mathematics - m. selkirk confey collegemathematics higher level – paper 1 marking scheme (300...

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J.18/20

Pre-Junior Certificate Examination, 2016

Mathematics Higher Level

Marking Scheme

Paper 1 Pg. 2 Paper 2 Pg. 36

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Mathematics

Higher Level – Paper 1 Marking Scheme (300 marks)

Structure of the Marking Scheme

Students’ responses are marked according to different scales, depending on the types of response anticipated. Scales labelled A divide students’ responses into two categories (correct and incorrect). Scales labelled B divide responses into three categories (correct, partially correct, and incorrect), and so on. These scales and the marks that they generate are summarised in the following table:

Scale label A B C D

No. of categories 2 3 4 5

5 mark scale 0, 3, 5 0, 2, 4, 5 0, 2, 3, 4, 5 10 mark scale 0, 4, 7, 10 0, 4, 6, 8, 10 15 mark scale

A general descriptor of each point on each scale is given below. More specific directions in relation to interpreting the scales in the context of each question are given in the scheme, where necessary.

Marking scales – level descriptors

A-scales (two categories) incorrect response (no credit) correct response (full credit)

B-scales (three categories) response of no substantial merit (no credit) partially correct response (partial credit) correct response (full credit)

C-scales (four categories) response of no substantial merit (no credit) response with some merit (low partial credit) almost correct response (high partial credit) correct response (full credit)

D-scales (five categories) response of no substantial merit (no credit) response with some merit (low partial credit) response about half-right (middle partial credit) almost correct response (high partial credit) correct response (full credit)

In certain cases, typically involving incorrect rounding, omission of units, a misreading that does not oversimplify the work or an arithmetical error that does not oversimplify the work, a mark that is one mark below the full-credit mark may also be awarded. Such cases are flagged with an asterisk. Thus, for example, scale 10C* indicates that 9 marks may be awarded.

The * for units to be applied only if the student’s answer is fully correct. The * to be applied once only within each section (a), (b), (c), etc. of all questions. The * penalty is not applied to currency solutions.

Unless otherwise specified, accept correct answer with or without work shown.

Accept a students’ work in one part of a question for use in subsequent parts of the question, unless this oversimplifies the work involved.

examsDEB

DEB 2014 (Paper 1)

Scale label A B CNo of categories 2 3 4

5 mark scale 0, 5 0, 3, 5 0, 2,

10 mark scale 0, 10 0, 5, 10 0, 3, 7

15 mark scale 0, 15 0, 10, 15 0, 10,

Name/version:

Printed: Whom:

Checked:

To: Ret’d:

Updated: Whom:

Name/version:

Complete (y/n): Whom: 2005 Print Stamp.doc

2016.1 J.18/20_MS 3/72 of 71 examsDEB

Summary of Marks – 2016 JC Maths (Higher Level, Paper 1)

Q.1 (a) 5C (0, 2, 4, 5) Q.8 (a) 5B (0, 3, 5) (b) (i) 5C (0, 2, 4, 5) (b) 5C* (0, 2, 4, 5) (ii) 5C (0, 2, 4, 5) 10 (c) 5D (0, 2, 3, 4, 5) 20 Q.9 (a) 5B (0, 3, 5) (b) 5D (0, 2, 3, 4, 5) Q.2 (a) (i) 5B (0, 3, 5) (c) 10D (0, 4, 6, 8, 10) (ii) 5B (0, 3, 5) 20 (b) (i) 5C (0, 2, 4, 5) (ii) 5C (0, 2, 4, 5) (iii) 5B (0, 3, 5) Q.10 (a) 10C (0, 4, 7, 10) (iv) 5C (0, 2, 4, 5) (b) (i) 5B (0, 3, 5) 30 (ii) 5B (0, 3, 5) (iii) 10D* (0, 4, 6, 8, 10) 30Q.3 (a) 5C (0, 2, 4, 5) (b) (i) 5C (0, 2, 4, 5) (ii) 5C (0, 2, 4, 5) Q.11 (a) (i) 5C (0, 2, 4, 5) (iii) 5C (0, 2, 4, 5) (ii) 5C (0, 2, 4, 5) (iv) 5C (0, 2, 4, 5) (b) (i) 5C (0, 2, 4, 5) (v) 5C (0, 2, 4, 5) (ii) 5C (0, 2, 4, 5) 30 20 Q.4 (a) 5C* (0, 2, 4, 5) Q.12 (a) Values10D (0, 4, 6, 8, 10) (b) 5D* (0, 2, 3, 4, 5) Graph 5D (0, 2, 3, 4, 5 ) 10 (b) 5B* (0, 3, 5) (c) 5C* (0, 2, 4, 5) (d) 5B* (0, 3, 5) Q.5 (a) 5C (0, 2, 4, 5) 30 (b) 10C* (0, 4, 7, 10) 15 Q.13 (a) 10D (0, 4, 6, 8, 10) (b) (i) 5C (0, 2, 4, 5) Q.6 (a) 5C (0, 2, 4, 5) (ii) 10D* (0, 4, 6, 8, 10) (b) 5C (0, 2, 4, 5) 25 (c) 5C (0, 2, 4, 5) (d) 5C (0, 2, 4, 5) 20 Q.14 (a) 10D (0, 4, 6, 8, 10) (b) 5C (0, 2, 4, 5) (c) 5C (0, 2, 4, 5) Q.7 (a) 5B (0, 3, 5) 20 (b) 5C (0, 2, 4, 5) (c) 5C (0, 2, 4, 5) (d) 5C (0, 2, 4, 5) 20

Assumptions about these marking schemes on the basis of past SEC marking schemes should be avoided. While the underlying assessment principles remain the same, the exact details of the marking of a particular type of question may vary from a similar question asked by the SEC in previous years in accordance with the contribution of that question to the overall examination in the current year. In setting these marking schemes, we have strived to determine how best to ensure the fair and accurate assessment of students’ work and to ensure consistency in the standard of assessment from year to year. Therefore, aspects of the structure, detail and application of the marking schemes for these examinations are subject to change from past SEC marking schemes and from one year to the next without notice.

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Mathematics


General Instructions

1. There are 14 questions on this examination paper. Answer all questions.

2. Questions do not necessarily carry equal marks.

3. Marks will be lost if all necessary work is not clearly shown.

4. Answers should include the appropriate units of measurement, where relevant.

5. Answers should be given in simplest form, where relevant.

Q.1 (Suggested maximum time: 10 minutes) (20)

1(a) During a discussion in Maths class, Adam said: “Prime numbers are a subset of Natural numbers (ℕ).”

Is Adam correct? Explain your answer. (5C)

Answer – yes

Explanation Any 1: – prime numbers are natural numbers greater than 1

that have no positive divisors other than 1 and itself, therefore prime numbers are a subset of natural numbers //

– natural numbers are positive whole numbers; prime numbers are positive whole numbers greater than 1 that have no positive divisors other than 1 and itself, therefore prime numbers are a subset of natural numbers // etc.

** Accept other appropriate answers.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. lists example(s) of prime numbers. – Correct answer but no explanation given.

High partial credit: (4 marks) – Correct answer, but incomplete or unsatisfactory explanation given.

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Q.1 (cont’d.)

1(b) Coins in the euro currency are issued in eight different denominations.

(i) Jennifer has one coin of each denomination in her pocket. How much money does she have in total? (5C)

Total = 1c + 2c + 5c + 10c + 20c + 50c + €1 + €2 = 1 + 2 + 5 + 10 + 20 + 50 + 100 + 200 = 388c or €3·88

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. identifies at least two coin denominations of euro currency.

High partial credit: (4 marks) – Identifies all coin denominations correctly, but fails to add or adds incorrectly.

* No deduction applied for omission of or incorrect use of units involving currency.

(ii) Jennifer would like the total value, in cent, of the coins she has in her pocket to be a prime number. What is the least amount of money she would have to add to her total to make it a prime number? (5C)

Smallest prime number > 388 = 389

Least amount of money = 389 – 388 = 1c or €0·01

** Accept students’ answers from part (i) if not oversimplified.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. identifies any prime number greater than 388 or greater than answer from part (i).

High partial credit: (4 marks) – Identifies correct prime number, but fails to finish or finishes incorrectly.


€

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Q.1 (cont’d.)

1(c) Express 232 and 368 as a product of prime numbers. Hence, or otherwise, find the highest common factor of 232 and 368. (5D)

232

232 ÷ 2 = 116 116 ÷ 2 = 58 58 ÷ 2 = 29 29 ÷ 29 = 1

232 = 2 × 2 × 2 × 29

368

368 ÷ 2 = 184 184 ÷ 2 = 92 92 ÷ 2 = 46 46 ÷ 2 = 23 23 ÷ 23 = 1

368 = 2 × 2 × 2 × 2 × 23

HCF of 232 and 368 = 2 × 2 × 2 = 8

Scale 5D (0, 2, 3, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes 232 and/or 368 as a product. – Divides correctly 232 and/or 368 by any prime number.

Middle partial credit: (3 marks) – Finds correctly 232 and/or 368 as a product of prime numbers.

High partial credit: (4 marks) – Finds correctly 232 and 368 as a product of prime numbers, but fails to find or finds incorrectly the HCF of both numbers.

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2(a) (i) On the Venn diagram below, shade in the region that represents A \ (B ∪ C). Use set notation to give an alternative description of your shaded region. (5B)

Shaded region

A

B

C

Alternative description

Answer – (A \ B) ∩ (A \ C)

** Accept set notation consistent with shaded region if not oversimplified.

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Correct region on Venn diagram shaded, but fails to name or incorrectly names alternative set notation. – Incorrect region on Venn diagram shaded, but correctly names alternative set notation for shaded region (if not oversimplified).

(ii) On the Venn diagram below, shade in the region that represents (A \ C) ∩ (B \ C). Use set notation to give an alternative description of your shaded region. (5B)

Shaded region

A

B

C

Alternative description

Answer – (A ∩ B) \ C

** Accept set notation consistent with shaded region if not oversimplified.

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Correct region on Venn diagram shaded, but fails to name or incorrectly names alternative set notation. – Incorrect region on Venn diagram shaded, but correctly names alternative set notation for shaded region (if not oversimplified).

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Q.2 (cont’d.)

2(b) A group of 100 students were asked which of the following social networking websites they used in the previous week: Facebook (F), Instagram (I) or Snapchat (S). These are the results:

60 had used Facebook. 45 had used Instagram. 31 had used Snapchat. 12 had used all three websites. 5 had used Facebook and Instagram, but not Snapchat. 8 had used Facebook and Snapchat, but not Instagram. 16 had used exactly two of the websites listed.

(i) Represent the above information on the Venn diagram. (5C)

U = [ 100 ] I = [45 ]

F = [60 ]

S = [31 ]

[ 5 ]

[60 5 12 8 ]= [ 35 ]� � �

[ 45 5 12 3 ]= [ 25 ]� � �

[ 31 8 12 3 ]= [ 8 ]

� � �

[ 16 5 8 ]= [ 3 ]� �[ 12 ]

[ 8 ]

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Three elements correctly identified.

High partial credit: (4 marks) – Between four and six elements correctly identified.

(ii) How many students did not use any of the social networking websites listed above? (5C)

# students = 100 − (35 + 5 + 12 + 8 + 25 + 3 + 8) = 100 − 96 = 4


Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. sum of three or more elements identified if not oversimplified. – Identifies universal set equals 100.

High partial credit: (4 marks) – Finds sum of elements equals 96 or equivalent, but fails to finish or finishes incorrectly.

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Q.2 (cont’d.)

2(b) (cont’d.)

(iii) Find the probability that a student chosen at random from the group used Snapchat only. (5B)

# students who used Snapchat only = 8

P(student used Snapchat only)

= 100

8

= 25

2 or 0·08


Scale 5B (0, 3, 5) Partial credit: (3 marks) – Some work of merit,

e.g. studentsTotal#

onlySnapchat used whoStudents#.

– Identifies # students who used Snapchat only equal to 8 or equivalent. – Finds any fraction with denominator equal to 100 or 25.

(iv) Find the probability that a student chosen at random from those who used the social networking websites listed above used at least two of the websites. (5C)

# students = # students who used 2 sites + # students who used 3 sites = (5 + 8 + 3) + (12) = 16 + 12 = 28

# students who used the social networking websites = 100 – 4 = 96

or

= 35 + 5 + 12 + 8 + 25 + 3 + 8 = 96

P(student used at least two websites)

= 96

28

= 24

7 or 0·291666...

