Cambridge International General Certificate of Secondary Education

0580 Mathematics November 2017

MATHEMATICS

Paper 0580/11
Paper 11 (Core)

Key messages

To succeed in this paper, candidates need to have completed the full Core syllabus, be able to remember
and apply formulae and to give answers in the form required. Candidates should be reminded of the need to
read all questions carefully, focussing on key words.

General comments

The paper was accessible to most candidates, with the majority attempting all questions. Candidates must
show all working to enable method marks to be awarded. This is vital in two or multi-step problems, in
particular with algebra, where each step should be shown separately to maximise the chance of gaining
marks. Rounding incorrectly and not using the appropriate degree of accuracy let down many candidates
particularly in Questions 5, 8, 9 and 16. The questions that presented least difficulty were Questions 1, 2,
10(a), 11, 14, and 19(a)(i). Those that proved to be the most challenging were Questions 8, limits, 15,
polygons, 20(b), surface area, and 21(b), identifying a region. Those questions that were occasionally left
blank were Questions 10(b), 15, 19(b) and 21.

Comments on specific questions

Question 1

This first question was answered well by the overwhelming majority of candidates, with most giving the
correct answer. A few added the given angles then subtracted 180. There were a small number of
candidates that made arithmetical errors, despite being able to use calculators for this paper.

Answer: 101

Question 2

There were many completely correct answers. However, some candidates reached the correct value but
then did further work in order to arrive at an answer with a decimal value, usually 9.944 or 99.44.
Occasionally, candidates used an incorrect method such as division or addition rather than multiplication.
This is a good example of where doing an approximate calculation would indicate the magnitude of the
expected answer.

Answer: 9944

Question 3

The most efficient method was to add the powers and this was often seen, and frequently evidenced as 21,
which was an acceptable answer. Others came to the correct answer by writing 0.0625 × 32 = 2, which was
a long method but nevertheless correct. Incorrect answers included 41, 2–9 and other various powers of 2,
either written as a power or evaluated, for example, 512.

Answer: 2

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 4

This was the first problem solving question on the paper where there was no scaffolding to aid candidates.
Many knew that the starting point was to find the total of Amber’s marks and those candidates frequently
went on to score full marks. Some got this correct with no working – showing no working in a two or more
mark question is a risky strategy. Some wrote (68 + 81 + 74 + 89 + x) / 5 = 80 but were then unable to solve
this, with 312x = 400, or similar, being commonly seen. Totally incorrect methods generally came from
finding the mean of the 4 given marks (78) or the marks being treated as a sequence. Thus, several times it
was seen that candidates wrote out the marks in ascending order and showed the term-to-term rule starting
with 68 + 6 = 74 and adding one more each time, finishing with 98 as her mark for the fifth test. This is an
incorrect approach as it is based on a complete misunderstanding of the scenario.

Answer: 88

Question 5

In both parts, a significant number of candidates gave answers that suggested that they had no
understanding of the basic concepts of rounding. Answers that were many orders of magnitude greater than
the given numbers were common and answers in which multiple digits had been changed were also seen in
many cases. Candidates were more confident with part (a), using decimal places, than answering part (b),
using significant figures. Some candidates gave an answer to part (a) of 18.7, a truncation to 3 significant
figures, or 18.77, rounding to 2 decimal places. In part (b), candidates often put a zero in each column of the
original number so gave 19.000.

Answers: (a) 18.8 (b) 19

Question 6

Many candidates did well with this question but the order of operations was a problem for some leading to
2 + 0.2
incorrect working of . A few got as far as 2 + 0.25 but this was not sufficient for a method mark.
0.8

Answer: 1.5

Question 7

Most candidates were able to factorise the given expression correctly and many were able to gain 1 mark for
either a partially correct or an incomplete factorisation. Some candidates did not recognise that there were
common factors and often seemed to be trying to factorise the expression as if it was a quadratic. A small
number of candidates identified a correct factor, but then divided every term rather than factorising and
arriving at the answer 4x + 5y – 3. A few candidates used an incorrect numerical factor leaving non-integer
coefficients in the bracket.

Answer: 3x(4x + 5y – 3)

Question 8

Candidates can find this area of the syllabus challenging and a number did not attempt this question. Very
many candidates realised that they had to go half a unit either side of the given measurement but were not
clear about the decimal places involved and what the half a unit was, so adding and subtracting 0.5 instead
of 0.05 was frequently seen. However, there were many answers that showed no understanding at all, with
numbers that were both greater than or both less than 14.3, or more than one unit from 14.3. A small number
gave two numbers that would round to 14.3 rather than the upper and lower bounds. Some candidates
earned 1 mark for one bound correct, often in the answer 14.25 and 14.34.

Answer: 14.25 14.35

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 9

In this question, one misconception was for candidates to use the circumference formula. Occasionally 2πr 2
22
was used as the formula. Of those who used the correct formula, some used 3.14 or for π even though
7
these approximation are insufficiently accurate. Of those that used the correct formula, some used the
incorrect value for the radius such as 3 cm or even the diameter, 9 cm.

Answer: 63.6

Question 10

Most candidates were able to identify the co-ordinates of point A but common incorrect answers included (3,
–2) (reversed co-ordinates) and occasionally (3, –1). Very occasionally, some candidates wrote
(x = –2, y = 3) which is not an acceptable form of the answer. In part (b), some answers were so inaccurate
that they implied that candidates did not have any understanding of the shape of a rhombus. In other cases,
the candidates’ plots were within half a unit of the correct position, suggesting that these candidates had
some understanding of what was expected, but were struggling to use the co-ordinate grid correctly.
Common errors included plotting D at (3, k) or (k, 3) or simply joining A to C to form a triangle.

Answers: (a) (–2, 3)

Question 11

Most candidates were able to identify the correct values in both parts. The common error in part (a) was not
cancelling completely. For part (b), the most common error was to give a decimal or percentage in which 9
did not have the correct place value.

5
Answers: (a) (b) 0.09 9%
9

Question 12

The biggest problem in this question proved to be in dealing with the mixed number, with a number of
candidates either omitting this step or making errors. Some felt that both fractions needed to be converted to
25 33
a common denominator, these often arrived at − . Only a tiny minority didn’t show any method at all; a
15 15
slightly larger number produced the correct answer, but showed calculations that were not completely
10 11
correct, such as − , presumably having arrived at the answer using their calculator. Others misread
15 15
11 11
as but were able to access a mark for the method of using common denominators. Candidates
15 5
should be clear that in fraction questions, a decimal answer is not acceptable.

14
Answer:
15

Question 13

Most candidates were able to gain some marks on this question, but there were few completely correct
1
answers. Both –7 and were suggested as the smallest number and some candidates thought that –7 and
3
–11 were natural numbers.

Answers: (a) 343 (b) –11 (c) 343

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 14

There were many completely correct answers that showed good understanding. The most common error
was to interpret the given vectors as fractions with some drawing a horizontal line in their answers. Inclusion
of such a line is not acceptable.

 2  2 8
Answers: (a)   (b)   (c)  
7 5  20 

Question 15

This was another problem solving question without the scaffolding to show candidates the first step. Also,
this was the question that was most often left blank; that could be a response to either the content or the
problem solving aspect. Many were able to perform a correct first step, either finding a value for the exterior
angle or the total of the interior angles. However, many seemed to be unclear what they had calculated, with
answers that implied that the interior angle of a regular pentagon is 72° being quite common. A minority
attempted to use the known angle of 90° in the quadrilateral formed by the line of symmetry, AB and the fact
that angle d added to 2 interior angles and 90° would total to 360°. However, these candidates were rarely
able to go on to solve their equation, since most did not identify the correct relationship, i.e. that
d + 2(2d) + 90° = 360°.

Answer: 54

Question 16

There were some excellent solutions but occasionally candidates gave the answer as 16 with no more
accurate value being stated. Candidates should know to always give a more accurate answer before
rounding. Those candidates who did not score often misapplied Pythagoras’ theorem or only completed one
step of a trigonometric argument.

Answer: 16.1

Question 17

For part (a), many candidates answered correctly but the most common incorrect answers included m3, m7,
m 25 and 2m 5 or only getting as far as m 5 × m 5. Again, many candidates did well with part (b), where the
errors included a mistake in one or more of the three elements, usually involving 4 + 5 = 9 or x2 × x3 = x6 or
y × y = y. A common misconception was to treat the question as though it had an addition sign and try to
factorise it, for example, x2y(4x + 5).

Answers: (a) m10 (b) 20x5y2

Question 18

To achieve full marks, the correct method must be seen as well as the values for x and y. Candidates need
to check for the most efficient way to tackle simultaneous equations so that they have as little opportunity as
possible to make numerical slips. Here, the simplest method to eliminate one variable is to multiply the first
equation by 2 so that there was a 6x term in each equation and then subtract the second from the first or to
multiply the second equation by 4 and add the two equations. A number of candidates re-arranged both
equations into x = ... (or y = ...), equated them and solved for x (or y). A few candidates used matrices but
those who did proved to be more likely to make sign errors than those who opted to use an algebraic
method. Many other methods, including substitution, will work but often have more opportunities for errors to
be made. Some candidates gained the mark for 2 values satisfying one of the original equations.

Answer: [x =] –2, [y =] 3

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 19

Part (a) was generally well answered with the overwhelming majority of candidates able to give the correct
three angles and most of these went on to draw the pie chart accurately. Occasionally, a mark for three
angles totalling 198 was awarded. It appeared that some candidates noticed that 180 – 18 (the number of
English speakers) is 162, so then went on to subtract each frequency in turn from 180, so showing a
complete misunderstanding of this area of the syllabus. With the pie chart, the most common errors were to
not label sectors, or to misread protractor scales. Part (b) proved to be more challenging. Answers that were
not fractions were common. Many gave the fraction that spoke Portuguese, rather than the fraction that did
not speak it. 70% was not acceptable as the answer, as the question asked for the fraction.

252
Answers: (a)(i) 99° 63° 36° (b)
360

Question 20

There were some excellent and completely correct answers to part (a). Some candidates showed
understanding but made errors in evaluating the area. In some cases candidates worked to an insufficient
degree of accuracy by rounding prematurely. Another approach was to divide up the shaded part in to
different rectangles and then combine them. This method had plenty of places where numerical errors could
and were made. One misconception was for candidates to assume the small rectangle was in the centre top
of the large rectangle which was not supported by the given information and it is not necessary to know
where the small rectangle is placed, in order to answer the question. There were a significant number of
candidates who did not know how to calculate the area of a rectangle; these found perimeter, or used
formulae involving doubling, halving or squaring the lengths of the sides. Candidates found part (b)
challenging. While some candidates did score full marks, a very common incorrect answer was 180 from
4 × (8 × 5) + 2 × (2 × 5) showing no logic to the understanding of the diagram. Examples of completely
incorrect answers included finding the volume or finding the total of all the edges or even using π.

Answers: (a) 71.48 (b) 132

Question 21

There were some excellent answers to this question but also many who omitted some or all parts. In part (a),
those who realised the need to use arcs in their accurate construction were usually able to do so to the
required degree of accuracy. The most common error was to draw the diagonal AC rather than the angle
bisector. A considerable number of candidates did not extend the line from C to meet AB which made it
impossible for them to answer part (b) correctly. Most were able to draw the correct bisector of DC, but not
all did so using an efficient method. The most common error was to only draw one pair of arcs. Again, a large
number of candidates didn’t extend their bisector as far as AB. If candidates had read the whole question
before starting part (a), they may well have drawn the bisectors long enough to enable them to go on to
identify the correct region in part (b). Even of those whose bisectors were long enough, very few were able
to identify the correct region. A small number of candidates produced shading for two or more regions, with
an overlap, but then didn’t give any indication as to which region they intended as their final answer.

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

MATHEMATICS

Paper 0580/12
Paper 12 (Core)

Key messages

Show working for all questions with more than one mark. Give answers to required or sufficient accuracy.

General comments

The standard of candidates’ responses was generally very good. However, it was noticeable that a
significant proportion of candidates either did not attempt some topics or their responses indicated lack of
experience of studying parts of the syllabus. It was also very common to see totally unrealistic answers to
questions. Triangle heights unrelated to given measurements and taxi fares in thousands of dollars were
examples of this. Just a few candidates appeared not to have a calculator, an essential item of equipment for
this paper.

Comments on specific questions

Question 1

Apart from the usual error of missing the zero or too many zeros, it was noticeable that some candidates
started with 40 instead of 14. Otherwise this question was well done.

Answer. 14 027

Question 2

While finding the temperature was well understood, the common error was to see the answer 3. Also a small
number of candidates added the data resulting in a response of 37.

Answer. −3

Question 3

Finding any number to the power 0 is quite often asked and most candidates knew the answer, or could use
their calculator. However, there were quite a number who gave 12 or 0.

Answer. 1

Question 4

Writing a number in standard form as an ordinary number seemed to be more of a challenge than the other
517
way round. Quite a variety of incorrect responses were seen, including 5170, 517, 51 700 and . Some
10 000
gave what seemed to be the correct answer but had no decimal point. Others had an incorrectly positioned
decimal point or even figures different to 517.

Answer. [0].00517

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 5

The ordering of the items was well done with much evidence of converting to decimals seen. The main errors
5 31 5
were before and last, presumably as it had 3 figures after the decimal point.
8 50 8

31 5
Answer. < < 0.63 < 64%
50 8

Question 6

This taxi journey question was one of the most challenging for candidates on the paper, with many totally
unrealistic responses given. Errors often resulted from reading the question incorrectly or ignoring the
mixture of cents and dollars in the data. 560 cents added to $4.50 ($564.50) and 4.5 added to 0.8 before
multiplication by 7 ($37.10) were regularly seen.

Answer. $10.10

Question 7

While there was quite a good response to this calculator question, poor use of the calculator and ignoring the
rounding instruction caused many to lose 1 mark or both. Simply entering 6.32 + 2.06 ÷ 4.15 – 0.12
produced the answer 6.696... An answer of 2.07 alone gained no mark while the often seen 2.08 indicated a
determination to give 3 significant figures regardless of the instruction of 1 decimal place.

Answer. 2.1

Question 8

While factors and multiples were generally well known, a considerable number of candidates did not gain the
marks in this question.

(a) 2 × 6, 3 × 4 was not a list of factors so did not score the mark. Other errors were missing a factor,
putting in incorrect factors and including multiples. Including 1 and 12 was not penalised.

(b) The question asked for multiples between 20 and 40, a fact ignored by quite a number of
candidates. Just one multiple was another error seen from a few candidates.

Answers: (a) 2, 3, 4, 6 (b) 27, 36

Question 9

The angle properties of straight lines and triangles were generally well known. Angle x was more often
correct than angle y while a few gained a mark for reversed answers or other cases of x + y = 100°. Negative
values were also seen.

Answer. [x =] 60 [y =] 40

Question 10

This percentage question was found challenging by many candidates. Many did not understand the
operation by multiplying 55 by 2.2 or dividing 2.2 by 55. Those who did understand how to tackle the
question often lost a mark by not changing or incorrectly changing both numbers to grams or kilograms. One
mark was quite common for figures 25 or 2200 or 0.055 seen.

Answer. 2.5

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 11

Candidates found finding the height of a triangle from a given area very challenging. The most common error
was to apply base × height = area rather than the correct formula. Some candidates did know the formula but
not how to change the subject to the height. Half the base, 16.5 was often seen as the answer.

Answer. 32

Question 12

A considerable number of candidates appeared not to be familiar with the topic due to the number of blank
responses and answers which were not correlation types.

(a) Those who were familiar with the topic nearly always gained this mark.

(b) In contrast, few candidates seemed able to cope with two unrelated properties so more blank
responses were seen. Many seemed to think that the only answer could be positive or negative.

Answers: (a) Positive (b) None

Question 13

Many candidates coped with finding the missing probability in the table but there were many who completely
misunderstood the method. Some added the probabilities and divided by 4 but also a significant number
added the three and subtracted it from 4 instead of 1. Other errors seen were simply the addition of the given
three, 0.65, as the answer, or an arithmetic error leading to 0.25. The latter would gain a mark if working was
shown but not without.

Answer. [0].35

Question 14

Upper and lower bounds is a topic candidates usually find challenging as evidenced by quite a number of
blank responses, but many gained the marks in this question. Some lost a mark by reversing the correct
limits or giving 362.4 as the upper limit. While many realised the limits had to be balanced around 362 there
was a great variety of responses for the limits.

Answer. 361.5 362.5

Question 15

Many candidates were successful in finding the angle in this right-angled trigonometry question. Pythagoras’
theorem was seen, although rarely leading to a correct long method. Nearly all who understood the topic
chose sine with only a few choosing cosine or tangent. Finding an angle seemed to pose more problems
than finding a side with a number not getting further than a correct expression for sin x. Insufficient accuracy
prevented some from gaining 2 marks with answers of 52 or 52.1 common.

Answer. 52.2

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 16

(a) Nearly all candidates coped with writing down the co-ordinates of a point with both numbers
positive. The only significant error was to reverse the co-ordinates.

(b) Again plotting was very well done with the same main error as in part (a). There were also a
number of careless mis-plots of the point C.

(c) Recognising the type of triangle was less well answered. Just the word ‘triangle’ was a common
response and had more joined the sides they may have recognised the type of triangle. It was
close to being ‘equilateral’ but again candidates would see, or realise from measuring, that only two
sides were equal. Other occasional incorrect responses were right-angled triangle and even
quadrilateral triangle.

Answers: (a) (2, 5) (c) Isosceles

Question 17

(a) Most candidates were able to give a correct response to the length of the line. The only noticeable
error was to misread the ruler to give the response 10 cm.

(b) While the mid-point was clearly identified by the vast majority, a point was not shown by quite a
number of candidates.

(c) Most candidates knew the term perpendicular but there were quite a number of parallel lines seen
as well as both parallel and perpendicular. Some candidates just drew any line at odd angles,
usually through the mid-point. Although the position of the perpendicular line could be anywhere on
the line, most drew (or constructed) the perpendicular bisector.

Answers: (a) 9

Question 18

Only a minority of the candidates found the correct angle for the hexagon. Many did not know that the
hexagon had 6 sides and 5, 7 and 8 sides were common. While 720 gained a mark, it was often seen as the
answer meaning many could not visualise the polygon or realise just the size of one angle was the question.
The step of finding the exterior angle, 60° gained a mark, but many did not take the final step.

Answer. 120°

Question 19

Many candidates found drawing a net of a cuboid with given dimensions rather challenging with a minority
gaining all 3 marks. Most had an idea of a net although quite a number of candidates drew a 3-D drawing or
showed disconnected rectangles. Common errors were to show 4 by 4 squares and adjacent edges not the
same length. Many lost a mark by drawing another 5 by 4 rectangle directly below the given one.

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 20

(a) Most candidates showed clear understanding of changing an improper fraction to a mixed number.
8
However, a significant number gave the answer as a decimal, 3.666.... or 3.7. Also seen was 2 .
3

(b) Basic fraction addition without the complication of mixed numbers was well done. Absence of
working by a few candidates meant no marks were awarded and it was quite common to lose a
mark by not reducing the answer to its lowest terms. This was quite common from those who chose
6 3
a common denominator of 48. Simply adding the numerators and denominators to give or
16 8
was seen from quite a significant number of candidates.

2 2
Answers: (a) 3 (b)
3 3

Question 21

Candidates found this question on finding the equation of a line challenging as they had to find both the
gradient and the y-intercept in one question rather than finding the gradient first. Consequently, only a
minority succeeded in gaining the 3 marks. Many attempted to find the gradient from the co-ordinates of two
points, but many who did this correctly did not know how to complete the equation. Some gained a mark
from realising that the constant was 2 but others lost that mark by just giving y = 2. Many candidates did not
appear to understand the meanings of m, x and c in the formula.

Answer. [y =] 0.5x + 2

Question 22

(a) (i) Most candidates worked out the common difference and were able to add it to 29 successfully.

(ii) This part was also answered well but quite a number of candidates did not interpret the question
correctly, usually by giving the expression for the nth term or n + 7. Another common error was to
write just 7 as the answer without any indication that it was added.

(b) Finding an expression for the nth term of a linear sequence wasn’t well answered with common
incorrect answers of n + 4, +4 or the next term, 18. Some candidates found the 4n part of the
expression but gave +2 or no constant. Others mixed up the parts of the expression to give the
response 2n – 4.

