Centre Number Candidate Number Edexcel GCSE … … · · 2014-03-27This might not be a good way to find out the types of film people like best. (b) ... Centre Number Candidate

Centre Number Candidate Number

Write your name hereSurname Other names

Total Marks

Paper Reference

Turn over

P40645A©2012 Pearson Education Ltd.

6/6/7/3/

*P40645A0128*

Edexcel GCSE

Mathematics APaper 1 (Non-Calculator)

Higher Tier

Monday 11 June 2012 – AfternoonTime: 1 hour 45 minutes 1MA0/1H

You must have: Ruler graduated in centimetres and millimetres, protractor, pair of compasses, pen, HB pencil, eraser. Tracing paper may be used.

Instructions Use black ink or ball-point pen. Fill in the boxes at the top of this page with your name,

centre number and candidate number. Answer all questions. Answer the questions in the spaces provided

– there may be more space than you need. Calculators must not be used.

Information The total mark for this paper is 100 The marks for each question are shown in brackets

– use this as a guide as to how much time to spend on each question. Questions labelled with an asterisk (*) are ones where the quality of your

written communication will be assessed.

Advice Read each question carefully before you start to answer it. Keep an eye on the time. Try to answer every question. Check your answers if you have time at the end.

2

*P40645A0228*

GCSE Mathematics 1MA0

Formulae: Higher Tier

You must not write on this formulae page.Anything you write on this formulae page will gain NO credit.

Volume of prism = area of cross section × length Area of trapezium = 12

(a + b)h

Volume of sphere = 43

�� 3 Volume of cone = 13

�� 2h

Surface area of sphere = 4�� 2 Curved surface area of cone = ��

In any triangle ABC The Quadratic Equation The solutions of ax2 + bx + c = 0

where ��0, are given by

xb b ac

a=

− ± −( )2 42

Sine Rule aA

bB

cCsin sin sin

= =

Cosine Rule a2 = b2 + c2 – 2bc cos A

Area of triangle = 12

ab sin C

length

sectioncross

b

a

h

rl

r

h

C

ab

c BA

3

*P40645A0328* Turn over

Answer ALL questions.

Write your answers in the spaces provided.

You must write down all stages in your working.

You must NOT use a calculator.

1 Sam wants to find out the types of film people like best.

He is going to ask whether they like comedy films or action films or science fiction films or musicals best.

(a) Design a suitable table for a data collection sheet he could use to collect this information.

(2)

Sam collects his data by asking 10 students in his class at school. This might not be a good way to find out the types of film people like best.

(b) Give one reason why.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(1)

(Total for Question 1 is 3 marks)

4

*P40645A0428*

2 The diagram shows a patio in the shape of a rectangle.

The patio is 3.6 m long and 3 m wide.

Matthew is going to cover the patio with paving slabs. Each paving slab is a square of side 60 cm.

Matthew buys 32 of the paving slabs.

(a) Does Matthew buy enough paving slabs to cover the patio? You must show all your working.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(3)

The paving slabs cost £8.63 each.

(b) Work out the total cost of the 32 paving slabs.

£ .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(3)


Diagram NOT accurately drawn

3.6 m

3 m

5


3 Bill uses his van to deliver parcels. For each parcel Bill delivers there is a fixed charge plus £1.00 for each mile.

You can use the graph to find the total cost of having a parcel delivered by Bill.

(a) How much is the fixed charge?

£ .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(1)

Ed uses a van to deliver parcels. For each parcel Ed delivers it costs £1.50 for each mile. There is no fixed charge.

(b) Compare the cost of having a parcel delivered by Bill with the cost of having a parcel delivered by Ed.

(3)


*

80

70

60

50

40

30

20

10

0 0 10 20 30 35 40 45 50

Distance (miles)

Cost (£)

5 15 25

6

*P40645A0628*

4 Here are the speeds, in miles per hour, of 16 cars.

31 52 43 49 36 35 33 2954 43 44 46 42 39 55 48

Draw an ordered stem and leaf diagram for these speeds.


7


5

You can work out the amount of medicine, c m�, to give to a child by using the formula

c ma=150

m is the age of the child, in months. a is an adult dose, in m��

A child is 30 months old. An adult’s dose is 40 m�.

Work out the amount of medicine you can give to the child.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . m�


Medicine

8

*P40645A0828*

6 Here are the ingredients needed to make 12 shortcakes.

Liz makes some shortcakes. She uses 25 m� of milk.

(a) How many shortcakes does Liz make?

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(2)

Robert has 500 g of sugar 1000 g of butter 1000 g of flour 500 m� of milk

(b) Work out the greatest number of shortcakes Robert can make.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(2)


Shortcakes

Makes 12 shortcakes

50 g of sugar 200 g of butter 200 g of flour 10 m� of milk

9


7 Buses to Acton leave a bus station every 24 minutes. Buses to Barton leave the same bus station every 20 minutes.

A bus to Acton and a bus to Barton both leave the bus station at 9 00 am.

When will a bus to Acton and a bus to Barton next leave the bus station at the same time?

... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


8 (a) Expand 3(2y – 5)

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(1)

(b) Factorise completely 8x2 + 4xy

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(2)

(c) Make h the subject of the formula

t gh=10

h = .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(2)


10

*P40645A01028*

9

Describe fully the single transformation that maps triangle A onto triangle B.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


–1 O 1 2 3 4 5 6 7

7

6

5

4

3

2

1

–1

–2

y

x–2

A

B

11


10 Railtickets and Cheaptrains are two websites selling train tickets.

Each of the websites adds a credit card charge and a booking fee to the ticket price.

Railtickets

Credit card charge: 2.25% of ticket price

Booking fee: 80 pence

Cheaptrains

Credit card charge: 1.5% of ticket price

Booking fee: £1.90

Nadia wants to buy a train ticket. The ticket price is £60 on each website. Nadia will pay by credit card.

Will it be cheaper for Nadia to buy the train ticket from Railtickets or from Cheaptrains?


*

12

*P40645A01228*

11

The diagram shows a parallelogram. The sizes of the angles, in degrees, are

2x 3x – 15 2x 2x + 24

Work out the value of x.

x = .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .



3x – 15

2x + 242x

2x

13


12 Jane has a carton of orange juice. The carton is in the shape of a cuboid.

The depth of the orange juice in the carton is 8 cm.

Jane closes the carton. Then she turns the carton over so that it stands on the shaded face.

Work out the depth, in cm, of the orange juice now.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cm


6 cm

10 cm

20 cm


14

*P40645A01428*

13

The diagram shows a regular hexagon and a regular octagon.

Calculate the size of the angle marked x. You must show all your working.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . °


x


15


14 The diagram shows the position of a lighthouse L and a harbour H.

The scale of the diagram is 1 cm represents 5 km.

(a) Work out the real distance between L and H.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . km(1)

(b) Measure the bearing of H from L.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . °

(1)

A boat B is 20 km from H on a bearing of 040°�

(c) On the diagram, mark the position of boat B with a cross (×). Label it B.

(2)


N

H

L

N

16

*P40645A01628*

15 Harry grows tomatoes. This year he put his tomato plants into two groups, group A and group B.

Harry gave fertiliser to the tomato plants in group A. He did not give fertiliser to the tomato plants in group B.

Harry weighed 60 tomatoes from group A. The cumulative frequency graph shows some information about these weights.

(a) Use the graph to find an estimate for the median weight.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . g(1)

60

50

40

30

20

10

0 140 150 160 170 180 190

Cumulative frequency

Weight (g)

17


The 60 tomatoes from group A had a minimum weight of 153 grams and a maximum weight of 186 grams.

(b) Use this information and the cumulative frequency graph to draw a box plot for the 60 tomatoes from group A.

(3)

Harry did not give fertiliser to the tomato plants in group B.

