2009 BECE Mathematics (Maths) Past Questions – Paper Two
1. (a) Simplify: and write your answer in standard form
(b) A plot of land measures 25m by 12m. A portion of this plot measuring 8m by 8m is used for the cultivation of vegetables. Find the area of the plot not cultivated
(c) The table below shows the performance of Aisha in her final examination.
Subject | Score |
English Language | 54% |
Mathematics | 36% |
Ga | 68% |
Science | 50% |
Social Studies | 32% |
Draw a pie chart to represent this information
1 (a) (1200×1260)/800 = (12×1260)/8
= 1890
= 1.89 × 103
1 (b)
Total area of land = 25 m × 12 m = 300 m2
Area cultivated = 8 m × 8 m = 64 m2
Area not cultivated = Total area – Area cultivated
= 300 m2 – 64 m2
= 236 m2
1 (c)
Subject | Score | Angles |
English Language | 54% | |
Mathematics | 36% | |
Ga | 68% | |
Science | 50% | |
Social Studies | 32% | |
TOTAL | 240 | 360 |
Pie chart showing the performance of Aisha in her final examination
2. The table below shows the scores of some students in an examination.
–
Scores | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
Frequency | 3 | 5 | 3 | 2 | 7 | 6 | 5 | 4 | 2 | 2 | 1 |
From the table find
(a) how many students wrote the examination
(b) the modal score
(c) the number of students that scored 7 or more
(d) the mean score correct to one decimal place
2 (a) 3 + 5 + 3 + 2 + 7 + 6 + 5 + 4 + 2 + 2 + 1 = 40
? 40 students wrote the examination
2(b) The modal score is the score that occurs most
(i.e., has the highest frequency)
? Modal score = 4
2(c) 4 + 2 + 2 + 1 = 9
? 9 students scored 7 or more.
2(d) Method 1
The Mean =
=
=
= = or 4.425
2(d) Method 2
Scores
(x) Frequency
(f)
f x
0 3 0
1 5 5
2 3 6
3 2 6
4 7 28
5 6 30
6 5 30
7 4 28
8 2 16
9 2 18
10 1 10
Sf=40 Sfx=177
The mean = = = or 4.425
3. (a) (i) Using a scale of 2cm 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 x-axis from -5 to 5 and y-axis from -6 to 6
(iii) Plot the points A(2,5), B(2,2) and C(4,2). Join the points A, B and C to form a triangle ABC
(iv) Using the y-axis as mirror line, draw the image triangle A1B1C1 of the triangle ABC such that A?A1, B?B1 and C?C1.
Write down the coordinates of A1, B1 and C1
(v) Draw the image triangle A2B2C2 of triangle ABC under anticlockwise rotation of 180o about the origin where A?A2, B?B2 and C?C2.
Write down the coordinates of A2, B2 and C2
(b) Given that
Evaluate 2a – 3c + b
3(a)
Approach 1 (By Inspection / Construction)
3(a)
Approach 2 (The rule / formula)
3 (a) (iv) Reflecting (x, y) in the y-axis
?
?
?
? Plot and join A1(-2,5), B1(-2,2) and C1(-4,2) as the image of triangle ABC under a reflection in the y axis as shown above.
3 (a) (v) Rotating (x, y) through 180° about the origin
? Plot and join A2(-2,-5), B2(-2,-2) and C2(-4,-2) as the image of triangle ABC under a rotation through 180° about the origin as shown above.
3 (b)
2 a – 3 c + b
? 2 – 3 +
? – +
? – +
?
?
?
4. (a) The ratio of sheep to goats on a farm is 4:7. If there are 1,428 sheep, find how many goats are on the farm.
(b) Using a ruler and a pair of compasses only, construct a triangle ABC with |AB| = 6cm, |AC| = 8cm and angle BAC = 30°.
Construct the bisector of angle ACB to meet line AB at D.
(i) Measure |AD| and |BD|
(ii) Write down the ratio |AD|:|BD|.
