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.
400.00m
692.82m
923.76m
1,600.99m
Correct answer is D
In \(\bigtriangleup\)KBC, sin 30 = \(\frac{800}{IKCI}\)
IKCL = \(\frac{800}{sin30^o}\)
= \(\frac{800}{0.5}\)
= 1600m
PQRT is square. If x is the midpoint of PQ, Calculate correct to the nearest degree, LPXS
53o
55o
63o
65o
Correct answer is C
In the diagram given,
tan\(\alpha\) = \(\frac{1}{0.5}\) = 2
\(\alpha\) = tan - 1(2) = 63.43o
= 63o
Given that tan x = \(\frac{2}{3}\), where 0o d" x d" 90o, Find the value of 2sinx.
\(\frac{2\sqrt{13}}{13}\)
\(\frac{3\sqrt{13}}{13}\)
\(\frac{4\sqrt{13}}{13}\)
\(\frac{6\sqrt{13}}{13}\)
Correct answer is C
tan x = \(\frac{2}{3}\)(given), is illustrated in a right-angled \(\Delta\)
thus m2 = 22 + 32
= 4 + 9 = 13
m = \(\sqrt{13}\)
Hence, 2sin x = 2 x \(\frac{2}{m}\)
2 x\(\frac{2}{\sqrt{13}}\)
= \(\frac{4}{\sqrt{13}}\)
= \(\frac{4}{\sqrt{13}} = \frac{\sqrt{13}}{\sqrt{13}}\)
= \(\frac{4\sqrt{13}}{13}\)
1
2
3
4
Correct answer is B
Ler: (x1, y1) = (0, 3)
(x2, y2) = (\(\frac{5}{4}, \frac{11}{2}\))
Using gradient, m = \(\frac{y_2 - y_2}{x_2 - x_1}\)
= \(\frac{\frac{11}{2} - 3}{\frac{5}{4} - 0}\)
= \(\frac{11 - 6}{2} + \frac{5}{4}\)
= \(\frac{5}{} + \frac{5}{4}\)
= \(\frac{5}{2} \times \frac{4}{5}\)
= 2
73o
67o
57o
37o
Correct answer is C
In the diagram given, < PRT = 3\(^o\) (Change in same segment)
< TPR = 90\(^o\) (angle in a semicircle)
Hence, < PTR = 180\(^o\) - (90 + 33)\(^o\)
= 180\(^o\) - 123\(^o\)
= 57\(^o\)