WAEC Mathematics Past Questions & Answers - Page 191

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

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

\(202^2_{three}\)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
\(\begin{matrix}
3 & 204\\
3 & 69 &R0\\
3 & 22 & R2\\
3 & 7 & R1\\
2 & 2 & R1\\
& 0& R2 \uparrow\\
\end{matrix} \\
=21120_3\)

Bearing of Q and P
\(180^{\circ} – (360^{\circ}-x)\\
180^{\circ}-360^{\circ}+x\\
-180^{\circ}+x\\
x-180^{\circ}\)

15% loss = N595

\(\implies\) (100 - 15)% of CP = N595

85% of CP = N595

CP = \(\frac{595 \times 100}{85}\)

= N700.00