2004 - WAEC Mathematics Past Questions and Answers - page 3

Mean \(= \frac{sum\hspace{1mm}of\hspace{1mm}all\hspace{1mm}the\hspace{1mm}numbers}{Total\hspace{1mm}Number}\
\frac{9+13+16+17+19+23+24}{7}\
\frac{121}{7}=17.2857=17.29\)

\(\frac{PQ}{PR}=\frac{XZ}{XY}; \frac{PQ}{8}=\frac{3.5}{5}\
PQ=\frac{8\times 3.5}{5}=\frac{28}{5}=5.6cm\)

27p²x² - 48y²; 3(3² p²x²-4²y²)
3[(3px) ² - (4y) ²]; 3[(3px-4y)(3px+4y)]

Volume of a solid cylinder = Base area x height
Base area (=\pi r^2 =\frac{22}{7} \times \frac{7}{2}\times \frac{7}{2} = \frac{77}{2})
Volume of a solid cylinder = (\frac{77}{2}\times \frac{7}{1} = \frac{539}{2}cm^3<br /> =269.5cm^3)

The general quadratic equation
x² - (sum of roots) x + (product of root) = 0
Comparing with the given equation
2x² +3x² - 9 = 0; (x^2 + \frac{3}{2}x - \frac{9}{2} = 0<br /> Sum \hspace{1mm} of \hspace{1mm}roots = -\frac{3}{2})

ξ = {1, 2, 3, . . . . . . ,10}, P= (2, 3, 5, 7) and Q = {1, 3, 5, 9},
P^l∩Q = {1, 4, 6, 8, 9, 10}∩{1, 3, 5, 7, 9} = {1, 9}

∠QRS = ∠RPQ (theorem: angle in the alternate segment are equal)
∠RPQ = 50^o
∠QRS = 50^o

(\begin{matrix}
x & x-m & |x-m| <br /> 6 & -2 & 2 <br /> 7 & -1 & 1 <br /> 8 & 0 & 0 <br /> 9 & 1 & 1 <br /> 10 & 2 & 2 <br /> & & 6
\end{matrix}
)
Mean deviation(=\frac{\sum|x-m|}{n}=\frac{6}{5} = 1.2)

If x is the bearing 342^ofrom a point y
= 360^o - 342^o = 18^o;
The bearing of y from x = 180^o - 18^o = 162^o

∠SOR = ∠ORQ alternate angle.
Also ∠OQR = ∠OQR (Base angle of isosceles)
∠QOR = 108 – 2(35^o) = 180 - 70^o - 110^o ∠QPR = ¹/₂QOR = ¹/₂(110^o) ∠QPR + 55^o (theorem angle at center is twice angle at circumference)