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(a) \(\sin 105^\circ = \sin(60 + 45)\) = \(\sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ\) = \(\frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2} + \frac{1}{2} \times \frac{\sqrt{2}}{2}\) = \(\frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}\) (b) a = 11, d = 2. \(S_{n} = \frac{n}{2} (2a + (n - 1) d\) \(551 = \frac{n}{2} (2(11) + (n - 1) 2\) \(551 = \frac{n}{2} (22 + 2n - 2)\) \(551 = \frac{n}{2} (2n + 20)\) \(1102 = 2n^{2} + 20n \implies 2n^{2} + 20n - 1102 = 0\) Divide through by 2. \(n^{2} + 10n - 551 = 0\) \(n^{2} - 19n + 29n - 551 = 0\) \(n(n - 19) + 29(n - 19) = 0\) \(\text{n = 19 or -29}\) Since n cannot be negative, n = 19 houses. (c)(i) \(T_{n} = ar^{n - 1}\) (terms of a G.P) \(a = 2\) \(T_{3} = 2 \times r^{3 - 1} = \frac{2}{9}\) \(r^{2} = \frac{1}{9} \implies r = \frac{1}{3}\) (ii) \(T_{5} = ar^{5 - 1}\) = \(2 \times (\frac{1}{3})^{4}\) = \(2 \times \frac{1}{81}\) = \(\frac{2}{81}\)

The numbers from 0 to 20 inclusive are: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20. Total numbers = 21 Odd prime numbers in this range: Primes must be greater than 1 and have no divisors other than 1 and themselves. Odd primes (exclude 2, which is even): 3, 5, 7, 11, 13, 17, 19 → There are 7 odd prime numbers. 3. Probability = \(\frac{\text{Number of favourable outcomes}}{\text{Total outcomes}}\) = \(\dfrac{7}{21} = \dfrac{1}{3}\)

Take half the coefficient of m and square it => (-8/2)2 = k = 16

q = 10\(^{2.7078}\) Antilog of 2.7078 = 510.2

= \(\frac{0.63954}{0.003}\) moving the decimal places, we have = \(\frac{639.54}{3}\) = 213.18 \(\approxeq\) 210 (to 2 s.f.)

log\(_{10}\) a = 4 a = 10\(^4\) = 10000

Let the actual length of the room = y m \(\therefore \frac{y - 3.99}{y} \times 100 = 5%\) \(100(y - 3.99) = 5y \implies 100y - 399 = 5y\) \(100y - 5y = 399 \implies y = \frac{399}{95}\) y = 4.2 m

\(\frac{6x-y}{x+2y}=2\) → \((6x-y) = (x+2y)2\) = 6x - y = 2x + 4y Collect like terms: 6x - 2x = 4y + y → 4x = 5y \(\frac{x}{y} = \frac{5}{4}\)

\(\frac{x}{x-2}-\frac{x+2}{x+3}\) \(\frac{[x][x+3] - ([x+2][x-2])}{[x-2][x+3]}\) = \(\frac{x^2 + 3x - (x^2 - 4)}{[x-2][x+3]}\) = \(\frac{3x + 4}{[x-2][x+3]}\)

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

r2 - r(2p + q) + 2pq = r2 - 2pr -qr + 2pq = r(r - 2p) - q(r - 2p) = (r - q)(r - 2p)

Using Pythagora's rule for △PSM Hyp\(^2\) = Opp\(^2\) + Adj\(^2\) 5\(^2\) = Opp\(^2\) + 4\(^2\) → Opp\(^2\) = 25 - 16 → Opp = \sqrt9 = 3 base SR = 18 + 3 = 21 base PQ = 18 side PS = 5 side QR = 4 perimeter of the trapezium PQRS = 21 + 18 + 5 + 4 = 48cm

diagonal of a square(x) = y\sqrt2 y in terms of x → y = \(\frac{x}{\sqrt2}\)

V = \(\pi r^2 h\) = 123.2 x 1000 = \(\frac{22}{7}\) x 28 x 28 x h = 123200 = 88 x 28h = 2464h ∴ h = \(\frac{123200}{2464}\) = 50cm N.B: 1 litre = 1 dm3 = 1000cm3

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