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Mauritius Examinations Syndicate
CPE 2015 Subject Report - Mathematics
April 2016
1
MATHEMATICS
(Subject Code No. 120)
General Comments
Candidates fared well in the CPE Mathematics Examinations 2015. The overall percentage pass
went up from 76.6 % in 2014 to 79.3 % in 2015.
A close examination of candidates’ scripts reveals that the majority of candidates had the ability
to carry out basic calculations accurately. In general, candidates were also able to solve simple
word problems of low procedural complexity.
The improvement observed in the overall performance of the candidates in 2015 is largely
attributed to the larger number of candidates who scored higher marks in Section B. Although
the difficulty level of the question paper was comparable to that in the previous years, it
comprised fewer unfamiliar situations. Questions such as 51, 52, 53 (a), 53 (b) and 54 (a), for
instance, were not uncommon. Arguably then, the improved results achieved in 2015 may be
ascribed to the fact that candidates had been exposed to and practised these types of
questions in the past. Consequently, they were more confident in tackling them.
However, it is worthwhile to note that, while candidates showed that they knew which
mathematical concepts or operations to apply in these situations, a good proportion of the
answers they offered revealed that they often applied the procedures inappropriately. In so
doing, they invariably obtained incongruous results that went unnoticed. A few telling
examples include the following:
getting Rs 104 as the original price of a shirt which cost Rs 520 in a sale (Qu. 52);
obtaining Rs 32 000 as the simple interest on a principal amount of Rs 8000 (Qu. 53 (a));
obtaining 4
31 as the fraction of blue marbles in a bag containing red, blue and yellow
marbles (Qu. 55).
2
Another shortcoming observed in 2015 relates to the loss of a considerable number of marks as
a result of the careless use of mathematical notations. For instance, candidates wrote 8000 ÷ 4
when they actually carried out a multiplication or they wrote 360o + 243o when in fact they
subtracted 243o from 360o or, yet again, they wrote 1035
45 and obtained 23 instead of
23
1.
In addition, the analysis of candidates’ performance suggests that pupils tend to be far too
subjected to repetitive problems of low procedural complexity that require little more than
using procedures during classroom instruction than being encouraged to think mathematically
and to adapt their mathematical understanding depending on the required application.
Whereas it is acknowledged that a certain degree of repetition is desirable in the teaching of
mathematics - repetition enables pupils ‘…to practise procedures so that they become a secure
part of their mathematical toolbox’ (Vincent & Stacey, 2008, p.103) - pupils also need to be
presented with sufficient problems that provoke their mathematical thinking or challenge their
mathematical understanding. Over-emphasis on repetition or limited exposure to complex
procedural problems may undermine pupils’ capability to apply practised procedures in novel
situations or to sustain a chain of reasoning when planning a solution pathway as testified by
candidates’ performance in Qu. 54 (b) and 55.
A final important point to which attention is strongly drawn regards candidates’ presentation of
their work. The presentation of the work of some of the candidates – not necessarily those
who achieved the highest marks – was of a very high standard. However, the general tendency
is for candidates to write their calculations messily all over the working space provided. In so
doing, they sometimes erased part of their essential work which, in their haste, they omitted to
replace. Sometimes, candidates did not show any intermediate working. It is to be highlighted
that the omission of essential working may lead to the unnecessary loss of marks regardless of
whether a candidate arrives at a correct answer or not. Examiners will often be able to give
partial credit where stages are shown in the working even if the final answer is incorrect.
3
Performance in Section A
Performance in Section A was marginally better than last year. Questions 1 to 15 were well-
answered by the majority of the candidates. A few stumbled on Qu. 6, 7, and 15 which
suggests that fractions and factors and multiples remain abstract concepts for many.
It was revealing to find that some of the high performing candidates lost marks in Section A
because they were not able to:
single out the shaded obtuse angle (Qu. 21);
state the L.C.M. of 6 and 8 (Qu. 22);
calculate the total surface area of a cube given its length (Qu. 33);
recognise a 3-D solid cylinder (Qu. 47(a));
decide where to place the decimal point on multiplying 4.3 by 2.5 (Qu. 47(b)(ii)).
