1994 - JAMB Mathematics Past Questions and Answers - page 1

\(\frac{1}{3} \div [\frac{5}{7}(\frac{9}{10} -1 + \frac{3}{4})]\)

\(\frac{1}{3} \div [\frac{5}{7}(\frac{9}{10} - \frac{10}{10} + \frac{3}{4})]\)

= \(\frac{1}{3} \div [\frac{5}{7}(\frac{-1}{10} + \frac{3}{4})]\)

= \(\frac{1}{3} \div [\frac{5}{7}(\frac{-2 + 15}{20})]\)

= \(\frac{1}{3} \div [\frac{5}{7} \times \frac{13}{20}]\)

\(\frac{1}{3} + [\frac{13}{28}]\) = \(\frac{1}{3} \times \frac{28}{13}\)

= \(\frac{28}{39}\)

\(\frac{0.36 \times 5.4 \times 0.63}{4.2 \times 9.0 \times 2.4}\)

= \(\frac{36}{420} \times \frac{54}{90} \times \frac{63}{240}\)

= \(\frac{6}{70} \times \frac{18}{30} \times \frac{21}{80}\)

= \(\frac{27}{2000}\)

= 0.0135

\(\approx\) = 0.013

\(\frac{log_5 0.04}{log_3 18 - log_3 2}\)

= \(\frac{log_5 0.04}{log_3(\frac{18}{2})}\)

= \(\frac{log_5 0.04}{log_3 9}\)

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

= -1

Let log₅ 0.04 = x

5x = 0.04

x = \(\frac{4}{100}\) = 5^-2

Let log₃ 9 = z

3² = 3²

z = 3

8x^-2 = \(\frac{2}{25}\)

= 200x^-2 = 2

= 100x^-2 = 1

x^-2 = \(\frac{1}{100}\)

x^-2 = 10^-2

x = 10

\(\sqrt{48}\) - \(\frac{9}{\sqrt{3}}\) + \(\sqrt{75}\)

Rearrange = \(\sqrt{48}\) + \(\sqrt{75}\) - \(\frac{9}{\sqrt{3}}\)

= (√16 x √3) + (√25 x √3) - \(\frac{9}{\sqrt{3}}\)

=4√3 + 5√3 - \(\frac{9}{\sqrt{3}}\)

Rationalize \(\to\) 9√3 = \(\frac{9}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\)

= \(\frac{9\sqrt{3}}{\sqrt{9}}\) - \(\frac{9\sqrt{3}}{\sqrt{3}}\)

= 3√3

\(\frac{1}{\sqrt{2}}\) = \(\frac{1}{\sqrt{2}}\) x \(\frac{\sqrt{2}}{\sqrt{2}}\)

= \(\frac{\sqrt{2}}{2}\)

= \(\frac{1.414}{2}\)

= 0.707

A \(\subset\) B means A is contained in B i.e. A is a subset of B(A \(\cap\) B)¹ = A¹

A(A \(\cap\) B)¹ = A \(\cap\) A¹

The intersection of complement of a set P and P¹ has no element

i.e. n(A \(\cap\) A¹) = \(\phi\)

\(\frac{(2m - u)^2 - (m - 2u)^2}{5m^2 - 5u^2}\)

= \(\frac{2m - u + m - 2u)(2m - u - m + 2u)}{5(m + u)(m - u)}\)

= \(\frac{3(m - u)(m + u)}{5(m + u)(m - u)}\)

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