Question Library

XTY = 50o = TXY = \(\frac{130^o}{2}\) = 65o

Area(S) = 2\(\pi r^2 + 2 \pi rh = 2\pi r(h + r)\) D = 14cm, 2r = D r = \(\frac{D}{2}\) = 7cm S = 2 x \(\frac{22}{7} \times [7 + 12] = 44 \times 19\) = 836cm2

\(\frac{1}{2}\)(PQ x 10 sin 30) = 35 PQ = \(\frac{35 \times 2}{10 \sin 30}\) = 14cm

Given:1.20, 1.00, 0.90, 1.40, 0.80, 1.20, 1.10 Sum of the numbers = 1.20 + 1.00 + 0.90 + 1.40 + 0.80 + 1.20 + 1.10 = 7.60 Number of terms = 7 Mean, m = 7.60 \div 7 = 1.0857 (to 4 d.p.) Highest value = 1.40 Lowest value = 0.80 Range, r = 1.40 − 0.80 = 0.60 m + r = 1.0857 + 0.60 = 1.6857 closest to 1.65.

(a) \(px - 2qx - 4qy + 2py\) = \(x(p - 2q) + 2y(p - 2q)\) = \((x + 2y)(p - 2q)\) (b) \(U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}\) \(P = {1, 2, 4, 6, 10} ; Q = {2, 3, 6, 9}\) \((P \cup Q) = {1, 2, 3, 4, 6, 9, 10}\) \((P \cup Q)' = {5, 7, 8}\) \(P' \cap Q' = {3, 5, 7, 8, 9} \cap {1, 4, 5, 7, 8, 10}\) = \({5, 7, 8}\) Hence, \((P \cup Q)' = P' \cap Q'\).

(a) \(y = k + \frac{c}{x^{2}}\) where c and k are constants. (b) When x = 1, y = 11 \(11 = k + \frac{c}{1^{2}} \implies 11 = k + c ... (1)\) When x = 2, y = 5 \(5 = k + \frac{c}{2^{2}} \implies 5 = k + \frac{c}{4}\) \(\equiv 20 = 4k + c ... (2)\) (2) - (1) : \(4k - k = 20 - 11 \implies 3k = 9\) \(k = 3\) Put k = 3 in (1), we have \(11 = 3 + c \implies c = 11 - 3 = 8\) \(\therefore y = 3 + \frac{8}{x^{2}}\) When x = 4, \(y = 3 + \frac{8}{4^{2}} = 3 + \frac{1}{2}\) = \(3\frac{1}{2}\)

(a) \(/QY/ = 2cm; /RS/ = 3cm\) \(\therefore /YZ/ = /RS/ = /PQ/ = 3cm\) (opp sides of a parallelogram) But \(/PZ/ = /PQ/ + /QY/ + /YZ/\) \(\therefore /PZ/ = 3cm + 2cm + 3cm = 8cm\) (b) Difference in longitude = 95^\circ - 25^\circ = 70^\circ Distance PQ = \(\frac{70}{360} \times 2\pi R \cos \theta\) = \(\frac{70}{360} \times 2 \times \frac{22}{7} \times 6400 \cos 56\) = \(\frac{7}{36} \times 2 \times \frac{22}{7} \times 6400 \times 0.5592\) = \(4374.19 km \approxeq 4374 km\) (to the nearest km)

(a) Total no of cards = 52; No of fives = 4; No of red nines = 2. Prob(of 5) = \(\frac{4}{52} = \frac{1}{13}\) Prob(red nine) = \(\frac{2}{52} = \frac{1}{26}\) \(\therefore\) Prob (of 5 or red nine) = \(\frac{1}{13} + \frac{1}{26} = \frac{3}{26}\) (b) \(\tan < RPQ = \frac{3}{5} = 0.6\) \(< RPQ = 30.96^\circ\) \(\therefore\) Bearing of R from P = 150^\circ - 30.96^\circ = 119.04^\circ \(\approxeq\) 119^\circ (to the nearest degree).

x 1 2 3 4 5 6 7 8 9 10 Total f 2 3 4 4 4 4 5 3 2 1 32 fx 2 6 12 16 20 24 35 24 18 10 167 (a)(i) Mean = \(\frac{\sum fx}{\sum f} = \frac{167}{32}\) = \(5.21875 \approxeq 5.22\) (ii) By inspection, the average of the numbers in the 16th and 17th position give the median = \(\frac{5 + 5}{2} = 5\) (iii) Mode = 7 marks having the largest frequency. (b) Students who scored at least 8 marks = Those who scored 8, 9, 10. Total number of students that scored at least 8 = 3 + 2 + 1 = 6. \(%age = \frac{6}{32} \times 100% = 18.75%\)

