Question Library

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)

x + 2x + 2x + (x + \(30^o\)) + (x + \(20^o\)) + (x - \(10^o\)) = (2n - 4) x \(90^o\) 8x + 50 \(^o\) - 10\(^o\) = (2 x 6 -4) x 90\(^o\) 8x + 40\(^o\) = 8 x 90\(^o\) = 720\(^o\) 8x = 720\(^o\) - 40\(^o\) = 680\(^o\) x = \(\frac{680^o}{8}\) = 85\(^o\)

|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

Using the two - point from \(\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}\) \(\frac{y - 2}{-2 - 2} = \frac{x - 4}{-8 - 4}\) \(\frac{y - 2}{-4} = \frac{x - 4}{-12}\) \(\frac{-12(y -2)}{-4}\) = x - 4 3(y -2) = x -4 3y - 6 = x - 4 3y = x - 4 + 6 3y = x + 2... By comparing the equations; 3y = px + , p = 1

Mean = \(\frac{\sum x}{n}\) 4 = \(\frac{1 + 3 + 5 + 7 + x}{5}\) 4 x 5 = 16 + x 20 - 16 = x 4 = x x = 4

Prob. (team scored 4 goals) = Prob. (team scored 7 goals) = \(\frac{3}{25}\) Hence, probability that a team selected at random scored either 4 or 7 goals; = \(\frac{6}{25} + \frac{3}{25}\) = \(\frac{9}{25}\)

No. of teams that scored at most 3 goals = 3 + 1 + 6 = 10 Hence, probability that a team selected at random scored at most 3 goals = \(\frac{10}{25}\) = \(\frac{2}{5}\)

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

\(2x^2 - x - 6\) = 0 \(2x^2 - 4x + 3x - 6\) = 0 2x(x - 2) + 3(x - 2) = 0 (2x + 3) (x - 2) = 0 Either; 2x + 3 = 0 or x - 2 = 0 x = \(\frac{-3}{2}\) or x = 2

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

2 x 3, 4 x 6, 8 x 9, 16 x 12,... 2\(^1\) x 3 x 1, 2\(^2\) x 3 x 2, 2\(^3\) x 3 x 3, 2\(^4\) x 3 x 4,.... 2\(^n\) x 3n

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

\(\frac{x}{y}\) = \(\frac{1}{4} \div \frac{3}{8}\) = \(\frac{1}{4} \times \frac{8}{3}\) = \(\frac{2}{3}\) \(\frac{y}{z}\) = \(\frac{1}{3} \div \frac{4}{9}\) = \(\frac{1}{3} \times \frac{9}{4}\) = \(\frac{3}{4}\) But, x = \(\frac{2}{5}T_1\), y = \(\frac{3}{5}T_1\) y = \(\frac{3}{7}T_2\), z = \(\frac{4}{7}T_2\) Using y = y \(\frac{3}{5}T_1\) = x = \(\frac{3}{7}T_2\) \(\frac{T_1}{T_2}\) = \(\frac{3}{7}\) x \(\frac{5}{3}\) = \(\frac{15}{21}\) \(T_1 = 15\) and \(T_2 = 21\) Thus , x = \(\frac{2}{5}\) x 15 = 6 y = \(\frac{3}{5}\) x 15 = 9 y = \(\frac{3}{7}\) x 21 = 9 z = \(\frac{4}{7}\) x 21 = 12 Hence; x : z = 6 : 12 = 1 : 2