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Edexcel GCSE

Mathematics 2540

Paper 5540H/4H

Mark Scheme

Summer 2008Mark Scheme

Mark Scheme ( Results)

matics

2540

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NOTES ON MARKING PRINCIPLES

1 Types of markM marks: method marksA marks: accuracy marksB marks: unconditional accuracy marks (independent of M marks)

2 Abbreviationscao correct answer only

ft follow throughisw ignore subsequent workingSC: special caseoe or equivalent (and appropriate)dep dependentindep - independent

3 No working

If no working is shown then correct answers normally score full marksIf no working is shown then incorrect (even though nearly correct)answers score no marks.

4 With workingIf there is a wrong answer indicated on the answer line always checkthe working in the body of the script (and on any diagrams), and

award any marks appropriate from the mark scheme.If it is clear from the working that the correct answer has beenobtained from incorrect working, award 0 marks. Send the responseto review, and discuss each of these situations with your TeamLeader.Any case of suspected misread loses A (and B) marks on that part, butcan gain the M marks. Discuss each of these situations with yourTeam 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If working is crossed out and still legible, then it should be given anyappropriate marks, as long as it has not been replaced by alternativework.If there is a choice of methods shown, then no marks should beawarded, unless the answer on the answer line makes clear themethod that has been used

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6 Ignoring subsequent workIt is appropriate to ignore subsequent work when the additional work

does not change the answer in a way that is inappropriate for thequestion: eg. incorrect cancelling of a fraction that would otherwisebe correctIt is not appropriate to ignore subsequent work when the additionalwork essentially makes the answer incorrect eg algebra.Transcription errors occur when candidates present a correct answerin working, and write it incorrectly on the answer line; mark the

correct answer.

7 ProbabilityProbability answers must be given a fractions, percentages ordecimals. If a candidate gives a decimal equivalent to a probability,this should be written to at least 2 decimal places (unless tenths).Incorrect notation should lose the accuracy marks, but be awardedany implied method marks.

If a probability answer is given on the answer line using bothincorrect and correct 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, or otherwise unambiguously indicated in working (withoutcontradiction elsewhere). Where the correct solution only is shownsubstituted, but not identified as the solution, the accuracy mark islost but any method marks can be awarded.

9 Parts of questionsUnless allowed by the mark scheme, the marks allocated to one partof the question CANNOT be awarded in another.

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5540H/4H

Question Working Answer Mark Notes

1 (a) 5

12

2M1 for

12

nor 12n or

543 ++

nor

)543( ++n where n is an integer12

A15

12or 0.41(6) or 41.6%

(b) 1 512

712

1 B1 ft 1- 512

provided the answer is

positive, or7

12or 0.58(3)

2 22.4 14.5 = 324.8

8.5 3.2 27.2

11.94117647 2M1 for 324.8 or 27.2 or

5

1624or

5

136

A1 11.941(17647) Accept17203 , 11

1716

3 (a) Points plotted 1 B1 points plotted 1 full smallest square

tolerance.

(b) Negative 1 B1

(c) lobf 1 B1 lobf that goes between (8,2000) and

(8,2400) and between (24,0) and (24,500)(d) 11-13 1 B1 11-13 or ft (tol 1 square) from single

straight line segment with a negative gradient

(e) 850-1150 1 B1 850-1150 or ft (tol 1 square) ) from

single straight line segment with a negative

gradient

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5540H/4H


4 58

Reason

2 B1 cao

B1 (dep) alternate or Z angle (oe)

5 Sketch 2 B2 complete 3-D sketch

(B1 for partial 3-D sketch e.g. pyramid only)

(B1 for partial 3-D sketch e.g. pyramid or

base only, or a shape with a box and 2pyramids either end)

Accept hidden lines as dashed or solid

NB: If more than one shape is shown:

For 2 marks there should be no choices or

alternatives other than those also worth 2

marks; if there are several diagrams of which

at least one is worth 1 or 2 marks, award B1.

2D diagrams get B0.

