2015.2 j.18 ms -14 sk-final -...

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J.18/20

Pre-Junior Certificate Examination, 2015

Mathematics Higher Level

Marking Scheme

Paper 1 Pg. 2 Paper 2 Pg. 36

2015.2 J.18/20_MS 2/72 of 71 examsDEB


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, 5 0, 3, 5 0, 2, 4, 5 0, 2, 3, 4, 5 10 mark scale 0, 5, 10 0, 4, 7, 10 0, 4, 6, 8, 10 15 mark scale 0, 7, 15 0, 6, 11, 15 0, 6, 10, 13, 15

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 (four 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 per question. The * penalty is not applied to currency solutions.

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

Accept a student’s 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,

2015.2 J.18/20_MS 3/72 of 71 examsDEB

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

Q.1 (a) (i) 5C (0, 2, 4, 5)

Q.9 (a) 10D (0, 4, 6, 8, 10)

(ii) (b) 5B (0, 3, 5) (b) (i) 10D (0, 4, 6, 8, 10) (c) 5C (0, 2, 4, 5) (ii) 5D (0, 2, 3, 4, 5) 20 20 Q.10 (a) (i)

5C (0, 2, 4, 5)

Q.2 (a) 5D (0, 2, 3, 4, 5) (ii) (b) (i)

10C (0, 4, 7, 10)

(b) 5C (0, 2, 4, 5) (ii) 10 (ii) (c) (i)

5C (0, 2, 4, 5)

(ii) Q.11 10C* (0, 4, 7, 10) 20 10 Q.3 (a) 5C (0, 2, 4, 5) Q.12 (a) 5C* (0, 2, 4, 5) (b) 5C (0, 2, 4, 5) (b) 5C (0, 2, 4, 5) (c) 5C (0, 2, 4, 5) (c) 5C* (0, 2, 4, 5) (d) 5C (0, 2, 4, 5) 15 20 Q.13 (a) 10C (0, 4, 7, 10) Q.4 (a) 5B (0, 3, 5) (b) 5B (0, 3, 5) (b) 5B (0, 3, 5) (c)

5C (0, 2, 4, 5)

(c) 5B (0, 3, 5) (d) 15 (e) 5C (0, 2, 4, 5) (f) 5B (0, 3, 5) 30Q.5 (a) 5C (0, 2, 4, 5) (b) 10C* (0, 4, 7, 10) 15 Q.14 (a) 5C (0, 2, 4, 5) (b) (i) 5B (0, 3, 5) (ii) 5B (0, 3, 5) Q.6 (a) 10C* (0, 4, 7, 10) (iii) 5C (0, 2, 4, 5) (b) 5C*(0, 2, 4, 5) (iv) 5B (0, 3, 5) 15 25 Q.7 (a) 5B* (0, 3, 5) Q.15 (a) 5C (0, 2, 4, 5) (b) 5B* (0, 3, 5) (b) 5C (0, 2, 4, 5) (c) 5B (0, 3, 5) (c) 5C (0, 2, 4, 5) 15 15 Q.8 (a) 10C (0, 4, 7, 10) Q.16 (a) Values 10D (0, 4, 6, 8, 10) (b) 5B (0, 3, 5) Graph 5D (0, 2, 3, 4,5 ) (c) 5C (0, 2, 4, 5) (b) 5B* (0, 3, 5) (d) 5B* (0, 3, 5) (c) 5B* (0, 3, 5) 25 (d) 5B* (0, 3, 5) 30

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.

2015.2 J.18/20_MS 4/72 of 71 examsDEB


Mathematics


General Instructions

1. There are 16 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. Question 1 (Suggested maximum time: 5 minutes) (20) 1(a) (i) What is an irrational number? (5C) Answer Any 1: – a real number that cannot be written / represented as

a simple fraction // – a number that cannot be expressed as a fraction p/q

for any integers p and q // – a real number that cannot be expressed as a ratio

of integers // – a number that has a decimal expansion that neither

terminates nor becomes periodic // – a real number that cannot be represented as a

terminating or a repeating decimal // etc.

** Accept other appropriate answers. (ii) π is one of the most identifiable irrational numbers.

The fraction 7

22 is commonly used to approximate the value of π.

To how many decimal places is this approximation correct?

π (approximation) ≅ 7

22

= 3·14285714...

π (exact) = 3·14159265... correct to two decimal places

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Lists example(s) of irrational number. – Incomplete or unsatisfactory explanation.

High partial credit: (4 marks) – Correct explanation, but only one value of π correctly evaluated.

examsDEB

2015.2 J.18/20_MS 5/72 of 71 examsDEB

1(b) (i) The columns in the table below represent the following sets of numbers: Natural Numbers (ℕ), Integer Numbers (ℤ), Rational Numbers (ℚ), Irrational Numbers (ℝ \ ℚ) and Real Numbers (ℝ).

Complete the table by evaluating each of the expressions using a = 3 and indicating with a tick () to show to which set(s) of numbers it belongs. (The first expression has been completed for you.) (10D)

Expression Evaluated Number Sets ℕ ℤ ℚ ℝ \ ℚ ℝ

a (3) () () () () (given)

a2 – a 6

a 1⋅732050...

a – a2 –6

2

1a

a −

9

2 / 0⋅222222...

5

a 0⋅346410...

a2 − a = 32 – 3 = 9 – 3 = 6

a = 3

= 1⋅732050...

a – a2 = 3 – 32 = 3 – 9 = –6

2

1a

a − =

2313 −

= 9

2 or 0⋅222222...

5

a =

5

3

= 5

...7320501⋅

= 0⋅346410...

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – One expression correctly evaluated and set(s) to which it belongs correctly identified

Middle partial credit: (6 marks) – Two or three expressions correctly evaluated and set(s) to which it belongs correctly identified [allow 1 error in identifying set(s) in each case].

High partial credit: (8 marks) – Four expressions correctly evaluated and set(s) to which it belongs correctly identified [allow 1 error in identifying set(s)]. – Five expressions correctly evaluated, but no more than one error in identifying set(s) in each case.

2015.2 J.18/20_MS 6/72 of 71 examsDEB

1(b) (ii) Place each of the numbers evaluated in part (i) in ascending order. (5D)

a2 − a = 32 – 3 = 9 – 3 = 6

a = 3

= 1⋅732050...

a – a2 = 3 – 32 = 3 – 9 = –6

2

1a

a − =

2313 −

= 9

2 or 0⋅222222...

5

a =

5

3

= 5

...7320501⋅

= 0⋅346410...

Order: = –6

9

2 / 0·222222...

5

3 / 0·346410...

3 / 1·732050...

6

** No penalty applied if student includes ‘3’ in ascending numbers correctly.

Scale 5D (0, 2, 3, 4, 5) Low partial credit: (2 marks) – One or two numbers in correct order.

Middle partial credit: (3 marks) – Three or four numbers in correct order.

High partial credit: (4 marks) – All numbers in the correct order, but not written out in ascending order.

2015.2 J.18/20_MS 7/72 of 71 examsDEB

Question 2 (Suggested maximum time: 10 minutes) (20)

U = {1, 2, 3, …, 15}. A is the set of prime numbers less than 15, B is the set of odd numbers less than 15 and C is the set of factors of 15.

2(a) Complete the Venn diagram. (5D)

A = set of prime numbers less than 15 = {2, 3, 5, 7, 11, 13}

B = set of odd numbers less than 15 = {1, 3, 5, 7, 9, 11, 13}

C = set of factors of 15 = {1, 3, 5, 15}

U

A

B

C

1

6

8 12 1410

4

9

15

35

2

7

1113

Scale 5D (0, 2, 3, 4, 5) Low partial credit: (2 marks) – One to five elements correctly placed.

Middle partial credit: (3 marks) – Six to ten elements correctly placed.

High partial credit: (4 marks) – Ten to fourteen elements correctly placed.

2(b) List the elements of each of the following sets: (10C)

(i) A \ (B ∪ C)

A \ (B ∪ C) = {2, 3, 5, 7, 11, 13} \ {1, 3, 5, 7, 9, 11, 13, 15} = {2}

(ii) (A ∪ B ∪ C)′, where (A ∪ B ∪ C)′ is the complement of the set A ∪ B ∪ C.

A ∪ B ∪ C = {1, 2, 3, 5, 7, 9, 11, 13, 15} (A ∪ B ∪ C)′ = {4, 6, 8, 10, 12, 14}

(iii) Find #(A ∪ B ∪ C)′.

(A ∪ B ∪ C)′ = {4, 6, 8, 10, 12, 14} #(A ∪ B ∪ C)′ = 6

** Accept answer for part (iii) consistent with part (ii) if not oversimplified.

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

High partial credit: (7 marks) – Two correct answers.

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2(c) (i) Ellen says that for all sets (A ∪ B) ∪ C = A ∪ (B ∪ C). Is she correct? Give a reason for your answer. (5C)

Answer – yes

Reason – the operation of union in sets is associative – (A ∪ B) ∪ C = {1, 2, 3, 5, 7, 9, 11, 13} ∪ {1, 3, 5, 15} = {1, 2, 3, 5, 7, 9, 11, 13, 15} A ∪ (B ∪ C) = {2, 3, 5, 7, 11, 13} ∪ {1, 3, 5, 7, 9, 11, 13, 15} = {1, 2, 3, 5, 7, 9, 11, 13, 15} ∴ (A ∪ B) ∪ C = A ∪ (B ∪ C)

(ii) Name another set in the Venn diagram above which displays the same property.

Answer – (A ∩ B) ∩ C = A ∩ (B ∩ C)

** Accept answer consistent with 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. – Gives further example only.

High partial credit: (4 marks) – Correct answer with satisfactory explanation, but no further example given.– Correct answer and further example given, but reason incomplete or unsatisfactory.

2015.2 J.18/20_MS 9/72 of 71 examsDEB


150 students went on a school tour to a ski resort in Switzerland. The resort offers three grades of mountain slopes: Easy (E) for beginners, Moderate (M) for more advanced skiers and Difficult (D) for experienced skiers. On their return, the students were asked on which slopes they had skied.

70 students said they had skied on the easy slopes. 60 students said they had skied on the moderate slopes.