Performance in Section B
Encouragingly, the trend observed since the last couple of years with respect to the growing
number of candidates who attempt Section B was maintained in 2015. Thus, fewer questions
in Section B were left unanswered.
4
Comments On Specific Questions
Very Short Answer Questions Questions 1 – 15
Question 6
Some candidates were not able to reduce 15
5 to its lowest terms. A fairly common mistake
observed was to give 5
1 as answer. This very likely occurred in haste as the answer suggests
that the candidates knew they had to divide the numerator and the denominator by 5 except
that they wrote 5 instead of 3 on dividing 15 by 5.
Question 7
This was the ‘very short answer question’ which recorded the highest number of incorrect
answers such as 2, 8, 12 and 24. The possible underlying misconceptions that can be picked out
from these invalid answers are that:
1. the H.C.F. of two numbers is the most common prime factor of the two numbers - 2 in
this case;
2. the H.C.F. of two numbers is the smaller number – reasonably extrapolating from prior
experience that the H.C.F. of, say, 2 and 4 is 2, that of 2 and 8 is 2, or the H.C.F. of 3 and
9 being 3, that of 4 and 8 being 4, or that 8 and 16 being 8 etc.; and
3. the H.C.F. of two numbers is the larger or higher number as the term H.C.F. itself
suggests.
The wrong answer, 24, stemmed from calculating the L.C.M. of 8 and 12 (2 × 2 × 2 × 3). The
steps to calculate the H.C.F. and the L.C.M. of two numbers are so subtly different from each
other that often candidates confuse between the two.
5
Question 15
Although performance in this question was slightly better than that in 2014 (Qu. 13), quite a
good number of candidates could not express 10
9 as a decimal. Common mistakes were to
misplace the decimal point giving 0.09, 9.0 or even 90 as wrong answers. In some instances,
10
9 was expressed as the ratio 9 : 10 or was incorrectly reduced to
5
3.
Multiple Choice Questions Questions 16 - 45
Question 21
A significant number of candidates were not able to recognise which of the shaded angles
represented an obtuse angle. Option D (visual representation of a reflex angle) was found to
be a quite strong distractor. It is important to note that, although some candidates correctly
wrote down what acute, obtuse, reflex and right-angles are on their scripts, they still could not
decide which of the diagrams represented an obtuse angle.
Question 31
Less than 50 % of the candidates were able to answer this question correctly. In general,
candidates realised that they had to add the duration of the movie (2 hours 35 minutes) to the
time at which it started (10 40). However, a large number of candidates omitted to add one
hour in the ‘hour column’ for subtracting 60 minutes from 75 minutes (35 + 40) in the ‘minutes
column’. Thus, they obtained 12 15 as their answer instead of 13 15. A quarter of the students
subtracted 2 hours 35 minutes from 10 40 and, consequently, chose either Option A or Option
B.
6
Question 33
This was the least well-answered multiple choice item. It should be noted that Qu. 33 is a very
common question that is set slightly differently from one year to another. In 2015, it required
candidates to calculate the total surface area of a cube of length 3 cm. Option C (18 cm3) was
the strongest distractor across all ability groups. This gives away the high proportion of
candidates who may not have read the question carefully mistaking the length of 3 cm for the
area of one face of the cube.
Question 35
A fairly small number of candidates was able to identify which of the numbers given was not a
prime number. Options B (31) and D (11) caused the most distraction.
Question 37
Performance in this question was rather fair. It is the multiple-choice item in which the high
performing candidates fared the least well possibly because, in their haste, they did not pay
enough attention to the question that was asked. The question read as follows:
The total mass of a box and 8 identical toys is 800 g. The mass of the box is 200 g.
What is the mass of one toy?
In most cases, these candidates chose Option A (600 g) omitting the fact that they were still
one step away from finding the mass of one toy.
Question 38
A quite small number of candidates attempted Qu. 38 successfully. As observed last year,
candidates wrongly, but, readily conceded that 225 minutes is equivalent to 2 hours 25
minutes. Option A was thus a very popular incorrect answer.
