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

Area of the quadrant = \(\frac{1}{4} \pi r^2 = \frac{1}{4} \pi (6)^2\) = \(\frac{36 \pi}{4} = 9 \pi \) Area of the triangle = \(\frac{1}{2} \times 6 = 18 \times \sin 90^o = 18\) Area of shaded portion =(9 \(\pi \) - 18)cm^2\) = 9(\(\pi - 2)cm^2\).

RY = YZ - RZ = 10 tan 45 - 10 tan 30 = 10(1 - \(\frac{1}{\sqrt{3}}\))

(a) \(9^{2x - 1} = \frac{81^{x - 2}}{3^{x}}\) \((3^{2})^{2x - 1} = \frac{(3^{4})^{x - 2}}{3^{x}}\) \(3^{4x - 2} = \frac{3^{4x - 8}}{3^{x}}\) \(3^{4x - 2} = 3^{4x - 8 - x}\) \(3^{4x - 2} = 3^{3x - 8}\) \(\therefore 4x - 2 = 3x - 8 \implies 4x - 3x = -8 + 2\) \(x = -6\) (b) \(\sqrt{\frac{0.81 \times 10^{-5}}{2.25 \times 10^{7}}}\) = \(\sqrt{\frac{0.81}{2.25}} \times \sqrt{10^{-5 - 7}}\) = \(\frac{0.9}{1.5} \times \sqrt{10^{-12}}\) = \(0.6 \times 10^{-6}\) = \(6 \times 10^{-7}\)

(a) \(2x + 5y = \frac{13}{2} \implies 4x + 10y = 13 ... (1)\) \(5x - 2y = 9 ... (2)\) \((2) \times 5 : 25x - 10y = 45 ... (2a)\) \((1) + (2a) : 29x = 58 \implies x = 2\) \(4(2) + 10y = 13 \implies 10y = 13 - 8 = 5\) \(y = \frac{5}{10} = 0.5\) (b) \(\log_{10} (2x + 1) - \log_{10} (3x - 2) = 1\) \(\log_{10} (\frac{2x + 1}{3x - 2}) = 1\) \(\frac{2x + 1}{3x - 2} = 10^{1} = 10\) \(2x + 1 = 10(3x - 2) \implies 2x + 1 = 30x - 20\) \(21 = 28x \implies x = \frac{3}{4}\)

(a) Length of arc AB = \(\frac{108^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 7 = \frac{132}{10} =13.2cm\) Perimeter of the sector = \(13.2cm + 7cm + 7cm = 27.2cm\) (b) Let T = top of cliff, C = bottom of cliff, B = the boat. Since angle of depression = 30^\circ then < CBT = 30^\circ (alternate angles TR//CB) \(\therefore \text{Height of the cliff} (CT) = 120 \times 0.5774m\) = \(69.288m \approxeq 69.3m\)

(a) < PQR = 180^\circ - 131^\circ = 49^\circ (angles on a straight line) \(\therefore < PQR = 49^\circ\) Also, \(< PQR = 180^\circ - 98^\circ = 82^\circ\) \(\therefore < QPR = 180^\circ - (49^\circ + 82^\circ) = 180^\circ - 131^\circ\) = \(49^\circ\) (b) Circumference of the circular track = \(2\pi r\) \(\therefore 2\pi r = 400 \) where \(\pi = \frac{22}{7}\) \(\therefore r = \frac{400 \times 1 \times 7}{22 \times 2} = \frac{700}{11}\) = \(63.63m \approxeq 64m\)

(a) P(obtaining a 3 from a die) = \(\frac{1}{6}\) P(obtaining a tail in 2 coins tossed at once) = \(\frac{1}{2}\). \(\therefore\) P(obtaining 3 and a tail = \(\frac{1}{6} \times \frac{1}{2} = \frac{1}{12}\) (b) No of black marbles = n(B) = 7 No of red marbles = n(R) = 3 Total marble = 10 (i) P(red then black) = P(1st red, then black) or P(1st black, then red) = \(\frac{3}{10} \times \frac{7}{9} + \frac{7}{10} \times \frac{3}{9}\) = \(\frac{7}{15}\) (ii) Prob (two blacks) = \(\frac{7}{10} \times \frac{6}{9} = \frac{7}{15}\)