Answers: (a)(i) 36 (ii) Add 7 (b) 4n – 2

Question 23

(a) This question was one of the most demanding types of solving linear equations by having the
variable on both sides as well as a negative term and a fraction answer. However, quite a number
of candidates did find the correct solution, either as a fraction or decimal to a minimum of 3 decimal
places. The error of 11n – 3n was very common as well as other incorrect rearrangements of the
terms. Many who reached the stage of 5 = 14n believed that a fraction less than 1 could not be the
correct answer and so divided 14 by 5 to reach 2.8.

(b) There was a better response to this linear equation question with many gaining at least 1 mark for
multiplying 3 by 5 as the first step. Many candidates then either subtracted 3 from 15 or divided by
p
3. Dividing just p or just −3 by 5 as a first step, resulting in = 6 or p – 0.6 = 3, was seen a
5
number of times.

5
Answers: (a) (b) 18
14

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 24

Many candidates were successful in this similar triangles question. The usual error of addition or subtraction
of lengths immediately meant no marks in both parts. The majority who did use a ratio of corresponding
sides were successful in gaining the marks.

(a) The most common error was to add 5 to (15 – 12.5). Some candidates gained a mark for a correct
fraction from corresponding sides but could not finish off correctly to find x.

(b) 9.5 from 12 – (15 – 12.5) was the common incorrect answer in this part.

Answers: (a) 6 (b) 10

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

MATHEMATICS

Paper 0580/13
Paper 1 Core

Key messages

Ensure answers are given to the required accuracy and avoid premature rounding in working.
Show working for questions worth more than 1 mark.
Understand how to measure a bearing.

General comments

The standard of candidates’ responses was generally very good. However checking answers would help to
eliminate errors, especially with directed numbers. Candidates should ensure they read the questions
carefully, for example when an estimated answer is required, marks will not be awarded for the exact
answer. Some candidates appeared not to have access to a calculator.

Comments on specific questions

Question 1

Although several candidates were able to give the correct answer, many used column subtraction and as a
result common incorrect answers were 2 hours 72 minutes and 3 hours 12 minutes.

Answer. 2 h 32 min

Question 2

Many correct answers were seen. A small number of candidates did not attempt the question.

Answer. 84

Question 3

This question was not well answered, with trapezium and parallelogram being the most common incorrect
answers.

Answer. Kite

Question 4

Most candidates knew they had to add the indices, but some multiplied giving y 20 as the most common
incorrect answer.

Answer. y 9

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 5

(a) A significant number of candidates were unable to read the scale correctly, with the most common
error being 0.13.

(b) Although several correct answers were seen, there were many candidates who could not correctly
order decimals. 0.42 and 0.5 were seen in the wrong order as were 0.078 and 0.06. A small
number ordered them by the number of figures rather than the value.

Answers: (a) 0.16 (b) 0.06, 0.078, 0.42, 0.5

Question 6

(a) This part was almost always correct but the probability, rather than the colour, was also seen.

(b) Again nearly all candidates were able to give the correct answer. A small number of candidates
simply wrote 3.

3
Answers: (a) Yellow (b)
16

Question 7

Whilst this was fully correct for most candidates, several gained only 1 mark. The most common reason for
this was writing 8 rather than 80 for the percentage.

8
Answers: 0.25, , 80
10

Question 8

Many candidates were able to give the correct answer to this two-step question. Including both multiplication
of a vector by a number and subtraction of vectors in one expression made this quite demanding. The rules
of directed numbers was also tested which led to some errors.

 11 
Answer.  
 −7 

Question 9

The majority of candidates were able to give the correct answer to this question. The most common errors
were adding, rather than subtracting 2x and 4.

Answer. 5

Question 10

Many candidates did not show rounding and just attempted the question as a straight calculation. Others did
as instructed but often 59.2 was left or rounded to 59, resulting in maximum 1 mark. Many who rounded
correctly gained the 2 marks.

Answer. 20

Question 11

Although some candidates were able to give the correct answer, this question caused problems for some. A
small number appeared to have seen the word estimate, then not used the 40 and 6. Several others divided
6 by 40 but did not know what the next step was.

Answer. 120

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 12

Many candidates did not earn any marks on this question. The main error was to use 0.26 rather than 0.026,
or to calculate 2.6% simple interest.

Answer. 1263.21

Question 13

(a) Almost all candidates gave the correct answer.

(b) Again, the majority of answers were correct.

(c) There were many correct answers seen. The most common error was −13 from subtracting rather
than adding 3 to −10.

Answers: (a) Moscow (b) 8 (c) –7

Question 14

(a) The majority of candidates had the frequencies correct, but a significant number lost a mark as they
converted them into probabilities or wrote the cumulative frequency in the frequencies column.

(b) Whilst most answers were correct, 80 to 89 was seen at times. Some candidates did not understand
‘modal group’ and gave a single value.

Answers: (a) 4, 5, 6, 3, 2 (b) 100 to 109

Question 15

Although finding the interior angle of a polygon is tested regularly, the correct answer was rarely seen. A
significant number of candidates gained 1 mark for dividing 360 by 12, but few knew what to do after this
step.

Answer. 150

Question 16

Many candidates were able to change a mixed number to an improper fraction and find a common
denominator. It was much more common to see the mixed numbers turned into improper fractions than
dealing with the whole number part and fraction part separately. A small number of candidates did not show
all working and so lost the method mark, and / or the accuracy mark for leaving their answer as an improper
fraction or for converting their answer to a decimal.

25
Answer. 1
28

Question 17

The majority of candidates were unsuccessful on this question. Pythagoras’ theorem was seen, although
rarely leading to a correct long method. Of those who used trigonometry many used cosine or tangent rather
than sine. Many candidates who correctly identified sine were unable to rearrange the formula correctly.
Insufficient accuracy prevented some from gaining full marks with an answer of 11 being common.

Answer. 10.9

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 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 18

(a) Many correct answers were seen. The most common incorrect answers were 1800 and 180 000.

(b) The majority of candidates were able to perform the calculation but had problems with the standard
form. Many gained 1 mark for figures 215. Others appeared not to understand that standard form
has only 1 digit before the decimal point.

Answers: (a) 18 000 (b) 2.15 × 106

Question 19

(a) Several candidates plotted the point (100, 60) but did not draw the line and as a result only gained
1 mark. A smaller number of candidates drew an incorrect line.

(b) (i) This part was mainly answered correctly.

(ii) Many correct answers were seen. The most common error was to give the answer 16 from
confusing the scales. The common answer of $33.33 implied calculations rather than use of the
graph.

Answers: (b)(i) 82 to 86 (ii) 31 to 35

Question 20

(a) (i) This part was almost always correct.

(ii) Many correct answers were seen. The most common incorrect answers were 6 and n + 6.

(b) There was a lack of understanding shown of the nth term expression. Some simply wrote +3 and a
small number of candidates gained 1 mark for 3n.

Answers: (a)(i) 34 (ii) add 6 (b) 3n + 8

Question 21

(a) Many correct answers were seen, with almost all candidates interpreting the scales correctly.

(b) This part was not well answered. The most common incorrect answer was 9.5 (the distance from
A to B in centimetres). Many candidates appeared not to understand how to measure a bearing.

(c) Few fully correct answers were seen. Many candidates had drawn a line which did not reach AB.
Many candidates drew arcs from all four corners, but few knew how to use them. Some simply
drew a line from A to C.

Answers: (a) 168 (b) 74

Question 22

Although a small number of candidates gained all 5 marks on this question, the majority of candidates made
the same error. The area of the square was calculated correctly but many then calculated the area of the
circle but did not halve their answer to find the area of the semi-circle. Therefore the most common answer
was 178.5. Many of those who did not know how to work out the area of the circle at all, generally gained a
mark for either the area of the square or for knowing the radius was 5. Several candidates gained the mark
for correct units.

Answer. 139 cm2

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

MATHEMATICS

Paper 0580/21
Paper 21 (Extended)

Key messages

To succeed in this paper candidates need to have completed full syllabus coverage, remember necessary
formulae, show all necessary working clearly and use efficient methods of calculation. They should be
encouraged to spend some time looking for the most efficient methods suitable in varying situations and
remember to check working and answers for avoidable mistakes such as arithmetic slips, copying previous
working incorrectly and misreading numbers in the question. Rounding values within working should be
avoided.

General comments

The level and variety of the paper was such that candidates were able to demonstrate their knowledge and
ability. There was no evidence that candidates were short of time, as most candidates attempted the whole
paper. Appropriate, thorough working was seen throughout the paper.

Candidates demonstrated a very good understanding of indices in Questions 2 and 13, dealt with the
fractions extremely well in Question 10 and could solve the data handling problem in Question 4.

Candidates particularly struggled with the bounds in Question 8, the vector problems in Question 14, set
notation in Question 15 and the 3D trigonometry problem in Question 21.

Comments on specific questions

Question 1

Almost all candidates gave a correct response in this question. The most common incorrect answer was 97,
using the rule for cyclic quadrilaterals.

Answer: 101

Question 2

The vast majority of candidates clearly understood indices or used their calculator efficiently to give the
correct value. 21 was seen regularly but was condoned in this instance. Of the few who did not gain the
mark, a common error was to multiply the powers and/or the 2s.

Answer: 2

Question 3

The majority of candidates demonstrated that they could use their calculator efficiently to obtain the correct
answer in part (a). Some lost the mark because they did not read the question and rounded prematurely.
Others showed a lack of understanding in order of operations and there were two common incorrect answers
as a result of this. The most common value of 1.305 came from (numerator ÷ 0.13) – 0.015 and the less
common answer of 1.01915 from 50.4 − 3 in the numerator. Most candidates were able to round their answer
to part (a) to 2 significant figures to gain the mark in part (b). The main reasons for the mark not to be
awarded were rounding to 2 decimal places rather than 2 significant figures and adding zeros after the 5.

Answers: (a) 1.49220 (b) 1.5

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 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 4

The vast majority of candidates understood the problem and gained both marks. Good, concise answers
included the full equation, then clear rearrangement to the answer. A common error was to divide by 2
instead of 5. Another error was to average the four given marks. Some gained the method mark but then
simplified to 312x rather than 312 + x. A few noticed that the four given numbers arranged in order gave an
increasing sequence, and gave the 5th value as 89 + 9 = 98.

Answer: 88

Question 5

Most candidates understood what was required in this factorisation although many were trying to factorise
into two sets of brackets and so did not score. Many candidates scored 1 mark for either a partial
factorisation or for making a slip and getting 2 out of the 3 terms inside the bracket correct.

Answer: 3x(4x + 5y – 3)

Question 6

Very few errors were made giving the co-ordinates in part (a), with just a few candidates reversing the
co-ordinates. Fewer marks were gained in part (b) for various reasons. Some candidates, who obviously
knew the properties of a rhombus, lost this mark due to inaccuracy, perhaps trying to construct rather than
use the co-ordinate grid. One of the most common incorrect answers was for candidates to simply draw a
line to connect A and C. Points (3, 3), (1, 1) and (4, 4) were often seen as point D, perhaps confusing the
shape with a kite. Others made AD horizontal or CD vertical.

Answer: (a) (–2, 3)

Question 7

Candidates understood the concept of acceleration and there were many completely correct graphs. There
were also a large proportion of candidates who gained 1 mark, almost always for a horizontal line following
an incorrect speed reached in the first 30 seconds. This speed was often 0.4, 4 or 6. Sometimes a horizontal
line at a speed of 0.4 was drawn for the whole width of the graph.

Question 8

This question proved to be among the most challenging on the paper. A small number of candidates worked
in millimetres and then converted back to centimetres; however some did not convert back and so lost the
final mark. The B1 mark was awarded fairly regularly, as many candidates gave the upper and lower bounds
in their working, but then continued their work incorrectly. While most candidates understood the need to
multiply by 3, the majority of answers which did not score any marks originated from applying a limit of
accuracy to the combined height of the stack of 3, rather than each individual cuboid. Working of
6.5 × 3 = 19.5 with answers of 20 or 19.55 were common. Those who did deal with individual cuboids often
gave an answer of 21 from 6.5 + 0.5. Candidates should understand that due to the nature of a bounds
question, it is incorrect to apply any rounding to the final answer, as many lost the final mark by rounding to
19.7 on the answer line.

Answer: 19.65

Question 9

The concept of decreasing exponentially caused problems for some. However, the majority scored at least 1
mark and there were many fully correct answers. Candidates need to be aware that exact answers should
not be rounded as some lost the answer mark by rounding to 7615 or 7620. The most common
misconception was to find 15% of the original price and to subtract three lots of this, leading to an answer of
6820. Others increased the value by 15% each year. Some candidates worked year by year which didn’t
cause too many problems in this instance as each value was exact. They should be encouraged to use the
formula and perform one calculation rather than carry out several calculations which often lead to errors.

Answer: 7615.15

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 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 10

Candidates had an excellent understanding of dealing with fractions, showing clear and concise working.
5
Most candidates converted to correctly and then used 15 as the common denominator. There were two
3
2
main issues for candidates losing marks. Firstly, 1 was converted incorrectly, but many did then get the
3
follow through mark for correctly finding a common denominator. The second common error was multiplying
11 25 33 8
the numerator in by 3, resulting in − =− . If candidates used a larger common denominator,
15 15 15 15
most cancelled down successfully for their final answer.

14
Answer:
15

Question 11

There were many fully correct answers to this angle problem and the majority scored at least 1 mark, usually
for showing a correct calculation to find either the sum of interior angles or an exterior angle. A large number
of candidates then did not know how to proceed and many simply halved the exterior angle of 72 to give 36
as their answer.

Answer: 54

Question 12

The majority of candidates answered this question well. Almost all were awarded at least 1 mark, with many
gaining all 3 marks. The vast majority were able to choose the cube number in part (a), the most common
1
incorrect choice being , presumably because of the digit 3. Slightly fewer could identify the lowest number
3
in part (b). A number of candidates seemed unaware that negative numbers were smaller than positive
1
fractions so was a common incorrect answer and some chose –7. Part (c) caused the most problems
3
where some may have been distracted by the 343 having already been used. This was obviously an
unfamiliar term to many who appeared to have a guess at this final part of the question. 5 was a common
incorrect choice, perhaps confusing with irrational.

Answers: (a) 343 (b) –11 (c) 343

Question 13

Candidates demonstrated an excellent knowledge of the laws of indices in this question. In part (a) the only
occasional error was to give the index as 7. There was a slightly lower success rate in part (b) where
sometimes the numbers were not combined or y did not have a power. A few candidates thought they had to
factorise, treating the multiplication sign as addition.

Answers: (a) m10 (b) 20x5y2

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 14

This was one of the most challenging questions on the paper and many candidates struggled with both parts
of the question, demonstrating a need for more understanding in the application of vectors. Candidates
should be encouraged to sketch diagrams in this type of question in order to understand the problem.
Part (a) was the better attempted part of the question. The main error here was to attempt to subtract the
values, giving the answers (5, 6) and (–5, –6) or a combination of these. Some candidates seemed to be
trying to find the midpoint of (2, –5) and (7, 1). A high proportion of candidates could not make an attempt at
part (b). The minority of candidates who answered correctly showed clear working and understood what was
required. Some understood that Pythagoras’ theorem was involved and made a correct start of
13 = t 2 + 122 or 132 = t 2 + 122 but were then unable to successfully re-arrange the equation to find a
solution, with many simply ‘cancelling’ all the squares and roots, resulting in 13 = t + 12. This is not usually
the case in a more familiar Pythagoras’ theorem setting, in which candidates are well-practised (indeed this
was not the case in Question 21). Others attempting Pythagoras’ theorem began with t 2 − 122 or 122 − t 2 . A
mark was often awarded for an answer of 5 where candidates had forgotten that the value of t was negative.
For the majority of candidates who did not understand the nature of the question, 25 and –1 were popular
incorrect answers, both resulting from simply adding or subtracting the 12 and 13.

Answers: (a) (9, –4) (b) –5

Question 15

Set notation appears to be an area which still requires more understanding. One of the most common
responses in part (a) was to include the whole of the original set and others added in negative values or
added the values together. Many candidates were clearly not familiar with the notation and did not attempt to
give an answer. Part (b) was more familiar to candidates. The right-hand diagram was better attempted than
the left where it was common to see the whole of N unshaded or the area outside of M and N unshaded.

Answer: (a) Fewer than 6 elements from {1, 2, 3, 4, 5, 6} or ∅

Question 16

The majority of candidates recognised the transformation as enlargement and gained the first mark. Some
didn’t earn this mark by stating that it was negative enlargement. The centre of enlargement was also
generally well done; most candidates who answered incorrectly here either gave the origin or (1, 2) although
a variety of points were given, both in between the triangles and to the right of triangle A. The scale factor
1
was the least well answered part of the question, with the most common answers of 3, –3 or sometimes −
3
1
or being given. The answer 3 may have been due to the candidate not reading the question correctly and
2
giving the enlargement of B to A. Combinations of transformations were only rarely seen; this was usually
when the candidate thought that a translation was also involved.

1
Answer: Enlargement, , (2, 1)
3

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 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 17

The majority of candidates understood the relationship and there were many fully correct answers.
Candidates should be aware that they may be asked to give an equation as a final answer as in part (a),
rather than a value of y, which they may be more familiar with. Many gained 1 mark for starting with a correct
k 1
equation and obtaining k = 72 but then reverted to a final answer of or sometimes . There
( x + 1) 2
( x + 1)2
were also those who found the correct equation and then substituted x = 0.2 back in to give a final answer of
y = 50. Where candidates did not score any marks, the most common approaches were to work with y
proportional to (x + 1)2 or inversely proportional to (x + 1). Virtually all candidates with the correct equation,
or a correct relationship containing k in part (a), gained the mark in part (b). Sometimes the value of k was
calculated in this part of the question. A follow through mark was awarded fairly frequently to those with an
incorrect relationship in part (a).

72
Answers: (a) (b) 32
( x + 1)2
Question 18

The majority of candidates were aware of the requirements to construct a bisector and an arc and attempted
this well. Most drew the bisector first, as usually just a very small arc marking the correct position of S was
seen. Good answers gave clear arcs and bisector, and clear indication of point S. A number of answers did
not fulfil all the requirements. Candidates should be aware that a construction question requires all
construction lines and arcs to be clearly shown; trial and error measurement with a ruler is not sufficient. A
bisector must show arcs and not be drawn using an angle measurer and a ruler and the majority of
candidates did this well. It was more common to either only go as far as drawing the bisector, or to give no
constructions and give S as a (correct) point, sometimes with the bird bath arc seen as well but sometimes
as a point with no other working on the diagram. Some candidates did give full constructions of the arc from
the bird bath and bisector, but then did not recognise that their intersection was S. Sometimes S was not
shown at all or it was placed in an incorrect position, usually somewhere on their bird bath arc, bisector
construction arcs or at the mid-point of the trees.

Question 19

Candidates appeared very competent in dealing with the two algebraic fractions and often scored 2 marks for
8 x + 26 1
reaching . Many then struggled to deal with adding and should understand that adding
( x − 3 )( x + 7 ) 2
numerical and algebraic fractions involves exactly the same process. The 1 and 2 were often just added to
the numerator and denominator. Many added (x – 3)(x + 7) to the numerator but forgot to multiply 8x + 26 by
2 or also multiplied (x – 3)(x + 7) by 2. Candidates sometimes ‘cancelled’ out the (x – 3)(x + 7) in the
numerator with the brackets in the denominator. Many attempted to deal with all three fractions together and
those who wrote clear, methodical working with the correct brackets were often successful in reaching the
correct answer. Those who had working in different parts of the working space and were careless with signs
and brackets almost always made errors while multiplying out and simplifying. Extra terms were often seen
as a result of multiplying a bracket by two values e.g. (2)(5)(x + 7) turned into 5x + 35 + 10 or 10x + 70 + 10.

x 2 + 20 x + 31
Answer:
2 ( x − 3 )( x + 7 )

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 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 20

The majority of candidates realised that a simple multiplication of the area of the cross section and height
was required to find the volume in part (a). Those who were unaware of this tried to use the formula for the
volume of a cylinder and often got into difficulties. Those who did successfully work backwards to find the
radius of the base circle to use the volume of a cylinder formula, usually ended up with a number slightly off
the correct answer because of rounding. Candidates had much more understanding of the relationship
between height and area scale factors than in previous sessions in part (b). Where the correct answer was
not given, it was usually 90, with the area scale factor used for height. Some gained a mark for using the
correct ratio but the wrong way round, finding the height of B to be less than 10 cm. A few used the cube root
rather than the square root.

