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Cambridge International General Certificate of Secondary Education
0444 Mathematics (US) June 2012
Principal Examiner Report for Teachers
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MATHEMATICS (US)
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Paper 0444/11
Paper 1
General comments
This paper covered a wide range of topics and produced a range of marks. Many candidates made a good
attempt at answering this paper and there were only a small number of very low marks. The total marks for a
good number of able candidates were in the 40’s and 50’s. However, there were no candidates who got full
marks.
This paper was fairly straight forward with many part questions worth only 1 mark. With those parts that
carried more marks, workings needed to be shown in order to access the method marks if the final answer
was incorrect. Along with this entreaty for workings to be shown, candidates must check their work for sense
and accuracy. Candidates must take notice that, if a question asks for the answer to a fraction calculation as
a mixed number then a vulgar fraction is not correct. Similarly, the answer to Question 22 was required to
be in terms of π. Candidates should be equipped with compasses for construction questions.
The Questions that presented least difficulty were 1, 2, 5, 8(b), 9, 20 and 21. The Questions that proved to
be the most difficult were 3, 8(a), 10, 18(b), 18(c) and 19(a)
Very few part questions were left blank so time does not appear to have been an issue over the whole paper.
Comments on specific questions
Question 1
The directed numbers caused some problems for a few with 6 being a common wrong answer from 18 – 12
but many candidates were correct.
Answer: 30
Question 2
The vast majority of the candidates were correct in part (a). Poor spelling was accepted if the meaning was
clear. The number of lines of symmetry was given as 1 or 2 with nearly ¾ of candidates giving the correct
answer of 3 lines. Some candidates matched their answer to part (a) of isosceles with 1 line of symmetry
but this was not given credit as the question stated the triangle had 3 equal sides.
Answers: (a)
Equilateral
(b)
3
Question 3
This time calculation was poorly done with some candidates using 100 minutes in an hour. Others simply
subtracted the figures in the question, 1827 – 319, giving 1508 as their answer. Another wrong method was
to say 24 – 18.37 + 3.19 = 8.92. Many who used the correct process to solve this question ended up with 9
hours and various minutes rather than 8 hours and 52 minutes. Sometimes candidates split the time period
into many sections rather than the time before midnight and the time after midnight. A step on from finding
the number of hours was the conversion to minutes. Sometimes candidates just wrote 852; whether this was
meant to be the number of hours and minutes or another use of 100 minutes to an hour was not clear.
Candidates need to better understand and use the fact that there are 60 minutes in an hour and 24 hours in
one day.
Answer: 532
1
© 2012
Cambridge International General Certificate of Secondary Education
0444 Mathematics (US) June 2012
Principal Examiner Report for Teachers
Question 4
Most candidates understood what was required of them but there was some confusion over the placement of
the decimal point leading to an incorrect number of zeros. After finding the increase in number of visitors, a
few candidates went on to give the total number of visitors in 2010 which was not asked for. In questions like
this, candidates should read the question carefully to make sure they understand exactly what is asked of
them. Underlining instructions of this sort may be beneficial.
Answer: 12 400
Question 5
This is another question about directed numbers, this time coupled with algebra, and was well answered.
Candidates who did not get the question correct gained credit for substitution into the expression or for
finding one of the two terms. Common wrong answers were –21 or 6 – 15 = –9.
Answer: 21
Question 6
This was reasonably well done but with wrong answers given such as ($)8.35 or b + n = 8.35. Some did not
change the bracelet cost from cents and gave 7.5n + 85b as their answer. A small number swapped the
costs over giving 7.5b + 0.85n. This is a simple example of where checking would have got a candidate 2
more marks.
Answer: 7.5n + 0.85b
Question 7
This question was not done well and was had one of the highest number of blank responses. It seems that
the term perpendicular bisector was not well understood as the wrong answers were not simply slightly
inaccurate bisectors but rather circles, lines parallel to or at 90° to AB in the wrong place. Compasses were
rarely used and arcs were added afterwards.
Answer: Ruled perpendicular bisector drawn with two correct pairs of arcs seen
Question 8
The correct answer, ‘Rhombus’ was generally not known as only about 25% of candidates got this correct.
In common with Question 2, candidates limited themselves to names of shapes with four sides with some
simply reiterating the word quadrilateral from the question. Some 3D shape names were seen. Far more
candidates got part (b) correct than part (a).
Answers: (a) Rhombus
(b)
131
Question 9
This algebra was generally well done but when candidates got to the answer of 2.25 they then proceeded to
round this to 2 or 2.3. It is not correct to round this type of answer that terminates.
Answer: 2.25 or equivalent
Question 10
As stated in the general comments, this was not well done. Candidates ignored the instruction to estimate
how many dollars John received and so did not gain any marks. The exchange rate should have been
rounded and then 600 / 0.6 lead to $1000. Many candidates used long division instead.
Answer: 1000
2
© 2012
Cambridge International General Certificate of Secondary Education
0444 Mathematics (US) June 2012
Principal Examiner Report for Teachers
Question 11
Many candidates got this correct or at least partially correct but others appeared not to be sure how to
proceed. This type of simplification needs to be tackled in stages with like terms being combined. The
numbers simplify to
2
, the index for a is 2 and the b2 term is left unchanged.
3
Answers:
2a
2
3b
2
2 −2
or
2a b
3
Question 12
This was not done particularly well. Many candidates did use the diagram to help them solve this question
but often signs were wrong.
Answers: (x =) –4 (y =) 9
Question 13
The common errors were to miss out a factor, most often it was 1 or 15, or to show 1 x 15 and 3 x 5. This
not a list of factors so it did not get any marks. Sometimes a list of multiples was given but this was not
common. Part (b) was not so well done as part (a) as the difficulty had been stepped up. About half of all
2
candidates got this correct but some tried to combine the correct answer into one term giving 39p t (from 3 x
2
[5 + 8]) or using the question to get 249pt (from 15 + 24) as their answer. Very few candidates only got as
few as taking out one factor.
Answers: (a)
1, 3, 5, 15
(b)
3p(5p + 8t)
Question 14
This question combined construction with applying a scale. Most candidates dealt well with the scale part of
the question but were reluctant to use compasses to draw the triangle. Some candidates reversed the
triangle so AC was 25 m instead of 35 m. Point C seemed to be found by trial and error using a ruler and
protractor with a few adding badly drawn arcs to try to add authenticity. Constructions such as this one, and
that in Question 7, must be drawn showing all construction lines and arcs.