Harry weighed 60 tomatoes from group B. He drew this box plot for his results.

(c) Compare the distribution of the weights of the tomatoes from group A with the distribution of the weights of the tomatoes from group B.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(2)


140 150 160 170 180 190Weight (g)

Group A

140 150 160 170 180 190Weight (g)

Group B

18

*P40645A01828*

16 (a) Simplify (m–2)5

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(1)

(b) Factorise x2 + 3x – 10

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(2)


17 (a) Write down the value of 100

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(1)

(b) Write 6.7 × 10–5 as an ordinary number.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(1)

(c) Work out the value of (3 × 107) × (9 × 106) Give your answer in standard form.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(2)


19


18

Triangle ABC is drawn on a centimetre grid. A is the point (2, 2). B is the point (6, 2). C is the point (5, 5).

Triangle PQR is an enlargement of triangle ABC with scale factor 12

and centre (0, 0).

Work out the area of triangle PQR.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cm2


A B

C

y

x

7

6

5

4

3

2

1

O 1 2 3 4 5 6 7 8

20

*P40645A02028*

19 Wendy goes to a fun fair. She has one go at Hoopla. She has one go on the Coconut shy.

The probability that she wins at Hoopla is 0.4 The probability that she wins on the Coconut shy is 0.3

(a) Complete the probability tree diagram.

Hoopla Coconut shy

(2)

(b) Work out the probability that Wendy wins at Hoopla and also wins on the Coconut shy.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(2)


Wendy does not win

Wendy wins

Wendy does not win

Wendy does not winWendy wins

Wendy wins

0.4

0.3

.. . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

21


20 Solve the simultaneous equations

5x + 2y = 114x – 3y = 18

x = .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

y = .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


22

*P40645A02228*

21

B, C and D are points on the circumference of a circle, centre O. AB and AD are tangents to the circle.

Angle DAB = 50°

Work out the size of angle BCD. Give a reason for each stage in your working.


*

D

O C

BA50°


23


22 The table gives some information about the speeds, in km/h, of 100 cars.

Speed (s km/h) Frequency

60 < s � 65 15

65 < s � 70 25

70 < s � 80 36

80 < s � 100 24

(a) On the grid, draw a histogram for the information in the table.

(3)

(b) Work out an estimate for the number of cars with a speed of more than 85 km/h.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(2)


60 70 80 90 100Speed (s km/h)

24

*P40645A02428*

23 (a) Simplify fully x xx x

2

23 4

2 5 3+ −− +

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(3)

(b) Write 42

32x x+

+−

as a single fraction in its simplest form.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(3)


24 Express the recurring decimal 0.28.1. as a fraction in its simplest form.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


25


25 The diagram shows a solid metal cylinder.

The cylinder has base radius 2x and height 9x.

The cylinder is melted down and made into a sphere of radius �.

Find an expression for � in terms of x.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .



9x

2x

26

*P40645A02628*

26 The graph of y = f(x) is shown on each of the grids.

(a) On this grid, sketch the graph of y = f(x – 3)

(2)

–1 O 1 2 3 4 5 6

6

5

4

3

2

1

–1

–2

y

x–2–3–4–5

–4

–3

7

27

*P40645A02728*

(b) On this grid, sketch the graph of y = 2f(x)

(2)


TOTAL FOR PAPER IS 100 MARKS

–1 O 1 2 3 4 5 6

8

7

4

3

2

1

–1

–2

y

x–2–3–4–5

–3

6

5

7

28

*P40645A02828*

BLANK PAGE

Centre Number Candidate Number

Write your name hereSurname Other names

Total Marks

Paper Reference

Turn over

P40647A©2012 Pearson Education Ltd.

6/6/7/3/

*P40647A0124*

Edexcel GCSE

Mathematics APaper 2 (Calculator)

Higher Tier

Wednesday 13 June 2012 – MorningTime: 1 hour 45 minutes 1MA0/2H

You must have: Ruler graduated in centimetres and millimetres, protractor, pair of compasses, pen, HB pencil, eraser, calculator. Tracing paper may be used.

Instructions Use black ink or ball-point pen. Fill in the boxes at the top of this page with your name,

centre number and candidate number. Answer all questions. Answer the questions in the spaces provided

– there may be more space than you need. Calculators may be used.

If your calculator does not have a � button, take the value of � to be 3.142 unless the question instructs otherwise.

Information The total mark for this paper is 100 The marks for each question are shown in brackets

– use this as a guide as to how much time to spend on each question. Questions labelled with an asterisk (*) are ones where the quality of your

written communication will be assessed.

Advice Read each question carefully before you start to answer it. Keep an eye on the time. Try to answer every question. Check your answers if you have time at the end.

2

*P40647A0224*

GCSE Mathematics 1MA0

Formulae: Higher Tier

You must not write on this formulae page.Anything you write on this formulae page will gain NO credit.

Volume of prism = area of cross section × length Area of trapezium = 12

(a + b)h

Volume of sphere = 43

�� 3 Volume of cone = 13

�� 2h

Surface area of sphere = 4�� 2 Curved surface area of cone = ��

In any triangle ABC The Quadratic Equation The solutions of ax2 + bx + c = 0

where ��0, are given by

xb b ac

a=

− ± −( )2 42

Sine Rule aA

bB

cCsin sin sin

= =

Cosine Rule a2 = b2 + c2 – 2bc cos A

Area of triangle = 12

ab sin C

length

sectioncross

b

a

h

rl

r

h

C

ab

c BA

3


Answer ALL questions.

Write your answers in the spaces provided.

You must write down all stages in your working.

1

ABC and DEF are parallel lines. BEG is a straight line. Angle GEF = 47�.

Work out the size of the angle marked x. Give reasons for your answer.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .�



AB

C

D E 47�F

G

x

4

*P40647A0424*

2 (a) Use your calculator to work out 38 5 14 218 4 5 9

. .. .

×−

Write down all the figures on your calculator display. You must give your answer as a decimal.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(2)

(b) Write your answer to part (a) correct to 1 significant figure... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(1)


5


3 Pradeep wants to find out how much time people spend playing sport.

He uses this question on a questionnaire.

How much time do you spend playing sport?

0 – 1 hours 1 – 2 hours 3 – 4 hours

(a) Write down two things wrong with this question.

1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(2)

(b) Design a better question for Pradeep’s questionnaire to find out how much time people spend playing sport.

(2)


6

*P40647A0624*

4 On the grid, draw the graph of y = 3x – 2 for values of x from –1 to 3


y

x

8

7

6

5

4

3

2

1

–1 O 1 2 3

–1

–2

–3

–4

–5

–6

7


5 Mr Weaver’s garden is in the shape of a rectangle.

In the garden there is a patio in the shape of a rectangleand two ponds in the shape of circles with diameter 3.8 m.

The rest of the garden is grass.

Mr Weaver is going to spread fertiliser over all the grass. One box of fertiliser will cover 25 m2 of grass.

How many boxes of fertiliser does Mr Weaver need? You must show your working.


17 m

9.5 m

2.8 m

3.8 mpond

patio grass


3.8 mpond

*

8

*P40647A0824*

6 Potatoes cost £9 for a 12.5 kg bag at a farm shop. The same type of potatoes cost £1.83 for a 2.5 kg bag at a supermarket.

Where are the potatoes the better value, at the farm shop or at the supermarket? You must show your working.


*

9


7 The scatter graph shows some information about 8 cars. For each car it shows the engine size, in litres, and the distance, in kilometres, the car travels on one litre of petrol.

(a) What type of correlation does the scatter graph show?

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(1)

A different car of the same type has an engine size of 2.5 litres.

