2007 - WAEC Mathematics Past Questions and Answers - page 4
31
A chord of length 6cm is drawn in a circle of radius 5cm. Find the distance of the chord from the centre of the circle.
A
2.5cm
B
3.0cm
C
3.5cm
D
4.0cm
correct option: d
Distance(d) = d2 = 52 - 32
= 25 - 9
= 16
d = \(\sqrt{16}\)
= 4cm
Users' Answers & Comments= 25 - 9
= 16
d = \(\sqrt{16}\)
= 4cm
32
The area of a square field is 110.25m2. Find the cost of fencing it round at N75.00 per metre
A
N1,575.00
B
N3,150.00
C
N4,734.30
D
N8,268.75
correct option: b
Area = 110.252
f2 = 110.25(square)
l = \(\sqrt{110.25}\) = 10.5m
Since N75.00 per metre
N75.00 = 1 metre; x = 10.5m
x = 787.5 for one side, then four sides = N787.5 x 4
= N3,150.00
Users' Answers & Commentsf2 = 110.25(square)
l = \(\sqrt{110.25}\) = 10.5m
Since N75.00 per metre
N75.00 = 1 metre; x = 10.5m
x = 787.5 for one side, then four sides = N787.5 x 4
= N3,150.00
33
A sector of a circle of radius 14cm containing an angle 60o is folded to form a cone. Calculate the radius of the base of the cone
A
5\(\frac{1}{2}cm\)
B
4\(\frac{2}{3}cm\)
C
3\(\frac{1}{2}cm\)
D
2\(\frac{1}{3}cm\)
correct option: d
Length of arc = circumference of the base of the
cone \(\frac{\theta}{360} \times 2\pi R = 2 \pi r\)
\(\frac{\theta R}{360}\) = r
r = \(\frac{60 \times 14}{360}\)
= \(\frac{7}{3} = 2\frac{1}{3}\)cm
Users' Answers & Commentscone \(\frac{\theta}{360} \times 2\pi R = 2 \pi r\)
\(\frac{\theta R}{360}\) = r
r = \(\frac{60 \times 14}{360}\)
= \(\frac{7}{3} = 2\frac{1}{3}\)cm
34
Find the volume of a solid cylinder with base radius 10cm and height 14cm.
A
2203
B
880cm3>
C
1400cm3
D
4400cm3
correct option: d
volume of a cylinder = \(\pi r^2h\)
r = 10cm
h = 14cm
= \(\frac{22}{7} \times 10 \times 10 \times 14\)
= 4400cm3
Users' Answers & Commentsr = 10cm
h = 14cm
= \(\frac{22}{7} \times 10 \times 10 \times 14\)
= 4400cm3
35
Each of the interior angle of a regular polygon is 162o. How many sides has the polygon?
A
8
B
12
C
16
D
20
correct option: d
\(\frac{(n - 2)180}{n}\) = 162 (where n = no. of sides)
162n = 180n - 360
162n - 180n = -360
-18n = - 360
n = \(\frac{360}{18}\)
n = 20
Users' Answers & Comments162n = 180n - 360
162n - 180n = -360
-18n = - 360
n = \(\frac{360}{18}\)
n = 20
36
In a , < PQR = < PRQ = 45o. which of the following statements is/are correct? i. \(\bigtriangleup\)PQR is an equalateral triangle ii. \(\bigtriangleup\)PQR is an isosceles triangle iii. \(\bigtriangleup\)PQR is a right-angled triangle
A
ii only
B
i and ii only
C
ii and iii only
D
i and ii only
correct option: c
Users' Answers & Comments37
The table above gives the distribution of the marks of a number of students in a test.
\(\begin{array}{c|c} Mark &1 & 2 & 3 & 4 & 5 & 6 \ \hline Frequency & 5 & 3 & 6 & 4 & 2 & 5\end{array}\), find the mode of the distribution.
\(\begin{array}{c|c} Mark &1 & 2 & 3 & 4 & 5 & 6 \ \hline Frequency & 5 & 3 & 6 & 4 & 2 & 5\end{array}\), find the mode of the distribution.
A
2
B
3
C
5
D
6
correct option: b
Users' Answers & Comments38
If the probability of an event occurring is x, what is the probability of the event not occurring?
A
1 - x
B
x - 1
C
0
D
\(\frac{1}{x}\)
correct option: a
If p is the probability of an event happening(occurring) and the probability of an event not occurring is p1 then p + p1 = 1
p1 = 1 - p
x1 = 1 - x
Users' Answers & Commentsp1 = 1 - p
x1 = 1 - x
39
The table shows the ages(in years) of twenty children chosen at random from a community. What is the mean age? \(\begin{array}{c|c} Age(years) & 1 & 2 & 3 & 4 & 5 \ \hline {\text {Number of children}} & 2 & 3 & 5 & 6 & 4 \end{array}\)
A
4.46 years
B
3.35 years
C
3.30years
D
3.00 years
correct option: b
Mean(age) = \(\frac{(1 \times 2) + (2 \times 3) + (3 \times 5) + (4 \times 6) + (5 \times 4)}{2 + 3 + 5 + 6 + 4}\)
= \(\frac{2 + 6 + 15 + 24 + 20}{20}\)
= \(\frac{67}{20}\)
= 3.35yrs
Users' Answers & Comments= \(\frac{2 + 6 + 15 + 24 + 20}{20}\)
= \(\frac{67}{20}\)
= 3.35yrs
40
The table shows the ages(in years) of twenty children chosen at random from a community. What is the median of the distribution? \(\begin{array}{c|c} Age(years) & 1 & 2 & 3 & 4 & 5 \ \hline {\text {Number of children}} & 2 & 3 & 5 & 6 & 4 \end{array}\)
A
3.5 years
B
3.0 years
C
2.5 years
D
2.0 years
correct option: a
Median = (\(\frac{n + 1}{2})^{th}\)
= \(\frac{20 + 1}{2} = \frac{21}{2}\)
= 10.5
= \(\frac{{\text{10th term and 11th term}}}{2}\)
= \(\frac{3 + 4}{2} = \frac{7}{2}\)
= 3.5yrs
Users' Answers & Comments= \(\frac{20 + 1}{2} = \frac{21}{2}\)
= 10.5
= \(\frac{{\text{10th term and 11th term}}}{2}\)
= \(\frac{3 + 4}{2} = \frac{7}{2}\)
= 3.5yrs