Processing math: 46%

WAEC Mathematics Past Questions & Answers - Page 201

1,001.

Given that $p = x-\frac{1}{x} and\hspace{1mm}q = x^2 + \frac{1}{x^2}$ express q in terms of p.

(p² + 2)

(p - 2) ²

(p + 2) ²

(p² - 2)

View answer & explanation

Correct answer is A

Given $p = x - \frac{1}{x}$ ; $q = x^2 + \frac{1}{x^2}$ .

$p^2 = (x - \frac{1}{x})(x - \frac{1}{x})$

$p^2 = x^2 + \frac{1}{x^2} - 2$

$p^2 = q - 2 \implies q = p^2 + 2$

1,002.

In the diagram O is the center of the circle, ∠SOR = 64° and ∠PSO = 36°. Calculate ∠PQR

100^o

96^o

94^o

86^o

View answer & explanation

Correct answer is D

< OSR = < ORS = $\frac{180° - 64°}{2}$ = 58°

< PSR = 36° + 58° = 94°

< PSR + < PQR = 180°

94° + < PQR = 180° $\implies$ < PQR = 180° - 94° = 86°

1,003.

In the diagram, $QR||TP and W\hat{P}T = 88^{\circ}$ . Find the value of x

92^o

68^o

67^o

23^o

View answer & explanation

Correct answer is C

Sum of the angles in a triangle = 180°

3x - 180° + 92° + x = 180°

4x - 88° = 180°

4x = 268°

x = 67°

1,004.

To arrive on schedule, a train is to cover a distance of 60km at 72km/hr. If it starts 10 minutes late, at what speed must it move to arrive on schedule?

60km/hr

80km/hr

90km/hr

108km/hr

View answer & explanation

Correct answer is C

$speed = \frac{distance}{time}\\ 72 = \frac{60}{time}\\ t = \frac{60}{72} = \frac{5}{6}hr$
time lost = 10mis $= \frac{10}{60}hr = \frac{1}{6}$
Time required for the journey
$=\frac{5}{6}-\frac{1}{6} = \frac{2}{3}\\ speed \hspace{1mm}=60 \div \frac{2}{3} = 90km/hr$