(b) Estimate the distance travelled on one litre of petrol by this car.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . kilometres(2)


16

14

12

10

8

6

Distance (kilometres)

Engine size (litres)

0 1 2 3 4 5

10

*P40647A01024*

8

(a) Rotate triangle A 90� clockwise, centre O.(2)

(b) Enlarge triangle B by scale factor 3, centre (1, 2).(3)


y

x

A4

2

–2–4 O 2 4

–2

–4

y

x

14

12

10

8

6

4

2

–2

–2

O 2 4 6 8 10 12 14

B

11


9 Linda is going on holiday to the Czech Republic. She needs to change some money into koruna.

She can only change her money into 100 koruna notes.

Linda only wants to change up to £200 into koruna. She wants as many 100 koruna notes as possible.

The exchange rate is £1 = 25.82 koruna.

How many 100 koruna notes should she get?

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


10 m is an integer such that –2 < m � 3

(a) Write down all the possible values of m.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(2)

(b) Solve 7x – 9 < 3x + 4

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(2)


12

*P40647A01224*

11 The equationx3 – 6x = 72

has a solution between 4 and 5

Use a trial and improvement method to find this solution. Give your answer correct to one decimal place. You must show all your working.

x = .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


13


12 The probability that a biased dice will land on a five is 0.3

Megan is going to roll the dice 400 times.

Work out an estimate for the number of times the dice will land on a five.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


13 Bob asked each of 40 friends how many minutes they took to get to work.

The table shows some information about his results.

Time taken (m minutes) Frequency

0 < m � 10 3

10 < m � 20 8

20 < m � 30 11

30 < m � 40 9

40 < m � 50 9

Work out an estimate for the mean time taken.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . minutes


14

*P40647A01424*

14 (a) Expand and simplify (p + 9)(p – 4)

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(2)

(b) Solve

5 83

4 2w w− = +

w = .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(3)

(c) Factorise x2 – 49

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(1)

(d) Simplify ( )9 8 312x y

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(2)


15


15 Henry is thinking about having a water meter.

These are the two ways he can pay for the water he uses.

Henry uses an average of 180 litres of water each day.

Henry wants to pay as little as possible for the water he uses. Should Henry have a water meter?


Water Meter

A charge of £28.20 per year

plus

91.22p for every cubic metre of water used

1 cubic metre = 1000 litres

No Water Meter

A charge of £107 per year

*

16

*P40647A01624*

16

LMN is a right-angled triangle. MN = 9.6 cm. LM = 6.4 cm.

Calculate the size of the angle marked x�. Give your answer correct to 1 decimal place.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . �


17 Liam invests £6200 for 3 years in a savings account. He gets 2.5% per annum compound interest.

How much money will Liam have in his savings account at the end of 3 years?

£ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .



N

x�

9.6 cm 6.4 cm

M

L

17


18 The diagram shows a quadrilateral ABCD.

AB = 16 cm. AD = 12 cm. Angle BCD = 40�� Angle ADB = angle CBD = 90��

Calculate the length of CD. Give your answer correct to 3 significant figures.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cm



B

A

C40�

16 cm

12 cm D

18

*P40647A01824*

19 p x yxy

2 = −

x = 8.5 × 109

y = 4 × 108

Find the value of p. Give your answer in standard form correct to 2 significant figures.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


20 Make t the subject of the formula 2(d – t) = 4t + 7

t = .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


19


21 Prove that (2n + 3)2 – (2n – 3)2 is a multiple of 8

for all positive integer values of n.


22 Solve 3x2 – 4x – 2 = 0 Give your solutions correct to 3 significant figures.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


20

*P40647A02024*

23 (a) Max wants to take a random sample of students from his year group.

(i) Explain what is meant by a random sample.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(ii) Describe a method Max could use to take his random sample.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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(2)

(b) The table below shows the numbers of students in 5 year groups at a school.

Year Number of students

9 239

10 257

11 248

12 190

13 206

Lisa takes a stratified sample of 100 students by year group.

Work out the number of students from Year 9 she has in her sample.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(2)


21


24

ABC is a triangle.

AB = 8.7 cm. Angle ABC = 49�. Angle ACB = 64�.

Calculate the area of triangle ABC. Give your answer correct to 3 significant figures.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .cm2



B

A

C

49�

64�

8.7 cm

22

*P40647A02224*

25 Carolyn has 20 biscuits in a tin.

She has

12 plain biscuits 5 chocolate biscuits 3 ginger biscuits

Carolyn takes at random two biscuits from the tin.

Work out the probability that the two biscuits were not the same type.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


23

*P40647A02324*

26

OAB is a triangle.

OA�

= a OB

� = b

(a) Find AB�

in terms of a and b.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(1)

P is the point on AB such that AP : PB = 3 : 1

(b) Find OP�

in terms of a and b. Give your answer in its simplest form.

.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(3)


TOTAL FOR PAPER IS 100 MARKS


B

P

AO

b

a

24

*P40647A02424*

BLANK PAGE

Mark Scheme (Results) Summer 2012 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, please visit our website at www.edexcel.com. Our website subject pages hold useful resources, support material and live feeds from our subject advisors giving you access to a portal of information. If you have any subject specific questions about this specification that require the help of a subject specialist, you may find our Ask The Expert email service helpful. www.edexcel.com/contactus 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 2012 Publications Code UG032625 All the material in this publication is copyright © Pearson Education Ltd 2012

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 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.

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 (a) Type of film

Tally Frequency

2

B2 for a table with all 3 aspects: Column/row heading ‘type of film’ or list of at least 3 film types Column/row heading ‘tally’ or tally marks (or key) Column/row heading ‘frequency’ or totals oe (B1 for a table with 2 of the 3 aspects)

(b) 1 B1 for acceptable reason eg. all same age, sample too small, biased, same school

1MA0_1H

Question Working Answer Mark Notes 2 (a)

360 ÷ 60 = 6 300 ÷ 60 = 5 6 × 5 =

Yes and 30 3 M1 for dividing side of patio by side of paving slab eg. 360 ÷ 60 or 300 ÷ 60 or 3.6 ÷ 0.6 or 3 ÷ 0.6 or 6 and 5 seen (may be on a diagram) or 6 divisions seen on length of diagram or 5 divisions seen on width of diagram M1 for correct method to find number of paving slabs eg. (360 ÷ 60) × (300 ÷ 60) oe or 6 × 5 or 30 squares seen on diagram (units may not be consistent) A1 for Yes and 30 (or 2 extra) with correct calculations OR M1 for correct method to find area of patio or paving slab eg 360 × 300 or 108000 seen or 60 × 60 or 3600 seen or 3.6 × 3 or 10.8 seen or 0.6 × 0.6 or 0.36 seen M1 for dividing area of patio by area of a paving slab eg. (3.6 × 3) ÷ (0.6 × 0.6) oe (units may not be consistent) A1 for Yes and 30 (or 2 extra) with correct calculations OR M1 for method to find area of patio or area of 32 slabs eg. 60 × 60 × 32 or 360 × 300 M1 for method to find both area of patio and area of 32 slabs eg. 60 × 60 × 32 and 360 × 300 (units may not be consistent) A1 for Yes and 115200 and 108000 OR Yes and 11.52 and 10.8 NB : Throughout the question, candidates could be working in metres or centimetres

1MA0_1H Question Working Answer Mark Notes

(b) 1726 25890 27616

8 6 3

2 2 4 1 8 9 37 1 6 1 2 6 2 6 1 6

24000+1800+90+1600+120+6 = 27616

800 60 3 30 24000 1800 90 2 1600 120 6

276.16 3 M1 for complete correct method with relative place value correct. Condone 1 multiplication error, addition not necessary. OR M1 for a complete grid. Condone 1 multiplication error, addition not necessary. OR M1 for sight of a complete partitioning method, condone 1 multiplication error. Final addition not necessary. A1 for digits 27616 A1 ft (dep on M1) for correct placement of decimal point after addition (of appropriate values) (SC: B1 for attempting to add 32 lots of 8.63)

