WAEC Mathematics Past Questions & Answers - Page 2

6.

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

A.

\(\frac{2}{5}\)

B.

\(\frac{5}{2}\)

C.

\(\frac{2}{7}\)

D.

\(\frac{7}{2}\)

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 = \(\frac{2}{5}\)

7.

Find the value of m in the diagram above.

A.

\(40^0\)

B.

\(50^0\)

C.

\(130^0\)

D.

\(140^0\)

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°

8.

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.

A.

\(72^0\)

B.

\(54^0\)

C.

\(36^0\)

D.

\(108^0\)

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°

9.

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

A.

\(36 cm^2\)

B.

\(48 cm^2\)

C.

\(72 cm^2\)

D.

\(18 cm^2\)

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 \(\frac{2}{3} and \frac{- 3}{4}\)

A.

\(12y^2 - y - 6 = 0\)

B.

\(12y^2 - y + 6 = 0\)

C.

\(12y^2 + y - 6 = 0\)

D.

\(y^2 + y - 6 = 0\)

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  \(12y^2 + y - 6 = 0\)