2001 - WAEC Mathematics Past Questions and Answers - page 3
The graph of the curve \(y = 2x^2 - 5x - 1\) and a straight line PQ were drawn to solve the equation \(2x^2 - 5x + 2 = 0\)
What is the equation of the straight line PQ?
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
Subtract (-y + 3x + 5z) from (4y - x - 2z)
(4y - x - 2z) - (-y + 3x + 5z)
= 4y + y - x - 3x - 2z - 5z
= 5y - 4x - 7z
If \(y \propto \frac{1}{\sqrt{x}}\) and x = 16 when y = 2, find x when y = 24
\(y \propto \frac{1}{\sqrt{x}}\)
\(y = \frac{k}{\sqrt{x}}\)
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,
\(24 = \frac{8}{\sqrt{x}}\)
\(\sqrt{x} = \frac{8}{24} = \frac{1}{3}\)
\(\therefore x = (\frac{1}{3})^2\)
\(x = \frac{1}{9}\)
\frac{2x}{x+1}=\frac{3}{2}\
∴ 4x = 3x + 3 x =3\)
The bearing S40°E is the same as
\(BC^2 = 13^2 - 5^2\
=169-25\
BC = \sqrt{144} = 12\
∴ cos P - tan P = \frac{12}{13} - \frac{5}{12}\
=\frac{79}{156}\)
Three observation posts P,Q and R are such that Q is due east of P and R is due north of Q. If |PQ| = 5km and |PR| = 10km, find |QR|
\((PR)^2 = (PQ)^2 + (QR)^2\)
\(10^2 = 5^2 + (QR)^2\)
\((QR)^2 = 100 - 25\)
\(QR = \sqrt{75}\)
= \(8.660\)
\(\approxeq\) 8.7 km
Express 25° 45' in decimal (Hint: 1° = 60')
A box contains 5 red, 3 green and 4 blue balls. A boy is allowed to take away two balls from the box. What is the probability that the two balls are red?
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) = \(\frac{5}{12} \times \frac{4}{11}\)
= \(\frac{5}{33}\)
A box contains 5 red, 3 green and 4 blue balls. A boy is allowed to take away two balls from the box. What is the probability that one is green and the other is blue?
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)
= \(\frac{3}{12} \times \frac{4}{11} + \frac{4}{12} \times \frac{3}{11}\)
= \(\frac{2}{11}\)