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2007 WAEC Mathematics Theory (a) Simplify : \(\frac{3\frac{1}{12} + \frac{7}{8}}{2\frac{1}{4} - \frac{1}{6}}\) (b) If \(p = \frac{m}{2} - \frac{n^{2}}{5m}\)...

Mathematics
WAEC 2007

(a) Simplify : \(\frac{3\frac{1}{12} + \frac{7}{8}}{2\frac{1}{4} - \frac{1}{6}}\)

(b) If \(p = \frac{m}{2} - \frac{n^{2}}{5m}\) ; 

(i) make n the subject of the relation ;  (ii) find, correct to three significant figures, the value of n when p = 14 and m = -8.

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Explanation

(a) \(\frac{3\frac{1}{12} + \frac{7}{8}}{2\frac{1}{4} - \frac{1}{6}}\)

\(3\frac{1}{12} + \frac{7}{8} = \frac{37}{12} + \frac{7}{8}\)

= \(\frac{74 + 21}{24}\)

= \(\frac{95}{24}\)

\(2\frac{1}{4} - \frac{1}{6} = \frac{9}{4} - \frac{1}{6}\)

= \(\frac{27 - 2}{12}\)

= \(\frac{25}{12}\)

\(\therefore \frac{3\frac{1}{12} + \frac{7}{8}}{2\frac{1}{4} - \frac{1}{6}} = \frac{95}{24} \div \frac{25}{12}\)

\(\frac{95}{24} \times \frac{12}{25} = \frac{19}{10}\)

= \(1.9\)

(b)(i) \(p = \frac{m}{2} - \frac{n^{2}}{5m}\)

\(\frac{n^{2}}{5m} = \frac{m}{2} - p\)

\(n^{2} = 5m(\frac{m}{2} - p)\)

\(n = \pm \sqrt{5m(\frac{m}{2} - p)}\)

(ii) When p = 14 and m = -8,

\(n = \sqrt{5(-8)(\frac{-8}{2} - 14)}\)

\(n = \sqrt{-40(- 4 - 14)}\)

\(n = \sqrt{720}\)

\(n = 12\sqrt{5}\) 

= \(\pm 26.83\)


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