Question Library

(a) \(3\sqrt{75} = 3 \times \sqrt{25 \times 3} = 3(5\sqrt{3}) = 15\sqrt{3}\) \(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\) \(\sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3}\) \(\therefore 3\sqrt{75} - \sqrt{12} + \sqrt{108} \equiv 15\sqrt{3} - 2\sqrt{3} + 6\sqrt{3}\) = \((15 - 2 + 6)\sqrt{3}\) = \(19\sqrt{3}\). (b) \(124_{n} = 232_{five}\) \(124_{n} = (1 \times n^{2}) + (2 \times n^{1}) + (4 \times n^{0}) \) = \(n^{2} + 2n + 4\) \(232_{five} = (2 \times 5^{2}) + (3 \times 5^{1}) + (2 \times 5^{0})\) = \(50 + 15 + 2\) = \(67\) \(\implies n^{2} + 2n + 4 = 67\) \(n^{2} + 2n + 4 - 67 = 0\) \(n^{2} + 2n - 63 = 0\) \(n^{2} - 7n + 9n - 63 = 0\) \(n(n - 7) + 9(n - 7) = 0\) \((n + 9)(n - 7) = 0\) \(\text{n = -9 or 7}\) Hence, \(n = 7\) \(124_{seven} = 232_{five}\)

(a) Area of the shaded segment = area of sector XOY - area of triangle OXY. \(\implies 504 = \frac{\theta}{360^\circ} \times \pi r^{2} - \frac{1}{2} r^{2} \sin \theta\) \(\implies 504 = \frac{90}{360} \times \frac{22}{7} \times r^{2} - \frac{1}{2} \times r^{2} \sin 90\) \(504 = r^{2}(\frac{22}{28} - \frac{1}{2})\) \(504 = \frac{2r^{2}}{7} \implies r^{2} = \frac{504 \times 7}{2} = 1764\) \(r = \sqrt{1764} = 42 cm\). (b) In the diagram, \(a_{1} = a_{2}\) (base angles of an isosceles triangle) \(\therefore a_{1} + a_{2} = 2a_{1}\) \(\therefore 2a_{1} + 109^\circ = 180^\circ \implies 2a_{1} = 180^\circ - 109^\circ = 71^\circ\) \(a_{1} = a_{2} = \frac{71^\circ}{2} = 35.5^\circ\) Hence, \(< RQS = 66^\circ + 35.5^\circ\) = \(101.5^\circ\)

(a) \(\frac{3}{2p - \frac{1}{2}} = \frac{\frac{1}{3}}{\frac{1}{4}p + 1}\) \(\implies 3(\frac{1}{4}p + 1) = \frac{1}{3}(2p - \frac{1}{2})\) \(\frac{3}{4}p + 3 = \frac{2}{3}p - \frac{1}{6}\) \(\frac{3}{4}p - \frac{2}{3}p = - 3 - \frac{1}{6}\) \(\frac{1}{12}p = - 3\frac{1}{6}\) \(p = \frac{-19}{6} \div \frac{1}{12}\) \(p = \frac{-19}{6} \times 12 = -38\) (b) \(5% \times ¢760.00 = \frac{5}{100} \times ¢760 = ¢38.00\) So the TV set was actually sold for ¢(760 - 38) = ¢722.00 Using, \(% profit = \frac{\text{SP - CP}}{CP} \times 100%\) \(20% = \frac{760 - x}{x} \times 100%\) \(20% x = (760 - x) \times 100%\) \(x = 5(760 - x) \implies x = 3800 - 5x\) \(6x = 3800 \implies x = ¢633.33\) Thus, the actual profit = actual SP - CP = ¢(722 - 633.33) = ¢88.67 Hence, actual %age profit = \(\frac{88.67}{633.33} \times 100%\) = 14.0006% = 14% (2 sig. figs)

(a) \(\otimes\) 1 5 9 10 1 1 5 9 10 5 5 3 1 6 9 9 1 4 2 10 10 6 2 1 (i) \((9 \otimes 5) \otimes (10 \otimes 10) = 1 \otimes 1 = 1\) (ii) (1) \(10 \otimes m = 2\) By comparison, \(10 \otimes 9 = 2\). (2) \(n \otimes n = 4\) From the table, \(9 \otimes 9 = 4\) Hence, n = 9. (b) Let the fraction = \(\frac{m}{n}\) \(\frac{m}{n} = \frac{3}{4}..... (1)\) \(\implies n = \frac{4m}{3} ....... (2)\) \(2m = n + 34 ...... (3)\) Put (2) in (3), \(2m = \frac{4m}{3} + 34\) \(2m - \frac{4m}{3} = 34 \implies \frac{2m}{3} = 34\) \(m = \frac{34 \times 3}{2} = 51\) \(n = \frac{4m}{3} = \frac{4 \times 51}{3} = 68\) Hence, the fraction = \(\frac{51}{68}\).

