Answer ALL questions - Maths Tallis -...

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 95 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.

Suggested Grade Boundaries (for guidance only)A* A B C D70 46 29 18 12

Practice Paper – Gold PlusThis publication may only be reproduced in accordance with Pearson Education Limited copyright policy.©2015 Pearson Education Limited.

Practice Paper – Gold

Gold Plus

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 = (a + b)h

Volume of sphere πr3 Volume of cone πr2h

Surface area of sphere = 4πr2 Curved surface area of cone = πrl

In any triangle ABC The Quadratic EquationThe solutions of ax2+ bx + c = 0where a ≠ 0, are given by

x =

Sine Rule

Cosine Rule a2 = b2+ c2– 2bc cos A

Area of triangle = ab sin C

Practice Paper: Gold Plus 2

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)___________________________________________________________________________


2. Work out an estimate for the value of 223

799.02.60

Give your answer as a decimal.

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

(Total 3 marks)___________________________________________________________________________


*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)

(Total 4 marks)___________________________________________________________________________


4. Jim did a survey on the lengths of caterpillars he found on a field trip.Information about the lengths is given in the stem and leaf diagram.

1 3 5 7 7 Key: 5|2 means 5.2 cm2 0 6 8 8 8 93 1 5 5 5 5 6 8 94 1 55 2

Work out the median.

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

(Total 2 marks)___________________________________________________________________________

5. Here are the first 5 terms of an arithmetic sequence.

3 9 15 21 27

(a) Find an expression, in terms of n, for the nth term of this sequence.

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

Ben says that 150 is in the sequence.

(b) Is Ben right?You must explain your answer.

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

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

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

(Total 3 marks)___________________________________________________________________________


6. 30 students took a test.The table shows information about how long it took them to complete the test.

Time (t minutes) Frequency0 < t 10 510 < t 20 720 < t 30 830 < t 40 640 < t 50 4

(a) On the grid, draw a frequency polygon for this information.

(2)

(b) Write down the modal class interval.

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

(Total 3 marks)


7.

(a) On the grid above, reflect shape A in the line x = –1.

(2)

(b) Describe fully the single transformation that will map shape P onto shape Q.

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

(Total 4 marks)___________________________________________________________________________


8. Margaret has some goats.The goats produce an average total of 21.7 litres of milk per day for 280 days.

Margaret sells the milk in 12

litre bottles.

Work out an estimate for the total number of bottles that Margaret will be able to fill with the milk.

You must show clearly how you got your estimate.

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

(Total 3 marks)___________________________________________________________________________


9. (a) A solid cube has sides of length 5 cm.

Work out the total surface area of the cube.State the units of your answer.

...............................................(4)

The volume of the cube is 125 cm3.

(b) Change 125 cm3 into mm3.

............................ mm3

(2)The weight of the cube is 87 grams, correct to the nearest gram.

(c) (i) What is the minimum the weight could be?

............................ grams

(ii) What is the maximum the weight could be?

............................ grams(2)

(Total 8 marks)


10. Use ruler and compasses to construct the perpendicular from point C to the line AB.You must show all your construction lines.

(Total 2 marks)___________________________________________________________________________


11. Peter, Tarish and Ben share £54.

Tarish gets three times as much money as Peter.Ben gets twice as much money as Tarish.

How much money does Ben get?

£ ..................................(Total 3 marks)

___________________________________________________________________________


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

(Total 3 marks)___________________________________________________________________________


13.

The diagram shows a square and 4 regular pentagons.

Work out the size of the angle marked x.

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

(Total 3 marks)___________________________________________________________________________


14. Use ruler and compasses to construct an angle of 30° at P.You must show all your construction lines.

(Total 3 marks)___________________________________________________________________________


15. Here is a scale drawing of a rectangular garden ABCD.

Scale: 1 cm represents 1 metre.

Jane wants to plant a tree in the garden

at least 5m from point C,nearer to AB than to AD

and less than 3m from DC.

