2005 - WAEC Mathematics Past Questions and Answers - page 4

Since 36km in 21 minutes
1km in (\frac{21}{36})minutes
60km in (\left(\frac{21}{36}\times 60\right))minutes = (\frac{70}{2})min
= 35 mins

(%Increase = \frac{Actual \hspace{1mm}increase}{initial \hspace{1mm}ratio}\times 100%)
Actual increase = 7; Initial ratio = 40
(=\left(\frac{7}{40} \times 100%\right)=\frac{35}{2}=17.5%)

Exterior angle (=\frac{360}{n}) where n = no of sides
(=\frac{360}{9}=40^{\circ})

(\frac{2}{3xy} - \frac{3}{yz} = \frac{2(4)-3(3x)}{12xyz} = \frac{8z-9x}{12xyz})

T.S.A of the wall (=7\time 5 + 7 \times x + 5 \times x +7 \times 5 = 96<br /> 70 + 7x + 5x = 96 \Rightarrow 12x = 96-70=26<br /> x = \frac{26}{12}=2)

The expression is not defines when y² - y - 6 = 0
=> y² - 3y + 2y - 6 = 0 => (y-3)(y+2) = 0 => y
-3 = 0 or y + 2 = 0 => y = 3 or -2

The two points of contact of the curve with the x-axis are x = 1 and x = 4. The roots of the equation are x = 1 and x = 4

The factor for the curve are (x-1)(x-4). The product off the curves gives (=x^2-5x+4=y)

∠QPT = ∠QTP = 70^o. Since ∠QTR = 140^o
∠PRS = TSR = 1/2(180-40) = 70^o

(tan 60^{\circ} = \frac{15}{x} \Rightarrow x = \frac{15}{tan60^{\circ}<br /> x = \left(\frac{15}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\right)m \Rightarrow 5\sqrt{3}m)