2012 - WAEC Mathematics Past Questions and Answers - page 2

x + y = 2y - x + 1 = 5

x + y = 2y - x + 1

x + x + y - 2y = 1

2x - y = 1....(i)

2y - x + 1 = 5

-x + 2y = 5 + 1

-x = 2y = 4

x - 2y = -4 .....(ii)

solve simultaneously (i) x 2x - y = 1

(ii) x x - 2y = -4

2x - y = 1

=2x - 4y = -8

3y = 9

y = (\frac{9}{3})

y = 3

substitute y = 3 into equation (i)

2x - y = 1

2x - 3 = 1

2x = 1 + 3

2x = 4

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

= 2

12 = (\frac{n}{3} - 2n = 1), multiply through by 3

36 + n - 6n = 3

-5n = 3 - 36

-5n = -33

-5n + 33 = 0

5n - 33 = 0

(\frac{-m}{2} - \frac{5}{4} \geq \frac{5m}{12} - \frac{7}{6})

= (\frac{-2m - 5}{4} \geq \frac{5m - 14}{12})

12(-2m - 5) (\geq) 4(5m - 14)

-24m - 60 (\geq) 20m - 56

-24m - 20m (\geq) -56 + 60

44m (\geq) 4

m (\leq \frac{4}{-44})

m (\leq \frac{-1}{11})

Curved surface area = 2(\pi h)

704 = 2 x (\frac{22}{7} \times 8 \times h)

704 = (\frac{352}{7})h

352h = 704 x 7

h = (\frac{704 \times 7}{352})

= (\frac{4928}{352})

h = 14cm

Let 0 = angle of the minor sector

angle of the major sector = 360 - (\theta)(angle at a point)

2 (\pi r) = 54 + 126(i.e circumference of minor and major arc)

2(\pi r = 180^o)

r = (\frac{180}{2\pi}) = (\frac{90}{\pi})

Lenght of ninor arc

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

54 = (\frac{\theta}{360} \times 3 \pi r)

(\theta = \frac{360 \times 54}{2 \pi r})

but r = (\frac{90}{\pi}) substituting (\frac{90}{\pi}) for r

(\theta = \frac{360 \times 54 \times \pi}{2 \times \pi \times 90})

(\theta = 2 \times 54 = 108^o)

angle of the major sector = 360 - 108^o

= 252^o

cos(x + 40)^o = 0.0872

x + 40 = cos^-10.0872

x + 40^o = 84.99^o

x = 84.99^o - 40^o

x = 44.99

x = 45^o

($$\text{Misplaced hline}$$)

(E(x - \bar{x})^2) = 150

N = 6

S.D = (\sqrt{\frac{(x - x)^2}{N}})

S.D = (\sqrt{\frac{150}{6}}) = 5

Prob(RB + BR) = Total balls = 4 + 6 = 10

= prob((\frac{4}{10} \times \frac{6}{9})) + prob((\frac{6}{10} \times \frac{4}{9}) = \frac{24}{90} + \frac{24}{90})

= (\frac{48}{90} = \frac{16}{30} = \frac{8}{15})