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})