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5540H/4H


6 3n+2 2 B2 for 3n+2 (oe, including unsimplified)

(B1 for 3n+k, k2)

7 2 12

3 33 2.5 20.(625)

2.1 13.(461) 2.6 22.(776)2.2 15.(048) 2.7 25.(083)

2.3 16.(767) 2.8 27.(5(52)

2.4 18.(624) 2.9 30.(189)

2.73 25.8(06) 2.74 26.0(508) or 26 2.75 26.2(96) 2.76 26.5(45)

2.7 4 B2 for trial between 2.7 and 2.8 inclusive

(B1 for trial between 2 and 3 inclusive)

B1 for different trial between 2.73 and 2.75inclusive

B1 (dep on at least one previous B1) for 2.7 only

NB trials wherex has 1 d.p should be rounded or

truncated to at least 2SF; trials wherex has 2 d.p.

or moreshould be rounded or truncated to at

least 3SF

8 Points at (5, 4), (15.5, 13), (25.5, 17), (35.5, 19),

(45.5, 7)

Polygon 2 B2 Complete polygon (ignore histograms and

any lines below a mark of 5 or above a mark of

45, but award B1 if there is a line joining the first

to last point)

(B1 One vertical or horizontal plotting error OR

incorrect but consistent error in placing the

midpoints horizontally OR correct plotting but

not joined. In this case ignore a line joining firstto last)

Plotting tolerance : half square

Points to be joined by lines (ruled or hand-

drawn, but not curves)

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5540H/4H


9 (a) 2x+2x+x+10+50=360 5x+60

=360

2 M1 3 or 4 out of 2x, 2 x,x + 10, 50 added

together

A1 2x+2x+x+10+50 = 360 oe includingx =60

(b) 5x+60=360

5x=300

60 3 M1 for isolating their terms inx

M1 for dividing their numerical term by the

coefficient of theirx term

A1 caoAll the marks in (b) may be given for work done

in answering (a) providing there is no

contradiction

Candidates can score full marks in (b)

independent of their answer in (a) (e. g. by

starting again)

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5540H/4HQuestion Working Answer Mark Notes

10 (a) 45 2 9 10 2 M1 for 45 "72" + or 45 2 or 5 seen, or 90seen or 10 seen as part of a ratio e.g 10:35

A1 cao

(b)

(80 17.5/100) + 80 = 14 + 80 =

94 3M2 for

100

5.11780 or 80 1.175 oe

A1 caoor

M1 for 80 0.175 or100

5.1780 oe or 14 seen

or 8 + 4 + 2 seen

M1(dep) 14 + 80 or 80 +100

5.1780 oe

A1 cao(c) 12000 0.82

OR

1st yr: 12000 0.2 = 2400; 12000 2400 = 9600

2nd yr: 9600 0.2 = 1920; 9600 1920 = 7680

[3rd year is 6144; 4th yr is 4915.20]

7680 3 M2 for 12000 0.82 or 12000 0.83

A1 cao

ORM1 12000 0.8 oe or 9600 or 2400 or 4800

or 7200 seen

M1(dep) 9600 0.8 oe

A1 cao

(if correct answer seen, ignore extra years)

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11 (a) 2a+4c 1 B1 2a+4c or 2(a +2c)

(b) (3)2 = 9 = 1.125 1.125 2 M1 for substitution: 32 oe

A1 1.125, 18

1,

8

9oe

(c) x(x-5) 2 B2 , accept )5( +xx

(B1 forx(linear expression inx) orx-5 seen)(d) x

2 + 3x + 4x + 12 x2+7x+12 2 B2 for fully correct

(B1 for 3 out of 4 terms correct in working

including signs, OR 4 terms correct, with

incorrect signs).

(e) (y+3)(y+5) 2 B2 for fully correct

(B1 for ))(( byay ++ with one ofab = 15,

a+b = 8)

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12 (a) 91 85 6100 100

85 85

= = 7.05882..

