Ordinary Level Marking Scheme - School of Maths · 2019. 4. 8. · 2017.1 L.16/19_MS 2/76 Page 2 of...

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L.16/19

Pre-Leaving Certificate Examination, 2017

Mathematics Ordinary Level

Marking Scheme

Paper 1 Pg. 2 Paper 2 Pg. 36

2017.1 L.16/19_MS 2/76 Page 2 of 75 examsDEB


Mathematics

Ordinary 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, 2, 5 0, 2, 4, 5

10 mark scale 0, 4, 7, 10 0, 4, 6, 8, 10

15 mark scale 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 (five categories) response of no substantial merit (no credit) response with some merit (low partial credit) response about half-right (mid 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 is to be applied only if the student’s answer is fully correct. The * is 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

DEB 2014 (LC-O1)

Scale label A B C D

No of categories 2 3 4 5

5 mark scale 0, 5 0, 2, 5 0, 2, 4, 5

10 mark scale 0, 5 10 0, 3, 7, 10 0, 2, 5, 8, 10

15 mark scale 0, 5, 10, 15 0, 4, 7, 11, 1


Summary of Marks – 2017 LC Maths (Ordinary Level, Paper 1)

Q.1 (a) (i) 5C (0, 2, 4, 5) Q.6 (a) (i) 5C (0, 2, 4, 5) (ii) 5C (0, 2, 4, 5) (ii) 5C (0, 2, 4, 5) (iii) 5C (0, 2, 4, 5) (b) 10D (0, 4, 6, 8, 10) (b) 10C* (0, 4, 7, 10) (c) 5B (0, 2, 5) 25 25 Q.2 (a) (i) 5B* (0, 2, 5) Q.7 (a) (i) 5C (0, 2, 4, 5) (ii)

5C (0, 2, 4, 5)

(ii) 10C (0, 4, 7, 10) (iii) (b) (i) 10C (0, 4, 7, 10) (b) (i)

15D* (0, 6, 10, 13, 15) (ii) 5C* (0, 2, 4, 5)

(ii) (c) (i) 5C* (0, 2, 4, 5) 25 (ii) 5C* (0, 2, 4, 5) (d) 10D (0, 4, 6, 8, 10) 50Q.3 (a) (i) 5C (0, 2, 4, 5) (ii) 5C (0, 2, 4, 5) (iii) 5C (0, 2, 4, 5) Q.8 (a) (i) 5C (0, 2, 4, 5) (b) 10D (0, 4, 6, 8, 10) (ii) 10D (0, 4, 6, 8, 10) 25 (iii) 5B (0, 2, 5) (iv) 5B (0, 2, 5) (v) 5B (0, 2, 5) Q.4 (a) (i) 5C (0, 2, 4, 5) (b) (i) 5B (0, 2, 5) (ii) 5C (0, 2, 4, 5) (ii) 5B (0, 2, 5) (b) 10D (0, 4, 6, 8, 10) (iii) 10D (0, 4, 6, 8, 10) (c) 5C (0, 2, 4, 5) 50 25 Q.9 (a) (i) 10D* (0, 4, 6, 8, 10) Q.5 (a) 5B (0, 2, 5) (ii) 10D* (0, 4, 6, 8, 10) (b) (i) 5C (0, 2, 4, 5) (iii) 5C* (0, 2, 4, 5) (ii) 5C (0, 2, 4, 5) (b) (i) 10C* (0, 4, 7, 10) (iii) 5C (0, 2, 4, 5) (ii)

5C* (0, 2, 4, 5)

(iv) 5C (0, 2, 4, 5) (iii) 25 (c) (i) 5C* (0, 2, 4, 5) (ii) 5C* (0, 2, 4, 5) 50

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.

General Instructions

There are two sections in this examination paper.

Section A Concepts and Skills 150 marks 6 questions Section B Contexts and Applications 150 marks 3 questions

Answer all questions.

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

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

Answers should be given in simplest form, where relevant.



Mathematics


Section A Concepts and Skills 150 marks

Answer all six questions from this section. (25 marks each)

Question 1 (25)

1(a) (i) Simplify:

5(4x + 3) – 3(2 – 4x). (5C)

5(4x + 3) – 3(2 – 4x) = 20x + 15 – 6 + 12x = 32x + 9

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. any correct attempt at simplifying equation, e.g. 20x + 15 or –6 + 12x, etc. – Simplifies one term correctly, i.e. 32x or 9.

High partial credit: (4 marks) – Simplifies fully to 20x + 15 – 6 + 12x, but fails to finish or finishes incorrectly. – Correct answer, but no work shown.

(ii) Solve for x:

6

15 +x +

2

1 x− = –

3

1, where x ∈ ℝ. (5C)

6

)1(3)15( xx −++ = –

3

1

6

3315 xx −++ = –

3

1

6

42 +x = –

3

1

2x + 4 = –3

6

= –2 2x = –2 – 4 = –6 x = –3

examsDEB


Question 1 (cont’d.)

1(a) (ii) (cont’d.)

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. identifies correct common multiple of 6, 2 and 3, and substitutes into equation correctly,

i.e. 6

)1(3)15( xx −++ = –

3

1.

– Multiplies terms by multiple of 6, 2 and 3,

i.e. 12

+

6

15x + 12

−

2

1 x = 12

−

3

1,

but fails to finish or finishes incorrectly.

High partial credit: (4 marks) – Finds 6

42 +x = –

3

1 or similar, but fails

to finish or finishes incorrectly. – Incorrect final answer, but no more than two errors in solving.

(iii) Verify your answer to part (ii) above. (5C)

6

15 +x +

2

1 x− = –

3

1

@ x = –3

6

15 +x +

2

1 x− =

6

1)3(5 +− +

2

)3(1 −−

= 6

115 +− +

2

31 +

= 6

14− +

2

4

= 3

7− + 2

= –23

1 + 2

= –3

1

as –3

1 = –

3

1, x = –3 is a solution (of the equation)

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some correct substitution of answer from part (ii) into equation.


15 +x +

2

1 x− = –

3

1 for x = –3,

but no conclusion given (or incorrect conclusion if not equal).

2017 LC Maths [OL] – Paper 1



1(b) Solve the equation x2 + 5x – 7 = 0 and give your answers correct to two decimal places. (10C*)

x = a

acbb

2

42 −±−

x = )1(2

)7)(1(4)5(5 2 −−±−

= 2

28255 +±−

= 2

535 ±−

= 2

...28010975 ⋅±−

x = 2

...28010975 ⋅−−

= 2

...28010912⋅−

= –6⋅140054... ≅ –6⋅14

x = 2

...28010975 ⋅+−

= 2

...2801092⋅

= 1⋅140054... ≅ 1⋅14

Scale 10C* (0, 4, 7, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down correct relevant ‘–b’ formula. – Some correct substitution into ‘–b’ formula, but fails to finish or finishes incorrectly. – Attempts to factorise quadratic equation.

High partial credit: (7 marks) – Substitutes correctly into ‘–b’ formula, but fails to finish or finishes incorrectly. – Finds one correct value of x only. – Finds incorrect solutions using appropriate method (allow up to three minor errors in total).

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



Question 2 (25)

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

2(a) Sophie is going on holiday to America and wants to exchange €500 for US dollars. She checks online and finds the exchange rate on that day to be $1 = €0·88.

