Diagrams for BECE 2016 Mathematics
The papers can be downloaded from the main Preppy site.
Available papers include
This diagram is from Question 31 in Paper 1.
Can you find x in the figure below?
Saturday 18 June 2016
Monday 13 June 2016
Preppy's BECE Question of the Day (June 13, 2016)
Preppy's BECE Question of the Day
June 13, 2016
In a class of 40 students, 28 like tea, 18 like coffee, and 10 like neither tea nor coffee. How many students like both tea and coffee?
A. 2
B. 14
C. 16
D. 30
E. None of the above
Friday 3 June 2016
Preppy's BECE Question of the Day (June 3, 2016)
Today's BECE Question of the Day is from Mathematics.
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?
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?
Wednesday 1 June 2016
BECE Question of the Day June 1, 2016
Today's BECE Question of the Day is from RME.
There are five pillars of Islam. What is the second pillar?
Sunday 14 June 2015
Timetable for the June 2015 BECE
The BECE is here again. This resit or remedial exam for this year has already taken place in February but the regular exam will start tomorrow, Monday, June 15 and end on Friday, June 19. The timetable is as follows:
Monday, June 15: English Language & Religious and Moral Education (RME)
Tuesday, June 16: Integrated Science & Basic Design and Technology (BDT)
Wednesday, June 17: Mathematics & Ghanaian Language and Culture
Thursday, June 18: Social Studies & French
Friday, June 19: Information and Communication Technology (ICT)
We wish all candidates the best of luck.
Monday, June 15: English Language & Religious and Moral Education (RME)
Tuesday, June 16: Integrated Science & Basic Design and Technology (BDT)
Wednesday, June 17: Mathematics & Ghanaian Language and Culture
Thursday, June 18: Social Studies & French
Friday, June 19: Information and Communication Technology (ICT)
We wish all candidates the best of luck.
Wednesday 10 June 2015
2015 May/June WASSCE Core Maths Objectives (5 interesting questions)
I would like to discuss some of the questions in the Core Maths paper of the just ended WASSCE. As usual, there were fifty objective questions for which students had one and a half hours to complete. I will discuss only five questions out of the fifty. These are the questions that I consider most interesting. They are interesting because they are difficult, or easy, or likely to trip up students. You can download the full list of questions from Preppy.
The five most interesting questions from the 2015 May/June WASSCE Core Maths Paper 1 are Questions 5, 20, 24, 31, and 50.
A. 37
B. 48
C. 50
D. 63
This question is not difficult but it requires students to apply an understanding of an important concept. I wish the examiners had asked for a more distant term, like the 70th term so that a true understanding of arithmetic progressions could be tested. However, this may be enough to see if students understand the pattern.
Though the sequence itself is not an arithmetic progression, the differences between the terms form an arithmetic progression:
from 2 we add 3 to get 5;
from 5 we add 5 to get 10;
from 10 we add 7 to get 17;
from 17 we add 9 to get 26, and so on.
3, 5, 7, 9, ... is an arithmetic sequence so it is easy to find its terms. In particular, the next two terms in the sequence are 11 and 13. Thus the sixth and seventh terms of the original sequence are 26 + 11 = 37 and 37 + 13 = 50.
In the diagram, O is the centre of the circle, RT is a diameter, angle PQT = 33° and angle TOS = 76°.
Find the size of angle PRS.
A. 76°
B. 71°
C. 38°
D. 33°
This question is interesting because the accompanying diagram looks scary. I suspect most students would run away from this question because the diagram looks cumbersome. Moreover, the circle theorems are not popular among test takers.
How do we find the size of angle PRS?
First, we note that PRS consists of two angles: PRT and TRS. Hence, we can find the sizes of each of those two angles and add them to get the answer. The size of angle PRT is 33° because it is subtended by chord PT at the circumference. We know this because chord PT subtends another angle (angle PQT) at the circumference whose size is 33°.
The size of angle TRS is half of 76° because it is subtended at the circumference by a chord which subtends an angle of 76° at the centre. Hence, the size of angle TRS is 38°. Adding the sizes of PRT and TRS gives 33° + 38° = 71°, which is the size of angle PRS.
A. 53 degrees
B. 55 degrees
C. 63 degrees
D. 65 degrees
This question is interesting because it is abstract. It is also short, masking its difficulty in a few words. Many students who take the Core Maths paper like to use a calculator to solve everything but this question does not easily yield to that. The student must use his imagination and maybe draw a diagram with which he can then answer the question.
An example of such a diagram may be the one shown on the right. The student must recognize that all the sides of a square are congruent. He must also understand what the midpoint is and how to construct angle PXS. Most importantly, the student must know how to label the square properly. A failure to do so will end in an error. For example, the square is given as PQRS. This means that in drawing it, one must label the vertices in order--whether clockwise or anticlockwise. Any other labeling would result in an error.
After drawing the square correctly and marking the important points, the student has to find the size of the angle in question. He can do so by using one of the trigonometric ratios. The tangent seems convenient here as we have all the information required to compute it. To find the tangent, we need to find the opposite and the adjacent. These are already obvious from the diagram. The opposite is k while the adjacent is k/2. Hence, the tangent is k/(k/2) = 2 and the angle itself is arctan(2), which is 63° to the nearest degree.
I can also imagine some students finding the hypotenuse of triangle PXS but there is no need to spend precious time on that.
