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(a) (i) (3 - x), 6, (7 - 5x). First term = a = (3 - x) common ratio = \(\frac{6}{3 - x} = \frac{7 - 5x}{6}\) \(36 = (3 - x)(7 - 5x)\) \(36 = 21 - 15x - 7x + 5x^{2}\) \(36 = 21 - 22x + 5x^{2}\) \(5x^{2} - 22x + 21 - 36 = 0\) \(5x^{2} - 22x - 15 = 0\) \(5x^{2} - 25x + 3x - 15 = 0 \implies 5x(x - 5) + 3(x - 5) = 0\) \((x - 5)(5x + 3) = 0\) \(x = -\frac{3}{5} \text{ or = } 5\) (ii) Constant ratio = \(\frac{6}{3 - x} = \frac{7 - 5x}{6}\) Using x = 5, \(r = \frac{7 - 5(5)}{6} = \frac{-18}{6}\) = \(-3\) (b) Considering \(\Delta\) AOC in the above diagram. \(b^{2} = 4^{2} + 3^{2}\) \(b^{2} = 16 + 9 = 25\) \(b = \sqrt{25} = 5 cm\) Considering \(\Delta\) ACD, Using cosine rule, \(\cos D = \frac{a^{2} + b^{2} - c^{2}}{2ab} = \frac{a^{2} + c^{2} - b^{2}}{2ac}\) = \(\frac{6^{2} + 7^{2} - 5^{2}}{2 (6)(7)}\) = \(\frac{36 + 49 - 25}{84}\) \(\cos D = \frac{60}{84}\) \(\cos D = 0.714\) \(D = \cos^{-1} (0.714)\) \(D = 44.4^\circ \approxeq 44^\circ\) (to the nearest degree)

(a) diagram above. (b) From the figure above, (i) |WZ| = 5.25 cm ; (ii) |ZN| = 4.86 cm.

(a) Length of the chord = 14 cm Radius = \(\frac{3}{2} \times 14 cm = 21 cm\) The chord AB = \(2r \sin \frac{\theta}{2}\) \(14 = 2(21) \sin \frac{\theta}{2}\) \(\sin \frac{\theta}{2} = \frac{14}{42} = 0.333\) \(\frac{\theta}{2} = \sin^{-1} (0.333)\) \(\frac{\theta}{2} = 19.469^\circ\) \(\theta = 19.469^\circ \times 2 = 38.938^\circ\) Length of the arc = \(\frac{\theta}{360^\circ} \times 2\pi r\) \(\frac{38.938}{360} \times 2 \times \frac{22}{7} \times 21\) = \(14.277 cm\) Perimeter of the remaining portion = 22 + 12 + 12 + 5 + 3 + 14.277 = 68.277 cm \(\approxeq\) 68.28 cm (2 decimal places). (b) \(\frac{3}{\sqrt{3}}(\frac{2}{\sqrt{3}} - \frac{\sqrt{12}}{6}\) = \(\frac{3}{\sqrt{3}}(\frac{12 - \sqrt{36}}{6\sqrt{3}}\) = \(\frac{3}{\sqrt{3}}(\frac{12 - 6}{6\sqrt{3}}\) = \(\frac{3}{\sqrt{3}}(\frac{6}{6\sqrt{3}}\) = \(\frac{18}{6\sqrt{9}}\) = \(\frac{18}{18}\) = 1.

Marks frequency cum freq lower bound upper bound 1 - 10 2 2 0.5 10.5 11 - 20 3 5 10.5 20.5 21 - 30 5 10 20.5 30.5 31 - 40 13 23 30.5 40.5 41 - 50 19 42 40.5 50.5 51 - 60 31 73 50.5 60.5 61 - 70 13 86 60.5 70.5 71 - 80 9 95 70.5 80.5 81 - 90 4 99 80.5 90.5 91 - 100 1 100 90.5 100.5 (a) see the diagram above. (b) (i) The 60th percentile is where the line from the cumulative frequency of 60 touches the curve i.e. 60th percentile = 58.5 (ii) If the pass mark was fixed at 35%, then the frequency will be 13 with the class mark 31 - 40.

(a) see diagram above. (b)(i) \(Speed = \frac{distance}{time}\) \(950 = \frac{d}{4} \implies d = 950 \times 4\) \(d = 3800km\) \(d = \frac{\theta}{360} \times 2\pi r\) \(3800 = \frac{\theta}{360} \times 2 \times \frac{22}{7} \times 6400\) \(3800 = \frac{\theta \times 281600}{2520}\) \(\theta = \frac{2520 \times 3800}{281600}\) \(\theta = 34.01^\circ\) \(\theta = 34^\circN\) (to the nearest degree) (ii) Distance between P and Q, correct to four significant figures. Longitude difference = 65^\circ - 15^\circ = 50^\circ Using, \(d = \frac{\theta}{360} \times 2 \pi r\) where \(r = R \cos \theta\) \(d = \frac{\theta}{360} \times 2 \pi R \cos \theta\) = \(\frac{50}{360} \times 2 \times \frac{22}{7} \times 6400 \cos 34.01^\circ\) = \(4631.53 km\) \(\approxeq 4632km\) (to 4 significant figures).