(i) Find, correct to the nearest cent, the amount that Sophie can expect to receive in dollars. (5B*)

Amount (in €) = 500

Amount (in $) = 880

500

⋅

= 568·181818... ≅ $568·18

Scale 5B* (0, 2, 5) Partial credit: (2 marks) – Some work of merit, e.g. indicates

division, i.e. 880

500

⋅, but fails to finish

or finishes incorrectly. – Finds 500 × 0·88 = €440 (with or without work shown).


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

(ii) Sophie goes into her local bank and exchanges her euro for US dollars. The exchange rate for the transaction is €1 = $1·0990.

Find the amount that Sophie receives in dollars. (5C)

Amount (in €) = 500 Amount (in $) = 500 × 1·0990 = $549·50

(iii) Suggest a reason why there is a difference between the two amounts.

Reason Any 1: – exchange rate in the bank is lower as it charges

a commission for exchanging currencies // – exchange rate fluctuates throughout the day

whereas the bank may offer a fixed rate on any specific day // etc.

** Accept any other appropriate reason.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. indicates multiplication, i.e. 500 × 1·0990, but fails to finish or finishes incorrectly.

– Finds 09901

500

⋅ = €454·959053...

(with or without work shown). – Correct reason given, but fails to find or finds incorrect amount in dollars.

High partial credit: (4 marks) – Finds correct amount in dollars, but reason incomplete or omitted.





2(b) Graham wishes to invest €3,000 for 3 years. His local bank offers him two options, Option 1 and Option 2, as shown in the table below.

Option 1 Option 2

2⋅75% compound interest per year, for 3 years

5% interest after 2 years

3% compound interest per year thereafter

No penalty applies if money withdrawn at the end of any year

Penalty applies if money withdrawn within the first 2 years

(i) Find, correct to the nearest cent, the value of Graham’s investment at the end of 3 years under both options. (15D*)

Option 1

F = P(1 + i)t = 3,000(1 + 0·0275)3 = 3,000(1·0275)3 = 3,000(1·0847895...) = 3,254·368640... = €3,254·37

or

Year 1 (end) = 3,000 × 1·0275 = 3,082·50

Year 2 (end) = 3,082·50 × 1·0275 = 3,167·26875

Year 3 (end) = 3,167·26875 × 1·0275 = 3,254·368640... = €3,254·37

Option 2

Year 2 (end) = 3,000 × 1·05 = 3,150

Year 3 (end) = 3,150 × 1·03 = 3,244·50

(ii) Which option would you recommend? Give a reason for your answer.

Option 1 Any 1: – greater return (€3,254·37 vs. €3,244·50) over 3 years // – can withdraw money from the investment at the end

of each year // etc.


Option 2 Any 1: – can withdraw money from the investment without

any penalty at any time after 2 years // – can only withdraw money from the investment under

Option 1 at the end of each year // etc.





2(b) (cont’d.)

Scale 15D* (0, 6, 10, 13, 15) Low partial credit: (6 marks) Option 1 – Some work of merit, e.g. writes down correct relevant formula and stops. – Some correct substitution into relevant formula.

Option 2 – Some work of merit, e.g. finds correct value of any amount after 1 year, i.e. any amount × 1·0275 correctly. – Finds correct value of investment at the end of Year 1 [ans. €3,082·50] and stops or continues incorrectly.

– Recommendation with reason given, but fails to finds or finds incorrect value of investment for both options.

Mid partial credit: (10 marks) – Finds correct value of investment for one option only.

High partial credit: (13 marks) – Finds correct value of investment for both options, but recommendation (and reason) incomplete or omitted.





Question 3 (25)

z1 = 3 + 2i and z2 = –1 + 2i are two complex numbers, where i2 = –1.

3(a) (i) Evaluate each of the following complex numbers in the form a + bi, where a, b ∈ ℤ. (5C)

2z1 = 2(3 + 2i) = 6 + 4i

z1 + z2 = (3 + 2i) + (–1 + 2i) = 3 + 2i – 1 + 2i = 2 + 4i

iz1 = i(3 + 2i) = 3i + 2i2 = 3i + 2(–1) = –2 + 3i

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. substitutes correctly z1 into iz1. – One correct answer.

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



(ii) Plot each of your answers in part (i) on the given Argand diagram and label each point clearly. (5C)

2z1 = 6 + 4i z1 + z2 = 2 + 4i iz1 = –2 + 3i

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – One point correctly plotted. – Two points correctly plotted, but without labels.

High partial credit: (4 marks) – Two points correctly plotted and labelled.– Three points correctly plotted, but without labels. – All points correctly plotted and labelled, but real and imaginary axes interchanged.

2

2

�2

4

6

�6

�4

4 6

Re( )z

Im( )z

��4�6

2z1 /6 4� i

z1 2� �z i/2 4

iz1 / 2+3� i



3(a) (cont’d.)

(iii) Using your answers to parts (i) and (ii) above, or otherwise, explain what happens under each of the transformations given in the table below. (5C)

Transformation Evaluation Explanation

z = 2z1 6 + 4i – z is twice as far from the origin as z1 // – modulus increases by a factor of 2

z = z1 + z2 2 + 4i – translation of the point z1 to z1 + z2

z = iz1 –2 + 3i – point rotated through 90° anti-clockwise

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. states that z = iz1 ‘rotates 90°’ or ‘rotates anti-clockwise’. – One correct explanation.

High partial credit: (4 marks) – Two correct explanations.

3(b) Find the complex number z3 such that z3 = 2

1

zz .

Give your answer in the form a + bi, where a, b ∈ ℤ. (10D)

z3 = 2

1

zz

= i

i

21

23

+−+

= i

i

21

23

+−+

× i

i

21

21

−−−−

= 2

2

4221

4263

iii

iii

−−+−−−−

= 2

2

41

483

i

ii

−−−−

= )1(41

)1(483

−−−−−− i

= 41

843

+−+− i

= 5

81 i− //

5

1 –

5

8i

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. correct substitution for z1 and/or z2. – Correctly identifies −

2z = –1 – 2i.

Mid partial credit: (6 marks) – Finds i

i

21

23

+−+

× i

i

21

21

−−−−

, but fails to

evaluate or evaluates incorrectly. – Some multiplication above and below, even if by wrong conjugate.

High partial credit: (8 marks) – Finds

2

2

41

483

i

ii

−−−−

or similar, but fails

to finish or finishes incorrectly.



Question 4 (25)

The function f : x ׀→ x3 – 9x2 + 15x + 2 is defined for x ∈ ℝ.

4(a) (i) Find the co-ordinates of the point where the graph of f cuts the y-axis. (5C)

f cuts the y-axis at x = 0

f (x) = x3 – 9x2 + 15x + 2 f (0) = (0)3 – 9(0)2 + 15(0) + 2 = 0 – 0 + 0 + 2 = 2

f cuts the y-axis at (0, 2)

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. states that f cuts the y-axis at x = 0. – Correctly substitutes ‘0’ into equation, i.e. f (0) = (0)3 – 9(0)2 + 15(0) + 2.

High partial credit: (4 marks) – Finds f (0) = 2, but fails to state co-ordinates of point where the graph of f cuts the y-axis. – Writes co-ordinates as (2, 0).

