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50\(^o\) + m = 105\(^o\) (exterior angle of \(\triangle\) m = 105\(^o\) - 50\(^o\) m = 55\(^o\)

Pr(green) = \(\frac{6}{5 + 6 + 7}\) = \(\frac{6}{18}\) = \(\frac{1}{3}\)

No of fruits = x \(\frac{80}{360}\) x \(x\) = 60 x = \(\frac{60 \times 360}{80}\) x 270 = 90

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

\(\frac{13 + 14}{2}\) = 13.5

(a) A(-2, 7), B(2, -3) m = \(\frac{y_2 - y_1}{x_2 - x_1} = \frac{-3 - 7}{2 - (-2)} = \frac{10}{4}\) \(\frac{-5}{2} = \frac{y - 7}{x + 2}\) -5(x + 3) = 2(y - 7) -5x - 10 = 2y - 14 2y - 14 + 10 + 5x = 0 2y - 14 + 10 + 5x = 0 2y + 5x - 4 = 0 (b) \(\frac{5b - a}{8b + 3a} = \frac{1}{5}\) 5(5b - a) = 1(1)(8b + 3a) 25b - 5a = 8b + 3a 25b - 8b = 3a + 5a 17b = 8a \(\frac{a}{b} = \frac{17}{8}\) \(\frac{a}{b}\) = 2.13 to 2 d.p

Total ratio = 3 + 5 + 8 = 26 Ali's share is \(\frac{3}{16}\) x N420,000 = N78,000 Yusuf's share is \(\frac{8}{16}\) x N420,000 = N210,000 Sum of Ali's share + Yusuf's share = N78,750 + N210,000 = N288,750 (b) 2(\(\frac{1}{8}\))\(^2\) = 32\(^{x - 1}\) 2\(^{1}\) x 2\(^{3x}\) = 2\(^{5x - 5}\) 2\(^{1 - 3x}\) = 2\(^{5x - 5}\) 1 - 3x = 5x - 5 -3x - 5x = -5 -1 -8x = -6 x = \(\frac{-6}{-8}\) = \(\frac{3}{4}\)

In \(\triangle\)PQR |PR|\(^2\) = |PQ|\(^2\) + |QR|\(^2\) = 3\(^2\) + 4\(^2\) |PR|\(^2\) = 9 + 16 |PR| = \(\sqrt{25}\) = 5cm In \(\triangle\)PRS |PS|\(^2\) = |PR|\(^2\) + |RS|\(^2\) 13\(^2\) = 5\(^2\) + |RS|\(^2\) 169 =25 + |RS|\(^2\) |RS|\(^2\) = 169 - 25 = 144 |RS| = \(\sqrt{144}\) = 12cm Area of quadrilateral PQRS = Area of \(\triangle\)PQR + Area of \(\triangle\)PQR + Area of \(\triangle\)PRS = (\(\frac{1}{2} \times 3 \times 4\)) + (\(\frac{1}{2} \times 12 \times 5\)) = 6cm\(^2\) + 30cm\(^2\) = 36cm\(^2\)

(a) Let number of blue balls = n Total number of balls = 3 + 5 + n = 8 + n p(red ball) = \(\frac{\text{no. of red balls}}{\text{Total no. of balls}}\) \(\frac{1}{6} = \frac{3}{8 + n}\) 8 + n = 18 n + 18 - 8 = 10 number of blue balls = 10 (b) P(green ball) = \(\frac{\text{No. of green balls}}{\text{Total no. of balls}}\) = \(\frac{5}{3 + 5 + 10}\) = \(\frac{5}{18}\) = 0.2778

F \(\infty\) \(\frac{m_1m_ 2}{d^2}\) F = \(\frac{km_1m_2}{d^2}\) 20 = \(\frac{k(25)(10)}{(25)}\) \(\frac{10k}{10} = \frac{20}{10}\) k = 2 (i) F = \(\frac{2m_1m_2}{d_2}\) (ii) Since F = \(\frac{2m_1m_2}{d_2}\) 30 = \(\frac{2(7.5)(4)}{d^2}\) 30 \(\frac{60}{d^2}\) d\(^2\) = \(\frac{60}{30}\) d\(^2\) = 2 d = \(\sqrt{2}\)m (b) Sum of angles = (n - 2) 180 = (5 - 2) 180 = 540\(^o\) x + x + 20 + x + 60 + x + 40 + x + 80 = 540 5x + 200 = 540 5x = 540 - 200 \(\frac{5x}{5} = \frac{340}{5}\) x = 68\(^o\)

Class Intervals Class Boundaries Frequency Cumulative Frequency 0 - 9 0.5 - 9.5 1 1 10 - 19 9.5 - 19.5 3 4 20 - 29 19.5 - 29.5 3 7 30 - 39 29.5 - 39.5 6 13 40 - 4921 39.5 - 49.5 8 21 50 - 59 49.5 - 59.5 12 33 60 - 69 59.5 - 69.5 6 39 70 - 79 69.5 - 79.5 3 42 80 - 89 79.5 - 89.5 4 46 90 - 99 89.5 - 99.5 4 50 \(\sum f = 50\) (b) (c) (I) Mean mark = 49.9 (ii) Percentage of at least 66 mark = \(\frac{13}{50} \times 100\)% = 26%