Answers: (a) 1480 (b) 30

Question 21

The main challenge in this question was to work out which angle was required and numerous incorrect
angles were suggested and calculated. If incorrect working followed, it was often difficult to award a mark as
candidates did not make it clear which angle they were attempting to find. Candidates should always be
encouraged to draw diagrams and label points, as this helps both the candidate in their further working and
the possibility of earning method marks. If the correct angle was identified, the main error was to give MC a
length of 6. Many candidates did use Pythagoras’ theorem, usually to find the length of AC which was then
halved for MC. There were a large number of candidates who lost the final answer mark because they
prematurely rounded 288 to 17 before halving to 8.5 which lead to an inaccurate answer. Candidates
should be encouraged to leave values in the calculator or to work to at least 4 decimal places in working.
This question was an example where inefficient methods were sometimes used; for example, finding MC and
then EC unnecessarily and using this length with sine or cosine rather than the given length of 9 and the
tangent ratio.

Answer: 46.7

Question 22

Candidates clearly understood how to interpret the graph and the majority gave the correct answer in part
(a). If full marks were not awarded, a method mark was frequently given for showing 120 with the candidate
then not completing the question by subtracting that value from 200. The graph was also drawn well in part
(b), with clear points plotted and clean curves drawn. There were very few candidates who lost marks
through drawing inaccurate curves through the points. The majority of marks lost were due to errors in
plotting, due to misreading the scale of the y-axis. Part (c) was less well answered. Many candidates did
earn full marks, and a method mark was awarded frequently for 130, perhaps retrieved from the table rather
than the graph. A very common incorrect answer was 25 which appeared to come from 3 different incorrect
methods. The most common appeared to be candidates using the mid-point between zero and the highest
value of each set of data i.e. 110 and 85. The second was using the mid-point of the range for each set of
data i.e. 160 and 135. The third was using 120 as the middle of the population rather than 100, perhaps
confusing with the line in part (a), resulting in values of 160 and 135.

Answer: (a) 80 to 84 (c) 26

Question 23

There were many correct inequalities given in part (a), although candidates often lost a mark for using strict
inequalities. There were a significant proportion of inequality signs which were reversed. Some candidates
did not understand what was being asked of them in part (a) but clearly understood the concept, as they
would often draw one, or even both lines correctly with correct shading in part (b), following no response, or
incorrect inequalities in part (a). It was common for candidates to score 2 marks for drawing and shading
x ⩾ 4 correctly in part (b). Lines were often drawn at y = 16, y = 12 and x = 8 rather than x + y = 16. In part
(c) the majority of candidates understood the need to choose a point within the region, and so even if they
did not comprehend where the optimal point would be, often scored 1 mark for showing correct working for a
point in the correct, or their, region.

Answers: (a) x + y ⩽ 16, x ⩾ 4 (c) 144

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

MATHEMATICS

Paper 0580/22
Paper 22 (Extended)

Key messages

To succeed in this paper candidates need to have completed full syllabus coverage, remember necessary
formulae, show all necessary working clearly and use a suitable level of accuracy.

General comments

The level of the paper was such that all candidates were able to demonstrate their knowledge and ability.
There was no evidence that candidates were short of time, as almost all attempted the last few questions.
Candidates showed evidence of good basic skills with the most success in Questions 1, 5, 13(b), 16(a) and
24(b). Most candidates were very good at showing their working although a few candidates showed just the
answers and missed the opportunity to gain method marks when answers were incorrect. Some candidates
did not work to an appropriate accuracy, losing marks due to not giving answers correct to three significant
figures; this was evident mostly in Question 24(a). Rounding or truncating part way through the working,
with inaccurate final answers as a consequence, was also evident, particularly in Questions 8 and 26.
Candidates found the locus question, the vectors question and the factional negative indices question
particularly challenging.

Comments on specific questions

Question 1

This question was well answered with nearly all candidates reaching the correct answer. Common errors
were finding 20 – 17 or 20 + 17 instead of 17 – 20.

Answer: –3

Question 2

This question was generally well answered. Occasionally candidates had the right idea but made an error
with the power, leading to answers such as 0.0517 or 0.000 517. Another error occasionally seen was to
517
express the answer as the fraction .
100 000

Answer: 0.005 17

Question 3

This question proved to be a good discriminator and was one of the most challenging questions on the
paper. A significant number of candidates thought two points were required rather than two lines, with the
most common incorrect answers being N and M or C and A. A small number stated BC with AC instead of
AB. It was also common to see the incorrect answer BN and BM.

Answer: BC AB

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 4

In part (a) most candidates gave the correct answers. A small number of candidates also listed 1 and 12,
which still gained the mark. Occasionally candidates listed their factors as product pairs; 2 × 6, 3 × 4 which is
not an acceptable answer. Part (b) was also answered successfully by many candidates, with most giving
the correct answers of 27 and 36. On a few occasions candidates also added an extra number, often 18. As
the question states the multiples should be between 20 and 40, including extra multiples meant the mark
could not be awarded. A common incorrect answer was 3, 4, where candidates were presumably indicating
that the third and fourth multiples were between 20 and 40.

Answers: (a) 2, 3, 4, 6 (b) 27, 36

Question 5

Nearly all candidates achieved full marks on this question. This was the best answered question on the
paper. Very rarely only 1 of the 2 marks was awarded which was usually for the answers reversed.

Answer: x = 60 y = 40

Question 6

This question was well answered with many candidates obtaining the correct answer of 2.5%. The most
common cause of error was incorrect conversion of 2.2 kg to grams or of 55 g to kilograms; in most cases
candidates had the misconception that there were 100 g in 1 kg rather than 1000 g. This usually led to an
answer with the correct digits so these candidates scored 1 mark as did those who forgot to multiply by 100
when expressing as a percentage, i.e. giving the answer 0.025.

Answer: 2.5

Question 7

This question was well answered by the majority of candidates. The most common error was to use
1
base × height to give the area of a triangle, instead of using × base × height, so a common incorrect
2
answer was 16.

Answer: 32

Question 8

Many candidates made a good attempt at this question and gained both marks. The more able candidates
55
who worked with fractions, i.e. 18 × usually earned full marks whereas for those who switched to
60
decimals, premature rounding or truncation sometimes meant that only 1 or no marks were earned. It was
common to see, e.g. 18 × 0.92 = 16.56 with no other working. There were occasional instances of division of
speed and time, rather than multiplication.

Answer: 16.5

Question 9

Many candidates successfully answered this question often with no working shown as calculators were
usually used. There was evidence of some working without a calculator, usually inefficiently re-writing as
normal numbers with lots of zeros rather than the method, e.g. 0.12 × 1041 + 1.2 × 1041. The most common
error was to work out 1.2 + 1.2 and then incorrectly deal with the powers of 10 in some way and the incorrect
answer 2.4 × 1081 was often seen. Occasionally candidates multiplied rather than added the two numbers
together.

Answer: 1.32×1041

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 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 10

This proved to be one of the more challenging questions on the paper although many candidates still scored
at least 1 mark. Some candidates lost a mark due to rounding their answer to three significant figures and
writing 20.8 as their final answer; candidates are advised that bounds should not be rounded. Common
errors were: subtracting 0.5 cm instead of 0.5 mm, i.e. 5.2 – 0.5 leading to a very common incorrect answer of
19.4; finding the perimeter of the triangle then subtracting 0.05 leading to an answer of 20.85 or finding the
perimeter of the triangle then subtracting 0.5, giving 20.4 as the answer. A small number of candidates
converted the original numbers given to millimetres and used this to find the lower bound. Of these
candidates the majority scored 1 mark instead of 2 as it was rare for the answer to be converted back into
centimetres. Very occasionally, candidates found the correct lower bounds, 5.15, 6.25 and 9.35, but then
used them incorrectly, for example by trying to find the area.

Answer: 20.75

Question 11

16
The response to this question was mixed. Many candidates obtained a correct fraction, usually , but were
33
unable to show the required working. Some candidates incorrectly thought the decimal was 0.488888 and
22
obtained a common incorrect answer of . The less able candidates ignored the recurring decimal
45
48 12
altogether and wrote down or . The more able candidates showed accurate working usually by
100 25
using x = 0.484848 and 100x = 48.484848 and subtracting these two equations to reach 99x = 48. On
some occasions candidates wrote out many different powers of 10x but were unable to correctly identify
which ones to subtract to cancel out the recurring decimals, e.g. it was common to see 100x–10x used.
0.48 3
Another common incorrect answer was = .
100 625

48
Answer:
99

Question 12

This question was generally well answered with most candidates correctly expanding the brackets. A
common error here was writing the final term as –n instead of –n2 or leaving that term out completely. The
most common errors on this question came from further working after successfully expanding the brackets,
either multiplying through by –1 to give n2 – 2n – 15 or re-factorising the expression. Most of these scored 1
mark for the original multiplication.

Answer: 15 + 2n – n2

Question 13

Nearly all candidates answered this question correctly, particularly part (b). A few did not understand the
meaning of mixed number. Sometimes an integer was followed by an improper fraction or a decimal answer
was given, usually 3.67. Most candidates handled part (b) very well. Most candidates tended to use the most
efficient 12 as the common denominator, although 48 and less commonly 24 were also seen. A small
8
number of candidates didn't cancel enough and left their answer as . Very rarely, with the less able
12
candidates, it was evident that two numerators were added together and then the two denominators were
6
added to reach instead of finding a common denominator.
16

2 2
Answers: (a) 3 (b)
3 3

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 Cambridge International General Certificate of Secondary Education
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Principal Examiner Report for Teachers

Question 14

This question produced a variety of responses and proved to be one of the more challenging questions.
More able candidates usually earned full marks. Of those who did not earn all 3 marks there were many who
reached the correct statement, –2 < n ⩽ 3, and did not list the integers. Some lists were incomplete, often
leaving out the 0, perhaps indicating the candidates were not sure whether 0 was an integer. Some lists
offered only the positive integers 1, 2 and 3. Some candidates earned 1 mark for making the required correct
progressive step in the solution of either the left or right hand side of the inequality, or both sides. The double
inequality proved too much for some; the less able candidates added 5 to the right hand side, leading to
2n – 1 ⩽ 10.

Answer: –1, 0, 1, 2, 3

Question 15

The majority of candidates recognised the need for a common denominator to gain at least 1 mark with many
showing sufficient working to gain a second for a correct first step with the numerators. There was a
significant number who were unable to deal correctly with the subtraction of the negative when simplifying
y−x
the numerator, leading to the common incorrect answer of which was almost as common as the
xy
correct answer. Some candidates showed poor understanding of efficient algebraic notation with
unnecessary brackets or missing brackets, e.g. a denominator of (x)(y) or having a numerator of
y × x + 1 – x × y – 1. A minority cancelled inappropriately i.e. not common factors, sometimes after a correct
first stage.

y+x
Answer:
xy

Question 16

Part (a) was answered very well with most candidates giving the correct answer of –1. A very small number
of candidates made an arithmetic error and some tried to work to the left rather than the right and found the
preceding value, 29. Part (b) was more challenging for candidates. Many gave the correct answer of
–6n + 29. Some candidates gave their answer in an unsimplified form, for example 23 – 6(n – 1) or
23 + (n – 1)(–6) after using the general formula for the nth term of an arithmetic series. A small number of
candidates left this in the form 23 + (n – 1) – 6 without the essential brackets which could not score any
marks. Another, fairly common, error was to give a final answer of 6n + 17, coming from use of a common
difference of +6 instead of –6, giving 23 + 6(n – 1).

Answers: (a) –1 (b) –6n + 29

Question 17

There were many good responses seen to this question, but some candidates struggled to make any
progress here. Those who identified that x + 29x = 180 were typically able to solve and find the correct value
of x. It was also quite common to see x + 29x = 360 leading to x = 12 and then often a final answer of 30.
Some candidates correctly wrote down x + 29x = 180 but then followed this by, e.g. 2x = 180 – 29 or
2x = 180 ÷ 29. A significant number of candidates quoted many formulae but were unable to solve them as
they involved both n and x. Some calculated that x = 6 but then divided a number other than 360 by this e.g.
540.

Answer: 60

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 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 18

This question was generally well answered with the most common, and most successful, method being the
x
substitution of y = or x = 2y into 2x – y = 1. The main error here came from mistakes when multiplying
2
through by 2. Other candidates re-arranged the equations in order to eliminate a variable by subtraction or
addition. Those who used this approach were more likely to make errors with algebraic manipulation,
particularly with negatives. However, when an incorrect value for x or y was found, candidates usually
evaluated the other value correctly, leading to a pair of numbers that satisfied one equation and therefore
had the opportunity to gain a mark. A further reason for losing marks was when candidates switched to
decimal answers rather than giving fractional answers. Often these were not correctly rounded to three
significant figures, for example x = 0.666 or 0.67 were quite common and then this often led to inaccuracies
in the y value also.

2 1
Answer: x = , y=
3 3

Question 19

Many candidates scored 2 or more marks on this question. Some made the error of simply transferring the
square root to the other side instead of using inverse operations. Some candidates got their operations in the
wrong order; it was common to see the 1 subtracted first then followed by square root. Some candidates
believed that a + b = a + b so a common incorrect first step was to write y = x + 1. A small number of
candidates did not have a long enough square root sign. A few candidates wrote some incorrect subsequent
work following the correct answer, e.g. attempting to cancel the square of y with the square root. A few
candidates made an error in the second step obtaining x2 = 1 – y2 but often they still scored 2 marks.

Answer: y2 −1

Question 20

This question was usually answered correctly with a roughly equal split between those using the formula for
the area of a trapezium and those using the area of two triangles and a rectangle added together. Usually
those who used the area of the trapezium formula were more successful as the most common error was
omitting the division by 2 when finding the area of one or both of the triangles.

Answer: 132

Question 21

This question proved to be one of the most challenging on the paper. Many candidates scored 0 or 1 mark
only. Correct notation is an area candidates need to work on, e.g. candidates are writing K or OK rather than
uuur
OK . There was evidence that many candidates did not appreciate the significance of the direction of the
uuur uuur
vector and who thought that AB was the same as BA . Some candidates gained a mark for writing
uuur uuur uuur uuur
AB = b – a or for writing OK without further correct working. Many attempted AK having found AB
uuur uuur uuur
without realising they needed to go further to find OK . Some who did go further attempted, e.g. OB + BK
uuur
but had a sign error, instead using a vector equivalent to KB . There were a number who worked through to
a correct unsimplified answer gaining 2 marks then made errors in their attempts to simplify.

1 2
Answer: a+ b
3 3

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 Cambridge International General Certificate of Secondary Education
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Principal Examiner Report for Teachers

Question 22

This question was not answered as well as many, with quite a few partly incorrect answers seen, although
few candidates scored 0. Many realised that w and x were opposite angles in a cyclic quadrilateral and thus
that w + x = 180 and many were also able to use the angles in quadrilateral ABCD giving w + x + y = 240.
Common errors included w = 108, presumably thinking that it was in the same segment as the angle at O,
which led to the fairly common response w = 108, x = 72 and y = 60, or writing w = 60, possibly thinking that
ADC and DCE are alternate angles, which led to the response w = 60, x = 120 and y = 60.

Answer: w = 54, x = 126, y = 60

Question 23

Full marks were not scored very frequently on this question. Many candidates were able to identify at least
30
one of the required formulae, usually × π × 62 , although this did not always result in a correct value for k
360
as candidates were not always leaving their answers in terms of π and it was quite common to see this
evaluated to 9.42. The more able candidates also knew the area of the triangle could be found by using
1 2
× 6 × sin30 . However it was quite common to see some candidates use a much more cumbersome
2
method. For example, by dividing the triangle in half with a line from O to the mid-point of the chord, then
using right-angled trigonometry to calculate the base and height of each triangle, and then 0.5 × b × h. It was
quite common for those that employed this method to prematurely round during their working resulting in an
inaccurate answer, such as c = 8.99. A few used the sine rule to calculate the base of the triangle and
Pythagoras’ theorem to find the perpendicular height of the triangle. Of those unable to find the area of the
30
sector, the most common incorrect calculation seen was × 2π × 6 . A few candidates were familiar with
360
radian measure and calculated the angle θ in radians then used the sector area formula 0.5 r2θ. However
1
some used × 62 × sin θ with their calculator still in degree mode. It was quite common to see c given as –9.
2
Many stopped at an answer of 0.425 and were unable to progress from this to relate it to (kπ – c), often
equating it instead.

Answer: k = 3, c = 9

Question 24

Part (a) was generally well answered with many candidates obtaining the correct answer expressed as a
fraction. Some candidates converted to decimals with some converting to a two significant figure version,
0.36, which, if they had not shown a correct version in the working, lost a mark. The other common error was
14
incorrect division of leading to a common incorrect answer of 2.8. However most candidates usually
5
obtained the method mark before this error as they generally showed their working. Part (b) was a very well
answered question with most candidates obtaining the correct answer of 18. The main error seen in this
question was, after they had secured a method mark by writing p – 3 = 15, they incorrectly subtracted 3
leading to an answer of 12. Sometimes this was a careless error made by some of the more able candidates.
Candidates are advised during their checking to substitute an answer back in to the original equation to
check that it works.

5
Answer: (a) (b) 18
14

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 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 25

Part (a) was well answered by many candidates; the most common error was to give the answer
( x − 11)( x + 12) and occasionally ( x − 66 )( x + 2) , ( x + 66 )( x − 2) , ( x − 132)( x + 1) or ( x + 132)( x − 1) .
Other errors included having the correct answer and then deciding to equate the expression to 0 and then
solve by giving answers of x = 12 and x = –11; some candidates didn't factorise and just used the quadratic
formula. Part (b) was more challenging with a very common answer being x(x2 – 4) which scored one of the
marks for the correct partial factorisation. The more able candidates were able to follow the hint in the
question, which asked the candidates to ‘factorise completely’ which is usually an indicator that taking out
just one factor and one bracket is not sufficient. Occasionally the incorrect answer x(x2 – 4x) was seen and
some candidates used inefficient notation, e.g. ( x + 0 )( x + 2 )( x − 2 ) . Some reached x(x2 – 4) and realised it
was the difference of two squares but incorrectly followed it with x ( x + 4 )( x − 4 ) .

Answers: (a) ( x − 12)( x + 11) (b) x ( x + 2 )( x − 2 )

Question 26

There were many correct and efficient solutions to this question with candidates correctly finding AC and
2
then finding tan−1   . Of those who did not succeed, many identified the wrong angle, often finding QAB,
5
QBC or AQC. Some good work was spoilt after correctly finding AC and/or AQ by going on to use an
2
incorrect trig ratio, e.g. sin−1   etc. It was quite common to see candidates using inefficient methods, e.g.
5
using sine or cosine ratios instead of tangent and therefore having to work out AQ, often having already
found AC. Usually these candidates prematurely rounded part way through their working using AQ as 5.38
and lost the final accuracy mark as a consequence. Some candidates made it even harder by opting to use
the cosine rule. It was less common for these to reach a successful outcome and again frequently,
premature rounding caused the loss of the final accuracy mark.

Answer: 21.8

Question 27

Part (a) was often correct although some candidates did not fully process their answers and wrote, e.g. 33.
Part (b) proved a little more challenging, although it was also often correct. Some candidates applied a
6
correct rule for the power but gave the answer as 2 rather than x2, or unsimplified as x 3 .
Part (c) was less successful and one of the more challenging questions on the paper with full marks not
being scored that often. A number correctly managed some simplification for 1 mark but very few fully
0.5 1
2−1
simplified their answer. Common was leaving it one stage away, often as −2 , 2−2 or −2 .
y y y
y2
Answers: (a) 27 (b) x2 (c)
2

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

MATHEMATICS

Paper 0580/23
Paper 23 (Extended)

Key messages

Generally answers should be given correct to three significant figures unless the answer is exact or if the
question asks for a different degree of accuracy.

General comments

Candidates should show all their working, particularly the calculations they do on their calculators.
Candidates need to read questions carefully and note the form the answer should be displayed in. Many
errors are made in converting between units; many candidates do not know how to convert between area
units and between volume units.

Comments on specific questions

Question 1

This question was usually answered correctly. The main error was to find the difference between the
numbers and give an answer of 3 (h) 28 (min).

Answer: 2h 32 min

Question 2

Most answers were correct; however some candidates added 9 to the cube root of 25 and then found the
square root, giving an answer of 3.45. Many candidates truncated the correct answer to 3.05.

Answer: 3.06

Question 3

Most answers were correct; some candidates gave 66 as their final answer without any supporting working.

Answer: 66.2

Question 4

The most usual answer was trapezium whilst other answers included rhombus and square.

