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.
(1,1) only
(1,1) and (1,2)
(0,0) and (1,1)
(√2,√2) only
Correct answer is A
x2 + y2 = (√2)2
x2 + y2 = 2
but y = x
Thus; x2 + x2 = 2
2x2 = 2
x2 = + or - 1
But x2 + y2 = 2
12 + y2 = 2
1 + y2 = 2
y2 = 2 - 1
y2 = 1
y = + or - 1
Thus point (x,y) = (1,1) only.
8
6
4
3
Correct answer is B
2x + x = 180°, => 3x = 180°, and thus x = 60°
Each exterior angle = 60° but size of ext. angle = 360°/n
Therefore 60° = 360°/n
n = 360°/60° = 6 sides
if P and Q are fixed points and X is a point which moves so that XP = XQ, the locus of X is
A straight line
a circle
the bisector of angle PXQ
the perpendicular bisector of PQ
Correct answer is D
No explanation has been provided for this answer.
-4/3
-3/4
3/4
4/3
Correct answer is A
Grad of 3y = 4x - 1
y = 4x/3 - 1/3
Grad = 4/3
Grad of Ky = x + 3
y = x/k + 3/4
Grad = 1/k
Since the two lines are perpendicular,
1/k = -3/4
-3k = 4
k = -4/3
An equilateral triangle of side √3cm is inscribed in a circle. Find the radius of the circle.
2/3 cm
2 cm
1 cm
3 cm
Correct answer is C
Since the inscribed triangle is equilateral, therefore the angles at all the points = 60°
Using the formula for inscribed circle,
2R = \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\)
where R = radius of the circle; a, b and c are the sides of the triangle.
⇒ 2R = \(\frac{\sqrt{3}}{\sin 60}\)
2R = \(\frac{\sqrt{3}}{\frac{\sqrt{3}}{2}}\)
2R = 2
R = 1cm