Mark Scheme Mock Paper - WordPress.com Scheme Mock Paper GCSE GCSE in Mathematics Specification A Higher Tier Paper 1 (Non-Calculator) Edexcel Limited. Registered in England and Wales

Mark Scheme

Mock Paper

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 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 schemeM1 – 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

Specification A: Paper 1 Higher Tier

1MA0/1H Question Working Answer Mark Additional Guidance1. 32 ÷ 80 × 100 40 2 M1 for 32 ÷ 80 × 100 oe

A1 cao

Total for Question 1: 2 marks2.

300 × 0.7 210 2 M1 for 300 × 0.7 A1 cao


(a) 2 × 2 × 2 × 3 × 5

2 M1 for correct method seenA1 cao

(b) 30 1 B1 caoTotal for Question 3: 3 marks

4. FE

(a) 24 ÷ 12 = 2 2 × 180

360 2 M1 for 24 ÷ 12 (= 2)A1 cao

(b) 18 ÷ 12 (=1.5) 1.5 × 200

300 2 M1 for 18 ÷ 12 (=1.5)A1 cao

Total for Question 4: 4 marks5. Shape enlarged

×3 in correct position

3 B3 shape enlarged × 3 in correct position(B2 shape enlarged ×3 but in wrong position or shape enlarged by a different scale factor correctly) (B1 shape enlarged by a different scale factor and in wrong position)


(a) 20 2 M1 for substitution into formulaA1 cao

(b) m13 1 B1 cao

(c) 1 1 B1 cao

(d) 4y3 2 B2 for 4y3

(B1 for ay3 or 4yn or 161/2(y3)1/2) Total for Question 6: 6 marks

7. FE

Question and response boxes

2 B1 for suitable questionB1 for response boxes

Total for Question 7: 2 marks

1MA0/1H Question Working Answer Mark Additional Guidance8. (i)

(ii)

0.39

0.41

3

B1 cao M1 for 1 – (0.2 + 0.16 + 0.23) A1 cao

Total for Question 8: 3 marks9. 49 4 M1 for 100 – 38 (=62)

M1 for 23 – 7 (-16) M1 for “62” – 18 – “16” A1 cao NB : working may be in a table or diagram

Total for Question 9: 4 marks10. FE

2 4 M1 for attempt to find LCM of any 2 of 12, 8 and 9M1 for attempt to find LCM of 8, 9 and 12 A1 for 72 A1 for 2


15000÷100×40 (=6000) 15000 – “6000” (=9000)

3000 4 M1 for 15000 – 15000÷100×40 oe (=6000)M1 for “9000” ÷ (3 + 1 + 2) (=1500) M1 for “1500” × 2 A1 cao


1MA0/1H Question Working Answer Mark Additional Guidance12. (a) 12x + 3y 2 M1 for 3×4x + 3×y or 12x or 3y

A1 cao (b) 5p2 – 15p 1 B1 cao

(c) y2 + 5y – 24 2 M1 for all 4 terms correct with or without signs or 3 out of no more than four terms correct with signs or y(y – 3) + 8(y – 3) or y(y + 8) – 3(y + 8) A1 cao

(d) 4t2 – 12t + 9 2M1 for all 4 terms correct with or without signs or 3 out of no more than four terms correct with signs or 2t(2t – 3) - 3(2t – 3) A1 cao

Total for Question 12: 7 marks13. m = (p – h)

÷ 6 2 M1 for p – h = 6m

A1 Total for Question 13: 2 marks

14. FE

Region shaded

4M1 for line parallel to AB, 2 cm ±2mm from AB M1 for circle, centre T, radius 3 cm ±2mm M1 for bisector of angle DCB ±2o A1 for correct region shaded within guidelines


2x +1 + 3x – 2 + 3x + 1 + 2x =

38 10x – 2 = 38 x = 4 7; 8; 13 ½ ×(7 + 13) × 10

80 5 M1 for 2x +1 + 3x – 2 + 3x + 1 + 2x = 38 M1 for correct method to solve linear equation A1 for x = 4 M1 for substitution of x = 4 into any expression for side A1 cao


1MA0/1H Question Working Answer Mark Additional Guidance16. 180 – (360÷ 5) oe (=108)

360 – “60” – 2×”108”

84 4B1 for 60o seen M1 for 180 – (360÷ 5) oe (=108) M1 for 360 – “60” – 2×”108” A1 cao


QWC FE

4000 × 1.032 Bank B 5 M2 for 4000 × 1.032 oe(M1 for 1.03 × 4000 oe or 120 seen) M1 for 3.2 × 4000 ÷ 100 oe A1 for 256 and 243.60 C1 for clear working conclusion following on from candidate’s working QWC : Working must be clearly laid out and conclusion must link to working


1MA0/1H

Question Working Answer Mark Additional Guidance18. (a) 6 ÷ 4 = 1.5

1.5 × 9 13.5 2 M1 for 6 ÷ 4 (=1.5) or 2 ÷ 3

A1 cao (b) 10.5 ÷ 1.5 7 2 M1 for 10.5 ÷ 1.5 oe

A1 cao

Total for Question 18: 4 marks19. x = 2,

y = -1.5 4 M1 for correct process to eliminate either x or y (condone one

arithmetic error) A1 for either x = 2 or y = -1.5 M1 (dep on 1st M1) for correct substitution of their found variable A1 cao for both x = 2 and y = -1.5


FE

(a) Points plotted and

cf graph drawn

2

B1 ft for at least 5 of 6 points plotted correctly ± ½ sq at end ofB1 ft (dep on previous B1) for points joined by curve or line segments provided no gradient is negative – ignore any part of graph outside range of their points (SC B1 if 5 or 6 pts plotted not at end but consistent within each interval and joined)

(b) Box plot drawn

3

B1 for median drawn correctly (ft from graph)B1 for UQ and LQ drawn correctly (ft from graph) B1 for whiskers correct

(c) Comparison

2 B2 ft for any comparison of spread in context(B1 ft for any comparison not in context)


1MA0/1H

Question Working Answer Mark Additional Guidance21. (14 – 2)/2 (=6)

“6” 3 (=18) “18” + 1

19 3 M1 for (14 – 2)/2 (=6) M1 for “6” 3 A1 cao or M1 for (k – 1)/12 = 3/2 M1 for 2(k – 1) = 12 × 3 A1 cao


96 4 M1 for Angle ABC = 0.5 × 168 (= 84)

M1 for Angle ADC = 180 – 0.5×168 A1 cao C1 for Angle at centre is twice angle at circumference and Opposite angles of a cyclic quadrilateral sum to 180o or M1 for reflex angle AOC = 360 – 168 (= 192) M1 for 0.5 × 192 A1 cao C1 for Angle at centre is twice angle at circumference and angles at a point add up to 360


1MA0/1H

Question Working Answer Mark Additional Guidance23.

7252

4

B1 for89ba

M1 for 84

93 or

82

93 or

83

94 or

82

94 or

84

92 or

83

92

M1 for 84

93 +

82

93 +

83

94 +

82

94 +

84

92 +

83

92

A1 for 7252

oe

or

B1 for89ba

M1 for 83

94 or

82

93 or

81

92

M1 for 1 – (83

94 +

82

93 +

81

92 )

C1 for 7252

oe


5 , - 0.5

5

M1 for common denominator on LHS or clearing fractionsM1 for multiplying out brackets A1 for 2x2 – 9x + 5 = 0 M1 for (2x ± 1)(x ± 5) or substitution into quadratic formula A1 for 5 and - 0.5


1MA0/1H

Question Working Answer Mark Additional Guidance25. (i)

(ii)

2b + a

½b + a

2 3

M1 for QRPQPR oe A1 cao

M1 for QR43

oe

M1 for QRPQSPSX43

oe

A1 for ½b + a oe


April 2010 For more information on Edexcel and BTEC qualifications please visit our website: www.edexcel.org.uk Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WC1V 7BH. VAT Reg No 780 0898 07

Mark Scheme

Mock Paper

GCSE

GCSE in Mathematics Specification A Higher Tier Paper 2 (Calculator)



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 schemeM1 – 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

Specification A Paper 2 Higher Tier

1MA0/2H Question Working Answer Mark Additional Guidance1. FE

7 ÷ 5 (=1.4) 2 × “1.4” (=2.8) 5.65 – 2.8 (=2.85) “2.85”÷3

£0.95 3 M1 for 7 ÷ 5 (=1.4) M1 for 5.65 – 2 × “1.4” (=2.85) A1 cao

Total for Question 1: 3 marks2. (a) 150774.1935… 2 M1 for 74.89.. or 0.0372

A1 for 159774.1…. (b) 151000 1 B1 ft


(a) Negative 1 B1 cao

(b) Line of best fit drawn

1 B1 St line between (15,50),(15,45) and (50,14),(50,9)

(c) 35 – 40 1 B1 ft


(a) 20 1 B1 cao

(b) Line from (10 10,10) to (10 40, 10) to

(11 20, 0)

3 B1 for line from (10 10,10) to (10 40, 10)M1 for 10 ÷ 15 or 40 minutes A1 for line from (10 40, 10) to (11 20, 0)

Total for Question 4: 4 marks5. 25 3 M1 for angle BAH = 28 or angle ABH = 180 – 53 (=127)

M1 for 180 – “127” – 28 A1 cao


1MA0/2H Question Working Answer Mark Additional Guidance6.

(a) 5 2 M1 for 5x – 2x = 17 – 2 A1 cao

(b)

67

x 2 M1 for 6x + 3 > 10

A1 cao


Stem and leaf

diagram (see end of

mark scheme)

+ key

3 B3 for fully correct diagram with correct key[B2 or ordered leaves (condone one error), key or no key OR unordered leaves (condone one error) + correct key] [B1 for unordered leaves (condone one omission), no key OR for a correct key (ignore diagram) OR for ordered leaves (no more than 2 errors with a correct key]


2 × 3.50 + 2.50 + 2.20 = £11.7010 ÷ 1.25 = 8 11.70 – 8

3.70 4 M1 for 2 × 3.50 + 2.50 + 2.20 (= £11.70)M1 for 10 ÷ 1.25 (= 8) M1 for “11.70” – “8” A1 cao or M1 for 2 × 3.50 + 2.50 + 2.20 (= £11.70) M1 for “11.70” × 1.25 = (14.625) M1 for “14.625” – 10 A1 cao


(a) 6y(y + 2) 2 M1 for any factor correctA1 cao

(b) (k + 10)(k + 3)

2 M1 for (k ± 10)(k ± 3) or (k + a)(k + b) where ab = 30A1 cao


1MA0/2H Question Working Answer Mark Additional Guidance10. (a) Proof 2 M1 for x × x × (x + 4) or equating an expression in x to 150

A1 for completion of proof (b) 4.2 144(.648…)

4.3 153(.467…) 4.4 162(.624…) 4.5 172(.125…) 4.25 149.0(15…)

4.3 4 B2 for trial 4.2 ≤ x ≤ 4.3 evaluated(B1 for trial 4 ≤ x ≤ 5 evaluated) B1 for different trial 4.25 ≤ x < 4.3 evaluated B1 (dep on at least 1 previous B1) for 4.3 Values evaluated can be rounded or truncated, but to at least 3sf when x has 1 dp and 4 sf when x has 2 dp NB allow 149 for evaluation using x = 4.25

Total for Question 10: 6 marks11. (a) 3 ≤ h < 4 1

B1 cao

(b) (1.5 + 12 + 17.5 + 35 +54) 40 = 120 40

3 4M1 for use of fx with x consistent within intervals (including end points) M1 (dep) for use of midpoints

M1 (dep on 1st M1) for use of ffx

A1 cao

Total for Question 11: 5 marks12. (a) Triangle at

(1,2) (1, -1) (3, -1)

1 B1 cao

(b) Rotation; 1800;

centre (0,0)

3 B1 for rotation B1 for 180o B1 for centre (0,0)



QWC Π × 62 × 15 (= 1696..)

15 × 1000 = 15000 15000 ÷ 1696 (=8.8…)

8 4 M1 for Π × 62 × 15 (= 1696..)B1 for 15000 M1 for “15000” ÷ “1696” C1 for reasoning how many bags for answer of 8 from 8.8… QWC : Working must be clearly set out with conclusion referring back to working


62 = x2 + 1.52

√33.75 (=5.809…) 5.8 3 M1 for 62 = x2 + 1.52

M1 for 25.236 A1 cao


52 ÷ 0.8 65 3 M1 for 0.8 or 80% seenM1 for 52 ÷ 0.8 oe A1 cao

Total for Question 15: 3 marks16. 6.32 3

M1 for tan 36 = 7.8BC

M1 for 8.7 × tan36 A1 for 6.32 – 6.325


(a) See table at end Table 2 B2 all 4 correct

(B1 any two correct) (b) Graph drawn (at end) 2 B1 at least 8 points plotted correctly (ft)

B1 smooth curve drawn through their 8 or 9 points (c) Graph drawn (at end) -1.5, -0.3,

1.9 1 B1 ft from graph


1MA0/2H Question Working Answer Mark Additional Guidance18. (a)

101

, 53

, 52

,

53

2B1 for

101

B1 for 53

,52

,53

(b)

259

2

M1 for 52

109

A1 for 259

oe


Region shaded 4 B1 y = - 4

B1 x = 2 B1 y = 2x+1 B1 region bounded by (2, 5), (2, -4), (2.5, -4) indicated

Total for Question 19: 4 marks20. (a) 3.7 × 10-4 1

B1 cao

(b) 8250 1B1 cao

(c) 1.26 × 104 2 M1 for 1260 or 1.26 × 10n or 126 × 310 A1 cao



(a) 18.5 1 B1 cao

(b) 564.25 2 M1 for “18.5” × UB (where 30 < UB ≤ 30.5)A1 cao

Total for Question 21: 3 marks22. 1490 3

M1 for 3734

21

(= 718.37…) or 15731 2

M1 for 3734

21

+ 15731 2

A1 1485 - 1490 Total for Question 22: 3 marks

23.