55 students said they had skied on the difficult slopes. 10 students said they had skied on the easy and the moderate slopes. 15 students said they had skied on the moderate and the difficult slopes. 12 students said they had skied on the easy and the difficult slopes. All students said they had skied.

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

U = [ 150 ] M = [60 ]

E = [70 ]

D = [55 ]

[ 10 ]�x

[ 70 (10 )(12 ) ]

= [ 48 ]

� �

� � �

�

x

x x

x

[ 60 (10 )(15 ) ]

= [ 35 ]

� �

� � �

�

x

x x

x

[ 55 (12 )(15 ) ]

= [ 28 ]

� �

� � �

�

x

x x

x

[ ]x

[ 12 ]�x

[ 15 ]�x

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

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

3(b) How many students skied on all three grades of mountain slopes? (5C)

(70) + (35 + x) + (15 – x) + (28 + x) = 150 148 + x = 150 x = 150 – 148 = 2

or

(60) + (28 + x) + (12 – x) + (48 + x) = 150 148 + x = 150 x = 150 – 148 = 2

or

(55) + (48 + x) + (10 – x) + (35 + x) = 150 148 + x = 150 x = 150 – 148 = 2

# students skied on all three grades = x = 2

2015.2 J.18/20_MS 10/72 of 71 examsDEB

3(b) (cont’d.)


Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit, e.g. identifies two correct elements. – Identifies sum of elements equals 150.

High partial credit: (4 marks) – Finds (70) + (35 + x) + (15 – x) + (28 + x) = 150 or equivalent, but fails to finish or finishes incorrectly.

3(c) How many students skied on only one grade of mountain slope? (5C)

# students skied on only one grade of slope = (48 + x) + (35 + x) + (28 + x) = 111 + 3x = 111 + 3(2) = 111 + 6 = 117

** Accept answer consistent with parts (a) and (b) if not oversimplified.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit, e.g. identifies two correct elements. – Sum of two or more elements identified in not oversimplified.

High partial credit: (4 marks) – Finds (48 + x) + (35 + x) + (28 + x) or equivalent, but fails to finish or finishes incorrectly.

3(d) Is it reasonable to say that students used this school tour to improve their skiing? Justify your answer. (5C)

Answer – no

Reason # students who were not experienced skiers = 150 – (28 – x) = 150 – (28 – 2) = 150 – 26 = 124

% students who took the opportunity to improve their skiing

= 124

33 ×

1

100

= 26⋅612903... ≅ 26⋅6%

only 26⋅6% of students who were not experienced skiers tried harder level(s)


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. – Correct answer with correct numerator or denominator, but fails to finish or finishes incorrectly.

2015.2 J.18/20_MS 11/72 of 71 examsDEB


(a) Factorise fully 3x2 – 27y2. (5B)

3x2 – 27y2 = 3(x2 – 9y2) = 3(x – 3y)(x + 3y)

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

(b) Factorise fully 6pr – 2qs – 3qr + 4ps. (5B)

6pr – 2qs – 3qr + 4ps = 6pr – 3qr – 2qs + 4ps = 3r(2p – q) + 2s(–q + 2p) = (3r + 2s)(2p – q)

or

6pr – 2qs – 3qr + 4ps = 6pr + 4ps – 2qs – 3qr = 2p(3r + 2s) – q(2s + 3r) = (2p – q)(3r + 2s)

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Any work of merit, e.g. any two terms correctly factorised.

(c) Use factors to simplify the following:

6

32

2

−++xx

xx. (5B)

6

32

2

−++xx

xx =

)2)(3(

)3(

−++xx

xx

= 2−x

x

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Any work of merit, e.g. factorises numerator or denominator correctly.

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** Deduct 1 mark off correct answer only if final answer is not rounded or incorrectly rounded or for the omission of or incorrect units for parts of question asterisked - this deduction should be applied only once throughout the question.

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

A company has a policy of renewing the company cars it supplies to its employees every two years. The company depreciates the value of the cars at a compound rate of 12⋅5% per annum in its accounts.

5(a) Mark receives a new company car which cost €40,000. How much will the car be worth at the end of two years? (5C)

F = P(1 + i)t = 40,000(1 – 0⋅125)2 = 40,000(0⋅875)2 = 40,000(0⋅765625) = €30,625

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

High partial credit: (4 marks) – Correctly substitutes in relevant formula, but fails to finish or finishes incorrectly.

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

or

Year 1 Depreciation = 40,000 × 12⋅5%

= 40,000 × 100

512⋅

= €5,000

or = 4,000 × 0·125

Value = 40,000 – 5,000 = €35,000


= 35,000 × 100

512⋅

= €4,375

or = 35,000 × 0·125

Value = 35,000 – 4,375 = €30,625

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit, e.g. find depreciation for first year, value at the end of first year.

High partial credit: (4 marks) – Finds depreciation for second year, but fails to find or finds incorrect value at the end of second year.


2015.2 J.18/20_MS 13/72 of 71 examsDEB

5(b) At the end of the second year, Mark may choose to keep the car for an additional year and after that he may purchase it from the company for two-thirds of its then value. If Mark takes this option, how much will it cost the company in the third year? (10C*)

Book value of car before it is sold

F = P(1 + i)t = 30,625(1 – 0⋅125)1 = 30,625(0⋅875) = €26,796⋅875 ≅ €26,796⋅88

or

F = P(1 + i)t = 40,000(1 – 0⋅125)3 = 40,000(0⋅875)3 = 40,000(0⋅669921875) = €26,796⋅875 ≅ €26,796⋅88

or


= 30,625 × 100

512⋅

= €3,828⋅125

or = 30,625 × 0·125

Value = 30,625 – 3,828⋅125 = €26,796⋅875 ≅ €26,796⋅88

Loss on sale of car

Loss = (1 – 3

2)(26,796⋅88)

= 3

1(26,796⋅88)

= 8,932⋅293333... ≅ €8,932⋅29

Cost to the company in the third year

Cost = (30,625 – 26,796⋅88) + 8,932⋅29 = 3,828⋅12 + 8,932⋅29 ≅ €12,760⋅41


Scale 10C* (0, 4, 7, 10) Low partial credit: (4 marks) – Any work of merit. – Finds book value of car before it is sold.– Finds loss on sale of car if book value incorrect.

High partial credit: (7 marks) – Finds book value of car before it is sold and loss on sale of car, but fails to finish or finishes incorrectly.

* Deduct 1 mark off correct answer only if not rounded or incorrectly rounded - apply only once throughout the question.


2015.2 J.18/20_MS 14/72 of 71 examsDEB



6(a) Solve for x:

5x2 – 7x – 10 = 0,

giving your answers correct to two decimal places. (10C*)

x = a

acbb

2

42 −±−

x = )5(2

)10)(5(4)7()7( 2 −−−±−−

= 10

200497 +±

= 10

2497 ±

= 10

...779733157 ⋅±

x = 10

...779733157 ⋅+, x =

10

...779733157 ⋅−

= 10

...77973322⋅ =

10

...7797338⋅−

= 2⋅277973... = –0⋅877973... ≅ 2⋅28 ≅ –0⋅89

Scale 10C* (0, 4, 7, 10) Low partial credit: (4 marks) – Any work of merit, e.g. writes down ‘–b’ formula.

High partial credit: (7 marks) – Full correct substitution into ‘–b’ formula,

i.e. x = )5(2

)10)(5(4)7()7( 2 −−−±−−.

– Both factors correct [not looking for ‘=0’].– Finds only one value of x (2⋅28 or –0⋅89).


6(b) Hence, or otherwise, solve 5(t2 – 2)2 – 7(t2 – 2) – 10 = 0. Give your answers correct to two decimal places. (5C*)

Let x = t2 – 2

t2 – 2 = 2⋅28 t2 – 2 = –0⋅89 t2 = 2⋅28 + 2 t2 = –0⋅89 + 2 t2 = 4⋅28 t2 = 1⋅11 t = ± 284⋅ t = ± 111⋅ = ±2⋅068816... = ±1⋅053565... ≅ ±2⋅07 ≅ ±1⋅05

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

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit, e.g. writes down x = t2 – 2 or equivalent. – Finds t2 = 4⋅28 or t2 = 1⋅11.

High partial credit: (4 marks) – Correct method with minor errors. – Finds only one value of t in each case or both values of t in one case only.


2015.2 J.18/20_MS 15/72 of 71 examsDEB



7(a) In the celebrity TV programme Splash!, Keith Duffy completed a dive from a platform, 30 m high. The speed of his dive was timed at 54 km/h. Calculate the time taken for Keith to complete the dive. (5B*)

Speed (km/h) = 54

Speed (m/s) = 6060

000,154

××

= 15 m/s

Speed = time

distance

Time = speed

distance

= 15

30

= 2 s

Scale 5B* (0, 3, 5) Partial credit: (3 marks) – Any work of merit, e.g. writes speed formula with or without manipulation. – Finds speed correctly in terms of m/s. – Some correct substitution into formula.

* Deduct 1 mark off correct answer only for the omission of or incorrect units - apply only once throughout the question.

7(b) Olympic diver, Tom Daly, repeated the same dive.

It took him 13

2 seconds from the instant he left the platform to hit the water.

Calculate the average speed of Tom’s dive in kilometres per hour. (5B*)

Speed = time

distance

=

3

21

30

= 5

330 ×

= 5

90

= 18 m/s

Speed (m/s) = 18

Speed (km/h) = 000,1

606018 ××

= 64⋅8 km/h

Scale 5B* (0, 3, 5) Partial credit: (3 marks) – Any work of merit, e.g. some correct substitution into speed formula with or without manipulation. – Finds speed correctly in terms of m/s, but fails to convert to km/h.


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7(c) Find the ratio of the average speeds of the two divers, giving your answer in its simplest form. (5B)

Ratio of speed = 54 : 64⋅8 (×5) = 270 : 324

270 = 3 × 3 × 3 × 2 × 5 324 = 3 × 3 × 3 × 2 × 6

Ratio of speed = 5 : 6

or


15 = 3 × 5 18 = 3 × 6


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

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Any work of merit e.g. finds the prime factors of 270, 324, 15 or 18. – Multiplies ratio by 5, but fails to factorise [ans. 270 : 324]. – Answer not given in its simplest form.

