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.
\(\frac{31}{15}\)
\(\frac{31}{5}\)
\(\frac{129}{5}\)
\(\frac{43}{5}\)
\(\frac{16}{5}\)
Correct answer is D
F = 1, 2, 3, 4, 5
x = 1, 2, 4, 8, 16
fx = 1, 4, 12, 32, 80, 3f = 15
(average size) = \(\frac{\sum fx}{\sum f}\)
= \(\frac{129}{15}\)
= \(\frac{43}{5}\)
\(\frac{59300}{625}\)
\(\frac{50900}{625}\)
200%
+100%
Correct answer is C
Let x rept. the size of the quantity
2x + \(\frac{116}{100}\) x \(\frac{-84}{100}\)
= 200%
What is log7(49a) - log10(0.01)?
\(\frac{49^a}{100}\)
\(\frac{a}{2}\) + 2
72a + 2
2a + 2
\(\frac{2a}{2}\)
Correct answer is D
log7(49a) - log10(0.01) = log7(72)a - log10100
log772a - log101 - log10 102
= 2a - 2
= 2a + a
\(\sqrt{2239cm}\)
\(\sqrt{2}\) x 2239cm
\(\frac{\sqrt{3}}{2}\) 2239cm
\(\sqrt{3}\) x 2239cm
4478cm
Correct answer is D
x = \(\sqrt{-2239^2 + 2239^2}\)
= -\(\sqrt{10026242}\)
= 3166.42
y = -\(\sqrt{10026242 + 5013121}\)
= -\(\sqrt{15039363}\)
= 3878
= \(\sqrt{3}\) x 2239
An arc of circle of radius 2cm subtends an angle of 60º at the centre. Find the area of the sector
\(\frac{2 \pi}{3}\)cm2
\(\frac{\pi}{2}\)cm2
\(\frac{\pi}{3}\)cm2
\(\pi\)cm2
Correct answer is A
Area of a sector \(\frac{\theta}{360}\) x \(\pi\)r\(^2\)
= \(\frac{60^o}{360^o}\) x \(\pi\)2\(^2\)
= \(\frac{2 \pi}{3}\)cm2