JAMB Mathematics Past Questions & Answers - Page 317

The first poster has 7 ways to be arranges, the second poster can be arranged in 6 ways and the third poster in 5 ways.

= 7 x 6 x 5

= 210 ways

or \(\frac{7}{P_3}\) = \(\frac{7!}{(7 - 3)!}\) = \(\frac{7!}{4!}\)

= \(\frac{7 \times 6 \times 5 \times 4!}{4!}\)

= 210 ways

(\(\sqrt2 + \frac{1}{\sqrt3})(\sqrt2 - \frac{1}{\sqrt3}\))

\(\sqrt4 - \frac {\sqrt2}{\sqrt3} + \frac {\sqrt2}{\sqrt3} - \frac {1}{\sqrt9}\)

= 2 - \(\frac {1}{3}\)

= \(\frac {16 - 1}{3}\)

= \(\frac{5}{3}\)

\(\frac{2 - \sqrt5}{3 - \sqrt5}\) x \(\frac{3 + \sqrt5}{3 + \sqrt5}\)

\(\frac{(2 - \sqrt5)(3 + \sqrt5)}{(3 - \sqrt5)(3 + \sqrt5)}\) = \(\frac{6 +2\sqrt5 - 3\sqrt5 - \sqrt25}{9 + 3\sqrt5 - 3\sqrt5 - \sqrt25}\)

= \(\frac{6 - \sqrt5 - 5}{9 - 5}\)

= \(\frac{1 - \sqrt5}{4}\)

log\(_{3}^{18}\) + log\(_{3}^{3}\) - log\(_{3}^{x}\) = 3

log\(_{3}^{18}\) + log\(_{3}^{3}\) - log\(_{3}^{x}\) = 3log₃3

log\(_{3}^{18}\) + log\(_{3}^{3}\) - log\(_{3}^{x}\) = log₃3³

log₃(\(\frac{18 \times 3}{X}\)) = log₃3³

\(\frac{18 \times 3}{X}\) = 3³

18 x 3 = 27 x X

x = \(\frac{18 \times 3}{27}\)

= 2

\((\frac{16}{81})^{\frac{1}{4}} \div (\frac{9}{16})^{-\frac{1}{2}}\)

\((\frac{16}{81})^{\frac{1}{4}} \div (\frac{16}{9})^{\frac{1}{2}}\)

\((\frac{2^4}{3^4})^{\frac{1}{4}} \div (\frac{4^2}{3^2})^{\frac{1}{2}}\)

\(\frac{2^{4 \times \frac{1}{4}}}{3^{4 \times \frac{1}{4}}} \div \frac{4^{2 \times \frac{1}{2}}}{3^{2 \times \frac{1}{2}}}\)

\(\frac{2}{3} \div \frac{4}{3}\)

\(\frac{2}{3} \times \frac{3}{4}\)

\(\frac{2}{4}\)

\(\frac{1}{2}\)