Answer: kite

Question 5

Most candidates gave a correct solution. The most common alternative answer was a partial factorisation
such as 3(6x + 9y).

Answer: 9(2x+3y)

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 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 6

Most answers were derived using a calculator rather than using indices and therefore, although the exact
answer is a fraction, many gave their answer as a decimal and some of them did not give enough accuracy
in their answer.

2
Answer:
3

Question 7

This question was answered well, although some used simple interest where $31.2 was often observed,
resulting in a final answer of $1262.40. The compound interest, $63.21, was sometimes presented as a final
answer.

Answer: 1263.21

Question 8

79
In many instances, appeared in the answer space with very little or no working. In the cases where
90
calculations were shown, 100x – 10x(=87.77 – 8.77 ) = 79 and 10x – x(=8.77 – 0.877 ) = 7.9 were
common, less so was the use of 100x – x. It was reasonably common to see 0.87& being interpreted as
0.878787.
79
Answer.
90

Question 9

Most candidates gained credit for correctly collecting the x’s and numbers together. If this resulted in
–10x ⩾ 12, in many cases x⩾ –1.2 was observed in the answer space. Common errors usually occurred in
the manipulation of the inequality and included 19 + 7(= 26) as well as 8x – 2x. Some candidates left their
answer as an unsimplified improper fraction.

Answer: x ⩽ –1.2

Question 10

The most common method was to not use volume scale factors and to multiply 2400 by 30 and divide by
100. Some attempts involved 2400 ÷ 30 as a first stage.

Answer: 64.8

Question 11

360
The most common method was to calculate = 30 before subtracting 30 from 180. A less common
12
method was to calculate (12 – 2) × 180 before dividing by 12. The most usual incorrect answer was 30 from
a partial method.

Answer: 150

Question 12

This question was usually well answered with the most popular method being 0.88 ÷ 0.8 = 1.1, although
occasionally the inverse method of 0.88 × 1.25 was seen, the 1.25 being the reciprocal of 0.8. The most
common incorrect answer was to add on 20% to 0.88 by calculating 0.88 × 1.2 = 1.056 or 1.06 which was
sometimes given as 1.1.

Answer: 1.1[0]

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 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 13

22 5 88 35 53 25
The most common correct method was − = − = = 1 . The most common error was not to
7 4 28 28 28 28
53
write the answer as a mixed number, instead leaving it as the improper fraction . The other main error
28
was not to convert the fractions to a common denominator before attempting to subtract them. Some tried to
deal with the fractions without involving the integers and occasionally this did lead to the correct answer.
There were a few who attempted to convert the fractions to decimals.

25
Answer: 1
28

Question 14

5
A very common error seen was the incorrect factorisation of (x + 4)(5x – 3). The answer of − was
3
frequently inaccurate with versions seen including −1.66 and −1.7. Many candidates tried to use the
quadratic formula suggesting that they were unable to factorise the expression or that they did not see the
request to factorise in the question.

5
Answer: 4, −
3

Question 15

The most common error was either calculating gf(x) as g(x) × f(x) or writing the first line of working as
(5x – 3)2 + 6x + 1. Those who started correctly sometimes expanded (5x – 3)2 incorrectly as 25x2 + 9 or
25x2 – 9.

Answer: 25x2 – 8

Question 16

The vast majority of candidates gaining full marks wrote their first line as 12m + 4 xy = xp . Those candidates
xp p 
who rearranged their first line to be 3m = − xy did not write their second line as 3m = x  − xy  . They
4 4 
usually attempted to multiply the equation by 4 and they usually did not multiply the xy term by 4.

The other common errors were making a sign error when attempting to rearrange their three term
expression, leaving an x term on both sides of their equation and choosing one ‘x’ term to make the subject
of their equation.

12m
Answer:
p − 4y

Question 17

Part (a) was answered correctly by most candidates. In part (b) most candidates started to solve the
equation 5n2 + 3 = 848 by subtracting 3 from 848. However instead of dividing by 5 and then taking the
square root, some tried to square root 845 first. A significant number did not solve this equation but they
would state ‘yes’ and then give an implicit form such as 5 × 132 + 3[= 848].

Answers: (a) 1, –4 and –9 (b) yes because 13 is an integer

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 18

Those reaching the correct answer used the sine rule to find angle ACB first then, using the property of the
angles of the triangle, they calculated angle ABC. The most common error was the incorrect rearrangement
of the correct implicit sine rule equation. Those who attempted to use the cosine rule usually found the length
of AC incorrectly.

Answer: 73.6

Question 19

In part (a) most answers were either completely correct or completely incorrect. The most common incorrect
6 0
answer was   . In part (b) the most common errors were to either not calculate the determinant at all
0 6
or to calculate 5 × 0 to be 5 in working out the determinant, thus reaching a determinant of 7 and not 12.
Some gave the answer in decimals correct to one decimal place without any working to support their answer.

 11 −6  1  −6 0 
Answers: (a)   (b)  
 −5 6  12  −5 −2 

Question 20

A large number of candidates were able to reach the correct answer. Almost all candidates correctly
calculated the area of the square and gave the correct units (cm2). Common errors were to work with 10 as
the radius in the formula for the area of the semi-circle or to use 5 as the radius but to find the area of the
whole circle. A few candidates wrote 10 × 10 for the area of the square but calculated it as 20. Some
attempted to convert the answer to square metres and most of these were attempted incorrectly.

Answer: 139 cm2

Question 21

In part (a) the majority of candidates were able to correctly calculate the mean number of hummingbirds
seen in Ali’s garden. A relatively common error was to misread the graph and calculate the mean number of
hummingbirds in Hussein’s garden. Some incorrectly multiplied the day number by the number of
hummingbirds before adding them together. In part (b) some candidates calculated the mean number of
hummingbirds seen in one of the two gardens rather than the median for Hussein’s garden. Where attempts
at listing numbers and finding the median were seen there was a variety of different errors observed, the two
most common being working with the data for Ali rather than for Hussein and not ordering the data before
choosing the middle one. Part (c) was well answered with sometimes 3 or 4 given as the answer.

Answers: (a) 3.4 (b) 5 (c) 10

Question 22

Part (a) was usually answered correctly. The most common incorrect answer was 23°. Part (b) was also
answered well, the most common error being to assume that both angles AEB and BEC were 19°. In part (c)
a good number of candidates gained credit for labelling angle EAB as 90°. Incorrect answers were often
based on calculations using differing combinations of 19° and 23°, for example 180° – 19° – 23°.

Answers: (a) 19 (b) 138 (c) 90

Question 23

In part (a) the common error with the first set was to shade B ′ only. In the second set the common error was
to shade A′ or ( A ∪ B )′ . In part (b)(i) the elements 60 and 70 were often included or some of the numbers
were omitted. The element 67 was sometimes in the incorrect place. In part (b)(ii) the elements were often
listed, rather than the number in the set. In part (b)(iii) 0, ‘nothing’ or an incorrect symbol was often written.

Answers: (b)(ii) 3 (iii) ∅ or { }

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

MATHEMATICS

Paper 0580/31
Paper 31 (Core)

Key messages

Centres should continue to encourage candidates to show formulas used, substitutions made and
calculations performed and emphasise that in show questions candidates must not start with the value they
are being asked to show. Attention should be paid to the degree of accuracy required in each question and
candidates should be encouraged to avoid premature rounding in workings. Candidates should also be
encouraged to process calculations fully and to read questions again once they have reached a solution so
that they provide the answer in the format being asked for and answer the question set.

General Comments

The difficulty of the paper was appropriate for the candidates intended. The majority of candidates were able
to access all questions and the presentation of their work was generally good. This paper gave all
candidates an opportunity to demonstrate their knowledge and application of mathematics. The vast majority
of candidates were able to complete the paper in the allotted time. Few candidates omitted part or whole
questions. The standard of presentation was generally good and there was evidence that most candidates
were using the correct equipment. Candidates should be encouraged to not write over previous wrong
answers when correcting their work, instead cross out wrong work and replace, as often answers are very
difficult to read.

Areas which proved to be important in gaining good marks on this paper were; using negative numbers in the
context of temperature, using money, calculate percentage increase, price after a percentage change and
calculating compound interest. Successful candidates were able to interpret a scatter diagram, calculate
mean and median from a set of data, plot and interpret a quadratic curve, measure, draw and calculate
bearings, use scale to draw an accurate diagram, use a calculator accurately, use and give times in 24 hour
format, understand probability, write a value as the product of its prime factors, draw and describe
transformations, simplify algebraic expressions, form and solving linear equations, calculate the area of a
triangle and continue sequences and describe the nth term of a sequence. Although this does not cover all
areas examined on this paper, these are the areas that successful candidates gained marks on.

Comments on specific questions

Question 1

(a) (i) Candidates were successful in interpreting the scale and showed that they understood each
division was equivalent to 2. Few candidates made errors on this question; the most common was
an answer of 10.6 from interpreting each division as 0.2.

(ii) Again candidates successfully interpreted this scale using each division as 5. Few errors were
seen; the most common was –18 or –19, using each division as 2 or 1 respectively.

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

(b) (i) The vast majority of candidates successfully identified the coldest day. Many candidates also gave
the temperature (–6) which was not penalised if they also gave the day.

(ii) Most candidates correctly calculated the difference, with a 50–50 split between those that gave the
answer of 6 and –6. Both answers however were acceptable.

(c) (i) Candidates found this time question more challenging. Most successful answers were given in 24-
hour time format but 4 05 pm was also accepted. However a large number of candidates lost the
mark because their time was given in the incorrect format, even though they clearly had found the
correct time, e.g. 16 05 pm and 4 05.

(ii) The vast majority of candidates successfully added 7 to –3. Some less able candidates gave
answers of 10 or –10 from subtracting 7 from –3 or adding 7 to 3.

Answers: (a)(i) 16 (ii) –15 (b)(i) Friday (ii) 6 (c)(i) 16 05 or 4 05 pm (ii) 4

Question 2

(a) This money question was successfully answered by nearly all candidates. It was the best answered
question of the whole paper. Very few candidates only gave the answer and lots of good quality
working was shown by most candidates. The few errors seen came when adding the cost of all the
items together. Most errors seen were due to candidates not using their calculators to do the final
addition.

(b) Most candidates attempted this question by finding the 8% first and then added it to $64. More able
candidates used the multiplication method, 64 × 1.08. Both methods were seen often and well
presented by successful candidates. The most common error seen was finding the 8% but not
adding it to find the new price.

(c) Calculating the percentage increase proved to be more challenging, with only the most able
candidates generally being successful on this question. The best answers calculated the increase
in price first and then expressed this as a percentage of the original price. The most common errors
were: expressing the increase in price as a percentage of the new price rather than the original
price, or dividing the original price by the new price. The most common incorrect answer seen was
30  250 
10.7% from × 100 or from  1 −  × 100 .
280  280 

(d) This question was well answered by more able candidates who showed clear working out. Very
good answers showed a clear calculation of area followed by a multiplication by 12 to give the cost.
Errors occurred generally because candidates did not calculate the area but found the perimeter
instead. Another common error was to multiply the length and width by 12 first and then to add
these together. Some candidates who used the correct method did not score full marks due to
premature rounding. Often candidates calculated the area as 46.75 but then rounded this figure to
47 or 46.8 before multiplying by 12. Candidates should be encouraged to use the full calculator
answer during the calculations and only round the final answer, if required.

(e) Calculating compound interest was a challenge for many candidates. However many correct
answers were given by correctly quoting and substituting into the formula for compound interest.
There were a number of errors seen; the most common was the calculation of simple interest.
Other common errors were down to poor rounding or truncation of the final answer. Another way in
which candidates lost a mark was rounding (1.06)3 to 1.19 or 1.2 and then multiplying by 3600 to
give an answer of 4284 or 4320.

Answers: (a) 180.5[0] (b) 69.12 (c) 12 (d) 561 (e) 4287.66

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 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 3

(a) (i) This was a very challenging question with only a minority of candidates giving a reason that was
acceptable. The majority of candidates gave the correct test but did not score the mark because
their reason wasn’t satisfactory. Correct answers compared the two test results for each candidate,
so an answer of ‘written test had higher scores’ gained the mark but an answer of ‘written test
because most of the class had high scores in the written test’ did not gain the mark as it does not
compare with the speaking test. Most candidates did not gain the mark for a variety of reasons; any
mention of the line of best fit did not gain the mark as it does not compare the two marks,
mentioning the highest mark for each test only compares one student’s results rather than all the
candidates. A number of candidates attempted to count how many scored more than, for example,
40 for each test. If they counted correctly this answer was accepted but many candidates
miscounted and therefore did not gain the mark. A small number of candidates put a lot of effort to
calculate the mean for each test, however if inaccurate they also did not gain the mark.

(ii) Most candidates correctly identified that the correlation shown was positive. Extra comments like
‘weak’ or ‘strong’ were ignored. Some less able candidates did not use the correct terminology and
‘direct proportion’ was a common incorrect answer as well as ‘negative’.

(iii) The vast majority of candidates correctly identified the point. The most common incorrect points
circled were (7, 45) and (58, 60).

(iv) Most candidates found the correct score. The most common error was 33 or 33.5 which was using
39 from the written test rather than the speaking test. Many less able candidates did not attempt
this question.

(b) (i) Few candidates found the mode or mean instead of the median. Good answers showed the whole
list of data in order before identifying 29 as the middle value. Common errors involved writing a list
of only 10 numbers and therefore finding the wrong middle value or finding the range instead of the
median.

(ii) Few candidates found the mode or median instead of the mean. Good answers clearly showed all
their working including an addition sum of all 11 values and a clear division by 11. Again some
errors in rounding caused candidates to lose marks. A significant number of candidates showed no
working and gave the answer of 27.4 which gained no marks as it is inaccurate and no working
given.

Answers: (a)(i) Written test and a valid reason (ii) Positive (iii) (45,10) indicated (iv) 42
(b)(i) 29 (ii) 27.5

Question 4

(a) (i) This was one of the best answered questions on the whole paper. The vast majority of candidates
correctly plotted the point (–4, 2) with only a few candidates plotting (2, –4) instead.

(ii) Giving the mathematical name of the triangle was much more challenging. Right angle(d) triangle
was the most common correct answer however scalene was seen. The most common incorrect
answers were ‘rectangle triangle’, ‘right triangle’ and ‘isosceles triangle’.

(iii) The correct vector was only given by more able candidates with many less able candidates not
answering or reversing the values or giving negative values instead of positive values.

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 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

(iv) (a) Calculating the gradient of the line was a very challenging question for many candidates. Few
correct answers were seen with a variety of incorrect answers given. These included calculations
involving run/rise, giving the common incorrect answer of 2. Many candidates attempted to use
co-ordinates on the line but few were successful in finding the change in y co-ordinates/change in x
co-ordinates.

(iv) (b) Candidates were able to gain a follow through mark if they used their gradient correctly but most
did not gain this mark as they did not use the y intercept as 0. Only the most able candidates gave
the correct equation with many candidates giving answers without the x variable. Candidates often
showed they needed to have an equation of the form y = mx + c, however did not put a value for c.
1
So common incorrect answers were y = x + c or y = mx + c .
2

(b) (i) Most candidates gained at least 2 marks for correctly finding the y values for x = 0, 2 and 4. The
most common errors were for y = –3 and –1 with x values of –17 and –7 respectively. This is
because candidates found (–3)² = –9 instead of 9 and (–1)² = –1 instead of 1.

(ii) There was good plotting of points and the follow through from part (b)(i) was seen often. Very few
straight lines joining points was seen and even fewer thick or feathered curves drawn. Some
candidates did not gain full marks as they drew a ‘flat bottomed curve’. It is important that
candidates draw a curved line between the points at x = 0 and x = –1 which goes below their
co-ordinates.

(iii) Candidates found this question extremely challenging. It was the question which most candidates
did not attempt in the whole of the paper. Those that did attempt it often did not use their graph to
find where their curve crossed the x-axis despite being told to use their graph in the question. Many
candidates attempted the quadratic formula but very few were successful. Candidates found the
scale difficult to use and often were not accurate enough in reading the x value from the graph.

8
Answers: (a)(ii) Right-angled or scalene (iii)   (iv)(a) 0.5 (iv)(b) [y =] 0.5 (b)(i) 1 –5 –5 1 15
4
(iii) –2.8 1.8

Question 5

(a) Candidates generally measured the distance accurately and multiplied by 12 correctly. Some less
able candidates measured to the nearest centimetre and therefore gave the answer of 48 km.

(b) The majority of candidates showed little understanding of bearings with the most common answer
being a measurement of length rather than an angle or 227° which was the bearing of A from B.
Very few candidates showed the ability to use a protractor accurately when measuring a bearing.

(c) Candidates found measuring a reflex angle even more challenging than the acute angle in part (b).
This part proved to be the most difficult in this question with candidates often not answering,
measuring in an anticlockwise direction from North or measuring C from B. Candidates should be
encouraged to draw a South line from town C and then measure, to split the reflex angle into 180
and 112, especially if not using a 360° protractor.

(d) (i) Good answers saw accurate arcs drawn using compasses from towns A and C and the town D
clearly marked where the arcs intersected. A significant number of candidates found the correct
position of town D but did not show any arcs, as instructed in the question.

(ii) Candidates generally showed they understood how to calculate speed by writing the correct
formula or attempting to divide the distance by the time. However only the most able candidates
were successful in gaining full marks. Most candidates lost marks because of premature rounding
or dividing by time in minutes instead of hours. The most common incorrect answer was 8.4, from
84
. Candidates who attempted to convert the time to hours usually lost a mark by prematurely
10
10
rounding their answer before dividing the distance by this time, e.g. = 0.16 or 0.17.
60

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

(e) Finding the bearing of town A from town E proved to be the most challenging question of the whole
paper. Those that attempted it often gave the incorrect answer of 242, from 360 – 118, or 62, from
180 – 118. The question asked for candidates to work out the bearing rather than drawing and
measuring, so candidates who did attempt this question using a diagram did not gain the marks
unless their answer was exactly 298.

Answers: (a) 51.6 (b) [0]47 (c) 292 (d)(ii) 504 (e) 298

Question 6

(a) (i) This was well answered with most candidates achieving at least 1 mark. The most common
omissions were 1 and/or 18. The most common incorrect inclusions were 4 or 8. A small minority
attempted prime factors.

(ii) Around half of the candidates correctly gave a multiple of 30, most commonly 60, although 30, 90,
150 and 900 were often seen. An equal number of candidates confused factors with multiples, so
2, 3, 5, 6, 10 or 15 were seen often.

(iii) Candidates showed good use of their calculators to find the square root.

(iv) This part was answered very well although many candidates gave a truncated or rounded answer,
usually 15.6.

(v) Again candidates showed good use of their calculators with very few errors seen. The most
common incorrect answer was –0.2.

(b) Many candidates showed a lack of understanding of the term ‘product of prime factors’ and only
factorised it into, e.g. 9 × 8. Most candidates chose to use a ladder or tree method but some left a
non-prime factor, e.g. 9 or 4. Many correct answers were seen, however a number of candidates
added the prime factors or listed the factors and did not write the answer as a product.

(c) Lists or ladders of prime factors were seen in many responses and more able candidates generally
went on to obtain the correct answer from this method. A significant number did not select the
correct factors, often missing a factor of 2 from their answer or giving 120. It was also common to
see the answer 2 following ladders of prime factors meaning candidates had identified the HCF
instead of the LCM. Many candidates gained 1 mark with this method by showing either 2 × 3 × 5
or 24, or multiplying all the prime factors to obtain 480. Others started lists of multiples; some
stopped well before 240 or made a slight error, while others led to the correct answer.

(d) Most candidates counted in 6s and 9s until they arrived at a corresponding time and many correct
answers resulted. Some incorrectly started their counting at 6 am and 9 am, instead of 2 am, or
added 2 am, 6 and 9 to achieve 5 pm. Confusion was seen often with 24-hour clock times having
am or pm added on, or 12-hour clock times not stating am or pm. The best answers followed lists
8, 14, 20 and 11, 20.

Answers: (a)(i) 1, 2, 3, 6, 9, 18 (ii) a multiple of 30 (iii) 46.2 (iv) 15.625 (v) 5 (b) 23×32 (c) 240
(d) 20 00 or 8 pm

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 7

(a) (i) Candidates showed good understanding of probability in all three parts of Question (a). The
majority of candidates gave their answers correctly as fractions with few giving their answers as
decimals or percentages. All correct equivalences were accepted for the marks.

