Mathematics questions and answers

Mathematics Questions and Answers

How good are you with figures and formulas? Find out with these Mathematics past questions and answers. This Test is useful for both job aptitude test candidates and students preparing for JAMB, WAEC, NECO or Post UTME.

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  \(12y^2 + y - 6 = 0\)