7
Question 39
The performance of candidates in Qu. 39 has only faintly improved from last year (Qu. 33,
2014). In this question, candidates were asked to calculate how many pieces they could cut
from a long rope of length 4 m if each piece cut were 20 cm long. This year again, a significant
number of candidates obtained the wrong answer by either multiplying 20 by 4 or by dividing
20 by 4, paying undue attention to the fact that the values, 20 and 4, had different units.
Consequently, these had to be converted to the same unit before they could be divided.
Question 42
Qu. 42 was a problem that related to the concept of average. It proved to be problematic for
many candidates. Very few gave the impression that they knew how to go about solving the
problem. It was the multiple-choice item that was most often left blank. There were quite a
good number of cases where candidates obtained the correct answer from an incorrect
working. For example, some candidates added the average mass of Rubina and Dolly (48.5 kg)
to the mass of Dolly (45 kg) in order to calculate the mass of Rubina. Since 48.5 kg added to
45 kg gave 93.5 kg, these candidates encircled the correct answer Option C (96 kg) which was
the nearest value to the answer they obtained in their working space.
8
Semi-Structured Questions Questions 46 - 50
Question 46
Less than 50 % of the candidates were able to identify the missing terms in all the three number
patterns given. In general, candidates easily recognised the sequence in part (a). However,
part (b) did not seem obvious to some. Occasionally, values were added to 110 arbitrarily.
In part (c), candidates had to identify the next square number in the sequence. While many did
not recognise the pattern of square numbers, a few noted that adding prime numbers to the
terms given would also yield the same pattern – that is, 1 added to 3 would give 4, 4 added to 5
would give 9, 9 added to 7 would give 16. Thus, they added 11 to 16 and gave 27 as their
answer. They were duly rewarded.
Question 46
(a) 52 , 56 , 60 , 64 , 68 .
(b) 50 , 60 , 80 , 110, 150 .
(c) 1 , 4 , 9 , 16, 25 .
1 , 4 , 9 , 16, 27 .
+ 4 + 4 + 4 + 4
+ 40 + 30 + 20 + 10
+ 11 + 7 + 5 + 3
9
Question 47
A fairly small number of the candidates were able to score full marks in Qu. 47. Whereas it was
felt that part (a) would be accessible to the majority of the candidates, few were able to state
that the 3-D solid shown was a cylinder. Those who found the correct answer were quite often
not able to spell the word ‘cylinder’ correctly: cilindre, cyclinder, cynder, cylindre, clylind are a
few incorrect spellings that were frequently seen. Some of the most able candidates gave
‘sphere’, ‘cube’ and ‘cuboid’ as answer. Occasionally, candidates associated the solid shown to
common objects and gave answers such as can, bottle, tin, tube etc.
Part (b) was generally well-answered. Mistakes often arose as a result of either omitting to
reduce the ‘3’ in 53.6 to ‘2’ after borrowing (which yielded 31.4 as answer) or neglecting to
consider the ‘6’ in 53.6 after subtracting 8 from the 10 that had been borrowed (which gave
30.2 as answer).
The main problem encountered by candidates in part (c) was that they could not decide with
confidence where to place the decimal point. The response 107.5 was seen time and again.
Question 48
Fewer than a third of the candidates were able to answer Qu. 48 successfully. Quite many
found the percentage increase in the price of the article to be 80 %. The mistake they made
was to divide the original price of the article (Rs 400) by its new price (Rs 500) to find the
percentage increase. A good number of candidates also obtained 20 % as the answer. In this
Question 47
(a) cylinder
(b)(i) 53.6 - 22.8 = 30.8 (b)(ii) 4.3 × 2.5 = 10.75
10
case, candidates rightly began by calculating the increase in the price of the article (Rs 500 – Rs
400 = Rs 100) but, instead of dividing 100 by the original price to find the percentage increase,
they divided it by the new price.
Question 48 was the least well-answered Short Answer Question in 2015.
Question 49
Question 49 was quite straightforward. It was the Short Answer Question in which the highest
number of correct answers was recorded. It is important to note, however, that a small
number of candidates were not able to convert 7 minutes into seconds. Sometimes this was
due to the fact that candidates did not recall their multiplication tables well. Quite often
answers such as 480 s or 540 s were seen in these cases. At other times, candidates wrongly
considered 1 minute to be equal to 100 seconds and obtained 700 s as their answer.
