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\(\frac{\sin y}{3} = \frac{\sin 120^o}{5}\) sin 120o = sin 60o 5 sin y = 3 sin 60o sin y = \(\frac{3 \sin 60^o}{5}\) \(\frac{3 \times 0.866}{5}\) = \(\frac{2.598}{5}\) y = sin-1 0.5196 = 30o 18'

\(\frac{2}{3} - \frac{1}{5}\) = \(\frac{10 - 3}{15}\) = \(\frac{7}{15}\) \(\frac{1}{3}\) Of \(\frac{2}{5}\) = \(\frac{1}{3}\) x \(\frac{2}{5}\) = \(\frac{2}{15}\) (\(\frac{2}{3} - \frac{1}{5}\)) - \(\frac{1}{3}\) of \(\frac{2}{5}\) = \(\frac{7}{15} - \frac{2}{15}\) = \(\frac{1}{3}\) 3 - \(\frac{1}{1 \frac{1}{2}}\) = 3 - \(\frac{2}{3}\) = \(\frac{7}{3}\) \(\frac{\frac{2}{3} - \frac{1}{5} \text{of} \frac{2}{15}}{3 - \frac{1}{1 \frac{1}{2}}}\) = \(\frac{\frac{1}{3}}{\frac{7}{3}}\) = \(\frac{1}{3}\) x \(\frac{3}{7}\) = \(\frac{1}{7}\)

\(\frac{0.0001432}{1940000}\) = k x 10n where 1 \(\leq\) k \(\leq\) 10 and n is a whole number. Using four figure tables, the eqn. gives 7.38 x 10-11 k = 7.381, n = -11

Let the selling price(SP from P to Q be represented by x i.e. SP = x When SP = x at 10% profit CP = \(\frac{100}{100}\) + 10 of x = \(\frac{100}{110}\) of x when Q sells to R, SP = N209 at loss of 5% Q's cost price = Q's selling price CP = \(\frac{100}{95}\) x 209 = 220.00 x = 220 = \(\frac{2200}{11}\) = 200 = N200.00

Let x represent the price of an orange and y represent the number of oranges that can be bought xy = 240k, y = \(\frac{240}{x}\).....(i) If the price of an oranges is raised by \(\frac{1}{2}\)k per orange, number that can be bought for N240 is reduced by 16 Hence, y - 16 = \(\frac{240}{x + \frac{1}{2}\) = \(\frac{480}{2x + 1}\) = \(\frac{480}{2x + 1}\).....(ii) subt. for y in eqn (ii) \(\frac{240}{x}\) - 16 = \(\frac{480}{2x + 1}\) = \(\frac{240 - 16x}{x}\) = \(\frac{480}{2x + 1}\) = (240 - 16x)(2x + 1) = 480x = 480x + 240 - 32x2 - 16 480x = 224 - 32x2 x2 = 7 x = \(\sqrt{7}\) = 2.5 = 2\(\frac{1}{2}\)k

Total yield = N3,700 Total amount invested = N 50,000 Let x be the amount invested at 6% interest and let y be the amount invested at 8% interest then yield on x = \(\frac{6}{100}\) x and yield on y = \(\frac{8}{100}\)y Hence, \(\frac{6}{100}\)x + \(\frac{8}{100}\)y = 3,700.........(i) x + y = 50,000........(ii) 6x + 8y = 370,000 x 1 x + y = 50,000 x 6 6x + 8y = 370,000.........(iii) 6x + 6y = 300,000........(iv) eqn (iii) - eqn(2) 2y = 70,000 y = \(\frac{70,000}{2}\) = 35,000 Money invested at 8% is N35,000

Number of boy's = 30 Number of girls = x Mean of boys' score = 6, for girls = 8 Total score for boys = 30 x 6 Total score for both boys and girls = 468 180 + 8x = 468 x = \(\frac{288}{8}\) = 36

Total cost of production = N120.00 Labour Cost = \(\frac{70}{120}\) x \(\frac{360^o}{1}\) = 210o

Numbers divisible by 4 between 1 and 300 include 4, 8, 12, 16, 20 e.t.c.To get the number of figures divisible by 4, We solve by method of A.P Let x represent numbers divisible by 4, nth term = a + (n - 1)d a = 4, d = 4 Last term = 4 + (n - 1)4 296 = 4 + 4n - 4 = \(\frac{296}{4}\) = 74 rn(Note: 296 is the last Number divisible by 4 between 1 and 300) Prob. of x = \(\frac{74}{298}\) = \(\frac{37}{149}\)

\(\frac{\log_{3} 27 - \log_{\frac{1}{4}} 64}{\log_{3} (\frac{1}{81})}\) \(\log_{3} 27 = \log_{3} 3^{3} = 3\log_{3} 3 = 3\) \(\log_{\frac{1}{4}} 64 = \log_{\frac{1}{4}} (\frac{1}{4})^{-3} = -3\) \(\log_{3} (\frac{1}{81}) = \log_{3} 3^{-4} = -4\) \(\therefore \frac{\log_{3} 27 - \log_{\frac{1}{4}} 64}{\log_{3} (\frac{1}{81})} = \frac{3 - (-3)}{-4}\) = \(\frac{6}{-4} = \frac{-3}{2}\)

P(x, y), P(0, 3) If x increases by 1 unit and y by 4 units, then ratio of x : y = 1 : 4 \(\frac{x}{1}\) = \(\frac{y}{4}\) y = 4x Hence the sign of the graph is y + 3 = 4x

Total cost of 1035 oranges at N145 each = 1035 x 145 = 20025 Total selling price at N245 each = (103)5 x 245 = 30325 Hence his gain = 30325 - 20025 = 10305

\(\frac{5\sqrt{7} - 7\sqrt{5}}{\sqrt{7} - \sqrt{5}}\) = \(\frac{5\sqrt{7} - 7\sqrt{5}}{\sqrt{7} - \sqrt{5}}\) x \(\frac{\sqrt{7} + \sqrt{5}}{\sqrt{7} + \sqrt{5}}\) = \(\frac{(5 \times 7) + (5 \sqrt{7} \times 5) - (7 \times \sqrt{5} \times 7) (-7 \times 5)}{(\sqrt{7})^2}\) = \(\frac{5 \sqrt{35} - 7\sqrt{35}}{2}\) = \(\frac{-2\sqrt{35}}{2}\) = - \(\sqrt{35}\)

\(\frac{3^n - 3^{n - 1}}{3^3 \times 3^n - 27 \times 3^{n - 1}}\) = \(\frac{3^n - 3^{n - 1}}{3^3(3^n - 3^{n - 1})}\) = \(\frac{3^n - 3^{n - 1}}{27(3^n - 3^{n - 1})}\) = \(\frac{1}{27}\)

P \(\alpha\) \(\frac{q^2}{r}\) P = \(\frac{kq^2}{r}\) k = \(\frac{pr}{q^2}\) = \(\frac{36 x 4}{(3)^2}\) p = \(\frac{16q^2}{r}\) = \(\frac{16 \times 25}{2}\) = 200