Bars of height 16, 14, 11,

4.5

3 M1 for use of frequency densityM1 for at least two bars of different widths drawn correctly or all correct heights seen A1 cao



QWC FE

24085sin

68sin

ADB

)68240

85sin(sin 1 ADB

½ × 240 × 68 × sin(180 – 85 – “16.39..”) + ½ × 240 × 95 × sin(136 – 16.39..)

17900 6M1 for

24085sin

68sin

ADB

M1 for )68240

85sin(sin 1 ADB

A1 for 16.39… M1 for ½ × 240 × 68 × sin(180 – 85 – “16.39..”) (7999.29…) or ½ × 240 × 95 × sin(136 – 16.39..) (= 9911.26…) M1 for ½ × 240 × 68 × sin(180 – 85 – “16.39..”) + ½ × 240 × 95 × sin(136 – 16.39..) C1 17850-17950 QWC : Working must be clearly set out with conclusion referring back to working

Total for Question 24: 6 marks25. (a) (3, 4) 1 B1 cao

(b) (5, 1) 1 B1 cao (c) (6, 1) 1 B1 cao


QWC 12 6 M1 for one correct expression for area

M1 for 2x(2x + 5) + (2x – 3)(x + 1) = 102 A1 for 2x2 + 3x – 35 = 0 or 6x2 + 9x – 105 = 0 M1 for (2x ± 7)(x ± 5) oe or substitution into quadratic formula C1 for 3.5 oe C1 ft for 12 QWC : Working must be clearly set out with conclusion referring back to working


Q 7

2 2 5 7

3 0 1 3

4 2 4 7 8

5 0 1 4 6

6 2 3 3 3 7 9

Plus appropriate key

Q 17 x -2 -1.5 -1 -0.5 0 0.5 1 1.5 2

y -3 0.125 1 0.375 -1 -2.375 -3 -2.125 1

1

2

-1

-2

-3

-4

1 2-1 -2 0 x

y

April 2010 For more information on Edexcel and BTEC qualifications please visit our website: www.edexcel.org.uk Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WC1V 7BH. VAT Reg No 780 0898 07

Mark Scheme

Sample Assessment Material

GCSE

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



• 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

75


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ect

expr

essi

on f

or x

inv

olvi

ng r

atio

s of

sid

es,

e.g.

42

5xoe

A1

cao

M1

695

xy

or6

59

xy

oe

A1

cao

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

1

0 6

8

1

0 1

2 8

10

1

2 14

10

12

14

1

6

OR 41

41

441

41

1643

M1

Att

empt

s to

lis

t al

l ou

tcom

e pa

irs

A1

all 16

fou

nd

A1

cao

OR

M2

441

41

(M1

2,1

4141

or 3

)

A1

164 o

e

(b

) Pr

ob A

li w

ins

= 166

Num

ber

of w

ins

= 80

166

30

3 B1

Pro

b Ali w

ins

= 166

oe

M1

80'

166'

A1

ft

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

cao

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 =

yAre

a of

rec

tang

le =

xy

Are

aAX

D =

4xy

Are

aC

YZ=

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

cao

OR

Dia

gram

M1

for

divi

ding

lef

t in

to 2

con

grue

nt t

rian

gles

f

or d

ivid

ing

righ

t in

to 4

con

grue

nt t

rian

gles

B1

lef

t =

2Aan

d 2A

or

sha

ded

= 21

of

21

=41

= 82

B1 r

ight

= 2

A an

d A

and

A or

shad

ed =

43

of21

= 83

A1

cao

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

int

erna

l sh

apes

B1

for

are

a of

DZX

B1 f

or a

rea

of Z

XBY

A1

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

int

erna

l sh

apes

B1

for

are

a AD

XB1

for

are

a ZC

YA1

cao

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

cao

(b

) O

X=

12b

, AX

=–4a

+ 1

2b=

4(–a

+ 3

b)

Cor

rect

re

ason

, w

ith

corr

ect

wor

king

3B1

OX

= 12

b

B1AX

=–4a

+ 1

2bC1

conv

inci

ng e

xpla

nati

on

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 =

720

120

720

120

+

8495

106 +

8594

106

720

360

4M

1 fo

r 85

96104

A1

for

720

120

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

cao

SC w

ith

repl

acem

ent

M1

106106

104

M1

106106

104×3

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 an

d co

here

ntm

anne

r

6 B2

the

nth

term

for

con

secu

tive

odd

num

bers

is

2n –

1 o

e (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

conc

lusi

on b

ased

on

corr

ect

alge

bra

QW

C: C

oncl

usio

n sh

ould

be

stat

ed,

wit

h co

rrec

t su

ppor

ting

alg

ebra

.

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

conc

lusi

on b

ased

on

corr

ect

alge

bra

QW

C: C

oncl

usio

n sh

ould

be

stat

ed,

wit

h co

rrec

t su

ppor

ting

alg

ebra

. Tota

l 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

0–1

0 40

4

40

10–2

0 60

6

100

20–4

0 90

4.

5 19

0 40

–80

60

1.5

250

>80

0 0

250

His

togr

amO

RCum

ulat

ive

Freq

uenc

ypo

lygo

n

82%

6 B1

Sca

les

labe

lled

and

als

o m

arke

d on

the

ver

tica

l ax

is w

ith

freq

uenc

y de

nsit

y or

wit

h cu

mul

ativ

e fr

eque

ncy

M1

freq

uenc

y de

nsit

ies

calc

ulat

ed,

at lea

st o

ne n

on-t

rivi

al o

ne c

orre

ct.

A1

all co

rrec

tly

plot

ted

(M1

cum

ulat

ive

freq

uenc

ies

corr

ect)

M1

Use

50

on t

he h

oriz

onta

l sc

ale

of C

F di

agra

m r

ead

off

vert

ical

axi

s (2

00-2

10)

or U

se 5

0 on

the

hor

izon

tal sc

ale

of a

his

togr

am 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 k

g

Mar

100

130.

13

13%

14.5

6 kg

14.

AB C

D

X

Y

Z

AB C

D

X

Y

Z


90

November 2009 For more information on Edexcel and BTEC qualifications please visit our website: www.edexcel.org.uk Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WC1V 7BH. VAT Reg No 780 0898 07

Mark Scheme

Sample Assessment Material

GCSE

GCSE in Mathematics Specification A Higher Tier Paper 2: (Calculator)



• 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


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

Cor

rect

ste

m

and

leaf

3 B3

Ful

ly c

orre

ct

(B2

All e

ntri

es c

orre

ct,

no k

ey)

(B1

corr

ect

entr

ies

unor

dere

d, k

ey o

r no

key

) O

R(B

2 Thr

ee r

ows

corr

ect,

key

or

no k

ey)

(B1

Tw

o ro

ws

corr

ect,

key

or

no k

ey)

(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

OR

M1

Red

oes

tabl

e (f

t) w

ith

each

val

ue inc

reas

ed b

y 5

and

atte

mpt

s to

fi

nd m

edia

n A1

cao

(c

)

The

sam

e +

reas

on1

C1

All t

he v

alue

s ha

ve inc

reas

ed b

y 5

min

utes

so

whe

n yo

u su

btra

ct t

he

5 m

inut

es w

ill c

ance

l ou

t.

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

litre

s,

200

gal

lons

= 9

08 lit

res

= 9

0800

0 cm

3

Vol

of

tank

60

2xπ

180

=

2035

752.

04..

cm3

9080

00<

1017

876.

02

OR

Vol

of

tank

60

2π

180

= 2

0357

52.0

4..c

m3

Hal

f vo

l of

tan

k

= 10

1787

6.02

cm

3

= 10

17.8

76…

litre

s

1017

.876

÷ 4

.54

= 22

4 ga

llon

s

224

> 20

0

No

5 Res

pons

e m

ay c

onve

rt int

o ga

llon

s, lit

res,

or

cm3

Cal

cula

tion

s m

ay b

e pe

rfor

med

in

diff

eren

t or

ders

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

or 1

0 lit

res

wit

hin

tole

ranc

e (4

.4 —

4.

6)A1

Ans

wer

in

cm3 ,

litre

s or

gal

lons

C1

Dec

isio

n an

d re

ason

QW

C: D

ecis

ion

shou

ld b

e st

ated

, w

ith

appr

opri

ate

supp

orti

ng s

tate

men

t

(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

770

— 7

72

Tota

l for

Que

stio

n: 8

mar

ks

7.12

055

225

245

150

4510

6.08

hou

rs

4 M

1 fo

r m

id in

terv

al v

alue

s

M1

for

mul

tipl

ying

fre

quen

cies

by

mid

-int

erva

l va

lues

M

1 fo

r ad

ding

(fr

eq

mid

-int

erva

l va

lues

) ÷

120

A1

cao

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

Cle

ar a

nd

cohe

rent

expl

anat

ion

1 C1

a cl

ear

and

cohe

rent

exp

lana

tion

(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

cao

OR

M1

collec

ts t

erm

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

cao

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 sa

les

of e

ach

shop

(263

5,

3530

)

Cor

rect

co

mpa

riso

ns

4 B1

, B1

, B1

for

any

4 o

f th

e fo

llow

ing

done

cor

rect

ly

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 sa

les

C1

for

com

men

ts o

n sh

ape

of t

he d

istr

ibut

ions

Q

WC:

Dec

isio

ns s

houl

d be

sta

ted,

and

all

com

men

ts s

houl

d be

cle

ar

and

follo

w t

hrou

gh f

rom

any

wor

king

or

diag

ram

s

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

26 —

30

(n

eed

21 s

q to

lera

nce

on e

ach)

FE(c

) 15

% o

f 80

= 1

2 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

corr

ect

conc

lusi

on

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

Rea

son

supp

orte

dby

calc

ulat

ion

4M

1 si

n 68

o =

5.8AC

M1

AC =

8.5

× s

in 6

8o

A1

7.88

(1…

C1

8.88

(1…

+ c

oncl

usio

n Q

WC:

Dec

isio

n sh

ould

be

stat

ed,

supp

orte

d by

cl

earl

y la

id o

ut w

orki

ng

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

A1

AT

4

A1

for

30.9

8… o

r 31

(.0)

OR

M2

for

40T =

10

060

oe

M1

for

100

6040

T oe

A1

for

30.9

8… 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

) Vol

=

2)

2(x

x=

51

Vol

=

xx

42

2–

51 =

0

Der

ives

giv

en

answ

er a

nd

cond

itio

n

4M

1 Vol

=

2)2

(xx

M1

expa

nds

brac

ket

corr

ectl

y A1

(E1)

set

s eq

ual 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

and

c) int

o qu

adra

tic

form

ula

M1

442

44

x

A1

6.14

(7…

, −

4.14

(7…

)

Tota

l for

Que

stio

n: 7

mar

ks

17.

Ang

le B

AC

= 18

0º —

47º

— 5

8º

= 75

º

)58

sin

(75

sin22

047

sin

ABAC

AC

=75

sin

47si

n22

0 =

166

.57.

.

Are

a=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

1550

0 m

2

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 iso

s tr

iang

les

5360

=72º

Base

ang

les

= (1

80 –

72)

2

=

54º

for

find

ing

equa

l si

des

of

isos

cele

s tr

iang

le;

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

…

Are

a 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 si

des

of iso

scel

es t

rian

gle;

x =

72si

n1054

sinx

A1

for

x =

8.50

6508

084…

M1

for

find

ing

area

of

isos

cele

s tr

iang

le =

72

sin

212 x

A1

for

34.4

0954

801…

(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

for

2065

cm

2(o

e)

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

6.88

19…

M1

for

find

ing

area

of

isos

cele

s tr

iang

le =

21

10

h

A1

for

34.4

0954

801…

(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

for

2065

cm

2(o

e)


167


168

9.

102030405060

1020

3040

5060

0C

Ds s

old

Freq

uenc

y


169

14.

2468

24

0x

y

October 2009 For more information on Edexcel and BTEC qualifications please visit our website: www.edexcel.org.uk Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WC1V 7BH. VAT Reg No 780 0898 07

Mark Scheme (Results) March 2013 GCSE Mathematics (Linear) 1MA0 Higher (Non-Calculator) Paper 1H

Edexcel and BTEC Qualifications Edexcel and BTEC qualifications come from Pearson, the world’s leading learning company. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information visit our qualifications websites at www.edexcel.com or www.btec.co.uk for our BTEC qualifications. Alternatively, you can get in touch with us using the details on our contact us page at www.edexcel.com/contactus. If you have any subject specific questions about this specification that require the help of a subject specialist, you can speak directly to the subject team at Pearson. Their contact details can be found on this link: www.edexcel.com/teachingservices. You can also use our online Ask the Expert service at www.edexcel.com/ask. You will need an Edexcel username and password to access this service. Pearson: helping people progress, everywhere Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We’ve been involved in education for over 150 years, and by working across 70 countries, in 100 languages, we have built an international reputation for our commitment to high standards and raising achievement through innovation in education. Find out more about how we can help you and your students at: www.pearson.com/uk March 2013 Publications Code UG035047 All the material in this publication is copyright © Pearson Education Ltd 2013

NOTES ON MARKING PRINCIPLES 1 All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as

they mark the last. 2 Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do

rather than penalised for omissions. 3 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.

4 Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and

exemplification may be limited. 5 Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response. 6 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 labeling 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.

7 With working If there is a wrong answer indicated on the answer line always check the working in the body of the script (and on any diagrams), and award any marks appropriate from the mark scheme. If working is crossed out and still legible, then it should be given any appropriate marks, as long as it has not been replaced by alternative work. If it is clear from the working that the “correct” answer has been obtained from incorrect working, award 0 marks. Send the response to review, and discuss each of these situations with your Team Leader. If there is no answer on the answer line then check the working for an obvious answer. Any case of suspected misread loses A (and B) marks on that part, but can gain the M marks. Discuss each of these situations with your Team Leader. If there is a choice of methods shown, then no marks should be awarded, unless the answer on the answer line makes clear the method that has been used.

8 Follow through marks

Follow through marks which involve a single stage calculation can be awarded without working since you can check the answer yourself, but if ambiguous do not award. Follow through marks which involve more than one stage of calculation can only be awarded on sight of the relevant working, even if it appears obvious that there is only one way you could get the answer given.

