2001 - WAEC Mathematics Past Questions and Answers - page 2

\(\frac{1}{2} + (\frac{3}{4} \text{ of } \frac{2}{5}) \div 1\frac{3}{5}\)

= \(\frac{1}{2} + (\frac{3}{4} \times \frac{2}{5}) \div \frac{8}{5}\)

= \(\frac{1}{2} + \frac{3}{10} \div \frac{8}{5}\)

= \(\frac{1}{2} + (\frac{3}{10} \times \frac{5}{8})\)

= \(\frac{1}{2} + \frac{3}{16}\)

= \(\frac{11}{16}\)

y = px + q

5 = 3p + q ... (i)

4 = 2p + q ... (ii)

(i) - (ii) : p = 1

\(\therefore\) 5 = 3(1) + q 

\(\implies\) q = 5 - 3 = 2

(p, q) = (1, 2)

\(\frac{x^2 + x - 2}{2x^2 + x - 3}\)

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

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

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

= \(\frac{x + 2}{2x + 3}\)

At x = -1, 

= \(\frac{-1 + 2}{2(-1) + 3}\)

= \(\frac{1}{1}\)

= 1

\(6x^2 + 7x - 20\)

= \(6x^2 + 15x - 8x - 20\)

= \(3x(2x + 5) - 4(2x + 5)\)

= \((3x - 4)(2x + 5)\)

\(\frac{2x - 1}{3} - \frac{x + 3}{2}\)

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

= \(\frac{4x - 2 - 3x - 9}{6}\)

= \(\frac{x - 11}{6}\)

\(\frac{y - 3}{2} < \frac{2y - 1}{3}\)

\(3(y - 3) < 2(2y - 1)\)

\(3y - 9 < 4y - 2\)

\(3y - 4y < -2 + 9\)

\(-y < 7\)

\(y > -7\)

\(\frac{4m + 3n}{4m - 3n} = \frac{5}{2}\)

\(5(4m - 3n) = 2(4m + 3n)\)

\(20m - 15n = 8m + 6n\)

\(20m - 8m = 6n + 15n\)

\(12m = 21n\)

\(\frac{21}{12} = \frac{m}{n}\)

\(m : n = 7 : 4\)

A quadratic equation is an equation of the second order. The highest power in the equation is 2.