BECE 2010 Mathematics – Paper One

1. Which of the following sets is well defined?
A. {Man, Kofi, Red, 14}
B. {Ink, Mango, Green, Nail}
C. {Car, Road, Glass, Book}
D. {Seth, Mary, Jacob, Evelyn}

D. {Seth, Mary, Jacob, Evelyn}

Solution

A well-defined set is a set whose members or elements show an identifiable uniform pattern or characteristic.

The set {Seth, Mary, Jacob, Evelyn} shows an identifiable uniform pattern or characteristic – they are all names of human beings.

The other sets are not uniform and contain different items.

Other examples of well-defined sets include:

{1, 2, 3, 4, 5, 6, …7} — counting numbers

{4, 6, 8, 10, 12, 14, 16} — even numbers between 2 and 18

{Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday} — days of the week

{J. J. Rawlings, J. A. Kuffour, J. E. A. Mills, John Mahama, Nana Addo-Dankwa Akufo Addo} — presidents of the 4th Republic of Ghana

 

2. If set B is a subset of set A, then
A. sets A and B have the same number of elements
B. some members of set B can be found in set A
C. no member of set B is in set A
D. all the members of set B are in set A

D. all the members of set B are in set A

Solution

Set B is a subset of set A if all the members of set B are also members of set A.

For example, if B = {2, 3, 4, 5} and A = (1, 2, 3, 4, 5, 6, 7, 8, 10}, then we say that B is a subset of A.

Note that every set is a subset of itself, and the null/empty set is a subset of every set.

 

3. The least common multiple (LCM) of 16, 30 and 36 is
A. 3
B. 6
C. 240
D. 720

D. 720

Solution

The least common multiple is the smallest positive integer that is divisible by both a and b.

Using factor trees, express the three numbers as products of prime factors:

 

 
$16 = 2\times2\times2\times2 = 2^4$
 
 

$30 = 2\times3\times5$
 

 

 

$36 = 2\times2\times3\times3 = 2^2 \times 3^2$
 

 

The prime factors of the three numbers are 2, 3 and 5.

For each of these of these prime factors, find out its biggest value or power.

For 2, we have 24, 2 and 22. The biggest value of 2 is 24.

For 3, we have 3 and 32. The biggest value of 3 is 32.

For 5, we have 5. The biggest value for 5 is 5.

To find the LCM of 16, 30 and 36, multiply the biggest values of their prime factors together.

=> LCM (16, 30, 36) = 24 × 32 × 5

= 16 × 9 × 5

= 720

 

4. The sum of 5 and $x$ divided by 4 is equal to 3.25. Find the value of $x$.
A. 8
B. 7
C. 2¼
D. – 3$4 \over 13$

A. 8

Solution

The sum of 5 and $x$ is (5 + $x$)

$=> \frac {5 + x}{4} = 3.25$
$=> 5 + x = 4 × 3.25$
$=> 5 + x = 13$
$=> x = 13 – 5$
$=> x = 8$

 

5. The numbers 32, 33, 34, …, …, 42 form a sequence in base 5. Find the missing numbers.
A. 35, 36
B. 30, 31
C. 40, 41
D. 31, 41

C. 40, 41

Solution

First, complete the sequence up to 40, as if we are counting in base 10

So we will have 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42

The new numbers are 35, 36, 37, 38, 39, 40, 41

But base 5 only uses the digits 0, 1, 2, 3 and 4. It does not include the digits 5 and above. This means that 35, 36, 37, 38 and 39 are out.

Therefore the missing numbers are 40 and 41 in base 5

 

6. Write down all the integers in the set A = {-10, -4, 0, ¼, 2½, 45, 100}.
A. {-10, -4, 0, 45, 100}
B. {-10, – 4}
C. {0, 45, 100}
D. {¼, 2½}

A. {-10, -4, 0, 45, 100}

Solution

An integer is a positive or negative whole number.

This means that integers do not include fractions. ¼ and 2½ are therefore out.