9 Ignoring subsequent work

It is appropriate to ignore subsequent work when the additional work does not change the answer in a way that is inappropriate for the question: e.g. incorrect cancelling of a fraction that would otherwise be correct It is not appropriate to ignore subsequent work when the additional work essentially makes the answer incorrect e.g. algebra. Transcription errors occur when candidates present a correct answer in working, and write it incorrectly on the answer line; mark the correct answer.

10 Probability Probability answers must be given a fractions, percentages or decimals. If a candidate gives a decimal equivalent to a probability, this should be written to at least 2 decimal places (unless tenths). Incorrect notation should lose the accuracy marks, but be awarded any implied method marks. If a probability answer is given on the answer line using both incorrect and correct notation, award the marks. If a probability fraction is given then cancelled incorrectly, ignore the incorrectly cancelled answer.

11 Linear equations

Full marks can be gained if the solution alone is given on the answer line, or otherwise unambiguously indicated in working (without contradiction elsewhere). Where the correct solution only is shown substituted, but not identified as the solution, the accuracy mark is lost but any method marks can be awarded.

12 Parts of questions

Unless allowed by the mark scheme, the marks allocated to one part of the question CANNOT be awarded in another.

13 Range of answers Unless otherwise stated, when an answer is given as a range (e.g 3.5 – 4.2) then this is inclusive of the end points (e.g 3.5, 4.2) and includes all numbers within the range (e.g 4, 4.1)

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 dep – dependent (on a previous mark or conclusion) indep – independent isw – ignore subsequent working

1MA0_1H Question Working Answer Mark Notes 1 183

× 47 1281 7320 8601 or

1 8 3 ×

4

3 2

1 2 4

8 7 5 6

2 1 7

6 0 1 or 4000 + 3200 +120 + 700 + 560 + 21 = 8601 or 183 × 100 = 18 300 183 × 50 = 18 300 ÷ 2 =9150 183 × 3 = 549 9150 – 549 = 8601

86.01 3 M1 for a complete method to multiply 183 by 47 and attempt at addition (condone one multiplication error)

A1 for digits 8601 given as the answer

B1 (dep on M1) for correctly writing their answer to 2 decimal places

100 80 3 4000 3200 120 40 700 560 21 7

1MA0_1H Question Working Answer Mark Notes 2 (a) Plot (2, 250) and (3.1, 190) Plot points 1 B1 for both points plotted accurately

(b) Relationship 1

B1 for “As the distance from the centre increases the monthly rent decreases” or the nearer you are to the centre the more you have to pay oe (accept negative correlation)

(c)

200 to 260 2

M1 for attempting a correct method, eg a line of best fit or any other indication, on a line that could be used as a line of best fit eg line to graph at x = 2.8or a mark on the line at 2.8 A1 for value in the range 200 to 260

3 (a) 2 reasons 2 B2 for 2 different reasons from given examples (B1 for 1 reason from given examples)

eg No time frame eg No box for less than £10 accept no box for zero or none or £0 eg Overlapping intervals or boxes or £30 and/ or £50 in two boxes

(b) 1 reason 1 C1 for reason why the sample is biased eg • they are only in the CD store, • the people in the store are more likely to buy CDs • she needs to ask people outside the store oe

1MA0_1H Question Working Answer Mark Notes 4 (a)

x –2 –1 0 1 2 y (1) 3 (5) 7 9

3, 7, 9 2 B2 for all three values correct in the table (B1 for 2 values correct)

(b)

graph of y = 2x + 5

2 (From their table of values) M1 ft for plotting at least 2 of their points (any points from their table must be correctly plotted) A1 for correct line from x = –2 to x = +2 (Use of y = mx + c) M1 for line drawn with gradient of 2 or line drawn with a y intercept of 5 and a positive gradient) A1 for correct line from x = –2 to x = +2

2

4

6

8

1

2O x

1MA0_1H Question Working Answer Mark Notes 5 (a) 6n – 3 2 M1 for attempt to establish linear expression in n with coefficient of

6 e.g. 6n + k where k is an integer (accept n = 6n –3 for one mark) A1 cao

(b) No + Reason 1 C1 ft from their answer to part (a) for decision and explanation eg “ stating no and because all the terms in the sequence are odd and 150 is even” or

“no and ‘6n – 3’ = 150, n = 153/6 ... so n is not an integer” or

Continuing the sequence to show terms 147 and 153 and state “no as 150 is not in the sequence” oe

1MA0_1H Question Working Answer Mark Notes 6 (a) 8 1 B1 for 8 (.00)

(b) 550

4 M1 for 600 – 200 ( = 400) M1 for correct method to convert ‘$400’ to £ M1 (dep on the previous M1) for 800 – ‘$400’ in £s A1 for value in the range 540 –560

OR M1 for correct method to convert $600 and $200 to pounds M1 for ‘375’–‘125’ M1 (dep on the previous M1) 800 –‘250’ A1 for a value in the range 540-560 OR M1 for correct method to convert £800 to dollars M1 for ‘1280’ + 200 – 600 M1 (dep on the previous M1) for attempt to convert ‘$880’ back to £ A1 for value in the range 540 – 560

7

(a) 6x – 3y 2 M1 for an attempt to combine terms in x or terms in y correctly eg 5x + x(= 6x), 4y – 7y(= – 3y) A1 for 6x – 3y oe

(b) 7x + 14 = 7 or x + 2 = 1 7x = –7

x = –1 2 M1 for correctly expanding the bracket or an attempt to divide both sides by 7 e.g. 7x +14 or x + 2 = 7 ÷ 7 oe A1 cao

1MA0_1H Question Working Answer Mark Notes 8 09 36 3 M1 for listing 9, 18, 27, 36, 45, ...(at least 3 correct multiples with

at most one incorrect) M1 for listing 12, 24, 36, 48, .... (at least 3 correct multiples with at most one incorrect) A1 for 09 36 or 9 36 (am) OR M1 for listing 9.09 9.18 9.27 9.36 ...(at least 3correct times with at most one incorrect) M1 for listing 9.12 9.24 9.36 ... (at least 3 correct times with at most one incorrect) A1 for 09 36 or 9 36 (am) OR M1 for 9 = 3 × 3 or 12 = 2 × 2 × 3 (could be in factor tree) M1 for 9 = 3 × 3 and 12 = 2 × 2 × 3 (could be in a factor tree) A1 for 09 36 or 9 36 (am) SC B2 for 9 36 pm or (after) 36 (minutes) on the answer line

9 (a) a 9 1 B1 for a 4 + 5 or a 9

(b)

9e5f 6 2

B2 cao (B1 for two of 9, e 6 – 1, f 8 – 2 as a product)

(c) 3 1 B1 (accept ± 3 but not just –3)

1MA0_1H Question Working Answer Mark Notes *10 Angle AED = 38 alternate angles are equal

Angle ADE = (180 – 38) ÷ 2 = 71 x = 180 – 71 base angles of an isosceles triangle are equal angles in a triangle add up to 180 angles on a straight line sum to 180 OR angle AEF = 142 allied angles/co-interior angles add up to 180 ADE = 142 ÷ 2 = 71 base angles of an isosceles triangle are equal exterior angle of a triangle is equal to the sum of the interior opposite angles , x = 180 – 71 angles in a straight line add to 180 OR Angle AED = 38 alternate angles are equal for angles BAE and AED and BAD and ADC (x) Angle DAE= (180 – 38) ÷ 2 = 71 base angles of an isosceles triangle are equal angles in a triangle add up to 180 Or Angle AED = 38 alternate angles are equal Angle ADE = (180 – 38) ÷ 2 = 71 base angles of an isosceles triangle are equal and angles in a triangle sum to 180 x = 38 + 71 alternate angles BAD and ADC(x) are equal

x = 109 4 B1 for angle AED = 38 or AEF = 142 M1 for a complete method to find one of the base angles of the isosceles triangle C2 (dep M1) for x = 109 with complete reasons (C1 (dep M1) for one reason correctly used and stated)

1MA0_1H Question Working Answer Mark Notes 11 730 5

M1 for 5 200

100× ( = 10) oe

M1 for 10 350100

× ( = 35) oe

M1 for 6 × ‘10’ or 4 × ‘35’ M1 (dep on M1 earned for a correct method for a percentage calculation) for “60” + “140”+ 530 A1 cao Or M1 for 6 × 200 (= 1200) or 4 × 350 (= 1400)

M1 for 5 "1200"( 60)

100× = oe

M1 for 10 "1400"( 140)100

× = oe

M1(dep on M1 earned for a correct method for a percentage calculation) for “60” + “140”+ 530 A1 cao

1MA0_1H Question Working Answer Mark Notes 12 240 4 M1 for 16 × 2 (= 32 girls)

M1 for 16 + ‘16 × 2’ (= 48) M1 (dep on the previous M1) for (16 + ‘32’ )× 5 or (16 + ‘32’) × (4 + 1) A1 cao OR M1 for 1 : 2 = 3 parts M1 for 5 schools × 3 parts (= 15 parts) M1 (dep on the previous M1) for ‘15’ parts × 16 A1 cao SC B2 for 176 given on the answer line

13 54

3 M1 for 180 – 360 ÷ 5 or 108 seen as the interior angle of a pentagon M1 (dep on previous M1) for 360 – 2 × ‘108’ – 90 A1 for 54 cao OR M1 for 180 × (5 – 2) (= 540) ÷ 5 or 108 given as the interior angle of a pentagon M1 (dep on previous M1) for 360 – 2 × ‘108’ – 90 A1 for 54 cao

1MA0_1H Question Working Answer Mark Notes 14 (a)

8, 23, 53, 70, 77, 80

1

B1 cao

(b)

graph

2

M1 ft from their table for at least 5 points plotted correctly at the ends of the intervals provided table values are cumulative, condoning one arithmetic error A1 cao for correct graph with points joined by curve or straight line segments [SC B1 if the shape of the graph is correct and 5 points of their points are not at the ends but consistently within each interval and joined.]

(c)

Readings at 60 and 20 420 to 440 – 280 to 295

120 – 160

2 M1 (dep on cf graph) for use of either cf = 20 or cf = 60 A1 ft from a cf graph

(d)

80 – 71 to 74

6 – 9

2

M1 (dep on cf graph) for evidence of reading off the cf axis from £530 0n the wages axis (could be the answer) A1ft for 6 - 9

1MA0_1H Question Working Answer Mark Notes

15

Required region 4 M1 arc radius 5 cm centre C M1 bisector of angle BAD M1 line 3 cm from DC A1 for correct region identified (see overlay)

16 (a) 820 000

1 B1 cao

(b) 3.76 × 10–4 1 B1 cao

(c)

5 × 108

2

M1 for 2.3 ÷ 4.6 × 1012 – 3 oe or 500 000 000 or 0.5 × 109 A1 cao (accept 5.0 × 108

17

1312 3 M1 for multiplying throughout by 10 oe or writing LHS as a single

fraction e.g 2(4x – 1) + 5(x + 4) = 3× 10 or

2(4 1) 5( 4)10

x x− + + or – +

M1 (dep) for a complete correct method to obtain linear equation of the form ax = b (condone one arithmetic error in multiplying out the bracket) A1 for

1312 oe (decimal equivalent is 0.923…)


18 Q at (– 3, 1), (– 6, 1) (–5, 3) (– 3, 3) R at (–3, – 1), (–6, – 1), (–5, – 3) (–3, –3)

Rotation 180° about (–1, 0)

3 M1 for showing R correctly on the grid without showing Q or for showing Q and R correctly on the grid A1 for rotation of 180° A1 for (centre) (–1, 0) Or M1 for showing R correctly on the grid without showing Q or for showing Q and R correctly on the grid A1 for Enlargement Scale Factor –1 A1 for centre (–1, 0) NB Award no marks for any correct answer from an incorrect diagram or any Accuracy marks if more than one transformation is given

19

68 3 M1 for angle OBC = 90° or angle OAC = 90° (may be marked on the diagram or used in subsequent working) M1 for correct method to find angle BOC or AOC or AOB e.g. angle BOC = 180 – 90 – 34 (= 56) or angle AOC = 180 – 90 – 34 (=56) or angle AOB = 180 – 2 × 34 (= 112) A1 cao NB (68 must be clearly stated as an answer and not just seen on diagram)


20

(a)(i)

(x – 9)(x – 3)

3

M1 for (x ± 9)(x ± 3) A1 for (x – 9)(x – 3)

(ii) x = 9 , x = 3 B1 cao

(b)

(y + 10)(y –10) 1 B1 for (y + 10)(y –10)

*21 (n + 1)2 – n2 = n2 + 2n + 1 – n2 = 2n + 1 (n + 1) + n = 2n + 1 OR (n + 1)2 – n2 = (n + 1 + n)(n + 1 – n) = (2n + 1)(1) = 2n + 1 (n + 1) + n = 2n + 1 OR n2 – (n + 1)2 = n2 – (n2 + 2n + 1) = –2n – 1 = – (2n + 1) Difference is 2n + 1 (n + 1) + n = 2n + 1

proof 4 M1 for any two consecutive integers expressed algebraically eg n and n +1 M1(dep on M1) for the difference between the squares of ‘two consecutive integers’ expressed algebraically eg (n + 1)2 – n2 A1 for correct expansion and simplification of difference of squares, eg 2n + 1 C1 (dep on M2A1) for showing statement is correct, eg n + n + 1 = 2n + 1 and (n + 1)2 – n2 = 2n + 1 from correct supporting algebra

22 Vertices at (–2, –4), (–4, –4), (–4, –6), (–2, –5)

Correct diagram 3 M1 for a similar shape in the correct orientation in the third quadrant M1 for an image in the correct orientation of the correct size A1 cao


23 75π 3 M1 for (4 × π × 52 ) ÷ 2 oe M1 for π × 52 oe A1 for 75π accept 235.5 Condone the use of π = 3.14…

