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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Leader.If working is crossed out and still legible, then it should be given anyappropriate marks, as long as it has not been replaced by alternative work.If there is a choice of methods shown, then no marks should be awarded,unless the answer on the answer line makes clear the method that has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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Paper 5525/05

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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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

7

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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

8

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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

10

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