# 2016 - WAEC Mathematics Past Questions and Answers - page 5

41

A bag contains 5 red and 4 blue identical balls. Id two balls are selected at random from the bag, one after the other, with replacement, find the probability that the first is red and the second is blue

A

\(\frac{2}{9}\)

B

\(\frac{5}{18}\)

C

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

D

\(\frac{5}{9}\)

**correct option:**c

n(red balls) = 5

n(blue balls) = 4

n(\(\iff\)) = 9

Hence, prob (R

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

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

Users' Answers & Commentsn(blue balls) = 4

n(\(\iff\)) = 9

Hence, prob (R

_{1}, B_{2})= \(\frac{5}{9} \times \frac{4}{9}\)

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

42

The relation y = x

^{2}+ 2x + k passes through the point (2,0). Find the value of kA

- 8

B

- 4

C

4

D

8

**correct option:**a

Y = x

y = o when x = 2

thus 0 = 2

0 = 4 + 4 + k

given k = -8

Users' Answers & Comments^{2}+ 2x + k (given)y = o when x = 2

thus 0 = 2

^{2}+ 2 x 2 + k0 = 4 + 4 + k

given k = -8

43

Find the next three terms of the sequence; 0, 1, 1, 2, 3, 5, 8...

A

13, 19, 23

B

9, 11, 13

C

11, 15, 19

D

13, 21, 34

**correct option:**d

44

Find the lower quartile of the distribution illustrated by the cumulative frequency curve

A

17.5

B

19.0

C

27.5

D

28.0

**correct option:**b

Lower quartile,

Q

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

= 150th term

= 15.5 + 3 + 0.5

= 19

Users' Answers & CommentsQ

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

= 150th term

= 15.5 + 3 + 0.5

= 19

45

The ratio of the exterior angle to the interior angle of a regular polygon is 1:11. How many sides has the polygon?

A

30

B

24

C

18

D

12

**correct option:**b

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

11e + e = 180

12e = 180

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

= 15

Hence, number of sides

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

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

= 24

Users' Answers & Comments\(\frac{e}{a}\) = \(\frac{l}{11}\) 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

46

Halima is n years old. Her brother's age is 5 years more than half of her age. How old is her brother?

A

\(\frac{n}{2} + \frac{5}{2}\)

B

\(\frac{n}{2}\) - 5

C

5 - \(\frac{n}{2}\)

D

\(\frac{n}{2}\) + 5

**correct option:**d

Halima's age = n years old

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

Users' Answers & CommentsHer brother's age = \(\frac{n}{2}\) + 5

47

In the diagram MN is a chord of a circle KMN centre O and radius 10cm. If < MON = 140

^{o}, find, correct to the nearest cm, the length of the chord MN.A

19cm

B

18cm

C

17cm

D

12cm

**correct option:**a

From the diagram

sin 70

x = 10sin 70

= 9.3969

Hence, length of chord MN = 2x

= 2 x 9.3969

= 18.7938

= 19cm (nearest cm)

Users' Answers & Commentssin 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)

48

An object is 6m away from the base of a mast. If the angle of depression of the object from the top of the mast is 50

^{o}, find, correct to 2 decimal places, the height of the mast.A

8.60m

B

7.83m

C

7.51m

D

7.15m

**correct option:**d

In the diagram give, \(\alpha\) 50

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

h = 6tan\(\alpha\)

= 6tan 50

= 6 x 1.1918

= 7.1508

= 7.15m (2d.p)

Users' Answers & Comments^{o}(alternative angles)tan\(\alpha\) = \(\frac{h}{6}\)

h = 6tan\(\alpha\)

= 6tan 50

^{o}= 6 x 1.1918

= 7.1508

= 7.15m (2d.p)

49

From the diagram, which of the following is true?

A

m + n + p = 180

^{o}B

m + n = 180

^{o}C

m = p + n

D

n = m + p

**correct option:**d

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

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

m + p = n

Users' Answers & Commentsm + \(\beta\) = n(sum of interior opp. angles)

m + p = n