24

EE + CC + HH Or EC+EH+CE+CH+HE+HC Or E,not E+ C,not C + H,not H

11076 5 M1 for use of 10 as denominator for 2nd probability

M1 for 4 3 5 4 2 111 10 11 10 11 10

or or× × ×

M1 for 4 3 5 4 2 1 3411 10 11 10 11 10 110

⎛ ⎞× + × + × =⎜ ⎟⎝ ⎠

M1 (dep on previous M1 for 1 – ‘11034 ’

A1 for 11076 oe

Or M1 for use of 10 as denominator for 2nd probability M1 for

114 ×

105 or

114 ×

102 or

115 ×

104 or

115 ×

102 or

112 ×

104 or

112 ×

105

M2 for 114 ×

105 +

114 ×

102 +

115 ×

104

+

115 ×

102 +

112 ×

104 +

112 ×

105

(M1 for at least 3 of these) A1 for

11076 oe

Or M1 for use of 10 as denominator for 2nd probability M1 for 4 7 5 6 2 9

11 10 11 10 11 10or or× × ×

M2 for 4 7 5 6 2 911 10 11 10 11 10× + × + ×

(M1 for two of these added) A1 for

11076 oe PTO for SC’s


SC: B2 for 12176

SC: B1 for × + × + × (= ) Or × + × + × + × + × + × Or

× + × + ×

25 (a) sketch

M1 for inverting the parabola, so maximum is at ( –2, 0 ) A1 for parabola passing through all three of the points ( –2, 0), (0, –4), ( –4, –4)

(b) y = f (x – 6) 1 B1 for y = f (x – 6) or y = (x – 4)2 oe


26 (a) 6b – 3a

1 B1 for 6b – 3a oe

(b) 4 M1 for AX =

13

AB or 13

’(6b – 3a)’ or ft to 2b – a

M1 for OY = OB + BY = 6b + 5a – b (= 5b + 5a ) oe M1 for OX = 3a + ‘2b – a’ = 2a + 2b oe Or

OX = 6b – ‘(6b – 3a)’ (= 2a + 2b) oe C1 for

52 OY =

52 ×5(a + b) = 2(a + b) = OX

4.

y 9 7 5 3 1 –1 1 –1

2

4

6

8

10

2 -2 O x

18.

y Q P R

1

2

3

4

5

6

7

1

-2

-3

-4

-5

-6

-7

1 2 3 4 5 6 7-1-2-3-4-5-6-7 O x

22

2

4

6

8

-

-

-

-

2 4 6 8---6- O x

y

25(a)

y

2

4

-2

-4

2 4-2 -4 0 x

Further copies of this publication are available from

Edexcel Publications, Adamsway, Mansfield, Notts, NG18 4FN

Telephone 01623 467467

Fax 01623 450481 Email [email protected]

Order Code UG035047 March 2013

For more information on Edexcel qualifications, please visit our website www.edexcel.com

Pearson Education Limited. Registered company number 872828 with its registered office at Edinburgh Gate, Harlow, Essex CM20 2JE

Mark Scheme (Results) March 2013 GCSE Mathematics (Linear) 1MA0 Higher (Calculator) Paper 2H

Edexcel and BTEC Qualifications Edexcel and BTEC qualifications come from Pearson, the world’s leading learning company. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information visit our qualifications websites at www.edexcel.com or www.btec.co.uk for our BTEC qualifications. Alternatively, you can get in touch with us using the details on our contact us page at www.edexcel.com/contactus. If you have any subject specific questions about this specification that require the help of a subject specialist, you can speak directly to the subject team at Pearson. Their contact details can be found on this link: www.edexcel.com/teachingservices. You can also use our online Ask the Expert service at www.edexcel.com/ask. You will need an Edexcel username and password to access this service. Pearson: helping people progress, everywhere Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We’ve been involved in education for over 150 years, and by working across 70 countries, in 100 languages, we have built an international reputation for our commitment to high standards and raising achievement through innovation in education. Find out more about how we can help you and your students at: www.pearson.com/uk March 2013 Publications Code UG035048 All the material in this publication is copyright © Pearson Education Ltd 2013

NOTES ON MARKING PRINCIPLES 1 All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as

they mark the last. 2 Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do

rather than penalised for omissions. 3 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.

4 Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and

exemplification may be limited. 5 Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response. 6 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








It is appropriate to ignore subsequent work when the additional work does not change the answer in a way that is inappropriate for the question: e.g. incorrect cancelling of a fraction that would otherwise be correct It is not appropriate to ignore subsequent work when the additional work essentially makes the answer incorrect e.g. algebra. Transcription errors occur when candidates present a correct answer in working, and write it incorrectly on the answer line; mark the correct answer.

10 Probability Probability answers must be given a fractions, percentages or decimals. If a candidate gives a decimal equivalent to a probability, this should be written to at least 2 decimal places (unless tenths). Incorrect notation should lose the accuracy marks, but be awarded any implied method marks. If a probability answer is given on the answer line using both incorrect and correct notation, award the marks. If a probability fraction is given then cancelled incorrectly, ignore the incorrectly cancelled answer.

11 Linear equations

Full marks can be gained if the solution alone is given on the answer line, or otherwise unambiguously indicated in working (without contradiction elsewhere). Where the correct solution only is shown substituted, but not identified as the solution, the accuracy mark is lost but any method marks can be awarded.





1MA0_2H Question Working Answer Mark Notes 1 1 7 8 8 9

2 0 0 1 2 3 5 9 3 3 7 7 4 2 1│8 represents 18

3 B2 for a fully correct ordered diagram (B1 for correct unordered diagram or ordered with at most two errors or omissions) B1 for a correct key Accept stem written as 10, 20 etc but key only acceptable if consistent with this

*2 No + comparison 3 M1 for a correct start to the process eg. or or or M1 for completion of a fully correct method that will lead to an appropriate comparison C1 (dep on M2) for a correct statement with conclusion with 500 g or 25g more needed or 19 cakes or 25g and 23.75g SC :If no working then B1 for a correct statement with correct figures and units

1MA0_2H Question Working Answer Mark Notes 3 (a) 30 1 B1 for 30 minutes

(b) 20 1 B1 cao

(c) graph completed 2 B1 for horizontal line from (5, 20) to (5.30, 20)

B1 for a single straight line with the correct gradient from ‘(5.30, 20)’ to the time axis

4 (a) 1 − 0.2 − 0.1 0.7 ÷ 2

0.35 3 M1 for correctly using total probability is 1 or 100% if percentages used M1 (dep) for complete correct method to complete the solution A1 for 0.35 or 35% or oe

(b) 20 2 M1 for 0.1 × 200 oe A1 cao SC : If M0 then award B1 for an answer of

5 π × 5 × 1.80 28.27 3 M1 for use of π × x (with x = 5 or x = 2.5) or 2 × π × x (with x = 5 or x = 2.5) M1 for π × 5 × 1.8(0) or 2 × π × 2.5 × 1.8(0) A1 for 28.26 or 28.27 or 28.28 or 28.3(0) or 28.8(0)


6 414.96 5 M1 for a correct method to work out the amount of oil required to fill the tank M1 for a correct method to find the cost of oil required before the discount M1 for a correct method of finding 5% of their calculated cost M1 (dep on previous M1) for a correct method to find the discounted cost A1 for correct answer of 414.96 or 41496p OR M1 for a correct method of finding 5% of the cost of 1 litre of oil M1 (dep on previous M1) for a correct method to find the discounted cost of 1 litre of oil M1 for a correct method to work out the amount of oil required to fill the tank M1 for a correct method to find the discounted cost of the oil required A1 for correct answer of 414.96 or 41496p OR M1 for a correct method to work out the amount of oil required to fill the tank M1 for a correct method of finding 5% of their calculated amount of oil M1 (dep on previous M1) for a correct method to find the reduced amount of oil M1 for a correct method to find the cost of the reduced amount of oil A1 for correct answer of 414.96 or 41496p

1MA0_2H Question Working Answer Mark Notes 7* (a) 2.5 2 M1 for 15 ÷ 6 oe

A1 for 2.5 or 2

*(b)

Yes + evidence

2

M1 for a correct method to change 15 miles into kilometres C1(dep M1) for 24 km and statement with correct conclusion [SC: B1 for “Yes” oe and 24 km shown if M0 scored] or M1 for a correct method to change 20 kilometres into miles C1(dep M1) for 12.5 miles and statement with correct conclusion [SC: B1 for “Yes” oe and 12.5 miles shown if M0 scored]

1MA0_2H Question Working Answer Mark Notes 8 x x³− 3x

2 2 2.1 2.(961) 2.2 4.(048) 2.3 5.(267) 2.4 6.(624) 2.5 8.(125) 2.6 9.(776) 2.7 11.(583) 2.8 13.(552) 2.9 15.6(89) 3 18

2.85 14.5(99...) 2.86 14.8(13...) 2.87 15.0(29...) 2.88 15.2(47...) 2.89 15.4(67...)

2.9 4 B2 for a trial 2.8 ≤ x ≤ 2.9 evaluated correctly (B1 for a trial evaluated correctly for 2 ≤ x ≤ 3 ) B1 for a different trial evaluated correctly for 2.85 ≤ x < 2.9 B1 (dep on at least one previous B1) for 2.9 NB For trials where x has one decimal place: x ≤ 2.6 trials must be evaluated to at least 1 sf truncated or rounded 2.6 < x < 2.85 trials must be evaluated to at least 2 sf truncated or rounded 2.85 ≤ x ≤ 2.9 trials must be evaluated to at least 3 sf truncated or rounded NB. Accept 15 or 15.0 for trial at x =2.87 No working scores 0 marks. If candidate is clearly working with x3 − 3x − 15 = 0 then use same scheme as above but subtract 15 from all evaluated values in the table

1MA0_2H Question Working Answer Mark Notes 9 1180 3 M1 for a correct method to find the area of the cross section

M1 (dep) for a complete correct method for the volume of the prism A1 cao OR M1 for a correct method to find the volume of one cuboid M1 (dep) for a complete correct method for the volume of the prism A1 cao

10 Translation; 2 B1 for translation B1 for

NB: B0 if more than one transformation given

11 (a) 3x + 12 + 10x − 2 13x + 10 2 M1 for correct method to expand one bracket eg 3 × x + 3 × 4 or 3x + 12 or 2×5x − 2×1 or 10x − 2 A1 for 13x + 10

(b) 2x² −8x + x −4 2x² − 7x − 4 2 M1 for all 4 terms (and no additional terms) correct ignoring signs or 3 out of no more than four terms correct A1 for 2x² − 7x − 4

(c) 3y(2y − 3x) 2 B2 for 3y(2y − 3x) (B1 for 3(2y² − 3xy) or y(6y − 9x) or 3y(2y + 3x) or 3y(2y − ax) where a is any positive integer except 3 or 3y(by − 3x) where b is any positive integer except 2)

1MA0_2H Question Working Answer Mark Notes 12 (a) −2, −1, 0, 1, 2 B2 for all 4 correct values; ignore repeats, any order

(B1 for 3 correct (and no incorrect values) eg. −2, −1, 0 or one additional value eg. −3, −2, −1, 0, 1)

(b) p>6 2 M1 for clear intention to add 7 to both sides or 3p > 11 + 7 or clear intention to divide all 3 terms by 3 as a first step or 3p > 18 or 3p = 18 or 3p < 18 or A1 for p > 6 as final answer NB: (p =) 6 on the answer line scores M1 A0

13 (a) 11.5 3 M1 for 13² − 6² or 169 − 36 or 133 M1 (dep on M1) for √"13 6 " or √133 A1 for answer in the range 11.5 − 11.6

(b)

47.2

3

M1 for cos (RPQ)= oe OR sin PQR = with PQR clearly identified

M1 for (RPQ =+) cos-1 oe OR PQR = sin-1 with PQR clearly identified A1 for answer in the range 47.1 − 47.2 SC : B2 for an answer of 0.823(033...) or 52.3(95...) or 52.4

1MA0_2H Question Working Answer Mark Notes 14 (a) 100 = 4 × 2 × c 12.5 2 M1 for correct substitution into formula

A1 for 12.5 oe

(b) 14

4 k + 1 k = 4m² − 1 or 2m = √( k + 1) 4 k + 1 k = 4m² − 1

k = 4m² − 1 3 M1 for correct method to clear fraction or remove square root sign M1 (dep) for a fully correct method to both clear fraction and remove square root sign A1 for k = 4m² − 1 or k = (2m + 1)(2m − 1)

15 (a) × (4 + 12) × 10 80 2

M1 for a fully correct method for area of QRST A1 cao

(b) For example 3

PT + 10 = 3PT 2PT = 10

5 3 M1 for a correct scale factor or ratio using two corresponding sides from two similar triangles or two sides within the same triangle (may be seen within an equation) eg. oe or 4 : 12 oe or or or etc. M1 for a correct equation with PT or PS as the only variable or complete correct method using scale factor A1 cao


16 (a) 154500 – 150000

× 100

3 3 M1 for 154500 – 150000 or 4500 M1 for 100 oe A1 cao OR M1 for (× 100) M1 for “ × 100 “ − 100 oe A1 cao

(b)

154500 × + 154500

160680 × + 160680 or 154500 × 1.04²

167107.20

3

M1 for 154500 × or 6180 or 12360 or 160680 or 166860 or 1.04 × 154500 M1 (dep) for (154500 + ‘6180’) × or 6427.2(0) or ‘ 160680’ × 1.04 A1 for 167107.2(0) as final answer OR M2 for 154500 × 1.04² (M1 for 154500 × 1.04) A1 167107.2(0) as final answer


17 2.73 …0.732 …

1.931851… 2 M1 for 2.73… or 0.732…or 3.73…or 1.931 or 1.932 or 1.93

or (1 +√3) or (√3− 1) or (2 +√3) or 1.65... or 0.855... A1 for 1.9318(5…) SC: B1 for 2.5127(17...)

