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INTERNATIONAL MATHEMATICS
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Cambridge International General Certificate of Secondary Education
0607 International Mathematics November 2011
Principal Examiner Report for Teachers
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Paper 0607/01
Paper 1 (Core)
Key message
In order to gain high marks in this paper, candidates need to have developed a good understanding of the
whole syllabus and be confident working without a calculator. They also need to ensure that they show all
relevant working in their answers.
General comments
The entry for this paper was small on this occasion and this should be taken into account when reading the
report. There was a wide range of marks with some candidates obtaining very high marks and others with
marks at the bottom end of the range. Most candidates were able to demonstrate an understanding of some
elements of the subject and it was rare to see a paper where a reasonable attempt had not been made to at
least some of the questions. Candidates answered Questions 1, 2(b), 4(a), 5(a), 8(b) and 10(a) particularly
well. Questions 5(c), 11(c) and 12 were found to be more challenging by some candidates. Working was
usually shown for those questions for which it was appropriate and the standard of presentation was good.
Candidates were usually able to carry out any necessary calculations accurately on this non-calculator
paper. Time did not appear to be a factor as the majority completed the paper with most candidates
attempting the final question.
Comments on specific questions
Question 1
Almost all the candidates identified the pattern in successive terms and gave the correct answer. Examples
of an incorrect answer were 34 and 36.
Answer: 35
Question 2
(a)
2
Most candidates answered this part correctly. The most common incorrect answer was 6.27 ×10 .
(b)
Virtually all the candidates answered this part correctly.
Answers: (a) 6.27 × 104 (b) 63 000
Question 3
(a)
Most candidates gave the correct factors. A few candidates included an incorrect factor in their list;
for example 7 or 8.
(b)
A small number of candidates gave 3 as the highest common factor but most gave the correct
answer.
Answers: (a) 3, 5, 9, 15 (b) 9
1
© 2011
Cambridge International General Certificate of Secondary Education
0607 International Mathematics November 2011
Principal Examiner Report for Teachers
Question 4
(a) (i)
A few candidates misunderstood the notation in this question and gave the answer as 6; the
majority of candidates, however, gave the correct answer.
(ii)
This part was quite well done. Candidates who attempted to expand the brackets, rather than
calculating 2 × 7 – 5, sometimes made an error; for example 5 + 6 – 5.
(b)
The majority of candidates answered this part correctly. A few candidates gave the square root of
4, rather than the square of 4.
Answers: (a) 8 (b) 9 (c) 16
Question 5
(a)
Most candidates showed a good understanding of the notation for the intersection of two sets and
gave the correct answer. A few candidates mistook the symbol for the intersection as that for the
union of two sets and so gave the answer as {p, q, r, s, t} or {q, r, s, t}.
(b)
There was a mixed response to this part. Some gave the answer as {q, r} assuming that A ′
represents the elements that are in A only and a number of candidates ignored the complement
sign and gave {p, q, r}, which is set A.
(c)
Most candidates found this part difficult. Some candidates attempted to give the elements in
{A ∪ B} rather than the number of elements in A ∪ B. The candidates who did give the number of
elements did not always give an answer of 5; answers of 0, 1, 3 and 4 were all given.
Answers: (a) p (b) s, t, u (c) 5
Question 6
This question was answered quite well with the majority of candidates giving the two correct lines of
symmetry. A number of candidates either omitted this question or only gave one line of symmetry. A small
number of candidates joined up the ends of two of the branches of the curve.
Answer: The two lines with equations y = x and y = –x drawn correctly.
Question 7
(a)
There was a mixed response to this part. The candidates who gave the correct answer generally
used the stem and leaf diagram to write down an ordered list of the numbers and then used this to
find the median. Those who did not do this were less successful and sometimes gave an answer
of 16, which did earn one mark. A few candidates omitted this part.
(b)
There was also a mixed response to this part with some candidates not giving an answer. Those
candidates who had an ordered list were often able to identify at least one of the quartiles which, in
most cases, was the lower quartile.
Answers: (a) 16.5 (b) 12
2
© 2011
Cambridge International General Certificate of Secondary Education
0607 International Mathematics November 2011
Principal Examiner Report for Teachers
Question 8
(a)
About half the candidates answered this correctly and there were others who did give the
x
. The most common
denominator as 12 but then gave an incorrect final answer, such as
12
x
obtained from subtracting the numerators and denominators.
incorrect answer was
−1
(b)
This was answered well by most candidates. A small number of candidates gave an answer of 6
with an incorrect power of c, for example 6c6, or gave an incorrect coefficient for c5, for example
5c5.
(c)
Most candidates answered this correctly. Some who did make an error, correctly divided 6 by 2 but
then gave an incorrect power such as 7 or 2.5. Other candidates left the answer as a fraction such
3x 5
3x 5
or 2 .
as
x
x
Answers: (a)
5x
(b) 6c5 (c) 3x3
12
Question 9
(a)
A few candidates answered this part correctly. Quite a number of candidates attempted to write
down the answer without showing any working and some of these gave an answer of 540°, the
sum of the angles of a pentagon. Other incorrect answers given were 120°, which is one interior
angle, and 360°.
(b)
This part was answered quite well. All the candidates that answered part (a) correctly went on to
earn full marks in this part. A number of those that had an incorrect answer for part (a) were also
able to score full marks in this part. Some did this by starting again and establishing that the sum
of the angles is 720° and then finding the correct value of x. Other candidates used a correct
method with their wrong answer to part (a) and then carried out the arithmetic accurately. A few
candidates made a false assumption such as assuming that all the angles were equal or that
x + x = 100 .
Answers: (a) 720 D (b) 160 D
Question 10
(a)
This part was answered very well with the majority of candidates understanding that the points A
and C were given by the co-ordinates (6, 2) and (1, 4) respectively and plotting the points
accurately.
(b) (i)
About half the candidates plotted the point B correctly. Those giving an incorrect answer did not
join O to A or O to C as the first step.
(ii)
Virtually all the candidates who had B correctly placed on the diagram gave the correct coordinates.
Answers: (a) Points correctly plotted
(b)(i) Parallelogram correctly drawn with vertices at (0, 0), (6, 2), (7, 6) and (1, 4)
(b)(ii) (7, 6)
3
© 2011
Cambridge International General Certificate of Secondary Education
0607 International Mathematics November 2011
Principal Examiner Report for Teachers
Question 11
(a)
This part was answered quite well with the majority of candidates using the formula for the midpoint
correctly.
(b)
There was a mixed response with about half the candidates able to quote the formula for the
gradient and then use it correctly with just a very small number making an arithmetic error. The
remaining candidates either omitted this part or gave an answer involving co-ordinates such as
(4, 10).
(c)
This part was often omitted. Some candidates used y = mx + c and substituted their answer from
part (b) for m but did not attempt to use either the co-ordinates of A, or the co-ordinates of B, to find
c.
