Mauritius Examinations Syndicate

CPE 2014 Subject Report - Mathematics

April 2015

1

MATHEMATICS

(Subject Code No. 120)

General Comments

The overall performance of candidates in the CPE Mathematics Examination 2014 was broadly

comparable to that of previous years. The overall percentage pass attained 76.6 %.

Achievement in Section B was seen to have slightly improved from that in the preceding year.

The ability to understand and apply mathematics in familiar and unfamiliar situations is a crucial

skill to acquire to be able to succeed in today’s world. For this reason, the question paper

aimed at offering a wide range of opportunities to assess candidates’ ability to solve problems

in different contexts in a graded fashion.

The statistical and qualitative analyses of candidates’ scripts indicate that, in general,

candidates had a good command of basic maths. However, they also revealed that a major

difficulty faced by some candidates was related to a limited capacity to recognise which

mathematical processes or ideas to apply in a given situation. In Section A, for instance,

candidates demonstrated they could add, subtract, multiply and divide whole numbers but

often applied those processes inappropriately. Such mistakes were observed in Questions 14,

26, 33, 48, and 49.

It was also noted that candidates still struggle with mathematical language. Terms such as

‘fewer’ in Qu. 26, ‘maximum’ in Qu. 33, ‘older than’ and ‘4 years ago’ in Qu. 48 were generally

misinterpreted by candidates. Questions 53 and 54 in Section B were two non-routine

problems set to stimulate pupils’ thinking and to assess their problem solving skills. Surface

understanding of key concepts made it difficult for some to draw on their mathematical

knowledge base to generate carefully planned strategies while dealing with these more

complex problems. They turned out to be the most challenging questions of the paper.

2

Performance in Section A

Performance in Section A was satisfactory on the whole. However, candidates performed less

well on a few questions which are frequently asked such as Qu. 29, 33, 35 and 37. Some

common errors observed occurred because candidates tended to work too hastily. For

example, slips in calculations such as ‘9 x 6 = 56’ in Qu. 12 were rather common. Sometimes,

candidates did not read the multiple choice items carefully enough and answered questions

that were not asked. A number of candidates of average ability were not able to:

convert kilograms into grams (Qu. 11);

express a fraction as a decimal number (Qu. 13);

convert minutes into hours and minutes (Qu. 37);

count the number of days on a calendar (Qu. 35).

It must be noted that some of the high performing candidates were not able to:

express 300 mL as a fraction of 1 L (Qu. 28);

divide a whole number by a fraction (Qu. 47(a)(ii));

recognise multiplication as a process of repeated addition (Qu. 47(b)).

Performance in Section B

A greater number of candidates, regardless of their acquired math competencies, attempted all

the questions in Section B. It was encouraging to observe the high proportion of candidates

who successfully solved parts (a) and/or (b) of Qu. 55.

3

Comments On Specific Questions

Very Short Answer Questions Questions 1 – 15

Question 11

A good number of candidates were unable to convert 15 kilograms into grams. The most

common mistake was to consider 1 kg as being equal to 100 g.

Question 13

There was a range of incorrect responses in this question. Candidates either took the reciprocal

of 10

2 or expressed

10

2 as 0.02 instead of 0.2. Some candidates also obtained 1.5 by first

reducing 10

2 to

5

1 and subsequently writing

5

1 as 1.5.

Question 14

Although performance on this question on proportion was satisfactory on the whole, it is

worthwhile to note that it was set the other way round from how it conventionally is. A non-

negligible number of candidates nevertheless multiplied 6 by 18 instead of dividing 18 by 6 to

find the cost of a banana. As a result, there were fewer correct answers for this question than

in the past.

4

Multiple Choice Questions Questions 16 - 45

Question 17

A number of candidates thought that a circle had no lines of symmetry and, consequently,

Option A proved to be a strong distractor in this case.

Question 23

Many candidates were not able to interpret the phrase ‘144 is 8 times _____’ correctly. In this

way, they could not decide which operation to use. One of the most common mistakes noted

was to multiply 144 by 8 instead of dividing 144 by 8.

