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\(\frac{0.24633}{0.03060}\) multiplying throughout by 100,000 = \(\frac{24633}{3060}\) = 8.05 = 8.1

Prime numbers between 1 and 6 are 2, 3, and 5 x = 2 + 3 + 5 = 10 H.C.F. of 6, 9, 15 = 3 ∴ y = 3 X x Y = 10 x 3 = 30

\(\frac{1\frac{1}{2}}{2 \div \frac{1}{4} \text{ of }32}\) = \(\frac{\frac{3}{2}}{2 \div \frac{1}{4} \times 32}\) \(\frac{3}{2} \times 4 = 6\)

% error = \(\frac{\text{actual error}}{\text{true value}}\) x 100 Where actual error = 5.1 - 5.0 = 0.1 true value = 5.0g % error = \(\frac{0.1}{5.0}\) x 100 = \(\frac{10}{5}\) = 2%

Let value of store = X Ratio of Taiwo's share to kehine's is 11:9 Keyinde sells \(\frac{2}{3}\) of her share to Taiwo for N720 \(\frac{2}{3}\) of 9 = 6 ∴ Sum of the ratio = 11 + 9 = 20 \{\frac{6}{20}\) of x = N720 \(\frac{6x}{20}\) = 720 ∴ x = \(\frac{720 \times 20}{6}\) x = N24,000

Let x represent total number of balls in the basket. If there are 10 blue balls, \(\frac{1}{8}\) of x = 10 x = 10 x 8 = 80 balls Green balls will be \(\frac{5}{8}\) x 80 = 50 and black balls = \(\frac{2}{8}\) x 80 = 20 Ratio = Green : black : blue 50 : 20 : 10 -10 : -10 : - ------------------ New Ratio 40 : 10 : 10 4 : 1 : 1

He pays tax on 1 - \(\frac{1}{8}\) = \(\frac{7}{8}\)th of his income 20% is 490, 100% is \(\frac{100}{20}\) x 490, N2,450.00 = \(\frac{7}{8}\) of his income = N2,450.00 \(\frac{1}{\frac{7}{8}}\) x 2450 = \(\frac{8 \times 2450}{7}\) = \(\frac{19600}{7}\) = N2800.00

\(\frac{8^{\frac{1}{3}} \times 5^{\frac{2}{3}}}{10^{\frac{2}{3}}}\) = \(\frac{(2^3)^{\frac{1}{3}} \times 5^{\frac{2}{3}}}{(2 \times 5)^{\frac{2}{3}}}\) = \(\frac{2 \times 5^{\frac{2}{3}}}{2^{\frac{2}{3}} \times 5^{\frac{2}{3}}}\) = \(2^{1 - \frac{2}{3}}\) x \(5^{\frac{2}{3} - \frac{2}{3}}\) = \(2^{\frac{1}{3}}\) x \(5^0\) = \(\sqrt[3]{2}\)

If \(log_{10} 2 = 0.3010\) and \(log_{10} 3 = 0.4771\), \(\log_{10} 4.5 = \log_{10} (\frac{3 \times 3}{2})\) \(log_{10} 3 + log_{10} 3 - log_{10} 2 = 0.4771 + 0.4771 - 0.3010\) = 0.6532

\(\frac{x + 2}{x + 1}\) - \(\frac{x - 2}{x + 2}\) = \(\frac{(x + 2)(x + 2) - (x - 2)(x + 1)}{(x + 1)(x + 2)}\) = \(\frac{(x^2 + 4x + 4) - (x^2 - x - 2)}{(x + 1)(x + 2)}\) = \(\frac{x^2 + 4x + 4 - x^2 + x + 2}{(x + 1)(x + 2)}\) = \(\frac{5x + 6}{(x + 1)(x + 2)}\)

\(\frac{1}{x^2 + 5x + 6}\) + \(\frac{1}{x^2 + 3x + 2}\) = \(\frac{1}{(x + 2)(x + 3)}\) + \(\frac{1}{(x + 1)(x + 2)}\) \(\frac{(x + 1)+ (x + 3)}{(x + 1)(x + 2)(x + 3)}\) = \(\frac{x + 1 + x + 3}{(x + 1)(x + 2)(x + 3)}\) \(\frac{2x + 4}{(x + 1)(x + 2)(x + 3)}\) = \(\frac{2(x + 2)}{(x + 1)(x + 2)(x + 3)}\) = \(\frac{2}{(x + 1)(x + 3)}\)

The volume of the pipe is equal to the area of the cross section and length. let outer and inner radii be R and r respectively. Area of the cross section = (R2 - r2) where R = 6 and r = 6 - 1 = 5cm Area of the cross section = (62 - 52)\(\pi\) = (36 - 25)\(\pi\)cm sq vol. of the pipe = \(\pi\) (R2 - r2)L where length (L) = 10 volume = 11\(\pi\) x 10 = 110\(\pi\)cm3

From the figure, 3x + x - 40o = 180o 4x = 180o + 40o 4x = 220o x = \(\frac{220}{4}\) = 55o

From the figure, PS = QR = YT = 7cm Area of parallelogram PQRS = 56cm 56 = base x height, where base = 7 7 x h = 56cm, h = \(\frac{56}{7}\) = 8cm Area of trapezium \(\frac{1}{2}\) (sum of two sides)x height where two sides are QT and PS but QT = QR + RY + YT = 7 +9 + 7 = 23cm Area of trapezium PQTS = \(\frac{1}{2}\)(23 + 7) x 8 \(\frac{1}{2}\) x 30 x 8 = 120cmsq

In the figure PS = RS = QS, they will have equal base QR = RP In angle SQR, angle S = 50º In angle QRP, 65 + 65 = 130º Since RQP = angle RPQ = \(\frac{180 - 130}{2}\) = \(\frac{50}{2} = 25^o\) QPR = 25^\circ