Part (b) required candidates to draw the lines of symmetry of a given shaded diagram. In
general, candidates fared well in this part. Occasionally, candidates drew 4 lines of symmetry
when there were only 2. In many instances, candidates did not use a ruler to draw the lines of
symmetry.
Question 48
Increase in price = Rs (500 -400)
= Rs 100
% increase = 100 × 100 %
400
= 25 %
Four year ago, Ravi was:
23 – 4 = 19 years old
11
Question 50
Performance in this question indicated that a considerable number of candidates did not
understand or, at least, could not apply their understanding of percentage in the context given.
It is not that many had overlooked the fact that the question required them to calculate the
number of white balls in the box. Rather, it appeared to them that calculating 60 % (the
percentage number of red balls in the box) of 25 balls would give the number of white balls in
the box. Quite often, candidates stopped at finding the percentage number of white balls in
the box (100 % - 60 % = 40 %) and gave their answer as 40 white balls. It is felt that oblivious
mistakes such as those could have been avoided if candidates had written down what ‘100 - 60
= 40’ stood for in the working space.
Question 49
(a) 7 × 60 = 420 seconds
(b)
12
Question 50
Method 1:
Total no. of balls = 25
No. of red balls in the box = 60 × 25
100
= 15
Therefore,
No. of white balls in the box = 25 – 15
= 10
Method 2:
% number of red balls in the box = 100 – 60
= 40
Hence,
No. of red balls in the box = 40 × 25
100
= 10
13
Structured/Open-ended Questions Questions 51 - 55
Question 51
This question was fairly well-attempted in general.
Part (a) proved to be highly scoring. Candidates, at large, recognised that they had to divide
376 by 8 to find the missing number. However, a good number of candidates were not able to
carry out the division successfully. A few cases where candidates had recourse to trial and
error were also noted.
Question 51 (a) 47
(b)
(i) Point A has coordinates ( 2 , 3)
(ii) Shown on graph.
B
14
In part (b), it was found that a considerable number of candidates did not know how to read
the coordinates of a point. Quite many gave the coordinates of point A as ( 3 , 2 ) instead of
( 2 , 3 ) and plotted point B at ( 1 , 5 ) on the graph.
Understandably, candidates who wrote the coordinates of point A as ( 3 , 2 ) also plotted point
B at ( 5 , 1 ). However, it was interesting to note that a good number of the candidates were
inconsistent in the mistakes they made in relation to parts (i) and (ii). They rightly wrote down
the coordinates of point A as ( 2 , 3 ). Yet, they plotted point B at ( 1 , 5 ) or, conversely, they
wrote the coordinates of point A as (3 , 2 ) and plotted point B at ( 5 , 1 ).
Question 52
A good number of candidates were able to solve part (a) which assessed candidates’
understanding of the angle properties of a quadrilateral. Mistakes arose mainly from incorrect
manipulations during addition and subtraction. Some candidates wrongly felt that angle x
would be equal to the sum of the three other angles of the quadrilateral. Consequently, they
did not proceed to subtract the sum of the three angles (243o) from 360o.
Question 52
(a)
The sum of the three interior angles
= (88o + 93o + 62o)
= 243o
Hence,
angle x = (360o – 243o)
= 117o
The sum of the four interior angles of a quadrilateral = 360o
15
Although the overall performance in parts (b) and (c) remained fairly low, it improved quite
significantly from the previous years.