Question 25

A small number of candidates attempted Qu. 25 successfully. Option A was chosen by the

majority. Some possibly dwelled on the emboldened word ‘shaded’ overlooking the fact that

they had to find the ‘percentage’ of the figure that was shaded. Some rightly found that the

shaded fraction was 9/20 but did not go further to express the fraction obtained as a

percentage.

Question 27

A significant number of candidates, even from the high ability group, mistook a straight angle to

be equal to 360 o.

Question 28

This question proved to be difficult for a large number of candidates. In many cases, they chose

an option at random. Only a few rightly recognised that they had to convert 300 mL into litres

to be able to find the required fraction.

5

Question 33

This was the least well-answered multiple choice item. Candidates were required to calculate

the maximum number of pieces that could be cut from a 6 m long rope so each piece would

have a length of 30 cm. Many multiplied 30 by 6 to obtain an answer. A good number of

candidates also divided 30 by 6, disregarding the fact that the two values that were being

divided had different units.

Question 35

The performance of candidates on this question on the calendar went down in 2014. Counting

days backwards was problematic for a very large number of candidates. Many drew calendars

in the working space to assist them in their count but the calendars drawn were often faulty.

Question 37

Many candidates chose Option A mistaking 210 minutes to be equal to 2 hours 10 minutes. It is

perhaps worthwhile to note that candidates fared better on this type of question in 2012.

6

Semi-Structured Questions Questions 46 - 50

Question 46

Most candidates were able to identify the missing terms in the sequences given. Part (a)(i) was

found to be the easiest sequence to work out. A few high performing candidates interpreted

this sequence in a number of unexpected ways. Although their approaches were more

complex, the responses which they gave were mathematically correct and they were rewarded

accordingly. These candidates started by determining the L.C.M of the denominators before

they could identify a familiar pattern. They consequently obtained the following answers:

2

1 ,

3

2 ,

4

3 , ____

12

6 ,

12

8 ,

12

9 ,

12

9

OR

12

6 ,

12

8 ,

12

9 ,

12

11

OR

2

1 ,

3

2 ,

4

3 ,

36

29

Performance in parts (a)(ii) and (a)(iii) was satisfactory although candidates fared much better

in part (a)(i). They encountered the most difficulty in part (a)(ii).

Question 46 (a)

(i) 2

1 ,

3

2 ,

4

3 ,

5

4

(ii) 190 , 180 , 160 , 90

(iii) 81 , 27 , 9 , 3

+ 1 + 2

+ 2 + 1 + 2

+ 12

1 +

18

1 +

6

1

2

+ 0

7

Question 47

A very small number of candidates were able to score full marks in Qu. 47 although the

majority was able to solve part (a)(i) that was related to the addition of two decimal numbers.

In part (a)(ii), a considerable number of candidates, across ability groups, obtained 5 after

dividing 10 by 2

1. This suggests a limited understanding of the process by which a number is

divided by a fraction. It was very common to find candidates replacing the division sign by the

multiplication sign. But, often, they did not take the reciprocal of 2

1. In a number of other

cases, candidates took the reciprocal of both, 10 and 2

1.

A significant number of candidates were not able to answer part (b) correctly. As noted in

previous reports, the notion of multiplication as a process of repeated addition does not seem

to have been acquired. As a result, they performed lengthy calculations when there was none

required. A great proportion of candidates began by multiplying 91 x 29 and then proceeded,

by trial and error, until they were able to find which missing number on the left-hand side of

the equation would equate to the value they had obtained on the right-hand side. Quite often,

they did not get to the end of their lengthy calculations.

Qu. 47 was the least well attempted Short Answer Question in 2014.

Question 47

(a)(i) 73.9 + 94.8 = 168.7

(a)(ii) 10 ÷ 1 = 10 x 2

2 1 = 20

(b) 90

8

Question 48

About one third of the candidates were able to answer Qu. 48 successfully. In most cases,

candidates misinterpreted or did not understand the common terms ‘older than’ and ‘4 years

ago’. As a result, many candidates subtracted 6 years from the age of Ashna to calculate Ravi’s

age when, in fact, Ravi was older than Ashna. Likewise, others also added 4 years to the age of

Ravi whereas they had to subtract 4 years.