Answer: Triangle drawn correctly with ruler and arcs
Question 15
This was done well, but sometimes the instruction to give the answer as a mixed number in its simplest form
was ignored or forgotten. Most candidates worked in sixtieths rather than thirtieths but either was
acceptable. It did not matter whether candidates included the 1 from the start or added it in at the end when
they converted to a mixed number as long as all steps of the method were given.
11
Answer: 2 15
Question 16
Parts (a) and (c) were done well. Here, the wrong answers of 39° or 129° were occasionally seen. In part
(b) candidates focused on the angle of 63° in the triangle when all that was needed was the fact the angle in
a semicircle is 90° regardless of the other two angles. Wrong answers included 117° and 63°. Quite a few
candidates left this part blank. With part (c), a few gave 132° (the sum of the base angles) or 84° (from
thinking that the equal angles were 48°)
Answers: (a)
51°
(b)
90°
(c)
66°
3
© 2012
Cambridge International General Certificate of Secondary Education
0444 Mathematics (US) June 2012
Principal Examiner Report for Teachers
Question 17
There were many correct responses to this question with about half of all candidate getting full marks
showing their confidence with the topic and methods of solving systems of linear equations. Many
candidates used the substitution method which can lead to sign errors which were penalised. However,
some produced a value for x or y that was not a simple number but still went on to the end of their method.
The answers to a system of linear equations will always be simple values (both positive and negative) and
candidates should realise this and recognise if their values are of a complex kind then they must have made
errors. This is one situation where candidates can check their work (by the method of substituting their
values into both equations) and then try to go through their working again.
Answers: (x =) –2 (y =) 3
Question 18
Part (a) was one of the best answered questions on the paper but occasionally reversed co-ordinates were
seen. Parts (b) and (c) were almost the worst answered questions on the paper. Part (b) had many sign
errors and reversed components. Some tried to give a 2x2 matrix made up of the co-ordinates for A and B.
Errors in understanding column vectors fed into part (c) but more got this correct than did part (b).
Answers: (a) (–1, 2)
⎛ 4 ⎞
⎜⎜ ⎟⎟
⎝ − 5⎠
(b)
(c) (1, 5)
Question 19
Part (a) was quite poorly done with many writing answers such as 33 or 326.41 (mixing up significant figures
with decimal places). Answers such as 330.000 are wrong – the extra trailing zeros are saying there are 6
significant figures in the number. Zeros are needed as place markers before the decimal point but not
afterwards. Part (b) was better done than part (a). With part (b) the problem was to decide how many
zeros in one million and to apply the fact that 10 x 10 = 100 so 100 x 100 = 10,000 and so on. One mark
was available for writing one million in figures.
Answers: (a)
330
(b)
1000
Question 20
Most were correct with this simplification, although some candidates reduced it further to a single term.
Quite a number of candidates gave the simplification of the terms involving q, as 10q or + − 4q. Part (b) was
y −g
y +g
generally well done but errors in the signs led to 2 or 2 .
Answers: (a)
(b)
9p – 4q
g −y
2
Question 21
For part (a) a common wrong answer was 16 (from 29 – 13). Sometimes the mean 19.2 or 19 were given
here or in part (b). Candidates were more successful with part (b) showing evidence of ordering but
sometimes this was careless. Difficulties finding the midpoint lead to 18 or 19 as answers. Only using one
of each value led to a median of 20 being chosen. Part (c) was a change of syllabus area to find a number
that is not a prime. 29 and 31 were the most common wrong answers.
Answers: (a)(i)
18
(ii)
17
(b)
21
4
© 2012
Cambridge International General Certificate of Secondary Education
0444 Mathematics (US) June 2012
Principal Examiner Report for Teachers
Question 22
The formula for the volume of a sphere is given on the formula page and only rarely was the wrong one was
chosen. Sometimes the fraction of 4/3 was inverted during the calculation. Only a few candidates failed to
find the correct radius before substituting into the formula. After substitution, the 30 must be cubed or some
cancelling done. If multiplying, candidates worked with the entire 30 and sometimes got confused with the
number of zeros rather than cubing the 3 and 10 separately. The answer was required to be given in terms
of π, but some went on to unnecessary multiply 36000 by 3.14 or 3.142. Those that gave 36000 alone as
their answer did gain some marks as their only error was to omit π.
Answer: 36 000π
5
© 2012
Cambridge International General Certificate of Secondary Education
0444 Mathematics (US) June 2012
Principal Examiner Report for Teachers
MATHEMATICS (US)
Paper 0444/21
Key Message
To succeed in this paper candidates need to have completed full syllabus coverage, 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 are showing evidence of good work in indices, working with the mean, adding fractions, and
solving equations. Candidates particularly struggled with geometrical drawing, vectors, trigonometry and arc
length. Not showing clear working and in some cases any working was occasionally a problem. Giving
answers to an incorrect degree of accuracy or an incorrect form was an issue with some candidates.
Comments on specific questions
Question 1
The majority of candidates correctly multiplied 300 by 0.619 with few arithmetic errors with most candidates
obtaining at least the method mark, a small minority divided. Many candidates ignored or missed the
instruction to give their answer to the nearest pound therefore 185.7 was a common incorrect answer.
Answer: 186
Question 2
Nearly all candidates correctly answered part (a) with only occasional incorrect answers which were usually
5
7
32 or 128 arising from 2 and 2 respectively. Whilst the majority of candidates also scored in part (b) they
1
1
found this part a little more challenging. Common incorrect answers were , -27 and −
.
9
27
Answers: (a) 64
(b)
1
27
Question 3
Many candidates were able to factor completely the expression with a large number obtaining full marks.
Common incorrect answers arose from partial factorization 3(5p2 + 8pt) and p(15p + 24t) both of which still
scored 1 mark. An occasional incorrect answer was 39p3t demonstrating both a lack of understanding of
how to add two algebraic terms and how to factor. Other incorrect answers seen were 3p(5p + 8pt) and
3p(5p + 6t).
Answer: 3p(5p + 8t)
6
© 2012
Cambridge International General Certificate of Secondary Education
0444 Mathematics (US) June 2012
Principal Examiner Report for Teachers
Question 4
This question proved to be a good discriminator. Most candidates were able to correctly expand (2x + 1)2 to
obtain 4x2 + 4x + 1 thus earning the method mark, although this was occasionally seen wrongly as 4x2 + 1
arising from the common misconception that (x + y)2 = x2 + y2. Many candidates demonstrated a lack of
understanding of where to proceed from this point toward finding the values of a and b. The most common
error was the following working: 4x2 + 11x – 2 = 4x2 + ax + b followed by 11x – 2 = ax + b followed by
11x − 2 − b
attempts to solve for a and b. Therefore the most common incorrect answers were a =
and
x
b = 11x − 2 − ax .
