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.
If \(\sin x = \frac{4}{5}\), find \(\frac{1 + \cot^2 x}{\csc^2 x - 1}\).
\(\frac{13}{2}\)
\(\frac{25}{9}\)
\(\frac{3}{13}\)
\(\frac{4}{11}\)
Correct answer is B
\(\sin x = \frac{opp}{Hyp} = \frac{4}{5}\)
5\(^2\) = 4\(^2\) + adj\(^2\)
adj\(^2\) = 25 - 16 = 9
adj = \(\sqrt{9}\) = 3
\(\tan x = \frac{4}{3}\)
\(\cot x = \frac{1}{\frac{4}{3}} = \frac{3}{4}\)
\(\cot^2 x = (\frac{3}{4})^2 = \frac{9}{16}\)
\(\csc x = \frac{1}{\sin x}\)
= \(\frac{1}{\frac{4}{5}} = \frac{5}{4}\)
\(\csc^2 x = (\frac{5}{4})^2 = \frac{25}{16}\)
\(\therefore \frac{1 + \cot^2 x}{\csc^2 x - 1} = \frac{1 + \frac{9}{16}}{\frac{25}{16} - 1}\)
= \(\frac{25}{16} \div \frac{9}{16}\)
= \(\frac{25}{9}\)
394 cm\(^3\)
425 cm\(^3\)
268 cm\(^3\)
540 cm\(^3\)
Correct answer is D
|AC| = |DF| = 13 cm
Using Pythagoras theorem,
|AC|\(^2\) = |AB|\(^2\) + |BC|\(^2\)
13\(^2\) = 12\(^2\) + |BC|\(^2\)
|BC|\(^2\) = 169 - 144 = 25
|BC| = \(\sqrt{25}\)
= 5 cm
Volume of triangular prism = \(\frac{1}{2} \times base \times length \times height\)
= \(\frac{1}{2} \times 5 \times 12 \times 18\)
= 540 cm\(^3\)
In the circle above, with centre O and radius 7 cm. Find the length of the arc AB, when < AOB = 57°
5.32 cm
4.39 cm
7.33 cm
6.97 cm
Correct answer is D
Length of arc = \(\frac{\theta}{360°} \times 2 \pi r\)
= \(\frac{57}{360} \times 2 \times \frac{22}{7} \times 7\)
= 6.97 cm
100
79
150
90
Correct answer is C
Number of students that scored above 40 = 55 + 45 + 30 + 15 + 5 = 150 students.
| Marks | 1 | 2 | 3 | 4 | 5 |
| Frequency | 2y - 2 | y - 1 | 3y - 4 | 3 - y | 6 - 2y |
The table above is the distribution of data with mean equals to 3. Find the value of y.
5
2
3
6
Correct answer is B
| Marks (x) | 1 | 2 | 3 | 4 | 5 | |
| Frequency (f) | 2y - 2 | y - 1 | 3y - 4 | 3 - y | 6 - 2y | 3y + 2 |
| fx | 2y - 2 | 2y - 2 | 9y - 12 | 12 - 4y | 30 - 10y | 26 - y |
Mean = \(\frac{\sum fx}{\sum f}\)
\(3 = \frac{26 - y}{3y + 2}\)
\(3(3y + 2) = 26 - y\)
\(9y + 6 = 26 - y\)
\(9y + y = 26 - 6\)
\(10y = 20 \implies y = 2\)