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.
A right circular cone has a base radius r cm and a vertical angle 2yo. The height of the cone is
r tan yo cm
r sin yo cm
r cot yo cm
r cos yo cm
r cosec yo cm
Correct answer is C
\(\frac{r}{h}\) = tan yo
h = \(\frac{r}{tan y^o}\)
= r cot yo
30
32
55
62
92
Correct answer is E
If y = cf(x)
cf(5) = 30 + 32 + 30
= 92
If pq + 1 = q2 and t = \(\frac{1}{p}\) - \(\frac{1}{pq}\) express t in terms of q
\(\frac{1}{p - q}\)
\(\frac{1}{q - 1}\)
\(\frac{1}{q + 1}\)
1 + 0
\(\frac{1}{1 - q}\)
Correct answer is C
Pq + 1 = q2......(i)
t = \(\frac{1}{p}\) - \(\frac{1}{pq}\).........(ii)
p = \(\frac{q^2 - 1}{q}\)
Sub for p in equation (ii)
t = \(\frac{1}{q^2 - \frac{1}{q}}\) - \(\frac{1}{\frac{q^2 - 1}{q} \times q}\)
t = \(\frac{q}{q^2 - 1}\) - \(\frac{1}{q^2 - 1}\)
t = \(\frac{q - 1}{q^2 - 1}\)
= \(\frac{q - 1}{(q + 1)(q - 1)}\)
= \(\frac{1}{q + 1}\)
If pq + 1 = q2 and t = \(\frac{1}{p}\) - \(\frac{1}{pq}\) express t in terms of q
\(\frac{1}{p - q}\)
\(\frac{1}{q - 1}\)
\(\frac{1}{q + 1}\)
1 + 0
\(\frac{1}{1 - q}\)
Correct answer is C
Pq + 1 = q2......(i)
t = \(\frac{1}{p}\) - \(\frac{1}{pq}\).........(ii)
p = \(\frac{q^2 - 1}{q}\)
Sub for p in equation (ii)
t = \(\frac{1}{q^2 - \frac{1}{q}}\) - \(\frac{1}{\frac{q^2 - 1}{q} \times q}\)
t = \(\frac{q}{q^2 - 1}\) - \(\frac{1}{q^2 - 1}\)
t = \(\frac{q - 1}{q^2 - 1}\)
= \(\frac{q - 1}{(q + 1)(q - 1)}\)
= \(\frac{1}{q + 1}\)
Simplify (1 + \(\frac{\frac{x - 1}{1}}{\frac{1}{x + 1}}\))(x + 2)
(x2 - 1)(x + 2)
x2(x + 2)
2 + 34
\(\frac{3x^2 - 1}{(x - 1)}\)
Correct answer is B
\((1 + \frac{x - 1}{\frac{1}{x + 1}})(x + 2)\)
\((x - 1) \div \frac{1}{x + 1} = (x - 1)(x + 1) = x^{2} - 1\)
\((1 + x^{2} - 1)(x + 2) = x^{2}(x + 2)\)