Question 49

Candidates had difficulties with Qu. 49. In general, they began to work out the problem

appropriately by calculating the total number of hours a week Diksha worked (5 x 7).

Subsequent workings, however, were improper. Occasionally, slips occurred while dividing 280

by 35. Average candidates often simply divided 280 by either 7 or 5 to get an answer.

Question 48

Ravi is 6 years older than Ashna.

So, Ravi is

17 + 6 = 23 years old

Four year ago, Ravi was:

23 – 4 = 19 years old

Question 49

Diksha works (7 x 5) = 35 hours per week

35 hours 1 week

280 hours 280 = 8 weeks

35

9

Question 50

For this Short Answer Question, the number of correct answers was the highest. However, a

number of candidates were not able to solve the problem completely. Partial marks were

scored for finding the total number of beads.

Question 50

Total no. of beads = 136 + 112

= 248

Number of beads in each box = 248

8

= 31

10

Structured/Open-ended Questions Questions 51 - 55

Question 51

This question was fairly well-attempted by the majority of candidates.

Part (a) proved to be scoring. A good number of candidates nonetheless read the coordinates

of point A as ( 4 , 5 ) instead of ( 5 , 4 ) and plotted point B on the x-axis instead of plotting it on

the y-axis.

A good number of candidates were able to solve part (b). Some candidates displayed firm

understanding of the properties of triangles. They worked out angle x from different

perspectives by projecting diverse lines in the diagram. Less successful candidates often

mistook the sum of angles of a triangle to be equal to 360 o or misused the properties of

isosceles and equilateral triangles.

Question 51 (a)

(a)(i) Point A has coordinates ( 5 , 4)

(a)(ii) Shown on graph.

B

11

Question 52

Performance in Qu. 52 was comparable to that of previous years. Parts (a) and (b) were well-

answered in general. However, in part (c), some candidates overlooked the fact that the bar

chart indicated the number of biscuits eaten and counted the number of shaded squares as

representing the number of biscuits left. Another common mistake was to wrongly interpret

the scale on the bar chart.

Question 51

(b)

Hence, Angle x = 360o – (90o + 60o + 45o)

= 165o

ABCD is a rectangle. So, angle ABC = 90o

1

A B

C D

E

F

x

AEB is a right-angled isosceles triangle. So, angle EAB = angle EBA = 45 o

45o

60o

BFC is an equilateral triangle. So, angle FBC = angle BFC

= angle FCB = 60 o

2

3

4

12

Question 53

A good number of candidates successfully attempted part (a).

Part (b), on the other hand, was challenging for a large number of candidates. The diagrams

given led a few to think that the volume had to be calculated (as per established practice).

Others had difficulty to connect the idea of volume to that of capacity which suggests that the

close relationship between the two concepts needs to be reinforced.

Question 52

(a) Brinda eats the greatest number of biscuits.

(b) Aleeyah & Divya eat (6 + 8) = 14 biscuits altogether

(c) Total number of biscuits left = ??

Total no. of biscuits at first = 15 x 4 = 60 Total no. of biscuits eaten = (6 + 12 + 4 +8) = 30

Hence,

Total no. of biscuits left = (60 – 30) = 30

13

A few candidates assumed that the total capacity of Container B would be 1000 ml and based

their calculations on that false assumption. A significant number of candidates were not able to

formulate an appropriate strategy to solve the problem.

Question 54

This was the most challenging question of the paper.

