2004 - WAEC Mathematics Past Questions and Answers - page 1

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

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

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
\(x = \frac{168}{8}=21^{\circ}\)

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

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 = \(\frac{x}{1}-\left(\frac{x}{4}+\frac{x}{3}\right)\
\frac{3x+4x}{12}; = \frac{12x-7x}{12} = \frac{5x}{12} = \frac{5}{12}\)

\(\frac{PR}{QR} = \frac{QR}{QY} = \frac{QP}{YR}\) (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) \(=\frac{Number\hspace{1mm} of\hspace{1mm} girls}{Number \hspace{1mm} of \hspace{1mm} boys \hspace{1mm} and \hspace{1mm} girls\hspace{1mm} in \hspace{1mm} class}\
= \frac{12}{m+12}\)

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