(a) \(17x = 375^{2} - 356^{2}\) \(17x = (375 + 356)(375 - 356)\) \(17x = (731)(19)\) \(x = \frac{731 \times 19}{17} = 817\) (b) \(4^{x} = 2^{\frac{1}{2}} \times 8\) \(2^{2x} = 2^{\frac{1}{2}} \times 2^{3}\) \(2^{2x} = 2^{3\frac{1}{2}}\) \(\implies 2x = 3\frac{1}{2} \implies x = \frac{7}{4}\) (c) \(S_{n} = \frac{n}{2} (2a + (n - 1)d)\) (sum of terms of an A.P) \(S_{9} = \frac{9}{2} [2a + (9 - 1) d] = \frac{9}{2} [2a + 8d]\) \(72 = 9(a + 4d) \implies 8 = a + 4d ... (1)\) \(S_{9 + 4} = S_{13} = \frac{13}{2} [2a + (13 - 1)d] = \frac{13}{2} [2a + 12d]\) \(72 + 71 = 143 = 13(a + 6d) \implies 11 = a + 6d ... (2)\) \((2) - (1) : 2d = 3 \implies d = \frac{3}{2}\) \(a + 4(1\frac{1}{2}) = 8 \implies a + 6 = 8\) \(\implies a = 8 - 6 = 2\) \(\therefore \text{The A.P is } 2, 3\frac{1}{2}, 5, 6\frac{1}{2}, 8,...\)

(c) |CP| = 6.7cm

(a) Area of the shaded segment = Area of sector - Area of triangle = \(\frac{\theta}{360^\circ} \times \pi r^{2} - \frac{1}{2} ab \sin \theta\) Area of sector = \(\frac{63}{360} \times \frac{22}{7} \times 10^{2}\) = \(55 cm^{2}\) Area of triangle = \(\frac{1}{2} \times 10 \times 10 \times \sin 63^\circ\) = \(50 \times 0.891 = 44.55 cm^{2}\) \(\therefore \text{Area of the shaded segment} = 55 - 44.55 = 10.45 cm^{2}\) (b) (i) Total surface area = \(2\pi r (r + h)\) = \(2 \times \frac{22}{7} \times 3 (3 + 6)\) = \(\frac{132}{7} \times 9 = \frac{1188}{7}\) = \(169.71 cm^{2}\) (ii) Volume of tin = \(\pi r^{2} h\) = \(\frac{22}{7} \times 3 \times 3 \times 6 = \frac{1188}{7}\) \(\approxeq 169.71 cm^{3}\)

(a) x -2.0 -1.5 -1.0 -0.5 0 0.5 1 1.5 2.0 y -15.5 -8.0 -2.5 1 2.5 2.0 -0.5 -5.0 -11.5 (b) see the diagram above (c) The maximum value of y = 2.5. (d) Obtaining the root of \(8x^{2} - 2x - 5 = 0\) Divide both sides by -2, i.e \(-4x^{2} + x + \frac{5}{2} = 0\) \(\therefore \text{Roots are } x = -0.7 ; x = 0.8.

(a) Sector angle represented by the officers: Total number of officers = 240 Clerks = \(\frac{78}{240} \times 360^\circ = 117^\circ\) Drivers = \(\frac{36}{240} \times 360^\circ = 54^\circ\) Typists = \(\frac{44}{240} \times 360^\circ = 66^\circ\) Messengers = \(\frac{52}{240} \times 360^\circ = 78^\circ\) Others = \(\frac{30}{240} \times 360^\circ = 45^\circ\) (b) Marks (x) 4 5 6 7 8 9 Total No of candidates (f) 5 8 5 6 4 2 30 \(fx\) 20 40 30 42 32 18 182 Mean score = \(\frac{\sum fx}{\sum f}\) = \(\frac{182}{30}\) = \(6.067\) marks \(\approxeq\) 6 marks to the nearest whole number.

< RPQ = 50^\circ + 75^\circ = 125^\circ By cosine rule, \(|QR|^{2} = |RP|^{2} + |PQ|^{2} - 2(|PR|)(|PQ|) \cos 125^\circ\) = \(90^{2} + 50^{2} + 2(90)(50) \cos 55^\circ\) = \(8100 + 2500 + 9000 \times 0.5736\) = \(10600 + 5162.4\) \(|QR|^{2} = 15762.4\) \(|QR| = \sqrt{15762.4} = 125.548km\) (b) By Sine rule, \(\frac{RQ}{\sin 125^\circ} = \frac{PR}{\sin

Source: Myschool Past Questions Archive (Record #54604)