2015.2 J.18/20_MS 17/72 of 71 examsDEB



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

The Doyle family of two adults and four children plans to visit the Titanic Belfast museum. The total cost of admission for the whole family is £60. On the same day, the Ryan family of five adults and two children will pay £92 to visit the museum.

8(a) Find the cost of admission for an adult and the cost of admission for a child to the museum. (10C)

Doyle f.: 2A + 4C = 60 Ryan f.: 5A + 2C = 92

2A + 4C = 60 (×1) 5A + 2C = 92 (×–2)

2A + 4C = 60 –10A – 4C = –184 –8A = –124

A = 8

124

−−

= £15⋅50

2A + 4C = 60 2(15⋅50) + 4C = 60 31 + 4C = 60 4C = 60 – 31 = 29 C = £7⋅25

or 5A + 2C = 92 5(15⋅50) + 2C = 92 77⋅50 + 2C = 92 2C = 92 – 77⋅50 = 14⋅50 C = £7⋅25

or

2A + 4C = 60 (×5) 5A + 2C = 92 (×–2)

10A + 20C = 300 –10A – 4C = –184 16C = 116

C = 16

116

= £7⋅25

2A + 4C = 60 2A + 4(7⋅25) = 60 2A + 29 = 60 2A = 60 – 29 = 31 A = £15⋅50

or 5A + 2C = 92 5A + 2(7⋅25) = 92 5A + 14⋅50 = 92 5A = 92 – 14⋅50 = 77⋅50 A = £15⋅50

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 A or C term. – Finds –8A = –124 or 16C = 116, but fails to finish or finishes incorrectly. – Finds either variable (A or C) correctly by trial and error, but fails to verify in both equations or verifies incorrectly.

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

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8(b) Verify your answers. (5B)

A = £15⋅50 C = £7⋅25

Doyle family 2A + 4C = 60

2A + 4C = 2(15⋅50) + 4(7⋅25) = 31 + 29 = 60

Ryan family 5A + 2C = 92

5A + 2C = 5(15⋅50) + 2(7⋅25) = 77⋅50 + 14⋅50 = 92

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

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Any work of merit, e.g. correct substitution into one equation. – Verifies one equation only.


8(c) The museum offers a family rate of £39 for two adults and two children. How much would each family save if they avail of this special family rate? (5C)

Doyle family 2A + 4C = (2A + 2C) + 2C = 39 + 2(7⋅25) = 39 + 14⋅50 = €53⋅50

Saving = 60 – 53⋅50 = €6⋅50

Ryan family 5A + 2C = (2A + 2C) + 3A = 39 + 3(15⋅50) = 39 + 46⋅50 = €85⋅50

Saving = 92 – 85⋅50 = €6⋅50


Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit, e.g. correct manipulation of one equation [e.g. 2A + 4C = (2A + 2C) + 2C]. – Finds reduced cost for one family only.

High partial credit: (4 marks) – Finds reduced cost for both families. – Finds savings for one family only.


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8(d) Emma Doyle availed of this special family rate and pays the admission fees for her family by credit card. Later, she notices the amount included on her credit card bill for the museum visit was €67⋅21. Assuming no commission was charged on the transaction, calculate the exchange rate on that day from sterling (£) to euro (€), correct to the nearest cent. (5B*)

Cost (in £) = 2A + 4C = (2A + 2C) + 2C = 39 + 2(7⋅25) = 39 + 14⋅50 = €53⋅50

Cost (in €) = €67⋅21

Exchange rate = £)(inCost

€)(inCost

= 5053

2167

⋅⋅

= 1⋅256261... ≅ 1⋅26 £1 (sterling) ≡ €1⋅26


Scale 5B* (0, 3, 5) Partial credit: (3 marks) – Any work of merit, e.g. writes formula to calculate exchange rate. – Finds inverse of exchange rate correctly,

i.e. 2167

5053

⋅⋅

= 0⋅796012... ≅ 0⋅80.

– Finds 5053

2167

⋅⋅

, but fails to finish or finishes

incorrectly.



2015.2 J.18/20_MS 20/72 of 71 examsDEB


9(a) Solve the following inequality and show the solution on the number line.

–14 < 2 – 4x < 10, x ∈ ℤ. (10D)

Solution

–14 < 2 – 4x 4x < 2 + 14 4x < 16 x < 4

and 2 – 4x < 10 –4x < 10 – 2 < 8

x > 4

8

−

> –2

–2 < x < 4 x = {–1, 0, 1, 2, 3}

or

–14 < 2 – 4x < 10 subtract 2 –14 – 2 < 2 – 4x – 2 < 10 – 2 –16 < 4x < 8

–16 < 4x < 8 divide by –4

4

16

−−

> 4

4

−x

> 4

8

−

4 > x < –2

–2 < x < 4 x = {–1, 0, 1, 2, 3}

Number line

0 1 2 3 4 5�1�2�3

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

Middle partial credit: (6 marks) – Finds x < 4 or x > –2 (accept with or without inequality sign).

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

9(b) Simplify (3x2 – 7)(5x – 2). (5B)

(3x2 – 7)(5x – 2). = 3x2(5x – 2) – 7(5x – 2) = 15x3 – 6x2 – 35x + 14

Scale 5B (0, 3, 5) Low partial credit: (3 marks) – Any work of merit, e.g. two terms correctly evaluated. – Correct solution, but with error(s) in signs.

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9(c) Divide 4x3 – 4x2 – 7x – 20 by 2x – 5. (5C)

2x2 + 3x + 4 2x – 5 ) 4x3 – 4x2 – 7x – 20 –4x3 + 10x2 6x2 – 7x – 20 –6x2 + 15x 8x – 20 –8x – 20 0

or

2x2 3x 4 2x 4x3 6x2 8x –5 –10x2 –15x –20

Answer – 2x2 + 3x + 4

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit, e.g. sets up division.– One term of the answer correctly identified. – Inserts the 4x3 or –20 term correctly in Method .

High partial credit: (4 marks) – Substantial work, but with one or more critical omissions.

2015.2 J.18/20_MS 22/72 of 71 examsDEB


10(a) A Pythagorean triple is a set of three positive integers that can form a right-angled triangle.

(i) Show that the three numbers 20, 21 and 29 form a Pythagorean triple. (5C)

| Opp |2 + | Adj |2 = | Hyp |2 ... Pythagoras’s theorem 202 + 212 = 292 400 + 441 = 841 841 = 841

(ii) List two other examples of Pythagorean triples.

Examples Any 2: – 3, 4, 5 // – 5, 12, 13 // – 6, 8, 10 // – 7, 24, 25 // – 8, 15, 17 // – 9, 12, 15 // – 9, 40, 41 // – 11, 60, 61 // – 12, 16, 20 // – 12, 35, 37 // – 13, 84, 85 // etc.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit, e.g. writes down formula for Pythagoras’s theorem. – Some correct substitution into formula for Pythagoras’s theorem. – Lists examples of Pythagorean triples only.

High partial credit: (4 marks) – Full correct substitution into formula for Pythagoras’s theorem, but fails to finish or finishes incorrectly. – First part fully correct, but no examples of Pythagorean triples given.

10(b) The diagram shows a right-angled triangle which has a base length of 4 + 5 and

a perpendicular height of 4 – 5 .

Calculate the value of h, giving your answer in surd form. (5C)

| Opp |2 + | Adj |2 = | Hyp |2 | h |2 = | 4 + 5 |2 + | 4 – 5 |2

= 16 + 8 5 + 5 + 16 – 8 5 + 25 = 62 | h | = 62

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit, e.g. some correct substitution into formula for Pythagoras’s theorem.

High partial credit: (4 marks) – Finds | h |2 = 62, but fails to find | h | or finds | h | incorrectly.

54 �

54 �

h

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Mo Farah is the current Olympic and World champion at both the 5,000 m and 10,000 m distance events.

He has accepted an invitation to run at a local athletics meeting on the track shown in the diagram which does not conform to international competition regulations.

How many laps of the track will the 10,000 m race need to be? Show the two approximate positions of the end of the race on the diagram. (10C*)

Perimeter = 2πr + 2(112⋅4)

= 2π(2

571⋅) + 224⋅8

= 71⋅5π + 224⋅8 = 224⋅623874... + 224⋅8 = 449⋅423874... m

Number of laps = ...423874449

000,10

⋅

= 22⋅250709... ≅ 22⋅25

Approximate positions of the end of the race

112 4 m·

71

m·5

Start

Scale 10C* (0, 4, 7, 10) Low partial credit: (4 marks) – Any work of merit, e.g. correct relevant formula for perimeter of track. – Finds perimeter of track (rounded or not).

High partial credit: (7 marks) – Finds perimeter of track (rounded or not).– Correct answer, but fails to indicate both possible end positions of the race.

* Deduct 1 mark off correct answer only if not rounding or incorrectly rounded - apply only once throughout the question.

112 4 m·

71

m·5

Start

2015.2 J.18/20_MS 24/72 of 71 examsDEB



Niamh is a young 100 m sprinter. During training and competition, her speed is regularly monitored and analysed using the formula:

d = 2

)( tvu +,

where d is distance, u is initial speed, v is end speed and t is the time interval.

12(a) When Niamh practises speed testing, her speed is recorded at five-second time intervals. During one time interval, Niamh’s initial speed was clocked at 4 m/s and her end speed was clocked at 7 m/s.

What distance did she run in that time interval? (5C*)

d = 2

)( tvu +

= 2

)5)(74( +

= 2

55

= 27⋅5 m

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit. – Some correct substitution into formula.

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


12(b) Write v in terms of d, u and t. (5C)

d = 2

)( tvu +

2d = (u + v)t = ut + vt vt = 2d – ut

v = t

utd −2

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

High partial credit: (4 marks) – Finds vt = 2d – ut and fails to finish or finishes incorrectly.

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12(c) Given that the 100 m event starts from rest, find the minimum end speed at which Niamh must cross the line to finish a race in under 12⋅5 seconds. (5C*)

v = t

utd −2

= 512

)512)(0()100(2

⋅⋅−

= 512

200

⋅

= 16 m/s

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

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit. – Some correct substitution into formula.



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Adam and Laura are brother and sister who receive pocket money each week. Adam receives €10 per week and Laura receives €15 per week.

The amount of pocket money each receives increases depending on the number of chores completed at home during that week, as shown in the table below.

