2015 - WAEC Mathematics Past Questions and Answers - page 2

Let the given positive number be x

Then 4 + x = x²

0 = x² - x - 42

or x² - x - 42 = 0

x² - 7x + 6x - 42 = 0

x(x - 7) + 6(x - 7) = 0

= (x + 6)(x - 7) = 0

x = -6 or x = 7

Hence, x = 7

Given; y = x² - x - 2, y = 2x - 1

Using y = y, gives

x² - x - 2 = 2x - 1

x² - 3x - 2 + 1 = 0

therefore, x² - 3x - 1 = 0

Using V = \(\frac{3}{1} \pi r^2h\),

so, 38\(\frac{1}{2} = \frac{1}{3} \times \frac{22}{7} \times r^2 \times 3\)

\(\frac{77}{2} = \frac{22}{7} \times r^2\)

r² = \(\frac{77 \times 7}{2 \times 22}\)

r² = \(\frac{49}{4}\)

Hence, r = \(\sqrt{\frac{49}{4}}\)

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

Volume of rectangular tank = L x B x H

= 2 x 7 x 11

= 154cm³

volume of cylindrical tank = \(\pi r^2h\)

154 = \(\frac{22}{7} \times r^2 \times 4\)

r² = \(\frac{154 \times 7}{22 \times 4}\)

= \(\frac{49}{4}\)

r = \(\sqrt{\frac{49}{4}} = \frac{7}{2}\)

= 3\(\frac{1}{2}\)m

In the diagram; 108° + x + x = 180° (sum of angle in a triangle)

108° + 2x = 180°

x = 180° - 108°

= 72°

x = \(\frac{72^o}{2}\)

= 36°

(Angle between tangent and a chord through the point of contact)

Hence, angle PTN = 90 + 36

= 126°

In the diagram; a = b = 48^o (alternate < S)

x = 180^o - b (angles on a str. line)

x = 180^o - 48^o

= 132^o

In the diagram given, < PRT = 3\(^o\) (Change in same segment) < TPR = 90\(^o\) (angle in a semicircle) Hence, < PTR = 180\(^o\) - (90 + 33)\(^o\) = 180\(^o\) - 123\(^o\) = 57\(^o\)

In the diagram,  < TRS = \(\frac{76^o}{2}\) = 38\(^o\) Hence, angle PRS = 33\(^o\) + 38\(^o\) = 71\(^o\)

Ler: (x₁, y₁) = (0, 3)

(x₂, y₂) = (\(\frac{5}{4}, \frac{11}{2}\))

Using gradient, m = \(\frac{y_2 - y_2}{x_2 - x_1}\)

= \(\frac{\frac{11}{2} - 3}{\frac{5}{4} - 0}\)

= \(\frac{11 - 6}{2} + \frac{5}{4}\)

= \(\frac{5}{} + \frac{5}{4}\)

= \(\frac{5}{2} \times \frac{4}{5}\)

= 2