September 2015 - Maths · PDF file(bi) 13-17 P1 for process to draw ... September 2015 ©...

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Pearson Edexcel Level 1/Level 2 GCSE (9-1) in Mathematics -Specimen Papers Set 2 - September 2015 © Pearson Education Limited 2015

145145Pearson Edexcel Level 1/Level 2 GCSE (9-1) in Mathematics -Specimen Papers Set 2 - September 2015 © Pearson Education Limited 2015

Pape

r 1M

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stio

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otes

1Tr

ansl

atio

n by�4 −3�

B1

for t

rans

latio

n

B1

�4 −3�

2(a

)Tr

end

desc

ribed

C1

for “

perc

enta

ge o

f peo

ple

who

use

the

shop

de

crea

ses”

oe

(bi)

13-1

7P1

for p

roce

ss to

dra

w tr

end

line

on g

raph

A1

for 1

3 -1

7

(bii)

No

+ re

ason

C1

for c

omm

ent,

eg “

no, b

ecau

se 2

020

is b

eyon

d th

e tim

e pe

riod

cove

red

by th

e gi

ven

data

”

3(a

)13

y−

1M

1fo

r exp

ansi

on o

f one

bra

cket

A1

for f

ull

sim

plifi

catio

n

(b)

35u3 w

7B

1fo

r 2 o

f 35,

u3an

dw

7co

rrec

tB

1ca

o

410

5P1

for p

roce

ss to

find

the

exte

rior a

ngle

or i

nter

ior

angl

e of

a h

exag

on o

r oct

agon

P1fo

r pro

cess

to fi

nd th

e bo

th e

xter

ior a

ngle

s or

bo

th in

terio

r ang

les

A1

for 1

05 f

rom

cor

rect

wor

king


146146 Pearson Edexcel Level 1/Level 2 GCSE (9-1) in Mathematics -Specimen Papers Set 2 - September 2015 © Pearson Education Limited 2015

Pape

r 1M

A1:

2H

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stio

nW

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nsw

erN

otes

5(a

)(i)

10, 1

2, 1

4, 1

5, 1

6,

18B

1ca

o

(ii)

12, 1

8B

1ca

o

(b)

7 10M

1fo

r 7 o

r ind

icat

ing

corr

ect r

egio

n or

for 1

0, 1

4,

16, 1

1, 1

3, 1

7, 1

9 lis

ted

A1

for 7 10

oe

66

:5 =

12

:10

2 : 1

= 1

0 : 5

70P1

P1 fo

r stra

tegy

to st

art t

o so

lve

the

prob

lem

eg

12

: 10

and

10: 5

C :

S : P

= 1

2 : 1

0 : 5

P1P1

for p

roce

ss to

solv

e th

e pr

oble

meg

10 27 ×

189

10 27 ×

189

A1

A1

cao

71 4

× π

× 4.

8²6.

58B

1fo

r use

of f

orm

ula

for a

rea

of a

circ

le

1 2×

4.8

× 4.

8P1

for c

ompl

ete

proc

ess t

o fin

d ar

ea o

f sha

ded

regi

on1 4

× π

× 4.

8² −

1 2×

4.8

× 4.

8A

1fo

r 6.5

6 –

6.58



Pape

r 1M

A1:

2H

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stio

nW

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nsw

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otes

8(a

)ex

plan

atio

nC

1fo

r “in

corr

ect e

xpan

sion

of b

rack

ets”

oe

(b)

expl

anat

ion

C1

for “

has n

ot o

btai

ned

both

solu

tions

” oe

9(a

)18

B1

cao

(b)

5(x

–1)

M1

for m

etho

d to

find

inve

rse

func

tion

A1

for 5

(x–

1) o

r 5x

− 5

(c)

9x–

48 sh

own

M1

for m

etho

d to

find

com

posi

te fu

nctio

nA

1fo

r wor

king

lead

ing

to 9

x–

48

10(a

)15

6000

0 ×

(1.0

52)2

1730

000

P1fo

r pro

cess

to fi

nd p

opul

atio

n in

201

6P1

for c

ompl

ete

proc

ess t

o fin

d po

pula

tion

in 2

017

A1

for 1

7250

00 -

1730

000

(b)(

i)20

20P1

for p

roce

ss to

find

whe

n po

pula

tion

will

exc

eed

2 00

0 00

0A

1fo

r 202

0

(ii)

C1

for c

orre

ct c

omm

ent o

n ho

w a

ssum

ptio

n w

ill

affe

ct th

e an

swer

, eg

if th

e pe

rcen

tage

gro

wth

is

high

er th

e po

pula

tion

may

exc

eed

2 00

0 00

0 ea

rlier

.



Pape

r 1M

A1:

2H

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stio

nW

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otes

11(a

)0.

