2009 - WAEC Mathematics Past Questions and Answers - page 3

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

= (\frac{60}{360} \times 2\pi \times 3cm)

= (\pi)cm

(\frac{2x + 1}{6} - \frac{3x - 1}{4}) = 0

(\frac{4(2n + 1) - 6(3x - 1)}{24}) = 0

-10x + 10 = 0

-10x = -10

x = (\frac{-10}{-10})

x = 1

3x - 4 < 6 = 3x < 6 = 4

3x < 10

x < (\frac{10}{3})

x < 3.33 and x - 1 = 0

n > 1 = 1< x; since x is an integer, and 1 < x3.33

x = {2, 3}

6 - x - x² = 0

a = -1; b = -1; c = 6

Sum of roots = c + k = -(\frac{-b}{a})

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

= -1

Each interior angle = 140

(\frac{(n - 2) \times 180}{n} = 140)

(n - 2) x 180 = 140n

150 - 360 = 140n

180m - 140n = 360

40n - 360

n = (\frac{360}{40})

n = 9

Sum of all interior angles = (n - 2) x 180

= (9 - 2) x 180

= 7 x 180

= 1260

No. of buckets of water = (\frac{\text{Capacity of reservoir}}{\text{Capacity of buckets}})

= (\frac{800 \times 700 \times 500}{10 \times 1000})

= (\frac{28000 0000}{10000})

= 28000

Total no of balls = 30

Let x = no. of red balls

Pr(red) = (\frac{x}{30})

Pr(black) = (\frac{3}{10} = \frac{9}{30})

Pr(white) = (\frac{2}{5} = \frac{12}{30})

No. of black balls = 9

No. of white balls = 12

9 = 12 + x = 30

x = 30 - 21

x = 9

No. of red balls = 9

(\frac{1}{2}\sqrt{32} - \sqrt{18} \sqrt{2}) = (\frac{1}{2} (4\sqrt{2}) - 3\sqrt{2} + \sqrt{2})

= 2(\sqrt{2} - 3\sqrt{2} + \sqrt{2})

= 3(\sqrt{2} - 3\sqrt{2} - 3\sqrt{2}) = 0

(x + 10)^o + (2x - 40)^o + (3x - 90)^o = 180

6x - 120 = 180

6x = 180 + 120

6x = 300

x = (\frac{300}{6})

x = 50

x + 10^o = 50^o + 10^o = 60^o

2x - 40 = 2(50^o) - 40 = 60^o

3x - 90 = 3(50^o) - 90^o = 60^o

Hence, it is an equilateral triangle

(x - 3y)² - (x + 3y)² = [(x - 3y) - (x + 3y)]

[(x + 3y + x + 3y)] = [-6y] [2x]

= -12xy