Question Library

N = \(\sum f\) = 15 Hence the median age is the \(\frac{N + 1}{2}\)th age, i.e. \(\frac{15 + 1}{2}\)th = 8th From the table, the age that falls on the 8th position when arranged in ascending order is 25years

Let \(\delta^2\) and \(\delta\) denote the variance and standard deviation of the distribution respectively. But \(\delta^2\) = 4 (given) Hence \(\delta\) = \(\sqrt{4}\) = 2

Range = Highest score - Lowest score = 10 - 3 = 7

A student can select 2 subjects from 5 subjects in; 5C3 ways, i.e. = \(\frac{5!}{2!(5 - 2)!}\) = \(\frac{5!}{2!3!}\)

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 \(\frac{50}{360} \times 80 \approx 11\)

Sum of angles in a quadrilateral = 360^\circ \(\therefore\) (3y + 10) + (2y + 30) + (y + 20) + 4y = 360 10y + 60 = 360 \(\implies\) 10y = 300 y = 30^\circ

Length of the tile = 30 cm = 0.3m Area of the tile = 0.3 \(\times\) 0,3 = 0.09 m\(^2\) Area of the room = (7.2 \(\times\) 4.2)m\(^2\) Number of tiles = \(\frac{7.2 \times 4.2}{0.09}\) = 336

Length of chord = \(2r \sin (\frac{\theta}{2})\) = \(2 \times 8 \times \sin (\frac{90}{2})\) = \(16 \times \frac{\sqrt{2}}{2}\) = \(8\sqrt{2} cm\)