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first work out the expression and then correct the answer to 4 s.f = 241.34..............(A) (3 x 10-\(^3\))\(^2\)............(B) = 3\(^2\)x\(^2\) = \(\frac{1}{10^3}\) x \(\frac{1}{10^3}\) (Note that x\(^2\) = \(\frac{1}{x^3}\)) = 24.34 x 3\(^2\) x \(\frac{1}{10^6}\) = \(\frac{2172.06}{10^6}\) = 0.00217206 = 0.002172(4 s.f)

_ Mean deviation = Σ|x - x| n _ x = 2.5 = |1 - 2.5| + |2 - 2.5| + |3 - 2.5| + |4 - 2.5| 4 = 4/4 = 1

\(\frac{0.00275 \times 0.0064}{0.025 \times 0.08}\) Removing the decimals = \(\frac{275 \times 64}{2500 \times 800}\) = \(\frac{88}{10^4}\) 88 x 10-4 = 88 x 10-1 x 10-4 = 8.8 x 10-3

2log \(\frac{2}{5}\) - log\(\frac{72}{125}\) + log 9 [\(\frac{2}{5}\))2 x 9] = log \(\frac{4}{25}\) x \(\frac{9}{1}\) x \(\frac{125}{72}\) = log \(\frac{72}{125}\) = log \(\frac{5}{2}\) = log \(\frac{10}{4}\) = log 10 - log 4 = log10 - log2\(^2\) = 1 - 2 log2

If log\(_9\)x = 1.5, 9\(^{1.5}\) = x 9^\(\frac{3}{2}\) = x (\sqrt9)\(^3\) = 3 ∴ x = 27

\(\sqrt{27}\) + \(\frac{3}{\sqrt{3}}\) = \(\sqrt{9 \times 3}\) + \(\frac{3 \times {\sqrt{3}}}{{\sqrt{3}} \times {\sqrt{3}}}\) = 3\(\sqrt{3}\) + \(\sqrt{3}\) = 4\(\sqrt{3}\)

\(\frac{2}{(2r - 1)}\) - \(\frac{5}{3}\) = \(\frac{1}{(r + 2)}\) \(\frac{2}{(2r - 1)}\) - \(\frac{1}{(r + 2)}\) = \(\frac{5}{3}\) The L.C.M.: (2r - 1) (r + 2) \(\frac{2(r + 2) - 1(2r - 1)}{(2r - 1) (r + 2)}\) = \(\frac{5}{3}\) \(\frac{2r + 4 - 2r + 1}{ (2r - 1) (r + 2)}\) = \(\frac{5}{3}\) cross multiply the solution 3 * 5 = (2r - 1) (r + 2) * 5 divide both sides 5 3 = 2r\(^2\) + 3r - 2 (when expanded) collect like terms 2r\(^2\) + 3r - 2 - 3 = 0 2r\(^2\) + 3r - 5 = 0 Factors are -2r and +5r 2r\(^2\) -2r + 5r - 5 = 0 [2r\(^2\) -2r] [+ 5r - 5] = 0 2r(r-1) + 5(r-1) = 0 (2r+5) (r-1) = 0 r = 1 or - \(\frac{5}{2}\)

use "T" to represent the total profit. The first receives \(\frac{1}{3}\) T remaining, 1 - \(\frac{1}{3}\) = \(\frac{2}{3}\)T The seconds receives the remaining, which is \(\frac{2}{3}\) also \(\frac{2}{3}\) x \(\frac{2}{3}\) x \(\frac{4}{9}\) The third receives the left over, which is \(\frac{2}{3}\)T - \(\frac{4}{9}\)T = (\(\frac{6 - 4}{9}\))T = \(\frac{2}{9}\)T The third receives \(\frac{2}{9}\)T which is equivalent to N12000 If \(\frac{2}{9}\)T = N12, 000 T = \(\frac{12 000}{\frac{2}{9}}\) = N54, 000

You can use any whole numbers (eg. 1. 2. 3) to represent all the mangoes in the basket. If the first child takes \(\frac{1}{4}\) it will remain 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\) Next, the second child takes \(\frac{3}{4}\) of the remainder which is \(\frac{3}{4}\) i.e. find \(\frac{3}{4}\) of \(\frac{3}{4}\) = \(\frac{3}{4}\) x \(\frac{3}{4}\) = \(\frac{9}{16}\) the fraction remaining now = \(\frac{3}{4}\) - \(\frac{9}{16}\) = \(\frac{12 - 9}{16}\) = \(\frac{3}{16}\)

I = \(\frac{PRT}{100}\) = \(\frac{10 \times 2 \times 4}{100}\) = \(\frac{4}{5}\) = 0.8 Total amount = N10.80 He pays N8.00 Remainder = 10.80 - 8.00 = N2.80

Interest I = \(\frac{PRT}{100}\) ∴ R = \(\frac{100 \times 1}{100 \times 5}\) = \(\frac{100 \times 7.50}{500 \times 5}\) = \(\frac{750}{500}\) = \(\frac{3}{2}\) = 1.5%

W \(\alpha\) \(\frac{\frac{1}{uv}}{u + v}\) ∴ w = \(\frac{\frac{k}{uv}}{u + v}\) = \(\frac{k(u + v)}{uv}\) w = \(\frac{k(u + v)}{uv}\) w = 8, u = 2 and v = 6 8 = \(\frac{k(2 + 6)}{2(6)}\) = \(\frac{k(8)}{12}\) k = 12 i.e 12 ( u + v) = uwv

-2 < 2x - 6 AND 2x - 6 < 4 -2 + 6 <2x AND 2x < 4 + 6 4 <2x AND 2x < 10 : 2

Source: WAEC / JAMB / NECO Classroom Archive (Record #60732)