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\(\bigtriangleup\) ABC, A + 153 = 180 (angles on straight line) o 9m = 180 - 27 9m = 153 m = \(\frac{153}{9}\) = 17

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

Arts = 90o business = 75o, science = 45o Technical 360o - (90o - 75o + 45o) = 360o - 210o Technical = 150o textbooks for technical class = \(\frac{150^o}{360^o}\) x 600 = 250

\(\frac{45^o}{360}\) x 100% = 12\(\frac{1}{2}\)%

PQRST is a regular polygon with 5 sides-pentagon n = 5, Each exterior angle = \(\frac{360^o}{n}\) = \(\frac{350}{5}\) = 70o Each interior angle = 180 - 72o (o Since the pentagon is regular, the exterior angles are equal o and o o - (o - (72o + 72o) = 180o - 144o = 36o

Sum of angles on straight line = 180º 80º + 50º + x + 10º = 180º 140º + x = 180º x = 180º - 140º x = 40º total sum of all angles = 360º 2yº + 2xº + 80o + 50º + xº + 10º = 360º 2yº + 2(40)º + 80º + 50º + 40º + 10º = 360º 2yº + 80º + 180º = 360º 2yº + 260º = 360º 2yº = 360º - 260º 2yº = 100º angle 2yº = 100º

R +198 = 360(angles at a point) R = 360 - 198 = 162o C = 72(alternate angle), b + c = 180(angles or straight line) b = 190 - c = 180 - 72 = 108 a + b = R = 162 a = R - b = 162 - 108 = 54, but a = y(alternate angles) y = 54o

(x \(\cap\) y2) = [a, b] n(x \(\cap\) y2) = 2

(a) \(\frac{\frac{3}{4}(3\frac{3}{8} + 1\frac{5}{8})}{2\frac{1}{8} - 1\frac{1}{2}}\) \(\frac{3}{4}(3\frac{3}{8} + 1\frac{5}{8}) = \frac{3}{4}(\frac{27}{8} + \frac{13}{8})\) = \(\frac{3}{4}(\frac{40}{8})\) = \(\frac{15}{4}\) \(2\frac{1}{8} - 1\frac{1}{2} = \frac{17}{8} - \frac{3}{2}\) = \(\frac{17}{8} - \frac{12}{8}\) = \(\frac{5}{8}\) \(\therefore \frac{\frac{3}{4}(3\frac{3}{8} + 1\frac{5}{8})}{2\frac{1}{8} - 1\frac{1}{2}} = \frac{15}{4} \div \frac{5}{8}\) = \(\frac{15}{4} \times \frac{8}{5}\) = \(6\) (b) \(\log_{10} 2 = 0.3010 ; \log_{10} 3 = 0.4771\) \(\log_{10} 1.125 = \log_{10}(\frac{1125}{1000})\) (Dividing through with 125) = \(\log_{10} (\frac{9}{8})\) = \(\log_{10} 9 - \log_{10} 8\) \(\log_{10} 9 = \log_{10} 3^{2} = 2\log_{10} 3 = 2 \times 0.4771 = 0.9542\) \(\log_{10} 8 = \log_{10} 2^{3} = 3\log_{10} 2 = 3 \times 0.3010 = 0.9030\) \(\therefore \log_{10} (\frac{9}{8}) = 0.9542 - 0.9030 = 0.0512\) \(\approxeq 0.051\) ( 2 significant figures)

(a) \(7x + 4 < \frac{1}{2}(4x + 3)\) \(7x + 4 < 2x + \frac{3}{2}\) \(7x - 2x < \frac{3}{2} - 4\) \(5x < -\frac{5}{2}\) \(\frac{5x}{5} < \frac{-\frac{5}{2}}{5}\) \(x < -\frac{1}{2}\). (b) When Salem gets N2 ------> Sunday gets 50 kobo (Recall N1 = 100 kobo) Salem gets 200 kobo ------> Sunday gets 50 kobo. Sunday gets N4 -----> Shaka gets N2. ====> Sunday gets 400 kobo ------> Shaka gets 200 kobo. When Sunday gets 400 kobo ------> Salem gets 1600 kobo. \(\therefore\) Salem : Sunday : Shaka = 1600 : 400 : 200 = 8 : 2 : 1 \(\therefore \text{Shaka's share} = \frac{1}{11} \times N1,100 = N100\)