** Accept students’ answers from parts (i) and (ii) if not oversimplified.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. identifies all correct elements and stops or continues. – Finds ‘# students who used at least two websites’ equal to 28 or equivalent. – Finds ‘# students who used websites’ equal to 96 or equivalent.

High partial credit: (4 marks) – Finds both 28 and 96 or equivalent if not oversimplified, but fails to finish or finishes incorrectly.

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Last summer, Roisin and Tom attended a Young Entrepreneurs camp for transition year students.

3(a) The ratio of girls to boys who attended the camp was 5:3. If there were 60 more girls than boys, find the total number of students who attended the camp. (5C)

Ratio of girls : boys = 5 : 3

35

35

+−

of students = difference between the number of girls and boys

8

2 of students = 60

8

1 of students =

2

60

= 30 Total students = 8 × 30 = 240

or

Number of girls = x Number of boys = x – 60 Ratio of girls : boys = x : x – 60 = 5 : 3

5

x =

3

60−x

3x = 5(x – 60) = 5x – 300 3x – 5x = –300 –2x = –300 2x = 300 x = 150

Total students = x + (x – 60) = 150 + (150 – 60) = 150 + 90 = 240

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. 3 + 5 = 8.

– Writes down number of girls = 8

5(total)

and/or number of boys = 8

3(total).

– Finds ratio x : x – 60 :: 5 : 3.

High partial credit: (4 marks) – Finds difference = 8

35 −(total) = 60

or equivalent, but fails to finish or finishes incorrectly.

– Finds 5

x =

3

60−x, but fails to finish

or finishes incorrectly. – Finds 30 (with work shown), but fails to multiply by 8. – Finds 150 (with work shown), but fails to finish or finishes incorrectly.

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Q.3 (cont’d.) 3(b) As part of the camp, Roisin and Tom set up a small business

enterprise selling cupcakes. They made 300 cupcakes and sold them at a local farmers’ market.

(i) Each cupcake cost 35 cent to make and they planned to sell them for €1·40 each.

Find the planned mark-up (profit as a percentage of cost price) on each cupcake. (5C)

Profit = 1·40 – 0·35 = €1·05

Planned mark-up = 350

051

⋅⋅

× 1

100

= 300%

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. finds correct profit on each cupcake (in cent or euro). – Finds margin (profit as a percentage

of selling price), i.e. 401

051

⋅⋅

× 1

100 = 75%.

High partial credit: (4 marks) – Correct expression for mark-up, but fails to finish or finishes incorrectly. – Incorrect profit, from substantially correct work, but finishes correctly.

* No deduction applied for the omission of ‘%’ symbol in final answer. (ii) A stall at the farmers’ market cost €40 to rent and the packaging for all the cupcakes

cost €15. Find the total cost of making and selling all the cupcakes. (5C)

Total cost = (0·35 × 300) + 40 + 15 = 105 + 40 + 15 = €160

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. finds 0·35 × 300 = 105 and stops.

High partial credit: (4 marks) – Finds three elements correctly, i.e. 105, 40 and 15, but fails to finish or finishes incorrectly.

* No deduction applied for omission of or incorrect use of units involving currency. (iii) On the day of the farmers’ market, Roisin and Tom sold 240 cupcakes at the planned price. Towards the end of the day, they sold the remaining cupcakes at a reduced price to ensure

that everything was sold. Given that the overall profit on the day’s enterprise was €226, find the total sale price of the remaining cupcakes. (5C)

Sales from 240 cupcakes = 1·40 × 240 = €336

Interim profit = Sales from 240 cupcakes – Total cost = 336 – 160 = €176

Overall profit = €226 Sales from (300 – 240) cupcakes = 226 – 176 Sales from 60 cupcakes = €50

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Q.3 (cont’d.)

3(b) (iii) (cont’d.)

** Accept students’ answers from part (ii) if not oversimplified.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. finds sales from 240 cupcakes, 1·40 × 240 = €336 and stops.– Subtracts ‘160’ or equivalent from incorrect sales.

High partial credit: (4 marks) – Finds correct interim profit, i.e. €176, but fails to finish or finishes incorrectly.


(iv) The remaining cupcakes were packed together in equal amounts and sold in bundles. In how many different ways could the cupcakes be packed together? (5C) Different ways – 1 × 60 // – 60 × 1 // – 2 × 30 // – 30 × 2 // – 3 × 20 // – 20 × 3 // – 4 × 15 // – 15 × 4 // – 5 × 12 // – 12 × 5 // – 6 × 10 // – 10 × 6

# ways = 12

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. lists 3 or more ways in which cupcakes could be packed.

High partial credit: (4 marks) – List all 12 ways in which cupcakes could be packed, but fails to state ‘# ways’.

(v) Roisin and Tom sold the bundles of cupcakes for €5 each. Find the number of cupcakes in each bundle. (5C)

Sales from 60 cupcakes = €50 Price for each bundle = €5

# bundles = 5

50

= 10 # cupcakes per bundle = 6

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. correct use in fraction form of either ‘50’ or ‘5’.

High partial credit: (4 marks) – Finds correct ‘# bundles’, i.e. 10, but fails to find or finds incorrect ‘# cupcakes per bundle’.

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** Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded or for the omission of or incorrect use of units in part(s) of question asterisked - this deduction should be applied only once per section (a), (b), (c), etc. of the question.

On 14 July 2015, the New Horizons space probe became the first spacecraft to perform a flyby study of Pluto. During its mission, the space probe took detailed measurements and observations of the planet and its moons.

It is estimated that the volume of Pluto is 6 970 000 000 km3.

4(a) The volume of the Earth is 1·08321 × 1012 km3. How many times greater than the volume of Pluto is the volume of the Earth? Give your answer correct to the nearest whole number. (5C*)

Volume of Pluto = 6,970,000,000 = 6·97 × 109 km3

# times = 9

12

10976

10083211

×⋅×⋅

= 91210976

083211 −×⋅

⋅

= 0·155410... × 103 = 155·410329... ≅ 155

or

Volume of the Earth = 1·08321 × 1012 = 1,083,210,000,000 km3

# times = 000,000,970,6

000,000,210,083,1

= 155·410329... ≅ 155

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. indicates correct

division, 000,000,970,6

10083211 12×⋅ or similar.

– Equates 6,970,000,000 to 6·97 × 109 or 1·08321 × 1012 to 1,083,210,000,000.

High partial credit: (4 marks) – Correct division with correct numerator

and denominator, i.e. 9

12

10976

10083211

×⋅×⋅

or

000,000,970,6

000,000,210,083,1 , but fails to finish or

finishes incorrectly.

* Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded - apply only once per section (a), (b), (c), etc. of question.

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Q.4 (cont’d.)

4(b) Find the difference between the diameters of the two planets, in kilometres. Give your answer in the form a × 10n, where 1 ≤ a < 10 and n ∈ ℤ, correct to three significant figures. (5D*)

Volume of Pluto = 3

4 πr3

= 6,970,000,000

3

4 πr3 = 6,970,000,000

r3 = 4

3000,000,970,6

×π×

= 1,663,964,930·025765... r = 3 ...025765930,964,663,1 ⋅

= 1,184·990102... ≅ 1,185 km

Volume of the Earth = 3

4 πr3

= 1·08321 × 1012

3

4 πr3 = 1·08321 × 1012

r3 = 4

310083211 12

×π××⋅

= 0·258597... × 1012

r = 3 1210...2585970 ×⋅

= 0·637100... × 104 = 6,371·006044... ≅ 6,371 km

Difference in diameters of the two planets = 2 × (6,371 – 1,185) = 12,742 – 2,370 = 10,372 km = 1·0372 × 104 ≅ 1·04 × 104 km

Scale 5D* (0, 2, 3, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. equates volume of either planet correctly, i.e. Pluto as

3

4 πr3 = 6,970,000,000 or Earth as 3

4 πr3

= 1·08321 × 1012.

Middle partial credit: (3 marks) – Finds correct radius or diameter of either Pluto or Earth.

High partial credit: (4 marks) – Finds correct radius or diameter of Pluto and Earth, but fails to find difference in diameters of the two planets or finds incorrect difference.


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** No deduction should be applied for the omission of or incorrect use of units involving currency.

Jim invests a sum of money for one year in a deposit account which offers an annual compound interest rate of 1·5%.

5(a) At the end of the year, the investment is worth €5,582·50. Find the sum of money that Jim invested. (5C)

F = P(1 + i)t P(1 + 0·015)1 = 5,582·50 P(1·015) = 5,582·50

P = 0151

50582,5

⋅⋅

= €5,500

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. identifies correct relevant formula from Tables.

High partial credit: (4 marks) – Substitutes correctly into relevant formula, i.e. P(1 + 0·015)1 = 5,582·50, but fails to finish or finishes incorrectly.


At the end of the year, Jim is offered a higher fixed rate of interest if he retains the money in the account for a further two years. Assuming he does not withdraw any money, Jim’s investment will be worth €5865·11 at the end of this period.

5(b) Calculate the annual rate of interest that Jim is offered on his investment. Give your answer as a percentage, correct to one decimal place. (10C*)

F = P(1 + i)t 5,865·11 = 5,582·50(1 + i)2

(1 + i)2 = 50582,5

11865,5

⋅⋅

= 1⋅050624... 1 + i = ...0506241⋅ = 1·024999... i = 1·024999... – 1 = 0·024999... = 2·499964... ≅ 2·5%

Scale 10C* (0, 4, 7, 10) Low partial credit: (4 marks) – Some work of merit. – Substitutes correctly into relevant formula, i.e. 5,582·50(1 + i)2 = 5,865·11 and stops or continues.

High partial credit: (7 marks) – Finds (1 + i)2 = 1⋅050624... or 1 + i = ...0506241⋅ / 1·024999... , but fails to finish or finishes incorrectly.


* No deduction applied for the omission of ‘%’ symbol in final answer.

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6(a) Express 32 as a power of 2. (5C)

32 = 2 × 16

= 21

2 × 4

= 21

2 × 22

= 212

2 or 25

2

or

32 = 21

)32(

= 21

5 )2(

= 25

2 or 212

2

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. 32 = 2 × 16

or 32 = 21

)32( and stops or continues.

High partial credit: (4 marks) – Finds 32 = 21

2 × 22 or 32 = 21

5 )2( ,

but fails to finish or finishes incorrectly.

6(b) Hence, solve the equation x2

25

= 32 . (5C)

x2

25

= 32

25 – x = 32

= 25

2

5 – x = 2

5

(5 – x)(2) = 5 10 – 2x = 5 –2x = 5 – 10 = –5 2x = 5

x = 2

5

** Accept students’ answers from part (a) if not oversimplified.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g.

x2

25

= 25 – x.

– Finds 25 – x = 32 or 25

2 or student’s own answer from (a) and stops or continues.

High partial credit: (4 marks) – Equates powers, i.e. 5 – x = 2

5 or power

of student’s own answer from part (a), but fails to finish or finishes incorrectly.

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Q.6 (cont’d.) 6(c) Verify your answer to part (b). (5C)

x2

25

= 32

Let x = 2

5

25

5

2

2 = 2

5

2

255

2−

= 25

2

25

2 = 25

2

x2

25

= 32 when x = 2

5

** Accept students’ answers from parts (a) and (b) if not oversimplified.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down

25 – x = 32 or x2

25

= 25

2 (one side

correctly evaluated). – Substitutes student’s answer from part (b)

correctly into equation, i.e. 25

5

2

2 = 2

5

2

and stops or continues.

High partial credit: (4 marks) – Correct substitution into equation and finishes, but values do not equate and no conclusion stated.

6(d) Use the properties of surds to show that 8

50128 − simplifies to a constant. (5C)

8

50128 − =

22

225264 −

= 2

2564 −

= 2

58 −

= 2

3

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down one or more of 128 = 64 2 or 8 2 ,

or 50 = 25 2 or 5 2

or 8 = 4 2 or 2 2 . – Uses decimal versions to find constant.

High partial credit: (4 marks) – Finds

2

2564 − or

2

58 −, but fails

to finish or finishes incorrectly.

2016.1 J.18/20_MS 18/72 of 71 examsDEB


7(a) Factorise fully 2x2 – 6x. (5B)

2x2 – 6x = 2x(x – 3)

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Some work of merit, e.g. any common factor identified.