The integers in the set are -10, -4, 0, 45, 100

 

7. Find the total cost of 25 pens and 75 books if each pen costs GH¢0.20 and each book costs GH¢0.30.
A. GH¢22.50
B. GH¢23.50
C. GH¢27.50
D. GH¢50.00

C. GH¢27.50

Solution

Number of pens = 25
Cost of each pen = GH¢0.20
=> Cost of pens = 25 × GH¢0.20

= GH¢5.00

Number of books = 75
Cost of each book = GH¢0.30
=> Cost of books = 75 × GH¢0.30

= GH¢22.50

∴ Total cost of 2 pens ad 75 book = GH¢5.00 + GH¢22.50

= GH¢27.50

 

8. Simplify – 27 + 18 – (10 – 14) – (– 2)
A. – 3
B. – 7
C. – 11
D. – 35

A. – 3

Solution

– 27 + 18 – (10 – 14) – (– 2)
= – 27 + 18 – (– 4) – (– 2)
= – 27 + 18 + 4 + 2
= -3

 

9. Arrange the following numbers from the lowest to the highest: 0.5, 3, -5, 0.
A. 0, 0.5, -5, 3
B. 0, -5, 0.5, 3
C. -5, 0, 0.5, 3
D. -5, 0.5, 0, 3

C. -5, 0, 0.5, 3

Solution

Picture a number line.

All negative numbers are less than zero, and all positive numbers are greater than zero.

The lowest number is -5 and the highest is 3.

So, from the lowest to the highest, we have: -5, 0, 0.5, 3

 

10. Find how many pieces of cloth 5½m long that can be cut from a roll of cloth 121m long.
A. 665½
B. 115½
C. 66
D. 22

D. 22

Solution

To find how many pieces of the 5½m cloth can be cut from the 121m cloth, we divide 121m by 5½m.

= $\frac{121}{5\frac12}$

= 121 $÷\,\, 5\frac12$

= $\frac{121}{1} ÷ \frac{11}{2}$

= $\frac{121}{1} × \frac2{11}$

= $\frac{\cancel{121}^{11}}{1} × \frac2{\cancel{11}_1}$

= 11 × 2

= 22

 

11. Find the value of 124.3 + 0.275 + 74.06, correcting your answer to one decimal place.
A. 198.6
B. 198.7
C. 892.0
D. 892.4

A. 198.6

Solution

 

12. Esi and Kwasi are 12 and 8 years old respectively. They share 60 mangoes in the ratio of their ages. How many mangoes does Esi get?
A. 42
B. 40
C. 36
D. 18

C. 36

Solution

Ratio of ages = Esi : Kwasi

= 12 : 8

= 3 : 2

Sum of ratio = 3 + 2

= 5

Number of mangoes to be shared = 60

Esi’s share = $\frac35 \times 60$

= $\frac3{\cancel{5}_{1}} × \frac{\cancel{60}^{12}}{1}$

= 3 × 12 = 36

 

13. It takes 6 students 1 hour to sweep their school compound. How long will it take 15 students to sweep the same compound?
A. 24 minutes
B. 12 minutes
C. 3 hours
D. 2 hours

A. 24 minutes

Solution

6 students = 1 hour (i.e. 60 minutes)

This is inverse proportion, because the higher the number of students, the smaller the time taken.

=> 15 students = $\frac{6}{15} × 60$ minutes

= $\frac{6}{\cancel{15}_{1}} × \frac{\cancel{60}^{4}}{1}$ minutes

= 6 × 4 minutes

= 24 minutes

 

14. A housing agent makes a commission of GH¢103,500 when he sells a house for GH¢690,000. Calculate the percentage of his commission.
A. 15.0%
B. 10.0%
C. 7.5%
D. 5.0%

A. 15.0%

Solution

Commission on sale = GH¢103,500

Amount at which house was sold = GH¢690,000

=> Percentage of commission = $\frac{1\,0\,3\,5\,\cancel{0\,0}}{6\,9\,\cancel{0\,0}\cancel{0\,0}} × 1\cancel{0\,0}$

= $\frac{\cancel{1035}^{15}}{\cancel{69}_1}$

= 15%

 

15. A simple interest of GH¢37,500.00 is earned on an amount of GH¢500,000.00 for 3 years. Find the rate of interest per annum.
A. 20.0%
B. 10.0%
C. 5.0%
D. 2.5%

D. 2.5%

Solution

Simple Interest (I) = $\frac{P × R × T}{100}$

=>R = $\frac{100\,I}{P\,T}$

=>R = $\frac{1\,0\,0\, × \,3\,7\,5\,0\,0}{5\,0\,0\,0\,0\,0\, × \,3}$

=>R = $\frac{1\cancel0\cancel0× \,3\,7\,5\cancel0\cancel0}{5\,0\cancel0\cancel0\cancel0\cancel0 × \,3}$

=>R = $\frac{\cancel{375}^{75}}{\cancel{50}_{10}\, × \,3}$

=>R = $\frac{\cancel{75}^{25}}{10\, × \cancel3_{1}}$

=>R = $\frac{25}{10}$

=>R = 2.5

 

