2012 Edexcel Higher C Paper 2 Mark Scheme

8/12/2019 2012 Edexcel Higher C Paper 2 Mark Scheme

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For Edexcel

GCSE Mathematics

Higher Tier

Paper 2C (Calculator)

Marking Guide

Method marks (M) are awarded for knowing and using a correct method.

ccurac! marks () are awarded for correct answers" ha#ing used a correct

method.

($) marks are independent of method marks.

(C) marks are for communication.

%ritten &! 'haun rmstrong

nl! to &e copied for use in the purchasers school or college

2*+2 E,C Paper 2 marks Page + - Churchill Maths imited


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Higher Tier Paper 2C Marking Guide

1 total spent = 315 + 120 = 435 B1total extra = 16 40 = 640 M1 A1it was a success as the extra he earned was oer 200 !orethan what he spent A1 "otal 4

2 #a$ positie B1

#%$ e&'& on lon'er (ourne)s she can use *aster roads !ore B1

#c$ B1

#d$ ,5 6-&5

200 40=

15&5

160= 0&0-. #2s*$ M1 A1

/#e$ it is the increase in aera'e speed in !h *or eachextra ! drien !ain' the delier) 2 "otal .

[ clear valid explanation ]

3 #a$ y10 B1

#%$ = 5m2 10m 2m2 3m M1= 3m2 13m A1

#c$ #3 +p$#3 p$ B1

#d$ 3x2= 4,x2= 16 M1

x= 16 = 4 A1

#e$ 3a#2a+ b$ M1 A1 "otal ,

2*+2 E,C Paper 2 marks Page 2 - Churchill Maths imited

istance #!$0 50 150100 200 250

Aera'e

speed

#!h$

.0

,0

.5

,5

-0


3/7

4 B = Blac 7 = 7hite 8 = 8re)9 = sa!e colour

B

7 8 7 B 8

A

B 97 9 9

7 9 9

8 9 9

7 9 9

25 outco!es - sa!e colour M2 A1

pro%a%ilit) =-

25A1 "otal 4

*5 :* ;aee! 'ies 3 to


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7 #a$ a*e ;ands 0&02 300 = 6 M1cost = , + 6 = 14 A1

@ast 7heels 0&03 300 = -cost = 4 + - = 13 #a%oe !ini!u!$

14 13 = 1 so @ast 7heels is 1 cheaper A1

/ #%$ e&'&

@ast 7heels is cheaper *or ite!s o* %etween 100 and 400 in alue&a*e ;ands is cheaper *or ite!s o* less than 100 or !ore than 400 in alue&

#"he) cost the sa!e *or ite!s o* exactl) 100 or 400 alue$&

'raphs or releant sets o* alues M2 A1report 2 "otal ,

[ coherent report summarising their findings (shown by graph / table etc ]

8 #a$ translation %) 15 B2

#%$ rotation -0 anticlocwise a%out #0 0$ B3

#c$ #2 3$ B1 "otal 6


0 100 300200 400

"otal cost #$

0

5

15

10

20

Calue o* ite! #$

a*e ;ands

@ast 7heels


5/7

9 #a$ = 0&3 320 = -6 M1 A1

#%$ x= 1 0&,5 = 0&15 B12x= 0&3

y= 1 #0&15 + 0&3 + 0&3$ = 1 0&.5 = 0&25 M1 A1 "otal 5

10 #a$ e&'& olu!e o* *ront cu%oid = 3 4 x= 12x M1olu!e o* rear cu%oid = 2xxx= 2x3

total olu!e = 2x3+ 12xso 2x3+ 12x= 200 M1hence x3+ 6x= 100 A1

#%$ e&'& x= 1 x3+ 6x= . too s!allx= 5 x3+ 6x= 155 too %i'x= 4 x3+ 6x= ,, too s!all M1x= 4&5 x3+ 6x= 11,&125 too %i' A1x= 4&3 x3+ 6x= 105&30. too %i'

x= 4&2 x3

+ 6x= --&2,, too s!allx= 4&25 x3+ 6x= 102&265&&& too %i' M1x= 4&2 #1dp$ A1 "otal .

11 #a$

M2 A1

#%$ see dia'ra! B1

#c$ 'radient =1

2yDintercept =

1

2M1

eEuation is y=1

2

x+1

2A1 "otal 6


y

x!

1

2

4

1 2 32 1

1

3

5


6/7

12 let radius o* circle %e rc!side o* sEuare = 2rr+ 2r= 12 M13r= 12r= 4 A1two se!icircles !ae a circle

area = F 4

2

+ ,

2

= 114 c!

2

#3s*$ M1 A1 B1 "otal 5

13 Gption 1 returns #1&035$3 500 = 554&36 M2 A1Gption 2 returns #1&033$2 1&041 500 = 555&42 M1Gption 2 is %est A1 "otal 5

14 #a$ 6, H wI .2 *reE dens = 16 J 4 = 4 M1so one lar'e sEuare on ertical axis = 1.2 H wI ,0 *reE dens = 16 J , = 2 M1,0 H wI -0 *reE dens = . J 10 = 0&.

A1

#%$ 40 H wI 60 *reE dens = 0&5 *reE = 0&5 20 = 10 M160 H wI 64 *reE dens = 3&5 *reE = 3&5 4 = 14 A1

7ei'ht #w'$ Ku!%er o*


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15 #a$ radius = 2 c!olu!e c)linder = F 22 , = 32F M1 A1

olu!e cone =1

3 F 22 4 =

16

3F M1

olu!e turret = 32F +16

3F =

112

3F or 11. c!3#3s*$ A1

#%$ = 1

2

3

112

3 F =

14

3 F or 14&. c!3

#3s*$ M1 A1 "otal 6

*16 104 -0 = 14 -0 53 = 3. 1,0 12- = 51 M1

cos 3. =1&,

"# so "#=

1&,

cos3.= 2&254 M1 A1

cos 14 =1&,

"$ so "$=

1&,

cos14= 1&,55

sin 51 =1

%&

so %&=1

sin51

= 1&2,. M1

sin 40 =1

'& so '&=

1

sin 40= 1&556

"$'&= 1&,55 + 2 + 1&556 = 5&411 ! M1"#%&= 2&254 + 2 + 1&2,. = 5&541 !5&541 5&411 = 0&130L oute"$'&%) 0&13 ! A1 1 "otal .

[ all parts of method clear and correct with a consistent conclusion, allow arithmetic errors ]

17 satellite distance H 3&65 104 B1!oon distance N 3&5 105 B1

satellite to !oon O 3&5 105 3&65 104 M1least possi%le = 3&135 105! P = 313500 ! Q A1 "otal 4

18 tan 21 =12

&M1

&=12

tan 21= 31&261 A1

&2=x2+x2= 2x2 M1

x2=&

2

2=

31&2612

2= 4,,&63 M1

x= 4,,&62. = 22&1 #3s*$ A1 "otal 5

TOTA !O" PAP#"$ 100 MA"%&


P

A

B

Q R

S

+ km

2 km

2 km

+.4 km0*5+05

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2012 Edexcel Higher C Paper 2 Mark Scheme

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