Answers: (a) (2, 5) (b) 3 (c) y = 3x – 1 (or in an equivalent form)
Question 12
Most candidates found this question difficult and only a few gave the correct answer. A small number of
3 4
x 3
=
or
which
=
candidates were able to use similar triangles to write down an equation, such as
x 6
6 4
they did not solve correctly. Most of the candidates gave an incorrect answer, such as 6 – 1 – 1 = 4, without
attempting to use similar triangles.
Answer: 4.5
4
© 2011
Cambridge International General Certificate of Secondary Education
0607 International Mathematics November 2011
Principal Examiner Report for Teachers
INTERNATIONAL MATHEMATICS
Paper 0607/02
Paper 2 (Extended)
Key message
In order to gain high marks in this paper, candidates need to have developed a good understanding of the
whole syllabus and be confident working without a calculator. They also need to ensure that they show all
relevant working in their answers.
General comments
All candidates appeared to have sufficient time to attempt all questions on this paper. Clear working was
shown on most papers and most candidates wrote legibly in pen. Method marks could be awarded for
correct working seen even when the answer was incorrect.
Marks obtained ranged from single figures up to full marks, with the majority gaining more than half marks.
Comments on specific questions
Question 1
(a)
Most candidates understood how to write 375 x 1012 in standard form. Occasional errors seen
included 3.75 x 1010 and inappropriate rounding to 3.8 x 1014.
(b)
Many candidates understood that the calculation 75 x 2.4 ÷ 100 could be used but made numerical
errors when calculating 75 x 2.4. The most successful candidates appreciated that 75% is
3
and calculated 2.4 ÷ 4 first before multiplying by 3.
equivalent to
4
(c)
x + 1 = 2 was not well understood with many candidates interpreting this as the pair of equations
x + 1 = 2 and x − 1 = 2 or the pair x + 1 = 2 and −x + 1 = 2 , leading to answers of 1 and 3, or 1 and
−1 . The correct equations of x + 1 = 2 and x + 1 = –2 were only seen occasionally.
Answers: (a) 3.75 x 1014 (b) 1.8(0) (c) –3, 1
Question 2
(a) (i)
(ii)
(b)
Almost all candidates understood and identified correctly n ( A ∪ B )′ .
Many candidates correctly found n( A ∩ B ) . Errors seen included an answer of 2, coming from the
equation 10 – x + 7 – x = 13 where the candidate had omitted the + x term from the left hand side.
Other candidates gave an answer of 3, from n(A) – n(B).
Shading the region A ∪ B ′ was commonly misunderstood. A significant number of candidates
shaded B ′ only. Others shaded A ∩ B ′ or ( A ∪ B )′ .
Answers: (a)(i) 7 (ii) 4 (b)
B
A
5
© 2011
Cambridge International General Certificate of Secondary Education
0607 International Mathematics November 2011
Principal Examiner Report for Teachers
Question 3
Many candidates successfully made a first step in rearranging the equation 3x + 4y = 12 to 4y = 12 – 3x and
then went on to complete the rearrangement correctly. After success with the first step some candidates
3
12 − 3 x
but either did not continue in order to reach y = 3 − x or,
then correctly progressed to y =
4
4
−3
x + 12 .
incorrectly, progressed to y = 3 − 3 x, or y =
4
Answer: −
3
x+3
4
Question 4
Many candidates correctly selected the formula for the volume of a sphere and gave a completely correct
4
π × 33
solution. The incorrect answer of k = 12 was common, coming from a correct substitution to give
3
but then using 33 = 9. Some candidates did not make use of the formulae printed on page 2 of the question
paper and thus used the wrong formula for the volume of a sphere. Others unnecessarily tried to calculate
using π as 3.14.
Answer: 36
Question 5
125 .
(a)
Candidates were competent at simplifying
(b)
Many candidates correctly identified the need to multiply by
provide a fully correct answer.
denominator of 6 + 3 .
Others expanded
6+ 3
, and some went on to
( 6 − 3 )( 6 + 3 ) incorrectly
A significant number of candidates incorrectly chose to multiply by
Answers: (a) 5 5 (b)
6+ 3
6− 3
6− 3
to give a
.
6+ 3
3
Question 6
(a) and (b) Candidates understood what was required and just a few numerical errors were seen.
(c) The expression for the nth term proved to be a challenging question for many candidates.
Candidates are familiar with quadratic sequences and having identified the non-linear sequence in
the question they often thought that the nth term should contain n2. To succeed in this type of
question, candidates need to be more aware of alternatives for non-linear sequences. The
candidates who recognised the power sequence gave a variety of correct expressions including,
3 x 2n, 6 x 2n-1, 2n+2 – 2n and 2n+1 + 2n.
Answers: (a) 192 (b) 768 (c) 3 x 2n
6
© 2011
Cambridge International General Certificate of Secondary Education
0607 International Mathematics November 2011
Principal Examiner Report for Teachers
Question 7
(a)
Candidates were competent at factorising x2 – 2x – 24 to (x – 6) (x + 4). Occasionally candidates
reversed the signs to give (x + 6) (x – 4) and some candidates unnecessarily went on to solve the
equation x2 – 2x – 24 = 0.
(b)
Almost all candidates achieved the first step in factorising xy2 – 4xz2 to get x(y2 – 4z2) and some
then recognised the difference of two squares and provided a fully correct factorisation. Many
candidates incorrectly factorised (y2 – 4z2) to (y – 2z)2 or alternatively did not recognise that
y 2 − 4z 2 could be factorised further. After successfully showing a correct first step in the
factorisation some candidates then spoilt their answer by ‘cancelling’ the x from outside the
bracket.
(
)
Answers: (a) (x – 6) (x + 4) (b) x(y – 2z) (y + 2z)
Question 8
Some candidates mistakenly found QP or disregarded the
(a)
Many correct answers were seen.
directional arrows and wrote p + q.
(b)
Candidates should be encouraged to write the step OR = OQ + QR both to clarify their thinking
and to gain method marks should they get the final answer incorrect. Candidates often incorrectly
1
started with OQ + PQ . To achieve full marks it was necessary to give a correct expression in its
4
simplest form.
Answers: (a) –p + q
(b)
1
3
p+ q
4
4
Question 9
This question highlighted the difficulties some candidates have when working with fractions and also some
misunderstanding of when it is, or is not, appropriate to use ‘without replacement’ methods in probability.
4 4 2
32
. Numerical errors in the multiplication may
The correct method × × sometimes led to answers of
6
6 6 6
be avoided if candidates are encouraged to simplify their fractions before attempting to multiply.
4 3 2 4 4 2
2 2 2
× × , + +
× × . Some candidates did not note that
and
6 5 4 6 6 6
6 6 6
5 5 1
zero appeared on two faces of the die and hence wrote × × .
6 6 6
Other incorrect answers included
Answer:
4
27
Question 10
The majority of candidates showed very good algebraic skills to solve the given equation and many correct
2(2 x + 1) 3( x + 1)
+
= 54 or a slip in
solutions were seen. Where an error was made it was often of the type
6
6
multiplying out the brackets correctly.