On the diagram, shade the region where Jane can plant the tree.

(Total 4 marks)___________________________________________________________________________


16. Colin took a sample of 80 football players.

He recorded the total distance, in kilometres, each player ran in the first half of their matches on Saturday.

Colin drew this box plot for his results.

(a) Work out the interquartile range............................................... km

(2)

There were 80 players in Colin’s sample.

(b) Work out the number of players who ran a distance of more than 5.6 km.

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

Colin also recorded the total distance each player ran in the second half of their matches.

He drew the box plot below for this information.

(c) Compare the distribution of the distances run in the first half with the distribution of the distances run in the second half.

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

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

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

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

(Total for Question 16 is 6 marks)___________________________________________________________________________


17.

In the diagram,

ABC is a triangle,angle ACB = 90°,P lies on the line AB,CP is perpendicular to AB.

Prove that the angles of triangle APC are the same as the angles of triangle CPB.

(Total 3 marks)___________________________________________________________________________


18. The diagram shows part of a pattern made from tiles.

The pattern is made from two types of tiles, tile A and tile B.

Both tile A and tile B are regular polygons.

Work out the number of sides tile A has.

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

(Total 4 marks)___________________________________________________________________________


19.

A, B, C and D are points on a circle, centre O.BC = CD.Angle BCD = 130°.

(a) Write down the size of angle BAD.Give a reason for your answer.

...................................°(2)

(b) Work out the size of angle ODC.Give reasons for your answer.

...................................°(4)

(Total 6 marks)___________________________________________________________________________


20.

ABCD is a square with a side length of 4x.M is the midpoint of DC.N is the point on AD where ND = x.

BMN is a right-angled triangle.

Find an expression, in terms of x, for the area of triangle BMN.Give your expression in its simplest form.

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

(Total 4 marks)___________________________________________________________________________


21. Here is a shape ABCDE.

AB, BC and CD are three sides of a square.BC = x cm.AED is a semicircle with diameter AD.

The perimeter, P cm, of the shape ABCDE is given by the formula

P = 3x + 2x

(a) Rearrange this formula to make x the subject.

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


The area, A cm2, of this shape is given by A = kx2 where k is a constant.

(b) Find the exact value of k.Give your answer in its simplest form.

.....................................(3)

(Total 5 marks)___________________________________________________________________________


22. (a) Find the value of 32

27 .

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

(b) Given that 2188

= a + b 2 , where a and b are integers,

find the value of a and the value of b.

a = ..............................

b = ..............................(3)

(Total 5 marks)___________________________________________________________________________


23. Express the recurring decimal 182.0 as a fraction in its simplest form.

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

(Total 3 marks)___________________________________________________________________________


24. Solve the equation 11

22

x

x.

(Total 4 marks)___________________________________________________________________________


25.

P, Q and T are points on the circumference of a circle, centre O.The line ATB is the tangent at T to the circle.

PQ = TQ.Angle ATP = 58°.

Calculate the size of angle OTQ.Give a reason for each stage in your working.

................................... °

(Total 5 marks)

TOTAL FOR PAPER IS 100 MARKS


1 (a) Type of filmTally

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 reasoneg. all same age, sample too small, biased, same school

2223

799.02.60

60 0.8 48 0.24200 200

0.24 3 B1 for any two of 60, 0.8, 200 seen or 48 seenM1 for at least one of 60, 0.8, 200 and a correct method to begin to evaluate eg. the numerator may be correctly evaluated ora correctly simplified fraction (NB. fraction may not be fully simplified)A1 for answer in the range 0.15 to 0.3 from correct working


3 (a) 10 1 B1 cao

(b)

Miles 0 10 20 30 40 50Ed 0 15 30 45 60 75Bill 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 oeC2 (dep on M1) for a correct full statement ft from grapheg. Ed cheaper up to 20 miles and Bill cheaper for more than 20 miles (C1 (dep on M1) for a correct conclusion ft from grapheg. cheaper at 10 miles with Ed ; eg. cheaper at 50 miles with Billeg. same cost at 20 miles; eg for £5 go further with Bill ORA general statement covering short and long distances eg. Ed is cheaper for shorter distances and Bill is cheaper for long distances)