(ii) Verify that the graph of f cuts the x-axis between x = 2 and x = 2·5. Explain your answer. (5C)

f (x) = x3 – 9x2 + 15x + 2

f (2) = (2)3 – 9(2)2 + 15(2) + 2 = 8 – 36 + 30 + 2 = 4

f (2·5) = (2·5)3 – 9(2·5)2 + 15(2·5) + 2 = 15·625 – 9(6·25) + 37·5 + 2 = 15·625 – 56·25 + 37·5 + 2 = –1·125

Explanation: – as the point (2, 4) lies above the x-axis and the point (2·5, –1·125) lies below the x-axis, the graph of f cuts the x-axis between x = 2 and x = 2·5

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. correctly substitutes ‘2’ or ‘2·5’ into equation i.e. f (2) = (2)3 – 9(2)2 + 15(2) + 2 or f (2·5) = (2·5)3 – 9(2·5)2 + 15(2·5) + 2. – Finds f (2) = 4 or f (2·5) = –1·125 only. – Correct explanation, but fails to find or finds incorrect co-ordinates.

High partial credit: (4 marks) – Finds f (2) = 4 and f (2·5) = –1·125, but explanation incomplete or omitted.




4(b) Find f ′(x), the derivative of f (x). Hence, find the co-ordinates of the local maximum turning point and local minimum

turning point of f. (10D)

f (x) = x3 – 9x2 + 15x + 2 f ′(x) = 3x2 – 18x + 15

f ′(x) = 0 at local max./min. turning points f ′(x) = 3x2 – 18x + 15 = 0 3x2 – 18x + 15 = 0 x2 – 6x + 5 = 0 (x – 1)(x – 5) = 0 x – 1 = 0 x = 1

x – 5 = 0 x = 5

@ x = 1 f (1) = (1)3 – 9(1)2 + 15(1) + 2 = 1 – 9 + 15 + 2 = 9 (1, 9) is the local maximum turning point

@ x = 5 f (5) = (5)3 – 9(5)2 + 15(5) + 2 = 125 – 225 + 75 + 2 = –23 (5, –23) is the local minimum turning point

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. states that f ′(x) = 0 at local max./min. turning points.– Differentiates at least one term correctly.

Mid partial credit: (6 marks) – Fully correct differentiation and stops. – Correct structure to f ′(x) = 0, but with errors and continues.

High partial credit: (8 marks) – Finds x = 1 and x = 5, but fails to finish or finishes incorrectly. – Finds x = 1 and f (1) = 9 or x = 5 and f (5) = –23 but fails to finish or finishes incorrectly.




4(c) Hence, sketch the graph of f on the axes below and indicate clearly both turning points. (5C)

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. sketches the correct shape of f , but with no reference to relevant points. – Two or three relevant points correctly plotted.

High partial credit: (4 marks) – Four relevant points correctly plotted, but sketch of f incorrect or omitted.


2 3 4 5 6 71�1

20

30

10

�10

�20

�30

x

y

(5, 23) / local min.turning point

�

(1, ) / local max. turning point�


Question 5 (25)

The first three patterns in a sequence of patterns formed by arranging square tiles are shown below.

5(a) Draw the fourth pattern in the sequence. (5B)

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Some work of merit, e.g. draws correct general shape of pattern. – Correct number of vertical tiles (5) or horizontal tiles (10).

5(b) (i) Complete the table below. (5C)

Pattern Number 1 2 3 4 5 6

Number of Tiles 5 8 11 14 17 20

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – One correct entry.

High partial credit: (4 marks) – Two correct entries.




5(b) (cont’d.)

(ii) Find, in terms of n, a formula for the number of tiles in the nth pattern in the sequence. (5C)

a = T1 = 5

d = T2 – T1 = 8 – 5 = 3

Tn = a + (n – 1)d = 5 + (n – 1)(3) = 5 + 3n – 3 = 3n + 2

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down correct relevant formula for Tn . – Identifies a or d correctly. – Some correct substitution into Tn .

High partial credit: (4 marks) – Identifies both a and d correctly and substitutes correctly into relevant formula, i.e. Tn = 5 + (n – 1)(3), but fails to finish or finishes incorrectly.

(iii) Using your formula, or otherwise, find the number of tiles in the 40th pattern. (5C)

Tn = 3n + 2

T40 = 3(40) + 2 = 120 + 2 = 122

** 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. writes down T40 and stops. – Some correct substitution into Tn formula.– Attempts to find T40 using list method (terms to at least T10.)

High partial credit: (4 marks) – Finds T40 correctly using list method. – Substitutes correctly into Tn formula, but fails to finish or finishes incorrectly.




5(b) (cont’d.)

(iv) Find the total number of tiles used in the first 40 patterns in the sequence. (5C)

Tn = 3n + 2 a = 5 d = 3

Sn = 2

n{2a + (n – 1)d}

S40 = 2

40{2(5) + (40 – 1)(3)}

= 20{10 + (39)(3)} = 20(10 + 117) = 20(127) = 2,540

** 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. writes down correct relevant formula for Sn. – Some correct substitution into Sn formula.

High partial credit: (4 marks) – Substitutes correctly into relevant formula,

i.e. S40 = 2

40{2(5) + (40 – 1)(3)}, but fails

to finish or finishes incorrectly. – Lists first forty terms, but fails to add or adds incorrectly.



Question 6 (25)

The diagram shows the graph of the function f (x) = x2 – 7x + 10 in the domain 0 ≤ x ≤ 7, x ∈ ℝ.

6(a) (i) The function g(x) is defined as

g : x 5 →׀ – x , where x ∈ ℝ.

Draw the graph of g(x) on the diagram. (5C)

g(x) = 5 – x

@ x = 0 g(0) = 5 – 0 = 5 (0, 5) ∈ g

@ x = 2 g(2) = 5 – 2 = 3 (2, 3) ∈ g

@ x = 3 g(3) = 5 – 3 = 2 (3, 2) ∈ g

@ y = 0 0 = 5 – x x = 5 (5, 0) ∈ g

** Accept co-ordinates of any two points on g.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. finds correctly y co-ordinate of any point on g. – Finds correctly co-ordinates (x, y) of one point on g. – Plots correctly co-ordinates (x, y) of one point on g (no work shown).

High partial credit: (4 marks) – Finds correctly co-ordinates (x, y) of two points on g, but fails to plot points or plots points incorrectly. – Plots correctly co-ordinates (x, y) of two points on g (no work shown), but fails to join points.


2 3 4 5 6 71

2

�2

4

g x( )

6

8

10y

x

f x( )



6(a) (cont’d.)

(ii) Use the graphs to find the two values of x for which g(x) = f (x). (5C)

Values of x for which g(x) = f (x)

Using graph: x = 1 x = 5

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

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. marks one or both points of intersection on graph. – Finds correctly x co-ordinate of one point of intersection, i.e. 1 or 5. – Finds correctly y co-ordinate of one point of intersection, i.e. 4 or 0.

High partial credit: (4 marks) – Finds correctly (1, 4) and (5, 0). – Finds correctly y co-ordinate of both points of intersection, i.e. 4 and 0.

6(b) Use algebra to solve g(x) = f (x). (10D)

g(x) = f (x) 5 – x = x2 – 7x + 10 x2 – 7x + 10 + x – 5 = 0 x2 – 6x + 5 = 0

(x – 1)(x – 5) = 0

x – 1 = 0 x = 1

x – 5 = 0 x = 5

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Any work of merit, e.g. states relevant ‘–b’ formula. – Equates 5 – x = x2 – 7x + 10 and stops or continues incorrectly.