Answer: a = 11 b = – 2
Question 5
This question was generally well answered by many candidates with the most success arising from those
with clear diagrams. Some candidates had a diagram clearly showing (3, 0) but went on to write (0, 3) on the
answer line. The most common errors were to find the midpoint of the line PQ or split the line in the ratio 1:2
instead of 2:1.
Answer: (3, 0)
Question 6
This was a well answered question with nearly all candidates scoring at least 1 mark. The majority
understood the method of multiplying the 2 and 7 and adding the indices. Common errors arose from
9
wrongly counting zeros (generally from those attempting 20000000 x 700) or wrongly converting 14 x 10 to
8
1.4 x 10 .
Answer: 1.4 × 1010
Question 7
The majority of candidates were able to correctly answer this question with the most success arising from
those who added the 1 then divided by the 2 before squaring. Some candidates squared at an earlier stage
and these tended to just square individual terms rather than entire expressions, for example common
incorrect first lines of working were 4y + 1 = 81 or 2y + 1 = 81. Other common incorrect starting points seen
were 2 y = 8 and 2 y = 10 followed by 2y = 100.
Answer: 25
Question 8
This question was very well answered with most candidates obtaining at least 1 mark. The best working
showed the equation
8+4+8+9+y
5
= 7.2 being correctly solved, although some candidates adopted a
‘trial and error’ approach. Some candidates misunderstood how to deal with the y in the equation with a
minority using 29y instead of 29 + y for the numerator.
Answer: 7
Question 9
This question was generally answered well with many candidates obtaining full marks. The best working
showed candidates awareness of the need to have the base number the same on both sides of the equation.
Those candidates writing 125 = 53 were generally successful. There were occasional attempts to solve the
equation using incorrect inverse operations 125 ÷ 5 x 3 = 75 was a common incorrect method and answer. It
was quite common to see no working for this two-mark question, which is a risky strategy for candidates
even when they feel confident in their answer.
Answer: 9
7
© 2012
Cambridge International General Certificate of Secondary Education
0444 Mathematics (US) June 2012
Principal Examiner Report for Teachers
Question 10
Candidates were able to demonstrate good understanding in this question. With the most success from
those who multiplied the first equation by 3 and then subtracted the second equation from it, eliminating x’s
and keeping all values positive. Those subtracting the other way round occasionally had problems dealing
with the negatives consequently this resulted in -16y = -48 not always being written or solved correctly. Few
candidates attempted to eliminate y. Candidates choosing to use the substitution method generally
recognised that it was easiest to made x the subject in the first equation and were usually successful. Those
candidates who checked the accuracy of their answer by substituting both figures into the unused equation
nearly always obtained full marks. Once common arithmetic slip seen was in finding the value of x, on quite
a few occasions x = 19 – 21 followed by x = -3 was seen.
Answer: x = -2 y = 3
Question 11
Candidates were generally able to use equivalent fractions to correctly convert the given fractions to those
with a common denominator. Most candidates obtained at least one mark on this question. Denominators of
30 or 60 were nearly always used and there were very few arithmetic slips in this step. The most successful
then went on to correctly add these and convert their fraction to a mixed number in its simplest form. Some
candidates either did not read the instructions in the question carefully enough or did not understand this
instruction and the most common incorrect answer was 41 along with other improper fractions or
15
unsimplified fractions.
Answer: 2 11
15
Question 12
Few incorrect answers were seen for all three parts of this question as candidates showed a clear
understanding of this work. It is worth noting in part (b)(i) there were a number of candidates writing
equations rather than expressions which was condoned. There were a small minority writing equations, such
T − 40
= h and then lost marks
as T = 40 + 35h who then went on to make h the subject of this equation i.e.
35
T − 40
for writing just
on the answer line demonstrating a lack of understanding of the phrase “in terms of
35
h”.
Answers: (a) 215
(b)(i)
40 + 35h
(b)(ii) 10
Question 13
In part (a) many candidates were correctly able to demonstrate they understood that rise/run gives the slope
y − y1
. Those
of the line. With the most success coming from those using correctly using the method: 2
x 2 − x1
candidates who chose to find the gradient by means of a sketch sometimes forgot the negative sign and so
½ was a common incorrect answer. There were a number of candidates who remembered the formula
y 2 − y1
wrongly, usually transposing the numerator and denominator but occasionally transposing any of
x 2 − x1
the figures. 2 and –2 were other common incorrect answers. In part (b) candidates generally realised that to
pass through the point (0, 2) they simply needed an equation of the form y = mx + 2 using the slope, m,
found in part (a). Occasionally 2 was used for the slope instead of the y-axis intercept.
Answers: (a)
−
1
2
(b)
y = − 21 x + 2
8
© 2012
Cambridge International General Certificate of Secondary Education
0444 Mathematics (US) June 2012
Principal Examiner Report for Teachers
Question 14
The candidates with the most success on this question began with y =
k
followed by correct substitution
x2
of x = 3 and y = 2 to find k = 18. A few candidates then lost the final mark because of the error of writing
18
18
y =
instead of y = 2 . The most common errors were using direct proportion or forgetting to square the
5
5
x. Some candidates attempted the method which does not require the constant k to be found directly first,
2
2
namely y 1 x 1 = y 2 x 2 however this method was less successful as it was often not correctly used as
2 × 3 2 = y × 5 2 with the most common error in this approach being
y1
x1
2
=
y2
x2
2
.
Answer: 0.72 or 18
25
Question 15
Less than half the candidates were able to correctly draw an angle of 67° in part (a). The most successful
candidates labelled C or made sure that their line BC did not extend beyond AD and had sharp pencils for
accurate diagrams. Sometimes the angle of 113° was drawn by those reading the incorrect scale on their
protractor or not reading the question carefully. Angles of 67° were also sometimes seen in various other
places on the diagram demonstrating a lack of understanding of the terminology ‘angle ABC = 67°’. There
were quite a few inaccurate drawings, candidates were either drawing with insufficient regard for accuracy or
did not have the correct equipment. Part (b) was more successful. Problems arose when candidates used a
ruler to measure half way, when the question specifies to use a ‘straight edge’ candidates need to
understand this means no measuring. Two sets of intersecting arcs are required to construct a
perpendicular bisector. With some candidates leaving this blank it is difficult to say whether this is due to
lack of understanding or lack of equipment. Some candidates did not read the question carefully and either
constructed the perpendicular bisector of AD or BC or bisected one of the angles in the triangle ABC.
