2008 - WAEC Mathematics Past Questions and Answers - page 3

\(\frac{\log\sqrt{8}}{\log 4 - \log 2} = \frac{\log 8\frac{1}{2}}{\log (\frac{4}{2})}\)

= \(\frac{8 \frac{1}{2}}{\log (\frac{4}{2})}\)

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

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

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

Average speed = \(\frac{Distance}{Time}\)

\(\frac{400km}{hr} = \frac{60km}{Time}\)

Time = \(\frac{60km}{400 km/hr}\)

= \(\frac{60hr}{400}\)

M = \(\frac{60hr}{400} \times \frac{60min}{1hr}\)

= 9 minutes

Length of an arc = \(\frac{\theta}{360} \times 2\pi r\)

18.33 = \(\frac{\theta}{360} \times 2 \times \frac{22}{7} \times 14\)

\(\theta = \frac{18.33 \times 360 \times 17}{2 \times 22 \times 14}\)

= 75^o (approx.)

Total surface area of cube = 6s²

6s² = x

s² = x = \(\frac{x}{6}\).....(1)

volume of a cube = s² = \(\frac{x}{2}\)

s² = \(\frac{x}{2}\)......(2)

put(1) into (2)

s(\(\frac{x}{6}\)) = \(\frac{x}{2}\)

s = \(\frac{x}{2} \times \frac{6}{x}\)

= 3cm

perimeter of minor segment = Length of arc xy + chord xy

where lxy = \(\frac{120}{360} \times 2x \times \frac{22}{7} \times 7\)

= 14.67cm

perimeter of minor segment = 14.67 + 8 = 22.67cm

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

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

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

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

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

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

1 + 2x = -2

2x = -2 - 1

2x = -3

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

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

Roots; x and -c

sum of roots = c + (-c) = 0

product of roots = c x -c = -c²

Equation; x² - (sum of roots) x = product of roots = 0

x² - (0)x + (-c²) = 0

x² - c² = 0

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

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

-2x < - \(\frac{4}{-6}\)

3x -2x < -4; -8x < -4

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

A = [2, 4, 6, 8}

B = {2, 3, 5, 9}

Pr = (Prime in B) = \(\frac{3}{4}\)