Mark Scheme Sample Assessment Material - TSSmathstssmaths.co.uk/Resources/smaths/ks4/1MA0Linear/spec/3983_GCSE … · Mark Scheme Sample Assessment Material GCSE GCSE in Mathematics

Mark Scheme

Sample Assessment Material

GCSE

GCSE in Mathematics Specification A Higher Tier Paper 1:(Non-Calculator)

Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WC1V 7BH

General Marking Guidance

All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark the last.

Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions.

All the marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should also be prepared to award zero marks if the candidates response is not worthy of credit according to the mark scheme.

Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may be limited.

Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.

Mark schemes will indicate within the table where, and which strands of QWC, are being assessed. The strands are as follows:

i) ensure that text is legible and that spelling, punctuation and grammar are accurate so that meaning is clear Comprehension and meaning is clear by using correct notation and labelling conventions. ii) select and use a form and style of writing appropriate to purpose and to complex subject matter Reasoning, explanation or argument is correct and appropriately structured to convey mathematical reasoning. iii) organise information clearly and coherently, using specialist vocabulary when appropriate. The mathematical methods and processes used are coherently and clearly organised and the appropriate mathematical vocabulary used.

Guidance on the use of codes within this mark scheme M1 method mark A1 accuracy mark B1 working mark C1 communication mark QWC quality of written communication oe or equivalent cao correct answer only ft follow through sc - special case

Edexcel GCSE Sample Assessment Materials Edexcel Limited 2009in Mathematics A

75


76

1MA

0/1H

Q

uest

ion

Wor

king

Ans

wer

Mar

kA

ddit

iona

l Gui

danc

e 3. FE

N

o of

tile

s ar

ound

roo

m

= 2

le

ngth

s of

roo

m =

8,

16,

16,

12

Tota

l num

ber

of t

iles

= 8

16

+ 8

1

2 =

224

Cost

= 4

2

24

OR

Area

of

the

room

=

4

8 +

4

6 =

56

Area

of

a ti

le

= 0.

5

0.5

= 0.

25

Num

ber

of t

iles

= 56

0.25

=

224

Cost

= 4

2

24

89

6 6

M1

for

doub

ling

each

leng

th t

o sh

ow n

umbe

r of

tile

s fo

r ea

ch s

ide

B1 f

or 8

, 16

, 16

and

12

M1

for

a fu

ll m

etho

d of

fin

ding

the

num

ber

of t

iles

(12

16

+ 8

4)

A1 f

or a

t le

ast

one

sec

tion

co

rrec

t M

1 fo

r 4

22

4

A1 c

ao

OR

M1

for

full

met

hod

for

find

ing

the

area

of

the

room

A1

at

leas

t on

e ar

ea c

orre

ct

B1 f

or a

rea

of t

ile =

0.2

5m2

or 2

500

cm2

or 4

tile

s =

1 m

2

M1

for

area

of

room

a

rea

of a

tile

M

1 fo

r 4

nu

mbe

r of

tile

s A1

cao

Tota

l for

Que

stio

n: 6

mar

ks

4.

(a)

5p=

20

p=

4 2

M1

add

16 t

o bo

th s

ides

A1

cao

(b

) 9

= 3

q q

=

32

M1

corr

ect

met

hod

to is

olat

e 3

qA1

cao

(c

) 6x

3 1

0

6x =

1

3 2

M1

at le

ast

one

expa

nsio

n co

rrec

t A1

13

or a

sta

tem

ent

that

the

ans

wer

is in

dep

of x

dep

endi

ng o

n co

rrec

t w

orki

ng

Tota

l for

Que

stio

n: 6

mar

ks


77


78

1MA

0/1H

Q

uest

ion

Wor

king

Ans

wer

Mar

kA

ddit

iona

l Gui

danc

e 8.