16. Simplify: $\left(8x^2y^3\right)\left(\frac{3}{8}xy^4\right)$
A. $3x^3y^7$
B. $3x^2y^7$
C. $3x^3y^4$
D. $3xy$

A. $3x^3y^7$

Solution

$\left(8x^2y^3\right)\left(\frac{3}{8}xy^4\right)$

= $\left(\cancel8^1x^2\,y^3\,\right)\left(\frac3{\cancel8}_1x\,y^4\right)$

= $(x^2y^3)(3xy^4)$

= $3\times{x^2}\times{x^1}\times{y^4}\times{y^3}$

= $3x^{2+1}y^{3+4}$

= $3x^3y^7$

 

17. The scores of 10 students in an examination are given as follows: 45, 12, 75, 81, 54, 51, 24, 67, 19 and 39. What is the median of the scores?
A. 39
B. 48
C. 51
D. 54

B. 48

Solution

45, 12, 75, 81, 54, 51, 24, 67, 19, 39

= 12, 19, 24, 39, 45, 51, 54, 67, 75, 81

= $\cancel{12}, \cancel{19}, \cancel{24}, \cancel{39}, 45, 51, \cancel{54}, \cancel{67}, \cancel{75}, \cancel{81}$

= $\frac{45 + 51}2$

= $\frac{96}2$

= $48$

 

18. A pie chart is to be drawn from the data in the following table:

Cassava20%
Yam17%
Plantain28%
Maize35%

What will be the value of the angle of the sector for maize?
A. 126.0°
B. 100.8°
C. 72.0°
D. 61.2°

A) 126.0°

Solution

Angle of sector for Maize = $\frac{35}{100}\times360°$

= $\frac{3\,5}{1\,0\cancel0}\times3\,6\cancel0°$

= $\frac{\cancel{3\,5}^7}{\cancel{1\,0}}_2\times3\,6°$

= $\frac{7}{\cancel{2}}_1\times\cancel{3\,6}^{18}°$

= $7\times18$

= 126°

 

19. Eighteen cards are numbered from 11 to 29. If one card is chosen at random, what is the probability that it contains the digit 2?

A. $3 \over 9$

B. $7 \over 18$

C. $5 \over 9$

D. $11 \over 18$

D. $11 \over 18$

Solution

$Probability (E) = {{Number \,of \,outcomes \,for \,E} \over {Total \,number \,of \,outcomes}}$

Outcomes for event that number chosen at radom contains the digit 2 = 12, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29

Number of outcomes for event = 11

Total number of outcomes = 18

$=> Probability (E) = {{11} \over {18}}$

 

20. Find the value of $x$, if $\frac{x}4 + 1 = 5$.
A. 24
B. 20
C. 19
D. 16

D. 16

Solution

$\frac{x}4 + 1 = 5$

$=> \frac{x}4 = 5 – 1$

$=> \frac{x}4 = 4$

$=> \frac{x}4 = \frac41$

$=> x = 4 \times 4$

$=> x = 16$

 

21. Factorize: xy + 5x + 2y + 10
A. (x + 5)(2y + 10)
B. (x + 2)(y + 10)
C. (x + 5)(y + 2)
D. (x + 2)(y + 5)

D. (x + 2)(y + 5)

Solution

xy + 5x + 2y + 10
= x(y + 5) + 2(y + 5)
= (y + 5)(x + 2)

 

22. If x ∈ {2, 3, 4, 5}, find the truth set of 2x + 1 < 8.
A. {2,3,4}
B. {2,3}
C. {3,4}
D. {4,5}

B. {2,3}

Solution

2x + 1 < 8

=> 2x < 8 - 1

=> 2x = 7

=>$\frac{\cancel2^1\text{x}}{\cancel2_1} < \frac72$

=>$\text{x} < \frac72$

=>$\text{x} < 3\frac12$

∴ Truth set = {x : x = 2,3}

 

23. Solve the inequality: 7x – (10x + 3) ≥ – 9
A. x ≥ 2
B. x ≤ 4
C. x ≥ 4
D. x ≤ 2

D. x ≤ 2

Solution

7x – (10x + 3) ≥ – 9

=> 7x – 10x – 3 ≥ – 9

=> 7x – 10x ≥ – 9 + 3

=> -3x ≥ – 6

=> $\frac{\cancel{-3}^1\text{x}}{\cancel{-3}_1} ≥ \frac{\cancel{-6}^2}{\cancel{-3}_1}$