Chores completed

Adam (€)

Laura (€)

0 10 15

1 13 17

2 16 19

3 19 21 4 22 23 5 25 25

13(a) Complete the table above and draw a graph to show the relationships between the number of chores completed and the amount of pocket money each person receives. (10C)

Table – see above

Graph

Chores Completed

Am

ount

of

Pock

etM

oney

Rec

eived

(�)

1 2 3 4 5

10

15

25

5

20Laura

Adam

2015.2 J.18/20_MS 27/72 of 71 examsDEB

13(a) (cont’d.)

** Accept graph for 20 ≤ x ≤ 80 for full credit - no necessity to extend lines.

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – One point correctly evaluated and correctly plotted. – Two points correctly evaluated. – Two points correctly plotted (including points given). – Co-ordinates reversed on graph and missing point(s).

High partial credit: (7 marks) – Six ‘new’ points correctly evaluated, but only 11 points correctly plotted (joined, incorrectly joined or not joined). – Six ‘new’ points correctly evaluated and only these points correctly plotted. – Six ‘new’ points correctly evaluated and end points plotted for each line. – Six ‘new’ points correctly evaluated and points for only one line correctly plotted.– Six ‘new’ points correctly evaluated and co-ordinates of all points reversed. – Treat any other type of graph (e.g. bar chart) as points plotted, but not joined.

13(b) Use your graph to find the number of chores each person must complete in that week in order that both receive the same amount of pocket money. (5B)

Adam – 5 (chores) Laura – 5 (chores)

** Accept answers based on fully plotted graph.

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Any work of merit, e.g. vertical line drawn from point on intersection of two graphs to horizontal axis, but not labelled.

13(c) Write down a formula to represent the amount of pocket money that Adam receives. State clearly the meaning of any letters you use in your formula. (5C)

Adam p = 10 + 3c p = pocket money

c = number of chores completed

13(d) Write down a formula to represent the amount of pocket money that Laura receives. State clearly the meaning of any letters you use in your formula.

Laura p = 15 + 2c p = pocket money

c = number of chores completed

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit, e.g. identifies rate of increase of one line [10 + 3c or 15 + 2c].

High partial credit: (4 marks) – One equation fully correct.

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13(e) Use your formulae from parts (c) and (d) to verify the answer that you gave to part (b) above. (5C)

Adam p = 10 + 3c 10 + 3c = 25 3c = 25 – 10 = 15

c = 3

15

= 5

Laura p = 15 + 2c 15 + 2c = 25 2c = 25 – 15 = 10

c = 2

10

= 5

** Accept answers from parts (c) and (d) if not oversimplified.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit, e.g. substitutes any value (excluding 25) from table into one correct formula.

High partial credit: (4 marks) – Substitutes 25 into one correct formula and gets correct value of c.

13(f) Which person do you think fares better? Give a reason for your answer. (5B)

Answer – Adam

Reason – he receives more money than Laura each week if both of them do more than five chores in that week // etc.

** Accept other appropriate answers.

or

Answer – Laura

Reason – she receives more money than Adam each week if both of them do fewer than five chores in that week // etc.


Scale 5B (0, 3, 5) Partial credit: (3 marks) – Incomplete or unsatisfactory reason.

2015.2 J.18/20_MS 29/72 of 71 examsDEB


14(a) The table below shows the growth, in cm, of a plant over a 4-week period.

Week 1 2 3 4

Growth (cm) 3 10 21 36

Is the pattern of growth shown in the table linear, quadratic or exponential? Explain your answer. (5C)

Answer – quadratic

Reason – second difference is constant

Week Growth (cm) 1st Diff. 2nd Diff.

1 3

7

2 10 4 11

3 21 4 15

4 36


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

14(b) The first three stages of a pattern are shown below. Each stage of the pattern is made up of square tiles.

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

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

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14(b) (ii) How many tiles are there in Stage 10 of the pattern? (5B)

Stage 1 2 3 4 5 6 7 8 9 10 # Tiles 4 6 8 10 12 14 16 18 20 22

Answer – 22

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Number of tiles in sixth or subsequent stage(s) identified.

(iii) Find a general formula for the number of tiles in Stage n of the pattern, where x ∈ ℕ. (5C)

a = 4

d = T2 – T1 = 6 – 4 = 2

Tn = a + (n – 1)d = 4 + (n – 1)2 = 4 + 2n – 2 = 2n + 2

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit, e.g. identifies first term or common difference, writes down general formula for nth term.

High partial credit: (4 marks) – Some correct substitution into correct general formula for nth term.

(iv) In which stage does the pattern consist of 250 tiles? (5B)

Tn = 2n + 2 = 250 2n + 2 = 250 2n = 250 – 2 = 248 n = 124 124th stage consists of 250 blocks

** Accept answers from part (iii) if not oversimplified.

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Any work of merit, e.g. formula for nth term equal to 250. – Minor error(s) in calculations.

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Let g be the function g : x ׀→ x2 – 3, where x ∈ ℝ.

15(a) Find the value of g(–2). (5C)

g(x) = x2 – 3 g(–2) = (–2)2 – 3 = 4 – 3 = 1

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


15(b) Express g(2t – 1) in terms of t. (5C)

g(x) = x2 – 3 g(2t – 1) = (2t – 1)2 – 3 = 4t2 – 4t + 1 – 3 = 4t2 – 4t – 2

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit, e.g. some correct substitution into given function.


15(c) Hence, find the values of t for which g(2t – 1) = g(–2). (5C)

g(2t – 1) = g(–2) 4t2 – 4t – 2 = 1 4t2 – 4t – 2 – 1 = 0 4t2 – 4t – 3 = 0 (2t – 3)(2t + 1) = 0 2t – 3 = 0 2t = 3

t = 2

3

or 2t + 1 = 0 2t = –1

t = –2

1

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

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit, e.g. correctly equates g(2t – 1) = g(–2) to find quadratic equation.

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

2015.2 J.18/20_MS 32/72 of 71 examsDEB



Irish golfer Rory McIlroy is currently the highest ranked player in the world.

A group of students have analysed his swing technique. They have come up with a function to predict the height of his golf ball above ground level, after he hits his tee shot.

The height, in metres, of the ball after t seconds is given by h : t 36 →׀t – 6t2, in the domain 0 ≤ t ≤ 6, where t ∈ ℝ.

16(a) On the grid below, draw the graph of y = h(t) in the domain 0 ≤ t ≤ 6. Values (10D)

Table (1st method)

t = 0 1 2 3 4 5 6 36t 0 36 72 108 144 180 216 – 6t2 0 –6 –24 –54 –96 –150 –216 y = h(t) 0 30 48 54 48 30 0

or Table (2nd method)

h(t) = 36t – 6t2 = y h(0) = 0 0 = 0 h(1) = 36 –6 = 30 h(2) = 72 –24 = 48 h(3) = 108 –54 = 54 h(4) = 144 –96 = 48 h(5) = 180 –150 = 30 h(6) = 216 –216 = 0

or Substitution method

h(t) = 36t − 6t2 h(0) = 36(0) – 6(0)2 = 0

h(1) = 36(1) – 6(1)2 = 30

h(2) = 36(2) – 6(2)2 = 48

h(3) = 36(3) – 6(3)2 = 54

h(4) = 36(4) – 6(4)2 = 48

h(5) = 36(5) – 6(5)2 = 30

h(6) = 36(6) – 6(6)2 = 0

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

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

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

2015.2 J.18/20_MS 33/72 of 71 examsDEB

16(a) (cont’d.)

Graph (5D)

Points (0, 0), (1, 30), (2, 48), (3, 54), (4, 48), (5, 30), (6, 0)

** Accept values calculated from previous work (six 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 point, (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) – Two or three points, (x, y) or (y, x), correctly plotted, labelled or not (joined or unjoined).

High partial credit: (4 marks) – Four or five points, (x, y) or (y, x), correctly plotted, labelled or not (joined or unjoined). – Six points, (x, y) or (y, x), correctly plotted, but joined together with inappropriate curve.

44 6·

1 75· 4 25·

6 4·

h t( )

Time in seconds, t

1 2 3 4 5 6

Hei

ght

of

Golf

Bal

lin

met

res

10

�10

�16

20

30

40

50

60

2015.2 J.18/20_MS 34/72 of 71 examsDEB

For parts (b), (c) and (d), you must show your working out on the diagram.

16(b) Use your graph to estimate the height of the ball after 1⋅75 seconds. How long will it take before the golf ball reaches this height again? (5B*)

Height – 44⋅6 m

Time – 4⋅25 s


Scale 5B* (0, 3, 5) Partial credit: (3 marks) – One correct answer. – Any work of merit, e.g. draws vertical line from 1⋅75 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.

* Deduct 1 mark off correct answer only for omission of or incorrect units - this deduction should be applied only once throughout the question.

16(c) Imagine Rory took the same tee shot again from an elevated position 16 m above ground level. By extending your graph, estimate the time it would take for the golf ball to hit the ground. (5B*)

Time – 6⋅4 s


Scale 5B* (0, 3, 5) Partial credit: (3 marks) – Any work of merit, e.g. extends graph to y = –16 or draws horizontal line at y = –16 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.

* Deduct 1 mark off correct answer only for omission of or incorrect units - this deduction should be applied only once throughout the question.

16(d) Write down a formula to represent the height of the golf ball above ground level after t seconds for this shot. State clearly the meaning of any letters you use in your formula. (5B*)

Answer – h(t) = 36t – 6t2 + 16

h = height of golf ball above ground level, in metres t = time after ball is hit, in seconds

Scale 5B* (0, 3, 5) Partial credit: (3 marks) – Any work of merit, e.g. identifies ±16 as constant in equation without writing down formula. – Gives final answer as h(t) = 36t – 6t2 – 16.

* Deduct 1 mark off correct answer only for omission of or incorrect meaning of letters used in formula - this deduction should be applied only once throughout the question.

2015.2 J.18/20_MS 35/72 of 71 examsDEB

Notes:

2015.2 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, 5, 10 0, 4, 7, 10 0, 4, 6, 8, 10 15 mark scale 0, 7, 15 0, 6, 11, 15 0, 6, 10, 13, 15

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 (four 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 per question. The * penalty is not applied to currency solutions.