Mid partial credit: (6 marks) – Finds correct quadratic equation, but fails to continue, i.e. x2 – 6x + 5 = 0.

High partial credit: (8 marks) – Finds correct quadratic equation, but finds only one correct value of x. – Finds correct quadratic equation and substitutes correctly into ‘–b’ formula, but fails to finish or finishes incorrectly.




6(c) Explain why the tangents to the curve y = f (x), at the points where the graphs of f (x) and g(x) intersect, are not parallel to each other. (5B)

Explanation:

@ x = 1, the function f (x) is decreasing slope of the tangent, m1, is negative

@ x = 5, the function f (x) is increasing slope of the tangent, m2, is positive

m1 ≠ m2 tangents are not parallel to each other

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Partial answer given - incomplete or unsatisfactory, but with some merit.



Section B Contexts and Applications 150 marks

Answer all three questions from this section. (50 marks each)

Question 7 (50)

The height of a diver after jumping off a platform into a diving pool is given by the formula:

h(t) = –5t2 + 5t + 10

where h is the height of the diver above the water line (the surface) of the diving pool in metres and t is the time in seconds after she jumps.

7(a) (i) Find the height of the diving platform. (5C)

h(t) = –5t2 + 5t + 10

@ t = 0 h(0) = –5(0)2 + 5(0) + 10 = 0 + 0 + 10 = 10 m

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down ‘height of diving platform @ t = 0’ or ‘h(0)’– Some correct substitution into h(t) equation and stops or continues. – Substitutes a value for t (≠ 0) into h(t) equation and evaluates correctly.

High partial credit: (4 marks) – Substitutes correctly into h(t) equation, i.e. h(0) = –5(0)2 + 5(0) + 10, but fails to finish or finishes incorrectly.

* No deduction applied for the omission of or incorrect use of units (‘m’) as units are mentioned in the question.

(ii) Find the length of time it takes before the diver enters the water. (10C)

h(t) = –5t2 + 5t + 10

diver enters water @ h = 0 0 = –5t2 + 5t + 10 5t2 – 5t – 10 = 0 t2 – t – 2 = 0

(t + 1)(t – 2) = 0 t + 1 = 0 t = –1 ... not applicable

t – 2 = 0 t = 2 s

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – Any work of merit, e.g. writes down diver enters water @ h = 0. – Substantially correct approach to solving quadratic equation.

High partial credit: (7 marks) – Solves quadratic equation correctly, i.e. finds t = –1 and 2, but fails to identify correct answer. – Correct answer, but no work shown.

* No deduction applied for the omission of or incorrect use of units (‘s’) as units are mentioned in the question.




7(b) (i) Find dt

dh. (10C)

h(t) = –5t2 + 5t + 10

dt

dh = –10t + 5

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – One term correctly differentiated.

High partial credit: (7 marks) – Two terms correctly differentiated with additional third incorrect term. – Correct co-efficient for both terms, but error(s) in indices. – Correct indices for both terms, but error(s) in co-efficient.

(ii) Hence, find the rate of change of height at the instant that the diver enters the water. (5C*)

Rate of change = dt

dh

= –10t + 5

@ t = 2 s Rate of change = –10(2) + 5 = –20 + 5 = –15 m/s

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

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

‘Rate of change = dt

dh ’ or ‘= –10t + 5’.

– Some correct substitution into h ′(t) equation and stops or continues. – Substitutes a value for t (≠ 2) into h ′(t) equation and evaluates correctly.

High partial credit: (4 marks) – Substitutes correctly into h′(t) equation, i.e. h ′(2) = –10(2) + 5, but fails to finish or finishes incorrectly.

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




7(c) (i) Find the length of time it takes the diver to reach the maximum height above the water line of the diving pool. (5C*)

Maximum height when dt

dh = 0

dt

dh = –10t + 5

= 0 –10t + 5 = 0 10t = 5

t = 10

5

= 2

1 s // 0·5 s

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


‘Maximum height at dt

dh = 0’ and stops.

– Substantially correct approach to solving linear equation.

High partial credit: (4 marks) – Equates linear equation correctly, i.e. h ′(t) = –10t + 5 = 0, but fails to find correct answer. – Correct answer, but no work shown.

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

(ii) Hence, find this maximum height. (5C*)

h(t) = –5t2 + 5t + 10

Maximum height @ t = 0·5 h(0·5) = –5(0·5)2 + 5(0·5) + 10 = –5(0·25) + 2·5 + 10 = –1·25 + 2·5 + 10 = 11·25 m

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

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down ‘Maximum height @ t = 0·5’ or ‘h(0·5)’.– Some correct substitution into h(t) equation and stops or continues. – Substitutes a value for t (≠ 0·5) into h(t) equation and evaluates correctly.

High partial credit: (4 marks) – Substitutes correctly into h(t) equation, i.e. h(0·5) = –5(0·5)2 + 5(0·5) + 10, but fails to finish or finishes incorrectly.

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



Question 7 (cont’d.) 7(d) The diver performs a different dive in competition. She reaches a maximum height of 0·75 m

above the diving platform after 0·5 s. After 1 second, she is at the same height again as her initial dive position. The height of the diver above the water line after t seconds is given by the formula g(t) = at2 + bt + 10, where a, b ∈ ℚ.

Find two expressions for the position of the diver in terms of a and b and hence, find the value of a and the value of b. (10D)

g(t) = at2 + bt + 10

@ t = 0·5 g(0·5) = a(0·5)2 + b(0·5) + 10 = 10 + 0·75 = 10·75 0·25a + 0·5b + 10 = 10·75 0·25a + 0·5b – 0·75 = 0 (× 4) a + 2b – 3 = 0

@ t = 1 g(1) = a(1)2 + b(1) + 10 = 10 a + b + 10 = 10 a + b = 0

a + 2b = 3 (× 1) a + b = 0 (× –1)

a + 2b = 3 –a – b = 0 b = 3

a + b = 0 a = –b = –3

or

a + 2b = 3 (× –1) a + b = 0 (× 2)

–a – 2b = –3 2a + 2b = 0 a = –3

a + b = 0 b = –a = –(–3) = 3

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down g(0·5) = 10·75 and/or g(1) = 10 and stops.– Finds one equation in terms of a and b, i.e. a + 2b – 3 = 0 or a + b = 0.

Mid partial credit: (6 marks) – Finds both equations in terms of a and b, and stops or continues incorrectly. – Multiplies equation(s) by appropriate constant(s) to facilitate the cancellation of a or b term. – Finds either variable (a or b) correctly by trial and error, but fails to verify in both equations or verifies incorrectly.

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



Question 8 (50)

8(a) The cost of hiring a well-known artist for a community music festival is €120 000. The cost will be shared equally by all the people who come to the event. The organisers are unsure of the number of people (n) who will attend.

(i) Complete the table below to show the cost per person (c) for each given value of n. (5C)

Number of people (n) 500 1,000 1,500 2,000 2,500 3,000

Cost per person (c) 240 120 80 60 48 40

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – One correct entry.

High partial credit: (4 marks) – Two correct entries.

(ii) On the grid below, draw a graph to show the relationship between n and c. (10D)

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

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. one point plotted correctly. – Draws correct general shape of graph.

Mid partial credit: (6 marks) – Two or three points plotted correctly with correct general shape of graph.

High partial credit: (8 marks) – Four or five points plotted correctly with correct general shape of graph. – All points plotted correctly, but not joined or joined by straight lines.