A. (30 - n) years
B. (20 - n) years
C. (25 - n) years
D. (30 + n) years
This question is not difficult but it requires students to be careful in their reasoning. If tom will be 25 years old in n years' time then he is (25 - n) years old now. If Tom is 5 years younger than Bade, then Bade is 5 years older than Tom. Hence, Bade is 25 - n + 5 = (30 - n) years old.
In the diagram, OX bisects angle YXZ and OZ bisects angle YZX. If angle XYZ = 68°, calculate the value of angle XOZ.
A. 68°
B. 72°
C. 112°
D. 124°
This is yet another geometry question. Students must do a bit of deduction to get the answer for this one. We are given only one figure and are required to find the others, including the answer. The problem that could arise here is that some students may want to find the sizes of angles OXZ and OZX individually, though that is impossible. What can be found from the information given is the sum of angles OXZ and OZX. This sum is half of the difference between 180° and 68° since OX and OZ bisect the two other angles of the triangle. Therefore, the sum of the sizes of OXZ and OZX is 1/2 of (180° - 68°) = 1/2 of 112° = 56°. We subtract this from 180° to get the size of XOZ as 124°.
The five most interesting questions from the 2015 May/June WASSCE Core Maths Objectives
The five most interesting questions from the 2015 May/June WASSCE Core Maths Paper 1 are Questions 5, 20, 24, 31, and 50.
Question 5
Find the 7th term of the sequence: 2, 5, 10, 17, 26, ...A. 37
B. 48
C. 50
D. 63
This question is not difficult but it requires students to apply an understanding of an important concept. I wish the examiners had asked for a more distant term, like the 70th term so that a true understanding of arithmetic progressions could be tested. However, this may be enough to see if students understand the pattern.
Though the sequence itself is not an arithmetic progression, the differences between the terms form an arithmetic progression:
from 2 we add 3 to get 5;
from 5 we add 5 to get 10;
from 10 we add 7 to get 17;
from 17 we add 9 to get 26, and so on.
3, 5, 7, 9, ... is an arithmetic sequence so it is easy to find its terms. In particular, the next two terms in the sequence are 11 and 13. Thus the sixth and seventh terms of the original sequence are 26 + 11 = 37 and 37 + 13 = 50.
Question 20
 |
| Diagram for Question 20 |
Find the size of angle PRS.
A. 76°
B. 71°
C. 38°
D. 33°
This question is interesting because the accompanying diagram looks scary. I suspect most students would run away from this question because the diagram looks cumbersome. Moreover, the circle theorems are not popular among test takers.
How do we find the size of angle PRS?
First, we note that PRS consists of two angles: PRT and TRS. Hence, we can find the sizes of each of those two angles and add them to get the answer. The size of angle PRT is 33° because it is subtended by chord PT at the circumference. We know this because chord PT subtends another angle (angle PQT) at the circumference whose size is 33°.
The size of angle TRS is half of 76° because it is subtended at the circumference by a chord which subtends an angle of 76° at the centre. Hence, the size of angle TRS is 38°. Adding the sizes of PRT and TRS gives 33° + 38° = 71°, which is the size of angle PRS.
Question 24
PQRS is a square. If X is the midpoint of PQ, calculate, correct to the nearest degree, angle PXS.A. 53 degrees
B. 55 degrees
C. 63 degrees
D. 65 degrees
 |
| Diagram for solving Question 24 |
An example of such a diagram may be the one shown on the right. The student must recognize that all the sides of a square are congruent. He must also understand what the midpoint is and how to construct angle PXS. Most importantly, the student must know how to label the square properly. A failure to do so will end in an error. For example, the square is given as PQRS. This means that in drawing it, one must label the vertices in order--whether clockwise or anticlockwise. Any other labeling would result in an error.
After drawing the square correctly and marking the important points, the student has to find the size of the angle in question. He can do so by using one of the trigonometric ratios. The tangent seems convenient here as we have all the information required to compute it. To find the tangent, we need to find the opposite and the adjacent. These are already obvious from the diagram. The opposite is k while the adjacent is k/2. Hence, the tangent is k/(k/2) = 2 and the angle itself is arctan(2), which is 63° to the nearest degree.
I can also imagine some students finding the hypotenuse of triangle PXS but there is no need to spend precious time on that.
Question 31
Tom will be 25 years old in n years' time. If he is 5 years younger than Bade, find Bade's present age.A. (30 - n) years
B. (20 - n) years
C. (25 - n) years
D. (30 + n) years
This question is not difficult but it requires students to be careful in their reasoning. If tom will be 25 years old in n years' time then he is (25 - n) years old now. If Tom is 5 years younger than Bade, then Bade is 5 years older than Tom. Hence, Bade is 25 - n + 5 = (30 - n) years old.
Question 50
 |
| Diagram for Question 50 |
A. 68°
B. 72°
C. 112°
D. 124°
This is yet another geometry question. Students must do a bit of deduction to get the answer for this one. We are given only one figure and are required to find the others, including the answer. The problem that could arise here is that some students may want to find the sizes of angles OXZ and OZX individually, though that is impossible. What can be found from the information given is the sum of angles OXZ and OZX. This sum is half of the difference between 180° and 68° since OX and OZ bisect the two other angles of the triangle. Therefore, the sum of the sizes of OXZ and OZX is 1/2 of (180° - 68°) = 1/2 of 112° = 56°. We subtract this from 180° to get the size of XOZ as 124°.
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