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).

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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.

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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.

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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.

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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.

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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.

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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

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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

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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

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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)

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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

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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

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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

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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

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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

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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

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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

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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.