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\(\sqrt{108} + \sqrt{125} - \sqrt{75}\) = \(\sqrt{3 \times 36} + \sqrt{5 \times 25} - \sqrt{3 \times 25}\) = \(6 \sqrt{3} + 5 \sqrt{5} - 5 \sqrt{3}\) = \(\sqrt{3} + 5\sqrt{5}\)

\([64^{\frac{1}{2}} + 125^{\frac{1}{3}}]^2\) = \([\sqrt{64} + \sqrt[3] {125}]^2\) \([8 + 5]^2\) = \([13]^2\) = 169

\(32_4 = 22_x\) \(3 \times 4^1 + 2 \times 4^o\) = \(2 \times x^1 + 2 \times x^o\) 12 + 2 x 1 = 2x + 2 x 1 14 = 2x + 2 14 - 2 = 2x 12 = 2x x = \(\frac{12}{2}\) x = 6

Population of school = 250 + 150 = 400 60% of 250 = \(\frac{\text{60%}}{\text{100%}}\) x 250 = 150 40% of 150 = \(\frac{\text{40%}}{\text{100%}}\) x 150 = 60 Total number of students who plays football; 150 + 60 = 210 Percentage of school that play football; \(\frac{210}{400}\) x 100% = 52.5%

\(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = 1 \(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = \(\log_{10}\)10 \(\log_{10}\)\(\frac{6x - 4}{2}\) - \(\log_{10}\)10 \(\frac{6x - 4}{2}\) = 10 6x - 4 = 2 x 10 = 20 6x = 20 + 4 6x = 20 x = \(\frac{24}{6}\) x = 4

F = \(\frac{9}{5}\)C + 32 When F = 98.6 98.6 = \(\frac{9}{5}\)C + 32 98.6 - 32 = \(\frac{9}{5}\)C 66.6= \(\frac{9}{5}\)C 66.6 x 5 = 9C C = \(\frac{66.6 \times 5}{9}\) = 37

y + 2x = 4 .....(1) y - 3x = -1 ......(2) Subtract (2) from (1) 2x - (-3x) = 4 - (-1) 2x + 3x = 4 + 1 5x = 5 X = \(\frac{5}{5}\) = 1 Substitute 1 for x in (1); y + 2(1) = 4 y + 2 = 4 y = 4 - 2 = 2 Hence, (x + y) = (1 + 2) = 3

\(\frac{2 - 18m^2}{1 + 3m}\) = \(\frac{2(1 - 9)m^2}{1 + 3m}\) = \(\frac{2(1 + 3m)(1 - 3m)}{1 + 3m}\) = \(2(1 - 3m)\)

Surface area of a sphere = \(4 \pi r^2\) \(4 \pi r^2\) = \(\frac{792}{7}cm^2\) 4 x \(\frac{22}{7}\) x \(r^2\) = \(\frac{792}{7}\) \(r^2\) = \(\frac{792}{7}\) x \(\frac{7}{4 \times 22}\) = 9 r = \(\sqrt{9}\) = 3cm Hence, volume of sphere = \(\frac{4}{3} \pi r^3\) = \(\frac{4}{3} \times \frac{22}{7} \times 3 \times 3 \times 3 \) = \(\frac{4 \times 22 \times 9}{7}\) \(\approx\) = 113.143 = 113\(cm^3\) (to the nearest whole number)

|MN| = \(\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\) = \(\sqrt{(-3 -5)^2 + (8 - 7)^2}\) = \(\sqrt{(-8)^2 + (8 + 7)^2}\) = \(\sqrt{64 + (15)^2}\) = \(\sqrt{64 + 225}\) = \(\sqrt{289}\) = 17 units

Total surface area of hemisphere is 3\(\pi r^2\) = 75\(\pi cm^2\) \(r^2\) = \(\frac{75 \pi}{3 \pi}\) \(r^2\) = 25 r = \(\sqrt{25}\) r = 5cm

The expression \(\frac{x - 5}{x(x - 1)}\) is undefined when x(x - 1) = 0 either x = 0 or x - 1 = 0 Hence, x = 0 or x = 1

\((x + 2)^2\) - \((2x + 1)^2\) = \((x^2 + 4x + 4) - (4x^2 + 4x + 1)\) = \(x^2 \) + 4x + 4 - 4 \(x^2 \) - 4x - 1 = -3 \(x^2 \) + 3 = 3 - 3 \(x^2 \) = 3(1 - \(x^2 \)) = 3(1 + x)(1 - x)

3x \(\equiv\) 4(mod 5) In modulo 5, multiples of 5 that give solution to the given equation are 5, 20, 35, 50,... but 5 will yield the leaast value of x. Thus; 3x = 4 + 5 = 9 x = \(\frac{9}{3}\) x = 3

\(Q_1 = \frac{1}{4}\) (N + 1)th \(\frac{1}{4} \times 12^{th}\) no. = 3rd no (\(\cong\) 4) \(Q_3 = \frac{3}{4}\) (N + 1)th = \(\frac{3}{4}\) x 12th no. = 9th no. (\(\cong\) 12) Hence, interquartile range = \(Q_3 - Q_1\) = 12 - 4 = 8