5
(ii) Most candidates correctly identified the probability that the counter was white was . This was
20
1
seen in equal frequency as or 0.25.
4

(iii) The vast majority of candidates identified the probability that the counter was yellow was 0. This
0 9
was equally seen as 0 or . A common incorrect answer was from calculating the remaining
20 20
fraction and therefore the probability that the counter was blue rather than yellow.

(b) This part was very well answered, although the method was rarely seen. 0.55 sometimes resulted
by assuming 0.3 was 0.03. Candidates should be reminded to show all working out. A small
number of candidates thought there was a pattern to the table as the values given decrease by
0.06 so the answer 0.36 was given.

8 6 2
(c) This part was very well answered. Many candidates showed that = = (or 40% or 0.4) to
20 15 5
achieve full marks. The common incorrect answer was to express red as a fraction of blue for each
8 6 2
bag, giving = = . Another error was to add the two fractions together. Cancelling and
12 9 3
conversion was generally of a high standard. A few candidates attempted to use a tree diagram,
however this was rarely successful.

6 5 8 6 2
Answers: (a)(i) (ii) (iii) 0 (b)(i) [0].28 (c) = =
20 20 20 15 5

Question 8

(a) This algebra question was well attempted by most candidates who generally scored at least 1 mark
for multiplying out at least one bracket correctly. –2x + 8 was a common incorrect answer when
multiplying out –2(x + 4). A significant number of candidates were able to expand both brackets
correctly but made an error when simplifying, e.g. 8x – 7, 8x + 23 or 8x – 23, even though +15 – 8
was clearly in the working. Another error was to change the sign as terms were moved around the
expression, e.g. 10x + 15 – 2x – 8 became 10x – 2x – 15 – 8.

(b) (i) This question was only answered successfully by the most able candidates. Many showed they
understood the question but did not give their answer in its simplest form. 2x + 2x + 2x, 3x2x, 2x3
were common incorrect answers.

(ii) As in part (b)(i) many candidates achieved an unsimplified answer which did not score the mark.

Many candidates formed equations here instead of an expression. The most common incorrect
answers were 20a ÷ 4, 20a ÷ 4 = 5, a = 5 and 20a × 4 = 80a.

(c) The best answers given in this part contained full working and correct simplification of algebraic
expressions. Few candidates gave the initial expression containing brackets but most candidates
did realise they needed to add 4 expressions together, although some candidates just added
together the two given sides. Others correctly multiplied each side by 2 with a view to adding them
but did not go on and gather like terms. Common errors were to confuse area with perimeter,
(3y + 1) × (2y + 5). Some, having formed a correct expression, then went on to attempt to solve an
equation, e.g. 3y + 1 = 2y + 5 or 6y + 2 = 4y + 10. Others achieved the correct answer 10y + 12
and then changed the + to = or set it = 0 and solved for y, thus losing a mark.

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

(d) Most candidates appeared to understand the problem but many found the algebraic skills required
to form a correct equation challenging and could only express a partial solution. Many candidates
therefore scored 1 of the first 2 marks in the first part for either m + 6 or for 3m + 7t = 182. Many
chose to omit the equation and use a trial and improvement approach. A common incorrect
equation was 3m + 7m = 182, but those that were able to form the correct equation
3m + 7(m + 6) = 182 generally then solved it correctly for full marks. Candidates who recognised
that the problem could be solved using simultaneous equations were often successful. The usual
approach was to substitute t = m + 6 into 3m + 7t = 182. Other successful, but lengthy approaches
(182 − 7t )
were to find t first using m = t – 6 or sometimes m = , and then find m. There was some
3
good use of elimination to solve for m (and t) as another alternative method, although this was
rarely seen.

Answers: (a) 8x + 7 (b)(i) 6x (ii) 5a (c) 10y + 12 (d) 14

Question 9

(a) (i) Most candidates correctly used the formula for finding the area of a triangle. Some answers
involved π, while others attempted to measure the height and base of the triangle instead of using
the grid and measured incorrectly. Counting squares proved very difficult for those that attempted it
with very few correct answers using this method.

(ii) This was well attempted with the correct triangle given in a variety of positions. This was the best
answered part of this question. A significant number of candidates however only enlarged one side
correctly while a few enlarged correctly but in the wrong orientation. The most common incorrect
answer saw candidates adding two squares to the height and base rather than multiplying by 2.

(b) (i) The transformation was generally described well. Most achieved at least 2 marks with the centre of
rotation being the usual omission. Some incorrectly called it reflection but very few gave more than
one transformation in their answer. A few lost marks by giving extra information, usually a vector.
Only a few used the full alternative transformation using enlargement.

(ii) Many candidates scored 2 marks for the correct reflection but the majority of responses were in an
incorrect horizontal line, usually y = 0. Some reflected in x = –1. Candidates who drew the mirror
line were more successful.

(iii) Many candidates achieved full marks by correctly translating the shape by the correct vector.
Candidates found the translation easier than reflection. The horizontal move was more successful
than the vertical, the common error being to move 1 down instead of up. A misconception was to
translate the triangle point (–3, 1) to (5, 1) using the same numbers in the vector.

Answers: (a)(i) 7.5 (b)(i) Rotation, 180°, [centre] (0, 0)

Question 10

(a) (i) This part was very well done with the vast majority of candidates correctly continuing the sequence
and giving the next term.

(ii) Several candidates tried to complicate this by attempting to give a formula for the nth term, when a
worded description of the term-to-term rule was required. n + 8 was the most common incorrect
answer. Others showed a sum to demonstrate how they answered part (i).

(iii) Giving an expression for the nth term proved to be the most challenging part of this question with a
large proportion of candidates not attempting this part. Many answers of –2(n – 1)8 and
–2(n + 1)8 were seen which was an attempt to use a + (n – 1)d but incorrectly quoting the formula.
Many candidates gained 1 mark by spotting the difference is 8 and writing 8n. Many correct
answers were seen, both simplified or not. The usual incorrect answer was n + 8.

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

(b) Many of those scoring 0 made no attempt to substitute 2 for n and consequently had answers
which included n. To be successful candidates needed to understand that ‘second term of the
sequence’ means n = 2. A common incorrect response was to expand the bracket, 5n + 5 – 6 was
seen regularly as the first step, often followed by 5n – 1 or just –1. The common incorrect answers
were 4 and –1 using n = 1 or n = 0. Many less able candidates chose not to attempt this question.

(c) Candidates who successfully found the next term showed evidence of working out the differences
between each term. Many responses were between 30 and 36 as a result of using incorrect
differences. The answer of 32, from using 13 as the next difference, was common. 23 was seen
frequently, being 19 plus the second difference of 4.

Answers: (a)(i) 30 (ii) add 8 (iii) 8n – 10 (b) 9 (c) 34

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

MATHEMATICS

Paper 0580/32
Paper 32 (Core)

Key messages

To succeed in this paper candidates need to have completed full syllabus coverage, remember necessary
formulae, show all working clearly and use a suitable level of accuracy. Particular attention to mathematical
terms and definitions would help a candidate to answer questions from the required perspective.

General comments

This paper gave all candidates an opportunity to demonstrate their knowledge and application of
mathematics. Candidates were able to complete the paper within the required time and most candidates
made an attempt at most questions. Although a number of questions have a common theme, candidates
should realise that a number of different mathematical concepts and topics may be tested within the
question. The standard of presentation and amount of working shown was generally good. Centres should
continue to encourage candidates to show formulae used, substitutions made and calculations performed.
Attention should be made to the degree of accuracy required and candidates should be encouraged to avoid
premature rounding in workings. Candidates should also be encouraged to read questions again to ensure
the answers they give are in the required format and answer the question set.

Comments on Specific Questions

Question 1

(a) (i) This part was generally answered well although the common error was 85 minutes, coming from
8.20 – 7.35 = 0.85, i.e. using time as a decimal value or over reliance on a calculator.

(ii) This part was generally answered well although a common error was in giving the answer as a time
period of 10 hours 10 minutes rather than the time of 10 10.

(iii) This part was generally answered reasonably well with the full method of (1.66 × 5) – 7.75 used,
resulting in the correct answer. Common errors included 7.75 ÷ 5 = 1.55, 1.66 – 1.55 = 0.11 and
7.75 – 1.66 = 6.09.

(b) (i) This part was generally answered well although a significant number of candidates did not
appreciate the definition of range and gave a variety of other statistical values.

(ii) This part on completing the frequency table was generally answered well although a small number
of candidates misplaced one or more values.

(iii) This part on drawing the bar chart was generally answered well with correct heights and widths of
the required bars. The common error was in not completing the scale used despite this requirement
being stated in the question.

(iv) This part was generally answered well although a significant number incorrectly answered in a way
that illustrated a lack of understanding of the term ‘modal class interval’.

Answers: (a)(i) 45 (ii) 10 10 (iii) 0.55 (b)(i) 50 (ii) 2, 7, 4, 5, 6, 6 (iv) 10 to 19

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 2

(a) (i) This part was generally answered correctly although a common error was writing ‘eighty’ instead of
‘eight’.

(b) (i) This part was generally answered correctly.

(ii) This part was generally answered correctly.

(c) (i) This part was generally answered well particularly by those candidates who used a factor tree or
factor table. However a significant number did not write their answer as a product of prime factors
and gave answers of 2, 7, 7 or 2, 7 or 2, 7, 14. Another common error was in writing down a list of
factors such as 2 × 49, 7 × 14 and 1 × 98.

(ii) This part was generally answered well again, particularly by those candidates who used a factor
tree or factor table for both 98 and 182. Common errors included 2, 7 and 1274 (LCM).

(d) (i) This part was generally answered correctly although a common error was 24.

(ii) This part was generally answered correctly although common errors were 156.2 (the square root)
and 468.5 (the square root × 3).

(iii) This part was generally answered correctly although common errors were 0 and 1.

1
(iv) This part was generally answered correctly. Common errors included –125, –15, and 0.005.
53

Answers: (a) eight thousand and forty-five (b)(i) 64 (ii) 61 or 67 (iii) 68 (c)(i) 2 × 72 (ii) 14
(d)(i) 1296 (ii) 29 (iii) 14 (iv) 0.008

Question 3

(a) Most candidates scored at least 1 of the 2 available marks with the order of rotational symmetry for
the hexagon causing the most problems.

(b) (i) This part was generally answered well although common errors included reflecting in the y-axis, the
x-axis or y = –1.

(ii) This part was generally answered well although common errors included enlargements drawn from
a variety of incorrect centres.

(iii) The majority of candidates correctly identified the transformation as a translation although a
number of errors were seen in the description of the required vector.

−5 
Answers: (a) 2, 6 (b)(iii) translation  
3 

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 4

(a) (i) This part was generally answered well with the correct answer reached although the lack of
working was noted for this part and a number of candidates may have lost method marks as a
result. A significant number did not realise the complexity of this multi-stage question and the
required stages of 10 ÷ 1.45 = 6.89 giving 6 pens bought, 6 × 1.45 = 8.70, 10.00 – 8.70 = 1.30
was often not seen or appreciated. A small number used the alternative valid method of repeated
subtraction.

(ii) This part on percentage decrease was generally answered well although a very common error
was simply finding 15% as 0.84 and giving this as the answer. Other common errors included
5.60 + 0.84 = 6.44, 5.60 – 0.15 = 5.45 and 5.60 ÷ 0.15 = 37.33.

(b) This part on finding the median was generally answered well particularly by those candidates who
re-wrote the values as an ordered list. Common errors included 18.5 (from unordered original list),
19 (mode), 27.6 (mean), 50 (range), 25 – 19 = 6, omitting one value from the ordered list, and the
calculator error of 19 + 25 ÷ 2 = 31.5.

(c) This part on ratio was generally answered very well with the correct answer reached although the
lack of working was noted and a number of candidates may have lost method marks as a result.

(d) This part on money conversion was generally answered well although the common error of
1400 × 1.54 was often seen.

(e) This part was generally answered well with the majority of candidates successfully applying the
compound interest formula. Those candidates who did the work in stages often lost the accuracy
mark or gave their final answer as 54.74. A small but significant number spoilt their method and lost
accuracy marks by adding or subtracting 2000 from their calculated answer. The use of simple
interest was rarely seen.

Answers: (a) 6 pens and 1.30 (ii) 4.76 (b) 22 (c) 3000, 1500, 2500 (d) 909.09 (e) 2160.09

Question 5

90
(a) Many candidates were able to show that 225 students chose Science by using × 900 or
360
1
× 900 . However, a significant number were unable to attempt this part while others used the
4
given value of 225 in a circular argument which is not acceptable in a ‘show that’ question.

(b) This part was generally answered well with a good number of candidates correctly applying
18 18
the method of × 900 . Common errors included × 360 = 7.2, 900 ÷ 18 = 50,
360 900
180 – 18 = 162 and 18.

(c) This part on completing the pie chart was generally answered well with an accurate diagram drawn,
although the lack of working was noted for this part.

1
(d) (i) The majority of candidates were able to state the required probability as 0 although was a
900
very common error.

(ii) This part was generally answered well with a good number of candidates correctly giving their
18 1 45 1
answer as a fraction in its lowest terms. Common errors included = , and = .
900 50 360 8

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

(e) This part was less successful and many candidates were unable to correctly calculate an estimate
for the expected number of students. Many and varied incorrect calculations involving 125, 900, 20,
360 and 2520 were seen.

1
Answers: (b) 45 (d)(i) 0 (ii) (e) 350
20

Question 6

(a) (i) The majority of candidates were able to measure the required distance and convert to kilometres
using the given scale although common errors of 9.5 and 100 were seen.

(ii) Although the measurement of the required bearing was performed quite well, many candidates
were confused by this term with common errors of the distance 9.5 cm and angles of 45, 225, and
315 being often seen.

(b) (i) This part was generally answered well with a good number of candidates able to score both marks.
One mark was often awarded for the correct distance and less often for the correct bearing.

(ii) The majority of candidates were able to use the correct formula to find the required speed but
common errors included 78 ÷ 45, 78 ÷ 0.45 and 78 × 45.

(c) The required construction proved very challenging for many candidates and proved to be a good
discriminator. Many candidates did not appear to appreciate that the two required constructions
were the perpendicular bisector of the line AB and an arc of radius 7 cm from centre A. Although a
number of candidates were able to draw one of these constructions they were often too short to be
fit for purpose.

Answers: (a) 95 (ii) 135 (b)(ii) 104

Question 7

(a) (i) This part was generally answered well although common errors of octagon, heptagon, 5, hexagon
were seen, together with a variety of other mathematical names.

(ii) This part was generally answered well although common errors of trapezium, parallel, rectangle,
4 were seen, together with a variety of other mathematical names.

(iii) This part was generally answered well although common errors of acute, reflex, 120, 60, isosceles
were seen, together with a variety of other mathematical terms.

(b) (i) This part was generally answered well although the common errors included 25 + 12 + 8,
25 × 12 × 8 × 6, 0.5 × 25 × 12 × 8, and the incorrect use of squares or cubes.

(ii) The required conversion in this part was generally poorly done with few correct answers seen.
Common errors included ÷10, ÷100, ÷1000, ×100, ×1000, and taking the square or cube root.
Those few candidates who started again with 0.25 × 0.12 × 0.08 were usually successful.

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

(c) (i) This part was generally answered reasonably well with a good number of candidates able to
identify the given line as the radius. Common errors included diameter, straight line, tangent, right-
angled together with a variety of other mathematical terms.

(ii) The required explanation in this part was generally poorly done with few correct answers of ‘angle
in a semicircle’ seen. Common incorrect statements included ‘triangle in a semicircle’, ‘because
BAD is a right angle’, ‘triangle ABD is a right angle’, ‘the tangent touches the circle’ and ‘the angles
are touching the circumference of the circle’.

(iii) This part on finding the circumference of the circle was generally answered well although a
significant number lost the final accuracy mark due to rounding errors or premature approximation.
Common errors included the use of π × 8² and 0.5 × π × 8 with other errors of 2 × π × 4 and
0.5 × 8 × 14 also seen.

(iv) This part on finding the length of CD was generally answered well by those candidates who
recognised the need for Pythagoras’ theorem, although again a small number lost the final
accuracy mark due to rounding errors or premature approximation and some incorrectly used 142 +
82. Very few fully correct answers were seen from those that chose to use a long method involving
trigonometry. Less able candidates tended to use a variety of incorrect methods using the numbers
given in the question such as 6 (from 14 – 8), 22 (from 14 + 8), 22 (from (8 × 14) – 90), 68
(from 90 – (14 + 8)) or 68 from (180 – (90 + 14 + 8)).

Answers: (a)(i) pentagon (ii) parallelogram (iii) obtuse (b)(i) 2400 (ii) 0.0024 (c)(i) radius (ii) angle in
a semicircle (iii) 50.3 (iv) 11.5

Question 8

(a) (i) This part was generally answered well. Many of the incorrect answers involved a change of sign
for some terms when re-ordering leading to answers such as 12p – 11r, 4p – 11r, 12p + 7r and
12p + –7r . Other common errors included 5pr and 12p2 + 7r2.

(ii) This part was generally answered well although common errors included 2304, 2304x , 24x, 24x6
and 10x5.

(b) This part was generally poorly answered with many candidates unable to write an acceptable
algebraic expression. Common errors included x + y = 165, xy = 165, x = 90 and y = 75, 90x – 75y,
90x × 75y, 165xy, 15xy, 165, 15 and 6750.

(c) The majority of candidates understood the method involved in this part and either gained full marks
or 1 mark for a partial factorisation with 2p(6p – 4) being the most common. Common
misconceptions seen included 12p2 – 8p = 4p2, p2 – p = p, 12p2 = 144p.

(d) This part was generally answered well with many candidates able to score full marks for solving the
given equation or at least 1 method mark for a correct step. Common errors included 7r – 3 = 4r,
28r – 12 =16r, 28r = 128 – 12 after the correct 28r – 12 = 128 and r = 140 – 28 after the correct
28r = 140.

(e) A significant number of candidates scored full marks on this simultaneous equations question,
showing a clear and succinct method. Most candidates used the elimination method and were able
to demonstrate an understanding that they needed to multiply both equations to make the
coefficients of one of the variables equal. There were many different types of errors seen and these
included: sign errors, arithmetic errors, incorrect coefficients in one or both equations, subtracting
the equations when they should have been added and vice-versa. Those who used the substitution
method were less successful although many scored the method marks but were then unable to
solve their resulting linear equation.

Answers: (a)(i) 12p – 7r (ii) 24x5 (b) 90x + 75y (c) 4p(3p – 2) (d) 5 (e) x = 2.5, y = 11

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 9

(a) (i) The table was generally completed very well with the majority of candidates giving 3 correct values
although a common error was calculating y = –8 when x = –1.

(ii) This was well answered by many candidates who scored full marks for accurate, smoothly drawn
curves. Most others scored 3 marks, the fourth mark being most commonly lost for one point being
plotted out of tolerance, or for just plotting the points without drawing the curve through them or for
joining the points with ruled lines.

(b) (i) This part was generally answered well. Common errors included drawing the lines of y = 4.8,
y = 5.2, y = 6 with x = 5, x + y = 5 and y = x + 5 also seen.

(ii) This part was generally answered well by candidates who had drawn the correct line in part (b)(i).
A significant number did not appreciate that the y co-ordinate of the intersection had to be 5. Other
common errors included reversed co-ordinates, misreading of the scale, and attempting to solve
the equation. Less able candidates were often unable to attempt parts (b)(i) and (ii).

Answers: (a)(i) –6, 6, 14 (b)(ii) 1.8 ⩽ x < 2, 5

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

MATHEMATICS

Paper 0580/33
Paper 33 (Core)

Key messages

To succeed in this paper candidates need to have completed full syllabus coverage, remember necessary
formulae, show all working clearly and use a suitable level of accuracy. Particular attention to learning
mathematical terms and definitions would help all candidates to answer questions giving the relevant name
or using the relevant process.

General comments

This paper gave all candidates an opportunity to demonstrate their knowledge and application of
mathematics. Candidates were able to complete the paper within the required time and most candidates
made an attempt at most questions. The standard of presentation and amount of working shown was
generally good, but candidates need to be aware that if they show no working and give a wrong answer they
cannot score any method marks. Centres should continue to encourage candidates to show formulae used,
substitutions made and calculations performed. Attention should also be made to the degree of accuracy
required, and candidates should be encouraged to avoid premature rounding in workings. A lack of
understanding of BIDMAS, whether performing arithmetic without a calculator, with a calculator or
manipulating algebra, was seen in Questions 2(a)(i), 2(c), 5(a)(ii), 5(a)(iii) and 6(a)(i). Marks were also
frequently lost in Questions 3(a) and 4(a) where candidates were not able to give correct mathematical
names. However, the graph in Question 7(a)(ii) was particularly well drawn by most candidates.

