Question Library

The bearing of X from Y = 360 - 120 = 240

Required distance = x where \(\frac{1}{50m}\) = sin 60 x 50m x 0.866 x = 43.3m

2(30 + x) = 360 - 200 60 + 2x = 160 2x = 160 - 60 2x = 100 x = \(\frac{100}{2}\) x = 50

\(\frac{\text{science}}{\text{graduates}}\) = \(\frac{\theta}{360}\) x 4320 where \(\theta\) + 100 + 80 + 108 = 360 \(\theta\) = 360 - 288 = 360 \(\frac{\text{science}}{\text{graduates}}\) = \(\frac{72}{360} \times 4320\) = 864

\(\frac{\text{education}}{\text{graduates}} = \frac{108}{360} \times 4320\) = 1296 Pr(Education) = \(\frac{1296}{4320} = \frac{1}{10}\) pr(not education) = 1 - \(\frac{3}{10}\) = \(\frac{10 - 3}{10}\) = \(\frac{7}{10}\)

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

(a) \((\sqrt{3} - 5\sqrt{2})(\sqrt{3} + \sqrt{2}) = \sqrt{3 \times 3} + \sqrt{3 \times 2} - 5\sqrt{2 \times 3} - 5\sqrt{2 \times 2}\) = \(3 + \sqrt{6} - 5\sqrt{6} - 5(2)\) = \(-7 - 4\sqrt{6}\) \(\therefore\) a = -7 and b = -4. (b) \(\frac{2^{1 - y} \times 2^{y - 1}}{2^{y + 2}} = 8^{2 - 3y}\) \(\frac{2^{1 - y + y - 1}}{2^{y + 2}} = (2^{3})^{2 - 3y}\) \(2^{0 - (y + 2)} = 2^{6 - 9y} \implies 2^{- y - 2} = 2^{6 - 9y}\) \(- y - 2 = 6 - 9y \implies - y + 9y = 6 + 2\) \(8y = 8 \implies y = 1\).

(a) \(9 \cos x - 7 = 1 \implies 9 \cos x = 1 + 7 = 8\) \(\cos x = \frac{8}{9}\) \(x = \cos^{-1} (0.8889)\) \(x = 27.26^\circ\) (b) \(7x < 2x - 13\) \(7x - 2x < - 13\) \(5x < - 13\) \(x < \frac{-13}{5}\) \([-2.6] = -3, -4, -5\) Note : The greatest value function rounds up a real number to an integer that is less or equal to the real number.

(a) (i) Mode = 2 (ii) Third quartile : Sum of frequency = 24 \(Q_{3} = \frac{3}{4} \times 24 = 18\) The 18th position falls at 3. (iii) P(at least 2 children) = \(\frac{7 + 4 + 3 + 2}{24}\) = \(\frac{16}{24}\) = \(\frac{2}{3}\) (b) The sector for 1 child = \(\frac{5}{24} \times 360\) = \(75^\circ\).

(a) Let B represent those with degree. Let D represent those with diploma. Let x represent those who have both degree and diploma. (17 - x) + x + (15 - x) + 4 = 30 36 - x = 30 x = 36 - 30 = 6. (b) \(< PNM = 75^\circ\) (corresponding angle) \(\therefore NQR = 125^\circ - 75^\circ \) = \(50^\circ\) \(< NPM = \frac{180^\circ - 125^\circ}{2}\) = \(\frac{55^\circ}{2}\) = \(27.5^\circ\). (Base angles of an isosceles triangle).

