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(a) (i) (ii) \(32 - x + x + 28 - x + 12 = 45\) \(72 - x = 45\) \(x = 72 - 45 = 27\) 27 students offered both Physics and Government. (iii) \(n(P \cup G) = 45 - 12 = 33\) (b) \(p = \frac{2u}{1 - u} ; q = \frac{1 + u}{1 - u}\) \(\frac{p + q}{p - q}\) \(p + q = \frac{2u}{1 - u} + \frac{1 + u}{1 - u}\) = \(\frac{2u + 1 + u}{1 - u}\) = \(3u + 1}{1 - u}\) \(p - q = \frac{2u - 1 - u}{1 - u}\) = \(\frac{u - 1}{1 - u}\) = \(\frac{-(1 - u)}{1 - u}\) = \(-1\) \(\frac{p + q}{p - q} = \frac{\frac{3u + 1}{1 - u}}{-1}\) = \(\frac{-(3u + 1)}{1 - u}\) = \(\frac{-(3u + 1)}{-(u - 1)}\) = \(\frac{3u + 1}{u - 1}\)

(a) (i) Length of sector = \(\frac{\theta}{360} \times 2\pi R\) Length of the base of the cone formed = \(2\pi r\) \(\frac{90}{360} \times 2 \times \pi \times 8 = 2\pi r\) \(4\pi = 2\pi r\) \(2r = 4 \implies r = 2 cm\) The base radius of the cone = 2 cm. (ii) Height of the cone \(h^{2} = 8^{2} - 2^{2}\) \(h^{2} = 64 - 4 = 60\) \(h = \sqrt{60}\) = \(7.746 cm\) (iii) volume of a cone = \(\frac{1}{3} \pi r^{2} l\) = \(\frac{1}{3} \times \frac{22}{7} \times 2 \times 2 \times 7.746\) = \(\frac{88 \times 7.746}{21}\) = \(32.46 cm^{3}\) (b)(i) \(100,000 cm = 1 km\) 1 cm represents 20,000 metres \(\therefore 4.5 cm = (4.5 \times 20,000) cm\) = 90,000 cm Converting to km, = \(\frac{90,000}{100,000}\) = 0.9km (ii) 1 metre represents 20,000 metres 16km = 16000m Distance on the map in metres = \(\frac{16000}{20000}\) = 0.8m (iii) Area of the forest on the map = \(\frac{85 \times (10^{5})^{2} cm^{2}}{(20000)^{2}}\) = \(\frac{85 \times 10^{10}}{4 \times 10^{8}}\) = \(2125 cm^{2}\)

IV is correct.

If 13m + 24m = 41m 1 * m1 + 3 0 + 2 * m1 + 4 * m0 = 4 * m1 + 1 * m0 1 * m + 3 * 1 + 2 * m + 4 * 1 = 4 * m + 1 * 1 m + 3 + 2m + 4 = 4m + 1 3m + 7 = 4m + 1 4m -3m = 7 - 1 m = 6

\(\frac{321_4}{23_4}\) = \(\frac{(3\times4^{2})+(2\times4^{1})+(1\times4^{0})}{(2\times4^{0})+(3\times4^{0})}\) = \(\frac{3\times16+2\times4+1\times1}{2\times4+3\times1}\) = \(\frac{48+8+1}{8+3}\) =\(\frac{57}{11}\) = 5\(\hspace{1mm}remainder\hspace{1mm}2\) ∴ r = 2\(_{10}\) Now\(\hspace{1mm}convert\hspace{1mm}2_{10} \hspace{1mm}to\hspace{1mm}base\hspace{1mm}4\frac{4}{2}\) = 2\(\frac{4}{0}\) = 0\(\hspace{1mm}or\hspace{1mm}2\) ∴ r = 2

3\(\frac{1}{2}\) - (2\(\frac{1}{3}\) x 1\(\frac{1}{4}\)) + \(\frac{3}{5}\)) = \(\frac{7}{2}\) - (\(\frac{7}{3}\) x \(\frac{5}{4}\)) + \(\frac{3}{5}\)) = \(\frac{7}{2}\) - \(\frac{35}{12}\) + \(\frac{3}{5}\) = L.C.M = 60 = \(\frac{(210 - 175 + 36)}{60}\) = \(\frac{71}{60}\) = 1\(\frac{11}{60}\)

Let x = the number of oranges The 1st received 1/3 of x = 1/3x ∴Remainder = x - 1/3x = 2x/3 The 2nd received 2/3 of 2x/3 = 2/3 * 2x/3 = 4x/3 The 3rd received 12 oranges ∴1/3x + 4x/9 + 12 = x (3x + 4x + 108)/9 = x 3x + 4x + 108 = 9x 7x + 108 = 9x 9x - 7x = 108 2x = 108 x = 54 oranges

\(\frac{81^{\frac{3}{4}}-27^{\frac{1}{3}}}{3 \times 2^3}\) = \(\frac{(3^{3-\frac{3}{4}}-3^{3-\frac{3}{4}})}{3\times 2^3}\) = \(\frac{3^3 - 3}{3 \times 8}\) = \(\frac{27-3}{24}\\=\frac{24}{24}\) =1

Log102 = 0.3010 and Log103 = 0.4771 Log104.5 = Log1041/2 = Log109/2 = Log109 - Log102 = log1032 - Log102 = 2Log103 - Log102 = 2(0.4771) - 0.3010 = 0.9542 - 0.3010 = 0.6532

\(\frac{(\sqrt{12}-\sqrt{3})}{(\sqrt{12}+\sqrt{3})}\) = \(\frac{\sqrt{4\times 3} - \sqrt{3}}{\sqrt{4\times 3}+\sqrt{3}}\) = \(\frac{2\sqrt{3}-\sqrt{3}}{2\sqrt{3}+\sqrt{3}}\) = \(\frac{\sqrt{3}}{3\sqrt{3}}\) = \(\frac{1}{3}\)

Two games played only = 5 + 7 + 3 = 15

m = 3, p = -3, q = 7 and r = \(\frac{5}{2}\) m(p+q+r) = 3(-3 + 7 + \(\frac{5}{2}\)) = 3(4 + \(\frac{5}{2}\)) = 3(4 + 2\(\frac{1}{2}\)) = 3 * 6\(\frac{1}{2}\) = 3 * \(\frac{13}{2}\) = \(\frac{39}{2}\) = 19.50

t = time taken and N = number of men t ∝ 1/N t = K/N K = Nt K = 30 * 6 K = 180 ∴t = 180/N 4 = 180/N 4N = 180 N = 180/4 45 men

7x - 3 > 25 + 3x 7x - 3x > 25 + 3 4x > 28 x > 28/4 x > 7

The graph of the function f(x) is a curve that starts at (-2,0) and ends at (2,0) with a peak at (0,1). The range of values of x for which f(x) \le 0 is the interval where the function's graph is above or touches the x-axis. From the given graph, we can see that this interval is [-2, 2]. Therefore, the correct answer is x \(\leq\) - 2, 0 \(\leq\) x \(\leq\) 2