Answer: 7
7
© 2011
Cambridge International General Certificate of Secondary Education
0607 International Mathematics November 2011
Principal Examiner Report for Teachers
Question 11
(a)
The majority of candidates understood how to evaluate p. The occasional error, p = 83, was seen.
(b)
However
Many candidates made good use of log ab = log a + log b and log a b = b log a .
candidates assumed that log 12 = q log 2 and log 9 = r log 3 which then led to confusion. Most
successful candidates used the approach log 12 + log 9 = log 108 and then used trial and error to
write 108 as 22 x 33 leading to log 22 + log33 and ultimately the correct solution. Others wrote both
log12 and log 9 in terms of log 2 and log 3 before deducing a value for r and hence realised that
log 12 contributed to the number of log 3’s.
Only a minority of candidates used log rules incorrectly and wrote log 12 = log(2 × 6 ) = 6 log 2 and
log 9 = log(3 × 3 ) = 3 log 3, hence giving q = 6 and r = 3.
Answers: (a) 2 (b) q = 2, r = 3
Question 12
(a)
The majority of candidates understood what was required and gave a correct solution. After having
identified the relationship F = kv2 and correctly evaluating k as 8, candidates should be urged to
then state explicitly that F = 8v2.
(b) (i)
This part was almost invariably answered correctly by candidates who had achieved a correct
formula in part (a) and also followed through correctly by candidates who had evaluated k
incorrectly.
(ii)
Again, very successfully completed by the majority. Some candidates did not recognise that 121 is
a square number and thus did not evaluate
121 .
Answers: (a) F = 8v2 (b) 32 (c) 11
8
© 2011
Cambridge International General Certificate of Secondary Education
0607 International Mathematics November 2011
Principal Examiner Report for Teachers
INTERNATIONAL MATHEMATICS
Paper 0607/03
Paper 3 (Core)
Key Message
To attain high marks in this paper, candidates need to
●
●
●
have a good knowledge of the whole syllabus
show all the relevant working in their answers
use a graphics calculator in the way outlined in the syllabus
General comments
Many candidates were able to attempt all the questions on the paper, demonstrating good syllabus
coverage. There were some candidates who could improve in certain areas of the syllabus as will be seen in
comments on specific questions. The paper proved to be accessible to all but the weakest candidates. The
performances were well spread out and the paper succeeded in discriminating amongst all grades.
The majority of candidates showed good working and clearly realised that answers without working could be
awarded with fewer marks. There are certain aspects of the graphics calculator that can be improved upon.
These include statistics functions and obtaining a domain and a range from a sketch. The presentation of
work was usually clear and appropriate levels of accuracy were generally seen.
Overall the paper should have been a positive experience for most candidates.
Comments on specific questions
Question 1
Although this question contained considerable context, it proved to be a friendly first question with many
candidates scoring high marks. Candidates are advised to read all the information given in a question and
should also realise that the paper allows sufficient time for careful and thorough interpretation.
(a)
This straightforward multiplication of a book width by the number of books was well answered.
(b)
Another straightforward multiplication which was usually correctly done.
(c) (i)
The simplification of a ratio was almost always correct.
(ii)
This part required the use of a ratio to calculate an amount and also required the use of the
answers to parts (b) and (c)(i). The question was much more challenging and candidates need to
be aware that longer questions often involve information found in previous parts.
(iii)
This part, to find a total cost, also involved looking back at earlier answers and again several
candidates demonstrated the need for more practice and experience in multi-step questions.
Answers: (a) 112 (b) 210 (c) (i) 2 : 3 (ii) 84 (iii) 1638
9
© 2011
Cambridge International General Certificate of Secondary Education
0607 International Mathematics November 2011
Principal Examiner Report for Teachers
Question 2
This question perhaps gave the impression that the statistics facilities on a graphics calculator would be
useful. This was true for certain parts but not others. Almost all the candidates chose not to use these
functions which may have been the best strategy considering that the question contained much more than
statistics.
A re-arrangement of the data in the list was often carried out and this proved to be useful in several parts.
(a)
The mean was usually found correctly.
(b)
The mode was also usually correct.
(c)
The range was often correct. A number of candidates were less familiar with this result. Several
omitted the part and another answer was 800 to 1500.
(d)
Most candidates demonstrated the ability to find a percentage and many scored full marks. Some
candidates showed the need to be more careful in counting from a list.
(e)
Most candidates demonstrated the ability to find a probability and many gained the mark. Again,
some candidates showed the need to be more careful in counting from a list.
(f)
The median from a shortened list was usually correct.
Answers: (a) 1090 (b) 900 (c) 700 (d) 30 (e)
6
10
(f) 950
Question 3
The algebraic techniques required in this question were generally evident and candidates appeared to be
well prepared in most areas.
(a)
Most candidates demonstrated their ability to expand brackets and then simplify.
(b)
This factorisation part proved to be a little more difficult. Candidates are advised to look for more
than one common factor in an expression.
(c)
The linear equation was usually correctly solved.
(d)
This substitution part was also well done with many candidates coping well with a negative number.
Answers: (a) 8x + 6 (b) 3x(x – 3y) (c) 3.5 (d) 12
Question 4
There appears to be the need for candidates to be more familiar with the uses of the graphics calculator.
Some improvement in this area from that of previous sessions was observed, but the evidence points to the
need for more practice and experience.
(a)
An improvement in graph sketching was seen. There continues to be the challenge of setting up a
suitable window and typing in an equation correctly. Candidates should be aware that many marks
are likely to depend on a good sketch.
(b)
Candidates who had a correct sketch were usually able to find the minimum point.
(c)
Candidates need to know that certain graphs have lines of symmetry. They also need to be able to
state equations of lines parallel to either axis.
(d)
Finding the range of a function is certainly one of the more challenging topics in the syllabus.
Candidates need to be aware that earlier parts may well lead towards the range. In this case it was
the y-co-ordinate of the minimum point.
(e)
The co-ordinates of the zeros of the function were usually correctly found.
10
© 2011
Cambridge International General Certificate of Secondary Education
0607 International Mathematics November 2011
Principal Examiner Report for Teachers
(f)
The sketch of the straight line was usually correct.
(g)
Most candidates were able to find the points of intersection by using the intersection facility of the
calculator. Those who traced along the graph were unable to give answers to the required
accuracy. Candidates are advised to read through a question and not overlook an instruction. In
this case it was the instruction on accuracy of answers.
Answers: (b) (0, – 4) (c) x = 0 (d) y ≥ −4 (e) (– 2, 0), (2, 0) (g) (– 2.21, 0.89), (2.71, 3.36)
Question 5
This question was written in a way to make it reasonably clear that it was a compound interest problem. The
majority of candidates realised this. A few candidates tried to make simple interest fit, indicating a need for a
little more work on distinguishing between the two situations.
(a)
Most candidates were able to find the interest for one year, perhaps because this did not depend
on the type of interest. Many candidates correctly treated this part as a straightforward percentage
question.