The most common mistake noted in part (a) was to read the sentence ‘His daughter received

1/3 of the amount of money received by his wife’ as meaning that the fraction representing the

amount of money received by the wife was 3

2. In fact, the sentence implied that the daughter

received 1 part and the wife received 3 parts of the total amount of money that was shared

between the two. So, the wife received 4

3 of this amount of money. It appeared that

Question 53

(a) Height of Container A : Height of Container B

5 : 4

5 parts 15 cm

4 parts 15 x 4 = 12 cm

5

(b) Height of water when Container B is one quarter full = ¼ x 12 cm = 3 cm

Increase in the height of water = (3 + 2) cm = 5 cm

Height of Container B which remains to be filled = (12 – 5) cm = 7 cm

2 cm of water 400 ml

7 cm of water 400 x 7 = 1400 ml

2

14

candidates had greater difficulty to interpret what the fraction ‘3

1’ stood for than to deal with

the concept of pie chart in this case.

Part (b) assessed candidates’ ability to solve a multi-step word problem on ‘numbers’.

Candidates’ scripts revealed that they were not very confident in tackling this question. The

question provided pieces of information which most candidates had difficulty in putting

together. Nevertheless, a few candidates solved the problem in a very coherent way.

Question 54

(a) METHOD 1: METHOD 2:

Angle representing ‘amount of money Angle representing ‘amount of money

shared between daughter and wife’ shared between daughter and wife’

= 360o - 144o = 360o - 144o

= 216o = 216o

Fraction of money received by wife 360o Rs 18 000

= 3/4 216o 18 000 x 162

360

Angle representing amount of money = Rs 10 800

received by wife

= 3/4 x 216o Amount of money shared between

= 162o daughter and wife = Rs 10 800

360o Rs 18 000 Fraction of money received by wife = 3/4

162o 18 000 x 162

360 Hence, wife received 3/4 x Rs 10 800

= Rs 8100 = Rs 8100

Hence, wife received Rs 8100.

15

Question 55

The most common mistake noted in parts (a) and (b) stemmed from the incorrect application of

the formula speed = distance/time. Since speed is quite an abstract concept, it is felt that

candidates would gain a better understanding if it were taught through candidates’ prior

knowledge of proportion rather than through the use of the formula. Pupils relate more easily

to notions of distance and time because these form part of their daily life experiences.

Expressing the speed of an object as the distance moved by the object every hour is thus more

likely to be understood than having to mechanically recall an imperceptible formula.

Part (c) was a little harder than parts (a) and (b). It was noteworthy that about a third of the

candidates managed to score partial marks in part (c). The main difficulty observed was for

candidates to keep different pieces of information in their minds whilst they were producing

new information. Often, in their attempt to find the distance that remained to be covered by

Shakeel, candidates multiplied the difference in the time taken by Shakeel and Tarun to

Question 54

(b) The cost of 4 pears = 4 x 8

= Rs 32

Implies,

Amount of money Nita had at first = cost of 16 oranges + Rs (32 + 3).

But, Nita & Asha had the same amount of money.

Implies,

The cost of 23 oranges = cost of 16 oranges + Rs 35.

So,

The cost 7 oranges (23 - 16) = Rs 35

The cost of 1 orange = 35/7

= Rs 5

Hence, Nita and Asha had Rs (23 x 5) = Rs 115 each at first.

16

complete the journey (half hour) by the speed of Tarun (36 km/h) instead of multiplying it by

the speed of Shakeel (30 km/h).

Question 55

(a) In 1 h, Shakeel travels 30 km

In 5 h, 30 x 5 = 150 km

Implies, journey is 150 km long.

(b) Tarun covers 32 km in 1 h.

To cover 96 km then, Tarun takes 1/32 x 96 = 3 h

Hence, Tarun covers the first 96 km in 3 h.

(c) Distance remaining for Tarun to complete the journey = (150 - 96) km

= 54 km

In second part of the journey, Tarun covers 36 km in 1 h.

So, to cover the remaining 54 km, Tarun takes 1/36 x 54 = 1 ½ h

Hence, time taken by Tarun to complete the journey = 1 ½ h

Total time taken by Tarun to complete journey = (3 + 1 ½ ) h

= 4 ½ h

Time difference between Shakeel and Tarun to complete the journey = (5 - 4 ½ ) h

= ½ h

Hence, Shakeel is (½ x 30) = 15 km behind Tarun at the moment Tarun completes

the journey.