1MA0_1H

Question Working Answer Mark Notes 3 (a) 10 1 B1 cao

(b)

Miles 0 10 20 30 40 50 Ed 0 15 30 45 60 75 Bill 10 20 30 40 50 60

Ed is cheaper up to 20

miles, Bill is cheaper for

more than 20 miles

3 M1 for correct line for Ed intersecting at (20,30) ±1 sq tolerance or 10 + x = 1.5x oe C2 (dep on M1) for a correct full statement ft from graph eg. Ed cheaper up to 20 miles and Bill cheaper for more than 20 miles (C1 (dep on M1) for a correct conclusion ft from graph eg. cheaper at 10 miles with Ed ; eg. cheaper at 50 miles with Bill eg. same cost at 20 miles; eg for £5 go further with Bill OR A general statement covering short and long distances eg. Ed is cheaper for shorter distances and Bill is cheaper for long distances) OR M1 for correct method to work out Ed's delivery cost for at least 2 values of n miles where 0 < n ≤ 50 OR for correct method to work out Ed and Bill's delivery cost for n miles where 0 < n ≤ 50 C2 (dep on M1) for 20 miles linked with £30 for Ed and Bill with correct full statement eg. Ed cheaper up to 20 miles and Bill cheaper for more than 20 miles (C1 (dep on M1) for a correct conclusion eg. cheaper at 10 miles with Ed; eg. cheaper at 50 miles with Bill eg. same cost at 20 miles; eg for £5 go further with Bill OR A general statement covering short and long distances eg. Ed is cheaper for shorter distances and Bill is cheaper for long distances) SC : B1 for correct full statement seen with no working eg. Ed cheaper up to 20 miles and Bill cheaper for more than 20 miles QWC: Decision and justification should be clear with working clearly presented and attributable

10

20

30

40

5 10 15 20 25 300 x

y

1MA0_1H

Question Working Answer Mark Notes 4 2 9

3 1 3 5 6 9 4 2 3 3 4 6 8 9 5 2 4 5 OR 20 9 30 1 3 5 6 9 40 2 3 3 4 6 8 9 50 2 4 5

2 9 3 1 3 5 6 9 4 2 3 3 4 6 8 9 5 2 4 5 Key: 2 9 = 29

3 B3 for fully correct diagram with appropriate key (B2 for ordered leaves, with at most two errors or omissions and a key OR correct unordered leaves and a key OR correct ordered leaves) (B1 for unordered or ordered leaves, with at most two errors or omissions OR key) NB : Order of stem may be reversed; condone commas between leaves

5 c =

30 40150×

8 2 M1 for

30 40150×

or 1200 seen

A1 cao

6 (a)

30 2 M1 for 25 ÷ 10 or 2.5 seen or 10 ÷ 25 or 0.4 seen or 12 + 12 + 6 oe or a complete method eg. 25 × 12 ÷ 10 oe A1 cao

(b) 1000 ÷ 200 × 12 60 2 M1 for 500÷50 or 1000÷200 or 500÷10 OR correct scale factor clearly linked with one ingredient eg. 10 with sugar or 5 with butter or flour or 50 with milk OR answer of 120 or 600 A1 cao

1MA0_1H

Question Working Answer Mark Notes 7 Acton after 24, 48, 72, 96, 120

Barton after 20, 40, 60, 80, 100, 120 LCM of 20 and 24 is 120 9: 00 am + 120 minutes OR Acton after 24, 48, 1h 12 m, 1h 36m, 2h Barton after 20, 40, 1 h, 1h 20m, 1h 40m, 2h LCM is 2 hours 9: 00 am + 2 hours OR Times from 9: 00 am when each bus leaves the bus station Acton at 9: 24, 9: 48, 10: 12, 10:36, 11:00 Barton at 9: 20, 9: 40, 10: 00, 10:20, 10:40, 11:00 OR 20 = 2 × 2 × 5 24 = 2 × 2 × 2 × 3 2 × 2 × 2 × 3 × 5 = 120

11: 00 am 3 M1 for listing multiples of 20 and 24 with at least 3 numbers in each list ; multiples could be given in minutes or in hours and minutes (condone one addition error in total in first 3 numbers in lists) A1 identify 120 (mins) or 2 (hours) as LCM A1 for 11: 00 (am) or 11(am) or 11 o'clock OR M1 for listing times after 9am when each bus leaves the bus station, with at least 3 times in each list (condone one addition error in total in first 3 times after 9am in lists) A1 for correct times in each list up to and including 11: 00 A1 for 11: 00 (am) or 11(am) or 11 o'clock OR M1 for correct method to write 20 and 24 in terms of their prime factors 2, 2, 5 and 2, 2, 2, 3 (condone one error) A1 identify 120 as LCM A1 for 11: 00 (am) or 11(am) or 11 o'clock

1MA0_1H

Question Working Answer Mark Notes 8 (a) 6y – 15 1 B1 cao

(b) 4x(2x + y) 2 B2 cao

(B1 for x(8x + 4y) or 2x(4x +2y) or 4(2x2 + xy) or 4x(ax + by) where a, b are positive integers or ax(2x + y) where a is a positive integer or 4x(2x – y))

(c) 10t = gh

h = 10tg

10tg

2 M1 for clear intention to multiply both sides of the equation by 10 (eg. ×10 seen on both sides of equation) or clear intention to divide both sides of the equation by g (eg. ÷g seen on both sides of equation)

or 10t = gh or 10

t hg= or

fully correct reverse flow diagram eg. ← ×10 ← ÷g ←

A1 for 10tg

oe

1MA0_1H

Question Working Answer Mark Notes 9 Rotation

180° Centre (3, 3)

or

Enlargement Scale factor -1 Centre (3, 3)

3 B1 for rotation B1 for 180° B1 for (3, 3) OR B1 for enlargement B1 for scale factor -1 B1 for (3, 3) B0 for a combination of transformations

1MA0_1H

Question Working Answer Mark Notes 10 2.25 × 60 ÷ 100 = 1.35

1.35 + 0.80 = 2.15 1.5 × 60 ÷ 100 = 0.90 0.90 + 1.90 = 2.80 OR

Railtickets with correct calculations

4 NB. All work may be done in pence throughout M1 for correct method to find credit card charge for one company eg. 0.0225 × 60(=1.35) oe or 0.015 × 60 (=0.9) oe M1 (dep) for correct method to find total additional charge or total price for one company eg. 0.0225×60 + 0.80 or 0.015×60 + 1.90 or 2.15 or 2.8(0) or 62.15 or 62.8(0) A1 for 2.15 and 2.8(0) or 62.15 and 62.8(0) C1 (dep on M1) for a statement deducing the cheapest company, but figures used for the comparison must also be stated somewhere, and a clear association with the name of each company OR M1 for correct method to find percentage of (60+booking fee) eg. 0.0225 × 60.8(=1.368) oe or 0.015 × 61.9(=0.9285) M1 (dep) for correct method to find total cost or total additional cost eg. '1.368' + 60.8(=62.168) or '1.368' + 0.8 (=2.168) or '0.9285' + 61.9 (=62.8285) or '0.9285' +1.9 (=2.8285) A1 for 62.168 or 62.17 AND 62.8285 or 62.83 OR 2.168 or 2.17 AND 2.8285 or 2.83 C1 (dep on M1) for a statement deducing the cheapest company, but figures used for the comparison must also be stated somewhere, and a clear association with the name of each company OR


