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.
The lengths of the sides of a right angled triangle are (3x + 1)cm, (3x - 1)cm and xcm. Find x
2
6
18
12
5
Correct answer is D
\((3x + 1)^{2} = (3x - 1)^{2} + x^{2}\) (Pythagoras's theorem)
\(9x^{2} + 6x + 1 = 9x^{2} - 6x + 1 + x^{2}\)
x2 - 12x = 0
x(x - 12) = 0
x = 0 or 12
The sides cannot be 0 hence x = 12.
Simplify \(\frac{x - 7}{x^2 - 9}\) x \(\frac{x^2 - 3x}{x^2 - 49}\)
\(\frac{x}{(x - 3)(x - 7)}\)
\(\frac{x}{(x + 3)(x - 7)}\)
\(\frac{x}{(x + 3)(x + 7)}\)
\(\frac{x}{(x - 3)(x + 7)}\)
Correct answer is C
\(\frac{x - 7}{x^2 - 9}\) x \(\frac{x^2 - 3x}{x^2 - 49}\)
= \(\frac{x - 7}{(x - 3)(x + 3)}\) x \(\frac{x(x - 3)}{(x - 7)(x + 7)}\)
= \(\frac{x}{(x + 3)(x + 7)}\)
y = \(\frac{10x^2}{31} + \frac{52}{31\sqrt{x}}\)
y = x2 + \(\frac{1}{\sqrt{x}}\)
y = x2 + \(\frac{1}{x}\)
y = \(\frac{x^2}{31} + \frac{1}{31\sqrt{x}}\)
Correct answer is A
y = kx2 + \(\frac{c}{\sqrt{x}}\)
y = 2when x = 1
2 = k + \(\frac{c}{1}\)
k + c = 2
y = 6 when x = 4
6 = 16k + \(\frac{c}{2}\)
12 = 32k + c
k + c = 2
32k + c = 12
= 31k + 10
k = \(\frac{10}{31}\)
c = 2 - \(\frac{10}{31}\)
= \(\frac{62 - 10}{31}\)
= \(\frac{52}{31}\)
y = \(\frac{10x^2}{31} + \frac{52}{31\sqrt{x}}\)
The value of (0.03)3 - (0.02)3 is
0.019
0.0019
0.00019
0.000019
0.000035
Correct answer is D
Using the method of difference of two cubes,
\(a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})\)
\((0.03)^{3} - (0.02)^{3} = (0.03 - 0.02)((0.03)^{2} + (0.03)(0.02) + (0.02)^{2}\)
= \((0.01)(0.0009 + 0.0006 + 0.0004)\)
= \(0.01 \times 0.0019\)
= \(0.000019\)
Make T the subject of the equation \(\frac{av}{1 - v}\) = \(\sqrt{\frac{2v + T}{a + 2T}}\)
T = \(\frac{3av}{1 - v}\)
T = \(\frac{1 + v}{2a^2v^3}\)
T = \(\frac{2v(1 - v)^3 - a^4v^3}{2a^3v^3 + (1 - v)^2}\)
\(\frac{2v(1 - v)^3 - a^4 v^3}{2a^3v^ 3 - (1 - v)^3}\)
Correct answer is D
\(\frac{av}{1 - v}\) = \(\sqrt{\frac{2v + T}{a + 2T}}\)
\(\frac{(av)^3}{(1 - v)^3}\) = \(\frac{2v + T}{a + 2T}\)
\(\frac{a^3v^3}{(1^3 - v)^3}\) = \(\frac{2v + T}{a + 2T}\)
= \(\frac{2v(1 - v)^3 - a^4 v^3}{2a^3v^ 3 - (1 - v)^3}\)