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.
F x varies inversely as y and y varies directly as Z, what is the relationship between x and z?
x \(\alpha\) z
x \(\alpha \frac{1}{z}\)
x \(\alpha z^2\)
x \(\alpha \frac{1}{z^2}\0
Correct answer is B
x \(\alpha \frac{1}{y}\)
y \(\alpha z\)
the relationship = x \(\alph \frac{1}{z}\)
Make u the subject of formula, E = \(\frac{m}{2g}\)(v2 - u2)
u = \(\sqrt{v^2 - \frac{2Eg}{m}}\)
u = \(\sqrt{\frac{v^2}{m} - \frac{2Eg}{4}}\)
u = \(\sqrt{v- \frac{2Eg}{m}}\)
u = \(\sqrt{\frac{2v^2Eg}{m}}\)
Correct answer is A
E = \(\frac{m}{2g}\)(v2 - u2)
multiply both sides by 2g
2Eg = 2g (\(\frac{M}{2g} (V^2 - U^2)\)
2Eg = m(V2 - U2)
2Eg - mV2 - mU2
mU2 = mV2 - 2Eg
divide both sides by m
\(\frac{mU^2}{m} = \frac{mV^2 - 2Eg}{m}\)
U2 = \(\frac{mV^2 - 2Eg}{m}\)
= \(\frac{mV^2}{m} - \frac{2Eg}{m}\)
U2 = V2 - \(\frac{2Eg}{m}\)
U = \(\sqrt{V^2 - \frac{2Eg}{m}}\)
10\(sqrt{2}\)
4\(sqrt{5}\)
5\(sqrt{2}\)
2\(sqrt{5}\)
Correct answer is D
p(4, 3) Q(2 - 1)
distance = \(\sqrt{(x_2 - x_1)^2 + (Y_2 - y_1)^2}\)
= \(\sqrt{(2 - 4)^2 + (-1 - 3)^2}\)
= \(\sqrt{(-2)^2 = (-4)^2}\)
= \(\sqrt{4 + 16}\)
= \(\sqrt{20}\)
= \(\sqrt{4 \times 5}\)
= 2\(\sqrt{5}\)
Find the truth set of the equation x2 = 3(2x + 9)
{x : x = 3, x = 9}
{x : x = -3, x = -9}
{x : x = 3, x = -9}
{x : x = -3, x = 9}
Correct answer is D
x2 = 3(2x + 9)
x2 = 6x + 27
x2 - 6x - 27 = 0
x2 - 9x + 3x - 27 = 0
x(x - 9) + 3(x - 9) = (x + 3)(x - 9) = 0
x + 3 = 0 or x - 9 = 0
x = -3 or x = 9
x = -3, x = 9
{0, 2, 6}
{1, 3}
{0, 6)
{9}
Correct answer is C
x = {0, 2, 4, 6}; y = {1, 2, 3, 4}; z = {1, 3}
u = {0, 1, 2, 3, 4, 5, 6}
y' = {0, 5, 6}
to find x \(\cap\) (Y' \(\cup\) Z)
first find y' \(\cup\) z = {0, 1, 3, 5, 6}
then x \(\cap\) (Y' \(\cup\) Z) = {0, 6}