18 (a) minimum = 5 lower quartile = 14 median = 25 upper quartile = 30 maximum = 44

box plot 3 B3 for fully correct box plot (B2 for at least 3 correct values plotted including box and tails or 5 correct values indicated) (B1 for at least 2 correct values plotted including box or tails or 3 or 4 correct values indicated)

(b) comparisons 2 B1 for a correct comparison (ft) of medians B1 for a correct comparison (ft) of ranges or IQRs

19 × π × 15² 58.8 2 M1 for a correct method to find the area of sector OAB A1 for answer in range 58.8 − 58.9125


20 15.0 3 M1 for 82 + 82 – 2 × 8 × 8 × cos 140 M1 (dep) for correct order of evaluation or 226.(05...) A1 for answer in range 15.0 − 15.04 OR M1 for

M1 for PR = × sin 140

A1 for answer in range 15.0 − 15.04 OR M1 for 8 × sin70 or 8 × cos20 M1 for 2 × 8 × sin70 or 2 × 8 × cos 20 A1 for answer in range 15.0 − 15.04


21 Total area = (0.12 × 40) + (0.36 × 20) + (0.7 × 20) + (0.56 × 20) + (0.18× 40) = 44.4 Area (140 < w < 200) = (0.36 × 20) + (0.7 × 20) + (0.56 × 20) = 32.4 32.4 ÷ 44.4

0.73 4 M1 for a method to find the frequency or the area of any one block M1 for a method (with correct values) to find total area of all blocks or 44.4 or 1110 or a correct method (with correct values) to find total area of middle 3 blocks or 32.4 or 810 M1 (dep on M2) for a correct method to find required proportion (could lead to a decimal or a percentage or a fraction) A1 for answer which rounds to 0.73 or 73% or or equivalent fraction


22

× π × 15² × 40

− × π × 7.5² × 20

8250 4 B1 for 15cm as diameter or 7.5 cm as radius of smaller cone (may be marked on diagram or used in a formula) M1 for a numerical expression for the volume of one cone eg. × π × 15² × 40 (=9424...) or × π × 7.5² × 20 (=1178...)

M1 for × π × 15² × 40 oe −

× × π × 7.5² × 20 oe A1 for answer in the range 8240 − 8250 OR B1 for 2³ M1 for a numerical expression for the volume of the large cone eg. × π × 15² × 40 (=9424...)

M1 volume of frustrum = × × π × 15² × 40 oe A1 for answer in the range 8240 − 8250

23

11 2 M1 for a × 50 oe A1 for 11 (accept 12)


*24

0.229 because the LB and UB agree to that number of

figures

5 B1 for 3.465 or 3.475 or 3.474999… B1 for 8.1315 or 8.1325 or 8.132499… M1 for √ .

. as UB OR √ .

. as LB

C1 (dep on all previous marks) for 0.2292… and 0.2288… both values must clearly come from working with correct values C1 for 0.229 from 0.2292… and 0.2288… and ‘both LB and UB round to 0.229’

25 1 + √5 5 M1 for × x × x × sin30° oe

M1 for (x − 2)( x + 1) oe or 2 1 sin90 M1 (dep on at least one previous M1) for formation of equation from equating areas with x as the only variable A1 for x² − 2x − 4 = 0 oe in the form ax2 + bx + c = 0 or ax2 + bx = c A1 cao





Order Code UG035048 March 2013



Mark Scheme (Results) Summer 2013 GCSE Mathematics (Linear) 1MA0 Higher (Non-Calculator) Paper 1H

Edexcel and BTEC Qualifications Edexcel and BTEC qualifications come from Pearson, the world’s leading learning company. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information visit our qualifications websites at www.edexcel.com or www.btec.co.uk for our BTEC qualifications. Alternatively, you can get in touch with us using the details on our contact us page at www.edexcel.com/contactus. If you have any subject specific questions about this specification that require the help of a subject specialist, you can speak directly to the subject team at Pearson. Their contact details can be found on this link: www.edexcel.com/teachingservices. You can also use our online Ask the Expert service at www.edexcel.com/ask. You will need an Edexcel username and password to access this service. Pearson: helping people progress, everywhere Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We’ve been involved in education for over 150 years, and by working across 70 countries, in 100 languages, we have built an international reputation for our commitment to high standards and raising achievement through innovation in education. Find out more about how we can help you and your students at: www.pearson.com/uk Summer 2013 Publications Code UG037223 All the material in this publication is copyright © Pearson Education Ltd 2013

NOTES ON MARKING PRINCIPLES 1 All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark

the last. 2 Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than

penalised for omissions. 3 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.

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

may be limited. 5 Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response. 6 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








It is appropriate to ignore subsequent work when the additional work does not change the answer in a way that is inappropriate for the question: e.g. incorrect canceling of a fraction that would otherwise be correct It is not appropriate to ignore subsequent work when the additional work essentially makes the answer incorrect e.g. algebra. Transcription errors occur when candidates present a correct answer in working, and write it incorrectly on the answer line; mark the correct answer.

10 Probability

Probability answers must be given a fractions, percentages or decimals. If a candidate gives a decimal equivalent to a probability, this should be written to at least 2 decimal places (unless tenths). Incorrect notation should lose the accuracy marks, but be awarded any implied method marks. If a probability answer is given on the answer line using both incorrect and correct notation, award the marks. If a probability fraction is given then cancelled incorrectly, ignore the incorrectly cancelled answer.

11 Linear equations Full marks can be gained if the solution alone is given on the answer line, or otherwise unambiguously indicated in working (without contradiction elsewhere). Where the correct solution only is shown substituted, but not identified as the solution, the accuracy mark is lost but any method marks can be awarded.





PAPER: 1MA0_1H Question Working Answer Mark Notes

1 (a) 331.705 1 B1 cao

(b) 179300 1 B1 cao

2 24 4 M1 for 0.15 × 240 ( = 36) oe M1 for × 240 ( = 180) oe M1 (dep on both prev M1) for 240 – “180” – “36” A1 cao OR M1 for 15(%) + 75(%) ( = 90(%)) M1 for 100(%) – “90(%)” ( = 10(%)) M1 (dep on both prev M1) for “ ” × 240 oe A1 cao OR M1 for 0.15 + 0.75( = 0.9) oe M1 for “0.9” × 240( = 216) oe M1 (dep on both prev M1) for 240 � “216” A1 cao OR M1 for 0.15 + 0.75( = 0.9) oe M1 for 1 – “0.9”( = 0.1) oe M1 (dep on both prev M1) for “0.1” × 240 oe A1 cao

PAPER: 1MA0_1H Question Working Answer Mark Notes 3 2| 4 7 8

3| 0 3 3 5 7 8 8 4| 1 1 2 4 4 5 Key,eg 4|1 is 4.1(kg)

3 B2 for correct ordered stem and leaf (B1 for fully correct unordered or ordered with one error or omission) B1 (indep) for key (units not required)

4 (a) 6 + 3t 1 B1 for 6 + 3t

(b) 6x2 + 15x 2 B2 for 6x2 + 15x (B1 for 6x2 or 15x)

(c) m2 + 10m + 3m + 30 m2 + 13m + 30 2 M1 for all 4 terms (and no additional terms) correct with or without signs or 3 out of no more than four terms correct with signs A1 for m2 + 13m + 30

5 5|525 5|105 3|21 7

3 × 5 × 5 × 7 3 M1 for continual prime factorisation (at least first 2 steps correct) or first two stages of a factor tree correct M1 for fully correct factor tree or list 3, 5, 5, 7 A1 3 × 5 × 5 × 7 or 3 × 52 × 7

PAPER: 1MA0_1H Question Working Answer Mark Notes 6 7 3

M1 for 4 × 10 or 40 or 12 6 15

4x+ + +

or a correct equation

M1 for a complete correct method A1 cao

7 (a) (4,0) (3, 0) (3, -1) (2, -1) (2, 2) (4, 2)

Correct position

2 B2 for correct shape in correct position (B1 for any incorrect translation of correct shape)

(b) Rotation 180° (0,1)

3 B1 for rotation B1 for 180° (ignore direction) B1 for (0, 1) OR B1 for enlargement B1 for scale factor -1 B1 for (0, 1) (NB: a combination of transformations gets B0)


8 20 3000.5

12000 3 B1 for 20 or 300 used M1 for “20” × “300” or " "

. or " "

. , values do not

need to be rounded A1 for answer in the range 11200 –13200 SC B3 for 12000 with or without working

9

LCM (80, 50) = 400 Matt 400 ÷ 50 = 8 Dan 400 ÷80 = 5 OR 50 = 2 × 5 (× 5) 80 = 2 × 5 (× 2 × 2 × 2)

Matt 8 Dan 5

3 M1 lists multiples of both 80 (seconds) and 50 (seconds) (at least 3 of each but condone errors if intention is clear, can be in minutes and seconds) or use of 400 seconds oe. M1 (dep on M1) for a division of "LCM" by 80 or 50 or counts up “multiples” (implied if one answer is correct or answers reversed) A1 Matt 8 and Dan 5 SC B1 for Matt 7, Dan 4 OR M1 for expansion of both 80 and 50 into prime factors. M1 demonstrates that both expansions include 10 oe A1 Matt 8 and Dan 5 SC B1 for Matt 7, Dan 4


10 1.5 4 M1 for correct expression for perimeter eg. 4 + 3x + x + 6 + 4 + 3x + x + 6 oe M1 for forming a correct equation eg. 4 + 3x + x + 6 + 4 + 3x + x + 6= 32 oe M1 for 8x = 12 or 12 ÷ 8 A1 for 1.5 oe OR M1 for correct expression for semi-perimeter eg. 4 + 3x + x + 6 oe M1 for forming a correct equation eg. 4 + 3x + x + 6 = 16 oe M1 for 4x = 6 or 6 ÷ 4 A1 for 1.5 oe


*11

× 60 = 75

Debbie + explanation

4 M1 for reading 24 (mins) and 30 (km) or a pair of other values for Debbie M1 for correct method to calculate speed eg. 30 ÷ 24 oe A1 for 74 – 76 or for 1.2 – 1.3 and 1.1 C1 (dep on M2) for correct conclusion, eg Debbie is fastest from comparison of “74 – 76” with 66 (kph) or “1.2 – 1.3” and 1.1 (km per minute) OR M1 for using an appropriate pair of values for Ian’s speed eg 66 and 60, 33 and 30, 11 and 10 M1 for pair of values plotted on graph A1 for correct line drawn C1 (dep on M2) for Debbie is fastest from comparison of gradients. OR M1 for reading 24 (mins) and 30 (km) or a pair other values for Debbie M1 for Ian’s time for same distance or Ian’s distance for same time. A1 for a pair of comparable values. C1 (dep on M2) for Debbie is fastest from comparison of comparable values.


12 x – 2 -1 0 1 2 3 4 y 4 4.5 5 5.5 6 6.5 7

y = x + 5 drawn

3 (Table of values/calculation of values) M1 for at least 2 correct attempts to find points by substituting values of x. M1 ft for plotting at least 2 of their points (any points plotted from their table must be plotted correctly) A1 for correct line between x = -2 and x = 4 (No table of values) M1 for at least 2 correct points with no more than 2 incorrect points M1 for at least 2 correct points (and no incorrect points) plotted OR line segment of y = x + 5 drawn A1 for correct line between x = -2 and x = 4 (Use of y=mx+c) M1 for line drawn with gradient 0.5 OR line drawn with y intercept at 5 M1 for line drawn with gradient 0.5 AND line drawn with y intercept at 5 A1 For correct line between x = -2 and x = 4 SC B2 for a correct line from x = 0 to x = 4

*13

Yes with explanation

3 M1 for bearing ± 2 ° within overlay M1 for use of scale to show arc within overlay or line drawn from C to ship’s course with measurement C1(dep M1) for comparison leading to a suitable conclusion from a correct method


14 (a) Line joins an empty circle at – 2 to a solid circle at 3

diagram 2 B2 cao (B1 for line from – 2 to 3)

(b) 2x ≥ 7 x ≥ 3.5 2 M1 for correct method to isolate variable and number terms (condone use of =, >, ≤, or <) or (x =) 3.5 A1 for x ≥ 3.5 oe as final answer

*Q15

No + explanation

3 M1 for 500 × 9 × 10-3 oe A1 for 4.5 C1 (dep M1) for correct decision based on comparison of their paper height with 4 OR M1 for 4 ÷ 500 oe A1 for 0.008 C1 (dep M1) for correct decision based on comparison of their paper thickness with 0.009 OR M1 for 4 ÷ (9 × 10-3) oe A1 for 444(.4...) C1 (dep M1) for correct decision based on comparison of their number of sheets of paper with 500

16 £500 3 M1 for 70% = 350 or

M1 for × 100 oe A1 cao

PAPER: 1MA0_1H

Question Working Answer Mark Notes 17 1 hour 45 mins 6 M1 for method to find volume of pond,

eg 21

(1.3 + 0.5) × 2 × 1 (= 1.8)

M1 for method to find the volume of water emptied in 30 minutes, eg 1 × 2 × 0.2 (= 0.4), 100 × 200 × 20 (= 400000) A1 for correct rate, eg 0.8 m³/hr, 0.4 m³ in 30 minutes M1 for correct method to find total time taken to empty the pond, eg “1.8” ÷ “0.8” M1 for method to find extra time, eg 2 hrs 15 minutes − 30 minutes A1 for 1.75 hours, 1 hours, 1 hour 45 mins or 105 mins OR M1 for method to find volume of water emptied in 30 minutes,.eg. 1 × 2 × 0.2 (= 0.4), 100 × 200 × 20 (= 400000) M1 for method to work out rate of water loss eg. “0.4” × 2 A1 for correct rate, eg 0.8 m³/hr M1 for correct method to work out remaining volume of water eg. (1.1 + 0.3) × 2 × 1 (= 1.4) M1 for method to work out time, eg “1.4” ÷ “0.8” A1 for 1.75 hours, 1 hours, 1 hour 45 mins or 105 mins NB working could be in 3D or in 2D and in metres or cm throughout


18 12x + 21y = 3 12x + 40y = 60 19y = 57 y= 3 3x + 10× 3 = 15 3x = – 15 Alternative method x =

3 + 10y = 15 3 – 21y +40y = 60 19y = 57 x =

x= -5, y = 3 4 M1 for a correct process to eliminate either x or y or rearrangement of one equation leading to substitution (condone one arithmetic error) A1 for either x = −5 or y = 3 M1 (dep) for correct substitution of their found value A1 cao

