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Transcript of 5525_05_rms_20050603
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GCSE
Edexcel GCSE
Mathematics A 1387
Paper 5525/05
Summer 2005
Mark Scheme (Results)
Mathem
aticsA1387
EdexcelGCSE
Paper5525/05
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NOTES ON MARKING PRINCIPLES
1 Types of mark
M marks: method marks A marks: accuracy marks B marks: unconditional accuracy marks (independent of M marks)
2 Abbreviationscao correct answer onlyft follow throughisw ignore subsequent workingSC: special caseoe or equivalent (and appropriate)dep dependentindep - independent
3 No workingIf no working is shown then correct answers normally score full marksIf no working is shown then incorrect (even though nearly correct) answersscore no marks.
4 With workingIf there is a wrong answer indicated on the answer line always check theworking in the body of the script (and on any diagrams), and award anymarks appropriate from the mark scheme.If it is clear from the working that the correct answer has been obtained
from incorrect working, award 0 marks. Send the response to review, anddiscuss each of these situations with your Team Leader.Any case of suspected misread loses A (and B) marks on that part, but cangain the M marks. Discuss each of these situations with your Team Leader.If working is crossed out and still legible, then it should be given anyappropriate marks, as long as it has not been replaced by alternative work.If there is a choice of methods shown, then no marks should be awarded,unless the answer on the answer line makes clear the method that has beenused.If there is no answer on the answer line then check the working for anobvious answer.
5 Follow through marksFollow through marks which involve a single stage calculation can beawarded without working since you can check the answer yourself, but ifambiguous do not award.Follow through marks which involve more than one stage of calculation canonly be awarded on sight of the relevant working, even if it appears obviousthat there is only one way you could get the answer given.
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6 Ignoring subsequent workIt is appropriate to ignore subsequent work when the additional work doesnot change the answer in a way that is inappropriate for the question: eg.incorrect cancelling of a fraction that would otherwise be correctIt is not appropriate to ignore subsequent work when the additional work
essentially makes the answer incorrect eg algebra.Transcription errors occur when candidates present a correct answer inworking, and write it incorrectly on the answer line; mark the correctanswer.
7 ProbabilityProbability answers must be given a fractions, percentages or decimals. If acandidate gives a decimal equivalent to a probability, this should be writtento at least 2 decimal places (unless tenths).Incorrect notation should lose the accuracy marks, but be awarded anyimplied method marks.
If a probability answer is given on the answer line using both incorrect andcorrect notation, award the marks.If a probability fraction is given then cancelled incorrectly, ignore theincorrectly cancelled answer.
8 Linear equationsFull marks can be gained if the solution alone is given on the answer line, orotherwise unambiguously indicated in working (without contradictionelsewhere). Where the correct solution only is shown substituted, but notidentified as the solution, the accuracy mark is lost but any method marks
can be awarded.
9 Parts of questionsUnless allowed by the mark scheme, the marks allocated to one part of thequestion CANNOT be awarded in anoth
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No Working Answer Mark
1 28742 + xxx (a) 2832 + xx 2 M1 for 4 terms corre
terms with correct si
A1 cao
(b) 24 2yy + 2 B2 caoB1 for or
4y
22y
(c) p(p+6) 2 B2 forp(p+6) orp(B1 forp(ap+b) whe
own, or part of an ex
(d) )32(3 yxx 2 B2 (B1 for 32(3 2x
2 question + response
boxes oe
2 1st
aspect: One quest
ignore other question
2nd
aspect: Response
3rd
aspect: Some men
question or response