7.06% 3M2 100

85

8591

(M191 85

85

or sight of

6

85or 0.0705

0.071 or 85

91or 1.0705 1.071)

A1 7.05 7.06

Or

M1 10085

91 (=107.05)

M1 (dep) 107.05 -100A1 7.05-7.06

T&I methods must lead to an answer 7.05

7.06 for full marks, otherwise 0 marks

(b) (64+73+85)/3 = 222/3 = 74

(73+85+91)/3 = 249/3 = 83

74, 83 2 M1 for (64+73+85)/3 or (73+85+91)/3 or

222/3 or 249/3 or 74 or 83 (condone missingbrackets)

A1 both answers in the correct order cao

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13 (a) 42 10 = 502.65(502-503)

503 2 M1 42 10 (=502.65)A1 502-503

SC B1 82 10(b) 502.65 0.6 = 301.59 302 2 M1 502.65 0.6

A1 300 302 ft on 502.65 to an answer

which would be correct on ft if rounded ortruncated to 3SF

14 (a) 7 8 = 56 = 28 28 2M1 7 8 or

2

1 7 8 sin 90o

A1 cao

(b) 82 + 72

64 + 49 = 113113 = 10.630145

10.63 3 M1 82 + 72 or 64+49 or 113

or 8

2

+ 7

2

90cos872

M1 (64+49) or113 where it is clearthat the 8 and 7 have been squared

A1 Any answer in 10.63 10.631 inclusive

SC B1 10.6 with no working with or without

a scale drawing

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(c) tany = 32/46 = 0.6956

tan-1 0.6956 = 34.82

34.8 3M1 tan (y = )

46

32

M1 tan-1 0.695(6) or tan-1

46

32or tan-1

46

32oe

(e.g.shift tanor inv tanfor tan-1)

A1 34.79 34.85o

Or

M1 for )4632( 22 + (=56.03(5..)) and either

32

sin

..)0.(56

90sin y= or

ysin

32

90sin

..)0.(56=

M1

..)06(571.0(sin(...)0.(56

90sin32sin)( 11 =

=y

A1 34.79 34.85o

SC1 B2 Radians 0.607-0.608

B2 Gradians 38.65 38.7

(both using tan)

Alternative methods using Pythagoras andthen sin or cos must have a fully correct

method for Pythagoras and sin/cos before

they score the first M1. The trigonometry

could be SOHCAHTOA or Sine rule/Cosine

rule

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15

B at ( 2, 1), ( 4, 1)

( 2, 4)

C at (4, 1), (6, 1), (4, 4)

Rotation

180

about (1,0)

3 B1 for rotation

B1 for 180

B1 for centre (1,0)

OR

B1 Enlargement

B1 Scale Factor1 Accept 1 on its own if it

is clear candidate is describing an enlargement

B1 Centre (1,0)

Ignore diagram unless no marks scored, in

which case

SC B1 for showing both B and C correctly

NB Award no marks for the description if more

than one transformation is given

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16 (a) 2)252

2)126

3) 63 or factor trees

3) 21

7) 7

1

22337 3 M1 for attempt at continual prime factorisation(at least 2 correct steps); could be shown as a

factor tree

OR sight of at least one each of 2, 3, 7 as

factors of 252

A1 for a fully correct factor tree or 2, 2, 3, 3, 7which may include 1, but no other numbers

A1 22337 or 22327 oe(b) HCF: The numbers must be 3n

and 3m where n and m are

coprime and at most one is a

multiple of 3

LCM: Factors of 45 are 1, 3, 5,

9, 15, 45

9 and 15

or 3, 45

3 B3 cao

(B2 for 2 numbers with HCF of 9 or LCM of

15)

(B1 for any attempt to list any 4 factors of 45 or

any 4 multiples of 3).

17 20 1.51 1026 3.021027 2 M1 20 1.51 1026 or 3.02 10n

or 30.21026

where n is a positive integer

A1 cao

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5540H/4H

Question Working Answer Mark Notes18 Region

indicated

3 M1 Bothx=2 drawn from at least (2,1) to (2,4)

and y= 1 drawn from at least (2,1) to (5,1)

M1 forx+y = 6 drawn from at least (2,4) to

(5,1)

A1 Correct region indicated by shading or

clearly labelled. Boundaries of the region may

be solid or dashed.

19 2150 13360

= 9291585.530641.0.