(b)
Many candidates realised that, if the interest is not removed, then the situation is compound
interest. These candidates went on to show that the value at the end of two years was that stated
in the question.
(c) (i)
The candidates who were successful in part (b) were usually able to extend the method to four
years without much difficulty.
(ii)
Most candidates showed an understanding of how to find the interest from the amount.
Answers: (a) 150 (c)(i) 5627.54 (ii) 627.54
Question 6
(a)
This simple algebraic expression was correctly stated by most candidates.
(b)
This part involved writing an equation in terms of two variables and proved to be more challenging.
The candidates well experienced in algebra were successful. Others showed, often by not
attempting this part, a need for more practice in setting up this type of equation.
(c)
The comments for part (b) apply to this part, which was to find a second equation in the two
variables.
(d)
The candidates who were successful were able to continue and solve the simultaneous equations
from parts (b) and (c). It is important to be aware that questions involving algebraic equations can
involve context, and experience in problem solving is needed.
Answers: (a) 6x (b) 6x + 4y = 27 (c) 2x + 3y = 14 (d) 2.50, 3
Question 7
(a)
Most candidates were able to divide the distance by the time.
(b)
The stronger candidates were able to understand the meaning of “the distance north” and apply
trigonometry. Many candidates found the situation difficult to interpret and were perhaps unable to
apply some basic skill in trigonometry that they possessed.
(c)
The reverse bearing was often correctly answered. Bearings is one of the more difficult topics in
the syllabus and some candidates gave angles relative to the south line and others omitted this
part.
Answers: (a) 20 (b) 38.3 (c) 220°
11
© 2011
Cambridge International General Certificate of Secondary Education
0607 International Mathematics November 2011
Principal Examiner Report for Teachers
Question 8
(a)
This question involving parallel lines and a triangle was found to be quite straightforward. Many
candidates gave correct answers for both angles.
(b)
This question was equally successful, with candidates demonstrating knowledge of tangents to a
circle and the angle sum of a quadrilateral.
(c)
This question involved angles around a point and angles on a straight line and again was usually
well done.
(d)
The length of the chord of a circle, given the radius and the distance of the chord from the Centre,
was perhaps more challenging as it involved a reverse Pythagoras calculation and then doubling
the answer. Stronger candidates found further success and scored highly in Question 8. A
number of candidates treated the half of the chord as the hypotenuse.
Answers: (a) 140, 80 (b) 90, 150 (c)(i) 60 (ii) 120 (iii) 80 (d) 16
Question 9
(a)
The total of the frequencies from the given cumulative frequency curve was usually correctly
stated. A few candidates gave the value at the top of the grid, overlooking the fact that the highest
point on the graph was 10 units lower. Candidates who made this error were able to earn all the
remaining marks by following through their answers in parts (b) and (c).
(b)
The frequency for below a particular value was usually correctly found.
(c) (i)
Most candidates were able to find a percentage of the total frequency.
(ii)
This part required reading the cumulative frequency from the top and did cause a few more
problems but many candidates found the correct value on the horizontal axis
Answers: (a) 150 (b) 129 to 131 (c)(i) 15 (ii) 64 to 66
Question 10
(a)
Most candidates were able to name the quadrilateral correctly.
(b)
There were three possible answers to this part. Most candidates gave the expected answer of a
reflection in the x-axis.
Others gave the rotation or enlargement, although these two
transformations required more details.
(c)
The translation was usually recognised together with the corresponding column vector. Translation
was the only acceptable word and so words such as “shift” or “translocation” did not receive the
mark.
(d)
The 90° rotation was often correctly drawn. Some candidates used the correct centre but the
incorrect direction and others used the bottom vertex of the object quadrilateral as the centre.
(e)
The enlargement was often correctly drawn, usually with the correct centre. A few candidates had
only three of the four vertices correctly plotted. Such candidates should be aware that an
enlargement will produce an image the same shape as the object and a slight distortion should
indicate an incorrect point.
Answers:
reflection, x-axis or rotation, (4, 0) 180° or enlargement, factor, – 1, (4, 0)
⎛ − 12 ⎞
⎟⎟
(c) translation ⎜⎜
⎝ − 10 ⎠
(a)
kite
(b)
12
© 2011
Cambridge International General Certificate of Secondary Education
0607 International Mathematics November 2011
Principal Examiner Report for Teachers
Question 11
(a) (i)
Most candidates understood that the length of the rectangle is the diameter of the semi-circle and
gave a correct radius.
(ii)
Most candidates realised that the width of the rectangle was the difference between the height of
the tunnel and the radius of the semi-circle.
(b)
Almost all candidates successfully calculated the area of the rectangle.
(c)
Many candidates were successful in finding the area of the semi-circle. A few candidates gave the
area of the full circle and a few others used the diameter instead of the radius. The requirement of
an answer to 2 significant figures was occasionally overlooked.
(d)
This volume of a prism was more complicated than many similar questions as the length needed to
be changed from kilometres into metres. There were many correct answers and there were many
answers that used the area multiplied by the length but did not convert the units correctly.
(e) (i)
Many candidates correctly divided the distance by the speed and many of these candidates went
on to convert their answer in hours into minutes. A few candidates did not cope with the time
conversion and a few others divided the speed by the distance.
(ii)
This part required some clear reasoning and candidates needed to distinguish between the time of
entering the tunnel and the time leaving the tunnel. This was overlooked by quite a considerable
number of candidates.
Answers: (a)(i) 3 (ii) 4 (b) 24 (c) 14 (d) 1 330 000 (e)(i) 20 (ii) 32
Question 12
(a)
This simple probability was almost always correctly answered.
(b)
Many candidates understood the “without replacement” situation and gave a correct probability for
the second event.
(c)
The completion of the tree diagram is quite a challenging task at the core level. There were some
good answers but there were also candidates whose knowledge of probability appeared to be in
need of attention and extension.
(d)
This final part of the last question on the paper was very discriminating and it gave the stronger
candidates a chance to show considerable ability in probability, as a sum of two products was
required. Many candidates did not attempt this part and many could only offer a single fraction
which had little relation to the situation. A few gained a mark by giving one of the products
correctly.
Answers: (a)
8
12
(b)
7
11
(d)
64
132
13
© 2011
Cambridge International General Certificate of Secondary Education
0607 International Mathematics November 2011
Principal Examiner Report for Teachers
INTERNATIONAL MATHEMATICS
Paper 0607/04
Paper 4 (Extended)
Key Message
To succeed in this paper, candidates need to have completed full syllabus coverage, including a full
awareness of graphics calculator skills which can be tested in this paper. In the case of graphics calculators,
it is important to understand that there are expectations of good coverage in the examinations, including the
statistics facilities mentioned in the syllabus. It is also necessary to show clear methods and use appropriate
levels of accuracy both in the working and in final answers.
