Question Library

5 people can take 3 places in; 5P3 ways, = \(\frac{5!}{(5 - 3)!}\) = \(\frac{5!}{2!}\) = \(\frac{5 \times 4 \times 3 \times 2!}{2!}\) = 5 x 4 x 3 = 60 ways

\((1 \leq x \leq 25)\) = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25} Number N of x divisible by both 2 and 3 is 4. n(\(\varepsilon\)) = 25 = \(\frac{N}{n(\varepsilon)}\) = \(\frac{4}{25}\)

n(apples) = 9 n(bananas) = 8 n(oranges) = 7 n(\(\varepsilon\)) = 24 Hence Prob(not apple, nor orange) = Prob(banana) = \(\frac{8}{24}\) = \(\frac{1}{3}\)

The graph crosses the x-axis at x = -1 and x = 2 Thus, x + 1 = 0 and x - 2 = 0 x\(^2\) - 2x + x - 2 = 0 x\(^2\) - x - 2 = 0 giving y = x\(^2\) - x - 2

In the diagram above, o (base o - (15 + 15)o = 180o - 30o = 150o But x + 110o = 150o (Sum of opp. interior o - 110o = 40o

Sum of the angles in a triangle = 180^\circ 180^\circ - x + 60 + 70 = 180 130^\circ - x = 180^\circ - 180^\circ = 0^\circ x = 130^\circ

80 + 160 + 200 + 80 + 128 + 72 = 720minutes

5x^\circ + (16x - 24)^\circ + 5x^\circ + (4x + 12)^\circ + (6x + 12)^\circ = 360^\circ 36x^\circ - 24 + 12 + 12 = 360^\circ 36x^\circ = 360^\circ x^\circ = \(\frac{360^\circ}{36}\) = 10^\circ Thus, the angle of sector representing Mathematics is 5 x 10^\circ = 50^\circ. Hence the number of students who offer mathematics is \(\frac{50}{360} \times 80 \approx 11\)

2.7 x 10-4 x 6.3 x 106 = 2.7 x 6.3 x 10-4 x 106 = 17.01 x 10-4 + 6 = 17.01 x 102 = 1.701 x 101 x 102 = 1.701 x 101 + 2 = 1.701 x 103

9(2 - x) = 3 32(2 - x) = 3 2(2 - x) = 1 4 - 2x = 1 -2x = 1 - 4 -2x = -3 x = \(\frac{-3}{-2}\) x = \(\frac{3}{2}\)

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

\(\frac{1\frac{7}{8} \times 2\frac{2}{5}}{6\frac{3}{4} \div \frac{3}{4}}\) from numerator \(1 \frac{7}{8} \times 2 \frac{2}{5}\) = \(\frac{15}{8} \times \frac{12}{5}\) = \(\frac{3 \times 3}{2 \times 1} = \frac{9}{2}\) from denominator \(6\frac{3}{4} \div \frac{3}{4}\) = \(\frac{27}{4} \div \frac{3}{4}\) = \(\frac{27}{4} \times \frac{4}{3}\) = \(\frac{9 \times 1}{1 \times 1} = \frac{9}{1}\) \(\frac{9}{2} \div \frac{9}{1} = \frac{9}{2} \times \frac{1}{9}\) = \(\frac{1}{2}\)

Un = n(n2 + 1) U5 = 5(2 + 1) = 5(25 + 1) = 5(26) = 130 U4 = 4(42 + 1) = 4(16 + 1) = 4(17) = 68 U5 - U4 = 130 - 68 = 62

\(\sqrt{50} - K\sqrt{8} = \frac{2}{\sqrt{2}}\) \(\sqrt{50} - \frac{2}{\sqrt{2}}\) = K\(\sqrt{8}\) = \(\sqrt{2} \times 25 - \frac{2}{\sqrt{2}}\) = K \(\sqrt{4 \times 2}\) \(\frac{5\sqrt{2}}{1} - \frac{2}{\sqrt{2}}\) = 2K\(\sqrt{2}\) \(\frac{5\sqrt{4} - 2}{\sqrt{2}} = 2K\sqrt{2}\) \(\frac{10 - 2}{\sqrt{2}} = 2K \sqrt{2}\) \(\frac{8}{\sqrt{2}} = \frac{2K\sqrt{2}}{1}\) = 2k\(\sqrt{2} \times \sqrt{2}\) = 8 2k \(\sqrt{4}\) = 8 2k x 2 = 8 4k = 8 k = \(\frac{8}{4}\) k = 2

% error = \(\frac{error}{\text{actual value}} \times 100\) error = N72 - N68 = 4 actual value = N72 %error = \(\frac{4}{72} \times 100\) = \(\frac{100}{18} = \frac{50}{9}\) = 5\(\frac{5}{9}\)%