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(a) \(\frac{0.09 \times 1.21}{3.3 \times 0.00025} = \frac{9 \times 10^{-2} \times 121 \times 10^{-2}}{33 \times 10^{-1} \times 25 \times 10^{-5}}\) = \(\frac{33 \times 10^{-4}}{25 \times 10^{-6}}\) = \(1.32 \times 10^{2}\) (b) Amount, A = Principal + Interest = GH¢(5,600 + 1,200) = GH¢6,800 Compound Interest formula : \(A = P(1 + \frac{r}{100})^{3}\) \(6,800 = 5,600(1 + \frac{r}{100})^{3}\) \((1 + \frac{r}{100})^{3} = \frac{6800}{5600} = 1.214\) \(1 + \frac{r}{100} = \sqrt[3]{1.214} = 1.06686\) \(\frac{r}{100} = 1.06686 - 1 = 0.06686\) \(r = 0.06686 \times 100 = 6.686 \approxeq 6.69\) (3 significant figures)

(a) In the diagram above, \(\alpha = 28^\circ\) (angles in alternate segment) \(46^\circ + \beta + \alpha = 180^\circ\) (angles on a straight line) \(46^\circ + \beta + 28^\circ = 74^\circ + \beta = 180^\circ\) \(\beta = 180^\circ - 74^\circ = 106^\circ\) \(x + \beta = 180^\circ\) (opposite angles of a cyclic quadrilateral) \(x + 106^\circ = 180^\circ\) \(x = 180^\circ - 106^\circ = 74^\circ\) Hence, < VUS = 74^\circ. (b) Total surface area = curved surface area + area of closed end = \(2\pi rh + \pi r^{2}\) = \(\pi r(2h + r)\) \(462 = \frac{22r}{7}((2 \times 7) + r)\) \(\frac{462 \times 7}{22} = r(14 + r)\) \(147 = 14r + r^{2}\) \(r^{2} + 14r - 147 = 0\) \(r^{2} - 7r + 21r - 147 = 0\) \(r(r - 7) + 21(r - 7) = 0\) \((r + 21)(r - 7) = 0\) \(r = \text{-21 or 7}\) The radius cannot be negative, therefore, r = 7 cm.

(a) In the diagram above, area of \(\Delta PSR = \frac{1}{2} \times |PR| \times 6\) \(36 = \frac{1}{2} \times |PR| \times 6\) \(36 = 3 \times |PR|\) \(|PR| = 12 cm\) But |PQ| = |TS| = 8cm (opp. sides of a parallelogram) Hence, |QR| = |PR| - |PQ| = 12 - 8 = 4cm. (b) In the diagram above, \(\beta = 60^\circ\) (alternate angles) \(\tan \beta = \frac{10.65}{|QB|}\) \(|QB| = \frac{10.65}{\tan 60}\) = \(\frac{10.65}{1.732} = 6.149m\) |PO| = |QB| = 6.149m (opp. sides of a rectangle) In \(\Delta POT, \tan 45 = \frac{|TO|}{6.149}\) \(|TO| = \tan 45 \times 6.149 = 1 \times 6.149 = 6.149m\) |OB| = |PQ| = 10.65m (opp sides of a rectangle) \(h = |OB| - |OT| \) = \(10.65 - 6.149\) = \(4.501 m \approxeq 4.50m\) (3 sig. figs) Hence, the height of the flagpole = 4.50m

(a) In the diagram above, \(\theta = y\) (vertically opposite angles) \(\alpha = 180^\circ - 110^\circ = 70^\circ\) (angles on a straight line) Thus, \(\beta = \theta + \alpha\) (sum of interior opp. angles) \(\beta = y + 70^\circ ..... (1)\) \(x + r + \beta = 180^\circ ..... (2)\) But 2x = r = y Hence, \(x + 2x + (2x + 70) = 5x + 70 = 180\) \(5x = 180 - 70 = 110\) \(x = 22^\circ\) (b) Let Nx represent the donation collected by the girls. Then, the donation collected by the boys = N(x + 600). Also, let y represent the average collection, \(x_{g}\), of the girls. Then the average collection, \(x_{b}\), of the boys = N(y + 100). Thus, \(x_{g} = \frac{x}{12} = y\) \(x = 12y ... (1)\) \(x_{b} = \frac{x + 600}{10} = y + 100\) \(x + 600 = 10(y + 100) .... (2)\) Substitute 12y for x in (2) : \(12y + 600 = 10y + 1000\) \(12y - 10y = 1000 - 600 \) \(2y = 400 \implies y = 200\) From (1), \(x = 12y\) \(\therefore x = 12 \times 200 = N2,400\). Thus, the girls collected N2,400 The boys collected N2,400 + N600 = N3000 Total = N(2,400 + 3,000) = N5,400

(a) Weight (kg) Frequency 62 1 63 5 64 12 65 14 66 10 67 6 68 2 (b) Using an assumed mean of 65kg \(x\) \(f\) \(fx\) \(d = x - A\) \(fd\) \(fd^{2}\) 62 1 62 -3 -3 9 63 5 315 -2 -10 20 64 12 768 -1 -12 12 65 14 910 0 0 0 66 10 660 1 10 10 67 6 402 2 12 24 68 2 136 3 6 18 \(\sum\) = 50 3253 3 93 (i) Mean \(\bar{x} = \frac{\sum fx}{\sum f}\) = \(\frac{3253}{50}\) = 65.06 kg (ii) \(SD = \sqrt{\frac{\sum fd^{2}}{\sum f} - (\frac{\sum fd^{2}}{\sum f})^{2}}\) \(SD = \sqrt{\frac{93}{50} - (\frac{3}{50})^{2}}\) = \(\sqrt{\frac{93}{50} - \frac{9}{2500}}\) = \(\sqrt{\frac{4650 - 9}{2500}}\) = \(\sqrt{\frac{4641}{2500}}\) = \(1.3625 \approxeq 1.36\) (2 d.p.)