\(\frac{1}{3}x - \frac{1}{4}(x + 2) \geq 3x -1\frac{1}{3}\) \(\frac{1}{3}x - \frac{(x + 2)}{12} \geq 3x - \frac{4}{3}\) \(\frac{4x - 3x - 6}{12} \geq 3x - \frac{4}{3}\) \(\frac{x - 6}{12} \geq \frac{3x}{1} - \frac{4}{3}\) \(\frac{x - 6}{12} \geq \frac{9x - 4}{3}\) 3(x - 6) \(\geq\) 12(9x - 4) 3x - 18 \(\geq\) 108x - 48 3x - 108x \(\geq\) -48 + 18 \(\frac{-105x}{-105} \geq \frac{-30}{-105}\) x \(\geq \frac{10}{35}\) x \(\geq\) \(\frac{2}{7}\) (b)(i) C\(B\)A = 108\(^o\) - (90\(^o\) + 35\(^o\)) = 180\(^o\) - 125\(^o\) = 55\(^o\) sin35\(^o\) = \(\frac{|AB|}{66}\) |AB| = 66 x sin30\(^o\) |AB| = 37.856m cos 35\(^o\) = \(\frac{|AC|}{66}\) |AC| = 66 x cos35\(^o\) |AC| = 55.06m Tan52\(^o\) = \(\frac{|AD|}{37.856}\) |AD| = 37.856 x Tan52\(^o\) = 48.45m the height of the tower to 2d.p (ii) Tan A\(C\)D = \(\frac{48.453}{54.06}\) =0.8962 A\(C\)D = tan\(^{-1}\)0.8962 = 41.867\(^o\) Angle of elevation of the tower from C = 35\(^o\) + 41.867\(^o\) = 76.87\(^o\) to 2d.p

(a) When x = 60\(^o\) y = 2cos 60\(^o\) + 3sin 60\(^o\) y = 3.6 When x = 120\(^o\) y = 2 cos 120\(^o\) + 3 sin 180\(^o\) y = 1.6 When x = 180\(^o\) y = 2 cos 180\(^o\) + 3 sin 180\(^o\) y = 2 When x = 300\(^o\) y = 2 cos 300\(^o\) + 3 sin 300\(^o\) y = 1.6 when x = 360\(^o\) y = 2 cos 360\(^o\) + 3 sin 360\(^o\) y = 2 (b) (c)(i) 1555.5\(^o\), 318\(^o\) (ii) When x = 3.42\(^o\), y = 0.9\(^o\)

(a) Weight of full tomatoes = 130kg Weight of half tomatoes = \(\frac{130}{2}\) = 65kg Cost price of half tomatoes = \(\frac{N52,000}{2}\) = N26,000 Profit = \(\frac{S.P - C.P}{C.P}\) x 100 30 = \(\frac{(S.P - 26,000)}{25,000}\) x 100 S.P = N7800 + N26,000 S.P = N7,800 + N26,000 S.P of half tomatoes = N33,800 Selling price per kg = \(\frac{33,800}{65}\) = N520 per kg Reduction in selling price = 12% = \(\frac{12}{100}\) x 520 = N62.40k per kg New selling price per kg = N520 - N62.40k = N457.60k (b) 5kg is thrown away remaining 60kg The selling price of entire 60kg = 60 x N457.60k = N27,456 The entire selling price N33,800 + N27,456 = N61,256 % profit = \(\frac{(61.256 - 50,000)}{52,000}\) x 100 = 17.8%

(a) (i) ar\(^2\) = \(\frac{1}{4}\) ........(i) ar\(^5\) = \(\frac{1}{32}\) ........(ii) \(\frac{ar^5}{ar^2} = \frac{1}{32} \div \frac{1}{4}\) r\(^3 = \frac{1}{32} \times \frac{4}{1}\) r\(^3 = \frac{1}{32} \times \frac{4}{1}\) r\(^3 = 2^{-3} = (\frac{1}{2})^3\) r = \(\frac{1}{2}\) Since ar\(^2\) = \(\frac{1}{4}\) a(\(\frac{1}{2}\))\(^2\) = \(\frac{1}{4}\) a = 1 The first term is 1 ad the common ratio is \(\frac{1}{2}\) (ii) T\(_7\) = ar\(^6\) = 1 x (\(\frac{1}{2}\))\(^6\) The seventh term = \(\frac{1}{64}\) (b) x = 2 and x = -3 a(2)\(^2\) + b(2) + c = 0 4a + 2b + c = 0 4a + 2b + c = 0.....(i) a(-3)\(^2\) + b(-3) + c = 0 9a - 3b + c = 0.......(ii) Solving eq.(i) and (ii) 4a + 2b + c = 0 -(9a - 3b + c) = 0 -5a + 5b = 0 (x - 2)(x + 3) = 0 x\(^2\) + 3x - 2x - 6 = 0 x\(^2\) + x - 6 = 0 a = 1, b = 1, c = -6