Co

stp

erp

erso

n(

)c

Pattern NumberNumber of people ( )n

1000

100

50

200

300

2000 2400 3000



8(a) (cont’d.)

(iii) Use your graph to estimate the minimum number of people required to attend the festival so that the cost per person is €50 or less. (5B)

Using graph: Minimum number = 2,400 (±125)

** Accept answer based on students’ graph in part (a)(ii) if not oversimplified. ** Accept tolerance of ±1 grid, i.e. 2,275 ≤ min. number ≤ 2,525.

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Some work of merit, e.g. marks clearly point on graph (with or without label), but no figure. – Work shown on student’s graph, but outside of tolerance. – Correct answer, but no work shown on graph.

(iv) Write a formula in n for the cost per person who attends the festival. (5B)

Cost (€) = n

000,120

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Some work of merit, e.g. use of n and 120,000 with incorrect operation.

– Writes down 000,120

n.

– Correct numerator or denominator

i.e. valueAny

000,120 or

n

valueAny.

(v) Verify your answer to part (iii) using your formula from part (iv). (5B)

Cost (€) = n

000,120

50 = n

000,120

n = 50

000,120

= 2,400

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

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Some work of merit, e.g. some correct substitution into student’s formula from part (a)(iv) and stops. – Correct substitution into student’s formula, but fails to finish or finishes incorrectly.




8(b) The organisers have sold 2000 tickets for the festival. They have only 3810 m2 available for parking of both cars and buses. The organisers plan to arrange buses to bring people to the event and plan to sell tickets for car parking spaces in advance.

The space allotted for each car is 12 m2 while the space allotted for each bus is 45 m2. It is assumed that the average occupancy per car is 2 and the average occupancy per bus is 53.

(i) Write down an expression for the total parking space available in terms of x (number of cars) and y (number of buses). (5B)

Total parking space:

x = number of cars y = number of buses

12x + 45y = 3,810

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Some work of merit, e.g. writes down one correct component, 12x or 45y.

(ii) Write down an expression for the number of people who have purchased tickets for the event in terms of x and y. (5B)

Total number of people:

2x + 53y = 2,000

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Some work of merit, e.g. writes down one correct component, 2x or 53y.

(iii) Hence, find the number of car parking spaces and bus parking spaces required in order to accomodate all ticket-holders at the event.. (10D)

12x + 45y = 3,810 (× –1) 2x + 53y = 2,000 (× 6)

–12x – 45y = –3,810 12x + 318y = 12,000 273y = 8,190

y = 273

190,8

= 30

12x + 45y = 3,810 12x + 45(30) = 3,810 12x = 3,810 – 1,350 = 2,460

x =

12

460,2

= 205




8(b) (iii) (cont’d.)

or

12x + 45y = 3,810 (× 53) 2x + 53y = 2,000 (× –45)

636x + 2,385y = 201,930 –90x – 2,385y = –90,000 546x = 111,930

x = 546

930,111

= 205

12x + 45y = 3,810 12(205) + 45y = 3,810 45y = 3,810 – 2,460 = 1,350

y = 45

350,1

= 30

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

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. identifies appropriate constant(s) by which to multiply equation(s) to facilitate the cancellation of x or y term.

Mid partial credit: (6 marks) – Multiplies equation(s) correctly by appropriate constant(s) to facilitate the cancellation of x or y term. – Finds 273y = 8,190 or 546x = 111,930, 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: (8 marks) – Finds first variable (x or y), 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) correctly by graphical means. – Finds both variables (x and y) by trial and error, but does not verify in both equations or verifies incorrectly.



Question 9 (50)

9(a) Bob works as a computer engineer. He pays income tax, a universal social charge (USC), and pay-related social insurance (PRSI) on his gross salary. His gross monthly salary is €4,250.

(i) The standard rate of income tax is 20% and the higher rate is 40%. Bob has tax credits of €3,300 per annum and a standard rate cut-off point of €32,800 per annum. How much income tax does Bob pay per month, correct to the nearest cent? (10D*)

Annual salary = 4,250 × 12 = €51,000

Tax @ 20% = 32,800 × 20%

= 32,800 × 100

20

= €6,560

Tax @ 40% = (51,000 – 32,800) × 40%

= 18,200 × 100

40

= €7,280

Gross tax per annum = Tax @ 20% + Tax @ 40% = 6,560 + 7,280 = €13,840

Net tax per annum = Gross tax – Tax credit = 13,840 – 3,300 = €10,540

Net tax per month = 12

540,10

= 878·333333... ≅ €878·33

Scale 10D* (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down annual salary is 4,250 × 12 or similar. – Finds tax correctly @ 20% or 40% of some relevant figure, but fails to continue.

Mid partial credit: (6 marks) – Finds Gross tax per annum correctly, but fails to find or finds incorrect Net tax per annum.

High partial credit: (8 marks) – Finds Net tax per annum correctly, but fails to find or finds incorrect Net tax per month.

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

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

or

Monthly tax credit = 12

300,3

= €275

Monthly standard rate cut-off point

= 12

800,32

= €2,733·333333...

Tax @ 20% = 2,733·333333... × 20%

= 2,733·333333... × 100

20

= €546·666666...




9(a) (i) (cont’d.)

Tax @ 40% = (4,250 – 2,733·333333...) × 40%

= 1,516·666666... × 100

40

= €606·666666...

Gross tax per month = Tax @ 20% + Tax @ 40% = 546·666666... + 606·666666... = €1,153·333333...

Net tax per month = Gross tax – Tax credit = 1,153·333333... – 275 = 878·333333... ≅ €878·33

Scale 10D* (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down Monthly tax credit is 3,300 ÷ 12 and/or Monthly SRCOP is 32,800 ÷ 12 or similar. – Finds Monthly tax credit and/or Monthly SRCOP, but fails to continue.

Mid partial credit: (6 marks) – Find tax @ 20% or 40% of some relevant figure, but fails to find or finds incorrect Gross tax per month.

High partial credit: (8 marks) .– Finds Gross tax per month correctly, but fails to find or finds incorrect Net tax per month.



(ii) Bob also pays USC on his gross salary every month. He pays USC at the rates of 0·5% on the first €1,001 per month, 2·5% on earnings between €1,001 and €1,564 per month and 5% on the balance.

Calculate the total amount of USC that Bob pays per month, correct to the nearest cent. (10D*)

USC @ 0·5% = 1,001 × 0·5%

= 1,001 × 100

50⋅

= €5·005

USC @ 2·5% = (1,564 – 1,001) × 2·5%

= 563 × 100

52⋅

= €14·075

USC @ 5% = (4,250 – 1,564) × 5%

= 2,686 × 100

5

= €134·3

Total USC = USC @ 0·5% + USC @ 2·5% + USC @ 5% = 5·005 + 14·075 + 134·3 = €153·38




9(a) (ii) (cont’d.)

Scale 10D* (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. finds 0·5%, 2·5% or 5% of some relevant figure. – Finds USC correctly at one rate.

Mid partial credit: (6 marks) – Finds USC correctly at two rates.

High partial credit: (8 marks) – Finds USC correctly at all three rates, but fails to finish or finishes incorrectly.