7(b) Factorise fully 3a2c – 9a + 6c – 2ac2. (5C)

3a2c – 9a + 6c – 2ac2 = 3a2c – 9a – 2ac2 + 6c = 3a(ac – 3) – 2c(ac – 3) = (ac – 3)(3a – 2c)

or

3a2c – 9a + 6c – 2ac2 = 3a2c – 2ac2 – 9a + 6c = ac(3a – 2c) – 3(3a – 2c) = (3a – 2c)(ac – 3)

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. one common factor or group indicated.

High partial credit: (4 marks) – Finds one correct factorisation - both inside and outside bracket, e.g. 3a(ac – 3).– Finds two correct factorisations, but with sign errors.

7(c) Use factors to simplify the following: 94

12522

2

−−+

x

xx. (5C)

94

12522

2

−−+

x

xx =

)32)(32(

)4)(32(

−++−

xx

xx

= 32

4

++

x

x

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. correct factors of –12 and/or –9 identified.

High partial credit: (4 marks) – Correct factorisation of numerator or denominator. – Correct factorisation of both numerator and denominator, but with sign errors.

2016.1 J.18/20_MS 19/72 of 71 examsDEB

Q.7 (cont’d.)

7(d) Multiply out and simplify (x2 – 3x + 5)(x – 7). (5C)

(x2 – 3x + 5)(x – 7) = x2(x – 7) – 3x(x – 7) + 5(x – 7) = x3 – 7x2 – 3x2 + 21x + 5x – 35 = x3 – 10x2 + 26x – 35

or

(x2 – 3x + 5)(x – 7) = x(x2 – 3x + 5) – 7(x2 – 3x + 5) = x3 – 3x2 + 5x – 7x2 + 21x – 35 = x3 – 10x2 + 26x – 35

or

x2 –3x2 +5x

x x3 –3x2 5x

–7 –7x2 21x2 –35x

(x2 – 3x + 5)(x – 7) = x3 – 7x2 – 3x2 + 21x + 5x – 35 = x3 – 10x2 + 26x – 35

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. any correct multiplication, i.e. one correct term (sign, number and x coefficient). – Sets up multiplication correctly, i.e. x2(x – 7) – 3x(x – 7) + 5(x – 7) or x(x2 – 3x + 5) – 7(x2 – 3x + 5). – Sets up grid correctly.

High partial credit: (4 marks) – Finds four correct (sign, number and x coefficient).

2016.1 J.18/20_MS 20/72 of 71 examsDEB

Q.8 (Suggested maximum time: 5 minutes) (10) ** Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded or for the

omission of or incorrect use of units in part(s) of question asterisked - this deduction should be applied only once per section (a), (b), (c), etc. of the question.

** No deduction should be applied for the omission of or incorrect use of units involving currency.

8(a) EuroMillions is a lottery game operated in several European countries. The winners of the largest lottery jackpot to date are Adrian and Gillian Bayford from Great Britain, who won €190 million on 10th August, 2012.

Given that the exchange rate was €1 = £0·7824 sterling on that day, calculate the value of the jackpot in sterling. (5B)

Jackpot = €190 million = 190,000,000 × 0·7824 = £148,656,000 stg.

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Some work of merit, e.g. writes formula to calculate value using exchange rate. – Incorrect use of exchange rate, i.e.

78240

0190,000,00

⋅ = 242,842,535·787...

– Finds 190,000,000 × 0·7824, but fails to finish or finishes incorrectly.


8(b) This was not the largest sterling amount won, as Chris and Colin Weir, also from Great Britain, scooped a €185 million jackpot on 12th July, 2011 but received £12·997 million more than the Bayfords due to a more favourable exchange rate.

Calculate the exchange rate on that day in the form €1 = £·. (5C*)

Exchange rate = €)(inValue

£)(inValue

= 000,000,185

000,997,120148,656,00 +

= 000,000,185

0161,653,00

= 0·8738 ≅ 0·87 €1 ≡ £0·87


Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. equates €185,000,000 ≡ £161,653,000 stg. – Finds total value of jackpot in sterling, i.e. 148,656,000 + 12,997,000 = 161,653,000 or answer from part (a) + 12,997,000.

High partial credit: (4 marks) – Finds inverse of exchange rate correctly,

i.e. 0161,653,00

000,000,185 = 1·144426... ≅ 1·14.

– Indicates division with correct numerator

and denominator, i.e. 000,000,185

0161,653,00, but

fails to finish or finishes incorrectly.



2016.1 J.18/20_MS 21/72 of 71 examsDEB


The first three stages of a pattern are shown below. Each stage of the pattern is made up of bowling pins.

9(a) Draw the next two stages of the pattern. (5B)

Scale 5B (0, 3, 5) Partial credit: (3 marks) – One stage correctly drawn.

9(b) Show that the numbers of bowling pins in each stage form a quadratic sequence. (5D)

Stage # of pins 1st Diff. 2nd Diff.

1 1

2

2 3 1 3

3 6 1 4

4 10 1 5

5 15

∴ as the first differences are not all the same, but the second differences are constant numbers of bowling pins in each pattern form a quadratic sequence

Scale 5D (0, 2, 3, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. finds one first difference.

Middle partial credit: (3 marks) – Finds one second difference. – Finds at least three first differences and stops or continues.

High partial credit: (4 marks) – Finds at least two second differences correctly, but fails to explain or explains incorrectly significance of second difference.

9(c) Find a general formula for the number of bowling pins in stage n of the pattern, where n ∈ ℕ. (10D)

Tn = an2 + bn + c

a = 2

difference2nd

= 2

1 // 0·5

2016.1 J.18/20_MS 22/72 of 71 examsDEB

Q.9 (cont’d.) 9(c) (cont’d.)

T1 = 2

1(1)2 + b(1) + c

= 1

2

1(1)2 + b(1) + c = 1

2

1 + b + c = 1

b + c = 1 – 2

1

= 0·5

T2 = 2

1(2)2 + b(2) + c

= 3

2

1(2)2 + b(2) + c = 3

2 + 2b + c = 3 2b + c = 3 – 2 = 1

T1: b + c = 0·5 T2: 2b + c = 1

b + c = 0·5 (×–1) 2b + c = 1 (×1)

–b – c = –0·5 2b + c = 1 . b = 0·5

b + c = 0·5 0·5 + c = 0·5 c = 0·5 – 0·5 = 0

or 2b + c = 1 2(0·5) + c = 1 1 + c = 1 c = 0

or

b + c = 0·5 (×–2) 2b + c = 1 (×1)

–2b – 2c = –1 2b + c = 1 . –c = 0 c = 0

b + c = 0·5 b + 0 = 0·5 b = 0·5

or 2b + c = 1 2b + 0 = 1 b = 0·5

Tn = an2 + bn + c = 0·5n2 + 0·5n

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down general formula for nth term, identifies coefficient of n2.

Middle partial credit: (6 marks) – Finds one or both equations for T1 and T2, i.e. b + c = 0·5 and/or 2b + c = 1. – Finds either variable (b or c) correctly by trial and error, but fails to verify in both equations or verifies incorrectly.

High partial credit: (8 marks) – Finds first variable (b or c), but fails to find second variable or finds incorrectly.– Finds both variables (b and c) correctly with no work shown or by graphical means.– Finds both variables (b and c) by trial and error, but does not verify in both equations or verifies incorrectly..

2016.1 J.18/20_MS 23/72 of 71 examsDEB

Q.10 (Suggested maximum time: 15 minutes) (30) ** Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded or for the

omission of or incorrect use of units in part(s) of question asterisked - this deduction should be applied only once per section (a), (b), (c), etc. of the question.

10(a) Write the following as a single fraction in its simplest form. (10C)

12

3

−x –

23

4

+x

12

3

−x –

23

4

+x =

)23)(12(

)12(4)23(3

+−−−+

xx

xx

= )23)(12(

4869

+−+−+

xx

xx

= )23)(12(

10

+−+

xx

x or 26

102 −+

+xx

x

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – Some work of merit, e.g. identifies (2x – 1)(3x + 2) as the denominator with some (incorrect) numerator. – Correct numerator, i.e. multiplies 3(3x + 2) – 4(2x – 1), but no denominator.

High partial credit: (7 marks) – Finds )23)(12(

)12(4)23(3

+−−−+

xx

xx or similar,


10(b) x is a real number. Number A is equal to 1 less than x. Number B is equal to 2 greater than seven times x.

(i) Write down the numbers A and B, in terms of x. (5B)

A = x – 1

B = 7x + 2

Scale 5B (0, 3, 5) Partial credit: (3 marks) – One correct answer.

(ii) The product of these two numbers is 2. Use this information to write an equation in x. (5B)

A × B = 2 (x – 1)(7x + 2) = 2

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Some work of merit, e.g. something = 2 or exhibits understanding that product means multiplication.

2016.1 J.18/20_MS 24/72 of 71 examsDEB

Q.10 (cont’d.)

10(b) (cont’d.)

(iii) Solve this equation to find the two possible values of x. Give each of your answers correct to two decimal places. (10D*)

(x – 1)(7x + 2) = 2 7x2 + 2x – 7x – 2 = 2 7x2 – 5x – 2 – 2 = 0 7x2 – 5x – 4 = 0

x = a

acbb

2

42 −±−

x = )7(2

)4)(7(4)5()5( 2 −−−±−−

= 14

112255 +±

= 14

1375 ±

= 14

...704699115 ⋅±

x = 14

...704699115 ⋅+, x =

14

...704699115 ⋅−

= 14

...70469916 ⋅ =

14

...7046996⋅−

= 1·193192... = –0·478907... ≅ 1·19 ≅ –0·48

Scale 10D* (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down ‘–b’ formula and identifies a, b and c. – Finds 7x2 – 5x – 4 = 0 correctly, but not does not proceed to use ‘–b’ formula.

Middle partial credit: (6 marks) – Full correct substitution into ‘–b’ formula,

i.e. x = )7(2

)4)(7(4)5()5( 2 −−−±−−.

– Two of the following steps correct: correct formula and correctly identifies a, b and c, fully correct substitution, works out solution to surd form, finishes correctly to two decimal places.

High partial credit: (8 marks) – Three of the above steps correct. – Finds only one value of x (1⋅19 or –0⋅48), but fails to find or finds incorrectly other value of x.


2016.1 J.18/20_MS 25/72 of 71 examsDEB


11(a) Let g be the function g : x 2 →׀x2 – 7, where x ∈ ℝ.

(i) Find the value of g(–2). (5C)

g(x) = 2x2 – 7 g(–2) = 2(–2)2 – 7 = 2(4) – 7 = 8 – 7 = 1

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. some correct substitution into given function, i.e. 2(–2)2, 2(4) or 8. – Correctly solves g(x) = –2.

High partial credit: (4 marks) – Full correct substitution into formula, but fails to finish or finishes incorrectly.

(ii) Find the values of k for which g(k) = 25. (5C)

g(x) = 2x2 – 7 g(k) = 2(k)2 – 7 = 25 2(k)2 – 7 = 25 2k2 = 25 + 7 = 32

k2 = 2

32

= 16 k = ± 16

= ±4

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. correctly equates g(k) = 25 to find correct equation, i.e. 2(k)2 – 7 = 25 or similar.

High partial credit: (4 marks) – Finds k2 = 16, but fails to find values for k or only finds one value of k.

2016.1 J.18/20_MS 26/72 of 71 examsDEB

Q.11 (cont’d.)

11(b) Let F = 12

)4(4

+−

C

C.

(i) Calculate the value of F when C = 2

1. (5C)

F = 1

2

12

2

144

+

−

= 11

216

+−

= 2

14

= 7

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. simplifies given

equation, i.e. F = 12

416

+−

C

C .

– Some correct substitution into equation.

High partial credit: (4 marks) – Full correct substitution into formula, but fails to finish or finishes incorrectly.

(ii) Write C in terms of F. (5C)

F = 12

)4(4

+−

C

C

F(2C + 1) = 4(4 – C) 2FC + F = 16 – 4C 2FC + 4C = 16 – F C(2F + 4) = 16 – F

C = 42

16

+−

F

F or )2(2

16

+−

F

F or F

F

24

16

+− or

)2(2

16

F

F

+−

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. one correct manipulation.

High partial credit: (4 marks) – Both sides fully expanded with or without C being isolated, i.e. 2FC + 4C = 16 – F or C(2F + 4) = 16 – F, but fails to finish or finishes incorrectly.

