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(a) see figure above a \(< QTR = < TRP = 70^\circ\) (alternate angle) \(< MNR = < TRP = 70^\circ\) (alternate angle; MN || ST) \(< MNR + < PNM = 180^\circ\) \(70^\circ + < PNM = 180^\circ \implies < PNM = 110^\circ\) \(\therefore x = 110^\circ\) (b)see figure b above \(< BDC = < BCD = 50^\circ\) (BD = BC) \(\therefore \Delta BDC = isosceles\) \(< ABD = 2(50^\circ) = 100^\circ\) (exterior angle) \(55^\circ + < BDA + 100^\circ = 180^\circ\) \(< BDA = 180^\circ - 155^\circ = 25^\circ\)

(a) x -3 -2 -1 0 1 2 3 4 \(y = x^{2}\) 9 4 1 0 1 4 9 16 (b) \(y = 2x + 3\) x -3 -2 -1 0 1 2 3 4 2x -6 -4 -2 0 2 4 6 8 3 3 3 3 3 3 3 3 3 y -3 -1 1 3 5 7 9 11 see graph above (c)(i) The roots of the equation are -1 and 3, from the graph. (ii) The gradient of \(y = x^{2}\) at x = -2 : \(\frac{8}{-2.5} = -3.2\)

(a) Frequency Distribution Table Birth 45 - 54 55 - 64 65 - 74 75 - 84 85 - 94 95 - 104 105 - 114 Freq(f) 1 2 15 21 7 3 1 (b) see histogram above. (c) Modal class is 75 - 84 Mode = \(\frac{75 + 84}{2} = 79.5\) (d) Birth freq (f) class midpoint (x) \(fx\) 45 - 54 1 49.5 49.5 55 - 64 2 59.5 119 65 - 74 15 69.5 1042.5 75 - 84 21 79.5 1669.5 85 - 94 7 89.5 626.5 95 - 104 3 99.5 298.5 105 - 114 1 109.5 109.5 Total 50 3915 Mean \(\bar{x} = \frac{\sum fx}{\sum f}\) = \(\frac{3915}{50}\) = \(78.3\).

(a)(i) (ii) \(21 + 4 + 5 + 10 + y + 35 - y + 32 - y = 100\) \(107 - y = 100\) \(y = 107 - 100 = 7\) Therefore, 7 traders sell oranges and mangoes alone. (b) \(312_{four} + 52_{x} = 96_{ten}\) \(312_{4} = 3 \times 4^{2} + 1 \times 4^{1} + 2 \times 4^{0}\) = \(48 + 4 + 2 = 54_{10}\) \(52_{x} = 5 \times x^{1} + 2 \times x^{0}\) = \((5x + 2)_{10}\) \(\therefore 54 + (5x + 2) = 96\) \(56 + 5x = 96 \implies 5x = 96 - 56 = 40\) \(x = 8\) \(\therefore 312_{4} + 52_{8} = 96_{10}\)