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

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

examsDEB

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

Q.1 (a) (i) 5C* (0, 2, 4, 5) Q.6 (a) 5B (0, 3, 5) (ii) 5C* (0, 2, 4, 5) (b) 5B (0, 3, 5) (iii) 5C* (0, 2, 4, 5) (c) 10D* (0, 4, 6, 8, 10) (iv) 5C* (0, 2, 4, 5) 20 (b) (i) 5C* (0, 2, 4, 5) (ii) 5C* (0, 2, 4, 5) (iii) 5B (0, 3, 5) Q.7 (a) 5B (0, 3, 5) (c) (i) 10D (0, 4, 6, 8, 10) (b) 5B (0, 3, 5) (ii) 5C* (0, 2, 4, 5) (c) 5C* (0, 2, 4, 5) 50 (d) 5C* (0, 2, 4, 5) 20 Q.2 (a) 5B (0, 3, 5) (b) 5C (0, 2, 4, 5) Q.8 (a) 5C (0, 2, 4, 5) (c) 5C* (0, 2, 4, 5) (b) 5C (0, 2, 4, 5) (d) 5B (0, 3, 5) (c) 5C (0, 2, 4, 5) (e) 5C (0, 2, 4, 5) (d) 10C (0, 4, 7, 10) (f) 5C (0, 2, 4, 5) (e) 10C (0, 4, 7, 10) 30 (f) 5C* (0, 2, 4, 5) 40 Q.3 (a) (i) 5D (0, 2, 3, 4, 5) (ii) 5C (0, 2, 4, 5) Q.9 (a) (i)

5D (0, 2, 3, 4, 5)

(b) (i)

5B (0, 3, 5)

(ii) (ii) (iii) (c) 5B (0, 3, 5) (b) 10C (0, 4, 7, 10) (d) 5C (0, 2, 4, 5) 15 (e) 5C (0, 2, 4, 5) (f) 5C (0, 2, 4, 5) (g) 5C (0, 2, 4, 5) Q.10 (a) (i)

5B (0, 3, 5)

40 (ii) (b) 10D* (0, 4, 6, 8, 10) (c) 5B (0, 3, 5) Q.4 (a) (i)

10D* (0, 4, 6, 8, 10)

20 (ii) (b) (i) 5C* (0, 2, 4, 5) (ii) 5B (0, 3, 5) Q.11 (a) 15D (0, 6, 9, 12, 15) 20 (b) 5B* (0, 3, 5) 20 Q.5 (a) (i)

5B (0, 3, 5)

(ii) Q.12 10C (0, 4, 7, 10) (b) 10C* (0, 4, 7, 10) 10 15

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.

2015.2 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.




1(a) The diagram below shows the plan for a rectangular park with sides measuring 75 m and 30 m. The park will consist of a grass lawn, a skateboarding zone with a dyed tarmacadam surface, in the shape of an isosceles triangle, and a 3 m-wide footpath around the edge of the park.

75 m

30 m

SkateboardingZone

Grass LawnFootpath

(i) Find the area of the skateboarding zone. (5C*)

Area of Δ = 2

1 × | base | × | ⊥height |

= 2

1 × | 30 – 3 – 3 | × | 30 – 3 – 3 |

= 2

1 × 24 × 24

= 2

576

= 288 m2

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit, e.g. correct area formula for triangle (no numbers).

High partial credit: (4 marks) – Correct substitution into relevant formulae.


examsDEB

2015.2 J.18/20_MS 39/72 of 71 examsDEB

1(a) (ii) Find the area of the grass lawn. (5C*)

Area of grass lawn = [(75 – 3 – 3) × (30 – 3 – 3)] – 288 = [69 × 24] – 288 = 1,656 – 288 = 1,368 m2

or

Area of grass lawn = [(30 – 3 – 3) × (75 – 24 – 3 – 3)] + 288 = [(30 – 6) × (75 – 30)] + 288 = (24 × 45) + 288 = 1,080 + 288 = 1,368 m2

or

Area of grass lawn = 2

1(30 – 3 – 3)[(75 – 3 – 3) + (75 – 24 – 3 – 3)]

= 2

1(24)(69 + 45)

= 2

1(24)(114)

= 2

736,2

= 1,368 m2

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit, e.g. states correct area formula for grass lawn (no numbers).



(iii) Find the area of the footpath around the edge of the park. (5C*)

Area of footpath = (75 × 30) – (1,368 + 288) = 2,250 – 1,656 = 594 m2

or

Area of footpath = 2(75 × 3) + 2[(30 – 3 – 3) × 3] = 2(225) + 2(24 × 3) = 450 + 2(72) = 450 + 144 = 594 m2

or

Area of footpath = (75 × 30) – [(75 – 3 – 3) × (30 – 3 – 3)] = 2,250 – (69 × 24) = 2,250 – 1,656 = 594 m2

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit, e.g. states correct area formula for footpath (no numbers).



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1(a) (iv) A safety fence is to be installed around the perimeter of the skateboarding zone. Find the total length of fencing required. Give your answer in metres, correct to the nearest whole number. (5C*)

Perimeter = 24 + 24 + Hypotenuse

| Hyp |2 = | Opp |2 + | Adj |2 ... Pythagoras’s theorem = 242 + 242 = 576 + 576 = 1,152 | Hyp | = 152,1

= 33⋅941125...

Perimeter = 24 + 24 + 33⋅941125... = 81⋅941125... ≅ 82 m

or

Perimeter = 24 + 24 + Hypotenuse

cos = |Hyp|

|Adj|

cos 45° = |Hyp|

24 ... likewise sin 45°

| Hyp | = °45cos

24

= ...7071060

24

⋅

= 33⋅941125...

Perimeter = 24 + 24 + 33⋅941125... = 81⋅941125... ≅ 82 m

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

High partial credit: (4 marks) – Finds | Hyp | or | Hyp |2, but fails to finish (i.e. find perimeter) or finishes incorrectly.

* Deduct 1 mark off correct answer only if not rounded or incorrectly rounded for the omission of or incorrect units - apply only once throughout the question.

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1(b) The skateboarding zone is to include a half-pipe made from concrete with a rubber membrane fitted to its curved surface as well as the two top landing areas either side of the curved surface. The membrane will protect the surface of the half-pipe against weathering and wearing due to skateboards. The half-pipe comprises of a rectangular cuboid with a semi-cylinder removed, as shown. The dimensions are shown in the diagram below.

8 m

12 m

5 m

(i) Find the volume of concrete used in the construction of the half-pipe. Give your answer correct to one decimal place. (5C*)

Volume of concrete = Volume of block – volume of semi-cylinder

= (L × W × H) – 2

1(πr2h)

= (8)(12)(3) – 2

1 π(2⋅5)2(12)

= 288 – 2

75 π

= 288 – 117⋅809724... = 170⋅190275... ≅ 170⋅2 m3

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit, e.g. writes one correct volume formula, uses incorrect relevant formula correctly. – Some correct substitution into relevant formula.


* Deduct 1 mark off correct answer only if not rounded or incorrectly rounded for the omission of or incorrect units - apply only once throughout the question.

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1(b) (ii) Find the total surface area that is to be covered by the rubber membrane. Give your answer correct to one decimal place. (5C*)

Total surface area = 2(L × W) + 2

1(2πrh)

= 2[12 × 2

1(8 – 5)] +

2

1(2π)(2⋅5)(12)

= 2(12)(1⋅5) + 30π = 36 + 30π = 36 + 94⋅247779... = 130·247779... ≅ 130·2 m2

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit, e.g. writes correct surface area formula, uses incorrect relevant formula correctly. – Some correct substitution into relevant formula.


* Deduct 1 mark off correct answer only if not rounded, incorrectly rounded, for the omission of or incorrect units - apply only once throughout the question.

(iii) A scaled diagram of the skateboarding zone is shown on the grid below. Each grid box represents a 2 m × 2 m square.

On the diagram, indicate a suitable location for the half-pipe. (5B)

Answer – See diagram above

** Accept other possible answers.

** Accept area indicated on diagram without sides labelled if both shown and to scale.

Scale 5B (0, 3, 5) Partial credit: (1 mark) – Any work of merit, e.g. one side (8m or 12m) correctly drawn to scale and labelled within the triangular area. – Indicate suitable position for 5m × 8m area within the triangular area.

SkateboardingZone

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1(c) A building company wishes to put in a bid for the construction of the park. The company’s costs for the different elements of the project (inclusive of labour) are shown in the table below.

Item Cost Dyed tarmacadam for the skateboarding zone €17 per m2

Green lawn (including planting of trees and shrubs) €110 per m2

Tarmacadam for footpath €15 per m2

Concrete for half-pipe €45 per m3

Rubber membrane for half-pipe €22 per m2

Fencing for perimeter of the skateboarding zone €14 per m

(i) Using your answers from parts (a) and (b), calculate the company’s total cost to construct the park. (10D)

Dyed tarmacadam = [288 – (8 × 12)] × 17 = 192 × 17 = €3,264

Green lawn = 1,368 × 110 = €150,480

Footpath = 594 × €15 = €8,910

Half-pipe = 170⋅2 × 45 = €7,659

Rubber membrane = 130⋅2 × 22 = €2,864⋅40

Fencing = 82 × 14 = €1,148

Total cost = 3,264 + 150,480 + 8,910 + 7,659 + 2,864⋅40 + 1,148 = €174,325⋅40

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – One or two correct costs.

Middle partial credit: (6 marks) – Three, four or five correct costs.

High partial credit: (8 marks) – Six correct costs, but total cost omitted or incorrect. – Five correct costs, but added correctly.

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

(ii) The building company put in a successful bid of €210,000 for the construction of the park. Find the percentage profit that the company will make if it completes the construction for the cost calculated in part (c)(i). Give your answer correct to one decimal place. (5C*)

Profit = 210,000 − 174,325⋅40 = €35,674⋅60

% Profit = PriceCost

Profit ×

1

100

= 40325,174

60674,35

⋅⋅

× 1

100

= 0⋅204643... × 100 = 20⋅464372...% ≅ 20⋅5%

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit, e.g. finds profit without attempting to evaluate % Profit.

High partial credit: (4 marks) – Finds 40325,174

60674,35

⋅⋅

or 40325,174

60674,35

⋅⋅

× 1

100,

but fails to finish or finishes incorrectly.