121

4x

x

215

x,

101x

OR

Choo

ses

a su

itab

le n

umbe

r of

ba

lls (

say

10)

5 w

ill b

e re

d Th

e ot

her

5 ne

ed t

o be

sh

ared

out

in t

he r

atio

1:4

, H

ence

1 y

ello

w a

nd 4

blu

e

1043

M1

121

4x

x

A1101

x

A1104

oe

Tota

l for

Que

stio

n: 3

mar

ks


79

1MA

0/1H

Q

uest

ion

Wor

king

Ans

wer

Mar

kA

ddit

iona

l Gui

danc

e 9.

(a

) (i

)

(ii)

a2 6x4 y

3

3 B1

cao

B26x

4 y3

(B1

for

2 ou

t of

3 t

erm

s co

rrec

t in

a p

rodu

ct)

(b

) x2

+ 3

x +

7x +

21

x2 +

10x

+ 2

1 2

M1

3 or

4 t

erm

s ou

t of

4 c

orre

ct in

a 4

ter

m e

xpan

sion

A1

cao

(c

)

3p(q

4p

) 2

B2 c

ao

(B1

p(3q

12

p), 1

2p( 41

q

p), p

(aq

+ bp

) whe

re a

and

b a

re n

umbe

rs)

(d

)(i)

(ii)

)32

(1

)2(3

xx

OR

320

1012

123

2x

xx

=5

23

2x

x

(3y

1)(y

3

) )1)(5

3(x

x

4 B2

cao

(B

1 (3

y

m)(y

n

) whe

re m

n =

3 o

r m

+ n

=

10

M1

use

of t

he f

acto

rise

d fo

rm w

ith

y re

plac

ed t

wic

e by

3x

+ 2

A1 c

ao

OR

B15

23

2x

xB1

cao

Tota

l for

Que

stio

n: 1

1 m

arks


80

1MA

0/1H

Q

uest

ion

Wor

king

Ans

wer

Mar

kA

ddit

iona

l Gui

danc

e 10

.

Reds

6,

12,

18,

24,

30

G

reen

s 9,

18,

27

201

3 B1

list

of

red

and

gree

n m

ulti

ples

(bo

th t

o at

leas

t 18

) or

exp

licit

ly

stat

es

LCM

B1

wor

ks o

ut h

ighe

st n

umbe

r (9

0 se

en)

B1201

(ac

cept

1005

)

Tota

l for

Que

stio

n: 3

mar

ks

11.

425x

695

xy

or6

59

xy

x =

2.5

y =1

1.25

4M

1 a

corr

ect

expr

essi

on f

or x

invo

lvin

g ra

tios

of

side

s, e

.g.

425x

oe

A1 c

ao

M1

695

xy

or6

59

xy

oe

A1 c

ao

OR

495y A1

cao

Tota

l for

Que

stio

n: 4

mar

ks


81

1MA

0/1H

Q

uest

ion

Wor

king

Ans

wer

Mar

kA

ddit

iona

l Gui

danc

e 12

.

(a)

4

6

8

10

6

8 1

0

12

8 1

0 1

2 1

4 10

12

14

16

OR 41

41

441

41

1643

M1

Atte

mpt

s to

list

all

outc

ome

pair

s A1

all

16 f

ound

A1

cao

OR

M2

441

41

(M1

2,1

4141

or 3

)

A1164

oe

(b

) Pr

ob A

li w

ins

= 166

Num

ber

of w

ins

= 80

166

30

3 B1

Pro

b Al

i win

s =

166 o

e

M1

80'

166 '

A1 f

t

Tota

l for

Que

stio

n: 6

mar

ks


82

1MA

0/1H

Q

uest

ion

Wor

king

Ans

wer

Mar

kA

ddit

iona

l Gui

danc

e 13

. (a

) 7

104.3

1 B1

cao

(b)

1005

104.2

12(

610

)5

102.1

2M

1 10

0510

4.212

oe (

610

)

A1 c

ao

Tota

l for

Que

stio

n: 3

mar

ks


83

1MA

0/1H

Q

uest

ion

Wor

king

Ans

wer

Mar

kA

ddit

iona

l Gui

danc

e 14

.