2016.1 J.18/20_MS 27/72 of 71 examsDEB



An archer, standing on level ground, shoots an arrow into the air. The height, in metres, of the arrow above ground level, t seconds after the arrow is released, is given by h : t 6 + 2 →׀t – t2.

12(a) On the grid below, draw the graph of y = h(t) in the domain 0 ≤ t ≤ 6, where t ∈ ℝ.

Values (10D)

Table (1st method)

t = 0 1 2 3 4 5 6 2 2 2 2 2 2 2 2 6t 0 6 12 18 24 30 36 –t2 0 –1 –4 –9 –16 –25 –36 y = h(t) 2 7 10 11 10 7 2

or Table (2nd method)

h(t) = 2 6t –t2 = y h(0) = 2 0 0 = 2 h(1) = 2 6 –1 = 7 h(2) = 2 12 –4 = 10 h(3) = 2 18 –9 = 11 h(4) = 2 24 –16 = 10 h(5) = 2 30 –25 = 7 h(6) = 2 36 –36 = 2

or Substitution method

h(t) = 2 + 6t – t2 h(0) = 2 + 6(0) – (0)2 = 2

h(1) = 2 + 6(1) – (1)2 = 7

h(2) = 2 + 6(2) – (2)2 = 10

h(3) = 2 + 6(3) – (3)2 = 11

h(4) = 2 + 6(4) – (4)2 = 10

h(5) = 2 + 6(5) – (5)2 = 7

h(6) = 2 + 6(6) – (6)2 = 2

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – One or two points, (x, y) or (y, x), correctly identified.

Middle partial credit: (6 marks) – Three or four points, (x, y) or (y, x), correctly identified.

High partial credit: (8 marks) – Five or six points, (x, y) or (y, x), correctly identified.

2016.1 J.18/20_MS 28/72 of 71 examsDEB

Q.12 (cont’d.) 12(a) (cont’d.)

Graph (5D)

Points (0, 2), (1, 7), (2, 10), (3, 11), (4, 10), (5, 7), (6, 2)

** Accept values calculated from previous work (seven co-ordinates needed). ** If no points are worked out, but correctly graphed, award the marks for the graph in both parts.

Scale 5D (0, 2, 3, 4, 5) Low partial credit: (2 marks) – One or two points, (x, y) or (y, x), correctly plotted, labelled or not. – Points incorrectly calculated, but correctly plotted to form a line.

Middle partial credit: (3 marks) – Three or four points, (x, y) or (y, x), correctly plotted, labelled or not (joined or not).

High partial credit: (4 marks) – Five or six points, (x, y) or (y, x), correctly plotted, labelled or not (joined or not). – All points, (x, y) or (y, x), correctly plotted, but joined together with inappropriate curve.

11

0 76· 5 24·

6 74·

h t( )

Time in seconds, t

1 2 3 4 5 6

Hei

gh

to

fA

rro

win

met

res,

h

2

�2

4

�4

6

8

12

10

�3

2016.1 J.18/20_MS 29/72 of 71 examsDEB

Q.12 (cont’d.) For parts (b), (c) and (d), you must show your working out on the diagram on the previous page.

12(b) Use your graph to estimate the height at which the speed of the arrow is zero after it is shot. (5B*)

Speed of arrow is zero @ maximum height

From graph Maximum height = 11 m

** Accept students’ answers based on fully plotted graph.

Scale 5B* (0, 3, 5) Partial credit: (3 marks) – Some work of merit, e.g. draws vertical line from 3 s and intersects graph with or without horizontal line from point of intersection to the y-axis. – Answer outside tolerance (1 grid box), but inside tolerance of ±2 grid boxes. – Answer = 3 seconds.

* Deduct 1 mark off correct answer only for omission of or incorrect use of units - apply only once per section (a), (b), (c), etc. of question.

12(c) Use your graph to estimate the length of time that the arrow is more than 6 m above ground level. (5C*)

From graph Time (> 6m) = 5·24 – 0·76 = 4·48 seconds


Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. draws horizontal line from 6 m and intersects graph with or without vertical line(s) from points of intersection to the x-axis. – Answer outside tolerance (1 grid box), but inside tolerance of ±2 grid boxes. – One correct x-intercept.

High partial credit: (4 marks) – Two correct x-intercepts, but fails to find or finds incorrect time interval.


12(d) Imagine the archer shoots the same arrow standing on an elevated position 3 m above ground level. By extending your graph, estimate the time that it would take the arrow to hit the ground. (5B*)

From graph Time = 6·74 seconds


Scale 5B* (0, 3, 5) Partial credit: (3 marks) – Some work of merit, e.g. extends graph to y = –3 or draws horizontal line at y = –3 and intersects graph with or without vertical line from point of intersection to the x-axis. – Answer outside tolerance (1 grid box), but inside tolerance of ±2 grid boxes.


2016.1 J.18/20_MS 30/72 of 71 examsDEB



13(a) Solve the following inequality and graph your solution on the number line.

–6 < 4x + 2 ≤ 14, x ∈ ℝ. (10D)

Solution

–6 < 4x + 2 –4x < 2 + 6 –4x < 8 x > –2

and 4x + 2 ≤ 14 4x ≤ 14 – 2 ≤ 12

x ≤ 4

12

≤ 3

–2 < x ≤ 3

or

–6 < 4x + 2 ≤ 14 ... subtract 2 –6 – 2 < 4x + 2 – 2 ≤ 14 – 2 –8 < 4x ≤ 12

–8 < 4x ≤ 12 ... divide by 4

4

8− <

4

4x ≤

4

12

–2 < x ≤ 3

–2 < x ≤ 3

Number line

0 1 2 3 4 5�1�2�3�4

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. substitutes in value for x.

Middle partial credit: (6 marks) – Finds x > –2 or x ≤ 3 (accept with or without inequality sign).

High partial credit: (8 marks) – Solution to inequality fully correct or number line correctly graphed.

2016.1 J.18/20_MS 31/72 of 71 examsDEB

Q.13 (cont’d.) 13(b) The area of a swimming pool is given by 2x3 + 6x2 – 25x – 25.

(i) The width of the swimming pool is x + 5. Find the length of the pool, in terms of x. (5C)

2x2 − 4x − 5 x + 5 ) 2x3 + 6x2 − 25x − 25 –2x3 – 10x2 −4x2 – 25x – 25 4x2 + 20x −5x – 25 5x + 25 0

or

2x2 −4x −5 x 2x3 –4x2 –5x 5 10x2 –20x –25

Answer – 2x2 − 4x − 5

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. sets up division.– Identifies one term correctly in answer. – Inserts the 2x3 or –25 term correctly in Method .

High partial credit: (4 marks) – Substantial work, but with one or more critical omissions, e.g. x2 identified on top with solution down to −4x2 – 25x – 25 stage correct. – Finds correctly two terms in answer.

(ii) Given that the length of the swimming pool is 25 m, find the possible widths of the pool. Which answer do you think is most appropriate? Give a reason for your answer. (10D*)

Possible widths

2x2 – 4x – 5 = 25 2x2 – 4x – 5 – 25 = 0 2x2 – 4x – 30 = 0 x2 – 2x – 15 = 0 (x – 5)(x + 3) = 0 x – 5 = 0 x = 5 x + 5 = 5 + 5 = 10 m

or x + 3 = 0 x = –3 x + 5 = –3 + 5 = 2 m

Most appropriate answer

Answer – 10 m

Reason – other answer, i.e. width = 2 m, would make the swimming pool very narrow, etc.

** Accept students’ answers from part (ii) if not oversimplified.

Scale 10D* (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. correctly equates 2x2 – 4x – 5 = 25 to find quadratic equation.

Middle partial credit: (6 marks) – Correctly solves quadratic equation, but fails to find roots or only finds one root.

High partial credit: (8 marks) – Finds both roots, but fails to identify most appropriate answer or fails to give reason.


2016.1 J.18/20_MS 32/72 of 71 examsDEB


The diagram below shows part of the graph of the function

f : x ׀→ x2 + bx + c, where x ∈ ℝ and b, c ∈ ℤ.

x

y

�2 3

The graph of f intersects the x-axis at the points where x = –2 and x = 3.

14(a) Find the value of b and the value of c. (10D)

y = x2 + bx + c = x2 – (sum of roots)x + (product of roots) = x2 – (–2 + 3)x + (–2)(3) = x2 – x – 6 b = –1 c = –6

The graph intersects the x-axis at x = –2 and x = 3 y = 0 @ x = –2 and x = 3

@ x = –2 0 = (–2)2 + b(–2) + c 0 = 4 – 2b + c 2b – c = 4

@ x = 3 0 = (3)2 + b(3) + c 0 = 9 + 3b + c 3b + c = –9

2b – c = 4 3b + c = –9 5b = –5 b = –1

2b – c = 4 2(–1) – c = 4 –2 – c = 4 –c = 4 + 2 = 6 c = –6

or 3b + c = –9 3(–1) + c = –9 –3 + c = –9 c = –9 + 3 = –6

2016.1 J.18/20_MS 33/72 of 71 examsDEB

Q.14 (cont’d.)

14(a) (cont’d.)

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. states correctly ‘x2 – (sum of roots)x + (product of roots)’ and stops or continues. – Forms one correct simultaneous equation and stops or continues.

Middle partial credit: (6 marks) – Substitutes into ‘sum and product’ rule correctly, but fails to find or finds incorrect values for b and c. – Forms two correct simultaneous equations and stops or continues.

High partial credit: (8 marks) – Substitutes into ‘sum and product’ rule correctly, but only finds correct value for either b or c. – Solves simultaneous equations, but only finds correct value for either b or c.

14(b) Write down the co-ordinates of the point where the graph intersects the y-axis. (5C)

Graph intersects the y-axis x = 0 y = x2 – x – 6 = (0)2 – (0) – 6 = –6


Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. correctly identifies x = 0.

High partial credit: (4 marks) – Correctly substitutes 0 into equation, but fails to finish or finishes incorrectly. – Correctly identifies x = c, but fails to find or finds incorrect value of c.

14(c) (t, 4t) is a point on the graph, where t ∈ ℤ. Find the two possible values of t. (5C)

y = x2 – x – 6 4t = t2 – t – 6 t2 – t – 6 – 4t = 0 t2 – 5t – 6 = 0 (t – 6)(t + 1) = 0 t – 6 = 0 t = 6

or t + 1 = 0 t = –1


Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. correctly substitutes (t, 4t) into equation to find quadratic equation in terms of t.

High partial credit: (4 marks) – Correctly solves quadratic equation, but fails to find roots or only finds one root.

2016.1 J.18/20_MS 34/72 of 71 examsDEB

Notes:

2016.1 J.18/20_MS 35/72 of 71 examsDEB

Notes:

2016.1 J.18/20_MS 36/72 of 71 examsDEB


Mathematics


Structure of the Marking Scheme

Students’ responses are marked according to different scales, depending on the types of response anticipated. Scales labelled A divide students’ responses into two categories (correct and incorrect). Scales labelled B divide responses into three categories (correct, partially correct, and incorrect), and so on. These scales and the marks that they generate are summarised in the following table:

Scale label A B C D

No. of categories 2 3 4 5

5 mark scale 0, 5 0, 3, 5 0, 2, 4, 5 0, 2, 3, 4, 5 10 mark scale 0, 4, 7, 10 0, 4, 6, 8, 10 15 mark scale

A general descriptor of each point on each scale is given below. More specific directions in relation to interpreting the scales in the context of each question are given in the scheme, where necessary.

Marking scales – level descriptors

A-scales (two categories) incorrect response (no credit) correct response (full credit)

B-scales (three categories) response of no substantial merit (no credit) partially correct response (partial credit) correct response (full credit)

C-scales (four categories) response of no substantial merit (no credit) response with some merit (low partial credit) almost correct response (high partial credit) correct response (full credit)

D-scales (five categories) response of no substantial merit (no credit) response with some merit (low partial credit) response about half-right (middle partial credit) almost correct response (high partial credit) correct response (full credit)

In certain cases, typically involving incorrect rounding, omission of units, a misreading that does not oversimplify the work or an arithmetical error that does not oversimplify the work, a mark that is one mark below the full-credit mark may also be awarded. Such cases are flagged with an asterisk. Thus, for example, scale 10C* indicates that 9 marks may be awarded.

The * for units to be applied only if the student’s answer is fully correct. The * to be applied once only within each section (a), (b), (c), etc. of all questions. The * penalty is not applied to currency solutions.

