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0.612 = \(\frac{0.612}{1}\) x \(\frac{1000}{1000}\) = \(\frac{612}{1000}\) = \(\frac{153}{250}\)

\(\frac{|PN|}{|PQ|}\) = \(\frac{|PM|}{|PR|}\)

From the truth table above, p \(\to\) q would be false if p is true and q is false

x + 2 \(\geq\) 2x + 1 x - 2x \(\geq\) 1 - 2 -x \(\geq\) -1 \(\frac{-x}{-1}\) \(\geq\) \(\frac{-1}{-1}\) x \(\leq\) 1

In the diagram, < WOZ = 180\(^o\) (angle on a straight line) < WOX = < XOY = < YOZ (|WX| = |XY| = |YZ|) \(\frac{180^o}{3}\) = 60\(^o\) = 60\(^o\) M + m =2m (base angles of isosceles \(\bigtriangleup\), |OY| and |OZ| are radii) < YOZ + 2m (base angles of a \(\bigtriangleup\)) 60\(^o\) + 2m = 180\(^o\) (sum of a \(\bigtriangleup\)) 60\(^o\) + 2m = 180\(^o\) 2m = 180\(^o\) - 60\(^o\) 2m = 120\(^o\) m = \(\frac{120^o}{2}\) = 60\(^o\)

m + n = 110\(^o\), (n + r) = 130\(^o\) (m + n) = 120\(^o\) then, r = 130\(^o\) - n and; m + (130^o - n) = 120\(^o\) m - n = -10\(^o\) 2m + (n + r) = 110 + 120 = 230 2m + 130 = 230 2m = 230 - 130 m = \(\frac{100}{2}\) = 50\(^o\) n = 110\(^o\) - 50\(^o\) = 60\(^o\) r = 130\(^o\) - 60\(^o\) = 70\(^o\) Hence, the ratio m : n : r = 50 : 60 : 70 = 5 : 6 : 7

Total donation = 4 x 500 + 7 x 2000 + 20 x 1000 + 9 x 700 + 4 x 500 + 5 x 100 + 3 x 50 + 1 x 2 + 2 x 10 = 20000 + 14000 + 20000 + 6300 + 2000 + 500 + 150 + 2 + 20 = N62,972

Let n(M \(\cup\) N \) = x Then 20 - x + x + 30 - x = n(M \(\cup\) N) 50 - x = 40 50 - 40 = x 10 = x x = 10 Hence, n(M \(\cup\N)' = 8 + (20 - 10) + (30 + 10) = 8 + 10 + 20 = 38

y = x\(^2\) ....(1) y = x ......(2) y = y x\(^2\) - x x\(^2\) - x = 0 x(x - 1) = 0 x = 0 or x - 1 = 0 x = 0 or x = 1 when x = 0, y = 0\(^2\) = 0 when x = 1, y = 1\(^2\) = 1 Hence; the two graphs interest at (0, 0) and (1, 1)

Beginning of 1st year (March 2011 - February 2012) - N900,000 Depreciation (30%) = \(\frac{30}{100} \times N900,000\) = N270,000 Price at the end of 1st year = N(900,000 - 270,000) = N630,000. Beginning of 2nd year (March 2012 - February 2013) - N630,000 Depreciation (22%) = \(\frac{22}{100} \times N630,000\) = N138,600 Price at the end of 2nd year = N(630,000 - 138,600) = N491,400. Beginning of 3rd year (March 2013 - February 2014) - N491,400 Depreciation (22%) = \(\frac{22}{100} \times N491,400\) = N108,108 Price at the end of the 3rd year = N(491,400 - 108,108) = N383,292. Beginning of 4th year (March 2014 - February 2015) - N383,292 Depreciation (22%) = \(\frac{22}{100} \times N383,292\) = N84,324.24 Price at the end of the 4th year = N(383,292 - 84,324.24) = N298,967.76 \(\approxeq\) N299,000 (to the nearest hundred naira).

(a) \(\frac{m + n + s + p + q}{5} = 12 \) \(\implies m + n + s + p + q = 12 \times 5 = 60\) \(\frac{(m + 4) + (n - 3) + (s + 6) + (p - 2) + (q + 8)}{5} = \frac{m + n + s + p + q + 4 - 3 + 6 - 2 + 8}{5}\) = \(\frac{60 + 13}{5} = \frac{73}{5}\) = 14.6. (b) In the community, there are 500 people. The 75th percentile age = 65 years Number of people in the 75th percentile = \(\frac{75 \times 500}{100} = 375\) The 25th percentile = 15 years Number of people in the 25th percentile = \(\frac{25 \times 500}{100} = 125\) Number of people between 15 years and 65 years = 375 - 125 = 250 people.

(a) (b) From the Venn Diagram above, 60% passed implies that 40% passed : 40% of 240 = 96. 28 + 2x + x + 12 + 6 + 6 + 8 = 96 60 + 3x = 96 3x = 36 x = 12 (i) Faulty brakes = 12 + 12 + 8 + 6 =38 (ii) Cars with only one fault = 28 + 2x + x = 28 + 3x = 28 + 3(12) = 28 + 36 = 64

The Chief Examiner reported that in part (a), some Candidates' carried out the requirement of BODMAS correctly tile few couldn't. However, in part (b), most Candidates' could not sketch the diagram correctly and hence failed to earn the marks allocated. In part (a), expressing as improper fraction we have (\(\frac{1}{4} \times 9\frac{1}{7} + \frac{2}{5} (\frac{2}{2} + \frac{3}{4})) \div (\frac{2}{5} - \frac{1}{4})\) Next, using BODMAS to get (\(\frac{16}{7} + \frac{2}{5} \times \frac{17}{12}) \div \frac{17}{12}\). Simplifying and inverting yields \(\frac{1198}{420} \times \frac{20}{3} = 19\frac{1}{63}\) In part (b), they were expected to sketch Then, using trigonometric ratio to have sin 48\(^o\) = \(\frac{x}{250}\) or cos 42\(^o\) = \(\frac{x}{250}\) and simplifying the vertical distance moved to be 186 m.

(a) First joining XZ and WY, we have < XWY = < XZY = \(\frac{1}{2}\)(180^\circ — 80^\circ) = 50^\circ. This implies that < ZXY = 50^\circ(base angles of an isosceles triangle). So that < XWZ = 180^\circ - 80^\circ = 100^\circ. (b) The rate in every N1.00 = \(\frac{2}{100}\) so that the rate of interest charged per annum will be \(\frac{2 \times 12}{100}\) x 100 which gives 24%.

(a) let the original price of the shirt be x, then we have two percentages reduction leaving 60% and 70%. This implies that \(\frac{60}{100} \times \frac{70}{100} \times x\) = 35. Solving for x yields x = GH₵ 83.33. (b) (i), the first reduction = (\(\frac{40}{100}\)) x 180 = 72.00 and the second reduction = (\(\frac{30}{100}\) ) x 108 = 32.40. Then total reduction = (72.00 + 32. 40) = GH₵ 104.40. (b) (ii), the percentage reduction in the price of the article = (\(\frac{104.40}{180} \times 100\)) = 58%