About a third of the candidates answered part (b) correctly. This was a net improvement of
+ 4.8 % from candidates’ performance in 2013 (Qu. 52 (a)). The question was about calculating
the selling price of a watch for which a profit of 10 % was made. The most common mistake
was for candidates to dwell on the emboldened word ‘profit’ without reading the question till
the end. Surface reading of the question thus led many to calculate the profit made without
proceeding any further. Sometimes, candidates miscalculated 100
10 × 1500. They wrote 1500 ×
10
100 instead and obtained Rs 15 000 as answer. Even then, the improbability of their answer
went unnoticed. In addition, it was quite peculiar to note the considerable number of
candidates who calculated 10 % of Rs 1500 as follows:
10
1500 × 100 = Rs 150
Question 52 (b)
Method 1
Buying Price = Rs 1500
Profit = 10 × 1500 = Rs 150
100
Hence,
Selling Price = Rs (1500 + 150) = Rs 1650
Method 2
Profit = 10 %
Selling Price = (100 + 10) = 110 %
Therefore,
Selling Price = 110 × Rs 1500
100
= Rs 1650
% Rs
Original Price 100 1500
Profit 10 10/100 × 1500 =
150
New/Selling Price
100 + 10 1650
16
About a fifth of the candidates successfully answered part (c). This amounted to a net increase
of + 5.1 % in the candidates’ performance compared to that in 2012 (Qu. 53 (a)). The majority
of the candidates had difficulty to interpret the sentence: ‘In a sale, the original price of a shirt
is reduced by 20 %’. Very often, they mistook the percentage new price to be 20 % instead of
80 % and wrongly calculated the original price by working out 20 % or 80 % of Rs 520.
Question 53
A good number of candidates answered part (a) correctly. Script analyses indicate that many
could recall the formula to calculate simple interest (100
PRT) but applied the formula
inappropriately. A small number of candidates translated 100
PRT literally as
100100
1048000
. A
good number of candidates proceeded by multiplying 8000 by 4 and, subsequently, calculated
10 % of Rs 32 000. However, they then went on to add 3200 to 8000 and gave their answer as
Rs 11 200. It was curious to note how results obtained were inconsistent with what pupils
wrote in the working space. As seen in Q. 52 (b), candidates often wrote 80010010
8000 .
Others wrote 320004
8000 . In these cases, it was difficult to discern if the candidates were
Question 52 (c)
Percentage (%) Price (Rs)
Original Price 100 ?
Reduction 20 ?
New Price 80 520
Hence,
Original price = Rs 650
80 % Rs 520
100 % 520 × 100
80
= Rs 650
17
using a wrong method or if they had the right intentions but were using inappropriate ways of
presenting their intermediate working.
Candidates’ performance in part (b) was slightly better compared to the performance in Qu. 53
(a) in 2013. The main observation made was that a large number of candidates successfully
found the number of tiles that would fit along the length (23 tiles) and the width (900 tiles) of
the floor. However, some of the candidates went on to add the two resulting figures whereas
they had to multiply them to obtain the total number of tiles used. It is important to highlight,
again, that in a considerable number of scripts, candidates used incorrect notations such as
231035
45 tiles or
900
45 20 tiles which were wrong.
Question 53(a)
Method 1 Method 2
Interest per year = 10 × Rs 8000
100
= Rs 800
Therefore,
Accrued interest = Rs 800 × 4
= Rs 3200
Simple Interest = PRT
100
= 8000 × 10 × 10
100
= Rs 3200
18
Question 54
Candidates scored quite a good number of partial marks in this question.
The main difficulty which candidates encountered in part (a) pertained to the division of a
whole number by a fraction which they had to carry out. In general, candidates were able to
find the time which Jane took to finish the race which they obtained by subtracting 07 30 from
08 54. They had then to convert this time in either minutes or hours to be able to determine
the average speed. Most converted the time into hours but were not able to divide 14 km by
5
21 h. It is worthwhile to note that a handful of candidates lost marks in this particular
question because of the inappropriate use of notations or improper presentation.
Question 53(b)
Method 1 Method 2
No. of tiles along the
length of the floor = 1035
45
= 23
No. of tiles along the
width of the floor = 900
45
= 20
Hence,
Total no. of tiles used = 23 × 20
= 460
Area of floor = 1035 × 900
= 931500 cm2
Area of one tile = 45 × 45
= 2025 cm2
Therefore,
total no. of tiles used = 931500
2025
= 460
19
In part (b), most of the candidates were rewarded for calculating the volume of Container B.
Candidates also reckoned that to be able to find the height of container A, they needed to
calculate its area of base. Mistakes stemmed primarily from misinterpreting the statement
‘Container B is now three quarter full’. Also, as a result of the number of calculations that
appeared all over in the working space, candidates could not figure out which of the values they
calculated represented the volume and which represented the area or whether the volume
they had calculated was the volume of Container A or that of Container B. It is felt that if
candidates had presented their work by writing down what they set out to calculate initially
instead of just working out a number of calculations, it would have been easier for them to
follow and to sustain their line of reasoning.