Unless otherwise specified, accept correct answer with or without work shown.

Accept students’ work in one part of a question for use in subsequent parts of the question, unless this oversimplifies the work involved.

examsDEB

Name/version:

Printed: Whom:

Checked:

To: Ret’d:

Updated: Whom:

Name/version:

Complete (y/n): Whom: 2005 Print Stamp.doc

2016.1 J.18/20_MS 37/72 of 71 examsDEB

Summary of Marks – 2016 JC Maths (Higher Level, Paper 2)

Q.1 (a) 5C* (0, 2, 4, 5) Q.7 (a) 5C (0, 2, 4, 5) (b) 5C* (0, 2, 4, 5) (b) 5B (0, 3, 5) (c) 5C (0, 2, 4, 5) (c) 5C (0, 2, 4, 5) (d) 5D* (0, 2, 3, 4, 5) (d) 5C (0, 2, 4, 5) (e) 5D* (0, 2, 3, 4, 5) (e) 5C (0, 2, 4, 5) 25 (f) 5C (0, 2, 4, 5) (g) 5C (0, 2, 4, 5) (h) 5C (0, 2, 4, 5) Q.2 (a) 5B (0, 3, 5) 40 (b) 5B (0, 3, 5) (c) 10D (0, 4, 6, 8, 10) (d) (i)

5B (0, 3, 5)

Q.8 (a) 5C* (0, 2, 4, 5) (ii) (b) 5C* (0, 2, 4, 5) (e) (i) 5B (0, 3, 5) (c) 5C* (0, 2, 4, 5) (ii) 5C (0, 2, 4, 5) (d) 5C* (0, 2, 4, 5) (f) 5C (0, 2, 4, 5) 20 40 Q.9 (a) 5B (0, 3, 5) Q.3 (a) 5C (0, 2, 4, 5) (b) 5B (0, 3, 5) (b) 5B (0, 3, 5) (c) 5A (0, 5) (c) (i) 5C (0, 2, 4, 5) (d) 5C (0, 2, 4, 5) (ii) 5C (0, 2, 4, 5) 20 20 Q.10 (a) 5C (0, 2, 4, 5) Q.4 (a) (i) 5C* (0, 2, 4, 5) (b) 5C* (0, 2, 4, 5) (ii) 5C (0, 2, 4, 5) (c) 10C (0, 4, 7, 10) (b) (i) 5C (0, 2, 4, 5) 20 (ii) 5C* (0, 2, 4, 5) (iii) 5C (0, 2, 4, 5) 25 Q.11 (a) 5C (0, 2, 4, 5) (b) 10C (0, 4, 7, 10) (c) 10D (0, 4, 6, 8, 10) Q.5 5B (0, 3, 5) (d) 5C (0, 2, 4, 5) 5B (0, 3, 5) 30 10C (0, 4, 7, 10) 20 Q.12 (a) 5C (0, 2, 4, 5) (b) 5D (0, 2, 3, 4, 5) Q.6 (a) 10C (0, 4, 7, 10) (c) 10C (0, 4, 7, 10) (b) 5B (0, 3, 5) 20 (c) 5C* (0, 2, 4, 5) 20

Assumptions about these marking schemes on the basis of past SEC marking schemes should be avoided. While the underlying assessment principles remain the same, the exact details of the marking of a particular type of question may vary from a similar question asked by the SEC in previous years in accordance with the contribution of that question to the overall examination in the current year. In setting these marking schemes, we have strived to determine how best to ensure the fair and accurate assessment of students’ work and to ensure consistency in the standard of assessment from year to year. Therefore, aspects of the structure, detail and application of the marking schemes for these examinations are subject to change from past SEC marking schemes and from one year to the next without notice.

2016.1 J.18/20_MS 38/72 of 71 examsDEB


Mathematics


General Instructions

1. There are 12 questions on this examination paper. Answer all questions.

2. Questions do not necessarily carry equal marks.

3. Marks will be lost if all necessary work is not clearly shown.

4. Answers should include the appropriate units of measurement, where relevant.

5. Answers should be given in simplest form, where relevant.



A standard DVD is in the shape of a circular disc of diameter 12 cm. It has a circular hole of diameter 1·8 cm in its centre, as shown.

1(a) Find the surface area of one side of a DVD. Give your answer in terms of π. (5C*)

Surface area = π 21r – π 2

2r

= π(2

12)2 – π(

2

81⋅)2

= π(6)2 – π(0·9)2 = 36π – 0·81π = 35·19π cm2

** Accept students’ answers in mm2 (3519π mm2) or m2 (0·00003519π m2).

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down correct area formula from Tables.

– Finds r1 = 2

12 or 6 and/or r2 =

2

81⋅ or 0·9

and stops.

High partial credit: (4 marks) – Finds π(6)2 – π(0·9)2, but fails to finish or finishes incorrectly. – Finds area using incorrect diameters correctly, i.e. π(12)2 – π(1·8) = 140·76π.– Final answer not in terms of π.

* Deduct 1 mark off correct answer only for the omission of or incorrect use of units (cm2) - apply only once per section (a), (b), (c), etc. of the question.

examsDEB

2016.1 J.18/20_MS 39/72 of 71 examsDEB

Q.1 (cont’d.)

1(b) The case for a single standard DVD is 19 cm long, 13·5 cm wide and 1·4 cm high.

Find the volume of a single DVD case. (5C*)

Volume = L × W × H = 13·5 × 19 × 1·4 = 359·1 cm3

** Accept students’ answers in mm2 (3519π mm2) or m2 (0·00003519π m2).

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down correct volume formula from Tables. – Some correct substitution into relevant volume formula (not stated).

High partial credit: (4 marks) – Correct substitution into relevant volume formula, but fails to finish or finishes incorrectly.

* Deduct 1 mark off correct answer only for the omission of or incorrect use of units (cm3) - apply only once per section (a), (b), (c), etc. of the question.

1(c) A DVD cake box holder is in the shape of a cylinder, as shown. It has an internal height of 6⋅1 cm. Given that the thickness of a DVD is 1⋅2 mm, find the maximum number of DVDs it can hold. (5C)

1·2 mm = 10

21⋅ cm

= 0·12 cm

# DVDs = 120

16

⋅⋅

= 50·83333... maximum # DVDs = 50

or

6·1 cm = 6·1 × 10 mm = 61 mm

# DVDs = 21

61

⋅

= 50·83333... maximum # DVDs = 50

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. converts 1·2 mm to 0·12 cm or 6·1 cm to 61 mm.

High partial credit: (4 marks) – Finds 120

16

⋅⋅

or 21

61

⋅ or 50·83333..., but

fails to round down correctly to 50.

2016.1 J.18/20_MS 40/72 of 71 examsDEB

Q.1 (cont’d.) 1(d) The external volume of the DVD cake box holder is 1,384 cm3. It has an external height of 8·5 cm.

Find the external diameter of the DVD cake box holder. Give your answer correct to one decimal place. (5D*)

External volume = πr2h = 1,384 cm3 πr2(8·5) = 1,384

r2 = π⋅58

384,1

= 51·828339... r = ...82833951⋅ = 7·199190...

d = 2r d = 2 × 7·199190... = 14·398380... ≅ 14·4 cm

** Accept students’ answers in mm (144 mm) or m (0·0144 m).

Scale 5D* (0, 2, 3, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down correct volume formula from Tables. – Some correct substitution into relevant volume formula (not stated).

Middle partial credit: (3 marks) – Correct substitution into relevant volume formula, but fails to manipulate or manipulates incorrectly.

High partial credit: (4 marks) – Finds r = ...82833951⋅ or 7·199190..., but fails to find or finds incorrect diameter.

* Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded or for the omission of or incorrect use of units (‘cm’) - apply only once per section (a), (b), (c), etc. of question.

1(e) Find the volume saved by using a DVD cake box holder as a percentage of using single DVD cases to store the maximum number of DVDs that the DVD cake box holder can contain. Give your answer correct to two significant figures. (5D*)

Volume of 50 cases = 50 × 359·1 = 17,955 cm3

Volume saved = 17,955 – 1,384 = 16,571 cm3

% Saving = VolumeTotal

Saving × 1

100

= 955,17

571,16 × 1

100

= 0⋅922918... × 100 ≅ 92%

** Accept students’ answers from part (b) if not oversimplified.

Scale 5D* (0, 2, 3, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. finds volume of 50 standard cases (ans. 17,555) and stops.

Middle partial credit: (3 marks) – Finds volume saved correctly without attempting to evaluate % Saving.

High partial credit: (4 marks) – Finds 955,17

571,16 or 955,17

571,16 × 1

100 , but fails



2016.1 J.18/20_MS 41/72 of 71 examsDEB


A simple random sample was taken from all the votes cast in last year’s marriage referendum. The sample was taken from both urban and rural areas. The way in which votes were cast is shown in the two-way table below.

Vote Urban Area Rural Area Total

Yes 602 Y 1,014

No X 388 786

Total 1,000 800 1,800

2(a) Find the missing values for X and Y in the table above. (5B)

Value of X

602 + X = 1,000 X = 1,000 – 602 = 398

or

X + 388 = 786 X = 786 – 388 = 398

Value of Y

602 + Y = 1,014 Y = 1,014 – 602 = 412

or

Y + 388 = 800 Y = 800 – 388 = 412

Scale 5B (0, 3, 5) Partial credit: (3 marks) – One value correct.

2(b) Explain what is meant by the term Simple Random Sample in the context of this question. (5B)

Any 1: – a simple random sample is where every person who

voted in the referendum has an equal chance of being selected //

– a simple random sample is meant to be an unbiased representation of all voters in the referendum // etc.

** Accept other appropriate material.

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Incomplete or unsatisfactory explanation, e.g. writes down ‘every person who voted’ or ‘equal chance of being selected’, etc.

2016.1 J.18/20_MS 42/72 of 71 examsDEB

Q.2 (cont’d.)

2(c) Complete the pie chart to display the data from the sample above. Show all of your calculations clearly. (10D)

Urban

‘Yes’

Urban

‘No’

Rural

‘No’

Rural

‘Yes’

Total number of votes = 1,800

Angle per vote = 800,1

360°

= 0·2°

Urban ‘Yes’ vote = 800,1

602 × 360°

= 120·4°

or = 602 × 0·2° = 120·4°

Urban ‘No’ vote = 800,1

398 × 360°

= 79·6°

or = 398 × 0·2° = 79·6°

Rural ‘Yes’ vote = 800,1

412 × 360°

= 82·4°

or = 412 × 0·2° = 82·4°

Rural ‘No’ vote = 800,1

388 × 360°

= 77·6°

or = 388 × 0·2° = 77·6°

** Allow a tolerance of ±2° in pie chart. ** Accept students’ answers from part (a) if not oversimplified.

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. finds angle per vote (ans. 0·2°), indicates that angles are fraction of 360°. – Finds one angle or fraction correct.

Middle partial credit: (6 marks) – Finds correct values of angles and stops or continues.

High partial credit: (8 marks) – Finds correct values of angles and two angles correct in pie chart (with labels). – Finds correct values of angles and all angles correct in pie chart, but no labels or incorrect labels. – Values of angles correct and correct pie chart (with labels), but no work shown. – Finds correct values of angles and all angles correct in pie chart (with labels), but outside tolerance (±2°).

2016.1 J.18/20_MS 43/72 of 71 examsDEB

Q.2 (cont’d.)

2(d) (i) Do you think that this is a suitable type of graph in which to illustrate this data? Give a reason for your answer. (5B)

Answer – no

Reason Any 1: – 3 out of the 4 angles look very similar in the pie chart // – figures (on vertical axis) of a bar chart / histogram may

add weight to the point that the graph is being used to illustrate // etc.

or

Answer – yes

Reason Any 1: – easy to see that the Urban ‘Yes’ vote is the largest

sector in the pie chart // – not cluttered by figures which may not be required

for the point that the graph is being used to illustrate // – not cluttered by figures which are only a sample of the

overall vote // etc.

** Accept other appropriate answers.

(ii) Identify another type of graph that could be more suitable to illustrate the data.

Answer Any 1: – bar chart // – histogram

** Award no marks for ‘line graph’, ‘line plot’ or ‘stem-and-leaf plot’ in part (ii).

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Correct reason, but no alternative suitable type of graph given. – No reason or incorrect reason, but suitable type of graph given.

