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.

21.

The bar chart represents the distribution of marks scored by students in an economics examination. Use the bar chart to answer questions 30 to 32

If the failed mark was 4, what is the probability that a student selected at random passed?

A.

0.36

B.

0.74

C.

0.52

D.

0.64

Correct answer is D

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

22.

The truth set of \(8 + 2x - x^2\) = 0 is {p, q}. Evaluate p + q.

A.

4

B.

2

C.

-6

D.

-2

Correct answer is B

\(8 + 2x - x^2 = 0\)
\(8 + 4x - 2x - x^2 = 0\)
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

23.

A bag contains 4 white marbles and 3 red marbles. Another bag contains 5 red and 6 blue marbles. If a marble is picked at random from each bag, find the probability that they are of the same colour.

A.

\(\frac{9}{11}\)

B.

\(\frac{18}{77}\)

C.

\(\frac{1}{2}\)

D.

\(\frac{11}{12}\)

Correct answer is B

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) = \(\frac{3}{7}\times \frac{6}{11} = \frac{18}{77}\)
 

24.

A line L passing through the point (6, -13) is parallel to the line which passes through (7, 4) and (-3, 9). Find the equation of the line L.

A.

y = \(\frac{1}{2}x - 10\)

B.

y = \(\frac{-1}{2}x + 10\)

C.

y = \(\frac{-1}{2}x - 10\)

D.

y = \(\frac{1}{2}x +10\)

Correct answer is C

(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 = \(\frac{-1}{2}x - 10\)

25.

John was facing S35°E. If he turned 90° in the anticlockwise direction, find his new direction.

A.

S55°E.

B.

S35°W.

C.

N55°E.

D.

N35°W.

Correct answer is C

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.