WAEC Mathematics Past Questions & Answers

Find the gradient of the line passing through the points $(\frac{1}{2}, \frac{- 1}{3}) and ( 3 , \frac{2}{3})$

$\frac{2}{5}$

$\frac{5}{2}$

$\frac{2}{7}$

$\frac{7}{2}$

View answer & explanation

Correct answer is A

Gradient(slope) m = $\frac{ y_2 - y_1}{ x_2 - x_1}$

the points are $(\frac{1}{2}, \frac{- 1}{3}) and ( 3 , \frac{2}{3})$

m = $\frac{\frac{2}{3} - (\frac{-1}{3})}{3 - \frac{1}{2}}$

= $\frac{\frac{2}{3} + \frac{1}{3}}{3 - \frac{1}{2}}$

= $1 \div\frac{5}{2}$ = $1\times\frac{2}{5}$

Therefore, m = $\frac{2}{5}$

Find the value of m in the diagram above.

$40^0$

$50^0$

$130^0$

$140^0$

View answer & explanation

Correct answer is C

∠EHI = ∠DEH = 40° (alternate angles are equal)
∠BEH = 90° (given)
∠BED = 90° - 40° = 50°
∠BEF = 180° - 50° = 130° (sum of angles on a straight line is 180o)
∠BEF = ∠ABE = 130° (alternate angles are equal)
∴ m = 130°

In the diagram above, O is the centre of a circle NST. |NT| = |ST| and ∠NTS = 36°. Find the measure of the angle marked t.

$72^0$

$54^0$

$36^0$

$108^0$

View answer & explanation

Correct answer is C

t = ∠NTS (the angle between a tangent and a chord is equal to the angle in the alternate segment).

Therefore, t = 36°

The length of the diagonal of a square is 12 cm. Calculate the area of the square.

$36 cm^2$

$48 cm^2$

$72 cm^2$

$18 cm^2$

View answer & explanation

Correct answer is C

Let each side be l, then area = $l^2$

Using Pythagoras theorem

$l^2 + l^2 = 12^2$

$2l^2 = 144$

divide both sides by 2

$l^2 = 72$

Therefore, the Area of the square is $72 cm^2$

10.

Find the quadratic equation whose roots are $\frac{2}{3} and \frac{- 3}{4}$

$12y^2 - y - 6 = 0$

$12y^2 - y + 6 = 0$

$12y^2 + y - 6 = 0$

$y^2 + y - 6 = 0$

View answer & explanation

Correct answer is C

Let p = $\frac{2}{3}$ and q = $\frac{- 3}{4}$

using (y - p)(y - q) = 0

= ( y - $\frac{2}{3})$ ( y - ( $\frac{- 3}{4})) = 0$

= ( $y - \frac{2}{3})( y + \frac{3}{4})$ = 0

$y^2 + \frac{3}{4}y - \frac{2}{3}y - \frac{6}{12} = 0$

$y^2 + \frac{1}{12}y - \frac{1}{2}$ = 0

= multiply through by the l. c. m of 3 and 4 = 12

∴ the quadratic equation is $12y^2 + y - 6 = 0$

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