
2019 18

Lecture1.1

Lecture1.2

Lecture1.3

Lecture1.4

Lecture1.5

Lecture1.6

Lecture1.7

Lecture1.8

Lecture1.9

Lecture1.10

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Lecture1.12

Lecture1.13

Lecture1.14

Lecture1.15

Lecture1.16

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MATHEMATICS 29

Lecture2.1

Lecture2.2

Lecture2.3

Lecture2.4

Lecture2.5

Lecture2.6

Lecture2.7

Lecture2.8

Lecture2.9

Lecture2.10

Lecture2.11

Lecture2.12

Lecture2.13

Lecture2.14

Lecture2.15

Lecture2.16

Lecture2.17

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Lecture2.19

Lecture2.20

Lecture2.21

Lecture2.22

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Lecture2.25

Lecture2.26

Lecture2.27

Lecture2.28

Lecture2.29


ENGLISH LANGUAGE 29

Lecture3.1

Lecture3.2

Lecture3.3

Lecture3.4

Lecture3.5

Lecture3.6

Lecture3.7

Lecture3.8

Lecture3.9

Lecture3.10

Lecture3.11

Lecture3.12

Lecture3.13

Lecture3.14

Lecture3.15

Lecture3.16

Lecture3.17

Lecture3.18

Lecture3.19

Lecture3.20

Lecture3.21

Lecture3.22

Lecture3.23

Lecture3.24

Lecture3.25

Lecture3.26

Lecture3.27

Lecture3.28

Lecture3.29


INTEGRATED SCIENCE 29

Lecture4.1

Lecture4.2

Lecture4.3

Lecture4.4

Lecture4.5

Lecture4.6

Lecture4.7

Lecture4.8

Lecture4.9

Lecture4.10

Lecture4.11

Lecture4.12

Lecture4.13

Lecture4.14

Lecture4.15

Lecture4.16

Lecture4.17

Lecture4.18

Lecture4.19

Lecture4.20

Lecture4.21

Lecture4.22

Lecture4.23

Lecture4.24

Lecture4.25

Lecture4.26

Lecture4.27

Lecture4.28

Lecture4.29


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

Lecture5.13

Lecture5.14

Lecture5.15

Lecture5.16

Lecture5.17

Lecture5.18

Lecture5.19

Lecture5.20

Lecture5.21

Lecture5.22

Lecture5.23

Lecture5.24

Lecture5.25

Lecture5.26

Lecture5.27

Lecture5.28

Lecture5.29


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Bece Past Questions & Answers – 2015 (Maths)
2015 BECE Mathematics (Maths) Past Questions – Paper One
A. {2, 3, 5}
B. {2, 6, 10}
C. {3, 5, 15}
D. {3, 6, 15}
1. A. {2, 3, 5}
2. Given that set P = {m, n, o, p}, find the number of subsets of P.
A. 4
B. 8
C. 10
D. 16
2. D. 16
3. If M = {multiples of 4 between 10 and 25} and N = {even numbers between 11 and 23}, find M ∪ N
A. {12, 16, 20}
B. {14, 18, 22}
C. {12, 14, 16, 18, 22}
D. {12, 14, 16, 18, 20, 22, 24}
3. D. {12, 14, 16, 18, 20, 22, 24}
4. What is the place value of 7 in 24.376 ?
A. Unit
B. Ten
C. Tenth
D. Hundredth
4. D. Hundredth
5. Find the Highest Common Factor of 24, 42 and 72
A. 4
B. 6
C. 7
D. 12
5. B. 6
6. Express as a number in base 10
A. 25
B. 27
C. 32
D. 35
6. D. 35
7. If p × q × r = 1197, and p = 19, q = 3, find r
A. 21
B. 49
C. 57
D. 61
7. A. 21
8. How many integers are within the interval – 5 < x < 7 ? A. 10
B. 11
C. 12
D. 13
8. B. 11
9. Divide 1.612 by 0.4
A. 4.3
B. 4.03
C. 0.403
D. 0.43
9. B. 4.03
10. Arrange the following fractions in ascending order: 5/8, 11/20, 7/10
A. 5/8, 11/20, 7/10
B. 7/10, 5/8, 11/20
C. 11/20, 5/8, 7/10
D. 5/8, 7/10, 11/20
10. C. 11/20, 5/8, 7/10
11. Abena spent 1/5 of her money on sweets, 4/7 on provisions and the rest on gari. What fraction of her money did she spend on gari?
A. 27/35
B. 13/35
C. 8/35
D. 5/35
11. C. 8/35
12. If 5 boys took 14 days to cultivate a piece of land, how long will it take 7 boys working at the same rate to cultivate the land ?
A. 14 days
B. 12 days
C. 10 days
D. 8 days
12. C. 10 days