4 3.1 2 M1 sight of 11th value or digits 31A1 3.1

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

Practice Paper: Gold Plus

10

20

30

40

5 10

15

20

25

30

0 x

y

29

7 (a) Reflection 2 B2 for vertices of shape plotted at (-3, 2), (-3, 3), (-5, 3), (-6, 2.5), (-5, 2) (B1 for a reflection in any vertical or horizontal line)

(b)Translation;

16 2 B1 for translation

B1 (indep.) for 6 left and 1 down OR

16

Note B0 if more than one transfomation given


8 12000 3 B1 for 20 or 300 used

M1 for “20” × “300” or or , values do not need to be roundedA1 for answer in the range 11200 –13200

SC B3 for 12000 with or without working

10 construction 2 M1 for a pair of arcs or a single arc, centre C, that cut line AB and at least one pair of arcs not at C within guidelinesA1 for perpendicular within guidelines with appropriate construction arcs


11 P: T: B = 1 : 3 : 654 ÷ 10 × 6orT = 3P and B = 2T oeSo, B = 2 × (3P) = 6PP + T + B = P + 3P + 6P = 10PP = 54 ÷ 10 = £5.40B = 6 × £5.40

32.40 3 M1 for 1 : 3 : 6 or any three numbers, in any order, in the ratio 1 : 3 : 6M1 for 54 ÷ (1 + 3 + 6) × 6)A1 for 32.4(0)

12 6 × 10 × 8 = 480480 ÷ (6 × 20) =

4 3 M1 for 6 × 10 × 8 or 480 seenM1 (dep) for '480' ÷ (6 × 20) oeA1 cao

13 54 3 M1 for 180 – 360 ÷ 5 or 108 seen as the interior angle of a pentagonM1 (dep on previous M1) for 360 – 2 × ‘108’ – 90A1 for 54 cao

14 Construction 3 M1 for arcs construction of 60 degreesM1 (dep) for arcs bisector of '60 degrees' (not 90 degrees)A1 (dep on both M marks) for 30 degrees within guidelinesORM1 for arc construction of 90 degreesM1(dep) for arc construction of 60 degreesA1 (dep on both M marks) for 30 degrees within guidelines


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 ½ (12 + 8) × 6 = 60‘60’ × 20 = 12001200 × 5 = 60006000 ÷ 1000 = 6

6 5 M1 ½ (12 + 8) × 6 oe or 60 seen M1 (dep) ‘60’ × 20M1 (indep) ‘1200’ × 5A1 6000 caoA1 ft (dep on 1st or 3rd M1 scored) for 6

17 PBC = 90 − PAC

BCP = 90 – (90 – PAC)

Proof 3 M1 for PBC = 90 – PAC or PAC = 90 – PBC or ACP = 90 – PCB M1 for BCP = 90 – (90 – PAC) or PAC = 90 – (90 – BCP) oeA1 for PAC = PCB and PCA = PBC and APC = CPBB1 SC if M0 awarded for APC = BPC = 90º or statement matching the 3 equal sets of angles PAC = PCB and PCA = PBC and APC = CPB

18 12 4 B1 for 60 seenM1 for (360 – 60) ÷ 2 (=150)M1 for 360 ÷ (180 – 150) or 150×n=180(n-2) oeA1 cao


19 (a) 50o

reason2 B2 for Angle BAD = 50 and the sum of opposite angles in

a cyclic quadrilateral is 180(B1 for angle BAD = 50 or angle BAD = 180 - 130)

(b) Angle BOD = 100o

Angle OBD = angle ODCAngle ODC = (360o – 230o) ÷ 2 = 65

65o 4 M1 angle BOD = 100° or ft 2 × their answer to (a) (may be on diagram)M1 360° – (130° + “100°”) and ÷2A1 caoB1 The angle at the centre of a circle is twice the angle at the circumference and Angles in a quadrilateral (4 sided shape) add up to 360º or opposite angles of a kite are the same.

