2003 - WAEC Mathematics Past Questions and Answers - page 2

5(x - 7) = 7 - 2x

5x - 35 = 7 - 2x

5x + 2x = 7 + 35

7x = 42

x = \(\frac{42}{7}\)

= 6

\(\frac{log8}{log\frac{1}{4}}=\frac{log2^3}{log2^{-2}}=\frac{3log2}{-2log2}=-\frac{3}{2}\)

\(V= \frac{1}{2}\hspace{1mm}base\hspace{1mm}area\hspace{1mm}\times\hspace{1mm}height\
120=\frac{1}{3}\times 5 \times 6 \times h; h = \frac{120}{10}=12cm\)

< QPS = 75°

< QRS = 180° - 75° = 105° (opposite angles of a cyclic quadrilateral)

The sum of angles in a triangle is 180^o
2x^o + (2x -21)^o + ∠ZYP = 180^o
Also ∠ZYP = 180^o - (3x+14)^o
∴2x^o + (2x-21)^o + 180^o - (3x+14)^o = 180
x = 35^o

3x\(^2\) - 8x - 3 = 0

3x\(^2\) - 9x + x - 3 = 0

3x(x - 3) + 1(x - 3) = 0

(3x + 1)(x - 3) = 0

3x = -1 \(\implies\) x = \(-\frac{1}{3}\)

or x = 3.

Using the sine rule
\(\frac{|QR|}{sinP}=\frac{|PQ|}{sinR}\
\frac{x}{sin43^o}=\frac{5}{sin53^o}\
x = \frac{5\times sin43^o}{sin53^o}\
=\frac{5\times 0.6820}{0.7986}\
=4.3\)

\(4P4_5 = 119_{10}\)

\(4 \times 5^2 + P \times 5^1 + 4 \times 5^0 = 119\)

\(100 + 5P + 4 = 119\)

\(5P = 119 - 104 = 15\)

\(P = 3\)

Sin 53 = \(\frac {h}{6} \)
h = 6 sin 53 = 6 x 0.7986
= 4.792m

\(Sin\theta^{\circ} = \frac{opp}{hyp}\
Sin X^{\circ}=\frac{5}{10}=0.5000\SinX^{\circ}\
X^{\circ} = sin^{-1}0.5000\
X^{\circ}=30^{\circ}\)