(iii) Bob also pays PRSI of €109 per month. How much can Bob expect his annual take-home pay to be? (5C*)

Total monthly deductions = Net tax + Total USC + PRSI = 878·33 + 153·38 + 109 = €1,140·71

Monthly take-home pay = Gross pay – (Total monthly deductions) = 4,250 – 1,140·71 = €3,109·29

Annual take-home pay = 3,109·29 × 12 = €37,311·48

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

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down relevant formula. – Some correct substitution into formula, but fails to finish or finishes incorrectly.

High partial credit: (4 marks) – Finds correct Monthly take-home pay, but fails to find or finds incorrect Annual take-home pay.





Question 9 (cont’d.) 9(b) (i) Bob wants to find out how much money he regularly spends. He prepares a spreadsheet

of all his regular financial outgoings, as shown in the table below.

Complete the table to find the total cost of Bob’s regular financial outgoings each month, correct to the nearest cent. (10C*)

Description Unit Cost Usage Cost (€)

Electricity Standing Charge 14·50

Price per unit 15·33 cent / kWh 66 kWh 10·1178

Broadband + TV 60·00

Rent 800·00

Gas Standing Charge 6·50

Price per unit 4·589 cent / kWh 300 kWh 13·767

Bus Ticket €132·00 132·00

Mobile Phone €45·00 45·00

Total 1,081·88

Scale 10C* (0, 4, 7, 10) Low partial credit: (4 marks) – Some work of merit, e.g. finds one correct figure, i.e. €10·1178 or 1,011·78c, €13·767 or 1,376·7c.

High partial credit: (7 marks) – Finds two correct figures, but fails to find or finds incorrect total.



(ii) How much disposable income does Bob have left each month after he pays all his regular outgoings? (5C*)

Disposable income = 3,109·29 – 1,081·88 = €2,027·41

(iii) Find this disposable income as a percentage of his net income. Give your answer correct to one decimal place.

Disposable income as a % of net income

= 29109,3

41027,2

⋅⋅

× 100

= 65·204918... ≅ 65·2%

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

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down relevant % formula. – Finds correct Disposable income, but but fails to finish or finishes incorrectly.

High partial credit: (4 marks) – Finds correct Disposable income with some correct substitution into % formula, but fails to find or finds incorrect %.





9(c) At the start of January, Bob owed €1,700 on his credit card. Interest is charged monthly at an annual equivalent rate (AER) of 18% of the amount owed.

(i) Using the formula (1 + r)12 = 1 + i, where r is the monthly rate and i is the annual rate of interest, find the rate of interest charged monthly which is equivalent to an AER of 18%, correct to three decimal places. (5C*)

(1 + r)12 = 1 + i (1 + r)12 = 1 + 0·18 = 1·18

1 + r = 1·1812

1

= 1·013888... r = 1·013888... – 1 = 0·013888... r = 1·388843...% ≅ 1·389%

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. identifies i = 0·18 or 1 + i = 1 + 0·18 = 1·18. – Divides or multiplies 1·18 by 12. – Some correct substitution into formula and stops or continues.

High partial credit: (4 marks) – Finds 1 + r = 1·1812

1

or 1⋅013888..., but fails to finish or finishes incorrectly.


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

(ii) Find the amount that Bob will owe on his credit card after 9 months if he does not make any further purchases or repayments over that period of time. Give your answer correct to the nearest euro. (5C*)

Amount owed after 9 months F = P(1 + i)t = 1,700(1 + 0·01389)9 = 1,700(1·01389)9 = 1,700(1·132185...) = 1,924·715208... ≅ €1,925

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

High partial credit: (4 marks) – Substitutes correctly into formula, i.e. 1,700(1·01389)9 or 1,700(1·132185...), but fails to finish or finishes incorrectly.





Notes:



.


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, 2, 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 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 (five categories) response of no substantial merit (no credit) response with some merit (low partial credit) response about half-right (mid 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 is to be applied only if the student’s answer is fully correct. The * is 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.

DEB 2014 (LC-O1 & O2)

Scale label A B C D

No of categories 2 3 4 5

5 mark scale 0, 5 0, 2, 5 0, 2, 4, 5

10 mark scale 0, 5 10 0, 3, 7, 10 0, 2, 5

15 mark scale 0, 5, 10, 15 0, 4, 7,

20 mark scale

25 mark scale 0, 6, 12

examsDEB


Summary of Marks – 2017 LC Maths (Ordinary Level, Paper 2)

Q.1 (a) (i) 5D (0, 2, 3, 4, 5) Q.7 (a) (i) 5C* (0, 2, 4, 5) (b) (i) 10D* (0, 4, 6, 8, 10) (ii) 10D* (0, 4, 6, 8, 10) (ii) 5C* (0, 2, 4, 5) (b) (i) 10C (0, 4, 7, 10) (c) 5C (0, 2, 4, 5) (ii) 5C* (0, 2, 4, 5) 25 (c) (i) 5C* (0, 2, 4, 5) (ii) 10C* (0, 4, 7, 10) Q.2 (a) (i) 5C (0, 2, 4, 5) 45 (ii) 5C (0, 2, 4, 5) (b) 5C (0, 2, 4, 5) (c) (i) 5C (0, 2, 4, 5) Q.8 (a) 10C* (0, 4, 7, 10) (ii) 5D (0, 2, 3, 4, 5) (b) 10D* (0, 4, 6, 8, 10) 25 (c) 10C* (0, 4, 7, 10) (d) 5C* (0, 2, 4, 5) Q.3 (a) 5C (0, 2, 4, 5) 35 (b) (i) 5C (0, 2, 4, 5) (ii) 5D (0, 2, 3, 4, 5) (c) (i) 5C (0, 2, 4, 5) Q.9 (a) (i) 5C (0, 2, 4, 5) (ii) 5C* (0, 2, 4, 5) (ii) 5B (0, 2, 5) 25 (iii) 10D (0, 4, 6, 8, 10) (iv) 5B (0, 2, 5) Q.4 (a) 10C (0, 4, 7, 10) (b) 10C (0, 4, 7, 10) (b) 5C (0, 2, 4, 5) (c) (i) 10C* (0, 4, 7, 10) (c) 10C (0, 4, 7, 10) (ii) 5C* (0, 2, 4, 5) 25 (iii) 5C (0, 2, 4, 5) (iv) 5C (0, 2, 4, 5) Q.5 (a) 10C (0, 4, 7, 10) (d) 10D (0, 4, 6, 8, 10) (b) 15D* (0, 6, 10, 13, 15) 70 25 Q.6 (a) 5B (0, 2, 5) (b) (i) 5C (0, 2, 4, 5) (ii) 5C (0, 2, 4, 5) (iii) 5C (0, 2, 4, 5) (c) 5C (0, 2, 4, 5) 25

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.

General Instructions

There are two sections in this examination paper.

Section A Concepts and Skills 150 marks 6 questions Section B Contexts and Applications 150 marks 3 questions

Answer all questions.

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

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

Answers should be given in simplest form, where relevant.



Mathematics


Section A Concepts and Skills 150 marks

Answer all six questions from this section. (25 marks each)

Question 1 (25)

1(a) Construct a triangle ABC in which | AB | = 7 cm, | BC | = 4 cm and | ∠ABC | = 90°. Show all your construction lines clearly. (5D)

B

C

A7 cm

4 cm

** Actual size as per examination paper. ** Tolerance: ±0·25 cm, ±2·5°.

Scale 5D (0, 2, 3, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. draws any triangle (pilot diagram). – Draws one correct line or angle and stops.

Mid partial credit: (3 marks) – Draws two correct elements and stops.