General comments
The majority of candidates demonstrated a good knowledge of the whole syllabus with most questions being
attempted. Certain topics such as areas of similar figures, functions, domain and range, probability and line
of regression indicated a need for more work and experience to be able to succeed fully in the examination.
More details are at the end of this section and in the comments on specific questions.
The use of graphics calculators continues to improve and many candidates gained high marks in this area.
Most candidates are now able to set up a correct window from values given in questions, which proves to be
extremely valuable when interpreting sketches. Challenges remain in finding ranges for given domains,
solving equations and inequalities and finding a line of regression. Quadratic equations continue to be
solved using the formula which, although perfectly acceptable, is probably less efficient for many candidates
than using a graphics calculator. A simple sketch with the two appropriate points indicated is considered to
be sufficient method. Candidates are advised to be hesitant in using calculator facilities, such as equation
solvers (without sketches), which are outside the syllabus and give little chance to show working.
Most candidates presented their work very clearly and used suitable accuracy throughout the paper.
Improvements are still possible in using accurate values held in a calculator rather than rounding off at
stages during working. The requirements for accuracy of answers is on the front cover and it is important
that candidates are aware that 2 significant figure answers will usually lose an accuracy mark each time.
Answers without working will often be penalised as it is the responsibility of the candidate to communicate by
showing all relevant working as clearly stated on the front cover of the examination paper.
The paper proved to be very accessible and the overall performances were successful. There appeared to
be adequate working space throughout the paper and all candidates had sufficient time to complete the
examination. Very few supplementary sheets were needed and Centres seem to appreciate that extra
sheets simply for working tend to disadvantage the candidate.
Questions involving basic knowledge and skills in all areas were well answered. Candidates can improve in
the more challenging parts of the syllabus such as areas of similar shapes, solving a reverse compound
interest problem, compound functions, compound areas and volumes, histograms, Pythagoras and
trigonometry in three-dimensional shapes, interpreting graphs, writing down the line of regression from the
graphics calculator, conversion of units, probability and problem solving using algebra. This list does not
suggest that these were difficult areas for all candidates but that some candidates found some of these
topics demanding and improvements can be made.
14
© 2011
Cambridge International General Certificate of Secondary Education
0607 International Mathematics November 2011
Principal Examiner Report for Teachers
Comments on specific questions
Question 1
(a) (i)
(b)
This straightforward time question was well answered.
(ii)
Most candidates correctly converted the 57 minutes into 0.95 hours. A few errors were seen in the
division by 60 or by writing 1 h 57 min as 1.57 h.
(iii)
Most candidates succeeded in dividing the distance by their time in part (ii).
This reverse percentage question was well answered with most candidates recognising that the
given time was equivalent to the given percentage. A few treated the given time as 100% and
calculated 26% of this time and then added it.
Answers: (a)(i) 12 22 (ii) 1.95 (iii) 574 (b) 7 h 30 min
Question 2
(a)
The correct angle from angles in the same segment was usually stated. A few candidates used
alternate angles when there were no parallel lines in the diagram.
Candidates need to consider diagrams and written information carefully and to only use given data.
(b)
The question did not indicate that the two triangles were similar but the requirement to look at
angles in part (a) should have been a strong enough hint. The outcome was rather mixed with
many candidates using correct ratios and many using ratios of non-corresponding sides. The
actual calculation of the ratio was usually correctly completed, by cross multiplication in most
cases.
(c)
Areas of similar figures is a challenging topic and it proved to be no exception in this examination.
The need to square the ratio of the sides was recognised by the stronger candidates whilst many
equated the ratio of the areas to the ratio of the sides.
Some candidates, not recognising the use of similar figures, calculated perpendicular heights and
then used half base × height for the required area of the triangle. This longer method was often
successful, although long methods do tend to lose accuracy by the time an answer is reached.
Answers: (a) CBX (b) 10.5 (c) 10.7
Question 3
(a)
Most candidates successfully found the compound interest, either by using the formula for
5
compound interest or simply multiplying the $480 by the factor 1.026 . Some candidates gave a
correct final amount, overlooking the requirement to find the interest.
(b)
To find a number of years for an investment to reach a certain value is quite challenging, although
many candidates produced good methods and reached the correct answer. The two expected
methods were the use of logarithms or the use of a graphics calculator to solve (graphically) the
equation 480 × 1.026 x = 800 . In the latter a simple sketch was sufficient for the method marks.
Some candidates repeated the multiplication by 1.026 until reaching 800. This clearly took time
and often led to slips in the calculation or too much rounding in the working, leading to an incorrect
answer.
Answers: (a) 65.73 (b) 20
15
© 2011
Cambridge International General Certificate of Secondary Education
0607 International Mathematics November 2011
Principal Examiner Report for Teachers
Question 4
(a)
The sine rule was almost always correctly applied with good clear working and a final answer
correct to 3 significant figures.
(b)
The calculation of an angle using the cosine rule was also generally well answered. Many
candidates quoted the formula for the angle and many others re-arranged the formula given in the
question paper.
Answers: (a) 8.95 (b) 80.3
Question 5
(a)
Most candidates used the formula to solve this quadratic equation, usually correctly with the
answers correct to 2 decimal places as required.
A few candidates used their calculator, drew a simple sketch and indicated two intersections
(usually with the x-axis), and simply wrote down the answers, gaining full marks. Candidates
should realise that this method is acceptable and may be quicker and less prone to error.
(b)
2
This question requiring 2x – 3 to replace x in the function f(x) = 2x – 3x – 3 was more challenging.
There were many good attempts and many correct answers. The squaring of 2x – 3 and then
multiplying it by 2 was found to be difficult with working such as 8x2 + 18 and 8x2 – 12x + (or –) 9
frequently seen.
Answers: (a) – 0.69, 2.19 (b) 30
Question 6
(a)
The area of the cross-section of the tunnel involved the calculation of the area of a sector with a
reflex angle together with the area of a triangle using two sides and the included angle. This was
often correctly carried out with clear working and two accurate values added to give a total area to
the required 3 significant figure accuracy.
(b)
To obtain the volume, the answer to part (a) needed to be multiplied by 2400 as the area was in
square metres and the length of the tunnel was in kilometres. Many candidates carried out this
simple calculation correctly. Many other candidates overlooked the km with the 2.4 and simply
multiplied by 2.4. Candidates should consider the context of a question and be aware of the use of
appropriate and sensible units.
(c)
This part also involved a conversion of units, from kilograms into tonnes. Since tonnes was printed
in bold, candidates appeared to be more practical in this part and multiplied by 1530 and divided by
1000. Almost all candidates multiplied their answer to part (b) either correctly or by 1530.
Answers: (a) 61.0 (b) 146 000 (c) 220 000
Question 7
(a)
Many candidates were able to calculate the frequencies from the given histogram. There were
some misunderstandings seen in this discriminating topic, often the multiplication by each
frequency density by 70, which was the width of the first column of the histogram.
(b)
Almost all candidates were able to calculate the mean from their frequency table, although many
did not use the facility available on the graphics calculator. As there were only three columns in the
table this was not too time consuming on this occasion.
