mark scheme sample assessment material - kesh · pdf filemark scheme sample assessment...

Mark Scheme

Sample Assessment Material

GCSE

GCSE in Mathematics Specification A Higher Tier Paper 2: (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 candidate’s 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

155


156


157

1MA

0/2H

Q

uest

ion

Wor

king

Ans

wer

Mar

kA

ddit

iona

l Gui

danc

e 5.

(a

) 0:

8

1: 0

2357

8 2:

012

2233

3:

134

5 4:

456

Key

4 I

6 m

eans

46

min

utes

Corr

ect

stem

an

d le

af3

B3 F

ully

cor

rect

(B

2 Al

l ent

ries

cor

rect

, no

key

) (B

1 co

rrec

t en

trie

s un

orde

red,

key

or

no k

ey)

OR

(B2

Thre

e ro

ws

corr

ect,

key

or

no k

ey)

(B1

Two

row

s co

rrec

t, k

ey o

r no

key

)

(b

) O

ld m

edia

n =

22

New

med

ian

= 22

+ 5

27

min

utes

2

M1

find

s m

edia

n co

rrec

tly

for

orig

inal

dat

a an

d ad

ds 5

A1

cao

O

RM

1 Re

does

tab

le (

ft)

wit

h ea

ch v

alue

incr

ease

d by

5 a

nd a

ttem

pts

to

find

med

ian

A1 c

ao

(c

)

The

sam

e +

reas

on1

C1 A

ll th

e va

lues

hav

e in

crea

sed

by 5

min

utes

so

whe

n yo

u su

btra

ct t

he

5 m

inut

es w

ill c

ance

l out

.

Tota

l for

Que

stio

n: 6

mar

ks


158

1MA

0/2H

Q

uest

ion

Wor

king

Ans

wer

Mar

kA

ddit

iona

l Gui

danc

e 6. FE QW

Cii,

iii

(a)

1 ga

llon

= 4.

54 li

tres

,

200

gal

lons

= 9

08 li

tres

=

908

000

cm3

Vol

of

tank

60

2xπ

180

=

2035

752.

04..

cm3

9080

00<

1017

876.

02

OR

Vol o

f ta

nk

602

π 1

80 =

203

5752

.04.

.cm

3

Hal

f vo

l of

tank

=

1017

876.

02 c

m3

= 10

17.8

76…

litre

s

1017

.876

÷ 4

.54

= 22

4 ga

llons

224

> 20

0

No

5 Re

spon

se m

ay c

onve

rt in

to g

allo

ns,

litre

s, o

r cm

3

Calc

ulat

ions

may

be

perf

orm

ed in

dif

fere

nt o

rder

s

M1

Usi

ng f

orm

ulae

to

find

vol

ume

of t

ank

B1 C

onve

rts

betw

een

litre

s an

d cu

bic

cent

imet

res

M1

read

s of

f gr

aph

for

1l,

2l ,

4l,

5l o

r 10

litr

es w

ithi

n to

lera

nce

(4.4

—

4.6)

A1 A

nsw

er in

cm

3 , lit

res

or g

allo

ns

C1 D

ecis

ion

and

reas

on Q

WC:

Dec

isio

n sh

ould

be

stat

ed,

wit

h ap

prop

riat

e su

ppor

ting

sta

tem

ent

(b

) “

9080

00”

cm3 ×

0.8

5 g/

cm3

= 77

1800

g

771.

8 3

M1

“908

000”

× 0

.85

M

1(de

p) 7

7180

0÷10

00

A1 7

70 —

772

To

tal f

or Q

uest

ion:

8 m

arks

7.

120

5522

524

515

045

106.

08 h

ours

4

M1

for

mid

inte

rval

val

ues

M

1 fo

r m

ulti

plyi

ng f

requ

enci

es b

y m

id-i

nter

val v

alue

s M

1 fo

r ad

ding

(fr

eq

mid

-int

erva

l val

ues)

÷ 1

20

A1 c

ao

Tota

l for

Que

stio

n: 4

mar

ks


159

1MA

0/2H

Q

uest

ion

Wor

king

Ans

wer

Mar

kA

ddit

iona

l Gui

danc

e 8.