Comments on specific questions

Question 1

(a) (i) Almost every candidate answered this part correctly.

160
(ii) This part was answered very well. The most common errors were either to find or to misread
3
800
the question and calculate ×3 .
10

(iii) The majority of candidates answered this part correctly. However, the most commonly seen error
3
was to convert to an inaccurate decimal of 0.37 or 0.38 and these wrongly gave either 59.2 or
8
60.8 hats.

(b) (i) Almost every candidate was able to work out correctly the cost of the T-shirts bought. The few
errors that were seen were generally from misreading the number of each type of T-shirt bought.

(ii) The majority of candidates were able to work out 20% of $9.50 as $1.90. However not all of these
candidates recognised that they needed to subtract this from $9.50 to find the reduced price rather
than just finding the reduction in price.

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

(c) (i) Although most candidates attempted this question, it was rare to see any working or method to find
the degrees per T-shirt. Whilst many scored full marks, a very common misconception was to
incorrectly assume that because the sector angle for plain T-shirts was ‘20’ more than the number
sold, then the sector angle for striped T-shirts would be 85 + 20 = 105° and for logo T-shirts
115 + 20 = 135°.

(ii) This part was generally well answered with accurate and ruled sectors drawn. The pie chart was
followed through from their table, provided their three sector angles had a total of 360°.

(d) Whilst most candidates attempted this part, it was clear that the phrase ‘percentage profit’ was not
well understood . Whilst most candidates recognised that the profit was $9, common errors were
either to give answers of 140%, 1.4 or 0.4 or to incorrectly divide the profit by the selling price.

Answers: (a)(i) 800 (ii) 48 (iii) 60 (b)(i) 43.5 (ii) 7.6 (c)(i) 102°, 138° (d) 40

Question 2

(a) (i) Almost equal numbers of candidates gave the incorrect answer of 3 as gave the correct answer
showing that many candidates do not understand BIDMAS.

(ii) Most candidates successfully completed this question.

(b) (i) Most candidates successfully completed this question.

(ii) Most candidates successfully completed this question.

(c) This part was not answered well. The most common errors seen were with candidates who either
evaluated the sum correctly but did not round it correctly to the required accuracy or who evaluated
3.67
the numerator and denominator separately but got no further than, for example, . Others
86.1
attempted to calculate the sum in one go but did not understand the BIDMAS rules and the need to
insert brackets. Others gave answers of 0.04 or 0.042 but showed no other working and so could
not be awarded a method mark.

(d) This part was answered well. The most common errors came from either summing rather than
1
multiplying the numbers, forgetting to square the 4.5, misreading the values or writing as 0.3.
3

(e) Relatively few candidates were able to correctly identify the irrational number.

(f) (i) Most candidates were able to evaluate T correctly.

(ii) Although some candidates scored full marks on this question, many did not even attempt to use a
recognised method such as a factor tree or tree diagram. Frequently candidates who scored no
marks had just written down a list of products of 80, that is 80 × 1, 40 × 2, 20 × 4, 16 × 5 and
10 × 8.

(iii) The candidates scoring full marks on this question were usually able to give the HCF as 20 with
little or no working. Others were able to gain 1 mark for giving a smaller common factor such as 2,
4, 5 or 10.

Answers: (a)(i) 9 (ii) 4 (b)(i) 1.4 (ii) 4096 (c) 0.043 (d) 64.8 (e) 5 (f)(i) 300 (ii) 24 × 5 (iii) 20

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 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 3

(a) (i) Relatively few candidates were able to give the mathematical name for AB as chord. Whilst some
incorrect mathematical answers such as tangent, radius and perpendicular were seen, there were
also a number of candidates who gave the non-mathematical words string and rope and a large
number of candidates who did not give a response.

(ii) Only around a third of candidates correctly named PQ as a tangent. Again, there were a range of
incorrect mathematical words given and a large number of blank responses.

(b) (i) This part was answered well. However it was clear that some candidates, in this and the next two
parts, did not understand what, for example, ‘angle COQ’ meant and/or which angle it represented.

(ii) This part was answered reasonably well with most candidates recognising that triangle ABO was
isosceles. Some candidates, who showed working, were awarded a method mark despite making
arithmetic errors.

(iii) This part was answered well with many candidates recognising that angle COQ + angle OQC =90°.
Candidates who could progress no further than indicating that angle OCQ = 90°, either on the
diagram or otherwise, were awarded 1 mark.

Answers: (a)(i) chord (ii) tangent (b)(i) 48 (ii) 66 (iii) 42

Question 4

(a) The majority of candidates could not give the correct name of the triangle. Common errors of
isosceles, normal, irregular, obtuse-angled, image, object and translation were seen.

(b) Candidates were frequently able to obtain at least 1 mark when describing the transformation.
Common errors that were seen included not giving the correct word ‘translation’, miscounting the
horizontal and vertical displacements, mapping B onto A (instead of A onto B), omitting one or
more of the negative signs, inverting the vector or giving the vector as a co-ordinate.

(c) Candidates generally rotated the triangle 90° clockwise. Candidates were not so careful about
using the correct centre of rotation and it was reasonably common to see (1, 2) or (1, 1) being used
as the centre.

(d) (i) Candidates were reasonably successful in correctly working out the area of triangle A. Common
errors seen were 2, 3 and 6.

(ii) Candidates generally understood the concept of enlarging triangle A with scale factor 2. However,
candidates did not find it so easy to use P, the centre of enlargement, correctly and triangles were
often drawn with a vertex at P.

(iii) Candidates found this part quite challenging. Those few candidates who calculated the area of the
enlarged triangle tended to be successful, but generally candidates showed little evidence of
working or any method. A common incorrect answer was 2.

 −5 
Answers: (a) scalene (b) translation   (d)(i) 1.5 (iii) 4
 −4 

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 5

(a) (i) This part was generally well answered with most candidates giving the correct expression in terms
of n.

(ii) Most candidates understood the concept but were not accurate in their response as was evidenced
by their lack of understanding of the need for brackets and BIDMAS. The most common incorrect
answer was 2n + 10.

(iii) Most candidates who attempted this question were able to gain at least 1 mark. Marks were
awarded for equating their linear expression from part (a)(ii) to 52 and/or starting to solve it. Some
candidates gave the correct answer without writing down an equation.

(iv) Candidates were often able to obtain the correct answer, or if not, were rewarded for doubling their
answer in part (a)(iii) and then adding 10 to it.

(b) (i) This part was answered well. Marks lost were usually for not giving the fraction in its simplest form
or for giving the answer in the wrong form such as a decimal or a percentage.

(ii) Whilst some candidates answered this correctly, on the whole this was not well done. The most
1
common errors were to show the probability at or to draw an arrow that was not clearly pointing
4
3
to .
4

(c) This question was answered well with many candidates showing evidence of their ability to convert
accurately from grams to kilograms. The most common errors seen were to use a wrong
conversion or to either forget to multiply by 6 or to forget to convert the units.

(d) This was a relatively easy bounds question but it was not well answered. The most common errors
included 110 with 130 and 119.5 with 120.5.

1 3
Answers: (a)(i) n + 10 (ii) 2(n + 10) (iii) 16 (iv) 42 (b)(i) (ii) arrow at (c) 2.7 (d) 115, 125
4 4

Question 6

(a) (i) Many candidates were able to find the correct median of the list of values. The most common error
made was not re-ordering the list and attempting to find the median from the wrong middle pair, 2.3
and 4.3. Some scored 1 mark for re-ordering the list and/or identifying the correct middle pair, 4.3
4.3 + 4.7
and 4.7. Again a lack of brackets and knowledge of BIDMAS saw errors such as = 6.65 .
2
Other errors included finding the 5th number or giving 5.5 as the answer or finding a different
statistical value, usually the mean.

(ii) Many candidates found the range successfully. The error seen most commonly was the answer of
4.9, from 9.6 – 4.7, the end values of the original list.

(iii) This was well answered with the majority of candidates able to calculate the correct mean of the
data.

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

(b) (i) The majority of candidates scored full marks. Many others scored the method mark for calculating
the correct time of 1.5 hours taken to complete the walk. Some candidates did not then convert 1.5
hours to 1 hour 30 mins, but added 1.5 to the starting time of 14 20 leading to incorrect answers
such as 15 70 or 16 10. Candidates scoring no marks had usually multiplied, rather than divided,
the distance by the speed, thus 9 × 6 = 54 followed by a finishing time of 15 14 was reasonably
common.

(ii) A minority of candidates scored full marks. Most were able to score the method mark for converting
6 km to 6000 m although a number of candidates thought there were 100 m in a kilometre. Others
knew they needed to divide by 60 to convert the hours to minutes and this could also score the
method mark. It was common to see various multiples of 6 and 10 used incorrectly.

(c) (i) This was generally well answered with the majority of candidates recognising positive correlation
on the scatter graph. A small, yet significant number of candidates gave answers referring to
distance, time, speed or acceleration.

(ii) This question tested interpretation of the points on a scatter graph recording distances and time
and candidates found this quite challenging. Candidates usually knew that one of the extremities
should be chosen but they chose the correct answer, one of the bottom pair (2.5, 35) or (3, 28) or
the highest point (5.8, 69) in roughly equal quantities. It was fairly common to see 2 or more points
circled.

Answers: (a)(i) 4.5 (ii) 8 (iii) 5.18 (b)(i) 15 50 (ii) 100 (c)(i) positive (ii) (4, 68) indicated

Question 7

12
(a) (i) Nearly all candidates completed the table of values correctly for y = .
x

(ii) This was well answered by many candidates who scored full marks for accurate, smoothly drawn
curves. Most others scored 3 marks, the fourth mark being most commonly lost for one point being
plotted out of tolerance, or for just plotting the points without drawing the curve through them or for
joining the points with ruled lines. However, only a few candidates joined the points across the
y-axis, thus spoiling the shape of the graph of this reciprocal function.

(iii) Many candidates scored the mark for drawing the line y = –5 using a ruler. Some lines were vertical
through x = –5 or slanting through the point (0, –5). Other candidates drew the lines y = –5.2 or
y = –4.6 either by lack of accuracy or not interpreting the scale correctly.

(iv) Those who scored the mark in part (a)(iii) were usually able to read off the point of intersection of
the curve and the line y = –5, although some attempts were spoilt by candidates who read the
negative y-axis scale backwards; –2.6 being given as –3.4 for example. A few others scored the
mark for reading off from their incorrect curve/line.

(b) (i) Candidates did not do very well on any of the parts (b)(i),(ii) and (iii). Although there were many
good attempts at rise/run seen, this part proved to be a challenge for the majority of candidates.
1
Incorrect answers included , +2, –2, vectors or just co-ordinates of point(s) taken from the grid.
2
1
Some correct answers were spoiled by including the x, that is, giving the gradient as − x .
2

(ii) Only a minority of candidates were able to find the equation of the line drawn on the grid. Many had
only a vague idea how to use y = mx + c . Answers often appeared as constants, contained m or
were left blank.

(iii) This part required a good understanding of y = mx + c and usually only those who had scored the
previous mark were able to score marks in this question. Whilst some candidates correctly drew
the line on the grid many candidates did not give any response.

Answers: (a)(i) –3, –6, 6, 3 (iv) –2.5 to –2.3 (b)(i) –0.5 (ii) y = –0.5x + 2 (iii) y = –0.5x + 3

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 8

(a) (i) A large majority of candidates scored full marks for an accurate drawing of the trapezium. Many
candidates assumed DC = 5 cm.

(ii) Only around half of the candidates were able to measure the obtuse angle in their trapezium
correctly. The most common error was to measure the acute angle instead.

(iii) The majority of candidates scored the mark for measuring their DC correctly. Many had assumed it
to be 5 cm and drawn it accordingly.

(iv) The majority of candidates made a good attempt to find the area of their trapezium either by using
the trapezium formula or by splitting the shape into a rectangle and triangle. Errors were many and
varied including incorrect formulae, finding the perimeter or multiplying values together.

(b) (i) Only a minority of candidates knew the formula for the volume of a cylinder and these candidates
generally scored full marks. Many incorrect formulae were used and not all of these included π.
Some candidates were able to earn a mark for recognising that the radius was 15 cm but most did
not. Simply calculating 30 × 25 was a common error.

(ii) The majority of candidates struggled with this part because they could not connect the dimensions
of the cylinder to the dimensions of the cuboid. Just getting as far as recognising that the three
dimensions were 30, 30 and 25 would have earned a mark. Candidates often found the area of one
face, usually 30 × 25 = 750 and either left this as their final answer or multiplied it by 6. A significant
number of candidates left this question blank.

Answers: (a)(ii) 124 (iii) 4.7 (iv) 31.25 to 32.25 (b)(i) 17 700 (ii) 4800

Question 9

(a) The majority of candidates were able to factorise this expression correctly. Common incorrect
answers were usually from trying to combine the two terms to give answers such as 8y 3 or 9y 3 or
8y 2 .

(b) Many candidates scored full marks. Most other candidates were able to score 1 mark usually for
correctly expanding the first bracket 3 ( 2 x − 1) . The most common error arose from a sign error
when expanding the second bracket −4 ( x − 5 ) as −4 x − 20 . Other errors included, for example,
expanding 3 ( 2 x − 1) as 6 x − 1 or sign errors such as −3 + 20 = −23 or −17.

(c) Many candidates demonstrated a good understanding of how to rearrange a formula and the
correct answer was often seen. The most common errors were usually sign errors, such as
k + 5m = 7 p or 5m − k = 7 p , or trying to subtract the 7.

(d) A significant number of candidates scored full marks on this final question. Most candidates used
the elimination method and were able to demonstrate an understanding that they needed to
multiply both equations to make the coefficients of one of the variables equal. There were many
different types of errors seen and these included: arithmetic errors, incorrect coefficients in one or
both equations, subtracting the equations when they should have been added and vice-versa. A
mark was earned by less able candidates for finding two values that satisfied one of the equations.

k − 5m
Answers: (a) y ( y + 8 ) (b) 2 x + 17 (c) (d) x = 4 , y = −3
7

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

MATHEMATICS

Paper 0580/41
Paper 41 (Extended)

Key messages

To achieve well in this paper, candidates need to be familiar with all aspects of the extended syllabus.

The recall and application of formulae and mathematical facts in varying situations is required as well as the
ability to interpret situations mathematically and problem solve with unstructured questions.

Work should be clearly and concisely expressed with answers written to an appropriate accuracy.

Candidates should show full working with their answers to ensure that method marks are considered where
answers are incorrect.

General comments

Some questions allowed candidates to recall and demonstrate their skills and knowledge, others provided
challenge where problem solving and reasoning skills were tested. Solutions were often well-structured with
clear methods shown in the space provided on the question paper. A number of candidates were less
systematic and often gave random working sometimes with a choice of methods.

Candidates appeared to have sufficient time to complete the paper and omissions were due to lack of
familiarity with the topic or difficulty with the question rather than lack of time.
22
Most candidates followed the rubric instructions with respect to the values for π although a few used or
7
3.14 which may give final answers outside the acceptable answer range. There were a number of candidates
losing unnecessary accuracy marks by either approximating values in the middle of a calculation or by not
giving their answers correct to at least three significant figures.

The topics that proved to be accessible were simple ratio, reverse percentage, solving simple equations,
factorising and solving quadratics, simplifying algebraic fractions, quadratic graph drawing, finding the mean
from a grouped frequency table, interpreting histograms, linear and simple quadratic sequences, and matrix
manipulation.

More challenging topics included working accurately with currency, problem solving with angles and circles,
using graphs to solve related equations, problem solving with 3D shapes, harder probability, reasoning and
problem solving with general triangles, vectors.

Comments on Specific Questions

Question 1

(a) This part was nearly always answered correctly. A few candidates divided the given total by 13 or
5, rather than 13 + 5. Occasionally the total for fiction books, rather than non-fiction books, was
found.

(b) Most candidates were familiar with ratio questions in which they are given the value of one of the
items but many were unsure how to deal with this question, in which they were given the difference
in value of two terms. A minority were able to link 10 – 6 with 384 in the ratio to obtain the solution.

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

(c) (i) Most candidates were able to make an attempt to find the speed and there were many correct
answers. The main source of error was with accuracy. For example the time of 23 minutes was
often converted to either 0.38 or 0.383 hours and then this was used to find the average speed. In
both these cases this gave an answer outside the accepted range. It is important for candidates to
appreciate that they must use full decimal values in calculations where appropriate to ensure
accuracy.

(ii) There were few candidates who scored full marks. Those who chose to find the journey time using
32
the new speed of 32 km/h sometimes made an error, such as giving but most used the correct
20
method. Those who worked in minutes often gave the exact answer of 37.5 minutes whereas those
working in hours sometimes made the same error as described in the previous part, using a time
such as 0.38 hours. Some did not then calculate the time difference and others divided by the new
time rather than the original time.

(d) This was not well answered. Those that worked in dollars often overlooked that the final answer
should be given to the nearest cent and $0.1 was a common final answer. Some truncated their
answer to $12.92 when converting to dollars. Others chose to work in euros and multiplied $12.99
by 0.9276 but then very few converted their difference in euros back into dollars. Candidates
should note that the method steps in the conversions should be very clearly shown to ensure
method marks when values are inaccurate.

(e) There were a large number of correct answers. There were a number of candidates who applied
the correct method but used 88% instead of 78%, presumably from calculating 100–22 incorrectly
although very few showed the subtraction. Many candidates did not realise that this was a reverse
percentage situation so it was quite common to see 22% of 7605 or 122% of 7605.

Answers: (a) 2915 (b) 1056 (c)(i) 52.2 (ii) 63.0 (d) 0.06 (e) 9750

Question 2

(a) In general, candidates responded to this question well and although the work was often set out
quite randomly, many candidates reached the correct answer. There were many possible methods,
most involving the drawing of additional lines on the diagram. It was often difficult to identify the
angles that candidates had used, although in a small number of cases these were written on the
diagram. The most common method which did not involve drawing extra lines was to use the
formula for the sum of the interior angles. Although some errors were seen involving substituting an
incorrect value for the number of sides, most gave 1080° and often went on to find the correct
answer.

(b) Candidates found this part challenging and few were able to give the correct answer. By far the
most common error was to falsely assume that PQRO is a cyclic quadrilateral so 2y − 60 + y = 180
was seen very frequently. Those who gave a correct first step such as 360 − ( 2y − 60 ) sometimes
omitted the brackets and those who were able to give a completely correct method, such as
360 − ( 2y − 60 ) = 2y made a sign error either when removing the brackets or when solving their
equation.

Answers: (a) 122 (b) 105

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 3

(a) Candidates answered this part very well. Any errors were usually as the result of a sign error when
rearranging the equation.

(b) (i) This was also answered well. Some candidates gave x ( x + 9 ) − 22 and others made sign errors
leading to ( x + 2 )( x − 11) .

(ii) Most candidates gave the correct answer but some did not appear to see the connection with the
previous part and so it was common to see the quadratic formula being used. Although this method
normally resulted in the correct answers being obtained, a few errors associated with this method
were also seen.

(c) Although some candidates only scored 1 mark, most earned 2 marks by removing the brackets
correctly and removing the fraction by multiplying both sides by x to reach xy = 2x – 2a. Candidates
found the next step rather more challenging but some separated the terms containing x from the
–2a term. Those that did this usually went on to factorise and then divide by y – 2 or 2 – y to obtain
the correct answer.

(d) This part was generally well answered, although a number of candidates did not attempt any
factorisation but carried out some ‘false’ cancelling e.g. cancelling the x2 in the numerator with the
x2 in the denominator. Those who attempted to factorise usually did both parts correctly although
some made a sign error.

2a x
Answers: (a) –2.75 (b)(i) (x + 11)(x – 2) (ii) –11 and 2 (c) (d)
2−y x+6

Question 4

(a) Most candidates successfully found both required values but the occasional sign error led to a
value of y = 18 for x = –1.

(b) The points were generally plotted well, with only a small number misinterpreting the scale and
some plotting the first point at (–2, 9) instead of at (–2, −9). Most candidates drew the curve
correctly although feathering, double lines and ruled sections were quite common.

(c) Most candidates read off the correct value from their graph. The most common incorrect answer
was 15.