20 5x2 4 M1 for 4x × 4xM1 for (2x ×4x)/2 or (2x × x)/2 or(3x ×4x)/2M1(dep M2) for “16 x2” – “4 x2”– “x2” – “6 x2”A1 for 5x2

21 (a) 3 32 2

32

xP x x

Px

x =

23 π

P 2M1 for

23 x

A1 for

23

Pxoe

Practice Paper: Gold Plus

22 (a) 1327 = 3

3 –2 = 2

13

19

2 M1 for a correct cube root, reciprocal or square

A1 for 19

or 0.11(1…)

(b)

218

28

2188

218

22

28

32

28

a = −3b = 4

3 M1 for attempt to rationalise denominator, e.g.

8 2 18 22 2 2 2

or

22

2188

Or 8 – √18 = √2(a + b √2) oeA2 for –3 + 4√2(A1 for −3)(A1 for 4)SC B1 if M0 scored for –3 or 4 seen on either answer line

34

23 e.g.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 subtractione.g. 100x = 28.1818…, x = 0.28181… and subtracting

e.g. 1000x = 281.8181…, 10x = 2.8181… and subtracting

OR 27.999

or 279990

oe

A1 cao


242( 1) 2 2( 1) 1 2( 1)

2 1x x x x

x

1241 xxx2242 xxx

062 xx 023 xx

x = 3, -2 4 M1 for an attempt to multiply one term of the equation by 2 or x + 1or 12 x or 2×x + 1with or without cancelling or attempt to write LHS with a common denominator

M1for attempt to multiply all terms by 2(x + 1) with or without cancelling

e.g. 2( 1) 2 2( 1) 1 2( 1)

2 1x x x x

x

Or x (x + 1) – 4 = 2(x + 1)

A1 for x2 + x –4 = 2x + 2 or x2 – x – 6 = 0

A1 cao for 3 and – 2


Question 1

Surprisingly, part (b) was answered correctly more often than part (a). In part (b) the majority of candidates generally recognised that the sample was too small or the age range too narrow. In part (a), despite the fact that a data collection table was asked for in the question, a significant number of questions suitable for a questionnaire were still seen. The modal mark scored was one as either a column for tallies or the frequency column was often omitted.

Question 2

The majority of candidates realised the need to work out an estimate rather than the actual value. However, many candidates failed to round the given numbers to one significant figure and therefore ended up attempting a much harder calculation than necessary for example, the denominator was often rounded to 220 rather than 200. Those who did approximate to

2008.060

then failed to evaluate this correctly with careless errors such as 60 × 0.8 given as

46 or 480 and 48 ÷ 200 being given as 2.4 or 0.0024 . Some candidates tried to simplify

2008.060

incorrectly as 2000

8600 multiplying each term in the numerator by 10 presumably to

avoid the decimal but only multiplying the denominator by 10 . Some answers were left as a fraction rather than a decimal as required by the demand of the question. A small minority of candidates did, however, attempt to work out the actual answer rather than an estimate. This gained no marks.

Question 3

Part (b) differentiated well. It was also a question testing QWC so it was essential that a method was shown. The more able candidates realised that drawing a graph to show Ed’s costs was the most efficient method of solution. Candidates who took this approach then generally made a correct statement that referred to 20 miles (the break-even point). Less able candidates used the information given and the graph to find the delivery costs for a particular distance and then either made a comment or just left the calculations as their final answer. It was not uncommon to see calculations which failed to refer to distance or Bill or Ed. Some failed to gain any marks as they just focused on comparing the fixed charges or cost per mile or a combination of these in a general way. Others were confused by Bill’s £10 fixed charge and added it on twice, e.g. if he went 10 miles then they said that he charged £30 (£20 plus his £10 fixed charge).