13. A man invested GHC 800.00 in a bank at a simple interest rate of 5% per annum. Find his total amount in the bank at the end of one year.
A. GHC 840.00
B. GHC 860.00
C. GHC 900.00
D. GHC 960.00
13. A. GHC 840.00
14. John sold a car for GHC 60,000.00 and made a profit of 20%. What is the cost price of the car?
A. GHC 48,000.00
B. GHC 50,000.00
C. GHC 72,000.00
D. GHC132,000.00
14. B. GHC 50,000.00
15. What is the value of x if 10x = 1000?
A. 1
B. 2
C. 3
D. 4
15. C. 3
16. Express 625.13 in standard form
17. Find the median of the numbers 17, 12, 15, 16, 8, 18, 13 and 14
A. 8
B. 12
C. 14.5
D. 15.5
17. C. 14.5
18. The ages in years of 10 children at a party are 2, 3, 3, 3, 4, 4, 5, 5, 5 and 6. If a child is chosen at random, what is the probability that he / she is not less than 5 years old ?
A. 2/3
B. 2/5
C. 3/10
D. 1/2
18. B. 2/5
19. Expand (2x + y) (2x – y)
20. Find the value of n, if 25.003 = (2 × 10) + (5 × 1) + (3 × n)
A. 0.001
B. 0.011
C. 0.01
D. 0.1
20. A. 0.001
21. Evaluate when m = 2.
A. 12
B. 18
C. 20
D. 24
21. D. 24
22. A wrist watch is priced GHC 2,000.00. A shopkeeper allows a discount of 2% on the cost price. Find the discount on 20 of such wrist watches.
A. GHC 500.00
B. GHC 600.00
C. GHC 800.00
D. GHC 1,000.00
22. C. GHC 800.00
23. Find the value of m, if 4(m + 4) = – 8.
A. – 6
B. – 2
C. 2
D. 6
23. A. – 6
24. Find the rule for the following mapping
A. y→x+2
B. y→2x
C. y→x2
D. y→2x+2
24. C. y→x2
25. How many vertices has a cuboid?
A. 6
B. 7
C. 8
D. 14
25. C. 8
26. The circumference of a circle is 440 m. Find the area of the circle. [Take π = 22/7]
A. 14,400 m2
B. 15,400 m2
C. 16,400 m2
D. 18,000 m2
26. B. 15,400 m2
27. What name is given to a triangle which has all its sides equal?
A. Isosceles triangle
B. Scalene triangle
C. Equilateral triangle
D. Rightangle triangle
27. C. Equilateral triangle
28. At eight o’clock, which of the following is the angle between the hour and the minute hands of the clock?
A. 150°
B. 120°
C. 90°
D. 60°
28. B. 120°
29. A rectangular field 50 m wide and y m long requires 260 m of fencing. Find y.
A. 15 m
B. 40 m
C. 80 m
D. 105 m
29. C. 80 m
30. Which of the following best describes the statement: ‘The locus of a point which moves so that its distance from two fixed points are always equal’?
A. Bisector of an angle
B. Perpendicular bisector
C. Circle
D. Two parallel lines
30. B. Perpendicular bisector
31. The point K (1, 5) is rotated through 90° anticlockwise about the origin. Find the coordinates of the image of K.
A. (5, 1)
B. (5, 1)
C. (1, 5)
D. (1, 5)
31. B. (5, 1)
32. Kwame is facing west. Through how many degrees should he turn anticlockwise to face north?
A. 90°
B. 180°
C. 270°
D. 360°
32. C. 270°
33. Given that vectors , find 2v – u
A.
B.
C.
D.
D.
34.
What is the name of the figure above?
A. Cuboid
B. Kite
C. Triangle
D. Pyramid
34. D. Pyramid
13  12  17 
E  F  10 
11  16  G 
Use the magic square above to answer questions 35 to 37
35. Find the value of F
A. 14
B. 15
C. 18
D. 23
35. A. 14
36. Find the value of E.
A. 14
B. 15
C. 18
D. 23
36. C. 18
37. Evaluate E + G
A. 29
B. 30
C. 33
D. 38
37. C. 33
38. The hypotenuse and a side of a rightangled triangle are 13 cm and 5 cm respectively. Find the length of the third side.
A. 8 cm
B. 9 cm
C. 12 cm
D. 17 cm
38. C. 12
39. Find the missing number in the sequence below:
11, 16, 22, 29, __, 46, 56
A. 30
B. 36
C. 37
D. 39
39. C. 37
40. A hall which is 20 m long is represented on a diagram as 10 cm long. What is the scale of the diagram?
