Year : 
2009
Title : 
Mathematics (Core)
Exam : 
WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

11 - 20 of 46 Questions

# Question Ans
11.

The diameter of a bicycle wheel is 42cm. If the wheel makes 16 complete revolution, what will be the total distance covered by the wheel? [Take \(\pi \frac{22}{7}\)

A. 672cm

B. 1056cm

C. 2112cm

D. 4224cm

Detailed Solution

d = 42cm

Circumference = \(\pi d = \frac{22}{7} \times 42cm\)

= 22 x 6cm = 132cm

Total distance covered = circumference x number of revolutions

= 132cm x 16 = 2112cm
12.

The interior angles of a pentagon are (2x + 5)o, (x + 20)o, xo, (3x - 20)o and (x + 15)o. Find the value of x

A. 80o

B. 70o

C. 65o

D. 40o

Detailed Solution

(2x + 5)o + (xo + 20o) + x + (3x - 20)o + (x + 15)o = (n - 2) x 180

8n + 20 = (5 - 2) x 180

where n = 5(Pentagon)

8n + 20 = 3 x 180

8n + 20 = 540

8n = 540 - 20

8n = 520

x = 65o
13.

If M and N are two fixed points in a plane. Find the locus L = [P : PM = PN]

A. a line equal to MN

B. line parallel to MN

C. Perpendicular bisector of MN

D. A circle centre P, radius MN

Detailed Solution

Locus L = (P : PM = Pn}

For M and N being two fixed points

Since PM = PN, P is equidistant from Mand N, So L must be the perpendicular bisector of the line MN
14.

Given that sin 60o = \(\frac{\sqrt{3}}{2}\) and cos 60o = \(\frac{1}{2}\), evaluate \(\frac{1 - sin 60^o}{1 + cos 60^o}\)

A. \(\frac{2 + \sqrt{3}}{3}\)

B. \(\frac{1 - \sqrt{3}}{3}\)

C. \(\frac{1 + \sqrt{3}}{3}\)

D. \(\frac{2 - \sqrt{3}}{3}\)

Detailed Solution

Sin 60 = \(\frac{\sqrt{3}}{2}\); cos 60o = \(\frac{1}{2}\)

= \(\frac{1 - \sin 60}{1 + \cos 60} = \frac{1 - \frac{\sqrt{3}}{2}}{1 + \frac{1}{2}}\)

= \(\frac{2 - \sqrt{3}}{3}{2}\div \frac{2 + 1}{2}\)

= \(\frac{2 - \sqrt{3}}{2} \times \frac{2}{3}\)

= \(\frac{2 - \sqrt{3}}{3}\)
15.

In an examination, Kofi scored x% in Physics, 50% in Chemistry and 70% in Biology. If his mean score for the three subjects was 55%, find x

A. 40

B. 45

C. 55

D. 60

Detailed Solution

\(\frac{x + 50 + 70}{3} = 55\)

x + 120 = 3(55)

x + 120 = 165

= 165 - 120

x = 45%
16.

What is the median of the following scores: 22 35 41 63 74 82

A. 82

B. 52

C. 49

D. 22

Detailed Solution

22 35 41 63 74 82


Median = \(\frac{41 + 63}{2} = \frac{104}{2}\)

= 52
17.

Express 2.7864 x 10-3 to 2 significant figures

A. 0.28

B. 0.27

C. 0.0028

D. 0.0027

Detailed Solution

2.7864 x 10-3 = 0.0027864

= 0.0028(approx. 2 s.f)
18.

If 92x = \(\frac{1}{3}\)(27x), find x

A. 2

B. 1

C. -1

D. -2

Detailed Solution

92x = \(\frac{1}{3}\)(27x)

(22)2x = \(\frac{(3^{3x})}{3}\)

(34x) = 33x - 1

4x = 3x - 1

4x - 3x = -1

x = -1
19.

If 85% of x is N3230, what is the value of x?

A. N2745.50

B. N3714.50

C. N3800.00

D. N4845.00

Detailed Solution

85% of x = 3230

\(\frac{85}{100} \times x\) = N3230

x = \(\frac{100}{85}\) x N3230

x = N3800.00
20.

The capacity of a water tank is 1,800 litres. If the tank is in form of a cuboid with base 600cm by 150 cm. Find the height of the tank

A. 2cm

B. 20cm

C. 200cm

D. 2000cm

Detailed Solution

1 litre = 1000cm3

1800 litre = 1800 x 1000cm3

l x b x h = capacity

600 x 150 x h = 1800000

h = \(\frac{1800000}{600 \times 150}\)cm

h = 20cm
11.