Question 4

Many candidates were able to write down the correct answers for part (a), although a significant minority wrote down the coordinates reversed.

For part (b), many plotted where they thought the midpoint was and tried to read off the coordinates and others had an intuitive idea of what to do, often finding the x-coordinate correct but not the y-coordinate, which was often given as 1 instead of the correct 0.5.


Question 5

Questions on finding the nth term of an arithmetic sequence regularly appear on our papers so it is surprising to see so many answers of n + 6 instead of 6n − 3.

In part (b), a multitude of solutions fell short of the mark because they were incomplete, e.g.

‘they are all odd’, without mentioning that 150 was even or saying that the value of n or 6153

is not an integer without stating the equation 6n − 3 = 150. It was also commonly thought that, because 15 was in the sequence, 150 was as well.

Question 6

Answers to this question were disappointing. There were several sources of error: many candidates plotted the points at the end of the interval rather than at the middle; many candidates either did not join their points or joined them with a curve; many candidates joined the last point back to the first point; many candidates drew a bar chart.

Part (b) was answered well although some candidates gave the frequency (8) rather than the class interval itself.

Question 7

This reflection was poorly answered with many candidates reflecting the shape in the x or y axis or other vertical or horizontal lines. Only 40% of candidates were able to reflect the shape in the correct line, x = –1, but 37% of candidates were awarded a mark for a reflection in a line of the form y = m or x = n (n ≠ –1) those who were successful were frequently seen to have drawn the line x = –1 on the grid. Part (b) again was not very well answered with many candidates unable to describe the correct translation as a vector or even as a description in words. Very often the word translation was omitted and some gave a combination of transformations when a single transformation was asked for. Many candidates who attempted the vector notation often omitted the word “translation” but used instead, transferred, transformed or moved, none of which were acceptable. Vectors were often given as the reverse of the correct one. A significant number of candidates tried to describe the transformation of Q onto P rather than P onto Q. Only 16% of candidates scored both marks whilst 36% gained one mark.

Question 8

Candidates were presented with two challenges in this question. Firstly, they had to decide on the calculations needed to work out the number of bottles that could be filled with milk and secondly, to find an estimate of this. Most candidates gained some credit for their responses, usually for identifying an appropriate calculation. However, the number of candidates who took the easiest route to find an estimate, ie to round values correct to one significant figure

then work out 5.030020

, was relatively small. Instead many candidates either failed to round

any of the quantities or rounded only one of the quantities, usually 21.7 to 22. As a result they made calculations more onerous and prone to error. Division by 0.5 was confused with dividing by 2. This question clearly identified an area where candidates would benefit from more practice.


Question 9

There were too many candidates who either worked out the volume (125 cm3) of the cube or who worked out 6 times the perimeter of one of the faces. Conversion from cm3 to mm3 was even more poorly done, with a usual answer of 1250. Few candidates made the link between the 5 cm as the edge of the cube and its equivalent 50 mm. Part (c) was competently answered with the lower bound of 86.5 more often identified than the 87.5 at the top end where often 84.49 was written or some attempt at a recurring decimal.

Question 10

This question was not done well. Few students could construct the perpendicular from the given point to the line. When drawing the arcs at point C centre A and centre B, students should be advised to draw arcs below the line as well as at point C. It was evident that a significant number of students did not use compasses to draw their construction arcs. A common incorrect answer was to draw the perpendicular bisector of the line AB.

Question 11

This ratio question was answered correctly by 41% of candidates. 19% of candidates gained one mark for writing any three values in the ratio 1 : 3 : 6 thus showing that they had an understanding of the problem. Some candidates, 3%, scored two marks for showing that they were going to divide £54 by (1 + 3 + 6) and then multiply by 6 whilst others used a decomposition method to show £50 as £5 : £15 : £ 30 followed by £4 as 40p : £1.20 : £2.40.