(d) Although the majority of candidates correctly drew the tangent at x = 3.5, some drew it elsewhere
or not at all. The gradient of the tangent was correctly found by many. A common error was to
count squares for the rise and the run instead of taking into account the different scales on the two
axes.

(e) This part proved to be a challenge for most candidates. There were attempts at rearranging the
given equation to compare it with the original function, and where candidates successfully reached
y = 2x + 10, the correct line was usually drawn and usually all three values given correctly. Those
who did not reach this equation often ruled a line through (0, 5) and a few attempted to draw the
curve corresponding to the given equation.

Answers: (a) 10, 7 (c) –1.7 to –1.55 (d) 6.5 to 11 (e) –1.3 to –1.1, 1, 4.1 to 4.25

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 5

(a) This was very well answered, with most candidates getting all three answers correct, although
more errors appeared in the number of apples in the 140 g to 170 g interval than any other.

(b) This was answered successfully by many candidates. Even those who had incorrect values in part
(a) were usually able to use their values correctly in this part. There was a common theme to errors
in this part: a significant but small group of candidates used class widths rather than mid-interval
values; a small number of candidates showed mid-interval values correctly but then found their
sum and divided by 5; a small number of candidates found the sum of fm but divided by 5 or by an
incorrect total obtained by adding their frequencies.

Answers: (a) 54, 76, 96 (b) 187

Question 6

(a) There were many correct answers in this part. The numerical answers were generally correct, with
only the number of small squares in Diagram 5 causing some difficulty. Many candidates correctly
gave 4n + 2 for the number of crosses, sometimes in an unsimplified form. More candidates
struggled to find the number of small squares in Diagram n.

(b) Many candidates were successful here, giving the correct answer even after an incorrect or
missing algebraic answer in part (a).

(c) Fewer candidates were successful here, although some misread the question and found the
number of squares for Diagram 226 or solved 4n + 2 = 226.

(d) Fewer candidates were successful in this part, which was not attempted by a number of
candidates. Those who did attempt it often substituted n = 1 or n = 2 into the given formula.
However, they then often put this equal to 0, so that answers of 4 or –4 or sometimes 12 were
common.

Answers: (a) 18, 22, 4n + 2, 17, 26, n2 + 1 (b) 242 (c) 15 (d) 3

Question 7

(a) This part was answered very well.

(b) The majority of candidates were able to substitute h(3) into function g correctly. Those that made
errors usually found a product of the two functions.

(c) Most candidates were successful in finding the inverse function. A few gave answers such as
3−y
and did not ‘interchange’ the y with x. Others struggled with directed terms and from e.g.
2
x – 3 = –2y, the negative was lost in the final division stage. Some less able candidates interpreted
1
the notation as an index and gave .
3 − 2x

(d) Although many made a correct start on this ‘show that’ question by giving 43–2x , most did not show
the correct product or division step using the rules of indices that would have led to the answer.

(e) This was reasonably well answered. Those that did not obtain the correct answer often gained a
method mark for showing 4x=8 in the working.

4 3−x
Answers: (a) –7 (b) (c) (e) 1.5
64 2

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 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 8

(a) Many candidates made a reasonable attempt at this part. Most attempted to write down the surface
area of the cone and hemisphere although some used the volume of a sphere. As the value of the
slant height of the cone is exact then, in order to score full marks, it is necessary to work with exact
values and then correctly cancel or divide with these exact values. The majority of candidates,
having written down algebraic forms for the surface areas, then used their calculators giving
decimal forms for the surface areas. This did receive some credit for the method but not full marks.
Candidates must also understand that when they are asked to show a result, they should not start
by assuming that result in their method.

(b) Some candidates used an incorrect method by attempting to find the height using an area or
volume but the majority identified Pythagoras’ theorem as the appropriate method. Most applied
this correctly and scored full marks with just a small number using h2 = 6.52 + 2.52. Those
candidates who had incorrectly used a radius of 5 in parts (a) and (c) were given credit in this part
and in part (c) for a correct method. The answer to part (a) was given and all candidates should be
aware that they must use 6.5 and not an incorrect value that they have calculated.

(c) This part was answered well with the majority identifying the correct volume formulas to use. The
main source of error was for candidates to use the formula for a full sphere rather than a
hemisphere. Those who used an incorrect value for the height of the cone from part (b) were given
credit for a correct method in this part.

(d) Most candidates appreciated that they needed to use their answer from part (c) and that this
needed to be converted from mm3 to cubic centimetres. Some were also aware that
1 cm3 = 1000 mm3 and so divided their volume by 1000. There were however many who divided by
10 or by 100 and in a small number of cases multiplied by their conversion factor. The majority then
went on to complete the method multiplying their volume by 19.3 and 38.62. Some did not show
their multiplications clearly and made it difficult to award method marks as a result.

Answers: (b) 6 (c) 72.0 (d) 53.7

Question 9

(a) (i) This part was well answered. The most common errors came from candidates either only getting as
far as giving the probability of choosing a red bead as 0.65 or finding the number of green beads as
28 from 0.35 × 80.

(ii) This part was well answered. The most common errors were either just re-stating the probability of
84
a green as 0.35 or by giving a final answer of .
240
3
(b)(i) The majority of candidates recognised that the probability of choosing a yellow marble was .
9
Candidates often then went on to complete the question successfully. However, a number of
3 2 1
candidates did not read the question carefully and did not replace the marble, finding × × .
9 8 7
Other candidates incorrectly attempted to add the fractions rather than to multiply them.

(ii) Only the candidates with a clear understanding of probability scored full marks on this question.
2 3 4
Many candidates scored 1 mark for × × but most were unable to recognise that this should
9 9 9
be multiplied by 6 for the six possible arrangements. A significant number of candidates incorrectly
used 1 − [P(BBB) + P(YYY) + P(WWW)]. Other common errors included, as in part (b)(i), not
replacing the marbles or adding the probabilities.

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

(c) Candidates found this part most challenging. However, most candidates correctly used the idea of
not replacing the counters and they were usually able to show either the product for P(PPP) or the
product for P(PPG) or P(PGP) or P(GPP). A common error was not to recognise that 3×P(PPG)
was required. Candidates who chose not to show any products to support their working frequently
scored no method marks as it was often very difficult to know where their answers were coming
from.

27 144 42
Answers: (a)(i) 52 (ii) 84 (b)(i) (ii) (c)
729 729 60

Question 10

(a) Many candidates found this part challenging and rather than using the cosine rule to set up an
equation involving x, tried to calculate x by first of all finding angle ABC and then used the cosine
rule to find the value of x. Those that made a correct substitution into the cosine rule using cos 60
were usually able to complete correctly to the given equation.

(b) This was quite well answered. Some used the quadratic formula correctly but did not round
answers correctly to two decimal places. A few did not use essential brackets during substitution
and made errors when, for example, squaring –17. Others gave correct answers from their
calculators but were unable to show a correct method and scored only partial marks.

(c) For this part many candidates had two values of x, one from part (b) and one from part (a) and
these were often different values. To score in this part they had to use their positive value from
part (b). Most candidates correctly used the sine rule.

1
(d) Some candidates used the area formula ab sin C reasonably well and a number scored full
2
marks for this part. Others used incorrect lengths or the wrong combination of lengths for the
included angle.

Answers: (b) 14.35, –5.85 (c) 12.2 (d) 138

Question 11

(a) (i) Most candidates were successful and scored full marks. A few made sign or arithmetic slips but
were able to score 1 mark for having two or three of the elements correct. A common error was to
4 9 
give the matrix   , found from squaring the individual elements.
 1 16 

(ii) This part was completed correctly by the majority of candidates and they were able to demonstrate
competency with the process to find the inverse of a matrix.

(b) There were mixed responses to this part. Some gave reflection, fewer gave the correct line of
reflection. The most common incorrect answers involved rotations.

(c) Some candidates were able to work out, or recall, the correct matrix and scored full marks. Other
candidates were aware the matrix should contain a combination of zeros and +1/–1 but it was
evident they did not have a clear strategy as to how they could work out where to place them in the
matrix.

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 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

(d) (i) There were some correct answers seen to this part. Answers were required to be correctly
simplified for full marks. Some slips in signs or in the adding of fractions were often seen but many
uuur
candidates were able to score at least 1 mark for giving either a correct route for OP or, for
uuur
example, AB = b − a .

(ii) Most candidates found this part challenging and many did not attempt it. To make progress,
uuur uuur uuur
candidates had to recognise the need to use OC = OB + BC . Only a few candidates were able to
set up a vector equation and solve it correctly.

 1 −18  1  4 3  0 1 4 3
Answers: (a)(i)   (ii)   (b) Reflection, y-axis (c)   (d)(i) a+ b
 6 13  11  −1 2   −1 0  7 7
7 4
(ii) ,
3 3

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

MATHEMATICS

Paper 0580/42
Paper 42 (Extended)

Key messages

To achieve well in this paper, candidates need to be familiar with all aspects of the extended syllabus.

The recall and application of formulae and mathematical facts in varying situations is required as well as the
ability to interpret situations mathematically and problem solve with unstructured questions.

Work should be clearly and concisely expressed with answers written to an appropriate accuracy.

Candidates need to be aware that in drawing graphs, linear functions should be ruled and curves should be
drawn freehand with a sharp pencil.

Candidates should show full working with their answers to ensure that method marks are considered.

Candidates should take sufficient care to ensure that their digits from 1 to 9 can be distinguished.

Candidates should ensure that their calculator is set in degrees.

General comments

Although a few question parts proved to be a challenge to many candidates, most were able to attempt
almost all of the questions reasonably well. Solutions were usually well-structured with clear methods shown
in the space provided on the question paper, but a number of candidates did not show full working on the
questions that asked for this requirement and scored only partial marks as a consequence.

Candidates appeared to have sufficient time to complete the paper and omissions were due to lack of
familiarity with the topic or difficulty with the question rather than lack of time.
22
Most candidates followed the rubric instructions with respect to the values for π although a few used or
7
3.14, giving final answers outside the range required. Very few candidates lost accuracy marks by not giving
their answers correct to at least three significant figures but there were a number of candidates losing
unnecessary accuracy marks by approximating values in the middle of a calculation, particularly when their
chosen method is not the most efficient available. This was apparent for example in Question 3. The
requirement for accuracy to one decimal place in Question 1(d) and to two decimal places in Question 8(b)
was sometimes ignored.

The topics that proved to be accessible were ratio, percentage decrease, simple interest and the recall of the
compound interest formula, recall of and recognition of when it is appropriate to use the cosine rule,
translation, rotation, matrix representation of enlargement, plotting points and drawing curves, drawing a
tangent and finding its gradient, use of a cumulative frequency curve, finding the mean of grouped data,
drawing histograms, probability, use of the quadratic formula to solve a quadratic equation and using
functions including composite and inverse functions.

More challenging topics included manipulation of the formulae for volume of spheres and cones, total surface
area of a cone, similar volumes, identifying the shortest distance from a point to a line, translation of a point
by a vector and length of a vector, probability involving more than two events, creating and solving linear
simultaneous equations, creating and manipulating equations with algebraic fractions and area of a
composite shape.

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 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Comments on specific questions

Question 1

(a) (i) Candidates cancelled the ratio to its simplest form with very few errors seen.

(ii) Nearly all candidates obtained the correct ratio. Some gave the unsimplified ratio 16 : 20. A
1
minority thought that was an actual amount as opposed to a proportion and so subtracted 0.2
5
from the values giving an answer of 19.8 : 24.8.

(iii) Many correct answers were given. Some candidates ignored the bold then and subtracted the 4
from the original values leading to an answer of 16 : 21.

(b) (i) The majority of candidates were able to calculate the percentage decrease. Some found the new
amount as a percentage of the original but then did not subtract this from 100. Others divided by
the new amount instead of the original amount. A number of candidates in this part and in
subsequent questions used arrows to link for example, 15 600 to x and 100 to 11 420. Candidates
should be reminded of the importance of showing their method in calculations rather than
symbolically.

(ii) Many candidates correctly obtained the answer 16 000. There were two significant misconceptions
that arose with this part. Candidates either reduced 15 600 by 2.5% leading to an answer of 15 210,
or increased 15 600 by 2.5%, leading to an answer of 15 990. Those that recognised that 15 600
was 97.5% of the new amount almost always went on to obtain the correct answer.

(c) The majority of candidates scored well on this simple interest calculation. The two common errors
200 × x × 15
seen were either to make equal to 248 or to omit the division by 100.
100

(d) A large majority of candidates were able to begin correctly with the compound interest formula
10
 y 
256 = 200  1 +  . Errors were then seen when manipulating this equation to find y. A number
 100 
of candidates took the 10th root first; some moved the 200 inside the bracket and others subtracted
the 200 from 256 instead of dividing. Those that were able to rearrange usually went on to score
full marks but some did forget that the question asked for 1 decimal place and gave answers to 3
significant figures.

Answers: (a)(i) 4 : 5 (ii) 4 : 5 (iii) 3 : 4 (b)(i) 26.8 (ii) 16 000 (c) 1.6 (d) 2.5

Question 2

(a) (i) This was usually well answered. A few candidates did not use the correct formula for the volume of
the cylinder, whilst a few others did not use 2r for the height of the cylinder. Most candidates
subtracted the volumes correctly and only a few either added or only gave one volume.

(ii) This was a much more challenging part and proved to be a good discriminator. The difficulties were
often in setting up a correct expression in terms of r. The volume of the sphere was often equated
alone to the given 36; one of the volumes with r = 8 was occasionally used; the subtraction was
occasionally in reverse order and an incorrect formula for the cylinder was sometimes seen. A
significant number of candidates who reached a correct expression in terms of r had some difficulty
in rearranging it to obtain the correct cube root. A few candidates used the alternative method of
using similar volumes. These tended to be the more able candidates and usually they achieved full
marks.

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 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

(b) (i) This was another challenging question requiring several steps and an understanding of the
situation. Most candidates realised that an area was required although a few used volume. The
perpendicular height of the cone was often used as the slant height and this caused the loss of
most marks, as the Pythagoras’ theorem calculation was not seen. The area of the base of the
cone was often omitted. The candidates who used Pythagoras’ theorem usually obtained the
correct length for the slant height although 122−52 was occasionally seen. A few candidates divided
by 0.015, which gave a very expensive cost.

(ii) The most challenging aspect of this question part was to realise that volume was now required.
Many candidates used areas and 16 was a very common answer. The successful candidates
usually found the volumes of the two cones and divided. This usually gave an answer very close to
64 and most candidates realised that this would be the final integer answer. A few left the answer
as a decimal and a few thought they should round down to 63. Some of the more able candidates
realised that the cones were similar with the linear scale factor of 4 and then simply cubed it to give
64. This was the intention of the question.

Answers: (a)(i) 1070 (ii) 2.58 (b)(i) 4.24 (ii) 64

Question 3

(a) Many candidates gave fully correct answers and the vast majority were able to find the area of the
right-angled triangle ABC. For triangle ADC some candidates misquoted the area formula using
cos or tan instead of sin and others incorrectly assumed it must be right-angled and so calculated
1 1
AD in order to work out × 110 × AD , or × 100 × AD .
2 2

(b) The majority of candidates showed their working clearly and were able to demonstrate partially
correct methods towards finding the perimeter. Most used Pythagoras’ theorem correctly to find BC
and a few used trigonometry in a less efficient method. A large majority of candidates recognised
that the cosine rule was required to find AD and recalled the rule correctly. Some errors seen with
the formula included missing out the 2 or using sin instead of cos. After stating the rule correctly a
significant minority gave the value 76.6 for AD2 from treating the rule as
(1102+1002 − 2×110×100) cos 40. A few candidates included the internal measurement of 100 in
their perimeter calculations.

(c) This question part proved to be a challenge for many candidates who did not know where the
relevant distance was on the diagram. A common error was to assume AD was the shortest
distance. Others recognised that they needed a perpendicular from A to CD but incorrectly
assumed it would bisect CD, so performed Pythagoras’ theorem and/or trigonometry using a length
of 55 m. Some candidates who did identify the correct distance used inefficient methods such as
the sine rule to calculate angle CDA, then basic trigonometry with that angle and length AD.

(d) Many candidates used a correct method to find either angle ACB or angle ABC. Their chosen
method was not always the most efficient, for example some used the sine rule with their
calculated length for BC, which inevitably led to rounding inaccuracies. Where possible, candidates
should use given values for calculations rather than calculated ones. Many candidates found a
correct relevant angle but did not understand the concept of bearing. This was clear from some of
the arcs indicated on the diagram. Some working shown by candidates was ambiguous. It would be
helpful to candidates if they either labelled an angle on the diagram or used the standard three
letter angle convention.

Answers: (a) 7040 (b) 374 or 375 (c) 64.3 (d) 235

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 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 4

In part (a) candidates needed to indicate the correct position of both the triangle and the pole.

(a) (i) The majority of candidates drew a correct translation of the flag.

(ii) Many completed this 180° rotation correctly but there were some candidates who used an incorrect
centre or who only rotated through 90°.

(iii) This was the most challenging of the required transformations. Many correct reflections were seen
but common errors included reflecting in the x- or y-axis. Other candidates correctly identified y = x
(or y = –x) but could not then visualise the effect on the flag.

(b) (i) The vast majority of candidates identified enlargement with centre (0, 0) and many of these also
gave the correct scale factor. The most common errors seen were to give scale factor 2 or –2. A
few candidates did not understand the concept of a centre and spoiled their answer by describing a
translation as well as an enlargement.

(ii) Many candidates were able to recall the correct matrix form for an enlargement.

(c) Candidates found this to be one of the most challenging questions on the paper. A significant
number did not know how to begin. A further significant number of candidates understood what
was required but the omission of brackets within their application of Pythagoras’ theorem spoiled
their solution. It was very common to see 4u2 + 3u2 = 7u2. A few candidates applied Pythagoras’
theorem incorrectly and wrote (4u)2 – (3u)2. Of the candidates who gave a correct answer some
used their knowledge of Pythagorean triples for a very concise and efficient method.

1 
2 0
Answers: (b)(i) Enlargement, centre (0,0), scale factor 0.5 (ii)   (c) ± 2.5
0 1
 
 2

Question 5

(a) Invariably this part was correctly answered. A few candidates truncated their values to 3.1 and/or
5.1.

(b) Most candidates scored 3 or 4 marks in this part and only a few lost marks through incorrectly
plotted points. The main difficulty seen was in dealing with the part of the curve between x = 0.5
and x = 1. Some candidates started with a vertical line which was too long and a few started the
curve going to the left of x = 0.5. These candidates usually scored 3 marks for correctly plotted
points.

(c) This part was very well answered. A few candidates gave for example, 0.7 instead of 1.7.

(d) These parts required an understanding of the number of times a horizontal line crossed the curve
and was generally well answered. The most common error was to overlook the requirement of an
integer answer. A few gave a list of integers in part (ii) and this was accepted if they were all
correct. Some candidates gave their answer as an inequality which was not accepted as an
inequality does not exclude non-integer values. A small number of candidates omitted both parts.

(e) Most candidates answered this tangent question extremely well. Accurate tangents were seen and
the gradient was usually correctly calculated. A few candidates divided the change in x by the
change in y; a few misread the scales when finding co-ordinates and some made errors in the
calculation as a result of using negative co-ordinates. A very small number of candidates drew the
tangent at an incorrect point and a small number of candidates omitted this part.

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 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

(f) (i) This part was generally well answered as most candidates drew the correct straight line. Almost all
candidates who drew this line obtained an answer within the required range. A few candidates
omitted this part and a small number of others drew a line from (0, 6) with an incorrect gradient.

(ii) This part was more demanding as it required considerable algebraic manipulation. It did appear
that most candidates knew that rearranging and multiplication was necessary to end up with the
equation in the correct form. Many candidates carried out these operations carefully and gained full
marks. A few candidates omitted this part although the main challenges were in keeping correct
signs as well as powers and multiples throughout all the steps.

Answers: (a) 3.2 or 3.15, 5.2 or 5.19 (c) 1.7 to 1.8 (d)(i) Any integer k⩾–1 (ii) Any integer k<−1
(e) 2.5 to 4 (f)(i) 2.85 ⩽ x ⩽ 3 (ii) 8, –48, –16

Question 6

(a) (i) The median was usually correctly stated.

(ii) The upper quartile was usually correctly stated.

(iii) The inter-quartile range was usually found correctly.

(iv) The number of students who ran more than 350 m was usually found correctly. Occasionally
candidates stated the number who had run less than 350 m.