2.25 – 1.5 = 0.75 0.075 × 60 ÷ 100 = 0.45 0.80 + 0.45 = 1.25 1.25 < 1.90

M1 for correct method to find difference in cost of credit card charge eg. (2.25 – 1.5) × 60 ÷ 100 oe or 0.45 seen M1 (dep) for using difference with booking fee or finding difference between booking fees eg. 0.80 + “0.45”(=1.25) or 1.90 – “0.45” (=1.45) or 1.90 – 0.8 (=1.1(0)) A1 1.25 and 1.9(0) or 0.45 and 1.1(0) C1 (dep on M1) for a statement deducing the cheapest company, but figures used for the comparison must also be stated somewhere, and a clear association with the name of each company QWC: Decision and justification should be clear with working clearly presented and attributable

1MA0_1H

Question Working Answer Mark Notes 11 3x–15 = 2x+24

x = 39 OR 2x+3x–15 +2x+ 2x+24 = 360 9x + 9 = 360 9x = 351 x = 39 OR 2x + 2x+24 = 180 4x + 24 = 180 4x = 156 x = 39 OR 2x + 3x–15 = 180 5x – 15 = 180 5x = 195 x = 39

39 3 M1 for forming an appropriate equation eg. 3x – 15 = 2x + 24 OR 2x + 3x – 15 + 2x + 2x + 24 = 360 OR 2x + 2x + 24 = 180 OR 2x + 3x –15 = 180 OR 2x + 3x – 15 = 2x + 2x + 24 M1 (dep) for correct operation(s) to isolate x and non-x terms in an equation to get to ax = b A1 cao OR

M2 for 3519

oe or 195

5oe or

1564

oe

A1 cao

1MA0_1H

Question Working Answer Mark Notes 12 6 × 10 × 8 = 480

480 ÷ (6 × 20) = 4 3 M1 for 6 × 10 × 8 or 480 seen

M1 (dep) for '480' ÷ (6 × 20) oe A1 cao OR

M1 for 20 ÷ 10 (=2) or 10 ÷ 20 (=12

) or 820

oe or 208

oe

M1 (dep) for 8 ÷ '2' or 8 × 12

or 820

× 10 oe or

10 ÷ 208

A1 cao

SC : B2 for answer of 16 coming from 20 8 6

10 6× ××

oe

1MA0_1H

Question Working Answer Mark Notes 13 180 – (360 ÷ 6) = 120

180 – (360 ÷ 8) = 135 360 – 120 – 135 = OR 360 ÷ 6 = 60 360 ÷ 8 = 45 60 + 45 =

105 4 NB. Do remember to look at the diagram when marking this question. Looking at the complete method should confirm if interior or exterior angles are being calculated M1 for a correct method to work out the interior angle of a regular hexagon eg. 180 – (360 ÷ 6) oe or (6 - 2)×180 ÷ 6 oe or 120 as interior angle of the hexagon M1 for a correct method to work out the interior angle of a regular octagon 180 – (360 ÷ 8) oe or (8 - 2)×180 ÷ 8 oe or 135 as interior angle of the octagon M1 (dep on at least M1) for a complete method eg. 360 – “120” – “135” A1 cao OR M1 for a correct method to work out an exterior angle of a regular hexagon eg. 360 ÷ 6 or 60 as exterior angle of the hexagon M1 for a correct method to work out an exterior angle of a regular hexagon 360 ÷ 8 or 45 as exterior angle of the octagon M1 (dep on at least M1) for a complete method eg. “60” + “45” A1 cao SC : B1 for answer of 255

1MA0_1H

Question Working Answer Mark Notes 14 (a) 35 1 B1 for 34 – 36

(b) 110 1 B1 for 108 – 112

(c) Position of B

marked 2 B1 for a point marked on a bearing of 40° (±2°) from H or

for a line on a bearing of 40° (±2°) (use straight line guidelines on overlay) B1 for a point 4 cm (± 0.2cm) from H or for a line of length 4 cm (± 0.2cm) from H (use circular guidelines on overlay) NB. No label needed for point

15 (a) 170 1 B1 accept answers in range 170 - 170.5 inclusive

(b)

3 B3 for box plot with all 3 aspects correct (overlay) aspect 1 : ends of whiskers at 153 and 186 aspect 2 : ends of box at 165 and 175 aspect 3 : median marked at 170 or ft (a) provided 165<(a)<175 (B2 for box plot with two aspects correct) (B1 for one aspect or correct quartiles and median identified) SC : B2 for all 5 values (153, 165, '170', 175, 186) plotted

(c) Two correct comparisons

2 B1 ft from (b) for a correct comparison of range or inter-quartile range eg. the range / iqr is smaller for group B than group A B1 ft from (b) for a correct comparison of median or upper quartile or lower quartile or minimum or maximum eg. the median in group A is greater than the median in group B


16 (a) m-10

1 B1 for m–10 or 10

1m

(b) (x + 5)(x – 2)

2 M1 for (x ± 5)(x ± 2) or

x(x – 2) + 5(x – 2) or x(x + 5) – 2(x + 5) A1

17 (a) 1 1

B1 cao

(b) 0.000067 1 B1 cao

(c) 2.7 × 1014 2 M1 for 27 × 107 + 6 or 27 × 1013 oe or an answer of 2.7 × 10n where n is an integer or an answer of a × 1014 where 1 ≤ a < 10 A1 cao

1MA0_1H

Question Working Answer Mark Notes 18 1

2 × 4 × 3 = 6

212

⎛ ⎞⎜ ⎟⎝ ⎠

× 6 =

1.5 3 M1 for

12

× 4 × 3 oe

M1 for 21

2⎛ ⎞⎜ ⎟⎝ ⎠

× “6”

A1 cao OR

M2 for 12

× 2 × 1.5 oe

(M1 for triangle with all lengths 12

corresponding

lengths of triangle ABC seen in any position or vertices seen at (1, 1) (3,1) and (2.5, 2.5) or stated) A1 cao

19 (a)

0.6 0.7, 0.3, 0.7

2 B1 for 0.6 in correct position on tree diagram B1 for 0.7, 0.3, 0.7 in correct positions on tree diagram

(b) 0.4 × 0.3 = 0.12 2 M1 for 0.4 × 0.3 oe or a complete alternative method ft from tree diagram A1 for 0.12 oe

1MA0_1H

Question Working Answer Mark Notes 20 15x + 6y = 33

8x – 6y = 36 23x = 69 5× 3 + 2y = 11 OR

11 25

11 24 3 185

44 8 15 9046 23

2

yx

y y

y yy

y

−=

−⎛ ⎞× − =⎜ ⎟⎝ ⎠− − =

− == −

x = 3 y = – 2

4 M1 for coefficients of x or y the same followed by correct operation (condone one arithmetic error) A1 cao for first solution 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 for second solution OR M1 for full method to rearrange and substitute to eliminate x or y, (condone one arithmetical error) A1 cao for first solution 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 for second solution Trial and improvement 0 marks unless both x and y correct values found

1MA0_1H

Question Working Answer Mark Notes 21*

ABO = ADO = 90°

(Angle between tangent and radius is 90°) DOB = 360 – 90 – 90 – 50 (Angles in a quadrilateral add up to 360°) BCD = 130 ÷ 2 (Angle at centre is twice angle at circumference) OR ABD = (180 – 50) ÷ 2 (Base angles of an isosceles triangle) BCD = 65 (Alternate segment theorem)

65o 4 B1 for ABO = 90 or ADO = 90 (may be on diagram) B1 for BCD = 65 (may be on diagram) C2 for BCD = 65o stated or DCB = 65o stated or angle C = 65o stated with all reasons: angle between tangent and radius is 90o; angles in a quadrilateral sum to 360o; angle at centre is twice angle at circumference

(accept angle at circumference is half (or 12

) the angle at the centre)

(C1 for one correct and appropriate circle theorem reason) QWC: Working clearly laid out and reasons given using correct language OR B1 for ABD = 65 or ADB = 65 (may be on diagram) B1 for BCD = 65 (may be on diagram) C2 for BCD = 65o stated or DCB = 65o stated or angle C = 65o stated with all reasons: base angles of an isosceles triangle are equal; angles in a triangle sum to 180o; tangents from an external point are equal; alternate segment theorem (C1 for one correct and appropriate circle theorem reason) QWC: Working clearly laid out and reasons given using correct language

1MA0_1H

Question Working Answer Mark Notes 22 (a)

F 15 25 36 24 Fd 3 5 3.6 1.2

Correct histogram 3 B3 for fully correct histogram (overlay) (B2 for 3 correct blocks) (B1 for 2 correct blocks of different widths) SC : B1 for correct key, eg. 1 cm2 = 5 (cars) or correct values for (freq ÷ class interval) for at least 3 frequencies (3, 5, 3.6, 1.2) NB: The overlay shows one possible histogram, there are other correct solutions.