High partial credit: (4 marks) – Constructs 90° correctly. – Draws two correct elements and finishes correctly/consistently.

examsDEB



1(b) (i) Measure the length of [ AC ] constructed in part (a).

Use the Theorem of Pythagoras to calculate | AC |, correct to two decimal places. Explain any difference you found in your two answers. (10D*)

Length of [ AC ]:

From diagram: | AC | = 8·1 ± 0·25 cm

Theorem of Pythagoras:

| Opp |2 + | Adj |2 = | Hyp |2 | AC |2 = | AB |2 + | BC |2 = 72 + 42 = 49 + 16 = 65

| AC | = 65 = 8·062257... ≅ 8·06

Difference:

Any 1: – | AC | can be determined with greater accuracy using Pythagoras’ theorem

as the method does not depend on how well the triangle is drawn // – the naked eye cannot measure with the same level of accuracy // etc.

** Accept other appropriate answers.

Scale 10D* (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down correct formula for Pythagoras’ theorem.– Measures | AC | accurately and stops. – Some correct substitution into formula for Pythagoras’ theorem. – Explanation of difference, but no supporting work, i.e. does not measure | AC | or does not find length using Pythagoras’ theorem.

Mid partial credit: (6 marks) – Finds | AC | = 65, 8·062257.... or 8·06 and stops.

High partial credit: (8 marks) – Measures | AC | accurately and finds correct | AC | using Pythagoras’ theorem, but omits explanation or gives incorrect explanation.

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




1(b) (cont’d.)

(ii) Use a trigonometric ratio to find | ∠CAB |. Give your answer correct to the nearest degree. (5C*)

sin | ∠CAB | = |Hyp|

|Opp|

= 068

4

⋅

= 0·496277... | ∠CAB | = sin–1(0·496277...) = 29·754052... ≅ 30°

or

tan | ∠CAB | = |Adj|

|Opp|

= 7

4

= 0·571428... | ∠CAB | = tan–1(0·571428...) = 29·744881... ≅ 30°

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down correct relevant trigonometric ratio (sin or tan) and stops.

– Finds sin | ∠CAB | = 068

4

⋅ or 0·496277...

or tan | ∠CAB | = 7

4 or 0·571428...

and stops.

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

| ∠CAB | = sin–1(068

4

⋅) or tan–1(

7

4) or

sin–1(0·496277...) or tan–1(0·571428...), but fails to finish or finishes incorrectly. – Finds correct answer, but no work shown.

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




1(c) On the same diagram above, construct a circle with [ AC ] as a diameter. (5C)

B

C

A

** Actual size as per examination paper. ** Tolerance: ±0·5 cm, ±5°.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. attempt to bisect [ AC ] and stops. – Indicates and labels midpoint of [ AC ] (no constructions shown) and stops.

High partial credit: (4 marks) – Draws perpendicular bisector of [ AC ] and indicates point of intersection as centre of circle, but no circle drawn. – Construction correct, but 90° angle of perpendicular bisector to [ AC ] outside tolerance.



Question 2 (25)

The circle c1 has equation (x – 1)2 + (y – 1)2 = 25.

2(a) (i) Write down the co-ordinates of the centre of c1. (5C)

centre = (1, 1)

Write down the length of the radius of c1.

radius = 25 = 5 units

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down correct x or y co-ordinate of centre. – Writes down r2 = 25 and/or r = 25 and stops.

High partial credit: (4 marks) – Writes down correct centre or radius of c1.

(ii) Draw the circle c1 on the grid below. Each unit on the co-ordinate grid is 1 cm. (5C)


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

High partial credit: (4 marks) – Two of the three components correct: centre, radius or drawing.


2

�1

1

�2

2

�3

3

�4

4

�5

�6

3 4 5 6

x

y

1�1�2�3�4�5�6

5

6

(1, 1)



2(b) Does the point (5, 3) lie inside, outside or on c1? Justify your answer using algebra. (5C)

Equation of c1: (x − 1)2 + (y − 1)2 = 25

Point (5, 3): (5 − 1)2 + (3 – 1)2 = (4)2 + (2)2 = 16 + 4 = 20 < 25 (radius2 of c1)

(5, 3) lies inside c1

or

Centre of circle (1, 1) Distance from (1, 1) to (5, 3)

= 212

212 )()( yyxx −+−

= 22 )13()15( −+−

= 22 )2()4( +

= 416 +

= 20 = 4·472135... < 5 (radius of c1)

(5, 3) lies inside c1

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. identifies correct relevant formula. – Substitutes (5, 3) into relevant formula and stops or continues.

High partial credit: (4 marks) – Finds 20 [Method ], but omits or incorrect conclusion. – Finds 20 [Method ], but omits or incorrect conclusion.




2(c) (i) Find the equation of another circle c2 with centre (0, 0) that passes through the point (2, 2). (5C)

Centre of c2 = (0, 0) Radius = distance from (0, 0) to (2, 2)

= 212

212 )()( yyxx −+−

= 22 )02()02( −+−

= 22 )2()2( +

= 44 +

= 8

Equation of c2: x2 + y2 = r2 x2 + y2 = ( 8 )2

x2 + y2 = 8

or

Equation of c2: (x – h)2 + (y – k)2 = r2 (x – 0)2 + (y – 0)2 = ( 8 )2

x2 + y2 = 8

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down correct relevant formula for circle. – Some correct substitution of (2, 2) into relevant formula and continues. – Correct substitution into relevant formula and stops.

High partial credit: (4 marks) – Finds correct radius of c2, i.e. 8 . – Correct substitution of centre or radius into relevant formula for circle.




2(c) (cont’d.)

(ii) Find, using algebra, the points of intersection of c2 and the line y = –2x + 2. (5D)

Equation of line: y = –2x + 2

Equation of c2: x2 + y2 = 8 x2 + 4x2 –8x + 4 = 8 5x2 – 8x + 4 – 8 = 0 5x2 – 8x – 4 = 0 (5x + 2)(x – 2) = 0

5x + 2 = 0 5x = –2

x = –5

2

y = –2x + 2

y = –2(–5

2) + 2

= 5

4 + 2

= 25

4

point of intersection (–5

2, 2

5

4)

x – 2 = 0 x = 2

y = –2x + 2 y = –2(2) + 2 = –4 + 2 = –2 point of intersection (2, –2)

Scale 5D (0, 2, 3, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down ‘substitute –2x + 2 into x2 + y2 = 8 or c2’ or similar. – Attempt at forming quadratic equation.

Mid partial credit: (3 marks) – Finds 5x2 – 8x – 4 = 0 and stops or continues to attempt to find factors.

– Finds one correct factor, i.e. –5

2 or 2.

High partial credit: (4 marks) – Finds one point of intersection correctly.– Finds both factors, but fails to find or finds incorrect points of intersection.



Question 3 (50)

The line l contains the point A (0, 2) and has a slope 3.

3(a) Find the equation of l. Give your answer in the form ax + by + c = 0, where a, b, c ∈ ℤ. (5C)

ml , slope of l = 3

Equation of l y – y1 = m(x – x1) y – 2 = 3(x – 0) = 3x 3x – y + 2 = 0

or

ml , slope of l = 3

Equation of l y = mx + c 2 = 3(0) + c c = 2

y = 3x + 2 3x – y + 2 = 0

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down correct relevant formula for the equation of a line and stops. – Some correct substitution into relevant formula and stops.

