1994 - WAEC Mathematics Past Questions and Answers - page 1

T\(_{2}\) = ar = 6

T\(_{5}\) = ar\(^{4}\) = 48

\(\frac{T_5}{T_2}\) = \(\frac{ar^{4}}{ar}\) = \(\frac{48}{6}\)

= r\(^{3}\) = 8

⇒ r = 2

S\(_{n}\) = \(\frac{a((r^n) - 1)}{r - 1}\)

S\(_{4}\) = \(\frac{a((r^4) - 1)}{r - 1}\)

S\(_{4}\) = \(\frac{3((2^4) - 1)}{2 - 1}\)

= 3(16 -1)

= 45 

\(\sin \theta = \frac{k}{1}\)

\(\implies 1^2 = k^2 + adj^2\)

\(adj = \sqrt{1 - k^2}\)

\(\therefore \tan \theta = \frac{k}{\sqrt{1 - k^2}}\)

\(2\frac{1}{3} \div 2\frac{2}{3} \times 1\frac{1}{7}\)

= \(\frac{7}{3} \div \frac{8}{3} \times \frac{8}{7}\)

= \(\frac{7}{3} \times \frac{3}{8} \times \frac{8}{7}\)

= 1

\(\frac{72}{125} = \frac{8 \times 9}{5 \times 5 \times 5}\)

= \(\frac{2^3 \times 3^2}{5^3}\)

\(\frac{8.75}{0.025}\)

= \(\frac{8750}{25}\)

= \(350\)

= \(3.5 \times 10^2\)

\(\frac{27^{\frac{1}{3}}}{16^{-\frac{1}{4}}} \)

= \(\frac{(3^3)^{\frac{1}{3}}}{(2^4)^{-\frac{1}{4}}}\)

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

= 6

16\(^{\frac{5}{4}}\) x 2\(^{-3}\) x 3\(^0\)

= \((2^4)^{\frac{5}{4}} \times 2^{-3} \times 1\)

= \(2^5 \times 2^{-3} \times 1\)

= \(2^2\)

= 4.

2log\(_{3}\) 6 + log\(_{3}\) 12 - log\(_{3}\) 16

= \(\log_3 6^2 + \log_3 12 - \log_3 16\)

= \(\log_{3} (\frac{36 \times 12}{16})\)

= \(\log_{3} (27)\)

= \(\log_{3} 3^3\)

= \(3 \log_{3} 3\)

= 3

Loss = N(100,000 - 80,000)

= N20,000.

Loss percentage = \(\frac{20,000}{100,000} \times 100%\)

= 20%