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.

3,921.

P(-6, 1) and Q(6, 6) are the two ends of the diameter of a given circle. Calculate the radius.

A.

6.5 units

B.

13.0 units

C.

3.5 units

D.

7.0 units

Correct answer is A

PQ\(^2\) = (x2 - x1)\(^2\)  + (y2 - y1)\(^2\) 

= 12\(^2\)  + 5\(^2\) 
= 144 + 25
= 169

PQ = √169 = 13

But PQ = diameter = 2r, r = PQ/2 = 6.5 units

 

3,922.

Find the number of sides of a regular polygon whose interior angle is twice the exterior angle.

A.

6

B.

2

C.

3

D.

8

Correct answer is A

Let the ext. angle = x
Thus int. angle = 2x
But sum of int + ext = 180 (angle of a straight line).
2x + x = 180
3x = 180
x = 180/3 = 60

Each ext angle = 360/n
=> 60 = 360/n
n = 360/60 = 6

3,923.

Find the value of P if the line joining (P, 4) and (6, -2) is perpendicular to the line joining (2, P) and (-1, 3).

A.

4

B.

6

C.

3

D.

0

Correct answer is A

The line joining (P, 4) and (6, -2).

Gradient: \(\frac{-2 - 4}{6 - P} = \frac{-6}{6 - P}\)

The line joining (2, P) and (-1, 3)

Gradient: \(\frac{3 - P}{-1 - 2} = \frac{3 - P}{-3}\)

For perpendicular lines, the product of their gradient = -1.

\((\frac{-6}{6 - P})(\frac{3 - P}{-3}) = -1\)

\(\frac{6 - 2P}{6 - P} = -1 \implies 6 - 2P = P - 6\)

\(6 + 6 = P + 2P \implies P = \frac{12}{3} = 4\)

 

3,924.

Given an isosceles triangle with length of 2 equal sides t units and opposite side √3t units with angle θ. Find the value of the angle θ opposite to the √3t units.

A.

100°

B.

120°

C.

30°

D.

60°

Correct answer is B

Cos θ° = t2 + t2 -(√3t)2 2 x t x t

= 2t2 - 3t2 2t2
= -t2/2t2
= -1/2

Thus θ = cos-1 (-0.5) = 120°

3,925.

A straight line makes an angle of 30° with the positive x-axis and cuts the y-axis at y = 5. Find the equation of the straight line.

A.

y = (x/10) + 5

B.

y = x + 5

C.

√3y = - x + 5√3

D.

√3y = x + 5√3

Correct answer is D

Cos 30 = 5/x
x cos 30 = 5, => x = 5√3

Coordinates of P = -5, 3, 0
Coordinates of Q = 0, 5
Gradient of PQ = (y2 - y1) (X2 - X1) = (5 - 0)/(0 -5√3)
= 5/5√3 = 1/√3

Equation of PQ = y - y1 = m (x -x1)
y - 0 = 1/√3 (x -(-5√3))

Thus: √3y = x + 5√3