2(e) (i) A vote is chosen at random from the entire sample. Find the probability that it is a ‘Yes’ vote. (5B)

# ‘Yes’ vote = 1,014 # Total sample = 1,800

P(‘Yes’ vote) = 800,1

014,1

= 300

169 or 0·563333...

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Correct numerator or denominator.

2016.1 J.18/20_MS 44/72 of 71 examsDEB

Q.2 (cont’d.)

2(e) (cont’d.)

(ii) If one vote is chosen at random from both the urban sample and the rural sample, find the probability that both are ‘No’ votes. (5C)

P(urban ‘No’ vote) = 000,1

398

= 500

199

P(rural ‘No’ vote) = 800

388

= 200

97

P(both ‘No’ vote) = 500

199 ×

200

97

= 000,100

303,19 or 0·19303

** Accept students’ answers for ‘398’ from part (a) if not oversimplified.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 mark) – Finds 500

199 and/or 200

97 , but does not

attempt to multiply them. – Uses addition and finishes correctly

[ans. 500

199 + 200

97 = 000,1

883 or equivalent].

High partial credit: (4 mark) – Finds 500

199 × 200

97 , but fails to finish

or finishes incorrectly.

2(f) Paula says, “It is more likely that a person from an urban area voted ‘Yes’ than a person from a rural area”.

Based on the data in the sample, do you agree with Paula? Give a reason for your answer. (5C)

Answer – yes

Reason

From the random sample:

P(urban ‘Yes’ vote) = 000,1

602

= 0·602

P(rural ‘Yes’ vote) = 800

412

= 0·515

as 0·602 > 0·4975, it is more likely that a person from an urban area voted ‘Yes’ than a person from a rural area

** Accept students’ answers for ‘412’ from part (a) if not oversimplified.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Correct answer, but no reason or incorrect reason given.

High partial credit: (4 marks) – Correct answer with incomplete or unsatisfactory explanation.

– Finds 000,1

602 and 800

412 , or equivalent, but

no conclusion given.

2016.1 J.18/20_MS 45/72 of 71 examsDEB


3(a) Explain what is meant by the term Trial in the context of probability and give an example. (5C)

Explanation Any 1: – a ‘trial’ is the act of doing a random experiment

in probability // – a ‘trial’ is any particular performance of a random

experiment // – a ‘trial’ is any procedure that can be infinitely repeated

and has a well-defined set of possible outcomes, known as a sample space // etc.


Example Any 1: – tossing a coin // – rolling a die // – picking a card from a deck of 52 cards // etc.


Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Incomplete or unsatisfactory explanation with no example or incorrect example given.

High partial credit: (4 marks) – Correct explanation with no example or incorrect example given. – Incomplete or unsatisfactory explanation with suitable example given.

3(b) A cinema offers three different combi-meal sizes on its menu, as shown below. A combi-meal consists of a soft drink and popcorn of the same size.

Combi-Meal Menu

Size Soft Drink Popcorn

Small Orange Plain Medium Cola Salted

Large Lemonade Buttered Ginger Ale Caramel Cream Soda

Calculate the number of different combi-meal options that can be ordered. (5B)

# options = 3 × 5 × 4 = 60

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Some work of merit, e.g. mention of the ‘Fundamental Principle of Counting’. – Writes a list of possible options. – Finds 3 × 5 × 4, but fails to finish or finishes incorrectly. – Uses addition instead of multiplication, i.e. # options = 3 + 5 + 4 = 12.

2016.1 J.18/20_MS 46/72 of 71 examsDEB

Q.3 (cont’d.)

3(c) Below is part of a tree diagram showing the probabilities that a customer chooses a medium combi-meal with cola and popcorn.

Caramel Popcorn

Buttered Popcorn

Salted Popcorn

Plain Popcorn

9

2

3

1

2

1

12

1

4

1

6

1

ColaMedium

(i) A person is chosen at random in the cinema. Find the probability that this person bought a medium combi-meal with cola and plain popcorn. (5C)

P(medium, cola, plain popcorn)

= 9

2 ×

3

1 ×

2

1

= 54

2 or

27

1 or 0·037037...

Scale 5C (0, 2, 4, 5) Low partial credit: (2 mark) – Some work of merit, e.g. identifies

9

2 and/or

3

1 and/or

2

1, but does not

attempt to multiply them. – Uses addition and finishes correctly

[ans. 9

2 ×

3

1 ×

2

1 =

18

19 or equivalent].

High partial credit: (4 mark) – Finds 9

2 ×

3

1 ×

2

1, but fails to finish


(ii) A customer is chosen at random from those who bought a medium combi-meal. Find the probability that this customer did not order a cola with the meal. (5C)

P(not Cola) = 1 – 3

1

= 3

2 or 0·666666...

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. identifies 3

1


High partial credit: (4 marks) – Finds 1 – 3

1, but fails to finish or finishes

incorrectly.

2016.1 J.18/20_MS 47/72 of 71 examsDEB



4(a) M, N and P are points on a circle with centre O. | ∠NPM | = 58°, as shown.

(i) Find | ∠OMN |. (5C*)

| ∠NOM | = 2(58°) = 116°

| ∠OMN | = 180° – (| ∠NOM | + | ∠MNO |) but | ∠OMN | = | ∠MNO |

| ∠OMN | = 2

1(180° – | ∠NOM |)

| ∠OMN | = 2

1(180° – 116°)

= 2

1(64°)

= 32°

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down that the angle at the centre is equal to twice the angle at the circumference. – Writes down | ∠NOM | = 2 × | ∠NPM | or similar. – Finds | ∠NOM | = 116°, but does not attempt to find | ∠OMN |.

High partial credit: (4 marks) – Finds | ∠NOM | = 64°.

* Deduct 1 mark off correct answer only for the omission of or incorrect use of units (‘°’) - apply only once per section (a), (b), (c), etc. of the question.

(ii) Given that | ∠PMO | = 29°, show that the triangle PMN is isosceles. (5C)

| ∠OMN | = 32°

| ∠PMN | = | ∠PMO | + | ∠OMN | = 29° + 32° = 61°

| ∠MNP | = 180° – (| ∠NPM | + | ∠PMN |) = 180° – (58° + 61°) = 180° – 119° = 61° = | ∠PMN | ΔPMN is isosceles

** Accept students’ answers for | ∠OMN | from part (a) if not oversimplified.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down that the sum of the three angles in a triangle is equal to 180°. – Finds | ∠PMN | = 61°, but does not attempt to find | ∠MNP |.

High partial credit: (4 marks) – Finds | ∠MNP | = 180° – (58° + 61°) or similar. – Finds | ∠MNP | = 61°, but no conclusion (i.e. ‘| ∠MNP | = | ∠PMN |, ΔPMN is isosceles’) or an incorrect conclusion given.

P

N

O

M

58�

2016.1 J.18/20_MS 48/72 of 71 examsDEB

Q.4 (cont’d.)

4(b) In the diagram, [ AB ] is parallel to [ CD ]. [ AD ] and [ CB ] intersect at the point E.

(i) Prove that triangles ABE and CED are similar. Give a reason for each of the statements that you make in your proof. (5C)

In ΔABE and ΔCED,

| ∠BEA | = | ∠CED | ... vertically opposite angles

| ∠EAB | = | ∠EDC | ... alternate angles as [ AB ] is parallel to [ CD ]

() | ∠ABE | = | ∠DCE | ... third angle in two triangles are equal if the other two angles are equal

or ... alternate angles as [ AB ] is

parallel to [ CD ]

ΔABE and ΔCED are similar

** Must include a reason for all three steps for full credit.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. indicates that - vertically opposite angles are equal, - alternate angles are equal, - third angle in two triangles are equal if the other two angles are equal. – One step correct with valid reason.

High partial credit: (4 marks) – Two steps correct with one valid reason.– Three steps correct with no reasons or incorrect reasons given.

(ii) Given that | ∠BEA | = 119° and | ∠DCE | = 32°, find | ∠EAB |. (5C*)

| ∠DCE | = 32° | ∠ABE | = 32°

| ∠EAB | = 180° – (| ∠ABE | + | ∠BEA |) = 180° – (32° + 119°) = 180° – 151° = 29°

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down that the sum of the three angles in a triangle is equal to 180°. – States that | ∠EAB | = 180° – (| ∠ABE | + | ∠BEA |) or similar. – Finds | ∠ABE | = 32°, but does not attempt to find | ∠EAB |.

High partial credit: (4 marks) – Finds | ∠EAB | = 180° – (119° + 32°), but fails to finish or finishes incorrectly.

* Deduct 1 mark off correct answer only for the omission of or incorrect use of units (‘°’) - apply only once per section (a), (b), (c), etc. of question.

C

A B

E

D

2016.1 J.18/20_MS 49/72 of 71 examsDEB

Q.4 (cont’d.)

4(b) (cont’d.)

(iii) | AE | = 16·5, | ED | = 55 and | CD | = 90. Find | AB |. (5C)

as ΔABE and ΔCED are similar

||

||

CD

AB =

||

||

ED

AE

90

|| AB =

55

516⋅

| AB | = 55

90516 ×⋅

= 55

485,1

= 27

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit, e.g. correct relevant ratio, corresponding sides identified or indicates that corresponding sides are in proportion. – Some correct substitution into correct relevant ratio with minor errors. – Identifies one correct relevant ratio,

i.e. 90

|| AB or

55

516⋅.


|| AB =

55

516⋅ or equivalent, but

fails to finish or finishes incorrectly. – Finds | AB | = 27, but no work shown.

2016.1 J.18/20_MS 50/72 of 71 examsDEB


Prove that, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. (5B, 5B, 10C)

Diagram

Given ΔABC where | ∠ BAC | = 90°.

To prove | BC |2 = | AB |2 + | AC |2

Construction Draw AD ⊥ BC.

Proof | ∠CAB | = | ∠BDA | ... both 90° | ∠ABC | = | ∠ABD | ... common angle ∴ triangles ABC and ABD are similar.

|AB|

|BC| =

|BD|

|AB| ... corresponding sides

are proportional | AB |2 = | BC |.| BD |

Likewise, triangles ABC and ADC are similar.

|AC|

|BC| =

|DC|

|AC| ... corresponding sides

are proportional | AC |2 = | BC |.| DC |

| AB |2 + | AC |2 = | BC |.| BD | + | BC |.| DC | = | BC |.(| BD | + | DC |) = | BC |.| BC | = | BC |2

∗∗ Some steps may be indicated on the diagram.

or

Diagram

Given Right-angled triangle with length of sides a, b, c, where c is the hypotenuse.

To prove a2 + b2 = c2

Construction Construct a square ABCD of side a + b Construct the point E on [AB] such that | AE | = b (and hence | EB | = a). Similarly construct points F, G and H on the other sides, as shown. Join E, F, G and H to divide the square ABCD into a quadrilateral and four triangles. Label the angles 1, 2, 3 and 4, as shown.

Proof Each of the four inscribed triangles is congruent to the original triangle ... SAS

∴ Each side of the inner quadrilateral has length c | ∠1 | + | ∠2 | = 90° ... angle sum of triangle | ∠1 | = | ∠3 | ... corresponding parts in congruent triangles ∴ | ∠2 | + | ∠3 | = 90° ∴ | ∠4 | = 90° ... straight angle ∴ The inscribed quadrilateral is a square

B D C

A

b

ac

A

D

E

G

B

F

H

C

b

b

a

a

b

a

a

b

c

1

2

4

3

2016.1 J.18/20_MS 51/72 of 71 examsDEB

Q.5 (cont’d.)

Proof (cont’d.)

Area of large square = 4(area of one triangle) + area of inscribed square (a + b)2 = 4 (area of one triangle) + c2 (a + b)2 = 4(½ ab) + c2 a2 + 2ab + b2 = 2ab + c2 ∴ a2 + b2 = c2

∗∗ Some steps may be indicated on the diagram.

Diagram/Construction

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Right-angled triangle drawn only.

Given/To prove

Scale 5B (0, 3, 5) Partial credit: (3 marks) – ‘Given’ correct. – ‘To prove’ correct.

Proof

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – Any correct step.

High partial credit: (7 marks) – One step missing or incomplete.

2016.1 J.18/20_MS 52/72 of 71 examsDEB



6(a) Construct a right-angled triangle ABC, where:

| ∠BAC | = 36° | AB | = 8 cm | ∠CBA | = 90°. (10C)

8 cm

C

A B

99

cm

·

36�

** Allow a tolerance of ±0·2 cm or ±2°.

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – One side or one angle (36° or 90°) correctly drawn. – Sketch drawn with given measurements shown.

