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 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
\(\frac{av}{1 - v}\) = \(\sqrt{\frac{2v + T}{a + 2T}}\)
\(\frac{(av)^3}{(1 - v)^3}\) = \(\frac{2v + T}{a + 2T}\)
\(\frac{a^3v^3}{(1^3 - v)^3}\) = \(\frac{2v + T}{a + 2T}\)
= \(\frac{2v(1 - v)^3 - a^4 v^3}{2a^3v^ 3 - (1 - v)^3}\)
L = -6, k = -9
L = -2, k = 1
k = -1, L = -2
L = 0, k = 1
k = 0,L = 6
Correct answer is A
f(x) = Lx3 + 2kx2 + 24
f(-2) = -8L + 8k = -24
4L - 4k = 12
f(1):L + 2k = -24
L - 4k = 3
3k = -27
k = -9
L = -6
If 0.0000152 x 0.042 = A x 108, where 1 \(\leq\) A < 10, find A and B
A = 9, B = 6.38
A = 6.38, B = -9
A = -6.38, B = -9
A = -9, B = -6.38
Correct answer is B
0.0000152 x 0.042 = A x 108
1 \(\leq\) A < 10, it means values of A includes 1 - 9
0.0000152 = 1.52 x 10-5
0.00042 = 4.2 x 10-4
1.52 x 4.2 = 6.384
10-5 x 10-4
= 10-5-4
= 10-9
= 6.38 x 10-9
A = 6.38, B = -9
\(\frac{1 - L}{1 + L}\)
\(\frac{L^2 \sqrt{3}}{1 + L}\)
\(\frac{1 + L^3}{L^2}\)
\(\frac{L(L - 1)}{1 - L + 1 \sqrt{1 - L^2}}\)
Correct answer is D
Given Cos z = L, z is an acute angle
\(\frac{\text{cot z - cosec z}}{\text{sec z + tan z}}\) = cos z
= \(\frac{\text{cos z}}{\text{sin z}}\)
cosec z = \(\frac{1}{\text{sin z}}\)
cot z - cosec z = \(\frac{\text{cos z}}{\text{sin z}}\) - \(\frac{1}{\text{sin z}}\)
cot z - cosec z = \(\frac{L - 1}{\text{sin z}}\)
sec z = \(\frac{1}{\text{cos z}}\)
tan z = \(\frac{\text{sin z}}{\text{cos z}}\)
sec z = \(\frac{1}{\text{cos z}}\) + \(\frac{\text{sin z}}{\text{cos z}}\)
= \(\frac{1}{l}\) + \(\frac{\text{sin z}}{L}\)
the original eqn. becomes
\(\frac{\text{cot z - cosec z}}{\text{sec z + tan z}}\) = \(\frac{L - \frac{1}{\text{sin z}}}{1 + sin \frac{z}{L}}\)
= \(\frac{L(L - 1)}{\text{sin z}(1 + \text{sin z})}\)
= \(\frac{L(L - 1)}{\text{sin z} + 1 - cos^2 z}\)
= sin z + 1
= 1 + \(\sqrt{1 - L^2}\)
= \(\frac{L(L - 1)}{1 - L + 1 \sqrt{1 - L^2}}\)