(b) 34

× 24 18 2

M1 for 34

× 24 (=18) oe or 1 244× (=6) oe

A1 cao OR M1 ft histogram for 15 × “1.2” or 5 × "1.2" A1 ft

1MA0_1H

Question Working Answer Mark Notes 23 (a) ( 4)( 1)

(2 3)( 1)x xx x+ −− −

4

2 3xx+−

3 M1 for (x + 4)(x – 1)

M1 for (2x – 3)(x – 1) A1 cao

(b) 4( 2) 3( 2)( 2)( 2) ( 2)( 2)

x xx x x x

− ++

+ − + −

7 2

( 2)( 2)x

x x−

+ −

3 M1 for denominator (x + 2)(x – 2) oe or x2 – 4

M1 for 4( 2)

( 2)( 2)x

x x−

+ − oe or

3( 2)( 2)( 2)

xx x

++ −

oe

(NB. The denominator must be (x + 2)(x - 2) or x2 – 4 or another suitable common denominator)

A1 for 7 2

( 2)( 2)x

x x−

+ − or 2

7 24

xx−−

SC: If no marks awarded then award B1 for

2 2

4( 2) 3( 2)2 2

x xx x− +

+− −

oe

1MA0_1H

Question Working Answer Mark Notes 24 eg.

x = 0.28181... 100x = 28.181... 99x = 27.9

31110

3 M1 for 0.28181(...) or 0.2 + 0.08181(...) or

evidence of correct recurring decimal eg. 281.81(...) M1 for two correct recurring decimals that, when subtracted, would result in a terminating decimal, and attempting the subtraction eg. 100x = 28.1818…, x = 0.28181… and subtracting eg. 1000x = 281.8181…, 10x = 2.8181… and subtracting

OR 27.999

or 279990

oe

A1 cao

25 Vol cylinder = π × (2x)2 × 9x = 36πx3

3 34363

x rπ π=

r3 = 27x3

3x 3 M1 for sub. into πr2h eg. π × (2x)2 × 9x oe

M1 for ( )2 342 93

x x rππ =× × oe

A1 oe eg. 3

3

3643

x

NB : For both method marks condone missing brackets around the 2x

1MA0_1H

Question Working Answer Mark Notes 26 (a) Parabola through

(4, –1), (2, 3), (6, 3) (3, 0) (5, 0)

2 B2 for a parabola with min (4, –1), through (2, 3), (6, 3),(3, 0), (5, 0) (B1 for a parabola with min (4, –1) or a parabola through (2, 3) and (6, 3) or a parabola through (3, 0) and (5, 0) or a translation of the given parabola along the x-axis by any value other than +3 with the points (–1, 3) (0, 0) (1, –1) (2, 0) (3, 3) all translated by the same amount)

(b) Parabola through (1, –2), (0, 0), (2, 0)

2 B2 parabola with min (1, –2), through (0, 0) and (2, 0) (B1 parabola with min (1, –2) or parabola through (0, 0), (2, 0) (-1, 6) and (3, 6))

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 UG032625 Summer 2012

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) Summer 2012 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, please visit our website at www.edexcel.com. Our website subject pages hold useful resources, support material and live feeds from our subject advisors giving you access to a portal of information. If you have any subject specific questions about this specification that require the help of a subject specialist, you may find our Ask The Expert email service helpful. www.edexcel.com/contactus 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 2012 Publications Code UG032626 All the material in this publication is copyright © Pearson Education Ltd 2012

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 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.

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_2H Question Working Answer Mark Notes 1 180 – 47 133 3 M1 for 180 – 47

A1 for 133 C1(dep on M1) for full reasons e.g. angles on a straight line add up to 180o and alternate angles are equal OR corresponding angles are equal and angles on a straight line add up to 180o

OR vertically opposite angles (or vertically opposite angles) are equal and allied angles (or co-interior angles) add up to 180o

2 (a) 5.127.546 =

43.736 2 B2 for 43.736

(B1 for 546.7 or10

5467 or 1255467 or 12.5 or

225 or 43.7

or 43.8 or 43.73 or 43.74 or 40 or 44)

(b) 40 1 B1 for 40 or ft from their answer to (a) provided (a) is written to 2 or more significant figures

1MA0_2H

Question Working Answer Mark Notes 3 (a) reasons 2 1st aspect : time frame

2nd aspect : overlapping boxes 3rd aspect : not exhaustive (eg. no box for more than 4) B2 any two aspects (B1 any one aspect)

(b) How much time do you spend playing sport each

week/month

None 1 hr to 2 hrs 3 hrs to 5 hrs

More than 5 hrs

2 B1 for a suitable question which includes a time frame and unit (the time frame and unit could appear with the response boxes) B1 for at least 3 non-overlapping response boxes (need not be exhaustive) or at least 3 response boxes exhaustive for all integer values of their time unit (could be overlapping). [Do not allow inequalities in response boxes]

1MA0_2H

Question Working Answer Mark Notes 4

x –1 0 1 2 3 y –5 –2 1 4 7

OR Using y = mx + c gradient = 3 y intercept = –2

Straight line from (–1, –5) to (3, 7)

3 (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 plotted from their table must be correctly plotted) A1 for correct line between –1 and 3 (No table of values) M2 for at least 2 correct points (and no incorrect points) plotted OR line segment of y = 3x–2 drawn (ignore any additional incorrect segments) (M1 for at least 3 correct points plotted with no more than 2 incorrect points) A1 for correct line between –1 and 3 (Use of y = mx+c) M2 for line segment of y = 3x – 2 drawn (ignore any additional incorrect segments) (M1 for line drawn with gradient of 3 OR line drawn with a y intercept of –2 and a positive gradient) A1 for correct line between –1 and 3

1MA0_2H

Question Working Answer Mark Notes 5 (17 – 2.8) × 9.5 = 134.9

π × (3.8 ÷ 2)2 = 11.34... 134.9 – 2 × 11.34... = 112.21 112.21 ÷ 25 = 4.488

5 5 M1 for (17 – 2.8) × 9.5 (=134.9) or 17×9.5 – 2.8×9.5 ( = 161.5 – 26.6 = 134.9) M1 for π × (3.8 ÷ 2)2 (= 11.33 – 11.35) M1 (dep on M1) for ‘134.9’ – 2 × ‘11.34’ A1 for 112 - 113 C1(dep on at least M1) for 'He needs 5 boxes' ft from candidate's calculation rounded up to the next integer

6 Farm shop 4 M1 for 12.5 ÷ 2.5 (=5) M1 for ‘5’×1.83 or ‘5’ × 183 A1 for (£)9.15 or 915(p) C1 (dep on at least M1) for decision ft working shown OR M1 for 12.5 ÷ 2.5 (=5) M1 for 9 ÷ ‘5’ or 900 ÷ ‘5’ A1 for (£)1.8(0) or 180(p) C1 (dep on at least M1) for decision ft working shown OR M1 for 9 ÷ 12.5 (=0.72) or 1.83 ÷ 2.5 (=0.732) M1 for 9 ÷ 12.5 (=0.72) and 1.83 ÷ 2.5 (=0.732) A1 for 72(p) and 73.(2)(p) or (£)0.72 and (£)0.73(2) C1 (dep on at least M1) for decision ft working shown OR M1 for 12.5 ÷ 9 (= 1.388...) M1 for 2.5 ÷ 1.83 (= 1.366...) A1 for 1.38.... and 1.36... truncated or rounded C1 (dep on at least M1) for decision ft working shown

