2004 - WAEC Mathematics Past Questions and Answers - page 1

(202^2_{three})when converted to base ten (=(202_3)^2<br /> 202_3 = 2 \times 3^2 + 0 \times 3^1 + 2\times 3^0 = 18 + 0 + 2<br /> =20_{ten}; (202_3)^2 = (20)^2_{ten} = 400<br /> 112^2_{three})when converted to base ten (= (112_3)^2<br /> 112_3 = 1 \times 3^2 + 1 \times 3^1 + 2\times 3^0 = 9+3+2=14_{ten}<br /> (112_3)^2 = (14)^2_{ten} = 196_{ten}<br /> Evaluate \Longrightarrow 400-196 = 204)
Reconvert to base three
(\begin{matrix}
3 & 204\
3 & 69 &R0\
3 & 22 & R2\
3 & 7 & R1\
2 & 2 & R1<br /> & 0& R2 \uparrow<br /> \end{matrix} <br /> =21120_3)

(23_{five} = X_{ten}; X_{ten} = 2\times 5^1 + 3\times 5^0 = 10 + 3 = 13<br /> 101_{three}=P_{ten}; P_{ten} = 1\times 3^2 + 0\times 3^1 + 1\times 3^0=9+0+1=10_{ten}<br /> Y = 13+10=23_{ten});
Converting to base two
(\begin{matrix}
2 & 23\
2 & 11 &R1\
2 & 5 & R1\
2 & 2 & R1\
2 & 1 & R0<br /> & 0& R1 \uparrow<br /> \end{matrix} <br /> =y=10111_2)

Sin (5x – 28)^o = cos (3x - 50)^o
Since by the trigonometry relation
Sin(5x – 28)^o = cos[90 – (5x – 28)]^o
Hence cos(3x – 50)^o = cos[90 – (5x – 28)]^o
3x – 50 = 90 - (5x-28)
3x – 50 = 90 – 5x + 28
3x + 5x = 90 + 28 + 50
8x = 168
(x = \frac{168}{8}=21^{\circ})

(\frac{3}{4}t+\frac{1}{3}(21-t) = 11; \frac{3t}{4} + \frac{7}{1} - \frac{t}{3} = \frac{11}{1}<br /> \frac{3 \times 3t + 7\times 12 – 4 \times t = 11 \times 12}{12}<br /> 9t + 84 – 4t = 132; 5t = 132 – 84<br /> 5t = 48; t = \frac{48}{5} = 9\frac{3}{5})

Let the girls pocket money be rep. by x. The amount spent on books = (\frac{1}{4}of\hspace{1mm}x = \frac{x}{4})
The amount spent on dress (=\frac{1}{3} of \hspace{1mm}x=\frac{x}{3})
∴The fraction that remains = (\frac{x}{1}-\left(\frac{x}{4}+\frac{x}{3}\right)<br /> \frac{3x+4x}{12}; = \frac{12x-7x}{12} = \frac{5x}{12} = \frac{5}{12})

\(\frac{PR}{QR} = \frac{QR}{QY} = \frac{QP}{YR}\) (SSS Congruence)

\(\frac{QR}{4} = \frac{8}{5}\)

\(QR = \frac{4 \times 8}{5}\)

= 6.4 cm

The sum of angles at point = 360^o
(x+10)^o + (4x+50)^o + 20^o + 3x^o + 2x^o = 360^o
10x + 80^o = 360^o
10x = 360^o - 80^o = 280^o
(\frac{280}{10}=28^{\circ})

Prob. (a girl) (=\frac{Number\hspace{1mm} of\hspace{1mm} girls}{Number \hspace{1mm} of \hspace{1mm} boys \hspace{1mm} and \hspace{1mm} girls\hspace{1mm} in \hspace{1mm} class}<br /> = \frac{12}{m+12})

\(7\frac{1}{2}-\left(2\frac{1}{2}+3\right)\div16\frac{1}{2}; \frac{15}{2} - \left(\frac{5}{2}+\frac{3}{1}\right)\div \frac{33}{2}\
\frac{15}{2} - \left(\frac{5+6}{2}\right)\div \frac{33}{2}; \frac{15}{2} - \frac{11}{2} \div \frac{33}{2}\
\frac{15-11}{2}\div \frac{33}{2}; \frac{4}{2} \times \frac{2}{33} = \frac{4}{33} = 0.1212\)
The nearest whole number is 0

Where H is the height of the tower H = ?
(Tan 30^{\circ} = \frac{H}{36} \Rightarrow H = 36 \times tan30^{\circ}<br /> H = 36 \times 0.5774 = 20.79)