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

2015.2 J.18/20_MS 44/72 of 71 examsDEB


** Accept correct answers with or without work shown. ** Deduct 1 mark off correct answer only if final answer is not rounded or incorrectly rounded or for the omission

of or incorrect units for parts of question asterisked - this deduction should be applied only once throughout the question.

Claire carried out a survey in her school and then displayed the results in the two-way table below.

Right-handed Left-handed

Girls 186 34

Boys 203 37

2(a) Based on the above data, write the two questions Claire asked in the survey in a suitable form. (5B)

Question Any 1: – Are you male or female? // – Put a tick () in one box below to indicate your gender.

Male Female // – Are you a boy or a girl? // etc.


Question Any 1: – Are you right-handed or left-handed? // – Put a tick () in one box below to indicate whether you

are right-handed or left-handed. Right-handed Left-handed //

– Do you write with your left hand or your right hand? // etc.


Scale 5B (0, 3, 5) Low partial credit: (5 marks) – One question written in a suitable form. – Two questions written in an unsuitable form, i.e. with deficiencies.

2(b) What type of data did Claire collect in the survey? Put a tick () in the correct box below. (5C)

Numerical Discrete

NumericalContinuous

CategoricalNominal

Categorical Ordinal

Explain your answer.

– categorical: each set of data can be sorted into two distinct categories // etc.

– nominal: data can be counted but not ordered or measured // 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.

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2(c) What percentage of students surveyed are left-handed? Give your answer correct to one decimal place. (5C*)

Number of students who are left-handed = 34 + 37 = 71

Total number of students surveyed = 186 + 34 + 203 + 37 = 460

% students surveyed who are left-handed

= 460

71 ×

1

100

= 0⋅154347... × 100 = 15⋅434782... % ≅ 15⋅4%

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit, e.g. identifies 71 and/or 460 without attempting to evaluate percentage.

High partial credit: (4 marks) – Finds 460

71 or

460

71 ×

1

100, but fails

to finish or finishes incorrectly.

* Deduct 1 mark off correct answer only if not rounded or incorrectly rounded - this deduction should be applied only once throughout the question.


2(d) If one student is chosen at random from the survey, what is the probability that the student chosen is not a right-handed girl? (5B)

Total number of students surveyed = 186 + 34 + 203 + 37 = 460

P(not R-h girl) = 1 – P(R-h girl)

= 1 – 460

186

= 460

186460 −

= 460

274

= 230

137 or 0⋅595652...

or = 1 – 460

186

= 1 – 0⋅404347... = 0⋅595652...

or

P(not R-h girl) = P(L-h girl) + P(R-h boy) + P(L-h boy)

= 460

34 +

460

203 +

460

37

= 460

3720334 ++

= 460

274

= 230

137 or 0⋅595652...

Scale 5B (0, 3, 5) Partial credit: (3 mark) – Finds 1 –

460

186 or

460

3720334 ++, but

fails to finish or finishes incorrectly.

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2(e) Claire says: “It is more likely to find a left-handed boy than a left-handed girl in her school.” Based on the data collected, do you agree with Claire? Give a reason for your answer. (5C)

Answer – no / incorrect

Reason

P(L-h boy) = 37203

37

+

= 240

37

= 0⋅154166...

P(L-h girl) = 34186

34

+

= 220

34 or 0⋅154545...

difference in levels of probabilities is insignificant



– Finds 240

37 and

220

34, but no conclusion

given.

2(f) If one boy and one girl are chosen at random from the survey, what is the probability that both students chosen are left-handed? (5C)

P(L-h boy) = 37203

37

+

= 240

37

P(L-h girl) = 34186

34

+

= 220

34

P(L-h boy and L-h girl)

= 240

37 ×

220

34

= 800,52

258,1

= 400,26

629 or 0⋅023825...

Scale 5C (0, 2, 4, 5) Low partial credit: (2 mark) – Finds 240

37 and/or

220

34 without attempting

to multiplying them. – Uses addition and finishes correctly

[ans. 240

37 +

220

34 =

800,52

300,16 or equivalent].

High partial credit: (4 mark) – Finds 240

37 ×

220

34, but fails to finish

or finishes incorrectly.

2015.2 J.18/20_MS 47/72 of 71 examsDEB




The prices of twelve pairs of designer shoes to buy in-store are shown in the incomplete back-to-back stem-and-leaf plot below.

In-store Online

9 3

5 4 4

7 5 0 0 5

9 6 5 6

5 7

3 8

3(a) (i) Complete the stem-and-leaf plot to display the online prices of the same shoes shown in the table below. (5D)

Online Prices (€)

43 59 31

56 76 42

67 48 54

46 58 63

In-store Online

9 3 1

5 4 4 2 3 6 8

7 5 0 0 5 4 6 8 9

9 6 5 6 3 7

5 7 6

3 8

** Accept unordered stem and leafs for full marks.

Scale 5D (0, 2, 3, 4, 5) Low partial credit: (2 marks) – One or two prices correctly placed.

Middle partial credit: (3 marks) – Three, four or five prices correctly placed.

High partial credit: (4 marks) – Six, seven, eight or nine prices correctly placed.

Key: 6 5 = €65

Key: 6 5 = €65

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3(a) (ii) Use your diagram to identify whether shoes are cheaper to buy in-store or online. Explain your answer. (5C)

Answer – online

Explanation Any 1: – highest online price is €76 compared to the highest

in-store price of €83; also the lowest online price is €31 compared to the lowest in-store price of €39 //

– most of the online prices are at the cheaper end of the stem-and-leaf plot compared to the in-store prices // etc.

** Accept other appropriate material.

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


3(b) (i) What is the in-store modal price? (5B)

Modal price – €50

(ii) Aoife says that she cannot find the online modal price. Explain why this in the case.

Answer Any 1: – all prices are different // – same price does not occurs more than once // etc.

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Correct in-store modal price, but no explanation or incorrect explanation given.– Correct explanation, but no modal price or incorrect modal price given.

3(c) Find the median price in each case. (5B)

In-store

39, 44, 45, 50, 50, 55, 57, 65, 66, 69, 75, 79

Median = 2

5755 +

= €56

Online

31, 42, 43, 46, 48, 54, 56, 58, 59, 63, 67, 76

Median = 2

5654 +

= €55

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

* Accept correct answers with or without work shown. * No deduction applied for the omission of or incorrect units involving currency.

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3(d) Where does the price of designer shoes have the largest range, in-store or online? Give a reason for your answer. (5C)

Answer – online

Reason Range (in-store) = 83 – 39 = €44

Range (online) = 76 – 31 = €45

price range online is €45, which is greater than the price range in-store, which is €44

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

High partial credit: (4 marks) – Two ranges correct, but no conclusion.

* Accept correct answers with or without work shown. * No deduction applied for the omission of or incorrect units involving currency.

3(e) Find the interquartile range in each case. (5C)

In-store

39, 44, 45, 50, 50, 55, 57, 65, 66, 69, 75, 79

Quartile 1 = 2

5045 +

= €47⋅50

Quartile 4 = 2

6966 +

= €67⋅50

Interquartile range = Quartile 4 – Quartile 1 = 67⋅50 – 47⋅50 = €20

Online

31, 42, 43, 46, 48, 54, 56, 58, 59, 63, 67, 76

Quartile 1 = 2

4643 +

= €44⋅50

Quartile 4 = 2

6359 +

= €61⋅00

Interquartile range = Quartile 4 – Quartile 1 = 61⋅00 – 44⋅50 = €16⋅50

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Correct identification of one upper or lower quartile.

High partial credit: (4 marks) – One correct interquartile range.


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3(f) Compare the prices of designer shoes to buy in-store and online. 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 – online prices are cheaper as both the mean and median

are lower Any 1: – mean online price is €53⋅58 while mean in-store price

is €58⋅17 // – median online price is €55 while median in-store price

is €56 // etc.


Measure of variability Any 1: – the spread of prices are very similar for both online and

in-store as the two ranges and the two interquartile ranges are almost the same

Any 1: – the range (online) is €45 whereas the range (in-store)

is just slightly less at €44 – prices are slightly more spread out online compared to

in-store as the range for online prices is slightly greater, i.e. the range (online) is €45 while the range(in-store) is €44 //

– the interquartile range (online) is €16⋅50 whereas the interquartile range (in-store) is €20 //

– the prices are slightly more spread out for in-store compared to online as the interquartile range for in-store prices is slightly greater, i.e. the interquartile range (online) is €16⋅50 whereas the interquartile range (in-store) is €20// 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.


3(g) Laura says: “Every pair of designer shoes is cheaper online.” Do you agree with her? Give a reason for your answer. (5C)

Answer – no

Reason – can not say with certainty that every pair of shoes are cheaper as the prices for specific pairs of shoes are not given and there is an overlap in the online and in-store price ranges of shoes // etc.


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

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

2015.2 J.18/20_MS 51/72 of 71 examsDEB



4(a) Mark wishes to find x, the distance ‘as the crow flies’ from his home to the top of a nearby mountain. He uses a clinometer to measure the angle of elevation to the top of the mountain and records an angle of 23°.

He then looks at a map of his local area and finds that the top of the mountain is 764 m above the ground level of his house.

764 m

x

23�

(i) Use these measurements to calculate x, correct to one decimal place. (10D*)

sin | ∠ | = |Hyp|

|Opp|

sin 23° = ||

764

x

| x | = °23sin

764

= ...3907310

764

⋅

= 1,955⋅308764... ≅ 1,955⋅3 m

(ii) Suggest one possible source of error that may affect Mark’s calculation.

Answer Any 1: – difficulty in determining precise angle of elevation

using a clinometer // – clinometer is subject to ±2° error which may

significantly affect Mark’s calculations // – clinometer used from eye level - not reflected in

Mark’s calculations // etc.


Scale 10D* (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Any work of merit, e.g. correct trigonometric ratio. – Substitutes correctly into inverted

trigonometric ratio, i.e. | x | = 764

23sin °.

High partial credit: (6 marks) – Trigonometric ratio with full substitution,

i.e. | x | = °23sin

764 or | x | =

...3907310

764

⋅,


High partial credit: (8 marks) – Correct answer, but no source of error or incorrect source of error given.


