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