4
(a) Method 1 (Unitary Approach – the value of one item)
If the sheep ratio, 4, corresponds to 1428,
then ratio, 1, corresponds to
hence, goat ratio, 7, corresponds to = 2499 goats
4
(a) Method 2 (Equivalent Fractions)
Ratio Actual no.
Sheep 4 1428
Goats 7 g
From the table above,
?
?
?
? = 2499 goats
4 (a) Method 3 (The rule: If more, less divides …)
If 4 ? 1428
Then 7 ? ? (more)
If more, less (i.e., 4) divides; ? we have
? = 2499 goats
4
(a) Method 4 (Equivalent ratios)
4 : 7 = 1428 : g
?
?
? 2499 goats
4(b)
(i) |AD| ˜ 4 cm, |BD| ˜ 2 cm
(ii) |AD| : IBD| = 4cm : 2cm
= 2 : 1
5. (a) The diagram is a triangle ABC with the side AC produced to D.
Find:
(i) The value of x
(ii) angle ACB
(b) The simple interest formula gives the interest, I on a principal, P invested at a rate, R per annum for time, T years.
(i) Find the simple interest on GH¢3,600.00 at 15% per annum for 2 years.
(ii) Make R the subject of the simple interest formula.
(iii) At what rate per annum will GH¢6,000.00 earn GH¢2,400.00 simple interest in 2 years?
a)
(i) Method 1 (Exterior angle = two opposite interior angles)
From the diagram,
?
?
?
?
?
?
?
5 (a)(i) Method 2 (Sum of interior angles of triangle = 180°)
From the diagram, ?ACB + (9x)° = 180°
?ACB = 180° – (9x)°
? 180° – (9x)° + (x+17)° + (2x+7)° = 180°
? 180° -180° + (x+17)° + (2x+7)° = (9x)°
?
?
?
?
?
?
5 (a)(ii) ?ACB + (9x)° = 180°
?ACB + (9×4)° = 180°
?ACB + 36° = 180°
?ACB = 180° – 36°
Angle ACB = 144°
5 (b) (i) =
= 36 × 2 × 15 = 1080
Simple interest = GH¢1,080.00
5 (b) (ii) Method 1
?
?
?
?
5 (b) (ii) Method 2
?
?
?
?
5 (b) (ii) Method 3
?
?
?
5 (b) (iii) Approach 1 (Substitute values in the expression for R)
Using the result obtained in (ii) above
?
? = 20
? The rate = 20%
Approach 2 (Substitute values in the given expression for Simple Interest)
Substituting in , we have
?
? R = 20
? The rate = 20%
6. (a) Given that ¢10,000.00 = GH¢1.00.
Complete the following table relating cedis (x) to Ghana cedis (y)
¢ (x) 10,000 50,000 150,000 250,000 350,000 450,000
GH¢(y) 1 45
(b) (i) On a graph sheet draw two perpendicular axes OX and OY.
(ii) Using a scale of 2cm to ¢50,000.00 on the Ox axis and 2cm to GH¢5.00 on the Oyo axis. Mark the Ox axis from 0 to ¢450,000 and OY axis from 0 to GH¢50.00
(c) Plot the points and join them with a straight line
(d) From your graph find the value of
(i) GH¢8.00 in cedis (¢)
(ii) GH¢35.00 in cedis (¢)
(iii) ¢260,000.00 in Ghana Cedis (GH¢)
6(a) Using the idea of simple proportion, the table is completed as shown below:
¢ (x) 10,000 50,000 150,000 250,000 350,000 450,000
GH¢(y) 1 5 15 25 35 45
6 (b), (c) The straight line graph is shown below
(IMAGE NEEDED)
6 (d) From the graph, the value of
(i) GH¢8.00 = ¢ 80,000.00
(ii) GH¢35.00 = ¢350,000.00
(iii) ¢260,000.00 = GH¢26,000.00