2015.2 J.18/20_MS 52/72 of 71 examsDEB

4(b) To prevent dangerous avalanches near ski resorts in Switzerland, the Swiss authorities initiate sets of mini-avalanches when the gradient of snowfall on the top of mountains reaches 45° to the vertical.

(i) Calculate the angle between the gradient of the snowfall and the vertical, as shown in the diagram. Give your answer in degrees and minutes (correct to the nearest minute). (5C*)

Let = angle of the gradient to the vertical

tan | ∠ | = |Adj|

|Opp|

= 200

250

= 1⋅25 | ∠ | = tan–1(1⋅25) = 51⋅340191... = 53° 20⋅411507...′ ≅ 51° 20′

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


i.e. | ∠ | = tan–1

200

250 or tan | ∠ | =

200

250.

– Trigonometric ratio inverted, but finishes

correctly [ans. | ∠ | = tan–1

250

200 = 38° 40′].

– Final answer left as 51⋅340191... or 51⋅34 (not converted to degrees and minutes).


(ii) The majority of avalanches occur when the gradient of snowfall on the mountain top is between 25° and 45° to the vertical. Is the gradient of the snowfall in (i) in danger of causing an avalanche? Give a reason for your answer. (5B)

Answer – no

Reason Any 1: – gradient of the snow fall on the mountain top

is 51° 20′, which is greater than 45° to the vertical // – 51° 20′ does not lie between 25° and 45° // etc.


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

250 m

200 m

2015.2 J.18/20_MS 53/72 of 71 examsDEB



The jib of a mobile crane is made up of a series of scalene triangles as shown in the diagram below. The triangles ABC and CDE are similar and the triangles BCD and DEF are similar.

BD

EC

A

5(a) (i) Write down one pair of lines that are parallel. (5B)

Answer Any 1: – AB and CD // – BC and DE

(ii) Explain what is meant by a scalene triangle.

Answer Any 1: – a triangle in which the three sides have different lengths // – a triangle in which no two sides are equal // etc.



5(b) Given that | AB | = 1⋅2 m, | BC | = 1⋅05 m and | CD | = 0⋅9 m, find the length of [ DE ]. (10C*)

as ΔABC and ΔCDE are similar

||

||

CD

AB =

||

||

DE

BC

90

21

⋅⋅

= ||

051

DE

⋅

| DE | = 21

90051

⋅⋅×⋅

= 21

9450

⋅⋅

= 0⋅7875 m

Scale 10C* (0, 4, 7, 10) Low partial credit: (4 marks) – Any work of merit, e.g. correct relevant ratio, corresponding sides identified or indicates that corresponding sides are proportional. – Some correct substitution into correct relevant correct with minor errors.

High partial credit: (7 marks) – Finds

90

21

⋅⋅

= ||

051

DE

⋅ or equivalent, but



B D

F

A C

E

2015.2 J.18/20_MS 54/72 of 71 examsDEB



Aer Lingus operates scheduled flights between five airports in Ireland and seventeen airports in Britain.

6(a) How many different possible flight routes can Aer Lingus operate between Ireland and Britain? (5B)

Number of possible flight routes = 5 × 17 = 85

** Accept correct answers with or without work shown.

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Any work of merit, e.g. 5 × 17. – Uses addition instead of multiplication [ans. 5 + 17 =22].

6(b) The actual number of scheduled flights operated by the airline on an average weekday between Ireland and Britain is 210. Suggest two reasons why this number is different to your answer in part (i). (5B)

Reasons Any 2: – not all destinations in Britain are catered for from all

airports in Ireland // – not all destinations in Britain are catered for by

scheduled flights every day - flights could, for example, be weekly, twice weekly or only at weekends //

– there may be several scheduled flights daily between the same airports in Ireland and Britain, e.g. Dublin and London //

– some routes may only be seasonal, e.g. operate only during the summer // etc.



2015.2 J.18/20_MS 55/72 of 71 examsDEB

6(c) On a particular Monday, Aer Lingus only operated 165 flights between the two countries due to bad weather in Britain. Given that flights between Ireland and Britain represent two-thirds of the airline’s daily scheduled flights, calculate the overall percentage decrease in the number of flights operated by the airline on that day. Give your answer correct to one decimal place. (10D*)

Number of flights scheduled between Ireland and Britain on that day = 210

3

2(Total) = 210

3

1(Total) =

2

210

Total = 2

210 × 3

= 315 Overall number of Aer Lingus scheduled flights on that day = 315

% decrease in the overall number of flights on that day

= 315

165210 − ×

1

100

= 315

45 ×

1

100

= 0⋅142857... × 100 = 14⋅285714... ≅ 14⋅3%

Scale 10D* (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Any work of merit, e.g. writes 3

2 ≡ 210.

– Finds overall number of scheduled flights [ans. 315]. – Finds decrease in number of flights [ans. 210 – 165 = 45].

Middle partial credit: (6 marks) – Uses incorrect numerator or denominator (other correct) and finishes correctly to find incorrect % decrease. – Finds correct numerator and denominator, but fails to calculate % decrease.

High partial credit: (8 marks) – Finds

315

165210 − ×

1

100,

315

165210 −,

315

45 ×

1

100 or

315

45 ×

1

100, but fails to

finish or finishes incorrectly.

– Gives 0⋅142857... or 0⋅14 as final answer (not multiplied by 100).

* Deduct 1 mark off correct answer only if not rounded or incorrectly rounded - this deduction should be applied only once throughout the question.


2015.2 J.18/20_MS 56/72 of 71 examsDEB




The installation of solar panels on the roof of a house can reduce the energy bills of the household. The cross section of the roof of a house which is being fitted with solar panels is shown below. The total span of the roof is 9 m and its height is 3 m. The apex of the roof is directly above the midpoint of the roof span.

B

X

SolarPanels

3 m

9 m

A

7(a) In which direction should the solar panels face on the roof of the house to maximise the impact of the sun? Give a reason for your answer. (5B)

Direction – south / south facing

Reason Any 1: – a south-facing roof will get more sun during the day // – best to point solar panels in the direction that captures

the most sunshine // etc.



7(b) Identify the type of triangle which represents the cross section of the roof shown in the diagram above. Give a reason for your answer. (5B)

Answer – isosceles triangle

Reason Any 1: – two angles at the base of the triangle are equal // – two sides are equal // – two angles are equal // – the angles opposite the equal sides are equal // etc.



2015.2 J.18/20_MS 57/72 of 71 examsDEB

7(c) Calculate the measure of the angle X, correct to one decimal place. (5C*)

tan | ∠X | = |Adj|

|Opp|

= 54

3

⋅

= 0⋅666666... | ∠X | = tan–1(0⋅666666...) = 33·690067... ≅ 33·7°

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


i.e. | ∠X | = tan–1

54

3

⋅ or tan | ∠X | =

54

3

⋅.

– Trigonometric ratio inverted, but finishes

correctly [ans. | ∠X | = tan–1

3

54⋅ = 56·3°].


2015.2 J.18/20_MS 58/72 of 71 examsDEB

7(d) Hence, find | AB |, the maximum length of solar panels that can be fitted to the roof. Give your answer correct to two decimal places. (5C*)

sin | ∠X | = |Hyp|

|Opp|

sin 33·7° = ||

3

AB

| AB | = °⋅733sin

3

= ...5548440

3

⋅

= 5⋅406921... ≅ 5⋅41 m

or

cos | ∠X | = |Hyp|

|Adj|

cos 33·7° = ||

54

AB

⋅

| AB | = °⋅

⋅733cos

54

= ...8319540

54

⋅⋅

= 5⋅408952... ≅ 5⋅41 m

or

| Hyp |2 = | Opp |2 + | Adj |2 | AB |2 = | 3 |2 + | 4⋅5 |2 = 32 + 4⋅52 = 9 + 20⋅25 = 29⋅25 | AB | = 2529⋅

= 5⋅408326... ≅ 5⋅41 m

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

High partial credit: (4 marks) – Correct substitution into trigonometric ratio and correctly manipulated,

i.e. | AB | = °⋅733sin

3 or | AB | =

°⋅⋅

733cos

54

– Correct substitution into formula for Pythagoras, i.e. | AB |2 = | 3 |2 + | 4⋅5 |2.

– Finds | AB | = ...5548440

3

⋅ or

...8319540

54

⋅⋅

,


– Finds | AB |2 = 29⋅25 or | AB | = 2529⋅ , but fails to finish or finishes incorrectly.


2015.2 J.18/20_MS 59/72 of 71 examsDEB



Scientists at the European Space Agency (ESA) monitor the flight paths of comets and asteroids that may collide with Earth. The progress of one particular asteroid is tracked over a 24-hour period. A researcher decided to plot the course of the asteroid on a co-ordinate grid to see how close it came to a collision. At the start of the tracking period, the asteroid was located at point A(–3, 2) and at the end of the tracking period, it was located at point B(4, −3). Earth is located at the point (0, 0).

8(a) Plot the points A and B on the co-ordinate plane below. Hence, draw the line that represents the flight path of the asteroid. (5C)

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – One point correctly plotted. – Both points reversed, i.e. (y, x).

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

2

�1

1

�2

2

�3

3

�4

4

3 4 5

x

y

1�1�2�3�4�5

A

B

2015.2 J.18/20_MS 60/72 of 71 examsDEB

8(b) Find the slope of AB. (5C)

A(–3, 2), B(4, –3) (x1, y1) (x2, y2)

mAB = 12

12

xx

yy

−−

= )3(4

23

−−−−

= 34

5

+−

= −7

5

or

mAB = run

rise

= 34

)32(

++−

= −7

5

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some correct substitution into slope formula. – Indicates that slope = rise/run. – Correct slope formula from tables.

High partial credit: (4 marks) – Both x and y reversed in substitution

[ans. −5

7 (inverted solution)].

– Slope not negative if rise/run method

used [ans. 7

5].

8(c) Find the equation of the line AB. (5C)

mAB = –7

5, A(–3, 2)

y – y1 = m(x – x1)

y – (2) = –7

5(x – (–3))

7(y – 2) = –5(x + 3) 7y – 14 = –5x –15 5x + 7y + 1 = 0

or

mAB = –7

5, A(–3, 2)

y = mx + c

2 = –7

5(–3) + c

7(2) = –5(–3) + 7c

c = 7

1514 −

= –7

1

y = mx + c

y = –7

5x –

7

1

7y = –5x – 1 5x + 7y + 1 = 0

2015.2 J.18/20_MS 61/72 of 71 examsDEB

8(c) (cont’d.)

