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Total amount invested by A and B = N5,000 farm yield was sold for N15,000.00 profit = 15,000.00 - 5,000.00 = N10,000.00 Profit was shared in ratio 2 : 3 2 + 3 = 5 A received \(\frac{2}{5}\) of profit = \(\frac{2}{5}\) x 10,000 = N4,000.00 A receive \(\frac{3}{5}\) of profit = \(\frac{3}{5}\) x 10,000 = N6,000.00 Difference in profit received = N6,0000 - N4,000.00 = N2,000.00

I = \(\frac{PRT}{100}\) 100(A - P) = prt P = \(\frac{100A}{100 + RT}\) = \(\frac{100 \times 7000}{100 + (4 \times 3)}\) p = \(\frac{700,000}{112}\) = N6,250.00

profit 8% of N1.35 = \(\frac{8}{100}\) x N1.35 = N0.08 Cost price = N1.35 - N0.10 = N1.25 If he sells the 20 oranges for N1.10 now %loss = \(\frac{\text{actual loss}}{\text{Cost price}}\) x 100 \(\frac{1.25 - 1.10}{1.25}\) x 100 = \(\frac{0.15 \times 100}{1.25}\) = \(\frac{15}{1.25}\) = 12%

Let x represents number of boys that can work at \(\frac{5}{4}\) the rate at which the 10 girls work For 1hr. x boys will work for \(\frac{\frac{1}{5}}{4}\) x 10 x = \(\frac{4}{5}\) x 10 = 8 boys 8 boys will do the work of ten girls at the same rate 4 + 8 = 12 boys cut the field in 5 hrs For 3 hrs, \(\frac{12 \times 5}{3}\) boys will be needed = 20 boys

(0.008) -\(\frac{1}{3}\) x (0.16) - \(\frac{3}{2}\) = (8 x 10-3)10-3 x (16 x 10-2)-\(\frac{3}{2}\) = \(\frac{(2^3)}{10^3}\) - 3 x \(\frac{(2^4)}{10}\) - \(\frac{-3}{2}\) = \(\frac{625}{8}\)

log\(_2\)6 - log\(_2\)3 = log\(_2\) (\(\frac{6}{3}\)) = log\(_2\)2... A log\(_2\)8 - 2log\(_2\)\(\frac{1}{2}\) log\(_2\)8 - log\(_2\)\(\frac{1}{2}\)\(^2\) log\(_2\)8 - log\(_2\)\(\frac{1}{4}\) = log\(_2\) 32... B \(\frac{A}{B}\) = \(\frac{log_22}{log_232}\) N.B. log\(_2\)2 = 1 = \(\frac{1}{5}\)

\(\frac{2\sqrt{14} \times 3\sqrt{21}}{7 \sqrt{24} \times 2\sqrt{98}}\) = \(\frac{6\sqrt{14} \times 3 \times \sqrt{7} \times \sqrt{3}}{7 \times 2 \sqrt{6} \times \sqrt{7} \times \sqrt{14}}\) = \(\frac{3\sqrt{3}}{14\sqrt{6}}\) = \(\frac{3\sqrt{3}}{14\sqrt{2} \times \sqrt{3}}\) = \(\frac{3\sqrt{2}}{28}\)

If P = \(\frac{\frac{2}{3}({1 - r^2})}{n^2}\), find n when r = \(\sqrt{\frac{1}{3}}\) and p = 1 p = \(\frac{\frac{2}{3}({1 - r^2})}{n^2}\) when r = \(\sqrt{\frac{1}{3}}\) and p = 1 1 = \(\frac{\frac{2}{3}({1 - (\sqrt{\frac{1}{3}})^2})}{n^2}\) \(n^2\) = \(\frac{2}{3}(1 - \frac{1}{3}\)) \(n^2\) = \(\frac{2 \times 2}{3 \times 3}\) = \(\frac{4}{9}\) n = \(\sqrt{\frac{4}{9}}\) = \(\frac{2}{3}\)

a = u\(^2\) - 3v\(^2\) = (1)2 - 3(-1) = -2 b = b = 2uv + v\(^2\) = 2(1)(-1) + (-1)2 = -1 ∴ 2(2a - b)(a - b\(^2\)) = [2(-2) + 1](-2 - (-1)\(^2\)) = (-3)( - 3) = 9

To find the extra cost of feeding one additional person, we start with the formula for the cost \( Q \): \(Q = 1.5 + 0.5n\) Calculate the cost for \( n \) \(Q(n) = 1.5 + 0.5n\) Calculate the cost for \( n + 1 \) \(Q(n + 1) = 1.5 + 0.5(n + 1) = 1.5 + 0.5n + 0.5 = 2 + 0.5n\) Find the extra cost \(\text{Extra cost} = Q(n + 1) - Q(n) = (2 + 0.5n) - (1.5 + 0.5n) = 2 - 1.5 = 0.5\) The extra cost of feeding one additional person is \( 0.5 \) Naira. To convert this to kobo (since \( 1 \) Naira = \( 100 \) kobo): \(0.5 \text{ Naira} = 0.5 \times 100 = 50 \text{ kobo}\) Thus, the extra cost of feeding one additional person is \( 50 \) kobo.

P = \(\frac{1}{v}\) and vR2 = P = \(\frac{k}{v}\)......(i) and v KR2 .......(ii) (where k is constant) Subst. for v in equation (i) = p = \(\frac{1^2}{KR}\).....(ii) when r = 7, p = 2 2 = \(\frac{k}{7^2}\) k = 2 x 49 = 98 Subt. foe k in ....(iii) P = \(\frac{98}{R^2}\) PR2 = 98

Z = x\(^2\) + \(\frac{1}{y^3}\) Z - x\(^2\) = \(\frac{1}{y^3}\) y\(^3\) = \(\frac{1}{Z - x^2}\) y = \(\sqrt[3]{\frac{1}{Z - x^2}}\) ∴ y = \(\frac{1}{\sqrt[3]{Z - x^2}}\) y = \(\frac{1}{(Z - x^2)^{\frac{1}{3}}}\)

x2 + 2(m + 1)x + m + 3 from (A + B)2 = a2 + 2AB + b2 2B = \(\frac{2(m + 1)}{2}\) = m + 1 .......(i) Also B2 = m + 3..........(ii) Subst. for B in equation ..........(iii) = (m + 1)2 = m + 3 = m2 + 2m + 1 = m + 3 = m2 + 2m - m + 1 - 3 = 0 m2 + m - 2 = 0 (m + 2)(m - 1) = 0 or m - 1 = 0 m = -2 or 1

62x + 1 + 7(6x) - 5 Let 6x = y....(i) 6yx + 7y - 5 = 0 (3y + 5)(2y - 1) = 0.....(i) Subt. for y from eqn. (i) in eqn. (ii) = [3(6x) + 5][2(6x) - 1]

x + y = 5.......(i) x2 - 2y2 = 1.......(ii) x = 5 - y.........(iii) Subst. for x in eqn.(ii) = (5 - y)2 - 2y2 = 1 25- 10y + y2 - 2y2 = 1 25 - 1 = y2 + 10y y2 + 10y2 - 24 = 0 (y + 12)(y - 2) = 0 Then Either y + 12 = 0 or y - 2 = 0 = (-12, +2)