WAEC Mathematics Past Questions & Answers - Page 5

Total number of students = total frequency = 3 + 6 + 9 + 1 + 12 + 7 + 5 + 10 = 53
Since the failed mark was 4,
Total number of students that passed = 12 + 7 + 5 + 10 = 34

Pr( passed) = $\frac{34}{53}$ = 0. 64

$8 + 2x - x^2 = 0$
$8 + 4x - 2x - x^2 = 0$
4(2 + x) - x(2 + x) = 0
(4 - x)(2 + x) = 0
4 - x = 0 or 2 + x = 0
x = 4 or x =- 2
So, p = 4 and q = -2
∴ p + q = 2

1st bag → 4 W, 3 B = 7 marbles
2nd bag → 5 R, 6 B = 11 marbles
The only possible way of picking marbles of the same colour is to pick a blue marble from each bag

therefore, Pr( same colour) = $\frac{3}{7}\times \frac{6}{11} = \frac{18}{77}$
 

(7, 4) and (-3, 9) = ($y_1 , x_1)(y_2 , x_2)$

slope$m_1$ of the points(7, 4) and (-3, 9) = $\frac{ y_2 - y_1}{ x_2 - x_1}$

 $m_1 = \frac{ 9 - 4}{ -3 - 7} = \frac{5}{-10} = \frac{-1}{2}$

$m_1 = m_2$ ( condition for parallelism)
Since the line L passes through the point (6, -13) and is parallel to the points (7, 4) and (-3, 9)

$\frac{-1}{2} = \frac{ y - (- 13)}{x - 6}$

= $\frac{-1}{2} = \frac{ y + 13}{x - 6}$

= $\frac{-1}{2}( x - 6) = y + 13$

$\frac{-1}{2}x + 3 = y + 13$

= $\frac{-1}{2}x + 3 - 13 = y$

∴ y = $\frac{-1}{2}x - 10$

Consider |NS|
θ = 180° - (90° + 35°) (sum of angles on a straight line is 180°)
= 180° - 90° - 35°
= 90° - 35°
= 55°
∴ His new bearing is N55°E.