A. 1:200
B. 1:250
C. 1:400
D. 1:500
40. A. 1:200
2015 BECE Mathematics (Maths) Past Questions – Paper Two
1. (a) Find the difference between the product of 2.5 and 7.5 and the sum of 2.75 and 9.55.
(b) Solve
(c) A container is 24 m long, 9 m wide and 8 m high. How many books can it hold if each book is 20 cm long, 16 cm wide and 6 cm thick.
(a) Product of 2.5 and 7.5
Sum of 2.75 and 9.55
=
2.75
+ 9.55
__________
12.30
__________
Difference between 18.75 and 12.30
= 18.75
– 12.30
__________
6.45
__________
(b) Solving
(c) Volume of container = length × width × height
= 24m × 9m × 8m
= 2400cm × 900cm × 800cm
= 1728000000 cm3
Volume of each book = 20 cm × 16 cm × 6 cm
= 320 cm2 × 6 cm
= 1920 cm3
2. (a) In a test consisting of 90 questions, Ama answered 75% of the first 40 questions correctly. If she had to get a score of 80% in the test,
(i) how many questions did she answer correctly out of the first 40 questions?
(ii) how many questions should she answer correctly out of the 90 questions ?
(iii) what percentage of the remaining 50 questions should she answer correctly in order to get the 80%?
(b) Three interior angles of a pentagon are 100°, 120° and 108°. Find the size of each of the remaining two interior angles, if one of them is three times the other.
2(a) (i) No.of questions Ama answered correctly out of first 40 questions
*** QuickLaTeX cannot compile formula: = 75\text{% of first 40 questions} = \frac{75}{100}\times40 = \frac{75\times4}{10} = 30 \ \ questions *** Error message: File ended while scanning use of \text@. Emergency stop.
(ii) To score 80% in the test, then she needs to answer
*** QuickLaTeX cannot compile formula: = 80 \text{ % x 90 questions } = \frac{80}{100}\times90 = 8 × 9 \ = 72 \ questions \ correctly *** Error message: File ended while scanning use of \text@. Emergency stop.
(iii) No. of questions she must answer correctly in the remaining 50 questions
= 72 – 30 questions
= 42 questions
Percentage of 42 out of 50 questions
=
= 42 × 2%
= 84%
(b) Sum of interior angles of a pentagon (5sided polygon)
= (n – 2) × 180° , where n = no.of sides
= (5 – 2) × 180° [n = 5 sides]
= 3 × 180°
= 540°
Let size of smaller missing angle = x
then, size of bigger missing angle = 3x
Now, if sum of interior angles = 540°,
⇒ 100° + 120° + 108° + x + 3x = 540°
⇒ 328° + 4x = 540°
⇒ 4x = 540° – 328°
⇒ 4x = 212°
⇒ x = 212/4
⇒ x = 53°
Hence, the other missing angle = 3x
= 3 × 53°
= 159°
The sizes of the two remaining interior angles = 53° and 159°
3. (a) Given that vectors
(i) q if q – p = ;
(ii) the magnitude of the vector q – p
(b)
(c) In the diagram AB = AC, angle ADC = 30° and angle ACD = 7x – 25°. Find
the value of x;
(ii) angle DAC;
(iii) angle BAD.
(a)
(ii) Magnitude of vector q – p
(b) (i) Since AB = AC
⇒ angle ABC = angle ACB [Base angles of isosceles triangle equal]
Let angle ABC = angle ACB = y
Then, y + y + 50° = 180° [interior angles of a triangle =180°]
⇒ 2y = 180° – 50°
⇒ 2y = 130°
⇒ y = 130/2
⇒ y = 65°
Now, 65° + (7x – 25°) = 180° [angles at a point on a straight line=180°]
⇒ 7x + 65° – 25° = 180°
⇒ 7x + 40 = 180°
⇒ 7x = 180° – 40°
⇒ 7x = 140°
⇒ x = (140°)/7
⇒ x = 20°
(ii) Angle DAC + 7x – 25° + 30° = 180° [interior angles of a triangle =180°]
Let angle DAC = a
⇒ a + 7x – 25° + 30° = 180°
⇒ a + 7(20°) – 25° + 30° = 180°
⇒ a + 140° – 25° + 30° = 180°
⇒ a + 115° + 30° = 180°
⇒ a + 145° = 180°
⇒ a = 180° – 145°
⇒ a = 35°
⇒ angle DAC = 35°
