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 tan θ = 4/3, calculate sin\(^2\) θ - cos\(^2\) θ.
16/25
24/25
7/25
9/25
Correct answer is C
\(\tan \theta = \frac{opposite}{adjacent} = \frac{4}{3}\)
Hyp\(^2\) = 4\(^2\) + 3\(^2\)
Hyp = 5.
\(\sin \theta = \frac{4}{5}; \cos \theta = \frac{3}{5}\)
\(\sin^{2} \theta - \cos^{2} \theta = \frac{16}{25} - \frac{9}{25}\)
= \(\frac{7}{25}\)
4π cm2
32 π cm2
16 π cm2
8 π cm2
Correct answer is D
Diameter = 4\(\sqrt{3}\) cm<
radius = 2\(\sqrt{3}\) cm
Area of major sector = \(\frac{\theta}{360} \times \pi r^{2}\)
\(\theta = 360 - 120 = 240°\)
= \(\frac{240}{360} \times \pi \times 12\)
= \(8\pi cm^{2}\)
If x varies directly as √n and x = 9 when n = 9, find x when n = (17/9)
4
27
√3
√17
Correct answer is D
\(x \propto \sqrt{n}\)
\(x = k \sqrt{n}\)
\(9 = k \sqrt{9} \implies 9 = 3k\)
\(k = 3\)
\(x = 3 \sqrt{n}\)
When n = 17/9,
\(x = 3 \times \sqrt{\frac{17}{9}} = \sqrt{17}\)
The sum to infinity of the series: 1 + (1/3) + (1/9) + (1/27) + ... is
11/3
10/3
5/2
3/2
Correct answer is D
The series is geometric with common ratio \(\frac{1}{3}\).
\(S_{\infty} = \frac{a}{1 - r}\)
= \(\frac{1}{1 - \frac{1}{3}} \)
= \(\frac{1}{\frac{2}{3}}\)
= \(\frac{3}{2}\)
59
19
67
38
Correct answer is A
\(p \ast q = pq + p + q\)
\(2 \ast (3 \ast 4) \)
\(3 \ast 4 = 12 + 3 + 4 = 19\)
\(2 \ast 19 = 38 + 2 + 19 = 59\)