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.
If angle \(\theta\) is 135°, evaluate cos\(\theta\)
\(\frac{1}{2}\)
\(\frac{\sqrt{2}}{2}\)
\(-\frac{\sqrt{2}}{2}\)
\(-\frac{1}{2}\)
Correct answer is C
\(\theta\) = 135°
Cos 135° = Cos(90 + 45)°
= cos90° cos45° - sin90° sin45°
= 0cos45° - (1 x \(\frac{\sqrt{2}}{2}\))
= \(-\frac{\sqrt{2}}{2}\)
Find the equation of the line through the points (-2, 1) and (-\(\frac{1}{2}\), 4)
y = 2x - 3
y = 2x + 5
y = 3x - 2
y = 2x + 1
Correct answer is B
\(\frac{y - y_1}{x - x_1}\) = \(\frac{y_2 - y_1}{x_2 - x_1}\)
\(\frac{y - 1}{x - -2}\) = \(\frac{4 - 1}{-\frac{1}{2} + 2}\)
= \(\frac{y - 1}{x + 2}\) = \(\frac{3}{\frac{3}{2}}\)
y = 2x + 5
The distance between the point (4, 3) and the intersection of y = 2x + 4 and y = 7 - x is
\(\sqrt{13}\)
\(3\sqrt{2}\)
\(\sqrt{26}\)
\(10\sqrt{5}\)
Correct answer is B
P1 (4, 3), P2 (x, y)
y = 2x + 4 .....(1)
y = 7 - x .....(2)
Substitute (2) in (1)
7 - x = 2x + 4
7 - 4 = 2x + x
3 = 3x
x = 1
Substitute in eqn (2)
y = 7 - x
y = 7 - 1
y = 6
P2 (1, 6)
Distance between 2 points is given as
D = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
D = \(\sqrt{(1 - 4)^2 + (6 - 3)^2}\)
D = \(\sqrt{(-3)^2 + (3)^2}\)
D = \(\sqrt{9 + 9}\)
D = \(\sqrt{18}\)
D = \(\sqrt{9 \times 2}\)
D = \(3\sqrt{2}\)
The gradient of the straight line joining the points P(5, -7) and Q(-2, -3) is
\(\frac{1}{2}\)
\(\frac{2}{5}\)
\(-\frac{4}{7}\)
\(-\frac{2}{3}\)
Correct answer is C
PQ = \(\frac{y_1 - y_0}{x_1 - x_0}\) = \(\frac{-3 - (-7)}{-2 - 5}\) = \(\frac{-3 + 7}{-2 - 5}\) = \(\frac{4}{-7}\)
Calculate the volume of a cuboid of length 0.76cm, breadth 2.6cm and height 0.82cm.
3.92cm3
2.13cm3
1.97cm3
1.62cm3
Correct answer is D
Volume of cuboid = L x b x h
= 0.76cm x 2.6cm x 0.82cm
= 1.62cm3