1995 - WAEC Mathematics Past Questions and Answers - page 2

\(\frac{1}{3}(2x - 1) < 5\)

\(2x - 1 < 15\)

\(2x < 16\)

\(x < 8\)

x\(^2\) - 3x + 2 = 0

x\(^2\) - 2x - x + 2 = 0

x(x - 2) - 1(x - 2) = 0

(x - 2)(x - 1) = 0

x = 1 or 2. The smaller value of x = 1.

x(a - c) + y(c - a)

= x(a - c) - y(a - c)

= (x - y)(a - c)

\(y \propto \frac{1}{x^2}\)

\(y = \frac{k}{x^2}\)

\(4 = \frac{k}{3^2}\)

\(k = 4 \times 3^2 = 36\)

\(y = \frac{36}{x^2}\)

When x = 2,

\(y = \frac{36}{2^2} = 9\)

3x\(^2\) + 25x - 18 = 0

3x\(^2\) + 27x - 2x - 18 = 0

3x(x + 9) - 2(x + 9) = 0

(3x - 2)(x + 9) = 0

x = \(\frac{2}{3}\) or x = -9.

(x + 2)(x - 7) = 0

x + 2 = 0 \(\implies\) x = -2

x - 7 = 0 \(\implies\) x = 7

x = -2 or 7

x + y = \(\frac{3}{2}\) ... (i)

x - y = \(\frac{5}{2}\) ... (ii)

(i) - (ii):

2y = \(\frac{-2}{2}\) = -1

y = \(-\frac{1}{2}\)

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

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

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

= \(\frac{4}{2}\)

x = 2

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

= -1 + 2 = 1.

|AB| = 12 cm; |DC| = 9 cm; |AE| = 8 cm.

\(\Delta\) ABE and \(\Delta\) EDC are similar triangles. Hence,

\(\frac{|AB|}{|DC|} = \frac{|AE|}{|EC|}\)

\(\frac{12}{9} = \frac{8}{|EC|}\)

\(\therefore |EC| = \frac{9 \times 8}{12}\)

= 6 cm