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Let the slant height = L ; height = 24cm ; r = 7cm. \(\therefore L^{2} = 24^{2} + 7^{2} = 576 + 49 = 625\) \(L = \sqrt{625} = 25cm\) Surface area of the structure = Area of cone + Area of hemisphere = \(\pi r L + 2 \pi r^{2} = \pi r (L + 2r)\) = \(\frac{22}{7} \times 7 (25 + 2(7))\) = \(22 \times 39 = 858 cm^{2}\)

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(a) (i) Given \(\cos \alpha = 0.6717\) \(\alpha = \cos^{-1} 0.6717 = 47.8^\circ\) (From cosine of angle table) (ii) \(\sin 47.8^\circ = 0.7408\) (b) From the diagram, \(< BPT = < PTR = 23.6^\circ\) \(\therefore |BT| = |BP| \tan 23.6^\circ\) = \(50 \tan 23.6^\circ = 50 \times 0.4369\) = \(21.8469 m\) \(\therefore\) The height of the building = 22m (to the nearest metre).

(a) (i) Area of minor segment POQ = \(\frac{84}{360} \times \frac{22}{7} \times (3.2)^{2}\) = \(\frac{11 \times 10.24}{15}\) = \(7.5093 cm^{2} \approxeq 7.51 cm^{2}\) (2 decimal places) (ii) Area of \(\Delta\) in POQ = \(\frac{1}{2} \times 3.2 \times 3.2 \times \sin 84^\circ\) = \(\frac{127.296}{25} = 5.0918 cm^{2}\) Area of segment PQ (shaded part) = Area of sector POQ - Area of triangle = \(7.5093 cm^{2} - 5.0918 cm^{2} = 2.4175 cm^{2}\) \(\approxeq 2.42 cm^{2}\) (b) Area of sector POQ = Curved surface area of the cone \(7.5093 cm^{2} = \pi rl\) where r = base radius, L = slant height = 3.2cm \(7.5093 = \frac{22}{7} \times r \times 3.2 = \frac{70.4r}{7}\) \(70.4r = 7.5093 \times 7 \implies r = \frac{7.5093 \times 7}{70.4} = 0.7467 cm\) \(\approxeq 0.7cm\)

Table of values of \(d = 60t - 10t^{2}\) t 1 2 3 4 5 \(60t\) 60 120 180 240 300 \(- 10t^{2}\) -10 -40 -90 -160 -250 d 50 80 90 80 50 (a)(i) It takes 1.6 seconds to reach a height of 70 metres. (ii) Height of stone at 5 seconds was 50 metres but after 5 seconds, d < 50 metres. (iii) It reaches its maximum height after 2.9 seconds. (b) From the graph, the gradient of the slope = \(\frac{AB}{BC}\) = \(\frac{56}{2.8} = 20 m/s\) But it is a negative slope, hence the slope = -20 m/s.

(a)(i) Given a circle APB with centre O. To prove that < AOB = 2 < APB. Construction: Draw PO and produce it to Q. Proof : Referring to the lettering in the figure, |OA| = |OB| (radius) \(\therefore x_{1} = x_{2}\) (base angles of isosceles triangle) \(< AOQ = x_{1} + x_{2}\) (exterior angles of triangle AOP) \(\therefore < AOQ = 2x_{2} (x_{1} = x_{2})\) Similarly, \(< BOQ = 2y_{2}\) \(< AOB = < AOQ + < BOQ \) Hence, \(< AOB = 2x_{2} + 2y_{2} = 2(x_{2} + y_{2})\) = \(2 \times < APB\) (ii) From the figure, \(|OQ| = |QT|\) (radius) \(< OQT = 22^\circ = < QTO\) (base angles of isosceles triangle) \(\therefore < POQ = 44^\circ \) (exterior angle of \(\Delta\) QOT) \(< QOR = 180^\circ - (44^\circ + 98^\circ) = 180^\circ - 142^\circ = 38^\circ\) (Angles on a straight line) \(< QOR = 2 \times < QSR = 38^\circ\) \(\therefore < QSR = \frac{38^\circ}{2} = 19^\circ\) (b) \(< BHC = (180^\circ - 70^\circ) = 110^\circ\) (angle on a straight line) \(< DAC = 41^\circ = < DBC\) (angles on the same segment) But \(< DBC = < HBC\) \(< AHB = < HBC + < HCB \) (exterior angles of \(\Delta\) BHC) \(70^\circ = 41^\circ + < HCB \implies < HCB = 70^\circ - 41^\circ = 29^\circ\) \(< ACB = < HCB = 29^\circ\)

