2005 - WAEC Mathematics Past Questions and Answers - page 2

101101₂ = 1 x 2⁵ + 0 x 2⁴ + 1 x 2³ + 1 x 2² + 0 x 2¹ + 1 x 2⁰ = 32 + 0 + 8 + 4 + 0 + 1 = 45₁₀

(2^7 = 2^{3(5-x)}\Rightarrow 7 = 3^{5-x} \Rightarrow 7 15 - 3x<br /> \Rightarrow -8 = -3x \Rightarrow x = \frac{8}{3})

((2x-3y)(x-5y)=2x^2 - 10xy - 3xy + 15y^2<br /> =2x^2 - 13xy + 15y^2)

(v=u+ft \Rightarrow v-u=ft\Rightarrow f=\frac{v-u}{t})

(x^2+(x-1)^2 = (\sqrt{13})^2 \Rightarrow x^2 + x^2 - 2x + 1 = 13<br /> 2x^2 - 2x - 12 = 0) dividing through by 2
(x^2 - x- 6 = 0; (x-3)(x-2) = 0 \Rightarrow x = 3 or -2 )

Prob.(John pass)\(=frac{3}{4}\) prob.(John fail) \(=1-frac{3}{4}=\frac{1}{4}\)
Prob.(James pass) \(=frac{3}{5}\) Prob.(James fail) \(=1-frac{3}{5}=\frac{2}{5}\)
Prob(both boys fail)\(=frac{1}{4}\times \frac{2}{5}=\frac{1}{10}\)

The straight line MN = 180 = 2(x+30^o) + 86^o =>
180 = 2x + 60^o + 86^o => 180^o = 2x + 146^o => 2x = 180 - 46 = 34^o => x = 34/2 = 17^o

Since ∠QRS = 180^o - 30^o = 150^o and ∠SPQ = ∠QRS (theorem opp. angles in a parallelogram are equal) ∴ x = 150^o

(cos\theta = \frac{adj}{hyp}=\frac{300}{600}=\frac{1}{2}<br /> \theta = cos^{-1}(0.5000)=60^{\circ})
The bearing of P from (Q = \theta + 180 = 60 + 180 = 240^{\circ})

(A=\pi r^2 \Rightarrow 77 = \frac{22}{7} \times \frac{r^2}{1}\Rightarrow r^2 = \frac{77\times 7}{22} \Rightarrow r^2 = \frac{49}{2}<br /> \Rightarrow r = \sqrt{\frac{49}{2}}=\frac{7}{\sqrt{2}}. Since D = 2r ∴ D = 2 \times \frac{7}{\sqrt{2}} = \frac{14}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = 7\sqrt{2}cm)