Answers: (a) 150, 100 (b) 70.9
16
© 2011
Cambridge International General Certificate of Secondary Education
0607 International Mathematics November 2011
Principal Examiner Report for Teachers
Question 8
(a) (i)
Most candidates produced a very good sketch of the given quartic equation.
(ii)
The co-ordinates of the points where the graph meets the axes were usually correctly stated
indicating good use of the graphics calculator.
(iii)
The equation of the line of symmetry was also usually correct.
(iv)
The co-ordinates of the two minimum points were found to be a little more challenging. Many
candidates correctly used the minimum point facility on their calculator. Some candidates traced
along their graphs and usually gave answers well outside an acceptable accuracy.
(v)
This question asking for the range of the function for all real numbers was only successfully
answered by a few candidates who were able to connect the y-co-ordinates of their minimum
points together with their sketch. Many candidates omitted this part, indicating the need for more
work and experience on domain and range.
(b) (i)
(ii)
Most candidates produced a very good sketch of the exponential function.
The solution of the function in part (i) equal to zero was usually correctly stated.
(c) (i)
Many candidates were able to successfully apply the intersection facility on their graphics
calculator. A number of candidates again traced and again were well outside the required
accuracy.
(ii)
Clearly this inequality depended on the answers to part (i) and many good answers were seen.
There were many candidates who omitted this part, some from not having an answer to part (i) and
some from not being able to interpret fully from the answer to part (i) together with the sketch.
Answers: (a)(ii) (– 1, 0), (0, 0), (1, 0) (iii) x = 0 (iv) (– 0.7071, – 0.25), ( 0.7071, – 0.25) (v) f(x) ≥ – 0.25
(b)(ii) 0.6781 (c)(i) 0.4988, 1.221 (ii) 0.4988 < x < 1.221
Question 9
(a)
Many candidates successfully calculated the total surface area of the cuboid. A few found the
volume and a few omitted one or two of the six rectangles.
(b)
The calculation of a named angle was carried out correctly by many candidates using the tangent
ratio. A few candidates found the three-dimensional aspect of the problem rather challenging
indicating a need for more practice and wider applications of trigonometry.
(c)
The use of Pythagoras twice to calculate the diagonal of the cuboid was frequently correctly done.
As in part (b) a few candidates found the three-dimensional situation difficult and found the length
of the diagonal of one of the faces of the cuboid.
Answers: (a) 548 (b) 35.0 (c) 17.1
Question 10
(a) (i)
(ii)
(b)
The use of parallel lines was almost always correctly applied to find the required angle.
The best way to find this angle was to use angles in a pentagon and many candidates
demonstrated their knowledge that the angle sum is 540˚. Another method seen was the drawing
of a line through D parallel to AE together with finding angle BDC. A few candidates perhaps
lacked some basic knowledge of angle properties and working was confused or an incorrect
answer was given without any working.
The line BD was usually correctly drawn and the resulting isosceles triangle BCD usually led to a
correct angle ABD. Many candidates put values on the diagram and this is a good practice making
the questions much more visual.
17
© 2011
Cambridge International General Certificate of Secondary Education
0607 International Mathematics November 2011
Principal Examiner Report for Teachers
(c) (i)
The extending of two lines and the naming of the resulting shape was usually correctly done.
(ii)
The many candidates who had a parallelogram in part (i) were easily able to use the opposite
angles property to give the required angle.
(d) (i)
The extension of two more lines leading to the calculation of an angle in a triangle was also usually
correctly done.
(ii)
This discriminating part to state a reason why a circle could not go through 4 given points was very
challenging. The stronger candidates tried to use a circle property, such as two named angles in
the same segment were not equal and earned the mark. Many candidates simply re-stated the
question.
Answers: (a)(i) 96 (ii) 154 (b) 61 (c)(i) parallelogram (ii) 84 (d)(i) 26
Question 11
(a)
Most candidates drew good sketches of the 4 parabolas. A few omitted the required labels.
(b) (i)
The translation and its vector were usually correct.
(ii)
The stretch was usually stated. The invariant line and factor were often correct although a number
of candidates could state stretch but were not able to give these details.
(iii)
Most candidates answered this correctly by giving the reflection in the x-axis. Two other answers
(one a rotation and the other an enlargement) were accepted by allowing candidates to simply look
at the object parabola and the image parabola without considering how points were being mapped.
⎛ − 2⎞
⎟⎟ (ii) stretch x-axis invariant factor 2 (iii) reflection x-axis
Answers: (b)(i) translation ⎜⎜
⎝ 0 ⎠
Question 12
(a)
The tree diagram was usually correctly drawn with correct pairs of probabilities against the
branches. A few candidates appeared to need to gain more knowledge of this type of probability
question and this comment applies to the following parts.
(b) (i)
The product of one pair of probabilities was usually correctly completed by the candidates who had
a correct tree.
(ii)
This probability was the sum of the answer to part (i) and another product. Again, the stronger
candidates succeeded with this challenging question.
(c)
Many candidates recognised the need to multiply the probability from part (b)(ii) by the 15 days to
find an expected number of days.
Answers: (b)(i) 0.765 (ii) 0.81 (c) 12
Question 13
(a)
The equation of the line parallel to the x-axis was almost always correctly stated.
(b)
The equation of the line, given the intercepts with the axes was usually correctly found. Many
candidates knew the answer immediately whilst others found the gradient and used y = mx + c.
(c)
The use of y = mx + c was more relevant in this part and many candidates were able to simply
write the equation down as they had identified m and c.
(d)
The intersection of the lines in parts (b) and (c) was often correctly found, using the simple
simultaneous equations. A number of candidates gave answers to only 2 significant figures.
18
© 2011
Cambridge International General Certificate of Secondary Education
0607 International Mathematics November 2011
Principal Examiner Report for Teachers
(e)
The three inequalities were usually correct if parts (a), (b) and (c) were correct. There were some
errors between ≤ and ≥ .
Answers: (a) y = 3 (b) x + y = 4 (c) y = 2x – 4 (d) ( 2 32 , 1 31 ) (e) y ≤ 3, x + y ≥ 4, y ≤ 2 x − 4
Question 14
(a)
The scatter diagram was almost always correctly completed by plotting the four remaining points.
(b)
The type of correlation was also almost always correctly indicated.
(c) (i)
The mean of the 10 values was usually correct. At this stage, only a few candidates appeared to
use the 2 variables statistics facility on their graphics calculator.
(ii)
The equation of the line of regression was often correctly stated, occasionally with only 2 significant
figure accuracy. Many candidates found this question to be outside their experience in statistics
and did not attempt it.
(iii)
The line of regression was drawn by most candidates. The rate of success was not very high as
most lines were inaccurate. The scatter diagram was accurate and so a sketch was not expected.
Candidates need to know that an accurate line of regression must go through the mean point, as
indicated in the syllabus. Part (c)(i) should have been a strong hint that this was needed.