Frequent incorrect methods were the use of 1 : 2 : 3 to give 9, 18, 27. Surprisingly several candidates found all 3 correct amounts and then selected the wrong value, thus losing a mark. Many candidates used a trial and improvement method to solve the question but the extra £4 proved difficult to share correctly. Other ratios seen were 1 : 2 : 3 and 1 : 3 : 5. Some merely divided 54 by 3 and based their incorrect solutions around £18 and so missed the point of the question.

Question 12

It was good to see a whole range of methods being used to successfully answer this question. Some candidates chose to find the volume of drink in the carton and then divide by the area of the new face in contact with the table. However, more popular was the use of scale factors taking into consideration that the area of the new face in contact with the table was twice the area of the previous face in contact and therefore the height of drink in the carton would halve. A very few candidates got the faces the wrong way round and ended up with an answer of 16 cm. Provided this answer was supported by correct working two marks were awarded. However, many candidates started off by either working out the volume of the container and were then unsure how to proceed further.

Question 13

Although few candidates gave a fully correct answer to this question, there was much misunderstanding of the relevance of dividing 360 by 5. A small number of candidates found 108˚ as the interior angle in a regular pentagon but could make no further progress and those who understood the question but showed inaccurate calculations scored 2 marks.

It was also clear that many candidates did not use the diagram, as they did not appreciate that the interior angle of a regular pentagon was obtuse and could not be 72.


Question 14

This question was very poorly answered with only 10% of candidates gaining full marks for an accurate 30º construction. Many candidates used a protractor to draw the angle then attempted with spurious arcs to pretend they had constructed the 30º angle. The most successful attempts drew a 60º angle either using an equilateral triangle or using the standard construction and gained one mark (7% of candidates) whilst a further 0.3% of candidates then bisected their angle to gain the second mark. Significant numbers of candidates knew how to bisect an angle but had not managed to successfully construct the 60 angle first. 83% of candidates scored no marks at all.

Question 15

This question on loci was poorly answered with very few candidates scoring full marks. The modal mark awarded was zero; 1 mark was awarded for the quarter circle of radius 5 cm and a line parallel to CD and 3 cm away from it. The most common mistake was to misunderstand ‘nearer to AB than to AD’ as few bisectors of angle A were given with the diagonal AC often seen in its place.

Question 16

Many candidates split their cross-sectional area into triangles and a rectangle, some doing it successfully and completing the question. Few could remember or correctly apply the formula for the area of a trapezium, or multiplied all the numbers they could see (or a selection of) or found the total surface area. For some, the step by step requirements of the question prevented them from following any sort of logical process, with the cross-sectional area just being the first hurdle. This was evident in the written work which was often chaotic and lacked any methodical approach. Many gained the latter two marks for correctly multiplying their volume by 5 and then converting correctly to kg by dividing by 1000. However, there were equally as many candidates who tried to convert g to kg by dividing by 100 or 10, or who tried to find the mass by dividing by 5. Only 9% of candidates scored full marks on this question with 68% failing to score any marks.

Question 17

This is a similar question problem. The first part can either be done from 12

10 6BC

or from

1012 6BC

. Of the second method, most candidates used the equivalent scale factor of 106

but

often rounding to 1.7 or truncating to 1.6. There was some latitude allowed for this in part (a) where answers from 19.9 to 20.4 were accepted but in part (b) the answer had to be exact.

Most candidates used their scale factor again in part (b) by working out 18 divided by their scale factor.

Many candidates also thought that this was the ‘Pythagoras question’ and some thought that you added on a constant amount to the lengths of the small triangle to get the lengths of the large triangle.

Question 18

Candidates drew on a number of different methods in making progress with this question. Those who gained the most marks generally worked on, and with, the diagram, making clear


which angles were being found. When calculating angles, it was not always clear whether it was an internal or external angle that was being found. Sometimes an angle was calculated in the working but then shown to be a different angle on the diagram; in these cases there was a penalty since it was not clear the candidate understood what they were finding.

