Question Library

\(\frac{BC}{60}\) = \(\frac{tan 62}{1}\) BC = 60 tan 62 \(\frac{AC}{60}\) = \(\frac{tan 62}{1}\) AC = 60 tan 64 AB = AC - BC = 60(tan 64o - tan 62o)

\(\cos^{2} x (\sec^{2} x + \sec^{2} x \tan^{2} x)\) = \(\cos^{2} x \sec^{2} x + \cos^{2} x \sec^{2} x \tan^{2} x\) = \(1 + \tan^{2} x\) = \(1 + \frac{\sin^{2} x}{\cos^{2} x}\) = \(\frac{\cos^{2} x + \sin^{2} x}{\cos^{2} x}\) = \(\frac{1}{\cos^{2} x} = \sec^{2} x\)

cosx = \(\sqrt{\frac{a}{b}}\) y2 + \(\sqrt{(a)^2}\) = \(\sqrt{(b)^2}\) by pythagoras y2 = b - a ∴ y = b - a cosec x = \(\frac{1}{sin x}\) = \(\frac{1}{y}\) \(\frac{b}{y}\) = \(\frac{\sqrt{b}}{\sqrt{b - a}}\) = \(\sqrt{\frac{b}{b - a}}\)

tan \(\theta\) = \(\frac{60\sqrt{3}}{60}\) = \(\sqrt{3}\) \(\theta\) = tan \(\frac{1}{3}\) = 60o Bearing of y from x = 060o

Education = 11% = \(\frac{11}{100} \times 360^\circ = 39.6^\circ\) Food = 59% = \(\frac{59}{100} \times 360^\circ = 212.4^\circ\)

From the diagram above x = 60 + 70 = 130º (sum of angles on a straight line was used to obtain 60º and 70º, and alternate angles are equal to obtain the value of x)

\(\frac{PQ}{PN} = \frac{PM}{PR} = \frac{QM}{NR}\) \(\frac{4.8}{12} = \frac{5}{PR}\) PR = \(\frac{5 \times 12}{4.8} = \frac{50}{4}\) = 12.5 \(\frac{PQ}{PN} = \frac{PM}{PT} = \frac{TM}{NT}\) \(\frac{PT}{12} = \frac{5}{PR}\) PT2 = 60 PT = \(\sqrt{60}\) = 7.746 = 7.7

PQM = 110o(Ext. o - (110 + 55) = 180o - 165o = 15o

Since PQRS is quadrilateral 2y + 2x = QPS = 360o i.e. 2(y + x) + QPS = 360o QPS = 360o - 2(y + x) But x + y + 68o = 180o x + y = 180o - 68o = 180o x + y = 180o - 68o = 112o QPS = 360o - 2(112o) 360o - 224o = 136o

In △ UQV |UV|\(^2\) = 6\(^2\) + 4\(^2\) (Pythagora's theorem) |UV| = \(\sqrt{52}\) = 7.21cm Also, in △ WVR |WV|\(^2\) = 2\(^2\) + 2\(^2\) = 4 + 4 = 8 (same reason as above) |WV| = \(\sqrt{8}\) = 2.83 Therefore, the area of △ UVW = \(\frac{1}{2}\) x 2.83 x 7.21 = 10.21 ≈ 10cm\(^2\)

(i) Sum of interior angles of a regular polygon = \((2n - 4) \times 90^\circ\) For a pentagon, \((2(5) - 4) \times 90^\circ = 6 \times 90^\circ = 540^\circ\) Interior angle = \(\frac{540^\circ}{5} = 108^\circ\) (ii) < BXA = 108^\circ - 90^\circ = 18^\circ < ABX = 108^\circ (as shown above) \(\therefore < BXA = 180^\circ - (108^\circ + 18^\circ) = 54^\circ\)

Waiting time (in mins) Class boundaries No of customers Cum Freq 1.5 - 1.9 1.45 - 1.95 3 3 2.0 - 2.4 1.95 - 2.45 10 13 2.5 - 2.9 2.45 - 2.95 18 31 3.0 - 3.4 2.95 - 3.45 10 41 3.5 - 3.9 3.45 - 3.95 7 48 4.0 - 4.4 3.95 - 4.45 2 50 (b) (c) Interquartile range of the distribution = \(Q_{3} - Q_{1}\) \(Q_{3} = \frac{3 \times 51}{4} = 38.25\) \(Q_{1} = \frac{1 \times 51}{4} = 12.75\) From the graph, \(Q_{3} = 3.25 ; Q_{1} = 2.45\) \(\therefore \text{The interquartile range of the distribution } = 3.25 - 2.45 = 0.80\) (ii) Proportion of customers who must have waited for more than 3 minutes = \(\frac{16}{50} = \frac{8}{25}\)

(a) & (b)

\(\frac{0.00275 \times 0.0064}{0.025 \times 0.08}\) Removing the decimals = \(\frac{275 \times 64}{2500 \times 800}\) = \(\frac{88}{10^4}\) \(88 x 10^{-4} = 88 x 10^{1} x 10^{-4} = 8.8 x 10^{-3}\)

use "T" to represent the total profit. The first receives \(\frac{1}{3}\) T remaining, 1 - \(\frac{1}{3}\) = \(\frac{2}{3}\)T The seconds receives the remaining, which is \(\frac{2}{3}\) also \(\frac{2}{3}\) x \(\frac{2}{3}\) = \(\frac{4}{9}\) The third receives the left over, which is \(\frac{2}{3}\)T - \(\frac{4}{9}\)T = (\(\frac{6 - 4}{9}\))T = \(\frac{2}{9}\)T The third receives \(\frac{2}{9}\)T which is equivalent to N12000 If \(\frac{2}{9}\)T = N12, 000 T = \(\frac{12 000}{\frac{2}{9}}\) Total share[T] = N54, 000 The first receives \(\frac{1}{3}\) of T → \(\frac{1}{3}\) * N54, 000 = N18,000 The second receives \(\frac{4}{9}\) of T → \(\frac{4}{9}\) * N54, 000 = N24,000 The third receives \(\frac{2}{9}\)T which is equivalent to N12000. Adding the three shares give total profit of N54,000