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Marks 3 4 5 6 7 8 Frequency 5 x - 1 x 9 4 1 fx 15 4x - 4 5x 54 28 8 \(\sum fx = 15 + 4x - 4 + 5x + 54 + 28 + 8 = 101 + 9x\) \(\sum f = 5 + x - 1 + x + 9 + 4 + 1 = 18 + 2x\) \(\bar{x} = \frac{\sum fx}{\sum f}\) \(5 = \frac{101 + 9x}{18 + 2x} \implies 101 + 9x = 5(18 + 2x)\) \(101 + 9x = 90 + 10x \implies 101 - 90 = 10x - 9x\) \(11 = x \) (b)(i) Mode = 5. (ii) Median Frequency = 18 + 2(11) = 40. Median position = \(\frac{40}{2} = 20\) 20th position = 5. (c) No of students that scored at least 7 marks = 4 + 1 = 5. Probability of scoring at least 7 marks = \(\frac{5}{40} = \frac{1}{8}\)

(a)(i) Volume of outer pipe = \(\pi r^{2} h\) = \(\frac{22}{7} \times (5.0)^{2} \times 28 \times 100\) Volume of inner pipe = \(\frac{22}{7} \times (3.5)^{2} \times 28 \times 100\) Volume of metal used = Volume of outer pipe - Volume of inner pipe = \(22 \times 400 \times (5.0^{2} - 3.5^{2})\) = \(22 \times 400 \times (5.0 - 3.5) \times (5.0 + 3.5)\) = \(112,200 cm^{3}\) (ii) Vol of water the pipe can hold = vol of inner pipe = \(\frac{22}{7} \times (3.5)^{2} \times 28 \times 100\) = \(107,800 cm^{3}\) = 107.8 litres. (b) < PMN = < LNM (alternate angles) < PMN = < LMN (angles in alternate segment) \(\therefore\) < LNM = < NLM, then \(\Delta\) LMN is isosceles.

(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) \(T_{5} = ar^{5 - 1}\) = \(2 \times (\frac{1}{3})^{4}\) = \(2 \times \frac{1}{81}\) = \(\frac{2}{81}\)

(a) \(TQ = 10 \sin 70^\circ = 10 \times 0.9397 = 9.397m\) \(\approxeq 9m\) (b)(i) \(QF = 10 \cos 70^\circ = 10 \times 0.3420 = 3.42m\) \(QF^{1} = 3.42m + 1m = 4.42m\) \(QT^{1} = \sqrt{10^{2} - 4.42^{2}} = \sqrt{100 - 19.5364}\) = \(\sqrt{80.4636} = 8.97m\) \(TT^{1} = 9.397 - 8.97 = 0.43m\) (ii) \(\sin < QF^{1}T^{1} = \frac{8,97}{10} = 0.897\) \(< QF^{1} T^{1} = 63.768^\circ \approxeq 64^\circ\)

Let P represent the cost ; n, the number of pupils and k and c, constant terms. \(\therefore P = k + cn\) \(15705 = k + 50c ... (1)\) \(13305 = k + 40c .... (2)\) (1) - (2) : \(2400 = 10c \implies c = \frac{2400}{10} = 240\) Put c = 240 in (1), so that \(15705 = k + (50)(240)\) \(15705 = k + 12000 \implies k = 3705\) \(P = 3705 + 240n\) ------ formula connecting price and number of pupils. (a) When number of pupils, n = 44 \(P = 3705 + 240(44) = 3705 + 10560 = $14265\) (b) For school not to be at loss: \(360 \geq k + cn\) \(360n \geq 3705 + 240n\) \(360n - 240n \geq 3705 \implies 120n \geq 3705\) \(n \geq \frac{3705}{120}\) \(n \geq 30.875 \) \(n \geq 31\) The pupils should be 31 at least for the school not to run at loss.

Coordinate of R and S : R(-1.6, 6.5) and S(-1.6, 0.6). Area of rhombus PRQS : \(\frac{(5 + 2) \times (6.5 + 0.6)}{2} = \frac{49.7}{2}\) = \(24.85 cm^{2}\)

(a) Price with 20% gain = 100% = x Selling price = 100% - 10% = 90% i.e. 90% of x = 864 \(\therefore x = 864 \times \frac{100}{90} = N960\) Let cost price = y = 100% x = 20% of y + y = 120% of y. \(y = \frac{100}{120} x = \frac{100}{120} \times N960 = N800\) (b) (i) \((x + 6)^{2} = (x + 5)^{2} + (x - 2)^{2}\) \(x^{2} + 12x + 36 = x^{2} + 10x + 25 + x^{2} - 4x + 4\) \(x^{2} + 12x + 36 = 2x^{2} + 6x + 29\) \(2x^{2} - x^{2} + 6x - 12x + 29 - 36 = 0\) \(x^{2} - 6x - 7 = 0\) \(x^{2} - 7x + x - 7 = 0 \implies x(x - 7) + 1(x - 7) = 0\) \((x - 7)(x + 1) = 0 \implies \text{x = 7 or -1}\) Since measurements cannot be negative, then x = 7 is the suitable answer. (ii) Length of the lawn = (x + 5) metres = (7 + 5) = 12 metres. Width of the lawn = (x - 2) metres = (7 - 2) = 5 metres \(\therefore \text{The area of the lawn} = 12 \times 5 = 60 m^{2}\)

x 0^\circ 30^\circ 60^\circ 90^\circ 120^\circ 150^\circ 180^\circ 210^\circ y 9 10.3 8.8 5 -0.2 -5.3 -9 -10.3 (b) (c)(i) From the graph, \(9\cos x + 5\sin x = 0 \implies x = 119^\circ\) (d) Maximum value of y = 10.3.

< VSU = 90^\circ (angle in a semi-circle) \(\therefore\) < SUV = 180^\circ - 90^\circ - 50^\circ = 40^\circ < RSV = 40^\circ (angle which chord makes tangent = angle in alternate segment)

loot - /luːt/ boot - /buːt/ rout - /raʊt/ route - /ruːt/

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

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

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

Research /rɪˈsəːtʃ/ disturb /dɪˈstəːb/

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