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.
A force (10i + 4j)N acts on a body of mass 2kg which is at rest. Find the velocity after 3 seconds.
\((\frac{5i}{3} + \frac{2j}{3})ms^{-1}\)
\((\frac{10i}{3} + \frac{4j}{3})ms^{-1}\)
\((5i + 2j)ms^{-1}\)
\((15i + 6j)ms^{-1}\)
Correct answer is D
Recall, \(F = mass \times acceleration \implies acceleration = \frac{force}{mass}\)
= \(\frac{10i + 4j}{2} = (5i + 2j) ms^{-2}\)
= \(v = u + at \implies v \text{at 3 seconds} = 0 + (5i + 2j \times 3)\)
= \((15i + 6j) ms^{-1}\)
Given that a = 5i + 4j and b = 3i + 7j, evaluate (3a - 8b).
9i + 44j
-9i + 44j
-9i - 44j
9i - 44j
Correct answer is C
= \(3(5i+4j) - 8(3i+7j) = 15i + 12j - 24i -56j\)
= \(-9i - 44j\)
Face | 1 | 2 | 3 | 4 | 5 | 6 |
Frequency | 12 | 18 | y | 30 | 2y | 45 |
Given the table above as the result of tossing a fair die 150 times, find the mode.
3
4
5
6
Correct answer is D
The mode is the occurrence with the highest frequency which, from the table, is 45 (the occurrence of obtaining a 6).
Face | 1 | 2 | 3 | 4 | 5 | 6 |
Frequency | 12 | 18 | y | 30 | 2y | 45 |
Given the table above as the results of tossing a fair die 150 times. Find the probability of obtaining a 5.
\(\frac{1}{10}\)
\(\frac{1}{6}\)
\(\frac{1}{5}\)
\(\frac{3}{10}\)
Correct answer is C
Probability of obtaining a 5 = \(\frac{\text{frequency of 5}}{\text{total frequency}}\)
\(12+18+y+30+2y+45 = 150 \implies 105+3y = 150\)
\(3y = 45; y = 15\)
Probability of 5 = \(\frac{2\times 15}{150} = \frac{1}{5}\)
\(\begin{pmatrix} \frac{17}{4} \\ 7 \end{pmatrix}\)
\(\begin{pmatrix} \frac{17}{4} \\ 5 \end{pmatrix}\)
\(\begin{pmatrix} \frac{17}{4} \\ 3 \end{pmatrix}\)
\(\begin{pmatrix} \frac{17}{4} \\ 2 \end{pmatrix}\)
Correct answer is B
\(a = \begin{pmatrix} 2 \\ 3 \end{pmatrix}\); \(b = \begin{pmatrix} -1 \\ 4 \end{pmatrix}\)
\(\implies 2 \times a = \begin{pmatrix} 4 \\ 6 \end{pmatrix}\) and \(\frac{1}{4} \times b = \begin{pmatrix} -\frac{1}{4} \\ 1 \end{pmatrix}\)
\(\therefore 2a - \frac{1}{4}b = \begin{pmatrix} 4 - \frac{-1}{4} \\ 6 - 1 \end{pmatrix}\)
= \(\begin{pmatrix} \frac{17}{4} \\ 5 \end{pmatrix}\)