43M

1fo

r use

of g

raph

at 2

40 m

inut

esA

1fo

r 0.4

2 –

0.44

oe

(b)

com

paris

on

B1

for a

t lea

st o

ne m

edia

n (2

49 –

252

or 2

73 –

276)

B1

for l

east

one

inte

rqua

rtile

rang

e (6

9 –

73 o

r 67

-71)

C1

for c

omm

ent c

ompa

ring

aver

age

times

eg

fem

ales

take

long

er th

an m

ales

oe

C1

for c

omm

ent c

ompa

ring

spre

ads o

f tim

es fr

om

IQR

s, eg

the

spre

ad o

f tim

es is

abo

ut th

e sa

me

(NB

–at

leas

t one

of t

he c

omm

ents

mus

t be

in

cont

ext)

12(a

)25

× 2

460

0P1

for p

roce

ss to

find

num

ber o

f way

sA

1ca

o

(b)

12 ×

10

× 11

10

× 1

2 ×

913

20 +

108

0

2400

P1fo

r pro

cess

to fi

nd n

umbe

r of l

ists

with

boy

then

gi

rl th

en b

oy o

rthe

num

ber o

f lis

ts w

ith g

irl

then

boy

then

girl

P1fo

r com

plet

e pr

oces

s to

find

the

tota

l num

ber o

f lis

tsA

1ca

o



Pape

r 1M

A1:

2H

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stio

nW

orki

ngA

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otes

1311

9M

1fo

r 1.0

6 ×

100

oe

M1

for 1

.063

× 10

0oe

A1

acce

pt 1

19.1

016

14ex

plan

atio

nC

1fo

r a c

orre

ct e

valu

atio

n, e

g th

e va

lue

of D

shou

ld b

e m

ultip

lied

by 8

, she

has

use

d 2

× 3

inst

ead

of 2

³

15(a

)1.

0–

1.3

M1

for f

indi

ng g

radi

ent b

y dr

awin

g ta

ngen

tM

1fo

r met

hod

to c

alcu

late

gra

dien

tA

1Fo

r 1.0

–1.

3

(b)

C1

for a

ccel

erat

ion

C1

for e

g“

4 se

cond

afte

r the

star

t of t

he ra

ce”,

“w

hen

the

spee

d is

7.6

m/s

”, “

in m

/s2 ”

(c)

limita

tion

C1

for c

omm

ent,

eg d

epen

dent

on

accu

racy

of

cons

truct

ing

a ta

ngen

t

16(i)

200

B1

cao

(ii)

5.6

B1

For 5

.6(2

...)



Pape

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stio

nW

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otes

17�

8.35

²−6.

05²

5.75

4997

828

B1

for f

indi

ng b

ound

s of o

ne m

easu

rem

ent,8

.25

8.35

, 6.0

5 or

6.1

5P1

for p

roce

ss o

f cho

osin

g an

d us

ing

corr

ect

boun

dsP1

for p

roce

ss o

f Pyt

hago

ras’

rule

with

cor

rect

bo

unds

A

1fo

r 5.7

54(9

97…

)

18( √𝑎𝑎

+2 √𝑏𝑏)

( √𝑎𝑎−

2 √𝑏𝑏)

a–

4bM

1fo

r exp

ansi

on o

f bra

cket

s or

42

bb

=

√𝑎𝑎× √𝑎𝑎

−2 √𝑎𝑎 √𝑏𝑏

+2 √𝑏𝑏 √𝑎𝑎−

2 √𝑏𝑏

×2 √𝑏𝑏

M1

for a

or (–

4b)

A1

cao

19(a

)sk

etch

B1

for c

orre

ct sh

ape

for 0

⩽x⩽

360

B1

for f

ully

cor

rect

sket

ch w

ith la

bels

(b)(

i)sk

etch

B1

cao

(ii)

sket

chB

1ca

o



Pape

r 1M

A1:

2H

Que

stio

nW

orki

ngA

nsw

erN

otes

20⦟

TSU

= 36

0 ÷

5 (=

72)

Exte

rior a

ngle

s of a

pol

ygon

ad

d up

to 3

60o

proo

fM

1fo

r met

hod

to fi

nd in

terio

r or e

xter

ior a

ngle

of

regu

lar p

enta

gon

⦟Q

RO=⦟

OTP

= 90

The

tang

entt

o a

circ

le is

pe

rpen

dicu

lar (

90o )to

the

radi

us(d

iam

eter

)

M1

for u

sing

ang

le b

etw

een

tang

ent a

nd ra

dius

⦟RO

T=

540

–2

× 90

–2

× 10

8 (=

144

)M

1fo

r met

hod

to fi

nd a

ngle

RO

T

⦟RU

T=

144

÷ 2

(= 7

2)Th

e an

gle

at th

e ce

ntre

of a

ci

rcle

is tw

ice

the

angl

e at

th

e ci

rcum

fere

nce

C1

for m

etho

d to

find

ang

le R

UT

with

reas

on

Bas

e an

gles

of a

n is

osce

les

trian

gle

are

equa

lC

1fo

r ded

uctio

n th

at S

T=

UT

with

reas

ons

212𝑥𝑥

− 1

𝑥𝑥 −

4

=16𝑥𝑥 +

1

2𝑥𝑥 −

1–1 12

, 5P1

for p

roce

ss to

writ

e as

an

equa

tion

(2x

–1)

² = (1

6x+

1)(x

–4)

P1fo

r pro

cess

to c

lear

the

frac

tions

12x²

–59

x–

5 =

0P1

for p

roce

ss to

writ

e eq

uatio

n in

form

ax

² + b

x+

c =

0(1

2x+

1)(x

–5)

= 0

P1fo

r pro

cess

to so

lve

the

equa

tion

A1

cao

September 2015 - Maths · PDF file(bi) 13-17 P1 for process to draw ... September 2015 ©...

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Transcript of September 2015 - Maths · PDF file(bi) 13-17 P1 for process to draw ... September 2015 ©...