Question 19

There was plenty of confusion evident in this question as to which was the cyclic quadrilateral with many candidates incorrectly using BODC as a cyclic quadrilateral. An even more common error was the belief that BODC was a parallelogram and angle BOD was 130°.

Too often in part (b) numbers were just written down with no attempt to demonstrate which angles these referred to meaning that method marks could not always be awarded. When reasons are asked for in a question it is essential that these are given clearly. For example, writing ‘angles in a cyclic quadrilateral add up to 180’ was not sufficient to gain the mark for a correct reason in part (a), the fact that it is opposite angles that sum to 180 had to be made clear. Most candidates ignored the instruction to give reasons. Too many candidates still use single letters to represent angles, e.g. B = D.

Question 20

The best candidates gave clear and concise solutions to this question. However most candidates were unable to make much headway in giving accurate expressions for the area of the square or for the area of the unshaded triangles or for the sides of the shaded triangle. A large proportion of the algebra seen was spoiled by the omission of brackets, for example by expressing the area of the square as 4x × x or as 4x2 instead of 4x × 4x, (4x)2, or 16x2 or in attempts to use Pythagoras rule. The square root sign was often used wrongly or ambiguously. These errors led to many candidates failing to score any credit for their attempts. Most candidates used the method of finding the area of the square and subtracting the areas of the three unshaded triangles but there were some excellent solutions harnessing Pythagoras rule to find the lengths of the sides NM and BM and then the area of triangle BNM. A significant proportion of candidates did not attempt this question.

Question 21

A common error in part (a) was to forget to multiply all terms by 2 when attempting to clear the fraction. Candidates who made this error but then went on to make the x the subject successfully were awarded one mark.

In part (a) only a few candidates realised that to isolate x they had to factorise the expression.

Only a very small minority of candidates realised that they had to find the area of the given shape for part (b). Of those who did appreciate this, the common errors were to forget to

halve the area of a circle to find the area of the semi-circle or to omit brackets around 2x

and

therefore end up with 2

2x rather than 4

2x . This was the question that candidates found the

most demanding.

Question 22

In part (a) of this question on irrational numbers and fractional indices 12% of candidates were able to gain one mark for establishing a correct root or power or reciprocal and a further


16% gained both marks for the correct answer. A common error was to interpret ³√27 as 27 ÷ 3. This would often be followed either by squaring or multiplying by –2. In part (b) most candidates realised they had to rationalise the denominator of the fraction or equated the given fraction to a + b√2 and multiplied this by √2 and 14% gained one mark for doing this. Full marks were only gained by 3% of candidates. The absence of the use of brackets when multiplying by √2 led to errors in subsequent work and there was the usual inappropriate ‘cancelling’ in many scripts.

Question 23

Candidates who were able to recognise that the given recurring decimal was 0.28181... rather than 0.281281... gained a generous first method mark. In order to gain the second method mark a full correct method had to be seen. Unfortunately, many attempted the subtraction of

281.8181... and 0.28181... which is an incorrect method. Some got as far as 99

9.27 or

99279

but

were then unable to finish their solution correctly to arrive at the correct answer of 110

3.

There were many incorrect guesses of 1000281

and 999281

seen.

Question 24

Only 6% of candidates were able to find the correct solution to this fractional equation but 18% of candidates either wrote the correct common denominator or multiplied one term by 2 or (x + 1) or 2(x + 1) and then a further 3% gained two marks for attempting to multiply all the terms by 2(x + 1) most usually failed to obtain this mark because they had forgotten to multiply the right-hand side of the equation by 2(x + 1). Once again the absence of brackets led to errors.

Question 25

Very few candidates appeared to have any knowledge or understanding of the ‘alternate segment theorem’. Consequently many solutions attempted to find the size of the angle PQT by long-winded methods. Whilst many were able to find that 58o was the size of angle PQT, very few gave full explanations for each stage of the working and thus restricted their award to just 3 out of the possible 5 marks. The most common alternative approach was to find the size of angle POT and then use the ‘angle at the centre theorem’ Other methods included the use of congruent triangles POQ and TOQ but this was rarely proved and so never gained full credit. Weaker candidates mistook PQ and ATB as parallel or ATP and QTB as equal.


