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 x = 3 - \(\sqrt{3}\), find x2 + \(\frac{36}{x^2}\)
9
18
24
27
Correct answer is C
x = 3 - \(\sqrt{3}\)
x2 = (3 - \(\sqrt{3}\))2
= 9 + 3 - 6\(\sqrt{34}\)
= 12 - 6\(\sqrt{3}\)
= 6(2 - \(\sqrt{3}\))
∴ x2 + \(\frac{36}{x^2}\) = 6(2 - \(\sqrt{3}\)) + \(\frac{36}{6(2 - \sqrt{3})}\)
6(2 - \(\sqrt{3}\)) + \(\frac{6}{2 - \sqrt{3}}\) = 6(- \(\sqrt{3}\)) + \(\frac{6(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})}\)
= 6(2 - \(\sqrt{3}\)) + \(\frac{6(2 + \sqrt{3})}{4 - 3}\)
6(2 - \(\sqrt{3}\)) + 6(2 + \(\sqrt{3}\)) = 12 + 12
= 24
Simplify 5\(\sqrt{18}\) - 3\(\sqrt{72}\) + 4\(\sqrt{50}\)
17\(\sqrt{4}\)
4\(\sqrt{17}\)
17\(\sqrt{2}\)
12\(\sqrt{4}\)
Correct answer is C
5\(\sqrt{18}\) - 3\(\sqrt{72}\) + 4\(\sqrt{50}\) = 5(3\(\sqrt{2}\)) - 3(6\(\sqrt{2}\)) + 4(5\(\sqrt{2}\))
15\(\sqrt{2}\) - 18\(\sqrt{2}\) + 20\(\sqrt{2}\) = 35\(\sqrt{2}\) - 18\(\sqrt{2}\)
= 17\(\sqrt{2}\)
Simplify \(\frac{(1.25 \times 10^{-4}) \times (2.0 \times 10^{-1})}{(6.25 \times 10^5)}\)
4.0 x 10-3
5.0 x 10-2
2.0 x 10-1
5.0 x 10-3
Correct answer is A
\(\frac{(1.25 \times 10^{-4}) \times (2.0 \times 10^{-1})}{(6.25 \times 10^5)}\) = \(\frac{1.25 \times 2}{6.25}\) x 104 - 1 - 5
\(\frac{2.50}{6.25}\) x 10-2 = \(\frac{250}{625}\) x 10-2
0.4 x 10-2 = 4.0 x 10-3
What is the value of x satisfying the equation \(\frac{4^{2x}}{4^{3x}}\) = 2?
-2
-\(\frac{1}{2}\)
\(\frac{1}{2}\)
2
Correct answer is B
\(\frac{4^{2x}}{4^{3x}}\) = 2
42x - 3x = 2
4-x = 2
(22)-x
= 21
Equating coefficients: -2x = 1
x = -\(\frac{1}{2}\)
Evaluate \(\log_{b} a^{n}\) if \(b = a^{\frac{1}{n}}\).
n2
n
\(\frac{1}{n}\)
\(\frac{1}{n^2}\)
Correct answer is A
Let \(\log_{b} a^{n} = x\)
\(\therefore a^{n} = b^{x}\)
\(a^{n} = (a^{\frac{1}{n}})^{x}\)
\(a^{n} = a^{\frac{x}{n}} \implies n = \frac{x}{n}\)
\(x = n^{2}\)