(a) \(T_{n} = a + (n - 1) d\) (terms of an A.P) Given: \(T_{8} - T_{4} = 20\) \(T_{8} = 1\frac{1}{2} \times T_{4}\) \(\therefore a + 7d - (a + 3d) = 20\) \(4d = 20 \implies d = 5\) (ii) Put d = 5 in the equation \(a + 7d = 1\frac{1}{2} \times (a + 3d)\) \(a + 7d = 1.5a + 4.5d\) \(7d - 4.5d = 1.5a - a \implies 2.5d = 0.5a\) \(a = \frac{2.5 \times 5}{0.5} = 25\) (b) \(A = P(1 - \frac{r}{100})^{n}\) where P = N10,250.00 r = 5% n = 9 \(\therefore A = (10,250) (1 - \frac{5}{100})^{9}\) = \(10,250 \times (0.95)^{9} = 10,250 \times 0.6302\) = \(N6,460.056 \approxeq N6,460\) (to 3 sig. figs)

(a) Sample space for a pair of fair dice = 36 results + 1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12 Total with sum at least 9 = 10 P(sum at least 9) = \(\frac{10}{36} = \frac{5}{18}\) (b) P(civil servant owns a car) = \(\frac{1}{6}\) P(civil servant owns no car) = \(\frac{5}{6}\) (i) P(A and B own a car) = \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\) (ii) P(C and D, only one owns a car) = P(C owns, not D) + P(not C but D owns) = \((\frac{1}{6} \times \frac{5}{6}) + (\frac{5}{6} \times \frac{1}{6})\) = \(\frac{5}{36} + \frac{5}{36}\) = \(\frac{10}{36} = \frac{5}{18}\) (iii) P(of X, Y and Z, only one owns a car) = P(X owns a car, not Y, not Z) + P(not X, Y owns a car, not Z) + P(not X, not Y, Z owns a car) = \(3(\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6})\) = \(3 \times \frac{25}{216}\) = \(\frac{25}{72}\)

(a) From the diagram and by Pythagoras theorem, \(|PR|^{2} = |PQ|^{2} + |QR|^{2}\) \(|PR|^{2} = (3a)^{2} + (4a)^{2}\) \(|PR|^{2} = 9a^{2} + 16a^{2} = 25a^{2}\) \(|PR| = \sqrt{25a^{2}} = 5a\) (b) (ii) From the figure, Bearing of Y from X = 060^\circ Using cosine rule, \(\cos XYZ = \frac{24^{2} + 18^{2} - 30^{2}}{2 \times 24 \times 18}\) \(\cos XYZ = 0 \implies < XYZ = 90^\circ\) \(< RYZ = 90^\circ - 60^\circ = 30^\circ\) \(< N_{2} YZ = 180^\circ - 30^\circ = 150^\circ\) (angles on a straight line) The bearing of Z from Y is 150^\circ. (iii) \(\cos XZY = \frac{30^{2} + 18^{2} - 24^{2}}{2 \times 30 \times 18}\) = \(\frac{900 + 324 - 576}{1080}\) \(\cos XZY = \frac{648}{1080} = 0.6\) \(< XZY = \cos^{-1} (0.6) = 53.13^\circ\) \(\therefore\) The bearing of X from Y = 360^\circ - (53.13^\circ + 30^\circ) = 276.87^\circ.

(a) Roots of the equation = \(\frac{1}{2}\) and -7 \(\therefore (x - \frac{1}{2})(x + 7) = 0\) \(x^{2} - \frac{1}{2} x + 7x - \frac{7}{2} = 0\) \(x^{2} + \frac{13}{2} x - \frac{7}{2} = 0\) Multiply through by 2 to get integers as coefficients, \(2x^{2} + 13x - 7 = 0\) (b) Let the father and daughter's present ages be r and t respectively. Three years ago, father = (r - 3) years \((r - 3) = 4t \implies r = (4t + 3) ... (1)\) \(r \times t = 430 ... (2)\) Substitute (1) in (2), \((4t + 3) \times t = 430 \implies 4t^{2} + 3t = 430\) \(\implies 4t^{2} + 3t - 430 = 0\) \(4t^{2} - 40t + 43t - 430 = 0\) \(4t(t - 10) + 43(t - 10) = 0\) \(\therefore (4t + 43)(t - 10) = 0\) \(t = \frac{-43}{4}; t = 10\) Since t is age, it cannot be negative. Hence, t = 10 years. Recall : \(r = 4t + 3\) \(\therefore r = 4(10) + 3 = 40 + 3 = 43\) Therefore, the father's present age is 43 years while the daughter's age is 10 years.

(101.5)2 - (100.5)2;(101.5 + 100.5) 202 x 1 = 202

0.06 x 0.09 = 0.0054 = 5.4 x 10-3

36\(^\frac{1}{2}\) x 64\(^{-\frac{1}{3}}\) x 5\(^0\) = \(6 \times \frac{1}{4} \times 1\) = \(1\frac{1}{2}\)

Un = a + (n-1)d      = -3 + (8 - 1)2      = 11

The mode=4. It has the highest frequency of 7