(iv)
The reading from the line of regression was often correct if candidates had a reasonably accurate
scatter diagram or if candidates used their equation in part (c)(ii).
Answers: (b) positive (c)(i) 13.2 (ii) 0.879x + 1.07 (iv) 17
Question 15
(a) (i)
(ii)
(b)
The expression in terms of n was usually correct.
This second expression in terms of n was also usually correct.
This part was a real test of the use of algebra to solve a problem. The answers to part (a) were to
be put together to form an equation based on their difference being 4.
The question was more challenging since the eventual quadratic equation was not given as a midway stage.
It was hoped that candidates would then think of using their graphics calculator with their original
equation thus removing the need for manipulations with algebraic fractions. This method was
rarely seen and it is hoped Centres will use this question to demonstrate the use of the graphics
calculator as well as the traditional algebraic method.
The candidates with sound algebra were able to re-arrange the equation into the correct quadratic
equation and then go on to solve it by factorising or by using the formula.
Many candidates made errors in the algebra and many did not attempt the question.
A few candidates overlooked the requirement of an equation and used a trial and improvement
approach. In some cases the correct answer was found but full marks were not awarded.
Answers: (a)(i)
360
360
(ii)
(b) 15
n+3
n
19
© 2011
Cambridge International General Certificate of Secondary Education
0607 International Mathematics November 2011
Principal Examiner Report for Teachers
INTERNATIONAL MATHEMATICS
Paper 0607/05
Paper 5 (Core)
Key message
In order to succeed in this paper, candidates need to ensure that they provided reasons and explanations in
their answers, as marks are awarded for communication skills.
General comments
This paper, as its title suggests, was based on an investigation into finding the maximum perimeter that could
be made by joining identical shapes. Examples and explanations were given throughout the paper and all
candidates showed at least some evidence of knowledge of looking for patterns.
Most candidates tackled the whole of Question 1 and the majority answered Question 2, with a good
number also trying Question 3. The numerical sequence questions were well answered and the drawing
answers were, on the whole, answered better than last year. Many candidates still found it difficult to move
from numeric to algebraic notation and few gave any sign that they tested their algebraic answers. Of those
candidates who attempted the algebraic parts of the questions many wrote equations rather than the
expressions that were specifically asked for.
Those candidates who stumbled over the last parts of Question 1 still made a good attempt at Question 2
and at least tried Question 3.
Comments on specific questions
Question 1
This question involved drawing and calculations as well as leading onto an algebraic expression.
(a) and (b)
(c)
Parts (a) and (b) were well answered with the most common error in (a) being the drawing of
a rotation of the given figure and in (b) the drawing of an enlargement.
Part (c) was well answered with the majority of candidates using the given grid to work out, test or
show examples of their answers. This drawing is a good form of communication, which is
assessed on this paper.
Answers:
(i)
(d)
10 (ii)
12 (iii)
14
Some candidates who had made errors on the previous parts of this question recovered in part (d)
by noticing the pattern in the table, and several candidates changed their answers to the missing
perimeters in the table and went on to answer parts (ii) and (iii) successfully.
Answers:
(i)
Number of
squares
Greatest
perimeter (cm)
(ii)
36 (iii)
2
3
4
5
6
7
8
9
10
6
8
10
12
14
16
18
20
22
15
20
© 2011
Cambridge International General Certificate of Secondary Education
0607 International Mathematics November 2011
Principal Examiner Report for Teachers
(e)
Following on from part (d) even those candidates who initially had not been able to complete the
table correctly were now able to complete part (e) correctly.
Answers:
(i)
(f)
2
(ii)
2, 2
Candidates need to understand what is meant by an ‘expression’ and they should be able to
identify the difference between an equation and an expression. Although most had realised, from
the table in part (d)(i) that +2 featured here, it was quite common to see an equation especially
those beginning x =.
Answer: 2x + 2
Question 2
The answers to this question followed the same route as those for Question 1, scoring well for the first parts
and less well when algebra became involved.
(a)
Fewer candidates used the grid to show how they were finding the answer to this part than they did
in Question (1)(c). Candidates should be encouraged to take every opportunity to communicate
their thinking to the Examiner; this includes drawing. Most candidates calculated the answer
successfully.
Answer: 8
(b)
This question was well answered apart from a few candidates who had not seen any earlier
patterns and stopped at this stage. Many candidates were now looking for a pattern in the table
and this time only a few had to change their first answers so that they would fit with the given
values.
Answers:
(i)
Number of
equilateral triangles
Greatest perimeter
(cm)
(ii)
(c)
12 (iii)
2
3
4
5
6
7
8
4
5
6
7
8
9
10
16
Those who managed to find a correct expression for Question 1(f) had no problem in extending
their thinking to answer this part correctly. Similarly, those who had written equations before
continued to write them now and those who had not managed to answer Question 1(f) did not
attempt this part either.
Answer: x + 2
Question 3
Some candidates drew groups of hexagons and using their knowledge of sequences managed to find the
correct expression. Many wrote something down following on from their incorrect answers (mostly
equations) in Questions 1(f) and 2(c). Those who had not managed the earlier algebraic expressions did
not attempt this question. Candidates should be encouraged to attempt all questions as a demonstration of
communication at the very least.
Answer:
4x + 2
21
© 2011
Cambridge International General Certificate of Secondary Education
0607 International Mathematics November 2011
Principal Examiner Report for Teachers
INTERNATIONAL MATHEMATICS
Paper 0607/06
Paper 6 (Extended)
Key message
In order to succeed in this paper, candidates need to ensure that they provided reasons and explanations in
their answers, as marks are awarded for communication skills.
General comments
This paper presented candidates with an investigation and a modelling task, carrying equal marks.
Candidates were advised to spend an equal amount of time on each of the two tasks.
Working out and answer spaces were provided on the paper giving a clear structure for the candidates to
follow.
The overall performance of candidates was very good. The investigation was completed very well by most
candidates although the modelling section presented more of a challenge.
Comments on specific questions
Part A Investigation
This part was based on an investigation into finding the maximum perimeter that could be made by joining
identical shapes. It was necessary to be able to draw examples as well as to recognise and extend patterns
in sequences and to state connections algebraically. Examples and explanations were given throughout the
paper and all candidates showed through their answers that they had been well prepared for this type of
task.
Question 1
This question involved drawing and calculations as well as leading onto an algebraic expression.
(a) and (b) Parts (a) and (b) were well answered with errors seen on very few scripts. A misconception
made by a few candidates was that for a larger perimeter the different shape needed to be an
enlargement. This was despite the introduction explaining that the triangles had sides of 1 cm.
Candidates had a large choice of shapes for part (b) but mainly chose the row combination.
(c)
Most candidates noticed the pattern for the greatest perimeter in the table and were able to use this
pattern or sequence to calculate the correct answers for parts (ii) and (iii). The majority used the
grid, as suggested on the paper, to work out or to confirm their answers. Candidates have
obviously received good guidance in looking for sequences and communicating their work to the
Examiner.
