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n(S) = 6(odd) = {1, 3, 5} n(odd) = 3 pr(odd) = \(\frac{n(odd)}{n(S)}\) 3/6 =1/2

Area of trapezium = 1/2(a+b)h = 1/2(4 + 7)x4 = 22m2 volume = 22 x 6 = 132m3

The shape is a prism so volume of the tank = base area x height = \(\frac{1}{2}\)( a + b) x height = \(\frac{1}{2}\)(4 + 7) x 4 x 6 = 132cm\(^3\) But 1000cm\(^3\) = 1litres 132cm\(^3\) =

log106 + log1045 - log1027 log10 \(\frac{6 \times 45}{27}\) = log102 x 5 log1010 = 1

In GP, when you are given three consecutive terms, say f, g, h, then \(f \times h = g^2\) Given: \(x, \frac{3}{2}, \frac{6}{7}, y\), then \(\frac{6x}{7} = (\frac{3}{2})^2 \implies \frac{6x}{7} = \frac{9}{4} ... (i)\) Also, \(\frac{3y}{2} = (\frac{6}{7})^2 \implies \frac{3y}{2} = \frac{36}{49} ... (ii)\) From \(\frac{6x}{7} = \frac{9}{4} \implies x = \frac{9 \times 7}{6 \times 4}\) \(x = \frac{21}{8}\) Also, \(\frac{3y}{2} = \frac{36}{49} \implies y = \frac{2 \times 36}{3 \times 49}\) = \(\frac{24}{49}\) \(xy = \frac{21}{8} \times \frac{24}{49} = \frac{9}{7}\)

From the figure \(< ABC = < ADC = 37^\circ\) (angles on the same segment) \(< BCA = 90^\circ\) (angle in semi-circle) \(\therefore\) In \(\Delta ABC\), \(< ACB = 180^\circ - (37^\circ + 90^\circ)\) = 53^\circ

5y 2 + 5ay - 3ay - 3a 2 5y(y + a) - 3a(y + a) (y + a) (5y - 3a)

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

Produce ML to R. From the figure < LRM = 89^\circ (corresponding angles) In triangle LRM, < LMR = 180^\circ - (61^\circ + 89^\circ) = 30^\circ

θ/360* 2πr = Rcos4o 4/360* 40041/1= 445km

Number of oranges = X; Costing N5X. Number of mangoes = 2X; Costing N8X. \(\therefore 65 \leq 5X + 8X \leq 130\) \(5 \leq X \leq 10\)

From the diagram, < LMN = 180^\circ - (74^\circ + 39^\circ) (sum of angle in a triangle) = 67^\circ < LON = 2 x 67^\circ (angle subtended at the centre is twice the angle at the circumference) = 134^\circ

\(\cos^2 \theta + \frac{1}{8} = \sin^2 \theta\)..........(i) from trigometric ratios for an acute angle, where \(cos^{2} \theta + \sin^2 \theta = 1\) ........(ii) Substitute for equation (i) in (i) = \(\cos^2 \theta + \frac{1}{8} = 1 - \cos^2 \theta \) = \(\cos^2 \theta + \cos^2 \theta = 1 - \frac{1}{8}\) \(2\cos^2 \theta = \frac{7}{8}\) \(\cos^2 \theta = \frac{7}{2 \times 8}\) \(\frac{7}{16} = \cos \theta\) \(\sqrt{\frac{7}{16}}\) = \(\frac{\sqrt{7}}{4}\) but cos \(\theta\) = \(\frac{\text{adj}}{\text{hyp}}\) \(opp^2 = hyp^2 - adj^2\) \(opp^2 = 4^2 - (\sqrt{7})^{2}\) = 16 - 7 opp = \(\sqrt{9}\) = 3 \(\tan \theta = \frac{\text{opp}}{\text{adj}}\) = \(\frac{3}{\sqrt{7}}\) \(\frac{3}{\sqrt{7}}\) x \(\frac{\sqrt{7}}{\sqrt{7}}\) = \(\frac{3\sqrt{7}}{7}\)

Thickness of an 800 pages book = 18mm to meter 18 x 103m = 1.8 x 10-2m One leaf = \(\frac{1.8 \times 10^{-2}}{800}\) = \(\frac{1.8 \times 10^{-2}}{8 \times 10^{2}}\) = \(\frac{-1.8}{8}\) x 10-4 = 0.225 x 10-4 = 2.25 x 10-5m one leaf contains 2 pages : 2 * 2.25 x 10\(^{-5}\)m = 4.5 * 10\(^{-5}\)m

\(\frac{4a^2 - 49b^2}{2a^2 - 5ab - 7b^2}\) = \(\frac{(2a)^2 - (7b)^2}{(a + b)(2a - 7b)}\) = \(\frac{(2a + 7b)(2a - 7b)}{(a + b)(2a - 7b)}\) = \(\frac{2a + 7b}{a + b}\)