2002 - WAEC Mathematics Past Questions and Answers - page 2

The longest side of a right angle triangle is the hypotenuse.

\((x + 2)^{2} = x^{2} + (x + 1)^{2}\)

\(x^2 + 4x + 4 = x^2 + x^2 + 2x + 1\)

\(2x^{2} - x^{2} + 2x - 4x + 1 - 4 = 0\)

\(x^{2} - 2x - 3 = 0\)

\(x^{2} - 3x + x - 3 = 0 \implies x(x - 3) + 1(x - 3) = 0\)

\((x - 3)(x + 1) = 0 \implies \text{x = 3 or -1}\)

\(x > 0 \implies x = 3\)

The longest side = 3 + 2 = 5.

\(log2 = 0.3010\hspace{1mm}given\
log2^y = 1.8062\
∴ ylog2 = 1.8062\
y=\frac{1.8062}{logy}=\frac{1.8062}{0.3010}=6\)

ZY\(^2\) = 12\(^2\) + 6\(^2\)

ZY\(^2\) = 144 + 36 = 180

ZY = \(\sqrt{180}\)

= \(13.416\)

= 13.4 km

20 members in the ratio 3:1

Number of women = \(\frac{1}{4} \times 20\)

= 5

Let the number of women to be added = x

Total number of members in the committee = 20 + x

\(\frac{5 + x}{20 + x} = \frac{2}{5}\)

\(5(5 + x) = 2(20 + x)\)

\(25 + 5x = 40 + 2x \implies 5x - 2x = 40 - 25\)

\(3x = 15 \implies x = 5\)

((x+3) \propto y<br /> ∴x + 3 = ky\hspace{1mm}when\hspace{1mm}x = 3, y = 12<br /> 3+3 = 12k<br /> ∴ k = \frac{1}{2}<br /> \Longrightarrow x + 3 = \frac{1}{2}y) to find x when y = 8
(x + 3 = \frac{1}{2}\times 8<br /> x=4-3<br /> x = 1)

\(T_{1} = 4; T_2 = 10; T_3 = 16\)

\(T_{2} - T_1 = T_3 - T_1 = 6\)

\(T_n = a + (n - 1) d\)

= \(4 + (n - 1) \times 6\)

= \(4 + 6n - 6\)

= \(6n - 2\)

= 2(3n - 1)

If (-3, -4) is a point on the line then

-4 = -3m + 2

-4 - 2 = -3m

-6 = -3m

m = 2

From the graph, the zeros of the equation exist at x = -2 and x = 3

\(\therefore\) (x + 2) = 0 and (x - 3) = 0

\(\implies (x + 2)(x - 3) = 0\)

\(x^2 - 3x + 2x - 6 = 0\)

\(x^2 - x - 6 = 0\) is the equation represented on the graph.

< XYQ = 180° - (80° + 56°)

= 44°

< PYQ = 56° (alternate angles, XP||YQ)

< QPY = 90°

< PQY = 180° - (90° + 56°)

= 34°