1MA0_2H

Question Working Answer Mark Notes 7 (a) negative 1 B1 for negative

(b) 10.3 – 11.7 2 M1 for a single straight line segment with negative gradient that could be used as a line of best fit or an indication on the diagram from 2.5 on the x axis A1 for an answer in the range 10.3 – 11.7 inclusive

8 (a) Triangle with vertices (2,–1) (4, –1) (4, –4)

2 B2 for triangle with vertices (2,–1) (4, –1) (4, –4) (B1 for triangle in correct orientation or rotated 90o anticlockwise centre O

(b) Triangle with vertices (7, 2) (13, 2) (7, 11)

3 B3 for triangle with vertices (7, 2) (13, 2) (7, 11) (B2 for 2 vertices correct or enlargement scale factor 3 in wrong position or enlargement, centre (1,2), with different scale factor) (B1 for 1 vertex correct or enlargement, not from (1,2), different scale factor)

9 51 3 M1 200 × 25.82 (= 5164) A1 for 5164 or 5160 or 5100 or 5200 or 51.64 or 51.6(0) or 52 A1 for 51 cao OR M1 for 100 ÷ 25.82 (= 3.87…) and 200 ÷ ‘3.87…’ (= 51.64) A1 for 5164 or 5160 or 5100 or 5200 or 51.64 or 51.6(0) or 52 A1 for 51 cao

1MA0_2H

Question Working Answer Mark Notes 10 (a) –1, 0, 1, 2, 3 2 B2 for all 5 correct values; ignore repeats, any order.

(–1 for each omission or additional value)

(b) 7x – 3x < 4 + 9 4x < 13

x < 3.25 2 M1 for a clear intention to use a correct operation to collect x terms or non-x terms in an (in)equality A1 for x < 3.25 oe (SC: B1 for 3.25 oe seen if M0 scored)

11

x = 4 gives 40 x = 5 gives 95 x = 4.1 gives 44.(321) x = 4.2 gives 48.(888) x = 4.3 gives 53.(707) x = 4.4 gives 58.(784) x = 4.5 gives 64.(125) x = 4.6 gives 69.(736) x = 4.7 gives 75.(623) x = 4.8 gives 81.(792) x = 4.9 gives 88.(249) x = 4.61 gives 70.3(12..) x = 4.62 gives 70.8(91..) x = 4.63 gives 71.4(72..) x = 4.64 gives 72.0(57..) x = 4.65 gives 72.6(44..)

4.6 4 B2 for a trial 4.6 ≤ x ≤ 4.7 evaluated (B1 for a trial 4 ≤ x ≤ 5 evaluated) B1 for a different trial 4.6 < x ≤ 4.65 evaluated B1 (dep on at least one previous B1) for 4.6 Accept trials correct to the nearest whole number (rounded or truncated) if the value of x is to 1 dp but correct to 1dp (rounded or truncated) if the value of x is to 2 dp. (Accept 72 for x = 4.64) NB : no working scores no marks even if the answer is correct.

1MA0_2H

Question Working Answer Mark Notes 12 0.3 × 400 120 2 M1 for 0.3 × 400 oe

A1 cao

13 5×3+15×8+25×11+35×9+45×9 =1130 1130 ÷ 40

28.25 4 M1 for finding fx with x consistent within intervals (including the end points) allow 1 error M1 (dep) for use of all correct mid-interval values M1 (dep on first M1) for Σfx ÷ 40 or Σfx ÷ Σf

A1 for 28.25 or 2841

1MA0_2H

Question Working Answer Mark Notes 14 (a) p2 – 4p + 9p – 36 p2 + 5p – 36 2 M1 for all 4 terms correct (condone incorrect signs) or

3 out of 4 terms correct with correct signs A1 cao

(b) 5w – 8 = 3(4w + 2) 5w – 8 = 12w + 6 –8 – 6 = 12w – 5w –14 = 7w

–2 3 M1 for attempting to multiply both sides by 3 as a first step (this can be implied by equations of the form 5w – 8 = 12w + ? or 5w – 8 = ?w + 6 i.e. the LHS must be correct M1 for isolating terms in w and the number terms correctly from aw + b = cw + d A1 cao OR

M1 for 38

35

−w

= 4w + 2

M1 for isolating terms in w and the number terms correctly A1 cao

(c) (x + 7)(x – 7) 1 B1 cao

(d) 23

43 yx

2 B2 for 23

43 yx or 5.143 yx or 21

143 yx (B1 for any two terms correct in a product eg. 3x4yn )


*15 180 × 365 =65700 65700 ÷1000 =65.7 65.7 × 91.22 =5993.154 5993.154÷100 + 28.20 =88.13...

D U C T 366 65880 6010 88.30365 65700 5993 88.13 65000 5929 87.49 66000 6020 88.40364 65520 5976 87.96360 64800 5911 87.31336 60480 5517 83.37

Decision ( Should have a water meter installed)

5 Per year M1 for 180 × ‘365’ (= 65700) M1 for ‘65700’ ÷ 1000 (= 65.7 or 65 or 66) M1 for ‘65.7’ × 91.22 (= 5993...) A1 for answer in range (£)87 to (£)89 C1 (dep on at least M1) for conclusion following from working seen OR (per day) M1 for 107 ÷ ‘365’ (= 0.293…) M1 for 180 ÷ 1000 × 91.22 (= 16.4196) M1 for 28.2 ÷ ‘365’ + ‘0.164196’ (units must be consistent) A1 for 29 – 30(p) and 24 – 24.3(p) oe C1 (dep on at least M1) for conclusion following from working seen OR M1 for (107 – 28.20) ÷ 0.9122 (= 86.384..) M1 for ‘86.384..’× 1000 (= 86384.5…) M1 for ‘365’ × 180 (= 65700) A1 for 65700 and 86384.5... C1 (dep on at least M1) for conclusion following from working seen NB : Allow 365 or 366 or 52×7 (=364) or 12 × 30 (=360) or 365¼ for number of days

1MA0_2H


cos x = 6.94.6

x = cos–1

6.94.6

=

48.2 3 M1 for cos x =

6.94.6

or cos x = 0.66(6...) or cos x = 0.67

M1 for cos–1

6.94.6 or cos–1 0.66(6...) or cos–1 0.67

A1 for 48.1 – 48.2 OR Correct use of Pythagoras and then trigonometry, no marks until

M1 for sin x = 6.9

'155.7' or tan x = 4.6

'155.7'

or sin x = 6.9

'155.7' × sin 90

or cos x = 6.94.62

'155.7'6.94.6 222

××−+

M1 for sin–1

6.9'155.7' or tan–1

4.6'155.7'

or sin–1 ⎟⎠⎞

⎜⎝⎛ × 90sin

6.9'155.7'

or cos–1⎟⎟⎠

⎞⎜⎜⎝

⎛××−+

6.94.62'155.7'6.94.6 222

A1 for 48.1 – 48.2 SC B2 for 0.841... (using rad) or 53.5... (using grad)

1MA0_2H

Question Working Answer Mark Notes 17 6200 × 1.0253 =

OR

6200 + 100

5.2 × 6200 = 6355

6355 + 100

5.2 × 6355 = 6513.875

6513.875 + 100

5.2 × 6513.875 =

6676.72 3 M2 for 6200 × 1.0253 (= 6676.72...) (M1 for 6200 × 1.025n, n ≠ 3) A1 for 6676.72, accept 6676.71 or 6676.73 OR M1 for 6200 × 1.025

or for 6200 + 100

5.2 × 6200 oe

or for 6355 or 155 or 465 or 6665 M1 (dep) for a complete compound interest method shown for 3 years A1 for 6676.72, accept 6676.71 or 6676.73 [SC B2 for 476.71 or 476.72 or 476.73 seen ]

1MA0_2H

Question Working Answer Mark Notes 18 BD2 + 122 = 162 oe

BD= 144256 − (=10.58...)

sin 40 = CD

'58.10'

CD = 40sin

'58.10'

16.5 5 M1 for BD2 + 122 = 162 oe or 162 – 122 or 112 seen M1 for 144256 − or 112 (=10.58...)