Question 54 (a)
METHOD 1: METHOD 2:
Time taken = 08 54 - 07 30
= 01 24
= 7/5 h or equiv.
Speed = distance
time
= 14 ÷ 7/5
= 14 × 5/7
= 10 km/h
Time taken = 08 54 - 07 30
= 01 24
= 84 minutes
84 min 14 km
1 min 14
84
60 min 14 × 60
84
= 10 km/h
Hence, average speed = 10 km/h
20
Question 54 (b)
METHOD 1:
The height of water in Container B = 3 × 40 cm
4
= 30 cm
Vol. of water in Container B = (70 × 30 × 30) cm3
= 63 000 cm3
Area of base of Container A = (60 × 50) cm2
= 3000 cm2
Therefore, the height of Container A = 63 000 cm3 ÷ 3000 cm2
= 21 cm
METHOD 2:
Volume of Container B = (70 × 30 × 40) cm3
= 84 000 cm3
Vol. of water in Container B = 3 × 84 000 cm3
4
= 63 000 cm3
Area of base of Container A = (60 × 50) cm2
= 3000 cm2
Therefore, the height of Container A = 63 000 cm3 ÷ 3000 cm2
= 21 cm
Method 3:
Volume of Container B = (70 × 30 × 40) cm3
= 84 000 cm3
Area of base of Container A = (60 × 50) cm2
= 3000 cm2
Assumed height of Container A = 84 0000 cm3 ÷ 3000 cm2
= 28 cm
Hence, the height of Container A = 3 × 28 cm
4
= 21 cm
21
Question 55
This was expectedly the most challenging question of the paper.
The most common mistake noted in part (a) was to assume that the fraction of blue marbles in
the bag was equal to 5
1 or
12
5. Candidates who obtained the fraction of blue marbles as
5
1 set
out to calculate the fractional number of blue and yellow marbles in the bag (1 - 5
2 =
5
3).
Then, they reasoned that if the ratio of the number of blue marbles to the number of yellow
marbles was 5 : 7, then out of 5
3,
5
2 of the marbles would be yellow and
5
1 of the marbles
would be blue. These fractions were thus intuitively assigned to the yellow and blue marbles.
Those who obtained 12
5, on the other hand, interpreted the sentence ‘The ratio of the number
of blue marbles to the number of yellow marbles is 5 : 7’ in isolation. As a result, they
translated the sentence as meaning that the fraction representing the number of blue marbles
would be equal to 5 over the total ratio, i.e., 12
5 overlooking the proportion of red marbles in
the bag.
Achieving success in part (b) was greatly dependent on the answer obtained to part (a).
Notwithstanding the mistakes made in part (a), the work of the candidate in part (b) was
carefully examined. Where a correct approach was used and sound mathematical thinking was
displayed, candidates were rightfully rewarded.
22
Question 55
(a) METHOD 1: METHOD 2:
No. of blue and yellow
marbles in the bag = 1 - 2
5
= 3
5
Implies,
12 shares 3
5
1 share 3 ÷ 12
5
5 shares 3 ÷ 12 × 5
5
Hence,
Fraction of blue marbles = 1
4
(b) METHOD 1 METHOD 2
Fractional difference
= 2 - 1 = 3
5 4 20
3 54 marbles
20
1 54
20 3
20 18 × 20
20
= 360
Hence,
Total no. of marbles = 360
No. of blue and yellow marbles in the bag = 1 - 2
5 = 3 5 Blue : Yellow 5 : 7 = 12 Implies, Fraction of blue marbles = 5 × 3 12 5 = 1 4
Red : Blue
8 : 5
8 shares - 5 shares = 3 shares
3 54 marbles
1 54 = 18 marbles
3
Therefore,
Total no. of marbles
= (8 × 18) + (5 × 18) + (7 × 18)
= 360
23
Reference:
Vincent, S & Stacey, K 2008, ‘Do Mathematics Textbooks Cultivate Shallow Teaching? Applying the TIMSS Video Study Criteria to Australian Eighth-grade Mathematics Textbooks’, Mathematics Education Research Journal, vol. 20, No. 1, pp. 82-107.