Year : 
2007
Title : 
Mathematics (Core)
Exam : 
WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

31 - 40 of 67 Questions

# Question Ans
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

Detailed Solution

Distance(d) = d2 = 52 - 32

= 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

Detailed Solution

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
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\(\frac{1}{3}cm\)

Detailed Solution

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
34.

Find the volume of a solid cylinder with base radius 10cm and height 14cm.

A. 2203

B. 880cm3>

C. 1400cm3

D. 4400cm3

Detailed Solution

volume of a cylinder = \(\pi r^2h\)

r = 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

Detailed Solution

\(\frac{(n - 2)180}{n}\) = 162 (where n = no. of sides)

162n = 180n - 360

162n - 180n = -360

-18n = - 360

n = \(\frac{360}{18}\)

n = 20
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

C
37.

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.

A. 2

B. 3

C. 5

D. 6

B
38.

If the probability of an event occurring is x, what is the probability of the event not occurring?

A. 1 - x

B. x - 1

D. \(\frac{1}{x}\)

Detailed Solution

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
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

Detailed Solution

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
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

Detailed Solution

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
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

Detailed Solution

Distance(d) = d2 = 52 - 32

= 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

Detailed Solution

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
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\(\frac{1}{3}cm\)

Detailed Solution

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
34.

Find the volume of a solid cylinder with base radius 10cm and height 14cm.

A. 2203

B. 880cm3>

C. 1400cm3

D. 4400cm3

Detailed Solution

volume of a cylinder = \(\pi r^2h\)

r = 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

Detailed Solution

\(\frac{(n - 2)180}{n}\) = 162 (where n = no. of sides)

162n = 180n - 360

162n - 180n = -360

-18n = - 360

n = \(\frac{360}{18}\)

n = 20
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

C
37.

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.

A. 2

B. 3

C. 5

D. 6

B
38.

If the probability of an event occurring is x, what is the probability of the event not occurring?

A. 1 - x

B. x - 1

D. \(\frac{1}{x}\)

Detailed Solution

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
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

Detailed Solution

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
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

Detailed Solution

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