Let

AB =

x,A

D =

yAr

ea o

f re

ctan

gle

=xy

Area

AXD

=

4xy

Area

CYZ

=8xy

Shad

ed a

rea

= 85xy

854

M1

a fu

ll m

etho

d to

fin

d th

e un

shad

ed a

rea

and

subt

ract

ing

from

1

B1 a

rea

of A

XD =

are

a of

ABC

D4

B1 a

rea

of C

YZ=

are

a of

ABC

D8

A1 c

ao

OR

Dia

gram

M1

for

divi

ding

left

into

2 c

ongr

uent

tri

angl

es

f

or d

ivid

ing

righ

t in

to 4

con

grue

nt t

rian

gles

B1

left

= 2

Aan

d 2A

or

sha

ded

= 21

of

21=

41 =

82

B1 r

ight

= 2

A an

d A

and

A or

shad

ed =

43of

21 =

83

A1 c

ao

Subs

titu

tion

M1

for

deci

ding

upo

n su

itab

le s

ide

leng

ths

for

ADan

d AB

and

cal

cula

ting

di

men

sion

s of

inte

rnal

sha

pes

B1 f

or a

rea

of D

ZXB1

for

are

a of

ZXB

YA1

cao

OR

M1

for

deci

ding

upo

n su

itab

le s

ide

leng

ths

for

ADan

d AB

and

cal

cula

ting

di

men

sion

s of

inte

rnal

sha

pes

B1 f

or a

rea

ADX

B1 f

or a

rea

ZCY

A1 c

ao

Tota

l for

Que

stio

n: 4

mar

ks


84

1MA

0/1H

Q

uest

ion

Wor

king

Ans

wer

Mar

kA

ddit

iona

l Gui

danc

e 15

.

(a)

(i)

(ii)

BC =

O

BC

O

AQ =

BQ

OB

AO

=

4a +

4b

+ 41

(12a

4b

)

12a

4b

3b

a

4

M1

BC =

O

BC

OA1

cao

M1

4a

+ 4b

+ 41

(12

a

4b)

A1 c

ao

(b

) O

X=

12b

, AX

=4a

+ 1

2b=

4(a

+ 3

b)

Corr

ect

reas

on,

wit

h co

rrec

t w

orki

ng

3B1

OX

= 12

b

B1AX

=4a

+ 1

2bC1

con

vinc

ing

expl

anat

ion

Tota

l for

Que

stio

n: 7

mar

ks


85

1MA

0/1H

Q

uest

ion

Wor

king

Ans

wer

Mar

kA

ddit

iona

l Gui

danc

e 16

.

8596

104 =

720120

720

120

+

8495

106 +

8594

106

720

360

4M

1 fo

r 85

96104

A1 f

or 7

20120

oe

M1

720'

120

' +

2 c

orre

ct c

ases

(M

1 an

y 2

corr

ect

case

s)

or72

0'12

0'

X 3

A1 c

ao

SC w

ith

repl

acem

ent

M1

106106

104

M1

106106

1043

Tota

l for

Que

stio

n: 4

mar

ks

17.

)53)(5

3()7

)(53(

xx

xx

53

7xx

3B1

)7)(5

3(x

xB1

)53)(5

3(x

x

B15

37

xx

Tota

l for

Que

stio

n: 3

mar

ks


86

1MA

0/1H

Q

uest

ion

Wor

king

Ans

wer

Mar

kA

ddit

iona

l Gui

danc

e 18

. (a

)

211

B1

(b

) (2

+

3)

)31(

=9

33

22

33

52

M1

4 te

rm e

xpan

sion

wit

h 3,

4 t

erm

s co

rrec

t an

d si

ght

of 3

or

9A1

cao

Tota

l for

Que

stio

n: 3

mar

ks

19.