(a) (i) \((16 - 2x) + (6 + x) + (19 - 3x) + 5x + 7x + 8x + 4x + 4 = 125\) \(45 + 20x = 125 \implies 20x = 125 - 45 = 80\) \(x = \frac{80}{20} = 4\) (ii) \(n(P \cup Q \cap R') = (16 - 2x) + 5x + (6 + x)\) = \(22 + 4x\) = \(22 + 4(4) = 22 + 16 = 38\) (b) In the diagram above, < XYW = 90^\circ (angle in a semi-circle; XW is a diameter) < XWY + 90^\circ + 50^\circ = 180^\circ (sum of angles in a triangle) < XWY = 180^\circ - 140^\circ = 40^\circ (i) < WYZ = 40^\circ (alternate angles; \(\overline{XW} || \overline{YZ}\)) (ii) < WOZ = \(2 \times 40^\circ \) (angle at centre = twice angle at circumference) = 80^\circ \(\therefore < WEO = 180^\circ - (40^\circ + 80^\circ)\) = \(180^\circ - 120^\circ \) = \(60^\circ\) \(\therefore < YEZ = < WEO = 60^\circ\) (vertically opposite angles).

(a) \(f : x \to 2x^{2} - 1 ; g : x \to 3x + 2\) (i) \(f(x - 1) - 7 = 0\) \(f(x - 1) = 2(x - 1)^{2} - 1 = 2(x^{2} - 2x + 1) - 1\) = \(2x^{2} - 4x + 2 - 1\) \(f(x - 1) - 7 = 2x^{2} - 4x + 1 - 7 = 2x^{2} - 4x - 6 = 0\) \(2x^{2} - 6x + 2x - 6 = 0 \implies 2x(x - 3) + 2(x - 3) = 0\) \((2x + 2)(x - 3) = 0 \implies 2x = -2; x = 3\) \(x = -1 ; 3\) (ii) \(\frac{f(-\frac{1}{2}) . g(3)}{f(4) - g(5)}\) \(f(-\frac{1}{2}) = 2(-\frac{1}{2})^{2} - 1 = \frac{1}{2} - 1 = -\frac{1}{2}\) \(g(3) = 3(3) + 2 = 9 + 2 = 11\) \(f(4) = 2(4^{2}) - 1 = 32 - 1 = 31\) \(g(5) = 3(5) + 2 = 15 + 2 = 17\) \(\frac{f(-\frac{1}{2}) . g(3)}{f(4) - g(5)} = \frac{(-\frac{1}{2}) . (11)}{31 - 17}\) = \(\frac{-\frac{11}{2}}{14}\) = \(-\frac{11}{28}\) (b) \(m \ast n = \frac{-n}{m^{2} + 1}; m, n \in R\) \(-3 \ast -10 = \frac{-(-10)}{(-3)^{2} + 1}\) = \(\frac{10}{10} = 1\) \(-10 \ast -3 = \frac{-(-3)}{(-10)^{2} + 1}\) = \(\frac{3}{101}\) Hence, \(\ast\) is not commutative as \(m \ast n \neq n \ast m\).

Let the number be y, subtract y from 2 i.e 2 - y 2 - y = 4 < \(\frac{1}{5}\) y, 2 - y = \(\frac{y}{5}\) - 4 2 - y + 4 = \(\frac{y}{5}\) 6 = \(\frac{y}{5}\) + y 6 = \(\frac{y + 5y}{5}\) 6 = \(\frac{6y}{5}\) multiplying through by 5 6 * 5 = 6y \(\frac{30}{6}\) = y = 5

\(\begin{array}{c|c} 3 & \text{27 rem 0} \\ \hline 3 & \text{ 9 rem 0} \\ \hline 3 & \text{ 3 rem 0} \\ \hline 3 & \text{ 1 rem 1}\\ \hline & 0 \end{array}\) Hence the correct answer is 10003

The sum, S of ratio is S = 5 + 3 + 2 = 10. But highest share = \(\frac{5}{10} \times T\), where T is the total number of apples. Thus, \(40 = \frac{5}{10} \times T\), given 40 x 10 = 5T, \(T = \frac{40 \times 10}{5} = 80\) Hence the smallest share = \(\frac{2}{10} \times 80\) = 16 apples

\(\frac{1.25 \times 0.025}{0.05}\) \( = \frac{125 \times 10^{-2} \times 25 \times 10^{-3}}{5 \times 10^{-2}}\) = 125 x 5 x 10-3 = 625 x 0.001 = 0.625 = 0.6 Approx to 1 d.p.

\(\frac{3^{-5n}}{9^{1-n}} \times 27^{n + 1}\) \(\frac{3^{-5n}}{3^{2(1-n)}} \times 3^{3(n + 1)}\) \(3^{-5n} \div 3^{2(1-n)} \times 3^{3(n + 1)}\) \(3^{-5n - 2(1-n) + 3(n + 1)}\) \(3^{-5n - 2 + 2n + 3n + 3}\) \(3^{-5n + 5n + 3 - 2}\) \(3^{1}\) = 3

log1041/3 = 1/3 log104 = 1/3 x 0.6021 = 0.2007

\(\frac{\sqrt{5}(\sqrt{147} - \sqrt{12}}{\sqrt{15}}\) \(\frac{\sqrt{5}(\sqrt{49 \times 3} - \sqrt{4 \times 3}}{\sqrt{5 \times 3}}\) \(\frac{\sqrt{5}(7\sqrt{3} - 2\sqrt{3}}{\sqrt{5} \times \sqrt{3}}\) \(\frac{\sqrt{3} (7 - 2}{\sqrt{3}}\) = 5

P = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} Q = {-1, 3, 5, 7, 11, 13, 17, 19, 23} P \(\cap\) Q = {3, 5, 7, 11, 13, 17, 19}

Let f(x) = x2 - x - k Then by the factor theorem, (x - 4): f(4) = (4)2 - (4) - k = 0 16 - 4 - k = 0 12 - k = 0 k = 12