** Accept student’s slope from part (b). ** Not necessary to write AB in the form ax + by + c = 0.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some correct substitution into a line formula. – Correct line formula from tables.

High partial credit: (4 marks) – Correct substitution into a line formula but incomplete. – One incorrect substitution into a line formula but continues correctly. – Both x and y reversed in substitution but continues correctly.

8(d) Find the equation of the line through (0, 0), perpendicular to AB. (10C)

mAB × ⊥mAB = –1

mAB = –7

5

⊥mAB = 5

7

⊥mAB = 5

7, point (0, 0)

y – y1 = m(x – x1)

y – (0) = 5

7(x – 0)

5y = 7x 7x – 5y = 0

or

⊥mAB = 5

7, point (0, 0)

y = mx + c

0 = 5

7(0) + c

c = 0

y = mx + c

y = 5

7x

5y = 7x 7x – 5y = 0

** Accept student’s slope from part (b). ** Not necessary to write AB in the form ax + by + c = 0.

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – Perpendicular slope found and stops. – Some correct substitution into a line formula. – Correct line formula from tables.

High partial credit: (7 marks) – Error in slope, but continues correctly. – Correct slope with one incorrect substitution into a line formula, but continues correctly.

2015.2 J.18/20_MS 62/72 of 71 examsDEB

8(e) Let C be the point where this perpendicular line through (0, 0) intersects AB. Calculate the co-ordinates of C. (10C)

5x + 7y = –1 (×5) 7x – 5y = 0 (×7)

25x + 35y = –5 49x – 35y = 0 74x = –5

x = –74

5

5x + 7y = –1 7y = –1 – 5x

7y = –1 – 5(–74

5)

7y = –1 + 74

25

7y = –74

49

y = –74

7

or 7x – 5y = 0 –5y = –7x

–5y = –7(–74

5)

–5y = 74

35

y = –74

7

C(–74

5, –

74

7)

or

5x + 7y = –1 (×7) 7x – 5y = 1 (×–5)

35x + 49y = –7 –35x + 25y = 0 74y = –7

y = –74

7

5x + 7y = –1 5x = –1 – 7y

5x = –1 – 7(–74

7)

5x = –1 + 74

49

= –74

25

x = –74

5

or 7x – 5y = 0 7x = 5y

7x = 5(–74

7)

7x = –74

35

x = –74

5

C(–74

5, –

74

7)

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 74x = –5 or 74y = –7, 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 with 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.

2015.2 J.18/20_MS 63/72 of 71 examsDEB

8(f) The co-ordinate grid is drawn to a scale of 1 unit = 500 000 km. How close did the asteroid come to a collision with Earth? Give your answer correct to the nearest kilometre. (5C*)

E(0, 0), C(–74

5, –

74

7)

(x1, y1) (x2, y2)

| EO | = 212

212 )()( yyxx −+−

= 22 )074

7()0

74

5( −−+−−

= 476,5

49

476,5

25 +

= 476,5

74

= 74

1 or 0⋅116247...

Closest distance = | EO | × 500,000

= 74

1 × 500,000

= 0⋅116247... × 500,000 = 58,123⋅819371... ≅ 58,124 km

** Accept student’s answer for co-ordinates for point C from part (e).

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit. – Some correct substitution into relevant distance formula, including Pythagoras’s Theorem. – Identifies correct relevant formula, including Pythagoras’s Theorem.

High partial credit: (4 marks) – Finds | EO | correctly, but fail to find distance or finds incorrect distance.


2015.2 J.18/20_MS 64/72 of 71 examsDEB


** Accept correct answers with or without work shown.

9(a) Examine the letters of the word shown below.

MATHEMAT ICS

(i) Which letter has no axes of symmetry? (5D)

Answer – S

(ii) Identify one letter which has only one axis of symmetry.

Answer Any 1: – M // – A // – T // – E // – C

(iii) Which two letters which have more than one axis of symmetry?

Answers – H – I

Scale 5D (0, 2, 3, 4, 5) Low partial credit: (2 marks) – One correct answer.

Middle partial credit: (3 marks) – Two correct answers.

High partial credit: (4 marks) – Three correct answers.

Full credit: (5 marks) – Four correct answers (no excess).

2015.2 J.18/20_MS 65/72 of 71 examsDEB

9(b) Each of the three images X, Y and Z in the diagram below shows the image of the object W under a transformation.

For each of X, Y and Z, describe fully one transformation (translation, axial symmetry or central symmetry) that will map the object onto that image. (10C)

X: – translation in the x direction

Y: Any 1: – central symmetry in the origin / (0, 0) // – 180° rotation around the origin

Z: – axial symmetry in the x-axis

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – One transformation fully described, e.g. central symmetry in the origin. – One or two transformations partially described, e.g. central symmetry.

High partial credit: (7 marks) – Two transformations fully described. – Three transformations partially described.

y

x

W

Z

X

Y

2015.2 J.18/20_MS 66/72 of 71 examsDEB



Beth has been asked to organise the music selection for an upcoming disco for students at her school. She conducted a survey by asking a group of girls in her year group to indicate their preferred choice of music. Her results are shown in the two charts below.

Other

Dance

PopRap

Rock

Dance Pop Rap Rock Other

2

0

4

6

8

10

12

14

16

10(a) (i) Beth wants to show that pop music was the most popular choice. Which chart do you think she should use? Give a reason for your answer. (5B)

Answer – pie chart

Reason Any 1: – easy to see that pop music is the largest sector

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

for the point she wishes to show // etc.


or

Answer – bar chart

Reason Any 1: – easy to see which bar is the highest // – figures (on vertical axis) may add weight to the point

she wishes to make // etc.


(ii) Identify another suitable type of graph that Beth could use to display this data.

Answer – line plot // etc.


** No marks awarded for ‘pie chart’ or ‘bar chart’ in part (i).

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.

2015.2 J.18/20_MS 67/72 of 71 examsDEB

10(b) Calculate the measure of the angle in the above pie chart that represents the students who indicated rap music as their preferred choice of music in the survey. (10D*)

Total number of students = 7 + 15 + 11 + 10 + 2 = 45

Angle per student = 45

360°

= 8°

| ∠Rap | = 8° × 11 = 88°

or

Total number of students = 7 + 15 + 11 + 10 + 2 = 45

Fraction of students who prefer rap

= 45

11

| ∠Rap | = 45

11 × 360°

= 11 × 8° = 88°

Scale 10D* (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Any work of merit, e.g. indicates that angles are fraction of 360°. – Finds total number of students [ans. 45].

Middle partial credit: (6 marks) – Finds angle per student (8°).

– Finds fraction who prefer rap [ans. 45

11].

High partial credit: (8 marks) – Finds | ∠Rap | = 8° × 11 or 45

11 × 360°,



10(c) Beth claims that the results of her survey are representative of the views of the mixed student population in her school. Do you agree with Beth? Give a reason for your answer. (5B)

Answer – no / incorrect

Reason Any 1: – Beth’s data may be biased as her sample is probably

not representative of the entire student population // – Beth only surveyed students in her own year group

and not from all segments / year groups in the school // – Beth only surveyed girls within her own year group and

did not take the views of boys within this group // etc.



2015.2 J.18/20_MS 68/72 of 71 examsDEB




11(a) Prove that in a parallelogram, opposite sides are equal and opposite angles are equal. (15D)

Diagram

CD

BA

1

2

3

4

Given Parallelogram ABCD

To Prove | AB | = | DC | and | AD | = | BC |. and | ∠DAC | = | ∠BCD | and | ∠CDA | = | ∠ABC |.

Construction Join A to C. Label angles 1, 2, 3 and 4.

Proof In triangles ABC and ACD, | ∠1 | = | ∠3 | ... alternate angles and | ∠2 | = | ∠4 | ... alternate angles | AC | = | AC | ... common side ∴ triangle ABC and ACD are congruent ... ASA | DC | = | AB | and | AD | = | BC | ... opposite sides are equal and | ∠CDA | = | ∠ABC | ... opposite angles are equal

also | ∠1 | = | ∠3 | and | ∠2 | = | ∠4 | ... alternate angles | ∠1 | + | ∠2 | = | ∠3 | + | ∠4 | | ∠DAB | = | ∠BCD | similarly | ∠CDA | = | ∠ABC | i.e. opposite angles are equal

Scale 15D (0, 6, 9, 12, 15) Low partial credit: (6 marks) – Correct diagram drawn.

Middle partial credit: (9 marks) – Diagram, Given, To Prove and Construction completed only. – More than one step missing in proof or incomplete.

High partial credit: (12 marks) – One step missing in proof. – Fully correct, but with no reason (i.e. congruent triangles) given. – Completes up to | ∠CDA | = | ∠ABC | or equivalent, but fails to finish.

Full credit: (15 marks) – Given, To Prove and Construction may be indicated on the diagram. – Some steps in Proof may be indicated on the diagram.

2015.2 J.18/20_MS 69/72 of 71 examsDEB

11(b) ABCD is a parallelogram, as shown. BA has been extended to E. | ∠EAD | = 134°, as shown.

Find the measure of the angle x° and the measure of the angle y°. (5B*)

x°: AB || DC x° = 134° ... alternate angles

y°: | ∠DAB | = | ∠BCD | ... opposite angles in a parallelogram | ∠DAB | = 180° – 134° = 54° y° = | ∠BCD | = 54°

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


AE B

134�

y�x�

CD

2015.2 J.18/20_MS 70/72 of 71 examsDEB


Using only a compass and a straight edge, construct a line through the point A perpendicular to the line l below. Show all construction work. (10C)

Question

A

Solution

A

** Allow tolerance of ±0·2 cm or ±2°. ** All construction lines and arcs must be shown.

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – Any correct construction line or arc.

High partial credit: (7 marks) – Accurate construction, but without construction lines and/or arcs. – Correct construction, but outside allowable tolerances.

2015.2 J.18/20_MS 71/72 of 71 examsDEB

Notes:

2015.2 J.18/20_MS 73/72 Page 73 of 71 examsDEB

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