=> x ≤ 2

 

24. Find the rule of the mapping:

A. x→4x-3
B. x→3-4x
C. x→4x+3
D. x→4x+5

C. x→4x+3

Solution

By careful observation, you will realise that:

(1 × 4) + 3 = 7

(2 × 4) + 3 = 11

(3 × 4) + 3 = 15

(4 × 4) + 3 = 19

(5 × 4) + 3 = 23

Therefore, the rule of the mapping is: x→4x+3

 

25. Find the circumference of a circle whose area is equal to 64 π cm2.
A. 32 π cm
B. 16 π cm
C. 8 π cm
D. 4 π cm

B. 16 π cm

Solution

To find the circumference, we need the radius or diameter of the circle

We have been given the area of the circle as 64πcm2, so we can find the radius

Area of a circle = πr2

=> 64π = πr2

=> 64π = πr2

=> $\frac{64\cancel\pi}{\cancel\pi} = \frac{\cancel{\pi}r^2}{\cancel\pi}$

=> $r^2 = 64$

=> $r = \sqrt{64}$

=> $r = 8$cm

Circumference of a circle = 2πr

=> Circumference of the circle = 2 × π × 8cm = 16πcm

 

26. Which of the following geometric figures is the plane shape of a cube?
A. Circle
B. Rectangle
C. Square
D. Triangle

C. Square

 

27. How many lines of symmetry has a rectangle?
A. 4
B. 3
C. 2
D. 1

C. 2

 

28. A rectangular box has length 20 cm, width 6 cm and height 4 cm. Find how many cubes of side 2 cm that will fit into the box.
A. 120
B. 60
C. 30
D. 15

B. 60

Solution

To find how many of the cubes will fit into the rectangular box, we need to divide the volume of the rectangular box with the volume of the cube.

Volume of rectangular box = L × B × H

= 20cm × 6cm × 4cm

= 480cm3

Volume of cube = s3

= 2cm3

= 8cm3

=> Number of cubes that will fit into the box = $\frac{480cm^3}{8cm^3}$

= $\frac{\cancel{480}^{60}\cancel{cm^3}}{\cancel8_1\cancel{cm^3}}$

= 60

 

29. The interior angle of a regular polygon is 120°. How many sides has this polygon?
A. 3
B. 4
C. 5
D. 6

D. 6

Solution

Interior angle of regular polygon = $\frac{180°(n-2)}{n}$

=>120° = $\frac{180°(n-2)}{n}$

=>120°× $n$ = 180°$(n-2)$

=>120°$n$ = 180°$n$ – 360°

=>360° = 180°$n$ – 120°$n$

=>360° = 60°$n$

=>$\frac{360°}{60°}$ = $\frac{60°n}{60°}$

=>$\frac{\cancel{360°}^6}{\cancel{60°}_1}$ = $\frac{\cancel{60°}n}{\cancel{60°}}$

=>n = 6

 

30.

In the diagram above, length of PS = length of SQ and angle SQR = 112°. Find the value of $x$.
A. 68°
B. 56°
C. 46°
D. 44°

D. 44°

Solution

∠PQS + ∠SQR = 180°

=> ∠PQS + 112° = 180°

=> ∠PQS = 180° – 112°

=> ∠PQS = 68°

△PSQ is an equilateral triangle because |PS| = |SQ|. This means that ∠PQS = ∠SPQ = 68°.

For △PSQ,

∠PQS + ∠QSP + ∠SPQ = 180°

=> 68° + $x$ + 68° = 180°

=> $x$ + 136° = 180°

=> $x$ = 180° – 136°

=> $x$ = 44°

 

31. XYZ is a right-angled triangle with length of sides as shown.

Which of the following equations gives the value of z2?
A. z2 = (x2 + y2 )
B. z2 = (x – y)
C. z2 = (y2 – x2 )
D. z2 = (x2 – y2 )

C. z2 = (y2 – x2)

Solution

△XYZ is a right-angled triangle. Using the Pythagorean theorem,

z2 + x2 = y2

=> z2 = y2 – x2

 

32. Express 7 min. 30 sec. as a percentage of 1 hour.
A. 2.5%
B. 7.5%
C. 11.7%
D. 12.5%

D. 12.5%

Solution

Both the 7 min. 30 sec. and the 1 hour have to be converted to seconds

7 min. 30 sec. = (7 × 60 sec.) + 30 sec.