Answers: (a) angle ABC = 67° (b) perpendicular bisector of AB
Question 16
Nearly all candidates obtained at least one mark on this question with many obtaining at least 2 marks for
their region R. The most common incorrect answers seen arose from being unable to deal with the two
1
inequalities y ≤ x + 4 and x + y ≥ 6 . It is worth advising candidates to shade diagrams in pencil as many
2
shaded in pen and then could not easily correct errors in shading, there were a number of unclear regions
due to this. Occasionally candidates were seen to be labelling certain points on the graph as R, rather than
labelling a region.
Answer:
9
© 2012
Cambridge International General Certificate of Secondary Education
0444 Mathematics (US) June 2012
Principal Examiner Report for Teachers
Question 17
This question was answered well by about half of the candidates. Some candidates were unsure how to
deal with the fact that w appeared twice in the formula. Those candidates with the most success began by
multiplying c by w + 3 and then collecting terms in w on one side and terms without w on the other side. The
next step of factorizing the expression to ensure that w only appeared once was the most challenging and
some candidates proceeded with less success from here. A number of candidates effectively multiplied the
right hand side of the equation by (w + 3)2 as a quadratic in w was sometimes seen here. A common
incorrect answer was w = cw + 3c – 4.
Answer: w =
4 − 3c
c −1
Question 18
In part (a) many candidates realised they were looking for P(x, 0) Q(0, y) with varying degrees of success in
finding the x and y. The most successful candidates showed the working 0 = 12 – 4x and y = 12 – 4(0). The
most common incorrect answers were P(0, 3) Q(12, 0) and P(0, 12) Q(3, 0). Part (b) proved to be more
challenging. Many candidates still obtained full marks and of those obtaining just the method mark there was
1
about an even split between those obtaining it for y = kx + 0 or for y = x + c.
4
Answer: (a) P(3, 0) Q(0, 12)
(b)
y =
1
4
x
Question 19
Nearly all candidates obtained full marks on part (a) with the most success coming from those quoting A =
π rl and then correctly substituting in the values of r and l. A small minority used 3.14 instead of π . Part (b)
proved more of a challenge. There were two main errors, firstly in finding the lateral surface area of a sphere
instead of a hemisphere and secondly in working out the total surface area of the cylinder (i.e. including the
circular end faces and not just the lateral surface area). Arising from both of these errors was the most
common incorrect answer of 224. Another error seen a few times was, when working out the hemisphere,
writing 4π 16 followed by 2π 8 , i.e. incorrectly halving both the 4 and the 16.
2
Answer: (a)
π × 4 × 10
(b) 192
Question 20
This question proved to be a good discriminator with few candidates obtaining full marks. Candidates had
the most success in part (a) of this question with about a third of them obtaining the correct answer. To have
more success candidates need to be aware that the direction is vital to the sign of the vector since p + t and
p – t were common incorrect answers. A few candidates did not understand the concept of vectors and
angles of 60° and 120° were sometimes seen used. In parts (b) and (c) candidates with the best working
made it clear which route they were taking, for example by writing PR = PS + SR , or OR , respectively, for
their method. The most common misconception was that PQ and TS were both t, consequently 2t was a
common incorrect answer for part (b).
Answers: (a) - p + t
(b)
p + 2t
(c)
2p + 2t
10
© 2012
Cambridge International General Certificate of Secondary Education
0444 Mathematics (US) June 2012
Principal Examiner Report for Teachers
Question 21
This question proved to be a good discriminator with few candidates obtaining full marks although quite a few
obtained 3 or more marks. Candidates were generally good at recognising that tangents meet a radius at
right angles and that tangents from the same point are equal in length but some were unsure where to
proceed from here. For those obtaining 3 marks there was about an even split for obtaining them for
correctly finding the arc length or for correctly calculating the lengths PT and PR using trigonometry. The
best explanations involved the use of the 30, 60, 90 triangle using a method of enlargement from this to
demonstrate that PT was 5 3 . Fewer candidates were able to show that where the expression came from
with the three most common errors being to find the minor arc length RT, to use a fraction of the circle area
formula or to simply halve the circumference.
Question 22
This question proved quite challenging although some candidates obtained full marks for all three parts. Part
(a) was the most successfully answered. Often candidates wrote no working in this question, for example 3
9
or 1 , with no method, were common wrong answers for part (a).
3
Occasionally seen incorrect methods
throughout all three parts were to add probabilities instead of multiplying (and vice versa) and sampling with
replacement. Candidates very occasionally incorrectly added 3+2+4 to find the total number of pencils. Part
(c) was generally the most challenging part with the most success from those working out P(GG|) + P(G|G)
many candidates chose instead to work out P(RG) + P(BG) + P(GR) + P(GB) and in both cases it was
common to not have considered all of the correct events. Occasionally probability answers greater than 1
were seen, particularly in parts (b) and (c), candidates are advised to check the sense of their answers.
Answers: (a)
1
12
(b)
5
18
(c)
5
9
11
© 2012
Cambridge International General Certificate of Secondary Education
0444 Mathematics (US) June 2012
Principal Examiner Report for Teachers
MATHEMATICS (US)
Paper 0444/03
Paper 3
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, and many were able to
make an attempt at most questions. Few candidates omitted whole questions. The standard of presentation
was generally good. There were occasions where candidates did not show clear workings and so did not
gain the method marks available. Centres should encourage candidates to show formulas used,
substitutions made and calculations performed. In questions where candidates were asked to ‘explain’, they
should be encouraged to answer in sentences, give all required information, and show all workings in order
to fully answer the question. This was particularly important in Questions 3c, 4a and 6e where full
explanations and correct information was required to gain full marks. 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 fully process calculations 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. The use of correct equipment was evident and should be emphasised by Centres. Centres
should be aware that candidates’ understanding of transformation of graphs was weak in comparison to the
other topics covered by this paper.
Comments on specific questions
Question 1
The first three parts of this question was generally well answered by candidates of all abilities. The final part
required candidates to calculate compound interest which many candidates found challenging and a large
number calculated simple interest by mistake.
(a)
Most candidates attempted this question however many candidates divided by 19 or 21 and not by
the total of 40. Some candidates calculated Wendy’s share of $2000 instead of Vince’s.
Candidates should be encouraged to check they have answered the question set once a solution
has been found.
(b)
Very well answered by candidates of all abilities. All candidates attempted division by $37 or
repeated addition of $37.
(c)
Very well answered by most candidates, nearly all candidates gaining one mark for calculating 27%
correctly. Some candidates misread the question and calculated 2/5 of the remaining amount,
leading to common answers of $86.70.