Practice paper: Gold 4Mean score for students achieving Grade:

Spec PaperSessionYYMM Question

Mean score

Max score

Mean% ALL A* A % A B C % C D E

1MA0 1H 1206 Q01 1.58 3 53 1.58 1.85 1.68 56.0 1.61 1.54 51.3 1.38 1.181380 1H 1203 Q02 1.52 3 51 1.52 2.57 2.16 72.0 1.76 1.33 44.3 0.87 0.501MA0 1H 1206 Q03 1.68 4 42 1.68 3.16 2.43 60.8 1.88 1.30 32.5 0.67 0.381380 1H 1111 Q04 0.99 2 50 0.99 1.39 1.21 60.5 1.13 1.00 50.0 0.81 0.671MA0 1H 1303 Q05 1.24 3 41 1.24 2.56 2.13 71.0 1.60 1.07 35.7 0.67 0.321380 1H 911 Q06 1.35 3 45 1.35 2.34 1.85 61.7 1.48 1.13 37.7 0.84 0.581380 1H 1106 Q07 1.86 4 47 1.86 3.26 2.63 65.8 1.91 1.19 29.8 0.66 0.441MA0 1H 1306 Q08 1.22 3 41 1.22 2.15 1.72 57.3 1.40 1.08 36.0 0.74 0.431380 1H 911 Q09 3.82 8 48 3.82 7.18 5.84 73.0 4.61 2.94 36.8 1.57 0.821MA0 1H 1411 Q10 0.33 2 17 0.33 1.64 1.15 57.5 0.68 0.29 14.5 0.15 0.061380 1H 1106 Q11 1.47 3 49 1.47 2.75 2.23 74.3 1.52 0.81 27.0 0.34 0.151MA0 1H 1206 Q12 1.11 3 37 1.11 2.55 1.74 58.0 1.12 0.75 25.0 0.48 0.361MA0 1H 1303 Q13 0.60 3 20 0.60 2.66 1.86 62.0 0.92 0.27 9.0 0.05 0.011380 1H 1106 Q14 0.39 3 13 0.39 1.64 0.60 20.0 0.21 0.10 3.3 0.06 0.041MA0 1H 1303 Q15 0.95 4 24 0.95 2.84 1.98 49.5 1.38 0.73 18.3 0.26 0.071380 1H 1111 Q16 0.91 5 18 0.91 4.14 2.74 54.8 1.30 0.36 7.2 0.09 0.051380 1H 1106 Q17 0.32 3 11 0.32 0.84 0.50 16.7 0.29 0.14 4.7 0.06 0.031MA0 1H 1211 Q18 0.92 4 23 0.92 3.13 1.95 48.8 1.18 0.73 18.3 0.44 0.281380 1H 1203 Q19 0.98 6 16 0.98 3.52 1.86 31.0 1.01 0.54 9.0 0.31 0.231MA0 1H 1306 Q20 0.46 4 12 0.46 2.43 1.27 31.8 0.44 0.07 1.8 0.01 0.001380 1H 1203 Q21 0.11 5 2 0.11 1.07 0.25 5.0 0.03 0.00 0.0 0.00 0.001380 1H 1106 Q22 0.73 5 15 0.73 3.03 1.29 25.8 0.37 0.11 2.2 0.05 0.041MA0 1H 1206 Q24 0.37 3 12 0.37 1.74 0.74 24.7 0.27 0.08 2.7 0.03 0.011380 1H 1106 Q27 0.61 4 15 0.61 2.89 1.04 26.0 0.26 0.05 1.3 0.01 0.011380 1H 1006 Q27 0.74 5 15 0.74 3.48 1.29 25.8 0.28 0.07 1.4 0.03 0.02 26.26 95 27 26.26 66.81 44.14 46.46 28.64 17.68 18.61 10.58 6.68


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