1a)\) \(12y - 9x = 33 ... (2a)\) \((1a) - (2a) : 19x = -57 \implies x = -3\) \(4y - 3(-3) = 11 \implies 4y + 9 = 11\) \(4y = 2 \implies y = \frac{1}{2}\) (b) \(/BC/^{2} = 36^{2} + 48^{2}\) \(/BC/^{2} = 1296 + 2304 = 3600\) \(/BC/ = \sqrt{3600} = 60 cm\) \(/AC/^{2} = /BC/^{2} + /AB/^{2}\) \((y + 48)^{2} = 60^{2} + 36^{2} + y^{2}\) \(y^{2} + 96y + 2304 = 3600 + 1296 + y^{2}\) \(96y = 3600 + 1296 - 2304 = 2592\) \(96y = 2592 \implies y = \frac{2592}{96} = 27 cm\)

(a) \(\log_{10} 0.72 =\log_{10} (\frac{72}{100})\) = \(\log_{10} 72 - \log_{10} 100\) = \(72 = 2^{3} \times 3^{2}\) \(\log_{10} 72 = \log_{10} 2^{3} + \log_{10} 3^{2} = 3\log_{10} 2 + 2 \log_{10} 3\) = \(3(0.3010) + 2(0.4771) = 0.9030 + 0.9542 = 1.8572\) \(\therefore \log_{10} 0.72 \equiv \log_{10} 72 - \log_{10} 100\) = \(1.8572 - 2 = -0.1428\) (b) \(\tan 39^\circ = \frac{20}{AB} \) = \(\frac{20}{\tan 39^\circ}\) \(AB = 24.698 m \approxeq 25m\)

\(y \propto \frac{1}{x^2}\) \(y = \frac{k}{x^2}\) \(4 = \frac{k}{3^2}\) \(k = 4 \times 3^2 = 36\) \(y = \frac{36}{x^2}\) When x = 2, \(y = \frac{36}{2^2} = 9\)

log\(_{10}\) x = \(\bar{2}.3675\) => x = 0.023 log\(_{10}\) y = \(\bar{2}.9738\) => y = 0.094 x + y = 0.117

x\(^2\) = 12\(^2\) + 5\(^2\) x\(^2\) = 144 + 25 x\(^2\) = 169 x = \(\sqrt{169}\) = 13m.

\(0.0014 = 1.4 \times 10^{-3}\) \(0.011 = 1.1 \times 10^{-2}\) \(\therefore 0.0014 \times 0.011 = 1.4 \times 10^{-3} \times 1.1 \times 10^{-2}\) = \(1.54 \times 10^{-5}\)

\(\frac{(81)^{\frac{3}{4}} - (27)^{\frac{1}{3}}}{3 \times 2^3}\) = \(\frac{(3^{4})^{\frac{3}{4}} - (3^{3})^{\frac{1}{3}}}{3 \times 2^{3}}\) = \(\frac{3^{3} - 3^{1}}{24}\) = \(\frac{27 - 3}{24} = 1\)

\(\frac{\sqrt{12} - \sqrt{3}}{\sqrt{12} + \sqrt{3}}\) \(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\) \(\therefore \frac{\sqrt{12} - \sqrt{3}}{\sqrt{12} + \sqrt{3}} = \frac{2\sqrt{3} - \sqrt{3}}{2\sqrt{3} + \sqrt{3}}\) = \(\frac{\sqrt{3}}{3\sqrt{3}}\) = \(\frac{1}{3}\)

\(x^{3} + px^{2} + qx + 6 = 0\) f(x - 1) = 0; f(1) = 0. \(1^{3} + p(1^{2}) + q(1) + 6 = 0 \implies p + q = -7 ... (1)\) f(x + 1) = 0; f(-1) = 0. \((-1)^{3} + p(-1^{2}) + q(-1) + 6 = 0 \implies p - q = -5 ... (2)\) Subtract (2) from (1). \(2q = -2 \implies q = -1\) \(p - (-1) = -5 \implies p = -5 - 1 = -6\) \((p, q) = (-6, -1)\)