(d)
The majority of candidates were able to calculate the result of two years compound interest, using
the compound interest formula. Unfortunately a significant number did not find the amount of
interest, forgetting to minus $500. Answers of 540.8, 540.80 and 541 were therefore common.
Fewer candidates followed the instructions given, to give their answer to the nearest dollar, with the
majority leaving their answers to one or two decimal places. A large number of candidates
incorrectly calculated simple interest by mistake, giving an answer of 540 or 40.
Answers: (a) 950
(b) 7
(c) 66
(d)
41
12
© 2012
Cambridge International General Certificate of Secondary Education
0444 Mathematics (US) June 2012
Principal Examiner Report for Teachers
Question 2
Candidates demonstrated a good understanding of reflection and rotation. Candidates generally gave the
correct information required to fully describe a particular transformation. Candidates found enlargement
more challenging. Candidates were required to describe fully a transformation which was dependent on their
correct initial transformation. Although follow through marks were available few candidates who made
mistakes in part (a) gained full marks in part (b).
(a) (i)
Candidates could generally draw a correct reflected shape but some reflected in the x-axis instead
of the y-axis.
(ii)
Most candidates could rotate the shape through 180˚ although some did not use the correct centre
of rotation.
(iii)
Candidates could draw the shape the correct size but again many could not locate it correctly.
Most commonly candidates enlarged the shape by the correct scale factor of 2 but from centre of
enlargement (0,0).
(b) (i)
Most candidates understood that only a single transformation should be given and despite
reflecting in the wrong axis in part (a)(i) were able to give the correct description. In order to
improve candidates need to know the equation for the y and x-axes as many candidates gave x=0
for the x-axis and y=0 for the y-axis.
(ii)
A large number of candidates did not give the correct answer of ‘translation’, most common
response was ‘slide’ which was not acceptable. Of those candidates who did identify a translation
many did not give a correct vector, either giving no answer, co-ordinates or incorrect values.
Answers: (b)(i) Reflection, y=0
(ii)
⎛ 4⎞
Translation, ⎜⎜ ⎟⎟
⎝8⎠
Question 3
This statistics question was attempted well by all candidates. The probability part of the question was more
successfully answered than the ‘explain’ question which challenged all candidates. It was evident that the
majority of candidates were able to use protractors and rulers to construct the pie chart, although a
significant number of non ruled attempts were seen.
In part (a) probabilities were generally given in a correct format. There were very few candidates who gave
their answers as ratios or ‘out of’. Those candidates that gave their answers as percentages were generally
successful, however centres should emphasise the correct number of significant figures required in all
answers (three).
(a) (i)
(ii)
(iii)
This part was attempted by all candidates and very well done.
This part was attempted by all candidates but less successfully. It was clear that some candidates
0
was common.
were unaware that 2 is a prime number and an answer of 0 or
6
6
which did
All candidates attempted this question however many candidates gave the answer of
6
not score. The candidates were required to simplify this to 1 to gain the mark.
(b)
This was answered very well by candidates of all abilities.
(c)
This part of the question was the most challenging. Candidates were aware that they were
required to compare the probabilities of Jon and Felix, however the vast majority used the
probabilities given in the question without changing to common denominators or a common form
(decimals or percentages), which was required for a complete answer. Those candidates that used
percentages or decimals to compare probabilities were more successful.
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Cambridge International General Certificate of Secondary Education
0444 Mathematics (US) June 2012
Principal Examiner Report for Teachers
(d) (i)
Many correct angles for both sections of the pie chart were seen. A small number gained a single
mark for realising the sum of the two angles to be found was 120˚.
(ii)
The required pie chart was well drawn by most candidates although common errors were
inaccurate measurement and drawing of the angles. The 30˚ angle was achieved by the vast
majority of candidates.
1
2
(ii)
(iii) 1
6
6
(b) 4, 4, 4, 4, 5, 5, 7, 7, 9
3
4
(c) Felix has probability
and Jon has probability
12
12
(d)(i) 72˚, 48˚
Answers: (a)(i)
Question 4
Parts of this question proved challenging to less able candidates. Again the ‘explain’ question was the most
difficult, candidates not giving enough information to gain full marks. Parts (b) and (c) followed on from each
other, and candidates who were unable to answer (b) found difficulty in scoring in part (c). However some
candidates found success by restarting the question in part (c) following an incorrect attempt at part (b).
(a)
This was the most challenging part of the question. Most candidates were able to explain that a
side would be zero or negative but were not specific enough, and did not explicitly say which side
this would be. Identification of the side 11-x was required to score the mark. A large number of
candidates believed this side to be the hypotenuse, despite the triangle not being right angled.
(b)
Most candidates were aware that the sides of the triangle had to be added and most were able to
add the sides correctly. A common mistake was to add all x values with the result of 6x + 14. It
was clear that the vast majority of candidates understood the term perimeter.
(c) (i)
This part was the most successful of the question. Candidates who had incorrect answers to part
(b) were more successful by restarting this part and forming the correct equation and then solving.
The vast majority of candidates who formed an equation were able to correctly reduce it to the form
ax=b by choosing the correct operation.
(ii)
Candidates were aware that they had to substitute their value for x into one of the sides of the
triangle. However because the diagram had side 2x + 3 drawn the shortest many of candidates
substituted into this side only. Candidates who substituted into all three sides were more
successful in identifying the shortest side. Candidates demonstrated confidence in substituting and
showed their substitution and workings.
Answers: (b) 14 + 4x
(c)(i) 4.5
(ii) 6.5
Question 5
The first two parts of this sequence question were very well answered. The parts involving algebra proved
th
more challenging for candidates, particularly as the n term was a quadratic.
(a)
The diagram was drawn correctly by nearly all candidates.
(b)
The vast majority of candidates correctly identified the number of crosses.
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Cambridge International General Certificate of Secondary Education
0444 Mathematics (US) June 2012
Principal Examiner Report for Teachers
(c) (i)
Many candidates were not able to write down an expression for the number of crosses in n rows.
Common answers were n or +2.
(ii)
Some candidates were able to follow through from their previous answer, however most candidates
were unable to deal with the quadratic nature of this sequence. Of those that did multiply by n a
large number used incorrect algebraic notation and were unable to gain the mark e.g. n×n+2 or
n.n+2 instead of n(n+2).
(iii)
This part was more successful. Some candidates followed through correctly from their part (ii),
although the majority did start again and calculated 20 × 22. A large number of candidates
calculated it from the beginning and gained full marks even if they had not earned marks in part
(ii).
