2001 - WAEC Mathematics Past Questions and Answers - page 3

Let the straight line PQ = y

2x\(^2\) - 5x - 1 - y = 2x\(^2\) - 5x + 2

y = 2x\(^2\) - 5x - 1 - 2x\(^2\) + 5x - 2 

y = -3

(4y - x - 2z) - (-y + 3x + 5z)

= 4y + y - x - 3x - 2z - 5z

= 5y - 4x - 7z

\(y \propto \frac{1}{\sqrt{x}}\)

\(y = \frac{k}{\sqrt{x}}\)

When x = 16, y = 2.

\(2 = \frac{k}{\sqrt{16}} \implies 2 = \frac{k}{4}\)

\(k = 8\)

\(y = \frac{8}{\sqrt{x}}\)

When y = 24,

\(24 = \frac{8}{\sqrt{x}}\)

\(\sqrt{x} = \frac{8}{24} = \frac{1}{3}\)

\(\therefore x = (\frac{1}{3})^2\)

\(x = \frac{1}{9}\)

\(2x : (x + 1) = 3:2\
\frac{2x}{x+1}=\frac{3}{2}\
∴ 4x = 3x + 3 x =3\)

= 90° + 50°

= 140°

If \(sin P = \frac{5}{13}\) from right angled triangle from pythagoras theorem
\(BC^2 = 13^2 - 5^2\
=169-25\
BC = \sqrt{144} = 12\
∴ cos P - tan P = \frac{12}{13} - \frac{5}{12}\
=\frac{79}{156}\)

\((PR)^2 = (PQ)^2 + (QR)^2\)

\(10^2 = 5^2 + (QR)^2\)

\((QR)^2 = 100 - 25\)

\(QR = \sqrt{75}\)

= \(8.660\)

\(\approxeq\) 8.7 km

25° 45' = 25° + \(\frac{45}{60}\)°

= 25° + 0.75°

= 25.75°

Total number of balls = 5 + 3 + 4 

= 12 balls

P(first ball is red) = \(\frac{5}{12}\)

P(second ball is red) = \(\frac{4}{11}\)

\(\therefore\) P(both balls are red) = \(\frac{5}{12} \times \frac{4}{11}\)

= \(\frac{5}{33}\)

Total number of balls = 5 + 3 + 4 = 12 balls

P(one ball is green and the other is blue) = P(first ball is green and second blue) + P(first ball is blue and the second green)

 = \(\frac{3}{12} \times \frac{4}{11} + \frac{4}{12} \times \frac{3}{11}\)

= \(\frac{2}{11}\)