(iii) angle BAD = angle BAC + angle DAC
= 50° + 35°
= 85°
4. (a) The Value Added Tax (VAT) paid by a man on a deep freezer was GHC 90.00. If VAT was charged at 15%,
what was the price of the deep freezer?
How much did the man pay including VAT?
(b) The average of the numbers 5, 7, 2, 6, x, (x+1), 7 and 4 is 5. Find the value of x.
(c) Simplify:
(a) (i) If (VAT) 15% → GH¢ 90.00
Then (Original price) 100% → ? (more)
If more, less (15%) divides, hence
=
*** QuickLaTeX cannot compile formula: \frac{100%}{15%}\times GHc\ 90 *** Error message: File ended while scanning use of \frac . Emergency stop.
= 100 × GHc 6
= GHc 600
Original price = GHc 600.00
(ii) Total amount paid = Original price + VAT
= GHc 600.00 + GHc 90.00
= GHc 690.00
(b) If the average of 8 no.s: 5, 7, 2, 6, x, (x+1), 7 and 4 = 5, then
(c) Simplification of
5. (a) A cylinder which has a height of 90 cm and diameter 14 cm is closed at both ends.
Find:
(i) its total surface area;
(ii) the volume of the cylinder
[Take π = 22/7]
(b) (i) Using a ruler and a pair of compasses only, construct triangle PQR such that
PQ = 8cm, angle PQR = 120° and QR = 6 cm.
(ii) Measure:
(α) PR;
(β) angle QPR
5(a) (i) h = 90cm, d = 14 cm,
⇒ r = 14cm÷2
r = 7cm
Total Surface Area of closed cylinder
= 2πr2 + 2πrh , where r = radius, h = height
(ii) Volume of cylinder
(b) (i)
(ii) Measure:
(α) PR = 12.1 cm (± 0.1cm)
(β) angle QPR = 25° (± 1°)
6. The table shows the distribution of grades of candidates in an examination.
Grade  1  2  3  4  5  6 
Frequency  2  3  6  5  4  10 
(a) Using a graph sheet, draw a bar chart for the distribution
(b) If all candidates who obtained grades above grade 3 were awarded credit, find the probability that a candidate selected at random obtained credit.
(c) Calculate, correct to the nearest whole number, the mean grade of the candidates.
(a) Bar chart for the frequency distribution table below
G
Grade  1  2  3  4  5  6 
Frequency  2  3  6  5  4  10 
(b) Number of candidates who obtained credit (grades above grade 3 for the distribution)
= Frequencies of Grade 1 and Grade 2
= 2 + 3
= 5
Total number of candidates = 2 + 3 + 6 + 5 + 4 + 10
= 30
Probability of selecting a candidate who obtained credit
= (No.of candidates who obtained credit)/(Total no.of candidates)
= 5/30 = 1/6
(c)
JUNE 2015 (Second Sitting)
MATHEMATICS 2
ESSAY
1 hour
1. (a) In a class of 70 students, 40 belong to the Red Cross Society, 27 belong to the Girls’ Guide Society and 12 belong to both the Red Cross Society and the Girls’ Guide Society. The remaining students do not belong to any of the two societies.
Illustrate the information on a Venn diagram
How many students belong to the Red Cross Society only?
How many students do not belong to any of the two societies?
(b) A farmer uses 1/3 of his land to plant cassava, 2/5 of the remaining land to plant maize and the rest vegetables. If vegetables cover an area of 10 acres, what is the total area of the farmer’s land.
(a) (i) Let n(R) = Number of Red Cross Society members
n(G) = Number of Girls’ Guide Society members
U = Universal set (total number in class)
x = Remaining students (not belonging any of the two)
(ii) No. of students in Red Cross Society only
= 40 – 12
= 28
(iii) No. of students who do not belong to any of the two societies
= 70 – (28 + 12 + 15)
= 70 – 55
= 15
(b)
2. (a) Solve for x, if
(b) At a rally attended by 520 people, 30% were Fantes, 25% Ewes, 15% Nzemas, 20% Gas and the rest Gonjas
