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.

$2=\frac{k}{\sqrt{16}}⟹2=\frac{k}{4}$

$k=8$

$y=\frac{8}{\sqrt{x}}$

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<br /> \frac{2x}{x+1}=\frac{3}{2}<br /> ∴ 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<br /> =169-25<br /> BC = \sqrt{144} = 12<br /> ∴ cos P - tan P = \frac{12}{13} - \frac{5}{12}<br /> =\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)

 = $\frac{3}{12}\times \frac{4}{11}+\frac{4}{12}\times \frac{3}{11}$

= $\frac{2}{11}$