= 221.22

221 2M1 for 2

15013

360 or 4.2132 oe

A1 220 - 222

20 (a) q = k ; 8.5 = k

t2 42

k= 8.5 42; k= 136

q = 136

t2

3 M1 q = k, (k 1 )t2

M124

5.8k

=

A1 cao

NB q = k in the answer line followed by k

t2

being found correctly anywhere in (a) or (b)

earns all 3 marks

(b) q = 136 52 = 136 25 5.44 1

B1 ft for25

'136'oe

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21 (a)

Freq = FD int width = 0.018 1000 = 18

Or = 18 1 = 18

= 0 .010 2000 = 20 or 10 2 = 20

= 0.006 2000 = 12 or 6 2 = 12

OR

No of small squares = 200 Total freq = 16

So 1 small square = 16 200 = 0.08

9 25 0.08 = 18

10 25 0.08 = 20

6 25 0.08 = 12OR 8 cm2 = 16 so 1 cm2 = 2 etc

18,20,12 2 M1 use of Freq = FD int width

or attempt to find freq of 1 standard square

(or one answer correct)

A1 cao: all three

(b) FD = Freq int width = 16 2000 = 0. 008 so

4 sqs up

= 8 4000 = 0.002 so 1 sq up

OR

16 0.08 = 200 200 25 = 8 so 4 sqs up8 0.08 = 100 100 25 = 4 so 1 square up

OR 16 2 = 8 so 4 sqs up etc

4000-6000

4 cm high

8000-12000

1 cm high

2 B1 4000-6000; 4 cm high

B1 8000-12000; 1 cm high

or

if B0, M1 use of Freq = FD int width

or attempt to find freq of 1 standard square

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22 238 has an UB 238.5, a LB of 237.5

27.3 has an UB of 27.35, a LB of 27.25

Upper: 238.5 = 8.75229

27.25

8.75 3 B1 for one of 238.5, 237.5, 27.35, 27.25,

94.238 ,

934.27 seen

M1 for UB no of milesLB no of litresWhere 238< UB no of miles238.5 and27.25 LB no of litres

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23 (a) 5(x 1)=(4 3x)(x + 2)

5x 5 = 4x + 8 3x2 6x (= 2328 xx )(3x

2+ 6x + 5x 4x 5 8 = 0)

3x2 + 7x 13 = 0

Proof 3 M1 multiply through by )2)(1( + xx and

cancel correctlyM1 expand 5(x 1 ) and (4 3x)(x + 2)

correctly, need not be simplified

A1 rearrange to give required equation (dep

on both Ms and fully correct algebra)(b) a = 3, b = 7, c= 13

x=-7(72+4313) = -7(49+156) = -72056 6 6

x = 1.2196 or 3.55297.

Or

03

13

6

7

6

722

=

+x

3

13

6

7

6

72

+

=

+x

x = 1.2196 or -3.55297.

1.22

-3.55

3 M1 correct substitution in formula ofa = 3, b

= 7 and c = 13

M1 reduction to6

2057

A1 1.215 to 1.22 and 3.55 to 3.555

Or

M1

2

6

7

+x

M136

205

6

7

A1 1.215 to 1.22 and -3.55 to -3.555

SC T&I 1 mark for 1 correct root, 3 marks for

both correct roots

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24

10122

151012cos

222

+=x =

240

19

x = cos-1 0.079 = 85.459

OR

152 = 122 + 102 21210cosx

cosx = 152 122 102 = 122 + 102 152 = 19

-21210 21210 240

x = cos-1 0.079 = 85.459

85.5 3M2

10122

151012cos

222

+=A

A1 85.4 -85.5

OR

M1 correct substitution into

Abccba cos2222 += M1 correct rearrangement of

Abccba cos2222 += algebraically to

cb

acbA

+=

2)(cos

222

oe

or to

(cosA =)10122

151012 222

+

oe

These can be earned in either order

A1 85.4-85.5

SC B2 Radians 1.49 seen

B2 Gradians 94.89-95 seen

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25 7 = ka1 ; 175 = ka3

k= 7 , 175 = 7a3 , 175 = 7a

2

a a

a2 = 25, so a = 5, k=1.4

Or3333

175,7 kaak ==

175

732 =k , k=1.4, a = 5

k= 1.4

a = 5

3 M1 eithera2 = 25

or 7 = ka (or 7 = ka1) and 175 = ka

3

A1 k= 1.4 oe

A1 a = 5

SC Eithera = 5 ork= 1.4 oe gets B2

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5540H_4H_rms_20080814