High partial credit: (7 marks) – Triangle correctly drawn, but unlabelled or incorrectly labelled. – Triangle correctly drawn, but outside allowable tolerances.

6(b) On your diagram, measure the length of the side AC. Give your answer in cm, correct to one decimal place. (5B)

From diagram: | AC | = 9·9 cm (±0·2 cm)

** Accept students’ triangle from part (a) if not oversimplified.

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Wrong side correctly measured. – Triangle incorrect and unlabelled, but one side correctly measured.

* No deduction applied for the omission of or incorrect use of units (‘cm’).

2016.1 J.18/20_MS 53/72 of 71 examsDEB

Q.6 (cont’d.)

6(c) Using trigonometry, verify your answer to part (b) above. (5C*)

cos | ∠BAC | = |Hyp|

|Adj|

= ||

||

AC

AB

cos 36° = ||

8

AC

| AC | = °36cos

8

= ...8090160

8

⋅

= 9·888543... ≅ 9·9 cm

or

tan | ∠BAC | = |Adj|

|Opp|

= ||

||

AB

BC

tan 36° = 8

||BC

| BC | = 8 × tan 36° = 8 × 0·726542... = 5·812340...

| Hyp |2 = | Opp |2 + | Adj |2 ... Pythagoras’ theorem | AC |2 = | AB |2 + | BC |2 = 82 + (5·812340...)2 = 64 + 33·783298... = 97·783298... | AC | = ...78329897⋅ = 9·888543... ≅ 9·9 cm


Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down trigonometric ratio (cos or tan) or states formula for Pythagoras’ theorem. – Some correct substitution into correct trigonometric ratio (cos or tan). – Some correct substitution into formula for Pythagoras’ theorem. – Finds | BC | and stops (ans. 8 × 0·726542... or 5·812340...).

High partial credit: (4 marks) – Correct substitution into trigonometric ratio and correctly manipulated, e.g. | AC | =

°36cos

8 .

– Correct substitution into formula for Pythagoras, i.e. | AC |2 = 82 + 5·812340...2.

– Finds | AC | = ...8090160

8

⋅, but fails to

finish or finishes incorrectly. – Finds | AC |2 = 97·783298... or | AC | = ...78329897⋅ , but fails to finish or finishes incorrectly.


* No deduction applied for the omission of or incorrect use of units (‘cm’).

2016.1 J.18/20_MS 54/72 of 71 examsDEB


7(a) Plot the points A(–4, 1), B(–4, 3) and C(–1, 3) on the co-ordinate plane below. Join the points to form a triangle. (5C)

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – One or two points correctly plotted. – All points reversed, i.e. (y, x), with or without triangle drawn.

High partial credit: (4 marks) – Three points correctly plotted, but not joined.

7(b) What type of triangle does ABC represent? Give a reason for your answer. (5B)

Type of triangle – right-angled triangle

Reason Any 1: – ∠ABC is equal to 90° // – [ AB ] is perpendicular to [ BC ] // – using Pythagoras’ theorem, | AB |2 + | BC |2 =| AC |2 // etc.

Type of triangle – scalene triangle

Reason Any 1: – all three sides of triangle ABC have different lengths // – triangle ABC has no two sides which are equal // etc.

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Correct answer, but no reason or incorrect reason given.

2

�1

1

�2

2

�3

3

�4

4

3 4 5

x

y

1�1�2�3�4�5

A

B C

2016.1 J.18/20_MS 55/72 of 71 examsDEB

Q.7 (cont’d.)

7(c) Find the area of triangle ABC. (5C) A(–4, 1), B(–4, 3) | AB | = | y2 – y1

| = 3 – 1 = 2

or

| AB | = 212

212 )()( yyxx −+−

= 22 )13())4(4( −+−−−

= 22 )2()0( +

= 4 = 2

B(–4, 3), C(–1, 3) | BC | = | x2 – x1

| = | –4 | – | –1 | = 4 – 1 = 3

or

| BC | = 22 )33())4(1( −+−−−

= 22 )0()3( +

= 9 = 3

Area of the triangle = 2

1 × | Base | × | ⊥Height |

= 2

1 × 2 × 3

= 3 units2

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down correct area formula from Tables. – Finds correct length of one relevant side.– Some correct substitution into correct area formula (not stated).

High partial credit: (4 marks) – Finds correct lengths of two relevant sides with some correct substitution into area formula.

– Finds 2

1 × 2 × 3 or

2

1 × 4 × 9 , but

fails to finish or finishes incorrectly.

* No deduction applied for the omission of or incorrect use of units (‘units2’).

2016.1 J.18/20_MS 56/72 of 71 examsDEB

Q.7 (cont’d.)

7(d) Draw in the image of triangle ABC under central symmetry in the origin on the co-ordinate plane above. (5C)

** Accept students’ triangle from part (a) if not oversimplified.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – One or two image points correctly plotted.– All points reversed, i.e. (y, x), with or without triangle drawn. – Correct image of ABC under axial symmetry in the x-axis or y-axis.

High partial credit: (4 marks) – Three image points correctly plotted, but not joined. – Correct image of ABC under central symmetry not in the origin.

7(e) State whether triangle ABC and its image under the transformation above are similar or congruent. Give a reason for your answer. (5C)

Answer – congruent

Reason – SSS: | AB | = | A′B′ |, | BC | = | B′C′ |, | AC | = | A′C′ | //

or

– SAS: | AB | = | A′B′ |, | ∠CBA | = | C′B′A′ |, | BC | = | B′C′ | //

or

– RHS: | ∠CBA | = | C′B′A′ |, | AC | = | A′C′ |, | AB | = | A′B′ |

or | BC | = | B′C′ | // etc.



High partial credit: (4 marks) – Correct answer, but reason given insufficient or incomplete.

2

�1

1

�2

2

�3

3

�4

4

3 4 5

x

y

1�1�2�3�4�5

A

B C

B'C'

A'

2016.1 J.18/20_MS 57/72 of 71 examsDEB

Q.7 (cont’d.)

7(f) Is the area of the triangle ABC equal to the area of its image under the transformation above? Give a reason for your answer. (5C)

Answer – yes

Reason Any 1: – central symmetry preserves area // – as ΔABC and its image are congruent, the areas of both

triangle are the same // etc.



High partial credit: (4 marks) – Correct answer, but reason given insufficient or incomplete.

7(g) Find | AC |, giving your answer in surd form. (5C)

A(–4, 1), C(–1, 3)

| AC | = 212

212 )()( yyxx −+−

= 22 )13())4(1( −+−−−

= 22 )2()3( +

= 49 +

= 13 units

or

A(–4, 1), C(–1, 3) Using Pythagoras’ theorem | AC |2 = | AB |2 + | BC |2 = | 2 |2 + | 3 |2 = 4 + 9 = 13 | AC | = 13 units

** Accept students’ answers from part (c) if not oversimplified in method .

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down correct distance formula from Tables. – Some correct substitution into correct distance formula (not stated).

High partial credit: (4 marks) – Fully correct substitution into correct distance formula, but fails to finish or finishes incorrectly. – Final answer given in decimal form (not surd form).

* No deduction applied for the omission of or incorrect use of units (‘units’).

2016.1 J.18/20_MS 58/72 of 71 examsDEB

Q.7 (cont’d.)

7(h) Sean says that cos | ∠ACB | is equal to sin | ∠CAB |. Is he correct? Give a reason for your answer. (5C)

Answer – yes

Reason

cos | ∠ACB | = |Hyp|

|Adj|

= ||

||

AC

BC

= 13

3

sin | ∠CAB | = |Hyp|

|Opp|

= ||

||

AC

BC

= 13

3

cos | ∠ACB | = sin | ∠CAB |

** Accept students’ answers from parts (c) and (g) if not oversimplified.


High partial credit: (4 marks) – Correct answer, with one trigonometric ratio (cos | ∠ACB | or sin | ∠CAB |) correctly evaluated.

2016.1 J.18/20_MS 59/72 of 71 examsDEB



The ‘Monument of Light’, alternatively known as the ‘Spire’, is a stainless steel, cone-shaped monument located in Dublin.

The Spire is the world’s tallest sculpture at a height of 121⋅2 m. It was officially unveiled in 2003 at a total cost of €4,000,000 to Dublin City Council.

8(a) The circumference of the Spire at its base is 9·4 m. Find the radius of its base, correct to one decimal place. (5C*)

Circumference = 2πr = 9·4 m 2πr = 9·4

r = π⋅

2

49

= 1·496056... ≅ 1·5 m

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down correct circumference formula from Tables.

High partial credit: (4 marks) – Fully correct substitution into circumference formula, but fails to finish or finishes incorrectly.

* Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded or for the omission of or incorrect use of units (‘m’) - apply only once per section (a), (b), (c), etc. of question.

8(b) Using the diagram, find l, the slant height of the Spire. Give your answer in metres, correct to two decimal places. (5C*)

Using Pythagoras’ theorem | Hyp |2 = | Opp |2 + | Adj |2 l2 = h2 + r2 = (121·2)2 + (1·5)2 = 14,689·44 + 2·25 = 14,691·69 l = 69691,14 ⋅

= 121⋅209281... ≅ 121⋅21 m

or

tan = |Adj|

|Opp|

= 51

2121

⋅⋅

= tan–1 (80·8) = 89·290930...

l

2016.1 J.18/20_MS 60/72 of 71 examsDEB

Q.8 (cont’d.)

8(b) (cont’d.)

sin = |Hyp|

|Opp|

sin 89·290930... = l

2121⋅

l = ...29093089sin

2121

⋅⋅

l = ...9999230

2121

⋅⋅

= 121·209281... ≅ 121·21 m


Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down Pythagoras’ theorem. – Some correct substitution into Pythagoras’ theorem and stops or continues. – Substitutes correctly into relevant

trigonometric ratio, i.e. tan = 51

2121

⋅⋅ ,


High partial credit: (4 marks) – Finds l2 = 14,691·69 or l = 69691,14 ⋅ ,

but fails to finish or finishes incorrectly. – Correct substitution into second relevant trigonometric ratio, i.e. sin 89·290930...

= l

2121⋅ or cos 89·290930 = l

51⋅ , but fails



* No deduction applied for the omission of or incorrect use of units (‘m’). 8(c) Find the curved surface area of the Spire. Give your answer in m2, correct to one decimal place. (5C*) Curved surface area = πrl = π(1·5)(121·21) = 181·815π = 571·188668... ≅ 571·2 m2


Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down correct formula for curved surface area of a cone from Tables. – Some correct substitution into relevant formula (not stated) and stops or continues.

High partial credit: (4 marks) – Fully correct substitution into relevant formulae, but fails to finish or finishes incorrectly.


* No deduction applied for the omission of or incorrect use of units (‘m2’).

2016.1 J.18/20_MS 61/72 of 71 examsDEB

Q.8 (cont’d.)

8(d) The Spire cost the construction consortium thst were awarded the contract €3,675,421 to manufacture and erect. Find the percentage profit made by the consortium, correct to one decimal place. (5C*)

Cost to DCC = €4,000,000 Construction cost = €3,675,421 Profit = 4,000,000 – 3,675,421 = €324,579

% profit = 421,675,3

579,324 ×

1

100

= 0·088310... × 100 = 8·883106... ≅ 8·9%

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. finds correct profit (i.e. €324,579), but does not attempt to evaluate percentage.

High partial credit: (4 marks) – Finds

421,675,3

579,324 or

421,675,3

579,324 ×

1

100,



* No deduction applied for the omission of or incorrect use of units (‘%’).

2016.1 J.18/20_MS 62/72 of 71 examsDEB


The back-to-back stem-and-leaf plot below shows the systolic blood pressure of thirteen people before and two hours after taking a particular drug.

Before After 10 9 11 1 6 8 6 12 5 7 4 13 1 7 9 7 0 14 5 6 8 6 4 3 15 9 7 5 16 0 8 17 0 18 1

Key: 15 9 means 159 mm Hg

9(a) Find the median blood pressure both before and after taking the drug. (5B)

Median (before)

126, 128, 134, 140, 147, 149, 153, 154, 156, 165, 167, 178, 180

Median = 153 mm Hg

Median (after)

109, 111, 116, 125, 127, 131, 137, 145, 146, 148, 159, 160, 181

Median = 137 mm Hg


* Accept correct answers with or without work shown. * No deduction applied for the omission of or incorrect use of units (‘mm Hg’).

