1992 - WAEC Mathematics Past Questions and Answers - page 1

\(\begin{array}{c|c} 2 & 77 \ \hline 2 & 38 R1 \ 2 & 19 R0 \ 2 & 9 R1 \ 2 & 4 R1 \ 2 & 2 R0 \ 2 & 1 R0 \ & 0 R1\end{array}\)

77_ten = 1001101_two

Error = 4.25 - 4.10 = 0.15

% error = \(\frac{0.15}{4.25} \times 100%\)

= \(\frac{15}{\frac{17}{4}} = \frac{15 \times 4}{17}\)

= \(3\frac{9}{17} %\)

\(T_n = 3. 2^{n - 2} \

T_{1} = 3. 2^{1 - 2} = 3. 2^{-1} \

T_1 = \frac{3}{2} \)

\(T_2 = 3. 2^{2 - 2} \

T_2 = 3. 2^0  = 3\)

\(T_3 = 3. 2^{3 - 2} = 3. 2^1 \

T_3 = 6\)

The first 3 terms of the sequence are \(\frac{3}{2}\), 3 and 6.

\((\frac{16}{81})^{\frac{1}{4}}\)

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

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

\(\log_{10} 25 + \log_{10} 32 - \log_{10} 8\)

= \(\log_{10} (\frac{25 \times 32}{8})\)

= \(\log_{10} 100 \)

= 2

2y\(^2\) + xy - 3x\(^2\)

2y\(^2\) + 3xy - 2xy - 3x\(^2\)

y(2y + 3x) - x(2y + 3x)

= (y - x)(2y + 3x)

If x = \(-\frac{1}{2}\) and 2; then

\(x + \frac{1}{2} = 0\) and \(x - 2 = 0\)

\(\implies (x + \frac{1}{2})(x - 2) = 0\)

\(x^2 - 2x + \frac{1}{2}x - 1 = 0\)

\(x^2 - \frac{3}{2}x - 1 = 0\)

\(2x^2 - 3x - 2 = 0\)

\(\frac{1}{1 - x} + \frac{2}{1 + x}\)

= \(\frac{(1 + x) + 2(1 - x)}{(1 - x)(1 + x)}\)

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

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

x\(^2\) - 18x to be a perfect square.

\((\frac{b}{2})^2\) is added to ax\(^2\) + bx + c in order to make it a perfect square.

\(x^2 - 18x + (\frac{-18}{2})^2\)

= \(x^2 - 18x + 81\)

\(\frac{m}{3} + \frac{1}{2} = \frac{3}{4} + \frac{m}{4}\)

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

\(\frac{m}{12} = \frac{1}{4}\)

\(4m = 12 \implies m = 3\)