19 – 5, 0.2, 0.5, 1 -5, 5-1, 0.5 , 50 2 M1 for either 5-1 or 50 evaluated correctly A1 for a fully correct list from correct working, accept original numbers or evaluated (SC B1 for one error in position or correct list in reverse order)

PAPER: 1MA0_1H Question Working Answer Mark Notes 20 5x2 4 M1 for 4x × 4x

M1 for (2x ×4x)/2 or (2x × x)/2 or(3x ×4x)/2 M1(dep M2) for “16 x2” – “4 x2”– “x2” – “6 x2”

A1 for 5x2

OR M1 for 2x ² 4x ² (= √20 ² = √20 x) M1 for ² 2 ² (= √5 ² = √5 x)

M1(dep M2) for "√ " "√ " (= √ ² A1 for 5x2

21 (a) Cf table: 4, 9, 25, 52, 57,60

cf graph Correct Cf graph 3 B1 Correct cumulative frequencies (may be implied by

correct heights on the grid) M1 for at least 5 of “6 points” plotted consistently within each interval A1 for a fully correct CF graph

(b)(i)

(ii)

IQR = UQ – LQ

172

12 - 14

3 B1 for 172 or read off at cf = 30 or 30.5 from a cf graph, ft provided M1 is awarded in (a) M1 for readings from graph at cf = 15 or 15.25 and cf = 45or 45.75 from a cf graph with at least one of LQ or UQ correct from graph (± ½ square). A1ft provided M1 is awarded in (a)


22 1200 cm3 4 M1 for 10 × 2 × 2 and 15 × 2 M1 for “40” × “30” A1 for 1200 B1 (indep) for cm3

OR M1 for 10 × 15 or 23 or 8 indicated as scale factor M1 for 10 × 15 × 2 × 2 × 2 A1 for 1200 B1 (indep) for cm3

SC B2 for 600 cm3 (B1 for 600)

23 4 55 3

43

2 M1 for (x ± 5)(x±3) A1 for

24 12 ÷ 10 = 1.2 15 ÷ 5 = 3 13 ÷ 5 = 2.6 18 ÷ 10 = 1.8 3 ÷ 15 = 0.2

Histogram

3 B3 for fully correct histogram (B2 for 4 correct blocks) (B1 for 3 correct blocks) (If B0, SC B1 for correct key eg 1cm2 = 2 (calls) Or frequency ÷ class interval for at least 3 frequencies) NB Apply the same mark scheme if a different frequency density is used.


25 (a) a = 4, b = 5 3 M1 for sight of (x – 4)2

M1 for (x – 4)2 – 16 + 21 A1 for a = 4, b = 5 OR M1 for x2 – 2ax + a2 + b M1 for –2a = – 8 and a2 + b =21 A1 for a = 4, b = 5

(b) (4, 5) 1 B1 ft

26 50 1 1 1 50 1 1 1 50

126720

4 M1 for 3 fractions , , where a < 10, b < 9 and c < 8 M1 for or or (= M1 for + +

or 3 ×

A1 for oe. eg.

Alternative Scheme for With Replacement M1 for (=

M1 for × 3 (= M0 A0 No further marks


27 (a) a - b 1 B1 for a - b oe

(b) a + b 3 M1 for a correct vector statement for eg. ( =) NQ + QR or ( =) NS + SR M1 for SQ (+ QR) or QS (+ SR) (SQ, QR, QS, SR may be written in terms of a and b) A1 for (a b) + b oe or (b – a) + a oe

28 (a) (90, 0) 1 B1 for (90, 0) (condone ( , 0))

(b) Correct graph 1 B1 for graph through (0, 2) (90, 0) (180, -2) (270, 0) (360, 2) professional judgement





Order Code UG037223 Summer 2013



Mark Scheme (Results) Summer 2013 GCSE Mathematics (Linear) 1MA0 Higher (Calculator) Paper 2H

Edexcel and BTEC Qualifications Edexcel and BTEC qualifications come from Pearson, the world’s leading learning company. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information visit our qualifications websites at www.edexcel.com or www.btec.co.uk for our BTEC qualifications. Alternatively, you can get in touch with us using the details on our contact us page at www.edexcel.com/contactus. If you have any subject specific questions about this specification that require the help of a subject specialist, you can speak directly to the subject team at Pearson. Their contact details can be found on this link: www.edexcel.com/teachingservices. You can also use our online Ask the Expert service at www.edexcel.com/ask. You will need an Edexcel username and password to access this service. Pearson: helping people progress, everywhere Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We’ve been involved in education for over 150 years, and by working across 70 countries, in 100 languages, we have built an international reputation for our commitment to high standards and raising achievement through innovation in education. Find out more about how we can help you and your students at: www.pearson.com/uk Summer 2013 Publications Code UG037224 All the material in this publication is copyright © Pearson Education Ltd 2013















10 Probability







PAPER: 1MA0_2H Question Working Answer Mark Notes 1 40.5 3 M1 for 1.5×6 or 1.5 ×1.5

M1 for adding area of 5 or 6 faces provided at least 3 are the correct area A1 cao NB: anything that leads to a volume calculation 0 marks.

*2

Not enough mincemeat since

600<700

OR

Only able to make 38 mince pies since

insufficient mincemeat

4

M1 for 45 ÷ 18 (= 2.5) M1 for 2.5 used as factor or divisor A1 for ingredients as 562.5 and 875 and 250 and 700 and 2.5 (accept 2 or 3) OR for availables as 400, 400, 200 240, 2.4 (accept 2 or 3) C1 ft (dep on at least M1) for identifying and stating which ingredient is insufficient for the recipe (with some supportive evidence) OR M1 for a correct method to determine the number of pies one ingredient could produce M1 for a correct method to determine the number of pies all ingredient could produce A1 for 80 and 51 and 90 and 38 and 108 C1 ft (dep on at least M1) for identifying and stating which ingredient is insufficient for the recipe. (with some supportive evidence)

PAPER: 1MA0_2H Question Working Answer Mark Notes 3

(a)

Points plotted at (1,8200) and (3.5,5000)

1

B1 for points accurately plotted ±1/2 square tolerance

(b)

‘the older the car the lower the value’ ‘the greater the value the

newer the car’

1

B1 for an acceptable relationship eg. ‘the older the car the lower the value’ (accept ‘negative correlation’ but not just ‘negative’)

(c) 5200 to 6600 2 M1 for a single line segment with negative gradient that could be used as a line of best fit or a vertical line from 2.5 or a point at (2.5,y) where y is from 5200 to 6600 A1 for given answer in the range 5200 − 6600

4 126 3

M1 for 1 – 0.05 – 0.32 (= 0.63) M1 for ‘0.63’ × 200 A1 cao OR M1 for 0.05 × 200 (= 10) or 0.32 × 200 (= 64) or 0.37 × 200 (=74) M1 for 200 – ‘10’ – ‘64’ A1 cao OR M1 for 100 – 5 – 32 (= 63)

M1 for "63" 200100

×

A1 cao

SC: B2 for 126200

as the answer.


(a) Response boxes overlap and are not

exhaustive

2

B2 for TWO aspects from: No time frame given Non-exhaustive responses Response boxes over-lapping (B1 for ONE correct aspect)

(b) How many magazines do you buy each

month? 0-4 5-8 over 8

2

B1 for a question with a time frame B1 for at least 3 correctly labelled response boxes (non-overlapping, need not be exhaustive) or for a set of response boxes that are exhaustive (could be overlapping) [Do not allow inequalities in response boxes]

(c) One reason 1

B1 for ONE reason Eg. All the same age, may all be males, may all like same types of magazines, sample too small, biased

6

4.8 4 M1 for 60 × 60 (=3600) M1 for 15000÷ 20 (=750) or 20÷15000 (=0.00133..) or “3600”÷15000 (=0.24) or 15000÷”3600” (=4.16..) M1 for “3600” ÷ (15000÷20) or “3600”×20÷15000 oe A1 cao


28% or

4 M1 for 100 � 30 (= 70) or 1 M1 for “70” ÷ (3 + 2) (= 14) or " "÷ (3+2) (=

)

M1 for “14” × 2 or 2

A1 for 28% or oe OR M1 for a correct method to find (100-30)% of any actual sum of money M1 for “350” ÷ (3 + 2) (= 70) M1 for “70” × 2 A1 for 28% or oe OR M1 for starting with two numbers in ratio 3:2, eg 21 and 14 M1 for equating sum of their numbers to 100 – 30 (=70%), eg ‘21’ + ‘14’ (=35) M1 for scaling sum of their numbers to 100%, eg ‘35’÷70×100 (=50) A1 for 28% or oe SC: award B3 for oe answers expressed in an incorrect form eg .

PAPER: 1MA0_2H Question Working Answer Mark Notes 8 10752 4 M1 for splitting the pentagon (or show the recognition of the

“absent” triangle) and using a correct method to find the area of one shape M1 for a complete and correct method to find the total area M1 (dep on at least one prev M1) for multiplying their total area by 2.56 (where total area is a calculation involving at least two areas) A1 cao

9 55 4 M1 for a correct method to find a different angle using 35° M1 for setting up a complete process to calculate angle x A1 cao B1 states one of the following reasons relating to their chosen method: Alternate angles are equal; Corresponding angles are equal; Allied angles / Co-interior angles add up to 180; the exterior angle of a triangle is equal to the sum of the interior opposite angles.


x x3 + 2x 4 72

4.1 77.(121) 4.2 82.(488) 4.3 88.(107) 4.4 93.(984) 4.5 100.(125) 4.6 106.(536) 4.7 113.(223) 4.8 120.(192) 4.9 127.(449) 5 135

4.65 109.8(44625) 4.66 110.5(14696) 4.67 111.1(87563) 4.68 111.8(63232) 4.69 112.5(41709)

4.7 4

B2 for a trial 4.6 ≤ x ≤ 4.7 evaluated correctly (B1 for a trial evaluated correctly for 4 ≤ x ≤ 5 ) B1 for a different trial evaluated correctly for 4.65≤ x < 4.7 B1 (dep on at least one previous B1) for 4.7 [Note: Trials should be evaluated to at least accuracy shown in table, truncated or rounded] No working scores 0 marks

11

3.52 3 M1 for 1.352 + 3.252 M1 (dep) for √(1.352 + 3.252 ) (= √12.385) A1 for answer in the range 3.51 to 3.52


12

(a)

3x – 6 = x + 7 2x = 13

6.5

3

M1 for 3×x – 3×2 (=3x – 6) or seen M1 for correct method to isolate the terms in x or the number terms on opposite sides of an equation A1 for 6.5 oe

(b) 2 – y = 1 × 5 – 3 2 M1 for intention to multiply both sides by 5 (to give 2 – y = 1 × 5) A1 cao

13

(a)

(3, 3.5) oe

2 M1 for a correct method to find the value of either the x coordinate or the y coordinate of the midpoint or x = 3 or y = 3.5 A1 cao

(b) -1.8 oe 2 M1 for correct method to find the gradient OR (+)1.8 A1 for -1.8 oe


*14 The Friendly Bank

4 M1 for a correct method to find interest for the first year for either bank OR correct method to find the value of investment after one year for either bank OR use of the multiplier 1.04 or 1.05 M1 for a correct full method to find the value of the investment (or the value of the total interest) at the end of 2 years in either bank A1 for 2100.8(0) and 2110.5(0) (accept 100.8(0) and 110.5(0)) C1 (dep on M1) ft for a correct comparison of their total amounts, identifying the bank from their calculations OR M1 for either 1.04 × 1.01 or 1.05 × 1.005 M1 for 1.04 × 1.01 and 1.05 × 1.005 A1 for 1.0504 and 1.05525 C1 (dep on M1) ft for a correct comparison of their total multiplying factors identifying the bank from their calculations


15

(a) -2 -1 0 1 2 3 4 8 3 0 -1 0 3 8

2

B2 for 8, -1, 0, 8 (B1 for at least two of 8, -1, 0, 8)

(b) Correct curve 2 M1 (ft) for at least 5 points plotted correctly A1 for a fully correct curve

(c) x2 – 2x – 3 = 0 OR (x − 3)(x + 1) = 0

3 and −1 2 M1 for the straight line y = 3 drawn to intersect the “graph” from (a) A1 for both solutions OR M1 for identifying y = 3 from the table A1 for both solutions OR M1 for (x ± 3)(x ± 1) A1 for both solutions


*16 Angle POT = 180 – 90 – 32 = 58 (angle between radius and tangent = 90o and sum of angles in a triangle = 180o) Angle OST =angle OTS = 58÷2 (ext angle of a triangle equal to sum of int opp angles and base angles of an isos triangle are equal) or (angle at centre = 2x angle at circumference) OR Angle SOT = 90 + 32 = 122 (ext angle of a triangle equal to sum of int opp angles) (180 – 122) ÷ 2 (base angles of an isos triangle are equal)

29 5 B1 for angle OTP = 90o, quoted or shown on the diagram M1 for a method that leads to 180 – ( 90 + 32) or 58 shown at TOP M1 for completing the method leading to “58”÷2 or 29 shown at TSP A1 cao C1 for “angle between radius and tangent = 90o” and one other correct reason given from theory used NB: C0 if inappropriate rules listed OR B1 for angle OTP = 90o, quoted or shown on the diagram M1 for a method that leads to 122 shown at SOT M1 for (180 – “122”) ÷ 2 or 29 shown at TSP A1 cao C1 for “angle between radius and tangent = 90o” and one other correct reason given from theory used NB: C0 if inappropriate rules listed


17

(a) Box plot overlay

2

M1 for a box drawn with at least 2 correct points from LQ, Med and UQ A1 for a fully correct box plot

(b) Comparison of a measure of

spread plus a comparison of

medians (in context)

2 B1 for a correct comparison of a measure of spread (using either range or iqr) B1 for a correct comparison of medians For the award of both marks at least one of the comparisons made must be in the context of the question.