(a)

Sm

ooth

cu

rve

2 B1

cor

rect

plo

t of

the

ir v

alue

s B1

sm

ooth

cur

ve t

hrou

gh t

heir

poi

nts

(b

)

x =

3 y

= 0

3 M

1 at

tem

pts

to d

raw

cir

cle

at o

rigi

n M

1 us

es r

adiu

s 3

cm (

usin

g gr

aph

scal

e co

rrec

tly)

A1

cao

OR

B1 f

or s

ubst

itut

ing

a va

lue

of x

into

y =

x(x

3)

and

2

2r

yx

B1 f

or s

ubst

itut

ing

yin

tox

= 3

into

x(x

3

) and

2

2r

yx

B1 c

ao

Tota

l for

Que

stio

n: 5

mar

ks


87

1MA

0/1H

Q

uest

ion

Wor

king

Ans

wer

Mar

kA

ddit

iona

l Gui

danc

e 20

.Q

WC

ii, ii

i

22

)12(

)12(

nn

=)1

44(

14

42

2n

nn

n

=8n

OR

22

)12(

)12(

nn

=))1

21

2()12(

)12(

nn

nn

=2

4n

= 8

n

Fully

alge

brai

c ar

gum

ent,

se

t ou

t in

a

logi

cal a

nd

cohe

rent

man

ner

6 B2

the

nth

term

for

con

secu

tive

odd

num

bers

is 2

n

1 oe

(B

12n

+ k

,1

kor

n=

2n

1 o

r 2x

1B1

use

of

2n +

1 a

nd 2

n

1 oe

M

1 2

2)1

2()1

2(n

nM

1)1

44(

14

42

2n

nn

n

C1 c

oncl

usio

n ba

sed

on c

orre

ct a

lgeb

ra Q

WC:

Con

clus

ion

shou

ld b

e st

ated

, w

ith

corr

ect

supp

orti

ng a

lgeb

ra.

OR

B1 u

se o

f 2n

+ 1

and

2n

1 o

e M

1 2

2)1

2()1

2(n

nM

1))1

21

2()12(

)12(

nn

nn

C1 c

oncl

usio

n ba

sed

on c

orre

ct a

lgeb

ra Q

WC:

Con

clus

ion

shou

ld b

e st

ated

, w

ith

corr

ect

supp

orti

ng a

lgeb

ra. T

otal

for

Que

stio

n: 6

mar

ks


88

1MA

0/1H

Q

uest

ion

Wor

king

Ans

wer

Mar

kA

ddit

iona

l Gui

danc

e 21

.

L

F FD

CF

0

10

40

4 40

10

20

60

6 10

0 20

40

90

4.5

190

408

0 60

1.

5 25

0 >8

0 0

0 25

0

His

togr

amO

RCu

mul

ativ

e Fr

eque

ncy

poly

gon

82%

6 B1

Sca

les

labe

lled

and

also

mar

ked

on t

he v

erti

cal a

xis

wit

h fr

eque

ncy

dens

ity

or w

ith

cum

ulat

ive

freq

uenc

y M

1 fr

eque

ncy

dens

itie

s ca

lcul

ated

, at

leas

t on

e no

n-tr

ivia

l one

cor

rect

. A1

all

corr

ectl

y pl

otte

d (M

1 cu

mul

ativ

e fr

eque

ncie

s co

rrec

t)

M1

Use

50

on t

he h

oriz

onta

l sca

le o

f CF

dia

gram

rea

d of

f ve

rtic

al a

xis

(200

-210

)or

Use

50

on t

he h

oriz

onta

l sca

le o

f a

hist

ogra

m a

nd c

over

t ar

ea t

o th

e le

ft t

o a

freq

uenc

y M

1 co

nver

t to

a p

erce

ntag

e A1

80

85

Tota

l for

Que

stio

n: 6

mar

ks


89

2.

Frac

tion

D

ecim

al

%

kg

Jan

1010.

110

%N

ot k

now

n

Feb

810.

125

12.5

% 15

kg

Mar

100

130.

13

13%

14.5

6 kg

14.

AB C

D

X

Y

Z

AB C

D

X

Y

Z


90

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Documents

Mark Scheme Sample Assessment Material - TSSmathstssmaths.co.uk/Resources/smaths/ks4/1MA0Linear/spec/3983_GCSE … · Mark Scheme Sample Assessment Material GCSE GCSE in Mathematics