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Volume of sphere = Volume of cylinder i.e. \(\frac{4}{3} \pi r^3 = \pi r^2 h\) \(\frac{4}{3} \pi r^3 = \pi \times 3^2 \times 4\) r3 = \(\frac{\pi \times 9 \times 4 \times 3}{4 \pi}\) r = 3\(\sqrt{27}\) = 3

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

Area of \(\Delta\) QNP = \(\frac{1}{2} \times 9 \times 6 \) = 27cm2 Area of \(\Delta\) QMN = Area of \(\Delta\) QNP = Area of \(\Delta\) PNO (triangles between the same parallels) Hence, area of the trapezium 3 x area of \(\Delta\) QNP = 3 x 27 = 81cm2

Perimeter of a sector = 2r + \(\frac{\theta}{360^o}\) x 2 x \(\frac{22}{7}\) x 21 64 = 2 x 21 + \(\frac{\theta}{360^o}\) x 2 x \(\frac{22}{7}\) x 21 64 = 42 + \(\frac{\theta}{360^o}\) x 44 x 3 64 - 42 = \(\frac{\theta}{360^o}\) x 11 x 3 22 = \(\frac{33\theta}{90}\) \(\theta = \frac{22 \times 30}{11}\) = 60o

From the venn diagram given, M = (a, b, c), N = (c, f, g) U = (a, b, c, d, e, f, g) Thus M' \(\cap\) N = (e, f, g) \(\cap\) (c, f, g) = (f, g)

First, reduce 20(mod 9) to its simplest form in mod 9; 9 x 2 + 2 = 2(mod 9) If 2(mod 9) y(mod 6), then y = 2 by comparism

\(\frac{(p - r)^2 - r^2}{2p^2 - 4pr}\) = \(\frac{(p - r)(p - r) - r^2}{2p^2 - 4pr}\)\ = \(\frac{p^2 - 2pr + r^2 - r^2}{2p(p - 2r}\) = \(\frac{p^2 - 2pr}{2p(p - 2r)}\) = \(\frac{p(p - 2r)}{2p(p - 2r)}\) = \(\frac{1}{2}\)

In the diagram, o(angles in same segment) o - o o (same reason as above) o

n(red balls) = 5 n(blue balls) = 4 n(\(\iff\)) = 9 Hence, prob (R1, B2) = \(\frac{5}{9} \times \frac{4}{9}\) = \(\frac{20}{81}\)

Y = x2 + 2x + k (given) y = o when x = 2 thus 0 = 22 + 2 x 2 + k 0 = 4 + 4 + k given k = -8

0, 1, 1, 2, 3, 5, 8. 0 + 1 = 1 1+ 1 = 2 1+ 2 = 3 2 + 3 = 5 3 + 5 = 8 5 + 8 = 13 8 + 13 = 21 13 + 21 = 34

Lower quartile, Q1 = \(\frac{1}{4}\)Nth term = \(\frac{1}{4}\) x 600th term = 150th term = 15.5 + 3 + 0.5 = 19

Let a represent an interior angle; e represent an exterior angle. A section of the polygon is down in the diagram. \(\frac{e}{a}\) = \(\frac{l}{11}\) given a = 11e a + e = 180o(angles on a straight line) 11e + e = 180o 12e = 180o e = \(\frac{180^o}{12}\) = 15o Hence, number of sides = \(\frac{360^o}{\tect{size of one exterior angle}\) = \(\frac{360^o}{14^o}\) = 24

Halima's age = n years old Her brother's age = \(\frac{n}{2}\) + 5

From the diagram sin 70o = \(\frac{x}{10}\) x = 10sin 70o = 9.3969 Hence, length of chord MN = 2x = 2 x 9.3969 = 18.7938 = 19cm (nearest cm)