18

3p2 = y + 4

p2 = 4

3y +

43

3 M1 for clear intention to add 4 to both sides or divide

all terms by 3(with at least 3 terms) M1 for clear intention to find the square root from p2 = (expression in y)

A1 for oe (accept ± a correct root)

19 (a) 3(2 + 3x) 1 B1 for 3(2 + 3x)

(b) (y + 4)(y – 4)

1 B1 for (y + 4)(y – 4)

(c) (2p − 5)(p + 2)

2 M1 for (2p ± 5)(p ± 2) A1 for (2p − 5)(p + 2)

PAPER: 1MA0_2H Question Working Answer Mark Notes 20 cos y = 2.25 ÷ 6

y = cos-1 (2.25 ÷ 6) OR 6cos 75 = 1.55…

The ladder is not safe

because y is not near to 75

3 M1 for cos y = 2.25 ÷ 6 oe M1 for cos-1 (2.25 ÷ 6) C1 for sight of 67-68 and a statement eg this angle is NOT (near to) 75o and so the ladder is not steep enough and so not safe. OR M1 for cos 75 = x ÷ 6 M1 for 6cos 75 C1 for sight of 1.55(29…) and a statement eg that 2.25 NOT (near to) 1.55 and so the ladder is not steep enough and so not safe.

21

48 or 49

2 M1 for

460 100460 320 165

×+ +

(=48.67 ….) or 4609.5

or

4609.45

A1 for 48 or 49

22 1.33 3 M1 for 3.4 oe or 3.4 × 52 (=85) M1 for ‘3.4 × 52’ ÷ 82

A1 for answer in range 1.32 to 1.33 or 8564

23

d: UB = 54.5 (or 54.499), LB = 53.5 C: UB = 170.5 (or 170.499), LB = 169.5 170.5 ÷ 53.5 169.5 ÷ 54.5

3.19 3.11..

4 B1 for any one correct bound quoted M1 for 170.5 ÷ 53.5 or 169.5 ÷ 54.5 A1 for UB = answer in range 3.18 to 3.19 from correct working A1 for LB = 3.11.. from correct working


24

(a)

18.2

2

M1 for × 6 × 7 × sin60 A1 for answer in range 18.1 to 18.2

(b)

6.56 3 M1 for 62 + 72 – 2 × 6 × 7 × cos60 M1 for correct order of operation eg 36 + 49 – 42 (=43) A1 for answer in range 6.55 to 6.56

25

x = 2.87, y = −0.87

and x = −0.87, y = 2.87

6 M1 for x2 + (2 – x)2 = 9 M1 for 4 – 4x + x2 A1 for 2x2 – 4x – 5 = 0 oe 3 term simplified quadratic M1 for a correct method to solve their quadratic Eg x = 4 ± √(16 – 4×2×−5) 4 A1 for x = 2.87, y = −0.87 or better A1 for x = −0.87, y = 2.87 or better Award marks for equivalent algebraic expressions. Apply the same scheme as above for y first.





Order Code UG037224 Summer 2013



Mark Scheme (Results) November 2013 Pearson Edexcel GCSE in Mathematics Linear (1MA0) Higher (Non-Calculator) Paper 1H

Edexcel and BTEC Qualifications Edexcel and BTEC qualifications are awarded by Pearson, the UK’s largest awarding body. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information visit our qualifications websites at www.edexcel.com or www.btec.co.uk. Alternatively, you can get in touch with us using the details on our contact us page at www.edexcel.com/contactus. Pearson: helping people progress, everywhere Pearson aspires to be the world’s leading learning company. Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We’ve been involved in education for over 150 years, and by working across 70 countries, in 100 languages, we have built an international reputation for our commitment to high standards and raising achievement through innovation in education. Find out more about how we can help you and your students at: www.pearson.com/uk November 2013 Publications Code UG037492 All the material in this publication is copyright © Pearson Education Ltd 2013

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10 Probability








1 90 450 225 1.5 960

3 M1 for 6 ÷ 4 (= 1.5) or 4 ÷ 6 (= 0.66..) or ÷4 × 6 oe or sight of any one of the correct answers A1 for three correct A1 for all correct

2 (a) Plot (90,17) 1 B1 cao

(b) Positive 1 B1 Positive

(c) In range 16 to 20 2 M1 for a single straight line segment with positive gradient that could be used as a line of best fit or a vertical line from 110 or a point plotted at (110, y) where y is in the range 16 to 20 A1 for an answer in the range 16 to 20 inclusive

3 120 cm3 4 M1 for 1

2 × 3 × 4

M1 (dep) for ‘ 12

× 3 × 4’ × 20

A1 for 120 B1 (indep) for cm3


4 (a) 4y + 5x + 5 2 M1 5x or 5 seen A1 cao

(b) 3x(3x – 2y) 2 B2 for 3x(3x – 2y) (B1 for x(9x – 6y) or 3(3x2– 2xy) or 3x(ax – by) where a and b are integers not equal to zero)

(c) 4x + 8 1 B1 cao

(d) x2 – 2x – 15 2 M1 for 4 terms correct with or without signs or 3 out of no more than 4 terms correct with correct signs A1 cao

5 (a) 0.25 1 B1 oe

(b) 150 2 M1 for 0.75 × 200 oe A1 cao

6 (a) Shape with vertices at (–1, 3), (0, 6), (2, 6), (1, 3)

1 B1 for correct shape in correct position

(b) Rotation centre (0,0)

90˚ anticlockwise

3 B1 rotation B1 (centre) (0,0) B1 90˚ anticlockwise or 270˚ clockwise Note: award no marks if more than one transformation is given


7 (i) 20, 40, 60 12, 24, 36, 48, 60 20 = 4×5 = 2×2×5 12 = 4×3 = 2×2×3

3 and 5 or

any multiple of 3, 5

4 M1 attempts multiples of both 20 and 12 (at least 3 of each shown but condone errors if intention is clear) or identifies 60 or a multiple of 60 M1 (dep on M1) for a division by 20 or 12 or counts up ‘multiples’ or identifies a common multiple (implied if one answer is correct or answers reversed) A1 cheese slices (packets) 3, burgers (boxes) 5 or any multiple of 3, 5 OR M1 for expansion of either 20 or 12 into factors M1 for demonstration that both expansions include 4 (or 2 × 2) A1 cao for cheese slices (packets) 3, burgers (boxes) 5

(ii) 60 B1 for 60 or ft from their correct answer in (i) or ft ‘common multiple’

8 38 5 M1 3x − 5 = 19 – x M1 for a correct operation to collect the x terms or the number terms on one side of an equation of the form ax + b = cx + d A1 for x = 6 M1 for substituting their value of x in the three expressions and adding or substituting their value of x after adding the three expressions A1 cao


9 (a) Criticisms 2 B1 Qu 1 Overlapping boxes, no units B1 Qu 2 e.g. no time frame, non-specific responses, no number quantities, open to interpretation, no option for those who do not exercise

(b) Question given 2 B1 for a correct question with a time frame B1 for at least 3 correctly labelled non-overlapping response boxes (need not be exhaustive) or at least 3 response boxes that are exhaustive for all integer values of their time unit (could be overlapping) NB Units must be included in either question or response boxes to score full marks [Do not allow inequalities in response boxes]


*10 Not enough, needs £133

5 M1 for splitting the shape (or showing recognition of the “absent” rectangle) and using a correct method to find the area of one shape M1 for a complete and correct method to find the total area M1 for a complete method to find 70% of 19 (= 13.3) or 70% of their total cost or 70% of their area A1 114(m2) and (£)133 or 114(m2) and (£)13.3(0) and 108(m2) C1 (dep on M2) for a conclusion supported by their calculations OR M1 for a complete method for the number of tins required for one section of the area of the floor M1 for a complete method to find the number of tins for the whole floor M1 for a complete method to find 70% of their total number of tins and multiply by 19 A1 (£)133 C1 (dep on M2) for a conclusion supported by their calculations

11 164 5 M1 200 ÷ (3+2) (= 40) or an equivalent ratio seen M1 (dep) 3 ×‘40’ (= 120) or 2 ×‘40’ (= 80) or 120: 80 or 80:120 M1 a complete method to find 70% of their total number of large letters e.g. 0.7 × ‘80’ (=56) M1 multiplies their three totals by the correct unit price and adds, e.g. 60(p) × ‘120’ + (£)1 × ‘56’ + (£)1.50 × ‘24’ A1 164


12

x –2 –1 0 1 2 y –4 –1 2 5 8

y = 3x + 2 drawn

4 B1 for axes scaled and labelled (Table of values) M1 for at least 2 correct attempts to find points by substituting values of x M1 ft for plotting at least 2 of their points (any points from their table must be correctly plotted) A1 for correct line between x = –2 and x = 2 (No table of values) M1 for at least 2 correct points with no more than 2 incorrect points M1 for at least 2 correct points (and no incorrect points) plotted OR line segment of y = 3x + 2 drawn A1 for correct line between x = –2 and x = 2 (Use of y = mx + c) M1 for line drawn with gradient of 3 OR line drawn with y intercept at 2 M1 for line drawn with gradient of 3 AND with y intercept at 2 A1 for correct line between x = –2 and x = 2 SC B2 (indep of B1) for correct line segment between x = 0 and x = 2 (ignore any additional incorrect line segment(s))


13 35 × 10 = 350 33 × 11 = 363 363 − 350 = 13 OR 10×(35−33) =20 33 − 20 = 13

13 3 M1 35 × 10 (= 350) or 33 × 11 (= 363) M1 (dep) finding the difference in their totals e.g. ‘363’ – ‘350’ A1 cao OR M1 10×(35 − 33) (=20) or 11×(35 − 33) (=22) M1 (dep) 33 − ‘20’ or 35 − ‘22’ A1 cao

14 (a) 15

1 B1 oe

(b) 19

1 B1 cao

(c) 9×104×3×103 2.7 × 108 2 M1 27 × 107 oe or 9×3×104+3 A1 cao


15 6x + 8y = 10 6x − 9y =27 y = −1 3x − 4 = 5 3x = 9 x = 3 OR 9x + 12y = 15 8x − 12y = 36 x = 3 9 + 4y = 5 4y = −4 y = −1

x = 3, y = −1 4 M1 for a correct process to eliminate either variable (condone one arithmetic error) A1 cao for either x or y M1 (dep on M1) for correct substitution of found value into one of the equations or appropriate method after starting again (condone one arithmetic error) A1 cao OR M1 for full method to rearrange and substitute to eliminate either variable (condone one arithmetic error) A1 cao for either x or y M1 (dep on M1) for correct substitution of found value into one of the equations or appropriate method after starting again (condone one arithmetic error) A1 cao Trial and improvement scores 0 marks unless both x and y are correct

16 120 ÷ 20 = 6 62 = 36 36×300 = 10 800

10 800 3 M1 120 ÷ 20 (= 6) oe, can be implied by 1202 ÷ 202 M1 ‘6’2 × 300 A1 cao

17 (3,6,7) to (−2,2,5) (−5, −4, −2) (−2− 5, 2 − 4, 5 − 2)

(−7, −2, 3) 2 M1 for midpoint plus change or complete method for 2 out of 3 coordinates, can be implied by 2 correct values A1 cao


18 (a) 68 1 B1 cao

*(b) 120 ÷ 20 = 6 62 = 36 36×300 = 10 800

Yes as 28 > 20

or 35% > 25% or 53 < 60

3 M1 for reading a value from graph at time = 60 (=28, accept 27 to 28) M1 for ‘28’ ÷ 80 × 100 (= 35) or 25 ÷ 100 × 80 (= 20) C1 (dep on M2) for correct decision based on their figures OR M1 for 25 ÷ 100 × 80 (= 20) M1 for reading a value from graph at cf = 20 (=53, accept 52 to 54) C1 (dep on M2) for correct decision based on their figures

(c) 28, 53, 68, 76, 96 Box plot plotted 3 B1 for ends of whiskers at 28 and 96 with a box B1 ft for median at ‘68’ inside a box B1 for ends of box at 53 (accept 52 to 54) and 76

19 0.82 3 M1 for 1 – 0.7 (= 0.3) or 1 – 0.4 (= 0.6) M1for 1− ‘0.3’×’0.6’ A1 for 0.82 oe OR M1 for 1 – 0.7 (= 0.3) or 1 – 0.4 (= 0.6) M1 (0.7 × 0.4) + (0.7 × ‘0.6’) + (‘0.3’× 0.4) A1 for 0.82 oe


20 (a) 4 3 M1 for correct expansion to 32x − 8 or multiplying both sides by 3x or dividing both sides by 4 M1 for a compete and correct method to isolate the x terms and the number terms (condone one arithmetic error in multiplying out the bracket) A1 cao

(b) 2( 6) ( 3)( 3)( 6)y yy y− − ++ −

15( 3)( 6)

yy y

−+ −

3 M1 for common denominator of ( 3)( 6)y y+ −

M1 for 2( 6) 3( 3)( 6) ( 3)( 6)

y yy y y y

− +−

+ − + −oe

or 2( 6) ( 3)( 3)( 6)y yy y− − ++ −

oe

A1 for 15( 3)( 6)

yy y

−+ −

or 2

153 18

yy y

−− −

21 100 4 M1 y = kx2 oe or 36 = k×32

A1 k = 4 M1 (dep on M1) (y =) ‘4’×52

A1 cao


*22 360 − y 180 − 2y

4 M1 ADC =

2y

A1 180 − 2y

C2 (dep on M1) for both reasons Angle at centre is twice the angle at the circumference Opposite angles in cyclic quadrilateral add to 180˚ (C1 (dep on M1) for one appropriate circle theorem reason) OR M1 reflex AOC = 360 − y

A1 2

360 y− oe

C2 (dep on M1) for both reasons Angles around a point add up to 360˚ Angle at centre is twice the angle at the circumference (C1 (dep on M1) for one appropriate circle theorem reason)

23 Triangle with vertices at (−1,−4), (−1,−5),

(−3,−4.5)

2 M1 for correct shape and size and the correct orientation in the wrong position or two vertices correct A1 cao


24 (a) 𝐴𝐵��⃗ = −a + b

𝑂𝑁��⃗ = 𝑂𝐴��⃗ + 23𝐴𝐵��⃗

𝑂𝑁��⃗ = a + 23

(–a + b)

