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x\(^2\) = 172\(^2\) - 8\(^2\) x\(^2\) = 289 - 64 x\(^2\) = 225 x = \(\sqrt{225}\) x = 15 tan y \(\frac{8}{15}\) 1 + 2 tan y = 1 + 2(\(\frac{8}{15}\)) = \(\frac{1}{1} + \frac{6}{15}\) = \(\frac{15 + 16}{15} = \frac{31}{15}\) \(\frac{\tan y}{1 + 2 \tan y}\) = \(\frac{8}{15} \div \frac{31}{15}\) = \(\frac{8}{15} \times \frac{15}{31} = \frac{8}{31}\) (b) Otobo shares = N60,000 Remainder = N300,000 - N60,000 = N240,000 Ada's share = \(\frac{5}{12}\) x N240,0000 = N100,000 Adeola's share =; = 300,000 - N160,000 = N140,000 Ratio of the money is Otobo : Ada : Adeola 60,000 : 100,000 : 140,000 60 : 100 : 140 6 : 10 : 14 3 : 5 : 7

T\(O\)S = 68\(^o\) x 2 = 136\(^o\) (Angle at centre is twice angle at circumference) T\(R\) = 180\(^o\) - (90 + 68\(^o\)) = 180\(^o\) - 158\(^o\) = 22\(^o\) x = 2 x T\(R\)O = 2 x 22\(^o\) = 44\(^o\) (b) Let the capacity of Tank B = x litres Capacity of Tank A = x + 600 litres If 100 litres is pumoed out of each tank then, (x + 600) - 100 = 3(x - 100) x + 500 = 3x - 300 500 + 300 = 3x - x \(\frac{800}{2} = \frac{2x}{x}\) x = 400 litres Capacity of Tank A = 400 + 600 = 1,000 litres Capacity of Tank B = 400 litres

In the diagram, Thus, m = 2 x 55\(^o\) (is a bisector of obtuse m = 110\(^o\) n = \(\frac{1}{2}\) x 110\(^o\) (angle at centre = 2 x angle at circum) n = 55\(^o\) m + n = 110 + 55 = 165\(^o\)

Since the salary increases by #36 after every 6 months : 2 years and 8 months imply an increase of five times only: 36 * 5 → #180 His salary then = initial salary + increment = 540 + 180 = #720 Answer is C

2\(\frac{1}{4} \times 3\frac{1}{2} \div 4 \frac{3}{8}\) = \(\frac{9}{4} \times \frac{7}{2} \div \frac{35}{8}\) = \(\frac{9}{4} \times \frac{7}{2} \times \frac{8}{35}\) = \(\frac{9}{5}\) = 1 \(\frac{4}{5}\)

6x\(^2\) - 2x + 1 \(\frac{6x^{2+1}}{2+1} - \frac{2x^{1+1}}{1+1}+ 1x = \frac{6x^3}{3} - \frac{2x^2}{2} + x\) \(2x^3 - x^2 + x + c\)

Let the number of people that offer both Mathematics and Physics = y Then, \((32 - y) + y + (24 - y) + 4 = 40\) \(60 - y = 40 \implies y = 20\) \(\therefore\) 20 students offer both Mathematics and Physics.

\(\frac {x+2}{4}\) - \(\frac{2x - 3}{3} < 4\) \(\frac{3(x + 2) - 4(2x - 3)}{12} < 4\) \(3x + 6 - 8x + 12 < 48 \) \(18 - 5x < 48 \implies -5x < 30\) \(\therefore x > -6\)

Let the angle for school fees = x^\circ Then Housing = 2x^\circ 120^\circ + 90^\circ + x^\circ + 2x^\circ = 360^\circ 3x^\circ = 150^\circ \(\implies\) x^\circ = 50^\circ. Amount spent on housing = \(\frac{100}{360} \times 7200\) = N2000.

Profit (P) = 10\(_x\) − \(_x\)2 Maximum profit can be achieved when the differential of profit with respect to number of bags(x) is 0 i.e. \(\frac{dp}{dx}\) = 0 \(\frac{dp}{dx}\) = 10 - 2x = 0 10 = 2x Then x = \(\frac{10}{2}\) = 5 Answer is C

y = 23\(_{five}\) + 101\(_{three}\) 23\(_{five}\) = \(2 \times 5^1 + 3 \times 5^0\) = 13\(_{ten}\) 101\(_{three}\) = \(1 \times 3^2 + 0 \times 3^1 + 1 \times 3^0\) = 10\(_{ten}\) y\(_{ten}\) = 13\(_{ten}\) + 10\(_{ten}\) = 23\(_{ten}\) = 10111\(_{two}\)

The diagram shows angles at a point, the total angle at a point is 360 x - 10 + 4x - 50 + 2x + 3x + 20 = 360 10x - 40 = 360 10x = 360 + 40 10x = 400 x = \(\frac{400}{10}\) x = 40

\(\frac{3}{4}\) t + \(\frac{1}{3}\) (21 - t) = 11 Multiply through by the LCM of 4 and 3 which is 12 12 x(\(\frac{3}{4}\) t) + 12 x (\(\frac{1}{3}\) (21 - t)) = (11 x 12) 9t + 4(21 - t) = 132 9t + 84 - 4t = 132 5t + 84 = 132 5t = 132 - 84 = 48 t = \(\frac{48}{5}\) t = 9 \(\frac{3}{5}\) Answer is D

let the girls' initial pocket money be whole(1) \(\frac{1}{4}\) of 1 = books, \(\frac{1}{3}\) of 1 = dress fraction of her money left = 1 - \(\frac{1}{4}\) - \(\frac{1}{3}\) = \(\frac{5}{12}\)

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