(a)(i) \(\sin 63.35^\circ = 0.8938\) \(2 \sin 63.35^\circ = 2 \times 0.8938 = 1.7876\) (ii) \(\log \cos 44.74^\circ = \bar{1}.8514\) (b) \(\log K - \log (K - 2) = \log 5\) \(\log (\frac{K}{K - 2}) = \log 5\) \(\therefore \frac{K}{K - 2} = 5 \implies K = 5(K - 2)\) \(K = 5K - 10 \implies 10 = 5K - K = 4K\) \(K = 2\frac{1}{2}\) (c) \(\frac{(3.68)^{2} \times 6.705}{\sqrt{0.3581}}\) No Log \((3.68)^{2}\) \(0.5658 \times 2 = 1.316\) \(6.705\) \(0.8264 = 0.8264\) = \(1.9580\) \(\sqrt{0.3581}\) \(\bar{1}.5540 \div 2 = \bar{1}.7770\) \(151.7\) \(\gets 2.1810\) \(\therefore \frac{(3.68)^{2} \times 6.705}{\sqrt{0.3581}} = 151.7\)

(a) \(p = x + ym^{3}\) \(ym^{3} = p - x\) \(m^{3} = \frac{p - x}{y}\) \(m = \sqrt[3]{\frac{p - x}{y}}\) (b) \(x^{2} - 6x + 7 = 0\) \(\implies x^{2} - 6x = -7\) \(x^{2} - 6x + (-3)^{2} = -7 + (-3)^{2}\) \(x^{2} - 6x + 9 = 2\) \((x - 3)^{2} = 2 \implies x - 3 = \pm \sqrt{2}\) \(x = 3 \pm \sqrt{2} = 3 \pm 1.414\) \(x = 4.414 ; 1.586\) \(x \approxeq 4.4 ; 1.6\) (c) Let the first odd number be z. The next positive odd number = z + 2 \(\therefore z(z + 2) = 195\) \(z^{2} + 2z = 195\) \(z^{2} + 2z - 195 = 0 \implies z^{2} - 13z + 15z - 195 = 0\) \(z(z - 13) + 15(z - 13) = 0 \implies \text{z = 13 or -15}\) Since we are told the odd numbers are positive, then z = 13 is the suitable answer. Hence the second odd number = 13 + 2 = 15. The two odd numbers are 13 and 15.

(a) x 0^\circ 30^\circ 60^\circ 90^\circ 120^\circ 150^\circ 180^\circ \(y = 2\cos 2x - 1\) 1.0 0 -2 -3 -2 0 1.0 (b) (c) x 0^\circ 30^\circ 60^\circ 90^\circ 120^\circ 150^\circ 180^\circ \(y = \frac{1}{180} (x - 360)\) -2.0 -1.83 -1.67 -1.5 -1.33 -1.17 -1.0 (d)(i) \(2 \cos 2x + \frac{1}{2} = 0\) \(2 \cos 2x + \frac{1}{2} - 1\frac{1}{2} = -1\frac{1}{2}\) \(2 \cos 2x - 1 = -1\frac{1}{2}\) The values of x at \(2 \cos 2x + \frac{1}{2} = 0\) are x = 51^\circ and 129^\circ. (ii) \(2 \cos 2x - \frac{x}{180} + 1 = 0\) \(2 \cos 2x - 1 = \frac{1}{180} (x - 360)\) The values of x is where the two graphs intersect; which are at x = 54^\circ and 132^\circ.

(a - d)

(a) < XZY = 180^\circ - (30^\circ + 45^\circ) = 105^\circ \(\frac{20}{\sin 105^\circ} = \frac{YZ}{\sin 30^\circ}\) \(YZ = \frac{20 \sin 30^\circ}{\sin 105^\circ} = 10.35km\) \(\approxeq 10.4km \) (to 3 sig. figs) (ii) \(\frac{20}{\sin 75} = \frac{XZ}{\sin 45}\) \(XZ = \frac{20 \sin 45}{\sin 75} = 14.64km\) \(\approxeq 14.6km\) ( to 3 sig. figs) (b) Let P and Q be two airfields. Difference in latitude = 36^\circ + 36^\circ = 72^\circ (i) Distance between P and Q = \(\frac{72}{360} \times 2 \times \frac{22}{7} \times 6400\) = \(\frac{22}{7} \times 2560\) = \(8,045.71km \approxeq 8,050 km\) (3 sig. figs) (ii) Time taken = \(\frac{8,045.71}{800} = 10.06 hours\) \(\approxeq 10hours\) (to the nearest hour)

(a)(i) Marks (x) No of pupils (f) Cum Freq 11 - 20 68 68 21 - 30 184 252 31 - 40 294 546 41 - 50 402 948 51 - 60 480 1428 61 - 70 310 1738 71 - 80 164 1902 81 - 90 98 2000 (ii) (b)(i) Cut off mark = 68 marks for reading at 1700. (ii) Candidates scoring more than 45% = 2000 - 720 = 1280 \(\therefore\) P(scoring at least 45%) = \(\frac{1280}{2000} = 0.64\)