(a) Present ages of a father and his son are in the ratio 10 : 3. Let the sum of their ages be x. The son's age = 15 years = \(\frac{3}{13} \times x = 15\) \(x = \frac{15 \times 13}{3} = 65\) \(\therefore\) The Father's present age = 65 - 15 = 50 years. In z years time, the ratio of their ages = 2 : 1 \(\frac{50 + z}{15 + z} = \frac{2}{1}\) \(\implies 50 + z = 2(15 + z)\) \(50 + z = 30 + 2z \implies 50 - 30 = 2z - z\) \(20 = z\) Therefore, in 20 years' time, the ratio of their age will be 2 : 1. (b) \(\frac{x + y + z}{3} = 6 \implies x + y + z = 18 .... (1)\) \(\frac{x + y + z + l + u + v + w}{7} = 9 \implies x + y + z + l + u + v + w = 63 .... (2)\) Putting (1) into (2), we have \(18 + l + u + v + w = 63 \) \(\therefore l + u + v + w = 63 - 18 = 45\) Mean of l, u, v and w = \(\frac{45}{4} = 11.25\).

Since it is divided into 3 sectors then \(8x^\circ = 360^\circ\) \(x^\circ = 45^\circ\)The 3rd sector = 3 times the size of the other two sectors put together = \(3(45 + 45) = 3(90) = 270^\circ\) Perimeter of a sector = \(2r + l\) where r = radius of the circle ; l = length of arc = \(\frac{\theta}{360^\circ} \times 2\pi r\) Area of circle = \(\frac{22}{7} \times r^{2} = 154 cm^{2}\) \(r^{2} = \frac{154 \times 7}{22} = 49\) \(r = 7 cm\) \(l = \frac{270}{360} \times 2 \times \frac{22}{7} \times 7 = 33 cm\) Perimeter of sector = \(2(7) + 33 = 14 + 33 = 47 cm\).

(a) (b) \(\tan \theta = \frac{4.2}{6 + x} ; \tan \theta = \frac{1.2}{x}\) \(\tan \theta = \tan \theta\) \(\implies \frac{4.2}{6 + x} = \frac{1.2}{x}\) \(4.2 x = 1.2(6 + x) \) \(4.2 x = 7.2 + 1.2 x \implies 4.2x - 1.2x = 7.2\) \(3x = 7.2 \implies x = 2.4 m\) (length of the shadow cast by the lamp) (ii) \(\tan \theta = \frac{1.2}{2.4} = 0.5\) \(\theta = \tan^{-1} (0.5) \) = \(27^\circ\) (angle of elevation)

(a)(i) \(p > q .... (1)\) \(p + q = 3(p - q) ...... (2)\) From (2), \(p + q = 3p - 3q\) \(p - 3p = - 3q - q \implies - 2p = - 4q\) \(p = 2q\) (ii) \(\frac{p^{2} + q^{2}}{pq}\) = \(\frac{(2q)^{2} + q^{2}}{2q \times q}\) \(\frac{4q^{2} + q^{2}}{2q^{2}}\) = \(\frac{5q^{2}}{2q^{2}}\) = \(\frac{5}{2}\) (b) Selling price = \(\frac{100 \times 66}{25}\) = \(N264.00\) Using \(\frac{SP - CP}{CP} = % gain\) \(\frac{264 - CP}{CP} = \frac{32}{100}\) \(100(264 - CP) = 32CP\) \(26400 = 32CP + 100CP = 132CP\) \(CP = \frac{26400}{132}\) = \(N200\) When he sells 20 for N50, Selling price = \(\frac{100 \times 50}{20} = N250.00\) Hence, he made gain. \(% gain = \frac{250 - 200}{200} \times 100%\) \(\frac{1}{4} \times 100%\) = \(25% gain\)

(a) x -3 -2 -1 0 1 2 3 4 y 35 15 1 -7 -9 -5 5 21 (b) see graph above (c) (i) The roots of the equation is where the graph cuts the x- axis = -0.9, 2.6. (ii) Minimum value of y = -9.21 (iii) At x = 2, gradient = \(\frac{y_{2} - y_{1}}{x_{2} - x_{1}}\) = \(\frac{4 - (-8.5)}{3.3 - 1.5}\) = \(\frac{12.5}{1.8}\) = 6.9444 \(\approxeq\) 6.9