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2006 WAEC Mathematics Theory (a) Simplify : \(\frac{4\frac{2}{9} - 1\frac{13}{15}}{2\frac{1}{5} + \frac{4}{7} \times 2\frac{1}{3}}\) (b) By rationalising the denominator,...

Mathematics
WAEC 2006

(a) Simplify : \(\frac{4\frac{2}{9} - 1\frac{13}{15}}{2\frac{1}{5} + \frac{4}{7} \times 2\frac{1}{3}}\)

(b) By rationalising the denominator, simplify : \(\frac{7\sqrt{5}}{\sqrt{7}}\), leaving your answer in surd form.

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Explanation

(a) \(\frac{4\frac{2}{9} - 1\frac{13}{15}}{2\frac{1}{5} + \frac{4}{7} \times 2\frac{1}{3}}\)

\(4\frac{2}{9} - 1\frac{13}{15} = \frac{38}{9} - \frac{28}{15}\)

= \(\frac{190 - 84}{45}\)

= \(\frac{106}{45}\)

\(2\frac{1}{5} + \frac{4}{7} \times 2\frac{1}{3} = \frac{11}{5} + (\frac{4}{7} \times \frac{7}{3})\)

= \(\frac{11}{5} + \frac{4}{3}\)

= \(\frac{33 + 20}{15}\)

= \(\frac{53}{15}\)

\(\therefore \frac{4\frac{2}{9} - 1\frac{13}{15}}{2\frac{1}{5} + \frac{4}{7} \times 2\frac{1}{3}} = \frac{106}{45} \div \frac{53}{15}\)

= \(\frac{106}{45} \times \frac{15}{53}\)

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

(b) \(\frac{7\sqrt{5}}{\sqrt{7}}\)

= \(\frac{7\sqrt{5}}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}}\)

= \(\frac{7\sqrt{35}}{7}\)

= \(\sqrt{35}\)


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WAEC offline past questions - with all answers and explanations in one app - Download for free
WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995
WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
WAEC offline past questions - with all answers and explanations in one app - Download for free
WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995