(a)(i) P(A) = \(\frac{2}{3}\) ; P(B) = \(\frac{5}{8}\) ; P(C) = \(\frac{3}{4}\). \(P(A \cap B \cap C) = P(A) \times P(B) \times P(C)\) = \(\frac{2}{3} \times \frac{5}{8} \times \frac{3}{4} = \frac{5}{16}\) (ii) P(A') = \(1 - \frac{2}{3} = \frac{1}{3}\) P(B') = \(1 - \frac{5}{8} = \frac{3}{8}\) P(C') = \(1 - \frac{3}{4} = \frac{1}{4}\) \(P(A' \cap B' \cap C') = P(A') \times P(B') \times P(C')\) = \(\frac{1}{3} \times \frac{3}{8} \times \frac{1}{4} \) = \(\frac{1}{32}\) (iii) P(A and B pass, C fails) = \(\frac{2}{3} \times \frac{5}{8} \times \frac{1}{4} = \frac{5}{48}\) P( A and C pass, B fails) = \(\frac{2}{3} \times \frac{3}{4} \times \frac{3}{8} = \frac{3}{16}\) P(B and C pass, A fails) = \(\frac{5}{8} \times \frac{3}{4} \times \frac{1}{3} = \frac{5}{32}\) P(only two of the boys pass) = \(\frac{5}{48} + \frac{3}{16} + \frac{5}{32} = \frac{43}{96}\) (b) Let the cost price be x % gain = \(\frac{\text{actual gain}}{\text{cost price}} \times 100%\) \(20 = \frac{36,000 - x}{x} \times 100%\) \(\frac{36,000 - x}{x} = \frac{1}{5}\) \(5(36,000 - x) = x \implies 180,000 - 5x = x\) \(180,000 = x + 5x = 6x\) \(x = N30,000\) Discount (5%) : \(\frac{5}{100} \times N36,000 = N1,800\) Selling price : \(N(36,000 - 1,800) = N34,200\) Actual gain : \(\frac{34,200 - 30,000}{30,000} \times 100% \) = 14%

(a) \(2x^{2} + (p + 1)x + q = 0\) The roots of the equation are 1 and 3, therefore (x - 1)(x - 3) = 0. Equation : \(x^{2} - 3x - x + 3 = 0\) \(x^{2} - 4x + 3 = 0\) \(\therefore 2x^{2} + (p + 1)x + q \equiv x^{2} - 4x + 3\) \(\frac{2x^{2} + (p + 1)x + q}{2} = x^{2} + \frac{(p + 1)}{2} x + \frac{q}{2}\) \(\implies \frac{q}{2} = 3\) \(q = 6\) \(\frac{p + 1}{2} = -4 \implies p + 1 = -8\) \(p = -9\) (p, q) = (-9, 6). (b) \(W \propto \frac{1}{d^{2}} \) \(W = \frac{k}{d^{2}}\) \(k = Wd^{2}\) \(W = \frac{80 \times (6400)^{2}}{d^{2}}\) (equation of variation). \(W = \frac{80 \times (6400)^{2}}{(6400 + 800)^{2}}\) \(W = \frac{80 \times (6400)^{2}}{(7200)^{2}}\) = 63.21 kg

< PAB = < ABE = 53^\circ (alternate angles) < CBE = 180^\circ - < DBC = 180^\circ - 161^\circ = 19^\circ < ABC = < ABE + < CBE = 53^\circ + 19^\circ = 72^\circ (a) In \(\Delta ABC\), \(AC^{2} = AB^{2} + BC^{2} - 2(AB)(BC) \cos < ABC\) \(AC^{2} = 15^{2} + 18^{2} - 2(15)(18) \cos 72\) = \(225 + 324 - 540 \cos 72\) = \(549 - 166.869\) \(AC^{2} = 382.131\) \(AC = \sqrt{382.131} = 19.548m\) \(\approxeq 19.5m\) (b) \(\frac{\sin A}{18} = \frac{\sin 72}{19.548}\) \(\sin A = \frac{18 \times \sin 72}{19.548}\) \(\sin A = 0.8757\) \(A = \sin^{-1} (0.8757) = 61.13^\circ\) The bearing of C from A = 61.13^\circ + 53^\circ = 114.13^\circ \(\approxeq\) 114^\circ. (c) In \(\Delta ATB\), \(\frac{15}{BT} = \tan 58\) \(BT = \frac{15}{\tan 58}\) \(9.373 m \approxeq 9.37m\)

(a) \(d = 15 cm \implies r = \frac{15}{2} cm\) \(V = \frac{4\pi r^{3}}{3}\) \(V = \frac{4}{3} \times \frac{22}{7} \times (\frac{15}{2})^{3}\) \(V = \frac{1125 \times 22}{7 \times 2}\) \(V = 1,767.857 cm^{3}\) (b)(i) Longitude difference : 105^\circ + 35^\circ = 140^\circ Distance along the parallel of latitude = \(\frac{140}{360} \times 2 \times \frac{22}{7} \times 6400 \cos 28\) = \(\frac{140 \times 44 \times 6400 \times 0.8829}{360 \times 7}\) = \(13812.48 km\) \(\approxeq 13800 km\) (ii) Distance from the equator : \(\frac{28}{360} \times 2 \times \frac{22}{7} \times 6400\) = \(3128.89 km\) \(\approxeq 3130 km\)

