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1993 WAEC Mathematics Theory (a) Simplify, without using Mathematical tables: \(\log_{10} (\frac{30}{16}) - 2 \log_{10} (\frac{5}{9}) + \log_{10} (\frac{400}{243})\)...

Mathematics
WAEC 1993

(a) Simplify, without using Mathematical tables: \(\log_{10} (\frac{30}{16}) - 2 \log_{10} (\frac{5}{9}) + \log_{10} (\frac{400}{243})\)

(b) Without using Mathematical tables, calculate \(\sqrt{\frac{P}{Q}}\) where \(P = 3.6 \times 10^{-3}\) and \(Q = 2.25 \times 10^{6}\), leaving your answer in standard form.

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Explanation

(a) \(\log_{10} (\frac{30}{16}) - 2 \log_{10} (\frac{5}{9}) + \log_{10} (\frac{400}{243})\)

= \(\log_{10} (\frac{30}{16}) - \log_{10} (\frac{5}{9})^{2} + \log_{10} (\frac{400}{243})\)

= \(\log_{10} (\frac{30}{16} \times \frac{400}{243} \div \frac{25}{81})\)

= \(\log_{10} (\frac{30}{16} \times \frac{400}{243} \times \frac{81}{25})\)

= \(\log_{10} (10) = 1\)

(b) \(\sqrt{\frac{P}{Q}} = \sqrt{\frac{3.6 \times 10^{-3}}{2.25 \times 10^{6}}\)

= \(\sqrt{\frac{36 \times 10^{-4}}{2.25 \times 10^{6}}\)

= \(\frac{6 \times 10^{-2}}{1.5 \times 10^{3}}\)

= \(4 \times 10^{-2 - 3} = 4 \times 10^{-5}\)


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Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts