
2019 18

Lecture1.1

Lecture1.2

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MATHEMATICS 29

Lecture2.1

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ENGLISH LANGUAGE 29

Lecture3.1

Lecture3.2

Lecture3.3

Lecture3.4

Lecture3.5

Lecture3.6

Lecture3.7

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Lecture3.19

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Lecture3.21

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INTEGRATED SCIENCE 29

Lecture4.1

Lecture4.2

Lecture4.3

Lecture4.4

Lecture4.5

Lecture4.6

Lecture4.7

Lecture4.8

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SOCIAL STUDIES 29

Lecture5.1

Lecture5.2

Lecture5.3

Lecture5.4

Lecture5.5

Lecture5.6

Lecture5.7

Lecture5.8

Lecture5.9

Lecture5.10

Lecture5.11

Lecture5.12

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Lecture5.19

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Lecture5.24

Lecture5.25

Lecture5.26

Lecture5.27

Lecture5.28

Lecture5.29


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Bece Past Questions & Answers – 2012 (Maths)
2012 BECE Mathematics (Maths) Past Questions – Paper One
1. If P = {2, 3, 5, 7}and Q = {2, 4, 6, 8}, find P∩Q
A. {2}
B. {3}
C. {4}
D. {5}
A. {2}
2. Which of the following numbers is an integer?
A. 5/4
B. 2/3
C. 0.5
D. 1
D. 1
3. Find the Lowest Common Multiple (LCM) of
4. How many diagonals are in a rectangle?
A. 1
B. 2
C. 3
D. 4
B. 2
5. Simplify – 4 (3 – 5) + 10 – 3 (7 + 4) + 30
A. – 1
B. 15
C. 56
D. 65
B. 15
6. An iron rod 15 m long is divided into 12 equal parts. How long is each part?
A. 0.80 m
B. 1.25 m
C. 1.50 m
D. 3.00 m
B. 1.25 m
7. Convert 42 to a base two numeral.
8. Simplify (
9. A tank contains 400 litres of water. If 100 litres is used, what percentage is left?
A. 25%
B. 30%
C. 40%
D. 75%
D. 75%
10. Triangle ABC is a rightangled triangle. Find the length of AC.
A. 1 cm
B. 5 cm
C. 7 cm
D. 12 cm
B. 5 cm
11. Arrange the following fractions in descending order of magnitude:
A. 5/7 ,2/5 ,2/3 ,1/2
B. 5/7 ,2/3 ,1/2 ,2/5
C. 1/2 ,2/5 ,5/7 ,2/3
D. 1/2 ,5/7 ,2/3 ,2/5
B. 5/7 ,2/3 ,1/2 ,2/5
12. Find the image of 3 under the mapping, x → 10 – 2x
A. 4
B. 5
C. 8
D. 16
A. 4
13. Simplify
14. If 2x = 5(x – 2) + 7, find the value of x
The table below shows the day and night temperatures of a town during a week.
Use it to answer Questions 15 and 16
Week day  Temperatures (°C)  
Day  Night  
Monday
Tuesday Wednesday Thursday Friday Saturday Sunday 
33
29 32 34 32 30 30

24
25 23 26 24 24 25 
15. Find, correct to one decimal place, the average day temperature for the week
A. 24.4 °C
B. 30.2 °C
C. 31.4 °C
D. 32.2 °C
C. 31.4 °C
16. On which day was the change in temperature the least?
A. Monday
B. Saturday
C. Sunday
D. Tuesday
D. Tuesday
17. A box contains 30 identical balls of which 16 are white and the rest yellow. If a girl picks a ball at random from the box, what is the probability that it is a yellow ball?
A. 1/16
B. 7/15
C. 8/15
D. 7/8
B. 7/15
18. Find the truth set of
A. {x: x ≤ –3}
B. {x: x ≤ –1}
C. {x: x ≥ 1}
D. {x: x ≥ 3}
C. {x: x ≥ 1}
19. The perimeter of the figure below is 71 cm. Find the diameter of the semicircular portion.
[Take = π]
A. 1.0 cm
B. 3.5 cm
C. 7.0 cm
D. 14.0 cm
D. 14.0 cm
20 Simplify
21. Kojo is 20% heavier than Afua. If Kojo weighs 6 kg, what is Afua’s weight?
A. 4.8 kg
B. 5.0 kg
C. 6.0 kg
D. 7.2 kg
B. 5.0 kg
22. Find the volume of a cylinder of height 3 cm and radius 2 cm.
A. 6π cm3
B. 12π cm3
C. 18π cm3
D. 24π cm3
B. 12π cm3
23. Given the points S(5, 2) and T(3, 2), calculate the gradient of the line ST.
24. Kofi invested GH¢ 150,000 at 2.5% per annum simple interest. How long will it take this amount to yield an interest of GH¢11,250.00?
A. 2 years
B. 3 years
C. 4 years
D. 5 years
B. 3 years
25. Express 3.75 as a mixed fraction.
26. A map is drawn to the scale 1:100,000. What distance in kilometres is represented by 5 cm on the map?
A. 0.5 km
B. 5.0 km
C. 50.0 km
D. 500.0 km
B. 5.0 km
27. Given that , find r – 2s
28. Esi went to the market and bought 500 g of meat, 850 g of fish and 900 g of eggs. What is the total weight of the items she bought in kilograms?
A. 2.20 kg
B. 2.25 kg
C. 2.35 kg
D. 22.50 kg
B. 2.25 kg
29. A watch gains minutes per hour. What is the total time gained from 12 noon to 12 midnight in a day?
A. 9 minutes
B. 15 minutes
C. 18 minutes
D. 36 minutes
C. 18 minutes
30. A printing machine prints 600 books in 3 hours. How many books will the machine print in 5 hours?
A. 360 books
B. 1000 books
C. 1800 books
D. 3000 books
B. 1000 books
31. The bearing of Atoru from Busase is 275°. What is the bearing of Busase from Atoru?
A. 180°
B. 175°
C. 095°
D. 075°
C. 095°
32. In a class of 24 pupils, 10 study French only and 8 study English only. If each pupil studies at least one of the two subjects, how many study English?
A. 12
B. 14
C. 16
D. 18
B. 14
33. Convert 84 to a base five numeral.
34. In the diagrams below, triangle A1B1C1 is an enlargement of triangle ABC. Determine the scale factor.
A. 0.50
B. 0.75
C. 2.00
D. 4.00
C. 2.00
35. Find the least number that must be added to 308 to make it divisible by 19.
A. 4
B. 7
C. 15
D. 18
C. 15
36. In a school of 940 pupils, the number of girls exceeds the number of boys by 150. How many girls are there in the school?
A. 620
B. 545
C. 470
D. 395
B. 545
37. Which of the following fractions is equivalent to ?
In the diagram below, line PQ is parallel to RS and UV is a line drawn through PQ and RS.
Use the diagram to answer Questions 38 and 39.
38. Find angle a.
A. 35°
B. 55°
C. 135°
D. 145°
D. 145°
39. Angle b and angle c are
A. alternate angles
B. vertically opposite angles
C. corresponding angles
D. interior opposite angles
C. corresponding angles
40. Expand – x(3 – 2x)
2012 BECE Mathematics (Maths) Past Questions – Paper Two
1.
(a) Evaluate , leaving the answer in standard form.
(b) An amount of GH¢4,200.00 was shared between Aba and Kwame. If Aba had 5/7 of the amount,
(i) how much did Kwame receive?
(ii) what percentage of Aba’s share did Kwame receive?
(c) Find the value of x in the diagram below.
1. (a)
Method 1 (Converting to common fractions)
1. (a)
Method 2 (Converting to whole numbers)
Multiplying both numerator and denominator by 100000
(to convert to whole numbers)
1. (b) (i) Method 1
Therefore Kwame’s share = GHc 4,200 – GHc 3,000
= GHc 1,200
Method 2
Percentage of Aba’s share that Kwame received
1.(c) (4x – 30°) + (2x + 30°) + (x +10°) = 360°
4x + 2x + x – 30° + 30° + 10° = 360°
7x + 10° = 360°
7x = 360° – 10°
7x = 350°
7x/7 = 350/7
x = 50°
2. (a) A car consumes a gallon of petrol for every 30 km drive. The driver of the car set out on a journey of 420 km with 10 gallons of petrol in the fuel tank.
(i) How many more gallons of petrol will be needed to complete the journey?
(ii) Find the cost of the petrol used for the journey of 420 km if a gallon of petrol costs GH¢5.50
(b) The average number of spectators at a football competition for the first five days was 3,144. The attendance on the sixth day was 3,990. Find the
(i) The total attendance on the first five days
(i) The average attendance for the 6 days
(c) The area enclosed by a square garden is 121 m2. What is the distance around the garden?
2. (a) (i) Let g = no. of gallons needed for the entire journey
Approach 1 (Equivalent Fractions)
Distance (km)  Petrol (Gallon)  
30  1  
The entire journey  420  g 
He already has 10 gallons; therefore he needs 4 gallons more
2. (a) (i)
Approach 2 (The rule: If more, less divides …)
If 30 km → 1
Then 420 km → ? (more)
If more, less (i.e., 30) divides; ∴ we have
Amount needed for the entire journey = 14 gallons
He already has 10 gallons, therefore he needs 4 gallons more
2. (a) (i)
Approach 3 (Equivalent ratios)
He already has 10 gallons, therefore he needs 4 gallons more
2. (a) (ii) Let THE COST OF PETROL USED FOR THE JOURNEY = c
Approach 1 (Common Logic)
If 1 gallon → GHc 5.50
Then 14 gallons → 14 × GHc 5.50 = GHc 77.00
Approach 2 (Equivalent Fractions)
Petrol (gallons)  Cost (GHc)  
1  5.50  
The entire journey  14  c 
Cost of petrol needed for the journey = GHc 77.00
2. (b) (i) Total attendance on first five days = 5 × average attendance
= 5 × 3144
= 15720 spectators
(ii)
= 3285 spectators
2. (c)
The distance around the garden (perimeter) = 4 × 11 m
= 44 m
3. (a) The table below shows the number of students who scored more than 80% in the listed subjects
Subject  Number of students 
Biology
Physics Chemistry French Geography History 
26
30 32 38 24 30 
(i) Draw a pie chart for the distribution
(ii) What is the probability that a student chosen at random from the distribution offers Chemistry?
(b) A woman bought 210 oranges for GH¢7.50. She sold all of them at 3 for 15 Gp. Find the
(i) total selling price of the oranges
(ii) percentage profit
3. (a) (i)
Subject  No. of students  Angle of sector 
Biology 
26


Physics  30  
Chemistry 
32


French  38  
Geography  24  
History  30  
TOTAL 
180 
360° 
Pie chart showing the number of students who scored more than 80% in the listed subjects
3 (a) (ii)
Approach 2 (using angle of sector)
3 (b) (i)
Approach 1 (Unitary Approach – the value of one item)
If 3 oranges sell for 15 Gp,
then 1 orange sells for = 5Gp
hence, 210 oranges would sell for 210 × 5 Gp
= 1050 Gp = GHc 10.50
Approach 2 (Equivalent Fractions)
No. of oranges  SP (in Gp) 
3  15 
210  y 
Approach 3 (The rule: If more, less divides …)
If 3 → 15 Gp
Then 210 → ? (more)
If more, less (i.e., 4) divides; ∴ we have
= 1050 Gp = GHc 10.50
Approach 4 (Equivalent ratios)
3 : 15 = 210 : y
= 1050 Gp = GHc 10.50
3. (b) (ii) Profit = Total SP – Total CP
= GHc 10.50 – GHc 7.50
= GHc 3.00
4. The marks scored by some students in a Mathematics test are as follows:
3 3 5 6 3 4 7 8 3 4
5 4 7 4 3 7 4 6 4 8
4 5 6 3 8 4 5 6 4 5
(a) Construct a frequency distribution table for the scores.
(b) Using the table, find for the distribution, the
(i) mode;
(ii) mean, correct to one decimal place
(iii) median
4. (a) Frequency distribution table
Mark (x)  Tally  Frequency (f) 
3  ∕∕∕∕ ∕  6 
4  ∕∕∕∕ ∕∕∕∕  9 
5  ∕∕∕∕  5 
6  ∕∕∕∕  4 
7  ∕∕∕  3 
8  ∕∕∕  3 
4. (b) (i) Mode = The mark with the highest frequency = 4
(ii) Mean
Mark (x)  Tally  Frequency (f) 
fx 
3  ∕∕∕∕ ∕  6  18 
4  ∕∕∕∕ ∕∕∕∕  9  36 
5  ∕∕∕∕  5  25 
6  ∕∕∕∕  4  24 
7  ∕∕∕  3  21 
8  ∕∕∕  3  24 
Σf = 30  Σfx = 148 
4.(b) (iii) Median
Approach1
Listing all the given marks in ascending order, we have
3,3,3,3,3,3, 4,4,4,4,4,4,4,4, 4, 5, 5,5,55, 6,6,6,6, 7,7,7, 8,8,8
The middle marks are 4 and 5
Approach 2
The median
= the mark at the ½(Σf+1)th position
⇒ the ½ (30+1)th position
= ½ (31) = 15½th position
The mark at the 15½th position
= the average of the marks at the 15th and 16th positions
= = = 4.5 or 4
5. (a) (i) Find the least Common Multiple (L.C.M.) of 9, 18 and 16.
(ii) Arrange
in ascending order of magnitude
(b) Using a ruler and a pair of compasses only,
(i) construct a triangle PQR with length PQ = 10 cm, angles QPR = 45° and PQR = 60°.
(ii) Construct the perpendicular bisectors of PR and RQ to meet at T.
(iii) Measure the length of TP.
(a) (i) LCM of 9, 18 and 16
Approach 1 (Successive division by prime factors)
2  9  18  16 
2  9  9  8 
2  9  9  4 
2  9  9  2 
3  9  9  1 
3  3  3  1 
1  1  1 
The LCM = 2 × 2 × 2 × 2 × 3 × 3
= 16 × 9 = 144
(a) (i) Approach 2 (Listing multiples)
9 = {9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, 108, 117, 126, 135, 144, …}
18 = {18, 36, 54, 72, 90, 108, 126, 144, …}
16 = {16, 32, 48, 64, 80, 96, 112, 128, 144, …}
Hence the Least Common Multiple (LCM) = 144
5. (a) (ii) in ascending order of magnitude
Therefore in ascending order, we have
5. (b) (i), (ii)
The length of TP = 5.2 cm (approx.)
6. (a) (i) Using a scale of 2 cm to 1 unit on both axes, draw two perpendicular axes Ox and Oy on a graph sheet.
(ii) Mark on the same graph sheet, the xaxis from 5 to 5 and yaxis from 6 to 6.
(iii) Plot the points P(4, 2), Q(2, 5) and R(2, 2). Join the points P, Q, R to form a triangle PQR
(iv) Using the xaxis as a mirror line, draw the image P1Q1R1 of the triangle PQR such that P→P1, Q→Q1, R→R1
(v) Write down the coordinates of P1, Q1 and R1.
(vi) Translate triangle PQR by the vector such that P→P2, Q→Q2, R→R2
(vii) Label the vertices of triangle P2Q2R2
(a)
Approach 1 (By Inspection / Construction)