How many Gonjas were at the rally?
How many more Fantes than Nzemas were at the rally ?
Draw a pie chart to illustrate the information.
(a)
(b) (i) Percentage of Gonjas = Total percentage – Sum of other percentages
= 100% – (30%+25%+15%+20%)
= 100% – 90%
= 10%
Number of Gonjas at the rally = 10% × Total number at rally
= 10% × 520
= 10/100 × 520
= 52
(ii)
Number of Fantes = 30 % of 520
= 30/100 × 520
= 3 × 52
= 156
Number of Nzemas = 15% of 520
= 15/100 × 520
= 3/2 × 52
= 78
Number of Fantes more than Nzemas = No. of Fantes – No.of Nzemas
= 156 – 78
= 78
(iii)
3. (a) Mr. Mensah’s farm is 20 km from his house. He uses his car to travel y km of the distance from his house and then walks hours at the rate of 3 km per hour to get to his farm. Find y
(b) The perimeter of a square field is the same as that of a rectangular field. If the length of the rectangular field is 8 km and the width is 5 km, calculate the area of the square field.
(c) Find the gradient of the line which passes through the points.
M (2, 1) and K (3, 6)
(a) Distance travelled by car = y km
Distance walked = one and half hr × 3 km / h
= 4.5 km
Total distance = 20 km
Dist.by car + Dist. walked = Total distance
y + 4.5 = 20
y = 20 – 4.5
= 15.5
(b)
Perimeter of square = Perimeter of rectangle
= 2 × length + 2 × width
= 2 × 8 km + 2 × 5 km
= 16 km + 10 km
= 26 km
Hence perimeter of square = 26 km
Now perimeter of square = 4 ×Length
Length of square = 26 km ÷ 4
= 6.5 km
Now, area of square = Length × length
= 6.5 × 6.5
= 42.25 cm2
(c) Gradient of line =
4. (a) Using a ruler and a pair of compasses only, construct triangle PST with angle PST = 30°, ST = 9 cm and PS = 12 cm
(b) With T as centre, draw a circle of radius 6 cm
(c) Construct the perpendicular bisector (mediator) of line PS.
(d) Label the intersections of the circle and the mediator Q1 and Q2.
(e) (i) Measure Q1Q2
(ii) Measure angle Q1TQ2
4. (a)
(e) (i) Q1Q2 = 11.3cm (± 1cm)
(ii) angle Q1TQ2 = 145° (± 1°)
5. (a) Anita bought 51 tubers of yam at 3 for GHC 10.00. If she sold them and made a loss of 40%, how much did she sell each tuber of yam?
(b) The volume of a cylinder closed at one end is 1056 cm3. If its height is 21 cm, find its
(i) diameter
(ii) total surface area.
(a) Cost Price of the 51 tubers =
= 17 × GHc 10
= GHc 170
Selling Price percentage = 100% – 40% = 60%
If (Cost Price) 100% → GHc 170
Then (Selling Price) 60% → ? (less)
If less, then more (100) divides
⇒
⇒ 6 × GHc 17
⇒ GHc 102
Hence selling price of all 51 tubers = GHc 102
⇒ Selling price of each tuber = GHc 102 ÷ 51
= GHc 2.00
(b) (i)
(ii)
6. (a) (i) Copy and complete the following mapping:
x 1 2 3 4 5 6 7
↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓
y 5 7 9 – – – 17
(ii) Determine the rule for the mapping
(b) Draw two perpendicular axes Ox and Oy on a graph sheet.
(c) Using a scale of 2cm to 1 unit on the xaxis and 2cm to 2 units on the yaxis, mark the xaxis from 0 to 8 and the yaxis from 0 to 20.
(d) Plot the point for each ordered pair (x, y) and join them with a straight line.
(e) Find:
(i) y, when x is 0;
(ii) x, when y is 14
6. (a) (i)
x 1 2 3 4 5 6 7
↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓
y 5 7 9 11 13 15 17
(ii)
Constant = 3 (The image of 0)
Hence from general equation y = mx + c, where m = gradient, c = constant
Equation = y = 2x+3
Therefore Rule = x → 2x + 3
(b), (c), (d)
(e) (i) When x is 0, y = 3
(ii) When y is 14, x = 5.5