9(b) Find the range of blood pressures both before and after taking the drug. (5B)

Range (before) = 180 – 126 = 54 mm Hg

Range (after) = 181 – 109 = 72 mm Hg


* Accept correct answers with or without work shown. * No deduction applied for the omission of or incorrect use of units (‘mm Hg’).

2016.1 J.18/20_MS 63/72 of 71 examsDEB

Q.9 (cont’d.)

9(c) What other measure of variability (spread) could have been used when examining this data? (5A)

Answer Any 1: – interquartile range // – standard deviation

Scale 5A (0, 5) – Hit or miss.

9(d) Compare the blood pressure results both before and after taking the drug. Refer to at least one measure of central tendency and at least one measure of variability (spread) in your answer. (5C)

Measure of central tendency

– systolic blood pressures are lower two hours after taking the drug

Any 1: – the median systolic blood pressure before taking the drug

is 153 mm Hg, while median systolic blood pressure after taking the drug is 137 mm Hg //

– the mean systolic blood pressure before taking the drug is 152 mm Hg, while median systolic blood pressure after taking the drug is 138 mm Hg // etc.


Measure of variability (spread)

– the spread of systolic blood pressure is greater two hours after taking the drug

Any 1: – the range of systolic blood pressure before taking the

drug is 54 mm Hg, while range of systolic blood pressure after taking the drug is 72 mm Hg //

– the interquartile range of systolic blood pressure before taking the drug is 29 mm Hg, whereas the interquartile range of systolic blood pressure after taking the drug is 33 mm Hg // etc.


Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Mention of mean, mode, median, range or interquartile range in answer.

High partial credit: (4 marks) – Comparison using measure of central tendency or measure of variability only.

* No deduction applied for the omission of or incorrect use of units (‘mm Hg’).

2016.1 J.18/20_MS 64/72 of 71 examsDEB



In 1975, an American daredevil known as Evel Knievel attempted to jump over thirteen single-deck buses on his motorcycle in front of 90,000 people at Wembley Stadium in London.

To complete the jump, a launch ramp 3·5 m in height with a base 10·4 m in length was constructed, as shown below.

10 4 m·

3 m·5

10(a) For Evel not to undershoot or overshoot the jump, he required the angle of elevation of the launch ramp to be between 15° and 20°. Would you conclude that the ramp met this criteria? Use calculations to justify your answer. (5C)

Answer – yes

Reason

tan | ∠X | = |Adj|

|Opp|

= 410

53

⋅⋅

= 0⋅336538... | ∠X | = tan–1 (0·336538...) = 18·600065... ≅ 18·6° ramp meets the criteria as 15° ≤ | ∠X | ≤ 20°

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit, e.g. writes down correct trigonometric ratio (tan). – Some correct substitution into correct trigonometric ratio (sin or cos).

High partial credit: (4 marks) – Correct substitution into trigonometric ratio with some manipulation, e.g. finds

| ∠X | = tan–1

410

53

⋅⋅ or tan–1 (0·336538...),

but fails to finish or finishes incorrectly. – Trigonometric ratio inverted, but finishes

correctly [ans. | ∠X | = tan–1

53

410

⋅⋅ = 71·4°].

– Finds | ∠X | = 18·6°, but no conclusion (i.e. ‘meets criteria as 15° ≤ | ∠X | ≤ 20°’) or an incorrect conclusion given.

2016.1 J.18/20_MS 65/72 of 71 examsDEB

Q.10 (cont’d.)

10(b) Find the length of the ramp. Give your answer in metres, correct to two decimal places. (5C*)

Using Pythagoras’ theorem | Hyp |2 = | Opp |2 + | Adj |2 l2 = | 3·5 |2 + | 10·4 |2 = 12·25 + 108·16 = 120·41 l = 41120⋅ = 10·973149... ≅ 10·98 m

or

sin = |Hyp|

|Opp|

sin 18·6° = l

53⋅

l = °⋅

⋅618sin

53

l = ...3189590

53

⋅⋅

= 10·973186... ≅ 10·98 m

or

cos = |Hyp|

|Adj|

cos 18·6° = l

410⋅

l = °⋅

⋅618cos

410

l = ...9477680

410

⋅⋅

= 10·973144... ≅ 10·98 m

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down trigonometric ratio (sin or cos) or states formula for Pythagoras’ theorem. – Some correct substitution into correct trigonometric ratio (sin or cos). – Some correct substitution into formula for Pythagoras’ theorem.

High partial credit: (4 marks) – Correct substitution into formula for Pythagoras, i.e. l2 = | 3·5 |2 + | 10·4 |2. – Finds l2 = 120·41 or l = 41120⋅ , but fails to finish or finishes incorrectly. – Correct substitution into trigonometric ratio and correctly manipulated, e.g.

l = °⋅

⋅618sin

53 or l = °⋅

⋅618cos

410 .

– Finds l = ...3189590

53

⋅⋅

or l = ...9477680

410

⋅⋅

,



* No deduction applied for the omission of or incorrect use of units (‘m’).

2016.1 J.18/20_MS 66/72 of 71 examsDEB

Q.10 (cont’d.)

10(c) The width of each bus was 2⋅5 m. Evel estimated that he needed to travel at an average speed of 90 km/h in mid-air to complete the jump.

Find the minimum time he needed to travel in mid-air to complete the jump successfully. Give your answer in seconds. (10C)

Length of mid-air jump = 13 × 2·5 = 32·5 m

Average Speed = 90 km/h

= s6060

m000,190

××

= 3,600

000,90

= 25 m/s

Average speed = Time

Distance

Time = SpeedAverage

Distance

= 25

532⋅

= 1·3 seconds

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down correct speed formula. – Finds correct length of mid-air jump i.e. 13 × 2·5 or 32·5 m. – Finds average speed required in mid-air

i.e. s6060

m000,190

×× or

3,600

000,90 or 25 m/s.

High partial credit: (7 marks) – Substitutes correct length of mid-air jump and/or average mid-air speed into speed formula, but fails to finish or finishes incorrectly.

* No deduction applied for the omission of or incorrect use of units (‘seconds’).

2016.1 J.18/20_MS 67/72 of 71 examsDEB


The customers in a restaurant were surveyed to find out how long they each had to wait for a table on a particular night. The results are shown in the grouped frequency table below.

Time (minutes) 0 − 5 5 − 10 10 − 15 15 − 20 20 − 25 Frequency 20 11 9 7 3

Note: 10 − 15 means at least 10 minutes but less than 15 minutes, etc.

11(a) What type of data was collected in the survey? Put a tick () in the correct box below. Explain your answer. (5C)

Type of data

NumericalDiscrete

NumericalContinuous

CategoricalNominal

Categorical Ordinal

Explanation

– numerical: any data that can be represented by numbers // etc.

– continuous: can take any value between two points, e.g. height, weight, time, speed, temperature // etc.


Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Correct answer, but no explanation or incorrect explanation given.

High partial credit: (4 marks) – Correct answer, but explanation given insufficient or incomplete.

11(b) Using mid-interval values, estimate the mean waiting time for a table on that particular night. (10C)

Mean = 3791120

)3522()7517()9512()1157()2052(

++++×⋅+×⋅+×⋅+×⋅+×⋅

= 50

5675122511258250 ⋅+⋅+⋅+⋅+

= 50

435

= 8·7

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – Some work of merit, e.g. shows indication of division by 50. – Finds one correct mid-interval value.

High partial credit: (7 marks) – Finds correct numerator, i.e. 435 or (2·5 × 20) + (7·5 × 11) + (12·5 × 9) + (17·5 × 7) + (22·5 × 3). – Finds incorrect mean using consistent incorrect mid-interval values.

– Finds 50

435, but fails to finish or finishes

incorrectly.

2016.1 J.18/20_MS 68/72 of 71 examsDEB

Q.11 (cont’d.)

11(c) Display the above data on a histogram. (10D)

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. draws correctly labelled axes (with no bars inserted). – One or two bars correct (each of consistent width).

Middle partial credit: (6 marks) – Three or four bars correct (each of consistent width). – Bar chart correctly drawn, but both axes not numerated or labelled.

High partial credit: (8 marks) – Five bars correct (each of consistent width), but both axes not labelled. – Bar chart correctly drawn.

11(d) What percentage of customers were waiting 15 minutes or longer for a table? (5C)

# people waiting 15 minutes or longer = 7 + 3 = 10

% people = 50

10 ×

1

100

= 0·2 × 100 = 20%

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. finds correct # people (i.e. 10), but does not attempt to evaluate percentage.


10 or

50

10 ×

1

100, but fails to finish


* No deduction applied for the omission of or incorrect use of percentage symbol (‘%’).

15

5

0

10

20

Time (minutes)

Fre

qu

ency

0 10 205 15 25

2016.1 J.18/20_MS 69/72 of 71 examsDEB


12(a) The equation of the line k is 2x + y – 1 = 0. Find the value of t such that (2t, –3t) is a point on the line k. (5C)

k: 2x + y – 1 = 0 (2t, −3t) ∈ k

2(2t) + (–3t) – 1 = 0 4t – 3t – 1 = 0 t – 1 = 0 t = 1

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down (2t, −3t) ∈ k or similar. – Some correct substitution into equation, i.e. 2(2t) or (–3t) or inverts x and y values in equation, i.e. 2(–3t) +(2t) – 1 = 0 and stops or continues.

High partial credit: (4 marks) – Correct substitution into line equation, but fails to finish or finishes incorrectly.

12(b) The line l passes through the point (3, 2) and is perpendicular to the line k. Find the equation of the line l in the form ax + by + c = 0. (5D)

k: 2x + y – 1 = 0 y = –2x +1 mk = –2

ml = ⊥mk

= 2

1

ml = 2

1, point (3, 2)

y – y1 = m(x – x1)

y – 2 = 2

1(x – 3)

2y – 4 = x – 3 x – 2y – 3 + 4 = 0 x – 2y + 1 = 0

Scale 5D (0, 2, 3, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down k in the form y = –2x +1 and stops. – Finds correct slope for k, i.e. 2.

Middle partial credit: (3 marks) – Finds correct slope for l, i.e. 2

1, and stops.

– Some correct substitution into a line formula.

High partial credit: (4 marks) – Correct slope with one incorrect substitution into line formula, but continues correctly. – Correct slope, but both x and y reversed in substitution, but continues correctly. – Correct substitution into line formula, but fails to finish or finishes incorrectly. – Final answer correct, but not in the form ax + by + c = 0.

2016.1 J.18/20_MS 70/72 of 71 examsDEB

Q.12 (cont’d.)

12(c) Find the co-ordinates of the point of intersection of lines l and k. (10C)

k: 2x + y = 1 (×2) l: x – 2y = –1 (×1)

4x + 2y = 2 x – 2y = –1 5x = 1

x = 5

1

2x + y = 1 y = 1 – 2x

y = 1 – 2(5

1)

= 1 – 5

2

= 5

3

or x – 2y = –1 –2y = –1 – x

y = 2

1(1 + x)

= 2

1(1 +

5

1)

= 5

3

l ∩ k = (5

1,

5

3)

or

2x + y = 1 (×1) x – 2y = –1 (×–2)

2x + y = 1 –2x + 4y = 2 5y = 3

y = 5

3

2x + y = 1 2x = 1 – y

x = 2

1(1 – y)

= 2

1(1 –

5

3)

= 5

1

or x – 2y = –1 x = –1 + 2y

= –1 + 2(5

3)

= –1 + 5

6

= 5

1

l ∩ k = (5

1,

5

3)

** Accept students’ answers from part (b) if not oversimplified.

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – Any work of merit, e.g. multiplying equation by appropriate constant to facilitate cancellation of x or y term. – Finds 5x = 1 or 5y = 3, but fails to finish or finishes incorrectly. – Finds either variable (x or y) correctly by trial and error, but fails to verify in both equations or verifies incorrectly.

High partial credit: (7 marks) – Finds first variable (x or y) correctly, but fails to find second variable or finds incorrectly. – Finds both variables (x and y) correctly, but no work shown. – Finds both variables (x and y) by graphical means. – Finds both variables (x and y) by trial and error, but does not verify into both equations.

2016.1 J.18/20_MS 71/72 of 71 examsDEB

Notes:

2016.1 J.18/20_MS 73/72 Page 73 of 71 examsDEB

mathematics - m. selkirk confey collegemathematics higher level – paper 1 marking scheme (300...

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