The diameter of a bicycle wheel is 42cm. If the wheel makes 16 complete revolution, what will be the total distance covered by the wheel? [Take \(\pi \frac{22}{7}\)

A. 672cm

B. 1056cm

C. 2112cm

D. 4224cm

Detailed Solution

d = 42cm

Circumference = \(\pi d = \frac{22}{7} \times 42cm\)

= 22 x 6cm = 132cm

Total distance covered = circumference x number of revolutions

= 132cm x 16 = 2112cm
12.

The interior angles of a pentagon are (2x + 5)o, (x + 20)o, xo, (3x - 20)o and (x + 15)o. Find the value of x

A. 80o

B. 70o

C. 65o

D. 40o

Detailed Solution

(2x + 5)o + (xo + 20o) + x + (3x - 20)o + (x + 15)o = (n - 2) x 180

8n + 20 = (5 - 2) x 180

where n = 5(Pentagon)

8n + 20 = 3 x 180

8n + 20 = 540

8n = 540 - 20

8n = 520

x = 65o
13.

If M and N are two fixed points in a plane. Find the locus L = [P : PM = PN]

A. a line equal to MN

B. line parallel to MN

C. Perpendicular bisector of MN

D. A circle centre P, radius MN

Detailed Solution

Locus L = (P : PM = Pn}

For M and N being two fixed points

Since PM = PN, P is equidistant from Mand N, So L must be the perpendicular bisector of the line MN
14.

Given that sin 60o = \(\frac{\sqrt{3}}{2}\) and cos 60o = \(\frac{1}{2}\), evaluate \(\frac{1 - sin 60^o}{1 + cos 60^o}\)

A. \(\frac{2 + \sqrt{3}}{3}\)

B. \(\frac{1 - \sqrt{3}}{3}\)

C. \(\frac{1 + \sqrt{3}}{3}\)

D. \(\frac{2 - \sqrt{3}}{3}\)

Detailed Solution

Sin 60 = \(\frac{\sqrt{3}}{2}\); cos 60o = \(\frac{1}{2}\)

= \(\frac{1 - \sin 60}{1 + \cos 60} = \frac{1 - \frac{\sqrt{3}}{2}}{1 + \frac{1}{2}}\)

= \(\frac{2 - \sqrt{3}}{3}{2}\div \frac{2 + 1}{2}\)

= \(\frac{2 - \sqrt{3}}{2} \times \frac{2}{3}\)

= \(\frac{2 - \sqrt{3}}{3}\)
15.

In an examination, Kofi scored x% in Physics, 50% in Chemistry and 70% in Biology. If his mean score for the three subjects was 55%, find x

A. 40

B. 45

C. 55

D. 60

Detailed Solution

\(\frac{x + 50 + 70}{3} = 55\)

x + 120 = 3(55)

x + 120 = 165

= 165 - 120

x = 45%
16.

What is the median of the following scores: 22 35 41 63 74 82

A. 82

B. 52

C. 49

D. 22

Detailed Solution

22 35 41 63 74 82


Median = \(\frac{41 + 63}{2} = \frac{104}{2}\)

= 52
17.

Express 2.7864 x 10-3 to 2 significant figures

A. 0.28

B. 0.27

C. 0.0028

D. 0.0027

Detailed Solution

2.7864 x 10-3 = 0.0027864

= 0.0028(approx. 2 s.f)
18.

If 92x = \(\frac{1}{3}\)(27x), find x

A. 2

B. 1

C. -1

D. -2

Detailed Solution

92x = \(\frac{1}{3}\)(27x)

(22)2x = \(\frac{(3^{3x})}{3}\)

(34x) = 33x - 1

4x = 3x - 1

4x - 3x = -1

x = -1
19.

If 85% of x is N3230, what is the value of x?

A. N2745.50

B. N3714.50

C. N3800.00

D. N4845.00

Detailed Solution

85% of x = 3230

\(\frac{85}{100} \times x\) = N3230

x = \(\frac{100}{85}\) x N3230

x = N3800.00
20.

The capacity of a water tank is 1,800 litres. If the tank is in form of a cuboid with base 600cm by 150 cm. Find the height of the tank

A. 2cm

B. 20cm

C. 200cm

D. 2000cm

Detailed Solution

1 litre = 1000cm3

1800 litre = 1800 x 1000cm3

l x b x h = capacity

600 x 150 x h = 1800000

h = \(\frac{1800000}{600 \times 150}\)cm

h = 20cm