(a

) Fr

ed p

ays

3x a

nd J

im

pays

210x

Mal

colm

get

s £1

70 f

or F

red

and

Jim

, so

Mal

colm

get

s

170

2103

xx

Clea

r an

d co

here

ntex

plan

atio

n

1 C1

a c

lear

and

coh

eren

t ex

plan

atio

n

(b

) Fr

ed h

as

32x

left

, so

sol

ving

for

xus

ing

170

2103

xx 2x

+ 3(

x –

10) =

170

6

5x

= 1

050

x

= 2

10O

R

6)

10(3

2210

3x

xx

x

170

630

5x

5

x =

105

0

21

0x

£140

4

M1

mul

tipl

y th

roug

h by

6 a

nd c

ance

ls f

ract

ions

M

1 (d

ep)e

xpan

d 3(

x−

10)

M1

(dep

)col

lect

ter

ms

on e

ach

side

cor

rect

ly

A1 c

ao

OR

M1

colle

cts

term

s ov

er 6

M

1(de

p) e

xpan

d 3(

x−

10)

M1(

dep)

mul

tipl

y th

roug

h by

6 a

nd c

olle

ct t

erm

s

A1 c

ao

Tota

l for

Que

stio

n: 5

mar

ks


160

1MA

0/2H

Q

uest

ion

Wor

king

Ans

wer

Mar

kA

ddit

iona

l Gui

danc

e 9.

QW

Ci,

iii

FE

M

akes

a c

ompa

riso

n of

the

sh

ape

of t

he d

istr

ibut

ion

by

draw

ing

M

akes

a c

ompa

riso

n of

the

m

odal

cla

sses

(31—

40,

11—

20)

Mak

es a

com

pari

son

of t

he

clas

s in

terv

als

that

con

tain

th

e m

edia

ns.(

31—

40,

21—

30)

Wor

ks o

ut a

n es

tim

ate

of t

he

tota

l sal

es o

f ea

ch s

hop(

2635

, 35

30)

Corr

ect

com

pari

sons

4

B1,

B1,

B1 f

or a

ny 4

of

the

follo

win

g do

ne c

orre

ctly

Plot

s fr

eque

ncy

poly

gon

or p

rodu

ces

tabl

e co

mpa

res

mod

es

com

pare

s m

edia

ns

com

pare

s to

tal s

ales

C1 f

or c

omm

ents

on

shap

e of

the

dis

trib

utio

ns

QW

C: D

ecis

ions

sho

uld

be s

tate

d, a

nd a

ll co

mm

ents

sho

uld

be c

lear

an

d fo

llow

thr

ough

fro

m a

ny w

orki

ng o

r di

agra

ms

Tota

l for

Que

stio

n: 4

mar

ks


161


162

1MA

0/2H

Q

uest

ion

Wor

king

Ans

wer

Mar

kA

ddit

iona

l Gui

danc

e 11

. (a

)

28

1 B1

27

— 2

9

(b

) 68

– 4

2 26

2

M1

68 —

42

A1 2

6 —

30

(ne

ed 21

sq

tole

ranc

e on

eac

h)

FE(c

) 15

% of

80

= 12

Ye

s, w

ith

corr

ect

conc

lusi

on

2 M

1 lo

oks

up 6

8 or

40

min

on

cum

ulat

ive

freq

uenc

y A1

cor

rect

con

clus

ion

Tota

l for

Que

stio

n: 5

mar

ks

12.

QW

Cii,

iii

FE

sin

68o =

5.8AC

AC =

8.5

× s

in 6

8o = 7

.881

7.

881

+ 1

< 9

Reas

onsu

ppor

ted

byca

lcul

atio

n

4M

1 si

n 68

o =

5.8AC

M1

AC =

8.5

× s

in 6

8o

A1 7

.88(

1…

C1 8

.88(

1… +

con

clus

ion

QW

C: D

ecis

ion

shou

ld b

e st

ated

, su

ppor

ted

by

clea

rly

laid

out

wor

king

Not

e

90si

n5.868

sinAC

does

not

get

mar

ks u

ntil

in t

he f

orm

68si

n90

sin5.8

AC

Tota

l for

Que

stio

n: 4

mar

ks


163

1MA

0/2H

Q

uest

ion

Wor

king

Ans

wer

Mar

kA

ddit

iona

l Gui

danc

e 13

.A

kT

; 40

=

100

kk

= 4

AT

460

4T

31.0

4

M1

Ak

TM

1 40

=

100

k

A1A

T4

A1 f

or 3

0.98

… o

r 31

(.0)

OR

M2

for

40T =

10

060

oe

M1

for

100

6040

T o

e

A1 f

or 3

0.98

… o

r 31

.0

Tota

l for

Que

stio

n: 4

mar

ks


164


165

1MA

0/2H

Q

uest

ion

Wor

king

Ans

wer

Mar

kA

ddit

iona

l Gui

danc

e 16

. (a

) Vo

l =

2)

2(x

x=

51

Vol =

x

x4

22

– 51

=0

Der

ives

giv

en

answ

er a

nd

cond

itio

n

4M

1 Vo

l =

2)2

(xx

M1

expa

nds

brac

ket

corr

ectl

y A1

(E1

) se

ts e

qual

to

51

B12

x a

s th

e le

ngth

s of

the

cub

oid

have

to

be p

osit

ive.