Answers:
(i)
Number of
equilateral triangles
Greatest perimeter
(cm)
(ii)
22
(iii)
2
3
4
5
6
7
8
4
5
6
7
8
9
10
30
22
© 2011
Cambridge International General Certificate of Secondary Education
0607 International Mathematics November 2011
Principal Examiner Report for Teachers
(d)
The move into algebra was tackled smoothly. Most candidates knew the meaning of an expression
and answered this question correctly. Some candidates muddled expression with equation and
lost marks for writing x = x + 2.
Answer: x + 2
Question 2
(a)
Most candidates answered this correctly and many used the grid to work out or to test their answer.
Candidates should be encouraged to use every opportunity to communicate their working out to the
Examiner.
Answer: 14
(b)
Candidates have been well prepared in looking for patterns in sequences of numbers and those
who initially completed incorrect values in this table were able to use this skill to change their
answers to the correct ones. By using this correct sequence from the table most candidates were
able to answer parts (ii) and (iii) correctly.
Answers:
(i)
Number of
squares
Greatest
perimeter (cm)
(ii)
(c)
36
(iii)
2
3
4
5
6
7
8
9
10
6
8
10
12
14
16
18
20
22
15
Here candidates were not required to simplify their answer so any expression that simplified to the
correct one was acceptable. Candidates should be encouraged to write an algebraic expression
from the sequence to obtain a simplified version. For example, 2(x + 2) – 2 was acceptable as an
answer to this question but it was obtained by using a longer method rather than the most efficient.
Answer:
2x + 2
Question 3
(a)
Most candidates drew sketches of hexagons to help them complete this table. Those who did were
able to spot their mistake if they made one and correct it to fit the new pattern for hexagons. This
sequence proved more difficult for those candidates who relied on working without drawings.
Candidates should be encouraged to sketch/draw wherever necessary to help them in an
investigation.
Answer:
(b)
Answer:
Number of regular
hexagons
2
3
4
5
6
Greatest
perimeter (cm)
10
14
18
22
26
Most candidates who had the correct values for the sequence in part (a) were able to write down a
correct algebraic expression. Many answers could have been simplified indicating that the
candidates used less efficient methods than finding the nth term to obtain the expression. Those
candidates who had written equations rather than expressions earlier in the paper continued to do
so in this question and the following one.
4x + 2
23
© 2011
Cambridge International General Certificate of Secondary Education
0607 International Mathematics November 2011
Principal Examiner Report for Teachers
Question 4
Many candidates saw a pattern linking their answers to Questions 1(d), 2(c) and 3(b). They then correctly
followed this pattern through for the octagons. This includes most of those candidates who were working
with unsimplified versions for their algebraic answers.
Answer:
6x + 2
Question 5
(a)
Answer:
(b)
Candidates had obviously been well prepared in looking for connections in sequences and made
excellent attempts to follow through their previous algebraic answers, many with great success.
x(y – 2) + 2
Candidates had also been well prepared in testing and trialling values in such situations as this
question. Communication was excellent and many candidates who had found part (a) very testing,
persevered with this part with great success.
Answer:
Three pairs from the following:
x
24
8
4
2
y
3
5
8
14
x
12
6
3
1
y
4
6
10
26
Part B Modelling
The modelling task required the candidates to know and use the formulae for volume and surface area within
the scenario of covering different shaped cakes with chocolate. Explanations and leads were given on the
question paper and knowing how to change the subject of a formula was an important skill required to
answer this paper.
The graph sketching questions were successfully tackled by the majority of candidates. In later questions
some misread and assumed the volume of the chocolate covering to be the volume of the cake.
Question 1
(a)
This question asked the candidates to show that a connection between the length, width and height
of a square cake was true. Candidates should be encouraged to start such ‘proofs’ with basic
formulae such as ‘volume = length × width × height’ and then to replace the general terms with
those specific to the question, e.g. v = x × x × y. In this situation most candidates started with
2
x y = 4000 and thus lost marks for making assumptions that were not shown.
(b)
This question asked the candidates to find an expression for the surface area and use it and the
equation given in part (a) to eliminate y. Candidates found this question more accessible than part
(a) and succeeded by starting with the required surface area and performing the substitution to
arrive at the answer given. Step by step working out was shown by many candidates who were
well prepared in communicating their thinking steps to the Examiner.
(c)
Those candidates who put the correct range as given into their calculators drew very good
sketches. The general shape of most sketches was good and few graphs cut or touched the x-axis
since x ≥ 2 was given. Candidates do need to be encouraged to be as accurate as possible when
copying a graph from their calculator and should be shown not to make assumptions about the
graph outside the given range.
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© 2011
Cambridge International General Certificate of Secondary Education
0607 International Mathematics November 2011
Principal Examiner Report for Teachers
(d)
Answer:
Candidates need to improve on their use of their graphical calculator to find such information as the
co-ordinates of minimum points. This was an easy question for all those who knew how to find the
minimum S using the trace on their calculator. Others produced a great deal of working and were
either unsuccessful or muddled the x with the y values.
minimum surface area = 1200
x = 20
y =10
Question 2
(a)
Again, candidates can improve on their ‘proof’ and ‘show that’ skills. They need to go back to basic
formulae, in this case the volume and surface area (top and sides) of a cylinder and remember to
show every line of working. It is a good idea to assume that the Examiner has no algebraic
knowledge and that every step must be given.
(b)
As shown in Question 1 the candidates graph sketching skills were excellent. Good use of
graphical calculators meant that most candidates achieved good marks for this part of Question 2.
(c)
Following on from Question 1(d) many candidates need to learn how to make better use of some
of the functions on their graphical calculator. Some candidates managed to get close enough to
the rounded correct answers to score some marks. Most of these candidates did a great deal of
work which could be avoided by using the trace on the graph on their calculator.
Answer: minimum surface area = 1110
x = 11
y = 11
Question 3
(a) and (b) Some candidates explained in great detail how to find the volume of the chocolate by going back
to the basic formulae for area and volume instead of using the models as instructed. This only cost
them in time for the last question because their answers were still valid. Other candidates
confused the volume of the chocolate covering with the volume of the cake which was already
known. It is important that candidates recognise what assumptions are made when modelling and
that these must be taken into account when looking at the validity of the model.
Answers:
(a)
Multiply by the thickness of the chocolate
(b)
Missing element of volume in the models
Assumes uniformity of cake
Assumes consistency of chocolate
Question 4
Many candidates made an excellent attempt at this less guided question. Those who read the question
carefully and noticed that it was using the ‘minimum surface area’ were very successful. Most of these
candidates were logical in their working, splitting their work into 2 sections, one for each cake shape, and
writing their working out methodically. The best candidates drew their work together with a concluding
comment which gave a natural completion to their modelling task.
Answer:
Square based: top = 400 cm2: sides = 800 cm2
Circular based: top = 369 cm2: sides = 738 cm2
Yes, both in ratio
top : sides = 1 : 2
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