M1 for sin 40 = CD

'58.10' or cos 50 = CD

'58.10'

M1 for (CD =) 40sin

'58.10' or 50cos

'58.10'

A1 for 16.4 – 16.5 OR M1 for BD2 + 122 = 162 oe or 162 – 122 or 112 seen M1 for 144256 − or 112 (=10.58..)

M1 for (BC =) ‘10.58’× tan 50 or 40tan

'58.10' (=12.6…)

M1 for 22 ...'58.10''6.12' + A1 for 16.4 – 16.5

1MA0_2H


89

89

104105.8104105.8

××××−×

= 18

9

104.3101.8

××

= 910...3823529.2 −× OR

98 105.81

1041

×−

×

= 109 1017647.1105.2 −− ×−×

= 910...3823529.2 −×

4.9 × 10–5 3 B3 for 4.88 × 10–5 to 4.9 × 10–5

(B2 for digits 238(23529) or 24 or 488(09353) or 49) (B1 for digits 81 or 34) OR B3 for 4.88 × 10–5 to 4.9 × 10–5

(B2 for digits 238(23529) or 24 or 488(09353) or 49) (B1 for digits 25 or 117(647))

20

2d – 2t = 4t + 7 2d – 7 = 4t + 2t 2d – 7 = 6t

672 −d

672 −d

3 B1 for 2d – 2t or 2t +

27 oe

M1 for rearranging 4 terms correctly to isolate terms in t e.g. ‘2d’ – 7 = 4t + ‘2t’ or 2d – 7 = 6t or –6t = 7 – 2d seen

A1 for 6

72 −d oe

21

4n2 + 12n + 32 – (4n2 – 12n + 32) = 4n2 + 12n + 9 – 4n2 + 12n – 9 = 24n =8 × 3n

Proof 3 M1 for 3 out of 4 terms correct in expansion of either (2n + 3)2 or (2n – 3)2

or ((2n + 3) – (2n – 3))((2n + 3) + (2n – 3)) A1 for 24n from correct expansion of both brackets A1 (dep on A1) for 24n is a multiple of 8 or 24n = 8 × 3n or 24n ÷ 8 = 3n

1MA0_2H

Question Working Answer Mark Notes 22 a = 3, b = –4, c = –2

x = ( )

3223444 2

×−××−−±−−

= 6

24164 +± = 6

404 ±

= 1.72075922 or = – 0.3874258867 OR

032

342 =−− xx

032

32

32 22

=−⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛ −x

32

32

32 2

+⎟⎠⎞

⎜⎝⎛=−x

x = 9

1032±

1.72, –0.387

3 M1 for

( )32

23444 2

×−××−−±−−

(condone incorrect

signs for –4 and –2)

M1 for 6

404 ± or 3

102 ±

A1 for one answer in the range 1.72 to 1.721 and one answer in the range – 0.387 to – 0.38743 OR

M1 for 2

32⎟⎠⎞

⎜⎝⎛ −x oe

M1 for method leading to 9

1032± oe

A1 for one answer in the range 1.72 to 1.721 and one answer in the range – 0.387 to – 0.38743

1MA0_2H

Question Working Answer Mark Notes 23 (a)(i) Explanation : Each member of

the population has an equal chance of selection

Each member of the population has an equal

chance of selection

2 B1 for explanation

(ii) Description : Eg. number each student and use random select on a calculator

Valid method B1 for an acceptable description

(b) 239+257+248+190+206=1140

1140239 ×100

21 2 M1 for

'1140'239 × 100 oe or 20.96…

A1 cao

1MA0_2H


64sin7.8

49sin=

AC

49sin64sin7.8

×=AC

(= 7.305…)

21 × 8.7 × 7.305... × sin (180 – 64 – 49)

29.3 5 M1 for

64sin7.8

49sin=

AC oe

M1 for (AC =) 49sin64sin7.8

×

A1 for 7.3(05…)

M1 for 21 × 8.7 × ‘7.305’ × sin(180 – 64 – 49)

A1 for 29.19 – 29.3 OR

M1 for 64sin7.8

)4964180sin(=

−−BC oe

M1 for (BC =) '67sin'64sin7.8

×

A1 for 8.9(10...)

M1 for 21 × 8.7 × ‘8.910’ × sin 49

A1 for 29.19 – 29.3 OR (X is point such that AX is perpendicular to BC) M1 for AX = 8.7×sin 49 (= 6.565…) or XB = 8.7×cos 49 (= 5.707...) M1 for XB = 8.7×cos 49 (= 5.707...) and CX = ‘6.565’ ÷ tan 64 oe (= 3.202…) A1 for 8.9(10...) or 5.7(07...) and 3.2(02...) M1 for ½ × ‘6.565…’ × (‘5.707’ + ‘3.202’) oe A1 for 29.19 – 29.3

1MA0_2H


192

203

194

205

1911

2012

×+×+×

⎟⎠⎞

⎜⎝⎛ ×+×+×−

192

203

194

205

1911

20121

380222

4

B1 for 1912 or

195 or

193 (could be seen in working or on

a tree diagram)

M1 for 195

203

1912

203

193

205

1912

205

193

2012

195

2012

×××××× ororororor

M1 for 195

203

1912

203

193

205

1912

205

193

2012

195

2012

×+×+×+×+×+×

A1 for 380222

oe or 0.58(421...)

OR

B1 for 198 or

1915 or

1917

M1 for 1917

203

1915

205

198

2012

××× oror

M1 for 1917

203

1915

205

198

2012

×+×+×

A1 for 380222

oe or 0.58(421...)

OR (continued overleaf…)

1MA0_2H


contd

B1 for 1911 or

194 or

192

M1 for 192

203

194

205

1911

2012

××× oror

M1 for ⎟⎠⎞

⎜⎝⎛ ×+×+×−

192

203

194

205

1911

20121

A1 for 380222

oe or 0.58(421...)

NB if decimals used they must be correct to at least 2 decimal places SC : with replacement

B2 for 200111 oe

OR e.g. B0

M1 for 2017

203

2015

205

208

2012

××× oror

M1 for 2017

203

2015

205

208

2012

×+×+×

A0

1MA0_2H

Question Working Answer Mark Notes 26 (a) b – a 1 B1 for b – a or –a + b

(b) APOAOP +=

AP = 43 × (b – a)

OP = a + 43 × (b – a)

OR

BPOBOP +=

BP = 41 × (a – b)

OP = b + 41 × (a – b)

41 (a + 3b)

3 B1 for

43 × ‘(b – a)’

M1 for )( =OP APOA + or )( =OP ABOA43

+

or a ± 43 × ‘(b – a)’

A1 for 41 (a + 3b) or

41 a +

43 b

OR

B1 for 41 × ‘(a – b)’

M1 for )( =OP BPOB + or )( =OP BAOB41

+

or b ± 41 × ‘(a – b)’

A1 for 41 (a + 3b) or

41 a +

43 b

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 UG032626 Summer 2012

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

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