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a = 50o (Alt. ∠s) Bearing of Q from P = 180o + a = 180o + 50o = 230o

Tan60o = h/55 h = 55Tan60o h 55 x √3 ; h = 55√3m

(a) \(\log_{10} 0.72 =\log_{10} (\frac{72}{100})\) = \(\log_{10} 72 - \log_{10} 100\) = \(72 = 2^{3} \times 3^{2}\) \(\log_{10} 72 = \log_{10} 2^{3} + \log_{10} 3^{2} = 3\log_{10} 2 + 2 \log_{10} 3\) = \(3(0.3010) + 2(0.4771) = 0.9030 + 0.9542 = 1.8572\) \(\therefore \log_{10} 0.72 \equiv \log_{10} 72 - \log_{10} 100\) = \(1.8572 - 2 = -0.1428\) (b) \(\tan 39^\circ = \frac{20}{AB} \) = \(\frac{20}{\tan 39^\circ}\) \(AB = 24.698 m \approxeq 25m\)

(a) \(\frac{0.45 \times 0.91}{0.0117} = \frac{45 \times 91}{117}\) = \(5 \times 7 = 35\) (b) Halfway between \(1\frac{6}{7}\) and \(2\frac{11}{28}\) = \(\frac{1\frac{6}{7} + 2\frac{11}{28}}{2}\) = \((\frac{13}{7} + \frac{67}{28}) \times \frac{1}{2}\) = \(\frac{119}{28} \times \frac{1}{2} \) = \(\frac{119}{56} = 2\frac{7}{56} = 2\frac{1}{8}\) (c) Let the interior angle = 5x^\circ and the exterior angle = x^\circ. \(\therefore 5x^\circ + x^\circ = 180^\circ\) \(6x^\circ = 180^\circ \implies x = 30^\circ\) Exterior angle = \(\frac{360^\circ}{n} \implies 30^\circ = \frac{360^\circ}{n}\) \(n = \frac{360^\circ}{30^\circ} = 12\) The polygon has 12 sides. (d) Length of pipe = 20 cm Internal diameter = 6 cm ; radius = 3cm External diameter = 8 cm ; radius = 4cm Let the external radius = R and internal radius = r \(\therefore\) Volume of material used = \(\pi R^{2} l - \pi r^{2} l\) Where l = length. \(Volume = \pi l (R^{2} - r^{2}) cm^{3}\) = \(\frac{22}{7} \times 20 (4^{2} - 3^{2}) = \frac{22}{7} \times 20 \times 7 = 440 cm^{3}\)

(a) Let the number be N. \(\frac{N}{3} + 8 = 18 - \frac{N}{2}\) \(\frac{N}{3} + \frac{N}{2} = 18 - 8 = 10\) \(\frac{5N}{6} = 10 \implies 5N = 60\) \(N = 12\) (b)

(a) Difference in longitudes = 124^\circ + 56^\circ = 180^\circ Distance PQ = \(\frac{180^\circ}{360^\circ} \times 2\pi r\) where \(r = R \cos \theta\) \(/PQ/ = \frac{180}{360} \times 2 \times 3.142 \times R \cos 68.7^\circ\) \(/PQ/ = 3.142 \times 6400 \times 0.3633\) = \(7305.53 km\) \(\approxeq 7300 km\) (3 sig. figs) (b) Red balls = 4, white = 3 ; green = 5 Total = 4 + 3 + 5 = 12 balls. (i) P(green) = \(\frac{5}{12}\) P(not green) = \(1 - \frac{5}{12} = \frac{7}{12}\) (ii) P(picking red and white) = P( red then white) or P(white then red) = \(\frac{4}{12} \times \frac{3}{11} + \frac{3}{12} \times \frac{4}{11}\) = \(\frac{2}{11}\)

\(\begin{array}{c|c} \text{class intervals} & Fre(F) & \text{class-marks(x)} & Fx & (x - x)& (x - x)^2 & F(x - x)^2 \\ \hline 1 - 3 & 5 & 2 & 10 & -3 & 9 & 45\\ 4 - 6 & 8 & 5 & 40 & 0 & 0 & 0 \\ 7 - 9 & 5 & 8 & 40 & 3 & 9 & 45 \\ \hline & 18 & & 90 & & & 90 \end{array}\) x = \(\frac{\sum fx}{\sum f}\) = \(\frac{90}{18}\) = 5 S.D = \(\frac{\sum f(x - x)^2}{\sum f}\) = \(\frac{90}{18}\) = \(\sqrt{5}\)

Total number of students (y) using the subject French ( 90 / 360 )º x Y = 30 Y = 30 x 4 Y = 120 History- 90º + French- 90º + Economics- 150º + CRK = 360º CRK = 360º - 330º = 30º Number that offered CRK: ( 30 / 360 )º x 120 = 10

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

|AB| = 12 cm; |DC| = 9 cm; |AE| = 8 cm. \(\Delta\) ABE and \(\Delta\) EDC are similar triangles. Hence, \(\frac{|AB|}{|DC|} = \frac{|AE|}{|EC|}\) \(\frac{12}{9} = \frac{8}{|EC|}\) \(\therefore |EC| = \frac{9 \times 8}{12}\) = 6 cm

Area of rhombus = \(\frac{pq}{2}\) where p and q are the diagonals of the rhombus. \(\therefore A = \frac{16 \times 12}{2}\) = 96 cm\(^2\)

Considering the smaller and larger triangle, these two are similar triangles. Hence, If the height of the smaller triangle = h, \(\therefore \frac{h}{6} = \frac{h + 12}{12}\) \(12h = 6h + 72 \implies 6h = 72\) \(h = 12 cm\) \(\therefore\) The height of the cone = 12 + 12 = 24 cm

Perimeter of sector = \(\frac{\theta}{360^\circ} \times 2\pi r + 2r\) = \(\frac{108}{360} \times 2 \times \frac{22}{7} \times \frac{7}{2} + 2(\frac{7}{2})\) = \(6 \frac{3}{5} + 7\) = \(13 \frac{3}{5} cm\)

4x + 3x + 2x + 90 =360^\circ [angle at a point] 9x + 90 = 360^\circ 9x = 360^\circ - 90^\circ 9x = 270 x = 270/9 x = 30^\circ

a = 52o [vertical opp. ∠s are equal] a = b = 52o [Alternate ∠s] b = c = 52o [vertical opp ∠s] c + x + 35o = 180o [sum of angles in a triangle] 52o + x + 35o = 180o x + 87o = 180o x = 180o - 87o = 93o