(b

)

22

)51

(2

4)4

()4

(2

x

4424

4x

6.15

, -4

.15

both

to

3sf

3 M

1 co

rrec

t su

bsti

tuti

on (

allo

w s

ign

erro

rs in

a,

b an

d c)

into

qua

drat

ic

form

ula

M1

442

44

x

A1 6

.14(

7…, −

4.14

(7…

)

Tota

l for

Que

stio

n: 7

mar

ks

17.

An

gle

BAC

= 18

0º —

47º

— 5

8º

= 75

º

)58

sin

(75

sin22

047

sin

ABAC

AC

=75

sin

47si

n22

0 =

166

.57.

.

Area

=58

sin

'57.

166

'22

021

= 15

538

1550

0 m

2 5

B1 f

or 7

5º

M1

)58

sin

(75

sin22

047

sin

ABAC

M1

AC

=75

sin

47si

n22

0

M1

21 ×

220

× “

166.

57”

× si

n58

A1 1

5500

m2

Tota

l for

Que

stio

n: 5

mar

ks


166

1MA

0/2H

Q

uest

ion

Wor

king

Ans

wer

Mar

kA

ddit

iona

l Gui

danc

e 18

.

Pent

agon

= 5

equ

al is

os

tria

ngle

s

5360

=72º

Base

ang

les

= (1

80 –

72)

2

=

54º

for

find

ing

equa

l sid

es o

f is

osce

les

tria

ngle

;

72si

n1054

sin

x=

8.50

6508

084…

area

of

isos

cele

s tr

iang

le =

72si

n21

2x

= 34

.409

5480

1..

area

of

pent

agon

=

5 3

4.40

9548

01

= 1

72.0

4774

01

area

of

dode

cahe

dron

=

12

172.

0477

401

OR

Usi

ng r

ight

-ang

led

trig

onom

etry

; h

= 5t

an54

º

= 6.

8819

…

Area

of

isos

cele

s tr

iang

le =

21 1

0 h

= 34

.409

5480

1…ar

ea o

f pe

ntag

on

= 5

34.

4095

4801

=

172.

0477

401

area

of

dode

cahe

dron

=

12

172.

0477

401

2065

cm

2 9

B1 f

or

5360

= 72

º

B1 (

180

–72

) 2

= 5

4º

M1

for

find

ing

equa

l sid

es o

f is

osce

les

tria

ngle

; x

=72

sin10

54si

nx

A1 f

or x

= 8

.506

5080

84…

M1

for

find

ing

area

of

isos

cele

s tr

iang

le =

72

sin

212 x

A1 f

or 3

4.40

9548

01…

(ft)

B1

for

are

a of

pen

tago

n =

5 (

ft)

= 17

2.04

7740

1…(f

t)

B1 f

or a

rea

of d

odec

ahed

ron

= 12

(

ft)=

206

4.57

2881

… c

m2

A1 f

or 2

065

cm2

(oe)

OR

B1 f

or

5360

= 72

º

B1 (

180

– 72

) 2

= 5

4 º

M

1 fo

r us

ing

righ

t-an

gled

tri

gono

met

ry;

h =

5 ta

n54º

A1

for

6.8

819…

M

1 fo

r fi

ndin

g ar

ea o

f is

osce

les

tria

ngle

= 21

10

h

A1 f

or 3

4.40

9548

01…

(ft)

B1 f

or a

rea

of p

enta

gon

= 5

(ft

) =

172.

0477

401…

(ft)

B1

for

are

a of

dod

ecah

edro

n =

12

(ft

) =

2064

.572

881…

cm

2

A1 f

or 2

065

cm2

(oe)


167


168

9.

102030405060

1020

3040

5060

0C

Ds s

old

Freq

uenc

y


169

14.

2468

24

0x

y

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mark scheme sample assessment material - kesh · pdf filemark scheme sample assessment...

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