From the diagram provided, x lies between -7 and or equal to 2, i.e. -7 < x \(\leq\) -1

From the constraint y < 3 and y > 0, the possible values for y are 1 and 2 When y = 1, 3y + 5x \(\leq\) 15 3 + 5x \(\leq\) 15 ⇒ 5x \(\leq\) 12 ⇒ 2.4 since x must be an integer greater than 0 but less than or equal to 2.4, the possible values of x are 1 and 2 When y = 2, 3y + 5x \(\leq\) 15 ⇒ 6 + 5x \(\leq\) 15 ⇒ 6 \(\leq\) 15 - 6 = 11 ⇒ x = 2.2 Again, Since x must be an integer greater than 0 and less than or equal to 2.2, the only possible integer value of x is 1 So, the pairs (x, y) that satisfy all the conditions are ( 1, 1), (1, 2) and (2, 1).

\(\begin{vmatrix}-1 & -1 & -1\\3 & 1 & -1\\1 & 2 & 1\end{vmatrix}\) = -\(\begin{vmatrix}1 & -1 \\2 & 1\end{vmatrix}\) + \(\begin{vmatrix}3 & -1\\1 & 1\end{vmatrix}\) - \(\begin{vmatrix}3 & 1\\1 & 2\end{vmatrix}\) = -( 1 + 2) + (3 + 1) - ( 6 - 1) = -3 + 4 - 5 = 4 - 8 = -4.

K = \(\begin{pmatrix}2 & 1\\3 & 4\end{pmatrix}\) K\(^2\) = K = \(\begin{pmatrix}2 & 1\\3 & 4\end{pmatrix}\) x K = \(\begin{pmatrix}2 & 1\\3 & 4\end{pmatrix}\) = \(\begin{pmatrix}7 & 6\\18 & 19\end{pmatrix}\) I = \(\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}\) K\(^2\) + K + I = \(\begin{pmatrix}7 & 6\\18 & 19\end{pmatrix}\) + \(\begin{pmatrix}2 & 1\\3 & 4\end{pmatrix}\) + \(\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}\) = \(\begin{pmatrix}10 & 7\\21 & 24\end{pmatrix}\)

If P = \(\begin{pmatrix}3 & -2 & 4\\5 & 0 & 6\\7 & 5 & -1\end{pmatrix}\) Then -2P = -2 x \(\begin{pmatrix}3 & -2 & 4\\5 & 0 & 6\\7 & 5 & -1\end{pmatrix}\) = \(\begin{pmatrix}-6 & 4 & -8\\-10 & 0 & -12\\-14 & -10 & 2\end{pmatrix}\)

a = N 100.00, d = N 20.00, Un = N 5800.00, n = ? Un = \(\frac{n}{2}\)(2a + (n - 1)d) 5800 = \(\frac{n}{2}\)( 2 x 100 + ( 20n - 20) 11600 = 200n + 20n\(^2\) - 20n 20n\(^2\) + 180n - 11600 = 0 divide through by 20 n\(^2\) + 9n - 580 = 0 solve using the quadratic formula n = 20 0r - 29 n = 20 years Years cannot be calculated in - Ve

Length of arc = \(\frac{Ø}{360} \times 2\pi\) 5.5 = \(\frac{360 - 210}{360}\) x 2 x \(\frac{22}{7}\) \(\frac{ 5.5 \times 360\times7}{150\times2\times22}\) = 2.1 length of major arc = \(\frac{210}{360}\) x 2 x \(\frac{22}{7}\) x 2.1 7.7m ≈ 8m