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 two dice are thrown together, what is the probability of obtaining at least a score of 10?
\(\frac{1}{6}\)
\(\frac{1}{12}\)
\(\frac{5}{6}\)
\(\frac{11}{12}\)
Correct answer is A
The total sample space when two dice are thrown together is 6 x 6 = 36
\(\begin{array}{c|c} & 1 & 2 & 3 & 4 & 5 & 6\\ 1. & 1.1 & 1.2 & 1.3 & 1.4 & 1.5 & 1.6\\ 2 & 2.1 & 2.2 & 2.3 & 2.4 & 2.5 & 2.6\\3 & 3.1 & 3.2 & 3.3 & 3.4 & 3.5 & 3.6\\ 4 & 4.1 & 4.2 & 4.3 & 4.4 & 4.5 & 4.6\\ 5 & 5.1 & 5.2 & 5.3 & 5.4 & 5.5 & 5.6\\6 & 6.1 & 6.2 & 6.3 & 6.4 & 6.5 & 6.6 \end{array}\)
At least 10 means 10 and above
P(at least 10) = \(\frac{6}{36}\)
= \(\frac{1}{6}\)
18
15
1.0
5
Correct answer is C
Mean \(\bar{x}\) = \(\frac{156}{10}\) = 15.6
Median = \(\bar{y}\) = \(\frac{15 + 16}{2}\)
\(\frac{31}{2}\) = 15.5
\(\frac{x}{y}\) = \(\frac{15.6}{15.5}\) = 1.0065
1.0(1 d.p)
2.7
4.33
3.1
3.3
Correct answer is B
\(\bigtriangleup\)PUS is right angled
\(\frac{US}{5}\) = sin60o
US = 5 x \(\frac{\sqrt{3}}{2}\)
= 2.5\(\sqrt{3}\)
= 4.33cm
120\(\pi\)cm3
110\(\pi\)cm3
60\(\pi\)cm3
50\(\pi\)cm3
Correct answer is B
The volume of the pipe is equal to the area of the cross section and length.
let outer and inner radii be R and r respectively.
Area of the cross section = (R2 - r2)
where R = 6 and r = 6 - 1
= 5cm
Area of the cross section = (62 - 52)\(\pi\) = (36 - 25)\(\pi\)cm sq
vol. of the pipe = \(\pi\) (R2 - r2)L where length (L) = 10
volume = 11\(\pi\) x 10
= 110\(\pi\)cm3
If \(\cos^2 \theta + \frac{1}{8} = \sin^2 \theta\), find \(\tan \theta\).
3
\(\frac{3\sqrt{7}}{7}\)
3\(\sqrt{7}\)
\(\sqrt{7}\)
Correct answer is B
\(\cos^2 \theta + \frac{1}{8} = \sin^2 \theta\)..........(i)
from trigometric ratios for an acute angle, where \(cos^{2} \theta + \sin^2 \theta = 1\) ........(ii)
Substitute for equation (i) in (i) = \(\cos^2 \theta + \frac{1}{8} = 1 - \cos^2 \theta \)
= \(\cos^2 \theta + \cos^2 \theta = 1 - \frac{1}{8}\)
\(2\cos^2 \theta = \frac{7}{8}\)
\(\cos^2 \theta = \frac{7}{2 \times 8}\)
\(\frac{7}{16} = \cos \theta\)
\(\sqrt{\frac{7}{16}}\) = \(\frac{\sqrt{7}}{4}\)
but cos \(\theta\) = \(\frac{\text{adj}}{\text{hyp}}\)
\(opp^2 = hyp^2 - adj^2\)
\(opp^2 = 4^2 - (\sqrt{7})^{2}\)
= 16 - 7
opp = \(\sqrt{9}\) = 3
\(\tan \theta = \frac{\text{opp}}{\text{adj}}\)
= \(\frac{3}{\sqrt{7}}\)
\(\frac{3}{\sqrt{7}}\) x \(\frac{\sqrt{7}}{\sqrt{7}}\) = \(\frac{3\sqrt{7}}{7}\)