(a), (b) (c) < ZMY = 120^\circ.

(a) \(\frac{3p + 4q}{3p - 4q} = 2\) \(3p + 4q = 2(3p - 4q)\) \(3p + 4q = 6p - 8q\) \(3p - 6p = - 8q - 4q\) \(-3p = - 12q\) \(p = 4q\) \(\frac{p}{q} = \frac{4}{1}\) \(\therefore p : q = 4 : 1\). (b)(i) Perimeter of cross- section = 4 + x + 4 + 2 + y + 2 i.e 34 = 12 + x + y x + y = 22 ..... (1) From the diagram, |PQ| = |UR| (opp. sides of a rectangle) i.e. x = 2 + d + 2 x - 4 = d ..... (2) From (1), y = 22 - x is the circumference of the semi-circle. \(22 - x = \frac{2\pi r}{2} = \pi r\) \(r = \frac{d}{2} \) \(22 - x = \frac{22}{7} \times \frac{(x - 4)}{2}\) \(154 - 7x = 11x - 44\) \(154 + 44 = 11x + 7x \) \(198 = 18x\) \(x = \frac{198}{18} = 11 m\) Hence, |PQ| = 11 m. (ii) Area of cross section = Area of rectangle PQRU - area of semi-circle Area of rectangle = \(11 \times 4 = 44 m^{2}\) Area of semi-circle = \(\frac{\pi r^{2}}{2} \) = \(\frac{22}{7} \times (\frac{(11 - 4)}{2})^{2} \times \frac{1}{2}\) = \(19.25 m^{2}\) Area of cross section = \(44 - 19.25 = 24.75 m^{2}\).

In a school with students' population 810, the sectoral angle for a class of 72 students is = \(\frac{72}{810}\) x 360º = 32º

Given: \(x + y = 90^\circ ... (1)\) \((\sin x + \sin y)^{2} - 2\sin x \sin y = \sin^{2} x + \sin^{2} y + 2\sin x \sin y - 2\sin x \sin y\) = \(\sin^{2} x + \sin^{2} y ... (2)\) Recall: \(\sin x = \cos (90 - x) ... (a)\) From (1), \(y = 90 - x ... (b)\) Putting (a) and (b) in (2), we have \(\sin^{2} x + \sin^{2} y \equiv \cos^{2} (90 - x) + \sin^{2} (90 - x)\) = 1

\( A = P \left(1 + \frac{r}{100}\right)^n \) Where P = 126, r = 4,n = 2 A=126 \( \left(1 + \frac{4}{100}\right)^2 \text{Using LCM} \) =126 \( \left(\frac{100+4}{100}\right)^2 = 126 \left(\frac{104}{100}\right)^2 \) =126 \( \left(1.04^2 \right) \) = 126 * 1.04 * 1.04 =136.28 A = 136.30 (approx.) The Amount A, = N136.30k

\( [1 ÷ (x^2 + 3x + 2)] + [1 ÷ (x^2 + 5x + 6)] \) = \( 1 ÷ (x^2 + 3x + 2) + [1 ÷ (x^2 +5x + 6)]\) = \( [1 ÷ ((x^2 + x) + (2x + 2) )] + [1 ÷ ((x^2 + 3x) + (2x + 6) )] \) = [1 ÷ (x(x + 2) + 2(x +1))] + [1 ÷ (x(x + 3) +2(x + 3) )] = [1 ÷ (x + 1)(x + 2)] + [1 ÷ ((x + 3) + (x + 2))] =((x + 3) + (x + 1)) ÷ (x + 1)(x + 2)(x + 3) Using the L.C.M =((x + x + 3 + 1)) ÷ (x + 1)(x + 2)(x + 3) =(2x+4)/(x+1)(x+2)(x+3) =2(x+2)/(x+1)(x+2)(x+3) = \( \frac{2 }{(x + 1)(x + 3)} \)

The probability of drawing a blue ball from the first bag is \( \frac{3}{8} \), since there are 3 blue balls out of 8 total balls. The probability of drawing a blue ball from the second bag is \( \frac{2}{6} = \frac{1}{3} \), since there are 2 blue balls out of 6 total balls. Since the draws from each bag are independent, the probability that both balls are blue is the product of the individual probabilities: \( \frac{3}{8} \times \frac{2}{6} = \frac{3 \times 2}{8 \times 6} = \frac{6}{48} = \frac{1}{8}. \)

Cost Price: Selling price = 100 : (100 + 15) = 100 : 115 Cost Price = 100 \div 115 of selling price (i.e N420,000) = 100 \div 115 \times N420,000 = N365,217.4 ∴ Cost Price = N365, 217

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

[1 \div (x+1)] + [1 \div (x − 1)] = ((x − 1) + [(x + 1)) \div (x+1)(x − 1)] Using the L.C.M. = (x − 1 + x + 1) \div (x + 1)(x − 1) = (x + 2 − 1 + 1) \div (x + 1)(x − 1) = 2x \div (x + 1)(x − 1) =2x \div (x + 1)(x − 1)