Answers: (b) 35
(c)(i)
n+2
(ii)
n(n+2)
(iii) 440
Question 6
Candidates found this question more challenging. The ‘explain’ question proved difficult for most candidates,
and many could not identify the correct angle to measure. Candidates were more confident at identifying
co-ordinates.
(a)
The question on calculating the slope was challenging for many candidates. The position of the
line in the diagram led most candidates to use the triangle ABE and give some indication of change
in x / change in y. Some candidates managed to get to a correct fractional value but did not
simplify to 2, or a correct numerical figure but gave the answer as a negative instead of a positive.
(b)
This part also proved challenging to some candidates. Some candidates gave a purely numerical
answer, without an x term or no x present e.g. y=-0.5+6. A number of candidates confused ‘m’ and
‘b’ in the equation of a straight line.
(c)
Slightly more candidates were able to answer this ratio question. Common errors were 3:12, -1:4,
1:-4 (which did gain a method mark) and -3:12. A small number of candidates gave non numeric
answers e.g. AE:EC and some candidates measured the length inaccurately with a ruler instead of
using the scale on the axes.
(d)
Candidates were confused with the notation for the angle ABE and many candidates measured
ABC, with the common incorrect answer of 90˚ seen. Those candidates that identified the correct
angle were able to give an answer in the correct range. It was evident that candidates had the
correct equipment to attempt this question.
(e)
Again the ‘explain’ question proved challenging to most candidates. Candidates were unaware of
the amount of information or the level of description needed to gain full marks. Common correct
answers identified the ‘same angles’ but significantly fewer candidates could explain the ‘lengths in
the same ratio’ in enough detail or accuracy to gain the mark. Most common answer was ‘same
shape different size’ which did not score. Candidates should be encouraged to write as detailed
answer as possible.
(f)
Candidates who calculated the area of ABC in one calculation using AC and AB were more
successful than those who used two triangles ABE and BCE, or sides AB and BC. The vast
majority of candidates showed they understood how to calculate the area of a triangle. Candidates
should be encouraged to use information given in the question rather than measure their own
lengths as an exact answer of 45 was required to gain full marks.
(g) (i)
This part of the question was more successful with many candidates correctly identifying the
correct position on the diagram
(ii)
This part was the most successful of the question with candidates correctly giving the correct
coordinate or gaining a correct follow through from their point in part (i).
Answers: (a) 2
(b) -0.5x + 6
(c) 1:4
(d) 25˚-29˚
(e) Equal angles and lengths in same ratio
(f) 45
(g)(i) (9,-6) correctly marked
(ii) (9,-6)
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© 2012
Cambridge International General Certificate of Secondary Education
0444 Mathematics (US) June 2012
Principal Examiner Report for Teachers
Question 7
This question discriminated well between candidates of differing abilities. It gave more able candidates an
opportunity to demonstrate their knowledge of trigonometry but also gave less able candidates the
opportunity to demonstrate their understanding of Pythagoras’ theorem and speed. Only the most able
candidates were able to gain full marks on part (e).
(a) (i)
(ii)
This question was well answered by the vast majority of candidates.
Most candidates recognised this as a Pythagoras’ theorem question and was successfully
answered by the majority of candidates. The use of a²+b² was quoted and substituted into, by
candidates of all abilities, however some went on to double this value or incorrectly evaluate 50² or
120².
(b)
Candidates again showed good understanding of the relationship between speed, distance and
time with the best candidates writing the correct speed=distance/time formula. Most common
mistakes came from incorrect answers to part (a) or candidates who used the distance ran by Said
instead of Bill.
(c)
Following a correct answer to part (b) most candidates could identify Said as arriving first however
less candidates could calculate by how many seconds. This question highlighted the importance of
candidates showing their workings as some candidates could have scored method marks if they
had shown how they reached their answer. Some candidates also tried to calculate Bills time to
reach R, despite it being given in part (b), and made mistakes which led to an incorrect time in this
part.
(d) (i)
This question challenged the more able candidates to demonstrate their understanding of
trigonometry. Of those that attempted the question most used a trigonometric ratio and the
majority correctly chose the tangent ratio, however a significant number of candidates used adj/opp
instead of opp/adj.
(ii)
Most candidates found the calculation of the bearing challenging with many candidates quoting a
length instead of an angle. Common misunderstanding was to subtract the previous answer from
360˚ instead of 180˚. Many candidates did not attempt this question showing a common
misunderstanding of the word ‘bearing’.
(e)
This part was challenging to all candidates, however candidates of all abilities were generally able
to gain two of the four marks for correctly multiplying 50 and 120 to calculate the area of the field in
m². Only the most able candidates were able to convert this value to square kilometres, the most
successful candidates started with the values of 0.05 and 0.12. Most candidates understood
scientific form, the most common answer was 6×10³ from 6000, however very few candidates were
able to gain the final mark for conversion to scientific form as most did not have an answer that led
to a negative index value.
Answers: (a)(i) 170
(d)(i) 67.4˚
(ii) 130
(b) 5
(c) Said, 1.5
(ii) 112.6
(e) 6×10-3
Question 8
Candidates were challenged by this question and many did not attempt parts of it. Candidates were more
successful at quoting the formula for the volume of a cylinder than the relationship between circumference
and diameter. The question on surface area challenged those who did not read the question carefully and
who did not know the difference between surface area and volume. A significant proportion of candidates
did not attempt part (d) and of those that did few were able to gain marks.
(a) (i)
Candidates were able to quote the correct formula for the volume of a cylinder and most were able
to substitute the correct values. All values of π were seen and most answers fell within the range
of acceptable answers. Many candidates did not give the units despite being prompted in the
question. Correct units seen were cm³, ml and litres (where candidates had successfully divided by
1000). However units of ml³, cl, and cm² were common mistakes.
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Cambridge International General Certificate of Secondary Education
0444 Mathematics (US) June 2012
Principal Examiner Report for Teachers
(ii)
This question challenged the most able of candidates and some candidates were unable to attempt
it. Common mistakes included incorrect conversion of 1.5 litres to cm³, dividing by the answer to
part (i) and misinterpretation of the value 8.8419.., giving the final answer as 9 instead of 8
glasses. This question again highlighted the importance of candidates showing their working as
some candidates could have gained follow through marks if they had shown their workings.
(b)
Many candidates did not know the relationship between circumference and diameter. Many
candidates could quote C=2πr and found the radius believing this to be the diameter. A significant
number of candidates simply divided 16 by 2. This question highlighted the importance of giving
answers to the required number of significant figures. Some candidates gave answers of 5.1,
instead of 5.09, which only gained the method mark.