= 13

a + 23

b

OR

𝑂𝑁��⃗ = 𝑂𝐵��⃗ + 13𝐵𝐴��⃗

𝑂𝑁��⃗ = b + 13

(−b + a)

= 13

a+ 23

b

13

a + 23

b 3 M1 for correct vector equation involvingON , eg. ON OA AN= +

, may be written, partially or fully, in terms of a and b, e.g. ( ON =

)

a + 32 AB

M1 for showing answer requires =AN32 AB or =BN

31 BA

A1 13

a + 23

b oe

(b) 𝑂𝐷��⃗ =𝑂𝐴��⃗ +𝐴𝐶��⃗ +𝐶𝐷��⃗ = a + b + b = a + 2b

OD = 3(31

a+32

b )

OD = 3 ON

Proof 3 M1 for a correct vector statement for OD

or ND

in terms of a and

b, e.g. 𝑂𝐷��⃗ = a + b + b oe or ND

= 32

(−b + a) + b + b oe

A1 for correct and fully simplified vectors for ON (may be seen in

(a)) and for OD

(= a + 2b) or ND

(= 23

a + 43

b)

C1 (dep on A1) for statement that 𝑂𝐷��⃗ or ND

is a multiple of 𝑂𝑁��⃗ (+ common point)


Mark Scheme (Results) November 2013 Pearson Edexcel GCSE In Mathematics Linear (1MA0) Higher (Calculator) Paper 2H

Edexcel and BTEC Qualifications Edexcel and BTEC qualifications are awarded by Pearson, the UK’s largest awarding body. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information visit our qualifications websites at www.edexcel.com or www.btec.co.uk. Alternatively, you can get in touch with us using the details on our contact us page at www.edexcel.com/contactus. Pearson: helping people progress, everywhere Pearson aspires to be the world’s leading learning company. Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We’ve been involved in education for over 150 years, and by working across 70 countries, in 100 languages, we have built an international reputation for our commitment to high standards and raising achievement through innovation in education. Find out more about how we can help you and your students at: www.pearson.com/uk November 2013 Publications Code UG037493 All the material in this publication is copyright © Pearson Education Ltd 2013















10 Probability








1 (a) 18.75 2 M1 for 84 or 4.48 or

11225

or 18.7 or 18.8 or 19 or 20 or 754

A1 cao

(b) 20 1 B1 for 20 or ft from their answer to (a) provided (a) is written to 2 or more significant figures

2 (a) 12 2 M1 for 32÷8 (=4) or

3 328× oe

A1 for 12

(b)

36 2 M1 for correct method to find 45% of 80 A1 cao

3 4 2 M1 for 14 or

3 7 57n

+= or any fraction equivalent to

27

or 57

A1 cao

4

2 M1 for a 5cm by 5 cm square or a 5cm by 3 cm rectangle or a 5 cm by 2 cm rectangle A1 for correct elevation with dividing line NB: diagrams which appear to have a 3D element get 0 marks


5 £26.50 or

HK$325.95

3 M1 for 3179.55 ÷ 12.3 (=258.5) M1 (dep) for 285 - '258.5' A1 for £26.50 (correctly stated with currency) OR M1 for 285 × 12.3 (=3505.5) M1 (dep) for '3505.5' - 3179.55 (=325.95) A1 for HK$325.95 (correctly stated with currency)

6

19 4 M1 for 130 − 96 (=34) M1 for 73 − 55 (=18) M1 for '34' − 9 − '18' + 12 A1 cao OR M1 for 96 − 55 − 12 (=29) M1 for 9 + '29' (=38) M1 for 130 − 73 − '38' A1 cao


*7 Small with correct figures for comparison

4 M1 for one calculation eg 6.5 ÷ 30 (=0.216...) or 8.95 ÷ 40 (=0.22375) or 10.99 ÷ 50 (=0.2198) M1 for all three calculations eg 6.5 ÷ 30 (=0.216...) and 8.95 ÷ 40 (=0.22375) and 10.99 ÷ 50 (=0.2198) A1 for 0.216(6...) and 0.223(75) and 0.219(8...); can be rounded or truncated as long as they remain different C1 (dep on M1) for conclusion ft from three comparable figures [could use different figures relating to 30, 40, 50] OR M1 for one calculation eg 6.5 × 20 (=130) or 8.95 × 15 (=134.25) or 10.99 × 12 (=131.88) M1 for three calculations eg 6.5 × 20 (=130) and 8.95 × 15 (=134.25) and 10.99 × 12 (=131.88) A1 for 130 and 134(.25) and 131(.88); can be rounded or truncated as long as they remain different C1 (dep on M1) for conclusion ft from three comparable figures eg cost of 600 plants or comparing small and medium and small and large e.g. 120 plants and 150 plants separately] OR M1 for one calculation e.g 30 ÷ 6.5 (= 4.615…) or 40 ÷ 8.95 (= 4.469…) or 50 ÷ 10.99 (= 4.549…) M1 for three calculations e.g. 30 ÷ 6.5 (= 4.615…) and 40 ÷ 8.05 (= 4.469…) and 50 ÷ 10.99 (= 4.549…) A1 for 4.6(15…) and 4.4(69…) and 4.5(49…) can be rounded or truncated as long as they remain different C1 (dep on M1) for conclusion ft from three comparable figures [or any other calculations leading to comparable figures]


8 (a) 7n − 4 2 B2 for 7n − 4 (B1 for 7n + d where d is an integer)

(b) explanation 2 M1 for '7n − 4' = 150 or any other valid method, eg. counting on 7s (to get 150) A1 for a complete explanation eg. the 22nd term is 150 or n = 22 from solution of equation or a clear demonstration based on 22 or complete sequence

9 115 4 M1 for 360 − 4 × 25 (=260) M1 (dep) for '260'÷4 (=65) M1 for 180 − '65' or (360 −2× '65') ÷ 2 A1 for 115 with working OR M1 for 360÷4 (=90) M1 (dep) for '90' − 25 (=65) M1 for 180 − '65' or (360 − 2×'65') ÷ 2 A1 for 115 with working

10

Merit 3 M1 for

62 10080

× (=77.5)

A1 for 77.5% or 78% B1 ft (dep on M1) for 'Merit' OR M1 for calculating a percentage between 70 and 85% of 80 eg 0.7×80 (=56) or 0.84 × 80 (=67.2) or 0.85 × 80 (=68) A1 for 56 and 67(.2) or 68 or for two appropriate values which can be compared with 62 B1 ft (dep on M1) for 'Merit'


11 (a) x10 1 B1 cao

(b) m12 1 B1 cao

(c) 4 63a f− 2 B2 for 4 63a f−

or 6

4

3 fa

(B1 for any two from 3, a-4 or 4

1a

, f 6 in a product)

12 440 2 M1 for 140 × π oe or 439

A1 for 439.6 – 440

*13 Distance ÷ speed: 30 ÷ 70 (= 0.42-0.43); Distance ÷ time: 30 ÷ 26 (=1.15…); Speed × time: = 70 × 26 (=1820 mins); mph to miles/min =70÷ 60 (=1.16-1.17); Minutes to hours is 26 ÷ 60 (=0.43…)

No with correct figure

3 M1 for a calculation which uses the Time × Speed = Distance relationship OR a conversion of units eg between hours & minutes or between mph & miles per min M1 for a calculation involving both of the above C1 for “no” with a correct calculation, with units, from working: 25.2-25.8 minutes, 30.1-30.8 miles, 69-69.3 mph NB: 70 ÷ 26 × 30 as a single stage calculation gets 0 marks


14 (a) 20 < T ≤ 24 1 B1 for 20 < T ≤ 24

(b) 6×10 + 8×14 + 13×18 + 21×22 + 2×26 = 920 920 ÷ 50

18.4 4 M1 for finding fx with x consistent within intervals (including the end points) allow 1 error; implied by 820, 1020 M1 (dep) for use of all correct mid-interval values eg 920 M1 (dep on 1st M1) for ∑fx ÷ ∑f A1 for 18.4 oe

(c)

correct frequency polygon

2 B2 for fully correct frequency polygon - points plotted at the midpoint (B1 for all points plotted accurately but not joined with straight line segments) or all points plotted accurately and joined with last joined to first to make a polygon or all points at the correct heights and consistently within or at the ends of the intervals and joined (can include joining last to first to make a polygon) NB: ignore parts of graph drawn to the left of the 1st point or the right of the last point

15 80.1 3 M1 for 392 + 702 M1 for "1521" "4900"+ or "6421" A1 for 80.1 - 80.2


16 (a)

6.25 3 M1 for clear intention to expand bracket or divide both sides of the equation by 5 as first step M1 for correct method to isolate terms in f A1 for 6.25 oe

(b)

−0.75 4 M1 for correct method to clear fractions eg. multiply all terms by 6 M1 for expansion of brackets oe M1 (dep on M1) for isolating the terms in h and the constant terms A1 for −0.75 oe

17 (a) -15, 0, 3, 0 ,-3, 0, 15

2 B2 for all correct (B1 for any 2 or 3 correct)

(b) Correct graph 2 M1 for at least 5 points plotted correctly (ft from table if at least B1 awarded in (a)) A1 for a fully correct curve


18

49.5 4 M1 for tan54 =

height6

M1 for (height =) 6 × tan54 (=8.2-8.3)

M1 for 12

× '8.258..' × 12

A1 for 49.2 - 50 OR

M1 for cos54 = 6

AC

M1 for (AC = ) 6

cos54 (=10.2(07...))

M1 for 12

× 12 × '10.207' × sin54

A1 for 49.2 - 50 OR

M1 for 12

sin 54 sin 72AC

=

M1 for (AC =) 12 sin54

sin 72× (=10.2(07...))

M1 for 12

× 12 × '10.207' × sin54

A1 for 49.2 – 50

19 (a) 0.00078 1 B1 cao


(b) 9.56 × 107 1 B1 cao

20

116 3

M1 for 80% or 0.8 seen oe or 4640.8

(=580)

M1 for 464 4640.8

−

A1 cao OR M1 for 80% or 0.8 seen oe M1 for 464 ÷ 4 or 464 ÷ (80÷20) A1 cao

21 (a) (2x + 3)(2x − 3) 1 B1 cao

(b)

m =53

ga−+

3 M1 for correct processes to isolate terms in m from other terms

M1 for taking m out as a common factor

A1 for m = 53

ga−+

or 5

3gm

a−

=− −


22 (a)

'show' 2 M1 for

1 ( 4 5) 22

x x x× − + + × or

2x ×(x − 4) + 1 2 92

x× ×

A1 for completion with correct processes seen

(b)

13 3 M1 for

21 1 4 2 3512 2

− ± − × ×−×

condone incorrect sign for 351

M1 for 1 2809

4− ±

A1 for 13 NB for either M mark accept + only in place of ± OR M2 for (2x + 27)(x − 13) (M1 for (2x ± 27)(x ± 13) ) A1 for 13

23

15 2 M1 for

134 1201065

× or 15.098... oe

A1 cao


24

14.4 3 M1 for π × 6.52 × 11.5 (=1526.42...)

M1 (dep) for 2

'1526.42... '5.8π ×

A1 for 14.4 - 14.5 OR

M1 for 5.8 6.5or6.5 5.8

or 0.89(23…) or 1.12(06896…)

M1 for 2 25.8 6.511.5 or 11.5

6.5 5.8⎛ ⎞ ⎛ ⎞÷ ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

A1 for 14.4 - 14.5 *25 Yes with

explanation 3 M1 For Line A: writes equation as y = 1.5x + 4 or gives the gradient as 1.5 or

constant term of 4 OR for Line B: shows a method which could lead to finding the gradient or gives the gradient as 2 or constant term of 4 or calculates a sequence of points including (0,4) or writes equation of line as y = 2x + 4 M1 Shows correct aspects relating to an aspect of Line A and an aspect of Line B that enables some comparison to be made eg gradients, equations or points. C1 for gradients 1.5 and 2 and Yes with explanation that the gradients are different or states the lines intersect at (0,4) or explanation that interprets common constant term (4) from equations OR M1 for a diagram that shows both lines drawn and intersecting at (0,4) M1 for a diagram that shows both lines and their intersection point identified as (0,4) C1 for Yes and states the intersection point as (0,4)


26 180-136-“34.4” =9.504

3.73 5 M1 for

sin sin13612.8 15.7

L=

M1 for L = 1 sin136sin 12.815.7

− ⎛ ⎞×⎜ ⎟⎝ ⎠

or sin-10.566...

A1 for 34.4 - 34.5

M1 for 15.7

sin(180 136 '34.4') sin136LN

=− −

or 12.8

sin(180 136 '34.4') sin'34.4'LN

=− −

or

(LN 2 = ) 15.72 + 12.82 - 2×15.7×12.8×cos(180 - 136 - '34.4') A1 for 3.73 - 3.74

27 12×20 + 10.8×10 + 7×15 + 5×15 + 1.8×30 + 0.6×30 =240+108+105+75 +54+18 =528+72=600

12% 3 M1 for attempt to work out total area (eg =600) or area greater than 60 (eg =72) by using fd or counting squares

M1 (dep) for '72' 100

'600'× oe (=12)

A1 cao (must have % otherwise 2 marks)


*28 Proof 3 M1 for one pair of equal angles or sides with reason M1 for second pair of equal angles or sides with reason C1 for proof completed correctly with full reasons and reason for congruence Acceptable reasons: AD common (oe eg both same) Angle BAD = angle CDA (angles in a semicircle are 90o) Angle ABO = angle DCA (angles in the same segment are equal) Triangle ABD and triangle DCA are congruent - ASA OR BD = CA (diameters of the circle) Angle BAD = angle CDA (angles in a semicircle are 90o.) AD common Triangle ABD and triangle DCA are congruent - RHS OR BD = CA (diameters of the circle) AD is common Angle ADB = angle CAD (base angles of an isosceles triangle are equal.) Triangle ABD and triangle DCA are congruent - SAS

Question 6:

F S G W 12 55 96 M 7 18 9 34 19 73 130

F S G W 12 55 29 96 M 9 19 73 38 130





Order Code UG027294 June 2011

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