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.
| Age in years | 10 - 14 | 15 - 19 | 20 - 24 | 25 - 29 | 30 - 34 |
| Frequency | 6 | 8 | 14 | 10 | 12 |
Find the mean of the distribution.
23.4
23.6
24.3
24.6
Correct answer is B
| Age in years |
Classmark (x) |
Frequency (f) |
fx |
| 10 - 14 | 12 | 6 | 72 |
| 15 - 19 | 17 | 8 | 136 |
| 20 - 24 | 22 | 14 | 308 |
| 25 - 29 | 27 | 10 | 270 |
| 30 - 34 | 32 | 12 | 384 |
| Total | 50 | 1170 |
\(Mean = \frac{\sum fx}{\sum f}\)
= \(\frac{1170}{50} = 23.4\)
| Age in years | 10 - 14 | 15 - 19 | 20 - 24 | 25 - 29 | 30 - 34 |
| Frequency | 6 | 8 | 14 | 10 | 12 |
In which group is the upper quartile?
15 - 19
20 - 24
25 - 29
30 - 34
Correct answer is C
The upper quartile occurs at the \(\frac{3N}{4}\) position.
= \(\frac{3 \times 50}{4} = 37.5\)
This occurs in the class 25 - 29.
| Age in years | 10 - 14 | 15 - 19 | 20 - 24 | 25 - 29 | 30 - 34 |
| Frequency | 6 | 8 | 14 | 10 | 12 |
What is the class mark of the median class?
17
22
27
32
Correct answer is B
\(\text{Total frequency} = 50\)
Median occurs at \(\frac{50}{2} = \text{25th position}\)
= Class 20 - 24 with classmark \(\frac{20 + 24}{2} = 22\)
\(8\pi cm^{2}s^{-1}\)
\(16\pi cm^{2}s^{-1}\)
\(24\pi cm^{2}s^{-1}\)
\(48\pi cm^{2}s^{-1}\)
Correct answer is D
Surface area of sphere, \( A = 4\pi r^{2}\)
\(\frac{\mathrm d A}{\mathrm d r} = 8\pi r\)
The rate of change of radius with time \(\frac{\mathrm d r}{\mathrm d t} = 3cm s^{-1}\)
\(\frac{\mathrm d A}{\mathrm d t} = (\frac{\mathrm d A}{\mathrm d r})(\frac{\mathrm d r}{\mathrm d t})\)
= \(8\pi \times 2cm \times 3cm s^{-1} = 48\pi cm^{2}s^{-1}\)
\(6\)
\(\frac{11}{6}\)
\(\frac{11}{4}\)
\(\frac{5}{3}\)
Correct answer is B
\(h : x \to 2 - \frac{1}{2x - 3}\)
\(h(x) = \frac{2(2x - 3) - 1}{2x - 3} = \frac{4x - 7}{2x - 3}\)
Let x = h(y)
\(x = \frac{4y - 7}{2y - 3}\)
\(x(2y - 3) = 4y - 7 \implies 2xy - 4y = 3x - 7\)
\(y = \frac{3x - 7}{2x - 4}\)
\(h^{-1}(x) = \frac{3x - 7}{2x - 4}\)
\(\therefore h^{-1}(\frac{1}{2}) = \frac{3(\frac{1}{2}) - 7}{2(\frac{1}{2}) - 4}\)
= \(\frac{\frac{-11}{2}}{-3} = \frac{11}{6}\)