Further Mathematics Questions and Answers
Test your knowledge of advanced level mathematics with this aptitude test. This test comprises Further Maths questions and answers from past JAMB and WAEC examinations.
Test your knowledge of advanced level mathematics with this aptitude test. This test comprises Further Maths questions and answers from past JAMB and WAEC examinations.
0.0082N
0.0084N
0.082N
0.084N
Correct answer is B
\(F = mass \times acceleration\) but \(accl = \frac{v - u}{t}\)
\(\therefore F = m(\frac{v - u}{t})\)
\(Mass = 28g = 0.028kg\)
\(v = 5.4 ms^{-1}; u = 0; t = 18secs\)
\(\therefore F = 0.028(\frac{5.4 - 0}{18}) = 0.028 \times 0.3 = 0.0084N\)
\(\begin{pmatrix} 9 \\ -5 \end{pmatrix}\)
\(\begin{pmatrix} -23 \\ -5 \end{pmatrix}\)
\(\begin{pmatrix} 9 \\ 17 \end{pmatrix}\)
\(\begin{pmatrix} -23 \\ 17 \end{pmatrix}\)
Correct answer is D
\(\overrightarrow{OY} \equiv -\overrightarrow{YO}\)
Also, \(\overrightarrow{YO} + \overrightarrow{OX} = \overrightarrow{YX}\)
\(\therefore \overrightarrow{YO} = -\overrightarrow{OY} = - \begin{pmatrix} 16 \\ -11 \end{pmatrix} = \begin{pmatrix} -16 \\ 11 \end{pmatrix}\)
\(\overrightarrow{YX} = \begin{pmatrix} -16 \\ 11 \end{pmatrix} + \begin{pmatrix} -7 \\ 6 \end{pmatrix}\)
= \(\begin{pmatrix} -23 \\ 17 \end{pmatrix}\)
\(\frac{1}{12}\)
\(\frac{1}{3}\)
\(\frac{1}{2}\)
\(\frac{2}{3}\)
Correct answer is D
\(\text{p(a head and a six)} = \text{p(a head)} + \text{p(a six)}\)
= \(\frac{1}{2} + \frac{1}{6} = \frac{2}{3}\).
Hint: Probability of A and B occurring should be greater than probability A or B happening.
Given that \(a = i - 3j\) and \(b = -2i + 5j\) and \(c = 3i - j\), calculate \(|a - b + c|\).
\(\sqrt{13}\)
\(3\sqrt{13}\)
\(6\sqrt{13}\)
\(9\sqrt{13}\)
Correct answer is B
Given \(a = i - 3j; b = -2i + 5j; c = 3i - j\)
\(a- b + c = (1 - (-2) + 3)i + (-3 - 5 + (-1))j = 6i - 9j\)
\(|a - b + c| = \sqrt{6^{2} + (-9)^{2}} = \sqrt{36 + 81} = \sqrt{117}\)
\(= \sqrt{9 \times 13} = 3\sqrt{13}\)
The marks scored by 4 students in Mathematics and Physics are ranked as shown in the table below
| Mathematics | 3 | 4 | 2 | 1 |
| Physics | 4 | 3 | 1 | 2 |
Calculate the Spearmann's rank correlation coefficient.
0.2
0.5
0.6
0.7
Correct answer is C
| Maths (x) | Rank \(r_{x}\) | Physics (y) | Rank \(r_{y}\) | \(d = |r_{x} - r_{y}|\) | \(d^{2}\) |
| 3 | 2 | 4 | 1 | 1 | 1 |
| 4 | 1 | 3 | 2 | 1 | 1 |
| 2 | 3 | 1 | 4 | 1 | 1 |
| 1 | 4 | 2 | 3 | 1 | 1 |
| Total | 4 |
\(\rho = 1 - \frac{6\sum{d^{2}}}{n(n^{2} - 1)}\)
\( 1 - \frac{6 \times 4}{4(4^{2} - 1)} = 1 - \frac{24}{60}\)
= \(1 - 0.4 = 0.6\)