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.
5.0
4.0
3.0
2.5
Correct answer is A
Diagonal |AC| = (2x + 1)cm
In the diagram,area of \(\Delta\)ABC
is \(\frac{110}{2}\) = \(\frac{1}{2}\) x |AC| x |HB|
55 = \(\frac{1}{2}\) x (2x + 1) x 10
55 = (2x + 1)5
55 = 10x + 5
55 - 5 = 10x
50 = 10x
x = \(\frac{50}{10}\)
= 5.0
Which of the following is used to determine the mode of a grouped data?
Bar chart
Frequency polygon
Ogive
Histogram
Correct answer is D
No explanation has been provided for this answer.
Make K the subject of the relation T = \(\sqrt{\frac{TK - H}{K - H}}\)
K = \(\frac{H(T^2 - 1)}{T^2 - T}\)
K = \(\frac{HT}{(T - 1)^2}\)
K = \(\frac{H(T^2 + 1)}{T}\)
K = \(\frac{H(T - 1)}{T}\)
Correct answer is A
T = \(\sqrt{\frac{TK - H}{K - H}}\)
Taking the square of both sides, give
T2 = \(\frac{TK - H}{K - H}\)
T2(K - H) = TK - H
T2K - T2H = TK - H
T2K - TK = T2H - H
K(T2 - T) = H(T2 - 1)
K = \(\frac{H(T^2 - 1)}{T^2 - T}\)
35o
40o
55o
60o
Correct answer is A
In the diagram, QOR + 2m(vertically opposite angles)
So, m + 90° + 2m = 180°
(angles on str. line)
3m = 180° - 90°
3m = 90°
m = \(\frac{90^o}{3}\)
= 30°
substituting 30° for m in
2m + 4n = 200° gives
2 x 30° + 4n = 200°
60° + 4n = 200°
4n = 200° - 60°
= 140°
n = \(\frac{140°}{4}\)
= 35°
\(\frac{1}{3}\)
\(\frac{2}{3}\)
1\(\frac{1}{3}\)
1\(\frac{2}{3}\)
Correct answer is C
\(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3}}\) = \(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\)
= \(\frac{\sqrt{2} \times \sqrt{3} + \sqrt{3} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\)
= \(\frac{\sqrt{6} + 3}{3}\)
= \(\frac{3 + \sqrt{6}}{3}\)
= Hence, (m + n) = 1 + \(\frac{1}{3}\)
= 1\(\frac{1}{3}\)