= 420 sec. + 30 sec.

= 450 sec.

1 hour = 60 min.

= 60 × 60 sec.

= 3600 sec.

7 min. 30 sec. as a percentage of 1 hour = $\frac{4\,5\,0}{3\,6\cancel0\cancel0}$ × $1\cancel0\cancel0$

= $\frac{\cancel{450}^{50}}{\cancel{36}_4}$

= $\frac{\cancel{50}^{25}}{\cancel{4}_2}$

= $\frac{5}{2}$

= 2.5%

 

33. The point (4,5) is translated to the point (3,1). What is the translation vector?

A. $\begin{pmatrix}– 1 \\4 \\\end{pmatrix}$

B. $\begin{pmatrix}1 \\4 \\\end{pmatrix}$

C. $\begin{pmatrix}1 \\– 4 \\\end{pmatrix}$

D. $\begin{pmatrix}– 1 \\– 4 \\\end{pmatrix}$

D. $\begin{pmatrix}-1 \\*4 \\\end{pmatrix}$

Solution

Translation vector = $\begin{pmatrix}3 \\1 \\\end{pmatrix}$ – $\begin{pmatrix}4 \\5 \\\end{pmatrix}$

=$\begin{pmatrix}3 – 4\\1 – 5 \\\end{pmatrix}$

= $\begin{pmatrix}– 1 \\– 4 \\\end{pmatrix}$

 

34. In the diagram below, triangle QRT is the enlargement of QST.

Which side of triangle QRT corresponds to side QT of triangle QST?
A. TS
B. TR
C. QR
D. SR

B. TR

Solution

For △QST, the right-angle is at S. Therefore QT is the hypotenuse.

For △QRT, the right-angle is at Q. Therefore TR is the hypotenuse, and corresponds to QT.

 

35.

In the diagrams above Fig. I is an enlargement of Fig. II. Find the side EF of Fig. II
A. 20 cm
B. 5 cm
C. 4 cm
D. 3 cm

D. 3cm

Solution

Scale factor = $\frac{E’H’}{EH}$

= $\frac{20cm}{5cm}$

= $\frac{\cancel{20cm}^4}{\cancel{5cm}_1}$

= 4

To find EF,

4 = $\frac{E’F’}{EF}$

=> 4 = $\frac{12cm}{EF}$

=> 4EF = 12cm

=> $\frac{4EF}{4}$ = $\frac{12cm}{4}$

=> $\frac{\cancel4^1EF}{\cancel4_1}$ = $\frac{\cancel{12}^3cm}{\cancel4_1}$

=> EF = 3cm

 

36. Express 4037 in standard form.
A. 4.037 × 10-4
B. 4.037 × 10-3
C. 4.037 × 103
D. 4.037 × 104

C. 4.037 × 103

 

37. Which of the following angles can be constructed by using the arcs at point C in the diagram below?

A. 30°
B. 45°
C. 60°
D. 75°

B. 45°

 

38. Given that vector a = $\begin{pmatrix}– 5 \\12 \\\end{pmatrix}$ and b = $\begin{pmatrix}10x \\12 \\\end{pmatrix}$, find the value of x if a = b.
A. -2
B. -½
C. ½
D. 2

B. -½

Solution

From the question,

$\begin{pmatrix}– 5 \\12 \\\end{pmatrix}$ = $\begin{pmatrix}10x \\12 \\\end{pmatrix}$

=> $- 5$ = $10x$

=> $\frac{- 5}{10}$ = $\frac{10x}{10}$

=> $\frac{- \cancel5^1}{\cancel{10}_2}$ = $\frac{\cancel{10}^1x}{\cancel{10}_1}$

=> $x$ = $-\frac12$

 

39.

In the diagram above, the bearing of point B from A is
A. 340°
B. 220°
C. 140°
D. 50°

B. 220° Solution
The bearing of point B from A = a + b + c a = 90° b = 90° c = 40° (c and the 40° at point B are alternate angles and therefore equal) => The bearing of point B from A = 90° + 90° + 40°

= 220°

 

40. Ama is 9 years older than Kwame. If Kwame is 18 years old, find the ratio of the age of Kwame to that of Ama.
A. 3 : 2
B. 1 : 3
C. 2 : 3
D. 2 : 1

C. 2 : 3

Solution

Kwame’s age = 18 years

Ama is 9 years older than Kwame

=> Ama’s age = 18 years + 9 years = 27 years

Ratio = Kwame : Ama

= 18 : 27

= 2 : 3