(c)
Candidates who knew the difference between surface area and volume were able to score some
marks on this question however a significant number of candidates multiplied all values to calculate
volume. Many candidates were able to calculate the area of one face and score one mark by
understanding that there were two of these faces. However some candidates believed that four of
the six sides were identical. This question again highlighted the importance of candidates showing
their workings.
Part (d) proved to be challenging to candidates of all abilities. A large proportion of candidates did not
attempt it.
(d) (i)
Most candidates were confused by the units given in the question. Very common mistakes were to
4
include the units, leading to answers of vm³ or m×v×m³ or vm . Many candidates wrote their
answer as an equation instead of an expression.
(ii)
This part was more successfully answered as candidates appreciated that they had to multiply their
answer to part (i) by s. However very few candidates gave the correct answer of msv and most
candidates who scored on this question did so as a follow through. A large proportion of the
candidates did not attempt this part of the question
(iii)
This part of the question was not attempted by a significant number of candidates. Those who did
attempt it did not understand the relationship between grams and kilograms with the majority
dividing by 1000 instead of multiplying.
Answers: (a)(i) 226 cm³
(ii) 8
(d)(i) mv
(ii) msv
(b) 5.09
(c) 148
(iii) 1000 msv
Question 9
This question was the most challenging on the paper and proved difficult for a large number of candidates.
All abilities found it challenging with many candidates not attempting some or all parts of the question.
Transformation of graphs is clearly an area of work that candidates find difficult at this level.
(a) (i)
This part proved to be the hardest of the whole paper with very few correct equations seen. A
large proportion of candidates did not attempt this question. Of those who attempted this question
a wide range of answers were given including, intervals, single numerical values and very complex
quadratic and cubic equations and fractions.
(ii)
This part also proved too difficult for most although it was attempted by more candidates. Very few
were able to give the correct solution of x-1. The most common mistake was to miss out the x and
only write -1 or more commonly +1.
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Cambridge International General Certificate of Secondary Education
0444 Mathematics (US) June 2012
Principal Examiner Report for Teachers
(b) (i)
All candidates found this question challenging with only a very few able to gain a mark. The most
common mistake was to move the graph up or down 2 units instead of stretching vertically.
(ii)
This part again challenged all abilities of candidate however more candidates were able to gain a
mark for identifying the correct transformation as a translation. A common mistake was to call this
transformation a ‘slide’. Very few candidates used a vector to describe the translation fully with
most candidates describing it in words. An answer of ‘move down 2’ did not gain the mark as the
candidate had not identified that there was no horizontal movement and therefore not described the
translation fully.
Answers: (a)(i)
k
x
(ii) x-1
⎛ 0 ⎞
(b)(ii) Translation, ⎜⎜ ⎟⎟
⎝ − 4⎠
Question 10
This question discriminated well between candidates of differing abilities. Nearly all candidates could answer
some part of the question with very few candidates not attempting any of it. Able candidates could
understand range and domain and draw a correct graph. Less able candidates could identify the vales for a
and b in part (a) and substitute correctly in part (d).
(a)
This part was generally answered well by candidates of all abilities.
(b)
This part proved more challenging to candidates, showing that only the more able candidates
understood the term ‘domain’. Many candidates simply wrote their values from part (a) for this
part.
(c)
Only the more able candidates again were able to answer this part with many candidates choosing
not to attempt it. This again showed that only the very able candidates understood the word range,
with even fewer able to give the answer in the required format of an interval. Most common answer
was to correctly find 50 and 110 but then to find the arithmetic range, i.e. subtracting these values.
This method only scored one mark.
(d)
More candidates were able to attempt this question and many more from all abilities were able to
gain the mark. This part did allow for a follow through from incorrect answers in part (a) however
the most successful candidates were those that had part (a) correct. Very few candidates gained
the mark as a follow through.
(e)
The graph proved challenging for all candidates. The more able candidates were able to draw a
line with the correct slope but very few realised that the domain identified in part (b) affected the
graph. Most answers had a straight line across the whole grid, the correct answer required the line
to go between w=1 and w=3 only.
(f)
This question was more successful for all abilities however a significant number of candidates did
not attempt it. The answer could have been gained without the use of the graph, which a large
number of candidates did. The wording of the question ‘Use your graph to…’ put a number of
candidates off attempting this part if they had been unable to draw the graph in part (e). The most
successful alternative method was to solve the equation 30w+20=65.
Answers: (a) 30, 20
(b)
1, 3
(c)
50<t<110
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(d) 95
(f) 1.5
© 2012
Cambridge International General Certificate of Secondary Education
0444 Mathematics (US) June 2012
Principal Examiner Report for Teachers
MATHEMATICS (US)
Paper 0444/04
Extended
General comments
There were some excellent scripts submitted indicating much mathematical talent and thorough preparation
for this paper. Unfortunately many had not studied the specialised topics, transformational geometry,
Question 7; functions, Question 8; and curves, Question 10.
Comments on specific questions
Question 1
(a)
This was well done by a few who knew the cosine rule and could apply it to this triangle. Some
spoiled their work by carelessly writing 180 for 108.
(b) (i)
Many had difficulty interpreting “how far L is south of K”. To see this distance necessitated drawing
a right angled triangle with KL as the hypotenuse. Some drew the south line through K to meet ML
at P say, then assumed angle KPL was 90°, no credit was given for this work. Some thought KP
bisected angle MKL. Many gave the answer as 9, or made no attempt.
(ii)
Good candidates got this right; some gave the bearing 233 for which they received one mark.
Answers: (a) 10.9
(b)(i) 5.16
(ii)
053
Question 2
There were many good responses to these arithmetic problems, the majority of candidates scored marks.
(a)
This was very well done by many.
(b) (i)
When an answer is a sum of money it is usual to give the answer in dollars and cents to the
nearest cent. Here there is no need to give 468.72 as 469.
(ii)
Many worked out 23% of $64.68.
(c)
3
It was pleasing to see candidates using the factor (1.016) , those who did not struggled to get the
3
right result. 1.6 , 1.016 × 3 were seen as well as simple interest attempts.
(d)
Many missed the rule that profit is usually given as a percentage of the cost price. Many used
324/288 but left their answer as 112.5%.
Answers: (a) 1134
(b)(i) 468.72
(ii) 84.00
(c)
262.19
(d) 12.5
Question 3
Many plotted correctly and drew a beautiful curve through the points, this was very pleasing especially as the
numbers were decimals and the scales on the axes were different. The line y = 0.8 was often in the right
place and many read off the intersections to score in (c)(ii). Just a few knew how to draw a tangent and did
so well. Many did not know what a tangent was, or they drew it in the wrong place.
