2012 Edexcel Higher C Paper 1 Mark Scheme

5/24/2018 2012 Edexcel Higher C Paper 1 Mark Scheme

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For Edexcel

GCSE Mathematics

Higher Tier

Paper 1C (Non-Calculator)

Marking Guide

Method marks (M) are awarded or knowing and using a correct method!

"ccurac# marks (") are awarded or correct answers$ ha%ing used a correct

method!

(&) marks are independent o method marks!

(C) marks are or communication!

'ritten # haun "rmstrong

*nl# to e copied or use in the purchaser+s school or college

,1, E.C Paper 1 marks Page 1 / Churchill Maths 0imited


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Higher Tier Paper 1C Marking Guide

*1 each strip is 50 cm wide6 m = 12 strips so 6.5 m = 13 strips M1a 10 m roll will give 3 strips of 3m B14 rolls gives 4 3 = 12 strips

he does not have enogh rolls !1 "otal 3

[ clear reasoning leading to correct answer ]

*2 e.g. mst have # on "es and B on $at B12 staff for each of 6 da%s = 12 da%s wor&each mem'er of staff wor&ing 2 da%s = 10 da%s wor&2 (e)tra* da%s wor& M1e)tra mst 'e # or B as the% mst total at least 6 da%scheapest if B wor&s e)tra da%s #1

Mon # and +"e # and +,ed B and !"h B and !-ri B and $at B and / lots of other soltions possi'le !2 "otal 5

[ communication of how criteria are met by writing, table with days crossed out etc. 2 marks for meeting all criteria, 1 mark if one not met ]

3 a -alse B1

' !annot tell B1

c "re B1 "otal 3

4 area # = 5 1 1 = 35 = 2 cm2 M1 #1

area B = 6 5 1

2 6 1

1

2 6 2 M1

= 30 3 6 = 21 cm2

area # area B = 2 21 = 7 M1 #1 "otal 5

5 14

8 1 512

= 312

8 1 512

= 1 12

/ = 1 23

M1 #1

12

38 6

1

2= 1

4

68 6

3

6= 8 1

1

6=

1

6M1

fraction wal&ing = 12

39

1

6=

5

39

47

6M1

=5

3

6

47=

5

1

2

47=

10

47M1 #1 "otal 6

,1, E.C Paper 1 marks Page , / Churchill Maths 0imited


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6

constrcting perpendiclar 'isector ofPand Q M1 #160 &m = 3 cm on map B1circle: radis 3 cm: centreR M1showing possi'le positions of plane #1 "otal 5

7 a

B1

,1, E.C Paper 1 marks Page / Churchill Maths 0imited

P

Q

R

;lane will 'e in one ofthese two positions

6 cm


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' area of


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10 a smallest positive when n= M1vale = 2 6 = 14 #1

' difference = 40thterm =11 4 = nth = 4n8 M1 #1 "otal 4

11 ; =1

2: ;; =

1

3: ;> =

1

6B1

; =1

2

1

2=

1

4M1

;;; =1

3

1

3=

1

7: ;>> =

1

6

1

6=

1

36

;conter =1

48

1

78

1

36=

14

36/ =

1 M1 #1 "otal 4

12 a 6x? 15 M1x? 2.5 #1

@'

y= 0 and x= 6 B1y=x 2 M1A on correct side of lines #1 !1 "otal 6

[ diagram comleted to an accetable standard, allow one error ]


1

2

4

3

5

y

1 2 x1

4

2

3

10 3 4 5 6

A


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13 a = 64 40 = 24 B1

' 4: 13: 30: 51: 6: 4 B1

c M1 #1

d see graph M1 #2

@e e.g. the average performance had improved as themedian was 56: 4 more than on the first moc& !1the mar&s were slightl% less spread ot with theCDA 'eing 20: 4 less than on the first moc& !1 "otal 7

[ rele!ant comarisons "uoting !alues ]

14 a 4 8 5 8 2 = 1122000 9 11 = 2000 M1mortgage = 4 2000 = E000 #1

per month = 000 9 12 = 2000 9 3 = E66 nearest E #1

' 110 = 22000 B110 = 22000 9 11 = 2000 M1100 = 10 2000 = E20000 #1 "otal 6

15 a 2x x8 3 M1 #1

' =53 x 6 3x 3

35M1 #1

=15x 30 3x 7

15=

12x 21

15=

4x

5M1 #1 "otal 6


Mar& 20 4030 50 60 0

!mlative-reFenc%

0

20

40

60

0

0

Mar&

20 30 5040 60 0 0


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16 a p8 q reglar he)agon = 6 eFilateral triangles B1

' =#$ 81

2$%

= p81

2p8 q =

3

2p8

1

2q M1 #1

c e.g.

&'=

&(8

() 8 k

)% = p8 q8 p kq: where kis a fraction M1 = 2p8 1 kq

parallel to#*so a mltiple of3

2p8

1

2q M1

coefficient of q=1

3of coefficient of p

so 1 k =2

3M1

&'= 2p82

3q #1 "otal

17 e.g. 2n8 1 and 2n8 32n8 32 2n8 12 M1= 4n28 12n8 7 4n28 4n8 1 M1= n8 #1= n8 1 which is a mltiple of #1 "otal 4

18 a y= kx2 M17 = k 62= 36k M1

k= 7 9 36 =1

4

y=1

4x2 #1

' y=1

4 102= 25 B1 "otal 4

19 a let angle$#%=xangle#$+= angle#%+= 70G /tangent and radisangle$+%= 360 70 8 70 8x M1

= 10 xangle$+(= 10 10 x =x= angle$#% #1

' angle#$%=1

210 x = 70

1

2x /isosceles triangle M1

let dotted lines meet at

angle&$= 70 70 1

2x =

1

2x M1

angle(&+= angle#$%= 70 1

2x

angle$&= angle(&+= 70 1

2x /opposite

angle&$8 angle$&=1

2x8 70

1

2x= 70

angle$&= 10 70 = 70 M1hence:(&e)tended is perpendiclar to$% #1 "otal 6

TOT! "O# PP$#% 100 M#&'


2012 Edexcel Higher C Paper 1 Mark Scheme

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