(a) see diagram above (b) From the above diagram, \(\sin 37^\circ = \frac{5}{|LM|}\) \(|LM| = \frac{5}{\sin 37}\) \(|LM| = \frac{5}{0.6018}\) = \(8.308 m\) \(\approxeq 8.31 m\) \(\cos 37 = \frac{|AM|}{|LM|}\) \(|AM| = 8.308 \times \cos 37\) \(|AM| = 8.308 \times 0.7986\) \(|AM| = 6.6348 m\) (c) The new |AM| = 6.6348 m - 2 = 4.6348 m \(\cos y = \frac{4.6348}{8.308}\) \(\cos y = 0.5579\) \(y = \cos^{-1} (0.5579)\) \(y = 56.089^\circ\) \(y = 56^\circ\) (to the nearest degree).

Sum of angles in a quadrilateral = 360^\circ \(\therefore\) (3y + 10) + (2y + 30) + (y + 20) + 4y = 360 10y + 60 = 360 \(\implies\) 10y = 300 y = 30^\circ

Length of the tile = 30 cm = 0.3m Area of the tile = 0.3 \(\times\) 0,3 = 0.09 m\(^2\) Area of the room = (7.2 \(\times\) 4.2)m\(^2\) Number of tiles = \(\frac{7.2 \times 4.2}{0.09}\) = 336

Length of chord = \(2r \sin (\frac{\theta}{2})\) = \(2 \times 8 \times \sin (\frac{90}{2})\) = \(16 \times \frac{\sqrt{2}}{2}\) = \(8\sqrt{2} cm\)

Angle of major sector = 360^\circ - 120^\circ = 240^\circ Area of major sector : \(\frac{\theta}{360} \times \pi r^{2}\) r = \(\frac{4\sqrt{3}}{2} = 2\sqrt{3} cm\) Area : \(\frac{240}{360} \times \pi \times (2\sqrt{3})^{2}\) = \(8\pi cm^{2}\)

If you take any point that is equidistant from the line PQ, it lies at a constant distance above and below the line. Therefore, the correct answer is: a pair of parallel lines to PQ.

Midpoint of a line PQ where P has coordinates (\(x_1\), \(y_1\)) and Q has coordinates (\(x_2\), \(y_2\)) is given as \((\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})\). \(\therefore\) If Q has coordinates (r, s), then \(\frac{-2 + r}{2} = 2\) and \(\frac{1 + s}{2} = 3\) \(-2 + r = 4 \implies r = 6\) \(1 + s = 6 \implies s = 5\) Q = (6, 5)

Given P(2, -3) and Q(-5, 1) Midpoint = \((\frac{2 + (-5)}{2}, \frac{-3 + 1}{2})\) = \((\frac{-3}{2}, -1)\) Slope of the line PQ = \(\frac{1 - (-3)}{-5 - 2}\) = \(-\frac{4}{7}\) The slope of the perpendicular line to PQ = \(\frac{-1}{-\frac{4}{7}}\) = \(\frac{7}{4}\) The equation of the perpendicular line: \(y = \frac{7}{4}x + b\) Using a point on the line (in this case, the midpoint) to find the value of b (the intercept). \(-1 = (\frac{7}{4})(\frac{-3}{2}) + b\) \(-1 + \frac{21}{8} = \frac{13}{8} = b\) \(\therefore\) The equation of the perpendicular bisector of the line PQ is \(y = \frac{7}{4}x + \frac{13}{8}\) \(\equiv 8y = 14x + 13 \implies 8y - 14x - 13 = 0\)

Using the cosine rule, we have \(p^{2} = q^{2} + r^{2} - 2qr \cos P\) \(p^{2} = 8^{2} + 6^{2} - 2(8)(6)(\frac{1}{12})\) = \(64 + 36 - 8\) \(p^{2} = 92 \therefore p = \sqrt{92} cm\)

\(\tan \theta = \frac{opp}{adj} = \frac{3}{4}\) \(hyp^{2} = opp^{2} + adj^{2}\) \(hyp = \sqrt{3^{2} + 4^{2}}\) = 5 \(\sin \theta = \frac{3}{5}; \cos \theta = \frac{4}{5}\) \(\sin \theta + \cos \theta = \frac{3}{5} + \frac{4}{5}\) = \(\frac{7}{5} = 1\frac{2}{5}\)