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Total allocation = ₦25m + ₦20m + ₦35m + ₦20m = ₦100,000,000 Fraction for Agriculture = ₦25,000,000 / ₦100,000,000 = 0.25 (or 25%) Angle for Agriculture = 0.25 \times 360^\circ = 90^\circ Other sectors: - Education: (20/100) \times 360^\circ = 72^\circ - Women Affairs: (35/100) \times 360^\circ = 126^\circ - Commercial and Industries: (20/100) \times 360^\circ = 72^\circ

xz2 = xy2 + yx2 - 2(xy) (yz) cos 120o = 22 + 12 - 2(2) (1) cos 1202 = 4 + 1 - 4x - cos 60 = 5 - 4x - \(\frac{1}{2}\) 5 + 2 = 7 xz = \(\sqrt{7}\)m

degree of excellent in the pie-chart = 360o - (90o + 120o + 80o) = 360o - 290o = 70o = 360o = 36 students 70 = \(\frac{70^o}{360^o}\) x 36 = 7

(a) \((\frac{4}{25})^{-\frac{1}{2}} \times 2^{4} \div (\frac{15}{2})^{-2}\) = \((\frac{25}{4})^{\frac{1}{2}} \times 2^{4} \div (\frac{2}{15})^{2}\) = \(\frac{5}{2} \times 16 \times \frac{225}{4} = 2250\) (b) \(\log_{5} (\frac{3}{5}) + 3\log_{5} (\frac{15}{2}) - \log_{5} (\frac{81}{8})\) = \(\log_{5} 3 - \log_{5} 5 + 3\log_{5} 15 - 3\log_{5} 2 - \log_{5} 81 + \log_{5} 8\) = \(\log_{5} 3 - \log_{5} 5 + 3\log_{5} (3 \times 5) - 3\log_{5} 2 - \log_{5} 3^{4} + \log_{5} 2^{3}\) = \(\log_{5} 3 - \log_{5} 5 + 3\log_{5} 3 + 3\log_{5} 5 - 3\log_{5} 2 - 4\log_{5} 3 + 3\log_{5} 2\) = \(2\log_{5} 5 \) = \(2\)

(a)(i) A = {5, 10, 15, 20} (ii) B = {4, 8, 12, 16, 20} (iii) C = {3, 6, 9, 12, 15, 18} (b) (i) \(A \cap B = {20}\) (ii) \(A \cap C = {15}\) (iii) \(B \cap C = {12}\) (c) \((A \cap B) \cup (A \cap C) = A \cap (B \cup C)\) \((A \cap B) \cup (A \cap C) = {15, 20}\) \(A \cap (B \cup C) = {5, 10, 15, 20} \cap {3, 4, 6, 8, 9, 12, 15, 16, 18, 20}\) = \({15, 20}\) \(\therefore (A \cap B) \cup (A \cap C) = A \cap (B \cup C)\)

Let |PC| = x cm; Hence, |AC| = 2x cm \(\tan 37^\circ = \frac{x}{5}\) \(x = 5 \tan 37 \) \(x = 3.768 cm\) \(\therefore |AC| = 2 \times 3.768\) = \(7.536 cm\) \(\approxeq 7.54 cm\) (3 sig. figs) (b) From \(\Delta ABC\), \(\tan < ABC = \frac{7.536}{5} = 1.5072\) \(< ABC = \tan^{-1} (1.5072) = 56.436^\circ\) \(\therefore < PBA = < ABC - < PBC\) = \(56.436^\circ - 37^\circ\) = \(19.436^\circ \approxeq 19.44^\circ\)

(a) \(|AB| = \sqrt{|BD|^{2} - |AD|^{2}}\) = \(\sqrt{17^{2} - 15^{2}} = \sqrt{289 - 225}\) = \(\sqrt{64} = 8 cm\) (b) Area of triangle BCD = \(\frac{1}{2} \times 9 \times 8 \) = \(36 cm^{2}\) (c) \(|EC| = 6 cm\) \(|CD| = \sqrt{8^{2} + 6^{2}} = \sqrt{100}\) = 10 cm. (d) Perimeter of the trapezium = 15 cm + 8 cm + 9 cm + 10 cm = 42 cm.

(a) \(3y - 2x = 21 ... (1)\) \(4y + 5x = 5 .... (2)\) Multiply (1) by 4 and (2) by 3 so we have, \(12y - 8x = 84 ... (3)\) \(12y + 15x = 15 ... (4)\) (3) - (4) : \(-8x - 15x = 69 \implies -23x = 69\) \(x = \frac{69}{-23} = -3\) Put x = -3 in (1), \(3y - 2(-3) = 21\) \(3y + 6 = 21 \implies 3y = 21 - 6 = 15\) \(y = \frac{15}{3} = 5\) \(x, y = -3, 5\). (b)(i) Probability of losing Probability of winning = \(\frac{1}{6}\) \(\therefore\) Probability of losing = \(1 - \frac{1}{6} = \frac{5}{6}\). (ii) Losing $20.00 after two picks = picking the card numbered 1 twice. = \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\)

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 \(\therefore \text{The area of the lawn} = 12 \times 5 = 60 m^{2}\)