(b) (i) This question was well answered with most candidates showing clear and accurate working leading
to the correct answer. Only a small minority of candidates used the interval widths.

(ii) Many accurately drawn histograms were seen. It was rare to see calculations for the frequency
densities. Some candidates drew two of the blocks with incorrect widths but the more common
error was to draw the height of 0.44 two squares above 0.4.

(c) Many candidates gave the expected answer of ‘further’ or ‘faster’ and a further significant number
of candidates gave the answer ‘more’ which was accepted.

Answers: (a)(i) 280 (ii) 320 (iii) 90 (iv) 10 (b)(i) 250.2 (c) further

Question 7

(a) This question part was invariably correctly answered.

2 2
(b) Many correct answers were given. Some occurrences of + were seen and some candidates
6 6
2 1
got confused with the concept of replacement, leading to a response of × . A very small number
6 5
2 2
of candidates could not correctly process × .
6 6

(c) The majority of candidates answered this in the correct format with very few giving a relative
frequency as their answer. Some candidates felt this was a continuation of part (b) and found
1
of 60.
9

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 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

(d) (i) Many fully correct diagrams were seen. There were a few instances of addition rather than
multiplication. The most common error was five rows of 1113.

(ii)(a) The majority of candidates gave a correct answer or followed through correctly from their diagram.
Some candidates stated the number of outcomes instead of giving the probability.

(ii)(b) Consistent with part (d)(ii)(a) the majority of candidates gave a correct answer or followed through
correctly from their diagram. Some candidates stated the number of outcomes instead of giving the
probability.

(e) This proved to be one of the most challenging questions on the paper. Many candidates did not
know where to start. Of those that did, a number thought that they were still dealing with the grid
from part (d) so used the probability of scoring a 1 from the grid which led to the response
4 4
 30   6  5  1
 36  ×  36  . A number of candidates wrote the simplified fractions  6  ×  6  again using the
       
5
2 2
grid or perhaps reverting to using an ordinary six sided dice. Others calculated   ,   × 5 or
6 6
 
4 2
4 ×   × . Probabilities greater than 1 did not alert candidates to an error.
6 6

5 4 9 4 512
Answers: (a) (b) (c) 20 (d)(ii)(a) (ii)(b) (e)
6 36 36 36 7776

Question 8

(a) (i) The correct equation was usually stated but a few candidates used inequalities.

(ii) Candidates who used p = a + 2 invariably successfully solved the problem. A significant number of
candidates made the errors p = 2a or a = p + 2.

(b) (i) This question was usually correctly answered.

(ii)(a) Those candidates who had the correct three term equation rarely made errors in reaching the
5
required equation. Some initial equations included 5 or instead of 2 for the total time. Less able
2
candidates omitted this question part.

(ii)(b) The correct formula was usually used with both fraction lines and square roots long enough to
ensure that the calculation was accurately completed. Most candidates did select the correct value
for the final answer but sometimes the addition of the two values was given as the final answer.
Those who used completion of the square rarely reached the correct answer.

2
Answers: (a)(i) 7a + 9 p = 354 (ii) 21, 23 (b)(i) (ii)(b) 3.19
x

Question 9

(a) f(–1) was usually calculated correctly.

(b) Most candidates were able to set up a correct equation. A significant number of candidates made
2 5
sign errors when rearranging the initial correct equation and answers of − and − were often
3 2
seen.

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 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

(c) The composition of two functions was more challenging but many candidates answered this part
well. A few took the composition to be a product of the functions. Errors with expansion of brackets,
signs and combinations of terms were seen frequently. Some candidates completely ignored the
negative signs in their working.

(d) hh(2) offered similar challenges to part (c). The question was well answered by the more able
candidates either by finding h(2) followed by h(5) or by working out the algebraic expression for
hh(x). Many of those who found h(2) = 5 went on to work out 5 × 5 instead of h(5).

x −1
(e) This inverse of a linear function was well answered. The most common answer was and this
−2
was accepted for full marks. As in parts (b) and (c), sign errors when rearranging terms were seen
frequently. A number of candidates lost the final mark by leaving an otherwise correct answer in
terms of y.

(f) This was the most searching part of this question, involving the composition of three functions.
Compared to part (c), more candidates took this part to be the product of the three functions, even
though the form of the answer was given. The more able candidates realised that the safer way
was to work out gf(x) first and then simplify it to the form (ax+b) and then substitute this into h(x).
Those who worked out hg(x) first had more substitution to do and were generally less successful
than those who worked out gf(x) first. Many candidates earned the first two marks by reaching
(1 – 2x + 4)2 + 1. If they then went on to (5 – 2x)2 + 1, they usually scored full marks with a small
number forgetting to complete by adding the 1. If, however, they tried to square (1 – 2x + 4), they
rarely obtained a correct nine term expansion.

2 1− x
Answers: (a) 3 (b) − (c) –2x – 7 (d) 26 (e) (f) –20, 26
5 2

Question 10

This was the most demanding question of the paper requiring strategy and application of several topics of
the syllabus. It was clearly set to be a discriminating final question and this proved to be the case. There
were some excellent answers with well organised working. Some candidates rounded off the three separate
areas and gave a final answer out of range. Most candidates earned one or two marks, although quite a
number did not attempt this question.

(a) Many candidates obtained a correct expression for the area of the sector. The area of the triangle
was also often correctly found, sometimes by methods which were more complicated than
necessary. The area of the rectangle proved to be challenging with many candidates taking the
unknown side length to be x or 2x. The more able candidates were able to use Pythagoras’
theorem or trigonometry to find this unknown side length. The collection of the three areas was
usually attempted although a subtraction was occasionally seen. A few candidates only found the
sector and the rectangle, overlooking the first line of the question.

(b) This part was more straightforward but did depend on some success in part (a).

Answers: (a) 5.68 (b) 4.40

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

MATHEMATICS

Paper 0580/43
Paper 43 (Extended)

Key messages

Candidates need to use efficient methods of calculation, show their working and always check their final
answers. They should always work with more figures than the final answer requires. In most cases this would
need intermediate values written to at least four significant figures.

General comments

This paper gave all candidates an opportunity to demonstrate their knowledge and application of
mathematics. Most candidates were able to complete the paper in the allotted time. Few candidates omitted
whole questions although some did not attempt several parts. A majority of candidates showed their working
and gained method marks but in a significant number of scripts insufficient or no working was seen. In
questions requiring candidates to show a result many were unable to gain marks as they used the value they
had to show from the beginning. Presentation of work was often good with some scripts showing working
that was clearly set out. For less able candidates, working tended to be more haphazard and difficult to
follow making it difficult to award method marks. Candidates need to be encouraged to write their working
clearly. There were many examples where candidates miscopied their own figures leading to a loss of marks.
Similarly, many candidates overwrite their initial answer with a corrected answer. This is often very difficult to
read and is not clear what the candidates’ final answer is. Candidates should be reminded to re-write rather
than overwrite.

Comments on specific questions

Question 1

(a) (i) There were many correct responses seen with a significant proportion calculating all three angles.
The question required candidates to show a result and most incorrect answers involved incomplete
methods, usually by not showing the division by 10.

(ii) Most candidates used an efficient method, either 12 sin 36 or 12 cos 54, although a significant
number calculated the third side and then used Pythagoras’ theorem. Some calculated the third
side and went no further and a small minority could not get started as neither of the angles 36 and
54 had been found. A small number also took the sides to be in the same ratio as the angles.

(b) (i) Many correct answers were seen although some chose to calculate the perimeter and then
calculate the longest side from that.

(ii) Many correct answers were seen. However, candidates were generally less successful in this part
as a significant number took the ratio of the angles to be the same as the ratio of the sides leading
to a common incorrect answer of 45°.

Answers: (a)(ii) 7.05 (b)(i) 13 (ii) 36.9

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 Cambridge International General Certificate of Secondary Education
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Principal Examiner Report for Teachers

Question 2

(a) Almost all candidates gave the correct answer.

(b) (i) Apart from the common errors of 0 and x, the vast majority gave the correct answer.

(ii) Almost all candidates gave the correct answer. The common error was x21.

(iii) This proved more challenging and far fewer correct answers were seen. Most of the incorrect
answers involved errors in dealing with the negative index in the denominator and/or dealing with
the coefficient of 3 in the bracket. Common errors involved x8 in the final answer or coefficients of
3, forgetting to square the coefficient, 8 and 2 by reducing the coefficient by 1 when dividing by the
denominator.

(c) (i) Incorrect answers were about as common as fully correct answers. Some candidates earned credit
for a partial factorisation such as (2x + 6)(x – 3) but answers such as 2(x2 – 9) were very common.
These candidates did not seem to pick up on the inclusion of the word ‘completely’ in the question.

(ii) Candidates were more successful in this part, with many earning at least 2 marks for a correct
factorisation of the denominator. Some of those that only reached 2(x2–9) in the previous part were
able to complete the factorisation in this part. Many of the less able candidates simply cancelled
like terms without any attempt at factorisation.

2( x + 3)
Answers: (a) 343 (b)(i) 1 (ii) x10 (iii) 9x16 (c)(i) 2(x – 3)(x + 3) (ii)
x + 10

Question 3

(a) (i) A small majority of candidates coped well with the required change of unit for time that was needed
and gave a correct answer. However, a large minority either changed the unit incorrectly or made
no change at all. Attempting to apply Pythagoras’ theorem to calculate the length of the sloping line
was a common error.

(ii) Many candidates understood that calculating the area under the graph would give the distance
between the two stations. Some of these were successful, usually by dividing the area into two
triangles and a rectangle instead of using the efficient method for the area of a trapezium. Some
lost marks because of their inaccurate conversion of times to decimal equivalents, such as 44
minutes equated to 0.73 hours. Some gained partial credit for a correct method for the area but
without the conversion of minutes to hours.

(b) (i) A majority showed the correct conversion but many lost out by not showing a complete method.

(ii) Most candidates appreciated that they needed to divide a distance by the speed. Deciding which
distance to use proved challenging for many of the candidates. Common errors included using
1400 + 2(200) or 1400 – 200 or simply 1400. Others were confused by the units and divided the
distance by 126 km/h rather than 35 m/s.

(c) This part was not well answered and only a minority of candidates gave a fully correct answer.
Many attempted to divide distance by time but a wide variety of incorrect values in the ranges
210 to 220 and 72 to 74 were used. It was common to see candidates attempting to divide their
upper bound for the distance by their upper bound for their time, as well as forgetting to change the
units of time. However, some did use the correct bounds and reached an answer of 3 km/min but
then forgot to convert this to km/h.

Answers: (a)(i) 1890 (ii) 103.95 (b)(ii) 46.3 (c) 180

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 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 4

(a) Many correct intervals were seen, the most common error being the middle interval 70 < t ⩽ 80.

(b) Candidates have a good understanding of this topic and many fully correct answers were seen.
Incorrect answers usually resulted from the use of one or more incorrect midpoints, the use of end
points and in a few cases, the class widths were used. A few candidates set out the working
correctly but did not reach the correct mean, almost certainly from keying the calculation
incorrectly.

(c) (i) The most frequent incorrect response was to state that the intervals were unequal. Others stated
that the individual times were unknown but did not specifically relate this to the highest and lowest
times. Only a minority gave a complete explanation.

(ii) If a candidate knew what to do they usually went on to earn full credit. Many used the efficient
26 26
method, × 360 , but it was quite common to see candidates converting to a percentage,
150 150
usually 17.3%, but this then lost the final mark as the answer was often inaccurate. Others lost
marks by using the wrong intervals and 10 and 34 were sometimes used instead of 26.

(d) Almost all candidates gave the correct answer.

(e) (i) Fewer candidates were able to cope with two events but a reasonable proportion of them did
manage to write down the correct product and follow it with the correct answer. The most common
1 10
incorrect answers were from replacement and 2 × . The probability for the second girl was
225 150
10 10
often given as and a few candidates mistakenly added it to .
149 150

(ii) Only a small minority were successful in this part of the question as the second possibility was
frequently overlooked. Again, many calculated their probability 'with replacement'. Some attempted
to work in decimals or percentages but sometimes lost marks by not giving answers to 3 significant
figures.

(f) Candidates seemed unfamiliar with this style of question on the histogram and fully correct
answers were rare. The one height that was given happened to be half of the frequency for that
interval as well as being the width of the interval. Many of the incorrect responses followed one of
these two patterns. Evidence of any working was rare but some candidates did list the frequency
densities as their heights.

22 90 440
Answers: (a) 80 < t ⩽ 100 (b) 86 (c)(ii) 62.4 (d) (e)(i) (ii) (f) 13, 8.5, 7.25, 1.1
150 22350 22350

Question 5

(a) (i) Many correct answers were seen. Reflection in the x-axis was the most common incorrect answer.

(ii) Candidates were slightly less successful with the enlargement. Many with an incorrect answer
earned partial credit for an enlargement of the correct size and orientation but in the wrong
position. Many of those with the wrong position had the apex of the triangle at (0, 4).

(iii) Again, many correct translations were seen with some candidates earning partial credit for a
translation with a correct displacement in one direction. Several candidates treated the translation
 3
as   .
 −5 

(b) Many fully correct or partially correct answers were seen. The centre of rotation caused the most
difficulty.

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

(c) (i) This proved very challenging for many candidates who were unable to set out the working in the
correct format to allow correct multiplication by not converting the co-ordinates (1, –4) into a
column vector. Several candidates made no attempt.

(ii) As this part included a second matrix it proved more challenging than the previous part. Similar
errors were seen and a significant number of candidates used their answer from the previous part
instead of using (1, –4). A significant number of candidates made no attempt.

(iii) Candidates were more successful in this part. Some clearly recognised the matrix and gave a
correct description of the transformation while others showed attempts to transform the unit square
to find the correct description. Many of those that had struggled in the previous two parts made no
attempt.

Answers: (b) Rotation, 90° clockwise,(4, –1) (c)(i) (4, 1) (ii) (8, –1) (iii) Rotation, 90° anticlockwise, (0, 0)

Question 6

(a) (i) This was well answered by many candidates. The two most common errors involved using 10 as
the radius instead of 5 and using 2πr2h for the volume of a cylinder. Some candidates lost the final
mark by giving an answer of 25.4, a truncated version of the correct answer.

(ii) Fully correct solutions were given by a minority of candidates only with many others using the
formula for a whole sphere instead of a hemisphere. If the equation was set up correctly at the
start, the radius was usually calculated correctly. Some lost the final mark by using square root
instead of cube root, despite setting out their working correctly.

(iii) A majority of candidates showed the correct working and were able to give the correct answer.
Quite a few candidates gave the area of one face only instead of the total surface area. Others lost
the final mark by not maintaining sufficient accuracy, usually by giving the cube root of 2000 to only
three figures. Other common errors involved multiplication by 2 when trying to square the length of
a side and division by 3 when trying to find the cube root.

(b) (i) The majority of candidates made a successful attempt at this question. However, a number treated
1
one angle as a right angle and × 10 × 7 was a common error. Some used cosine in the area
2
formula instead of sine.

(ii) Many candidates demonstrated competency in identifying and using the cosine rule. Most were
able to find the correct length for the third side and show a correct method for finding the perimeter.
This was expected to be 23.46 or better but premature rounding of the third side resulted in a
significant number of candidates only giving the perimeter as 23.5. Some candidates used less
efficient methods, correctly drawing the perpendicular height and working their way through two
right-angled triangles.

(c) This part proved more challenging and fewer fully correct answers were seen. Many candidates
used the 28.2 as the arc length earning only partial credit; others used the formula for area instead
of arc length. A few used 10.2 as the third side of the triangle and used the cosine rule to calculate
the angle c.

Answers: (a)(i) 25.5 (ii) 9.85 (iii) 952 (b)(i) 22.5 (c) 64.9

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 7

(a) Most candidates completed the table correctly with many of the errors arising from incorrect
squaring of negative values of x.

(b) The plotting of the points was carried out accurately with many going on to draw a smooth curve.
When points were plotted correctly any loss of marks was the result of the curve missing one or
more points, the curve having straight line segments or occasionally excessive feathering.

(c) Most candidates understood what was required and were able to read off two values accurately.
For some, reading the scale led to errors.

(d) This proved challenging for all but the most able candidates. Those that attempted to complete the
square often started incorrectly with (x + 2.5)2. Others attempted to expand 2(x + a)2 + b with the
intention of equating coefficients but few were successful. Common errors tended to lead to terms
such as 2ax, a2 and 2b.

5 49
Answers: (a) 9, –6, 9 (c) –3.4, 0.9 (d) a = , b=−
4 8

Question 8

(a) (i) Many correct answers were seen in this part.

(ii) A majority of candidates gave the correct gradient but many didn’t appreciate that the equation
3
should be written in the form y = mx + c. Apart from the incorrect answer of , other common
2
2 2
errors included 3, –3, and − .
3 3

(b) A small majority gave a correct point on the x-axis. A wide variety of incorrect answers were given,
frequently without any working. The most common errors included (0, 0.8) and the intercept with
the y-axis.

(c) The process of finding the gradient of a perpendicular line was not understood by many of the
candidates and only a minority gave a correct equation. Many of the incorrect equations resulted
1
from an incorrect gradient, often given as 5, –5, and a variety of others. With or without the
5
correct gradient, not all candidates appreciated that (10, 9) needed to be substituted to find the
constant term. A significant number made no attempt.

(d) This proved quite challenging and a minority of candidates gave fully correct answers. Those that
found the co-ordinates usually solved the simultaneous equations algebraically. Several attempted
to sketch the two lines but this was rarely successful. Some attempted to find the point of
intersection of line A and its perpendicular line. Many gave an incorrect pair of co-ordinates without
showing any working and were unable to earn any method marks. A significant proportion of
candidates made no attempt.

(e) In addition to a very high proportion of candidates that made no attempt, a majority of those that
made an attempt struggled to make any progress. Many of those that were successful had drawn a
diagram showing the triangle with intercepts clearly labelled.

3
Answers: (a)(i) 5 (ii) − (b) (0.8, 0) (c) y = –0.2x + 11 (d) (2, 6) (e) 13
2

© 2017
 Cambridge International General Certificate of Secondary Education
0580 Mathematics November 2017
Principal Examiner Report for Teachers

Question 9

10 10
(a) Many correct expressions for the time were seen. Common errors included − 0.5 and .
x x + 0.5

(b) (i) Those candidates that had struggled in the previous part struggled again or made no attempt.
Many attempted to set up an equation but it was common to see the times subtracted in the wrong
order, and occasionally added. Algebraic fractions were often successfully combined but the
expansion of brackets often led to errors with the signs. Some tried unsuccessfully to work
backwards from the given equation, or use values found in the next part. Less able candidates
often attempted to solve the quadratic equation.

(ii) A majority made a good attempt and usually obtained two correct solutions, although not always
written to 2 decimal places as requested. When solving the equation some candidates made sign
errors and the use of –1 instead of –(–1) for –b and squaring –1 to obtain –1 were common errors.

(iii) Many candidates made no attempt to find the time. Many of those attempting the question earned
some credit but only a minority gave a fully correct solution. Some correctly calculated the time in
hours but did not go on to give their answer in hours and minutes. Others that did convert their
answers sometimes forgot to round their answer to the nearest minute. A significant number
attempted to calculate Alfredo's time instead of Luigi's.

10
Answers: (a) (b)(ii) –4.23 and 4.73 (iii) 2 h 7 min
x − 0.5

Question 10

(a) (i) Most candidates were successful in finding the prime factors, often by using a factor tree.
Occasionally not all factors were reduced to primes but at least some credit was earned. Some
gave an answer as the sum of prime factors rather than as a product.

(ii) The vast majority were able to give the correct LCM, the most common method being to list the
multiples of both numbers. A few showed the prime factors of each number in a Venn diagram.
Some gave 540 without any working. A common error was to think this was HCF giving an answer
of 18 or to give a higher multiple of both which was usually 9720.

(b) This proved challenging for many candidates and a very high proportion made no attempt. Many of
the candidates could not see the link between the given expressions for X and Y and the LCM and
HCF values resulting in no progress being made. Forming an expression for the HCF and equating
this to 1225 proved to be an efficient method. This led to a = 5 and enabled solutions for X and Y to
be found. Others started by finding prime factor products for 1225 and 42 875, again leading them
to deduce that a = 5. Some trialled values of prime numbers and if they reached a = 5 they were
usually successful. A few divided 42 875 by 5 and 7 to give the required answers.

Answers: (a)(i) 22 × 32 × 5 (ii) 540 (b) X = 8575, Y = 6125

© 2017