Answers: (a) 1, 0.98, 0, 0.98, −1
(c)(ii) ranges around −1.15, −0.45, 1.6
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(d) 4 to 5.5
© 2012
Cambridge International General Certificate of Secondary Education
0444 Mathematics (US) June 2012
Principal Examiner Report for Teachers
Question 4
The majority made very poor attempts at all parts of this question.
(a)
Many could see the 90° angle at B but chose to give 13 as their answer; they were not asked for
this angle.
(b)
TanC = 7/10 was the direct method to get 34.99. Many chose to find AC = 12.2 by Pythagoras and
then they used sine of angle C. Some chose to use angle CAB = 55, this method was usually not
valid.
(c)
Some candidates quoted a “bow tie theorem” or mentioned “butterflies”; splendid methods to help
understanding in class, but stuffy Examiners prefer “angles on the same arc” or even better “angles
in the same segment”.
In part (d) just a few knew the sine rule and the expression ½absinC, the majority did not.
In (d)(i) candidates needed to focus on triangle ADB and they would see that there are sufficient known
quantities to find AD.
In (d)(ii) they had sufficient information to get angle DAB = 68 and to use the area formula.
Here they had to see that their AD and 10 were corresponding sides in the similar triangles and so
the ratio of the areas was the square of 10:their AD. A few excellent mathematicians did this.
(e)
Answers: (a) 90
(d)(i)
11.9
(ii)
38.6
(e)
8.69
Question 5
Some very good candidates scored full marks for this; many just a few marks; many had no ideas and did
not score.
(a)(i)(ii)(iii) and (b) Those familiar with a cumulative frequency curve managed these questions easily.
(c) (i)
(ii)
(d)
Again, those in the know converted the information on the curve to complete this table.
Candidates had to follow the standard methods for dealing with a grouped frequency distribution,
just a few could do this.
There were some pleasing attempts at the histogram. Many candidates set up the axes and
produced the appropriate six bars. The common fault was to get the wrong heights for the first and
last columns.
Answers: (a)(i)
2.8
(ii)
3.8
(iii) 1.8
(b)
6
(c)(i) 9, 4, 4
(ii) 2.95
Question 6
(a) (i)
Many did not set up the appropriate equation as their first move; ½bh was a common opening as
they could see that 8x2 had to become 4x2; similarly (−1)(−7) became 6. There was a full range of
responses and all possible marks were scored.
(ii)
This was tackled extremely well by the majority of the entry; the only common error seen was 0.407
given as having two decimal places.
(iii)
This was usually correct.
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Cambridge International General Certificate of Secondary Education
0444 Mathematics (US) June 2012
Principal Examiner Report for Teachers
(b) (i)
The candidates were extremely well prepared for the difference of the two squares, very few
indeed could not do it.
(ii)
This fierce-looking equation was very well solved by many candidates. The temptation is to deal
with the denominator (x−4)(x2 − 16), but no, the majority took the more sensible line of multiplying
throughout by (x2 – 16). Careful handling of many terms led to a simple solution, −7, very pleasing.
Answers: (a)(ii)
1.84 0.41
(iii) 0.36
(b)(i) (x−4)(x+4)
(ii) x = −7
Question 7
(a) (i)
Many could describe this rotation, some missed the centre, some got the angle wrong.
(ii)
Some recognised an enlargement but very few could give the scale factor as −3.
(iii)
This is another enlargement, some got the scale factor as ⅓ and many got the centre (0,6).
In this part (a) many candidates missed the significance of the word “single” in the questions. In any part, if
more than one transformation is mentioned then all three marks are lost.
(b) (i)
(ii)
(c)
The translation was well done by some. A common fault was to reverse the x and y moves.
Again this was well done by many, the drawing of x = −1 was not always correct.
This image was hardly ever correct. Even the best candidates who had clearly studied this topic
could not draw a shear
Answers: (a)(i) rotation, 180, (0,0)
(ii) enlargement, −3, (0,−3)
(iii) enlargement, ⅓, (0,6)
(b)(i) (−1,0); (−2,−2); (−4,−2)
(ii) (−5, 5), (−2, 3), (−4, 3)
(c)(i) (0,3); (6,5); (4,3)
Question 8
(a) (i)
This was well done, very few got it wrong.
(ii)
Many did a correct interpretation, 7 – 2 (x – 3) but then spoiled their good work by removing the
brackets badly.
(iii)
Again many had (5x)2 – 8 in their working, 5x2 – 8 as their answer.
(b)
Many knew how to tackle this problem but ruined their work by making a sign error in transposing.
A few gave their answer as 1/(7 – 2x).
(c)
A few could do this but a common interpretation was (x2 – 8)(3x + 5).
(d)
Again (3x + 5)(3x + 5) was common.
(e)
Most candidates wrote down the inequality and many solved it correctly.
Answers: (a)(i) 14
(ii) 13 – 2x
(d) 7 (e) x < −3/8
(iii) 25x2 – 8
(b) (7 – x)/2 (c) 9x2 + 30x + 17
Question 9
(a)
Most could substitute into the given formula and nearly all of those correctly calculated the answer.
(b)
The first task was to find the radius of the missing top cone, this was not well done. Similar
triangles gets r = 3 almost immediately but many could not see this. A very inelegant method was
to work backwards from 1960 to get r = 3.01. However, many excellent candidates had no trouble
and some even employed the efficient method:- heights are in ratio 1:3 so volumes are in ratio 1:27
26
× part (a).
so volume of base =
27
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Cambridge International General Certificate of Secondary Education
0444 Mathematics (US) June 2012
Principal Examiner Report for Teachers
(c)
This part was well done and often scored full marks. Just a few did not know the volume of a
cylinder, ⅓πr2h, 2πr2h, 2πrh were seen.
Answers: (a) 2036
(b)
1960.35
(c) 2.88
Question 10
(a) (i)
A few candidates got this right. A few translated 2 to the right for one mark. Others translated
upwards or produced no recognisable curve.
(ii)
There were a few correct efforts. Many did not keep the origin invariant; they produced y = f(x) + 2,
a different curve altogether.
(b)
Just a few could sketch this graph, many drew only the middle sweep, many were totally wrong,
many made no response to the question.
(c)
The period was occasionally 180, the amplitude was often 2.
(d)
Most candidates managed to fit one, two or three of the functions to the drawings. All four right
3
was achieved a few times, as was all four wrong. y = x was the favourite to get right. The thinking
behind some of the choices was difficult to understand.
Answers: (c) 180, 2
(d)
4 – x2, 10/x, 3x, x3
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