2016 - WAEC Mathematics Past Questions and Answers - page 5

n(red balls) = 5

n(blue balls) = 4

n(\(\iff\)) = 9

Hence, prob (R₁, B₂)

= \(\frac{5}{9} \times \frac{4}{9}\)

= \(\frac{20}{81}\)

Y = x² + 2x + k (given)

y = o when x = 2

thus 0 = 2² + 2 x 2 + k

0 = 4 + 4 + k

given k = -8

Lower quartile,

Q₁ = \(\frac{1}{4}\)Nth term

= \(\frac{1}{4}\) x 600th term

= 150th term

= 15.5 + 3 + 0.5

= 19

Let a represent an interior angle; e represent an exterior angle. A section of the polygon is down in the diagram.

\(\frac{e}{a}\) = \(\frac{l}{11}\) given

a = 11e

a + e = 180^o(angles on a straight line)

11e + e = 180^o

12e = 180^o

e = \(\frac{180^o}{12}\)

= 15^o

Hence, number of sides

= \(\frac{360^o}{\tect{size of one exterior angle}\)

= \(\frac{360^o}{14^o}\)

= 24

Halima's age = n years old

Her brother's age = \(\frac{n}{2}\) + 5

From the diagram

sin 70^o = \(\frac{x}{10}\)

x = 10sin 70^o

= 9.3969

Hence, length of chord MN = 2x

= 2 x 9.3969

= 18.7938

= 19cm (nearest cm)

In the diagram give, \(\alpha\) 50^o(alternative angles)

tan\(\alpha\) = \(\frac{h}{6}\)

h = 6tan\(\alpha\)

= 6tan 50^o

= 6 x 1.1918

= 7.1508

= 7.15m (2d.p)

In the diagram, \(\beta\) = p(vertically opposite angles)

m + \(\beta\) = n(sum of interior opp. angles)

m + p = n