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 (R1, B2)
= \(\frac{5}{9} \times \frac{4}{9}\)
= \(\frac{20}{81}\)
Users' Answers & Commentsn(blue balls) = 4
n(\(\iff\)) = 9
Hence, prob (R1, B2)
= \(\frac{5}{9} \times \frac{4}{9}\)
= \(\frac{20}{81}\)
42
The relation y = x2 + 2x + k passes through the point (2,0). Find the value of k
A
- 8
B
- 4
C
4
D
8
correct option: a
Y = x2 + 2x + k (given)
y = o when x = 2
thus 0 = 22 + 2 x 2 + k
0 = 4 + 4 + k
given k = -8
Users' Answers & Commentsy = o when x = 2
thus 0 = 22 + 2 x 2 + k
0 = 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
Users' Answers & Comments44
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,
Q1 = \(\frac{1}{4}\)Nth term
= \(\frac{1}{4}\) x 600th term
= 150th term
= 15.5 + 3 + 0.5
= 19
Users' Answers & CommentsQ1 = \(\frac{1}{4}\)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 = 180o(angles on a straight line)
11e + e = 180o
12e = 180o
e = \(\frac{180^o}{12}\)
= 15o
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 = 180o(angles on a straight line)
11e + e = 180o
12e = 180o
e = \(\frac{180^o}{12}\)
= 15o
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 = 140o, 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 70o = \(\frac{x}{10}\)
x = 10sin 70o
= 9.3969
Hence, length of chord MN = 2x
= 2 x 9.3969
= 18.7938
= 19cm (nearest cm)
Users' Answers & Commentssin 70o = \(\frac{x}{10}\)
x = 10sin 70o
= 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 50o, 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, \(\alpha\) 50o(alternative angles)
tan\(\alpha\) = \(\frac{h}{6}\)
h = 6tan\(\alpha\)
= 6tan 50o
= 6 x 1.1918
= 7.1508
= 7.15m (2d.p)
Users' Answers & Commentstan\(\alpha\) = \(\frac{h}{6}\)
h = 6tan\(\alpha\)
= 6tan 50o
= 6 x 1.1918
= 7.1508
= 7.15m (2d.p)
49
From the diagram, which of the following is true?
A
m + n + p = 180o
B
m + n = 180o
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