Award B2 for all thr
* 0-1, 2-3, 4-5 is OK
3 reflection in xy = 2 B2 cao accept the wo(B1 any statement in
4 (a) 2235 += xx xx 3225 += 5
3 3 B1 for 22 +x seen O
M1 for correct rearra
A1 for5
3oe
(b) 2,1,0,1,2,3 2 B2 (B1 for 5 correct integers)
3
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Paper 5525/05
No Working Answer Mark
5 (a)
3
2
4
3=
12
6
2
1
2M1 for
12
6or
6
3or
A1 accept 0.5
(b)1+2+
12
8+
12
94
12
5
2 M1 for attempt to co
denominator e.g twoA1 correct conversio
or12
20and
12
33seen
A1 cao for12
54
OR
attempts to convert t
M1 0.66+0.75 (or 1.6
A1 4.41, 4.417, 4.41
A1 4.416 (recurring
6 (a)(i) 65 1 B1 accept 15625, 54+
(ii) 35 1 B1 accept 125, 59-6
(b) 10=+ yx and 4= yx x = 7y = 3
3 M1 for eitherx + y =
A2 for both values c
If M0, award B3 for
correct, otherwise B0
SC B2 forx = 3 ory
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7 8622
1 or 48
9109698 ++ or 72 +54 + 90
264
cm2
4 M1 attempt to find th
2
168 or (89) o
24 or 48M1 all five faces wit
A1 cao numerical an
B1 (indep) cm2
with
8
d
ab
3
3
3)(3 dc + 23 bc3 B3 (B1 for each one
Nb B1 for each of t
9 (a)
(b)
12.03.0 =+++ xx
2003.0
0.25
60
2
2
M1 for 2.03.0 ++xA1 oe
M1 0.3 200A1 cao Accept 60 ou
SC B1 for200
60
10 (a)
(b)
8(0)24)12(
5 points plotted
accurately
points joined with
smooth curve
3
2B3 for all correct [(B
B1 1 full (2mm) s5 points plotted)
B1 ft for any smooth
NB: curve must pass
11 m=3
n=5
2 B1 for 3
B1 for 5
(B2 for or523 2
SC: award B1 only i
5
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12
21
15
= -2
52 += xy 4 M1 for clear attempt-1,5 in numerator, 2,
2A1 for cao
B2 ft for "2" += xy
(B1 for 5+= mxy
13 (a) 4
1on LH branch 2 B1 cao
3
2&
3
1&
3
2on RH
branches
B1
(b)3
1
4
1
3
2
4
3+ =
12
1
12
6+
12
7 3 M1 for
4
3
3
2 or
4
1
M1 for sum of two p
A1 for 12
7
oe(c) 421=n or
4
1
6
1: oe
846
1 or
3
221
14 3M1 for either
3
1
4
3
tree diagram
M1 for 214 (=84) o
A1 for 14 cao
SC: B2 for 63 seen in
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14 (a)(i) 150 2 B1 accept 150 or 210
(ii) B1 for angle at the ce
circumference
(b)
"150"9090360 or 180 "150"
30 3 B1 identifies angle b90as (may be in w
9090M1 360 o
A1 ft from (a)(i) exc
Or
B1 for 90
M1 for 9180(2
A1 ft from (a)(i) exc
Or
B3 for 180 (a)
SC: 180 210 canquadrilateral
15 (a)
(b) egx = 0.3939 so 100x = 39.3939
99x = 39
so33
13
99
39==x
0.2727 1
3
B1 for 2.27 recurring
3.39100M1 for =xM1 dep for subtracti
A1 for33
13from cor
Alternative method
M1 for 13.00033M1 for remainders 3
A1 for 0.39 recurring
SC:B1 for99
39
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16 2ktd=(a)
(b)
(c)
2480 = k
2545 t=
25td=
245
3
3
1
2
M1 for or2
ktd= dM1 sub d=80 and t=
A1 for oe (25td=
B1 ft from (a) using
M1 ft from (a) for su
A1 for 3 cao (condon
17 (a)(i)
(ii)
(b)LHS =
+
4
)6416(100 2 xx
=
+
4
1636 2xx
RHS =
+
4
18236 2xxx= LHS
(0, 9)
(8, 25)
3
3
B1 cao
B1 for 8=x caoB1 for 25=y cao
SC: B1 for (25, 8)
M1 for expansion of
terms correct
M1 for common den4 or reducing each n
A1 for fully correct s
Alternative method
M1 for )2
)8(5(
x
M1 for2
(52(
x
A1 for4
)(18( + xx
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18 (a)
(b)
90
810or 9
9 or 3
33
or 27 or 2700
12
2700
3
2
M1 for
90
810or 9
M1 for
90
810or
A1 caoSC:B1 for answer of
M1 for or 27 or3"3"
A1 cao
19 (a)(i)
(ii)
(iii)
(b)
3
23
2
64
164 =
or123
2
)4(64
=
27 = 39 or 27 = 33 or27 = 33/2
1
8
16
1
2
5oe
1
1
2
2
B1 cao
B1 cao
M1 for knowing neg
3
1root is a cube ro
A1 cao for 16
1
M1 for 27 = 9
A1 for2
5oe (cao)
Alternative method
M1 for 927 = 32n
A1 for2
5oe (cao)
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20 (a)(i)
(ii)
(b)(i)
(ii)
(90, 1) B1 cao could be indi
(180, 0)
(45, 0)
(90, -3)
2
2
B1 cao could be indi
B1 cao could be indi
B1 cao could be indi
21 32 )2(3
4
3
1xhx =
32 84 xhx =
32x 3 M1 for substitution iM1 (dep.) for correc
2
3
3
1
)2(3
4
x
x
h
= oe or
A1 for 32x cao
22 (a)( )OM= a+ 2b ( ) 3ON= a or
6
2a
( )MN= a b + 3a2
2a 2b 2 B2
(B1 for either OM
SC: B1 for 2b2a
(b) ( )OX= 2a +b ( )OY= b + 4a 2=XY a 2 B1 for either OX o( QR =) 2ab or ( RQ =)b 2a (hence parallel) B1 forXY= 2a or
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