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 y = \(\frac{x}{x - 3}\) + \(\frac{x}{x + 4}\) find y when x = -2
-\(\frac{3}{5}\)
\(\frac{3}{5}\)
-\(\frac{7}{5}\)
\(\frac{2}{5}\)
Correct answer is A
y = \(\frac{x}{x - 3}\) + \(\frac{x}{x + 4}\) when x = -2
y = \(\frac{-2}{-5}\) + \(\frac{(-2)}{-2 + 4}\)
= \(\frac{2}{5}\) + \(\frac{-2}{2}\)
= \(\frac{4 -10}{10}\)
= \(\frac{-6}{10}\)
= -\(\frac{3}{5}\)
Factorize completely 8a + 125ax3
(2a + 5x2)(4 + 26ax)
a(2 + 5x)(4 - 10x + 25x2)
(2a + 5x)(4 - 10ax + 25x2)
a(2 + 5x)(4 + 10ax + 25x2)
Correct answer is B
\(8a + 125ax^{3} = (2^{3})a + 5^{3} ax^{3}\)
= \(a(2^3 + 5^3 x^3)\)
∴\(a[2^3 + (5x)^3]\)
\(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)
∴ \(a(2^3 + (5x)^3)\)
= \(a(2 + 5x)(4 - 10x + 25x^2)\)
If x varies directly as y3 and x = 2 when y = 1, find x when y = 5
2
10
125
250
Correct answer is D
x \(\alpha\) y3
x = ky3
k = \(\frac {x}{y^3}\)
when x = 2, y = 1
k = 2
Thus x = 2y3 - equation of variation
= 2(5)3
= 250
\(\frac{1}{3}\)
1
3
9
Correct answer is B
\(\frac{9^{\frac{1}{3}} \times 27^{-\frac{1}{2}}}{3^{-\frac{1}{6}} \times 3^{-\frac{2}{3}}}\) = \(\frac{(3^2)^{\frac{1}{2}} \times (3^3)^{-\frac{1}{3}}}{3^{-\frac{1}{6}} \times 3^{-\frac{2}{3}}}\)
= \(\frac{3^{\frac{2}{3}} \times 3^{-\frac{3}{2}}}{3^{-\frac{1}{6}} \times 3^{-\frac{2}{3}}}\)
= \(\frac{3^{-\frac{5}{6}}}{3^{-\frac{5}{6}}}\)
= 1
Find n if log\(_{2}\) 4 + log\(_{2}\) 7 - log\(_{2}\) n = 1
10
14
27
28
Correct answer is B
log\(_2\) 4 + log\(_2\) 7 - log\(_2\) n = 1
= log\(_2\) (4 x 7) - log\(_2\) n = 1
\(\therefore\) log\(_2\) 28 - log\(_2\) n = 1
= \(\frac{28}{n} = 2^1\)
\(\frac{28}{n}\) = 2
2n = 28
∴ n = 14