(a) From the diagram, \(|AC|^{2} = 24^{2} + 7^{2}\) = \(576 + 49\) \(|AC|^{2} = 625 \implies |AC| = \sqrt{625} = 25 cm\) \(Area = \frac{1}{2} \times b \times h\) = \(\frac{7 \times 24}{2}\) = \(84 cm^{2}\). (b) Total Surface Area of prism = Area of two right angles ( ABC and EFD) + Area of 3 rectangles (ACDF, ABEF and BCDE) = \(( 2 \times 84 cm^{2}) + (25 \times 40)cm^{2} + (24 \times 40)cm^{2} + (7 \times 40)cm^{2}\) = \((168 + 1000 + 960 + 280) cm^{2}\) = \(2408 cm^{2}\)

(a) \(\log (\frac{35 \times 49}{40 \div 56})\) Given \(\log 5 = 0.6990 , \log 7 = 0.8451 , \log 8 = 0.9031\) \(\log (\frac{35 \times 49}{40 \div 56} = \log (\frac{(7 \times 5) \times 7^{2}}{(8 \times 5) \div (8 \times 7)}\) = \(\log (\frac{7^{3} \times 5}{5 \div 7}\) = \(\log 7^{3} + \log 5 - (\log 5 - \log 7)\) = \(3 \log 7 + \log 5 - \log 5 + \log 7\) = \(4 \log 7\) = \(4 \times 0.8471\) = \(3.3804\) (b) (i) Since there are 60 more adults than children, then the number of adults = x + 60 \(\therefore D 5(x + 60) + D x(2) = D 1280\) \(5x + 300 + 2x = 1280\) \(7x + 300 = 1280 \implies 7x = 1280 - 300 = 980\) \(x = \frac{980}{7} = 140\) There were 140 children. (ii) Ratio of children to adults = \(x : x + 60\) = \(140 : (140 + 60)\) = \(140 : 200\) = \(7 : 10\) (iii) Total number of persons at the show : 140 + 200 = 340 persons. Total amount gotten = D 1280 Average paid per person = \( D \frac{1280}{340}\) = \(D \frac{64}{17}\) = D 3.765 (iv) Percentage profit Profit : D (1280 - 720) = D 560 % profit : \(\frac{560}{720} \times 100% = \frac{700}{9} %\) = \(77.78%\)

(a) (i) From \(\Delta\) TFO, \(\tan x = \tan \theta\) (alternate angles) \(\tan x = \frac{25}{30} = 0.8333\) \(x = \tan^{-1} (0.8333)\) = \(39.80^\circ \approxeq 40^\circ\) Angle of depression of the pole from the woman = 90^\circ - 40^\circ = 50^\circ. (ii) From \(\Delta\) TPQ, \(\tan 46 = \frac{25}{TQ}\) \(TQ = \frac{25}{\tan 46}\) \(TQ = \frac{25}{1.0355}\) = \(24.1429 m\) Height of the pole : \(TF - TQ\) = \(30 m - 24.1429 m\) = \(5.8571 m\) \(\approxeq 6m\)

(a) Marks frequency mid-point \(fx\) 21 - 30 2 25.5 51 31 - 40 10 35.5 355 41 - 50 12 45.5 546 51 - 60 15 55.5 832.5 61 - 70 8 65.5 524 71 - 80 3 75.5 226.5 \(\sum\) 50 2535 (b) Mean (x) = \(\frac{\sum fx}{\sum f}\) = \(\frac{2535}{50}\) = \(50.7\) (c) Number of students that scored above 60% : 8 + 3 = 11 %age : \(\frac{11}{50} \times 100% = 22%\)

(a) \(\frac{x + 2}{x - 2} - \frac{x + 3}{x - 1}\) = \(\frac{(x - 1)(x + 2) - (x - 2)(x + 3)}{(x - 2)(x - 1)}\) = \(\frac{x^{2} + 2x - x - 2 - (x^{2} + 3x - 2x - 6)}{(x - 2)(x - 1)}\) = \(\frac{x^{2} - x^{2} + x - x - 2 + 6}{(x - 2)(x - 1)}\) = \(\frac{4}{(x - 2)(x - 1)}\) (b) (1, 4) and (2, 10) are supposed roots of the equation When x = 1, equation becomes \(A(1)^{2} + B(1) = 4\) When x = 2, \(A(2^{2}) + B(2) = 10\) \(\implies A + B = 4 ...... (1)\) \(4A + 2B = 10 ....... (2)\) Solving for A and B, we have A = 1 and B = 3. (iii) \(y = x^{2} + 3x\) \(y = x(x + 3)\) x = 0 or x = 3.