High partial credit: (4 marks) – Substitutes correctly into equation of a line formula, but fails to finish or finishes incorrectly. – Substitutes almost correctly into equation of a line formula (allow one incorrect or omitted substitution) and finishes correctly.– Finds correct equation for l, but fails to give final answer in the desired form (ax + by + c = 0).

3(b) (i) Determine whether the line k: x + 3y – 26 = 0 is perpendicular to l. (5C)

Equation of k x + 3y – 26 = 0 3y = –x + 26

y = –3

1x +

3

26

y = mx + c

mk , slope of k = –3

1

mk × ml = –3

1 × 3

= –1 lines l and k are perpendicular




3(b) (i) (cont’d.)

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down correct relevant condition for ⊥ lines (i.e. mk × ml = 1) or equation of a line (i.e. y = mx + c) and stops.

High partial credit: (4 marks) – Finds correct slope for mk , but fails to finish or finishes incorrectly. – Finds mk × ml = –1, but omits or incorrect conclusion.

(ii) Find the co-ordinates of B, the point of intersection of the lines l and k. (5D)

l: 3x – y + 2 = 0 (× 3) k: x + 3y – 26 = 0 (× 1)

9x – 3y = –6 x + 3y = 26 10x = 20 x = 2

3x – y = –2 3(2) – y = –2 6 – y = –2 –y = –2 – 6 = –8 y = 8

x + 3y = 26 2 + 3y = 26 3y = 26 – 2 = 24 y = 8

or

l: 3x – y + 2 = 0 (× 1) k: x + 3y – 26 = 0 (× –3)

3x – y = –2 –3x – 9y = –78 –10y = –80 y = 8

3x – y = –2 3x – 8 = –2 3x = –2 + 8 = 6 x = 2

x + 3y = 26 x + 3(8) = 26 x + 24 = 26 x = 26 – 24

= 2


Scale 5D (0, 2, 3, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. multiplies equation(s) correctly by appropriate constant(s) to facilitate the cancellation of x or y term.

Mid partial credit: (3 marks) – Finds 10x = 20 or 10y = 80, 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: (4 marks) – Finds first variable (x or y), 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) correctly by graphical means. – Finds both variables (x and y) by trial and error, but does not verify in both equations or verifies incorrectly.




3(c) k intersects the x-axis at the point C.

(i) Find the co-ordinates of C. (5C)

k: x + 3y – 26 = 0

k ∩ x-axis: y = 0 x + 3(0) – 26 = 0 x = 26 co-ordinates of C = (26, 0)

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit, e.g. writes down y = 0 and stops. – Writes down y = 0 and continues.

High partial credit: (4 marks) – Correct substitution into equation, but fails to finish or finishes incorrectly. – Correct answer with no work shown. – Finds x = 26, but fails to show answer in (x, y) format.

(ii) Hence, or otherwise, find the area of the triangle ABC. (5C*)

A(0, 2) → (0, 0) 0 → 0: ↑ 0 2 → 0: ↓ 2

B(2, 8) → (2 + 0, 8 – 2) → (2, 6)

C(26, 0) → (26 + 0, 0 – 2) → (26, –2)

Area of ΔABC = 2

1| x1y2 – x2y1

|

= 2

1| (2)(–2) – (26)(6) |

= 2

1| –4 – 156 |

= 2

1| –160 |

= 80 units2

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

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit, e.g. writes down correct relevant formula for the area of a triangle and stops. – Correct translation of at least one point.

High partial credit: (4 marks) – Substitutes correctly into relevant area formula, but fails to finish or finishes incorrectly. – Substitutes almost correctly into relevant area formula (allow one incorrect or omitted value) and finishes correctly.

* Deduct 1 mark off correct answer only for the omission of or incorrect use of units (‘units2’) - apply only once to each section (a), (b), (c), etc. of question.



Question 4 (50)

The diagram shows two triangles, ACB and FCD. Both triangles have right angles at C.

The point D is the midpoint of [ AC ]. [ FD ] is extended such that it intersects [ AB ] at the point E. | AE | = | ED |.

4(a) Prove that triangles ACB and FCD are similar. Justify your answer by setting out a series of steps with geometrical statements supporting each step. (10C)

| ∠FCD | = | ∠ACB | ... both angles are 90° (given)

| ∠CDF | = | ∠ADE | ... vertically opposite angles are equal but | ∠EAD | = | ∠ADE | ... the angles opposite the equal sides in

an isosceles triangle are equal | ∠CDF | = | ∠EAD | i.e. | ∠CDF | = | ∠BAC |

| ∠DFC | = | ∠CBA | ... remaining angles in the two triangles are equal if the other two angles in the two triangles are equal

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down one or two correct steps without supporting geometrical statement. – Writes down one correct step with supporting geometrical statement.

High partial credit: (7 marks) – Writes down three correct steps without supporting geometrical statements. – Writes down two correct steps with supporting geometrical statements.

4(b) Hence, show that the triangle EFB is isosceles. (5C)

From step in part (a): | ∠DFC | = | ∠CBA |

but ∠DFC is ∠EFB and ∠CBA is ∠FBE | ∠EFB | = | ∠FBE | triangle EFB is isosceles

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down correct step from part (a) and stops. – Finds | ∠DFC | = | ∠CBA | by another method. – Identifies ∠DFC as ∠EFB and/or ∠CBA as ∠FBE.

High partial credit: (4 marks) – Finds | ∠DFC | = | ∠CBA | and identifies ∠DFC as ∠EFB and ∠CBA as ∠FBE, i.e. finds | ∠EFB | = | ∠FBE |, but omits or incorrect conclusion.


C

D

F B

A

E



4(c) Given that | AE | = 2 cm, and using your answers from parts (a) and (b), find | BE |. (Hint: Let | BE | = x and | DF | = y.) (10C)

| BE | = x | DF | = y

ΔEFB is isosceles | BE | = | FE | x = y + 2 y = x – 2

ΔACB and ΔFCD are similar

||

||

DF

AB =

||

||

DC

AC

y

x 2+ =

||

||2

DC

DC

x + 2 = 2y

Substitute y = x – 2 into equation: x + 2 = 2(x – 2) x + 2 = 2x – 4 2x – x = 2 + 4 x = 6 | BE | = 6 cm

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – Some work of merit, e.g. identifies ΔEFB as isosceles and | BE | = | FE | and stops. – Finds y = x – 2 [step ] and stops.

– Identifies ΔACB and ΔFCD as similar

triangles and ||

||

DF

AB =

||

||

DC

AC and stops.

– Finds y

x 2+ = 2 or x + 2 = 2y [step ]

and stops.

High partial credit: (7 marks) – Finds both y = x – 2 [step ] and

y

x 2+ = 2 or x + 2 = 2y [step ], but

fails to finish or finishes incorrectly.



Question 5 (50)

A game consists of rolling two fair, six-sided dice, one red and one black. The dice are thrown at the same time and the scores are then added. The red die has the numbers 1, 2, 2, 3, 4 and 5 on its sides. The black die has the numbers 1, 2, 3, 3, 4 and 6 on its sides.

5(a) Complete the table below to show all possible outcomes of the game. (10C)

Red Die

1 2 2 3 4 5

Bla

ck D

ie

1 2 3 3 4 5 6

2 3 4 4 5 6 7

3 4 5 5 6 7 8

3 4 5 5 6 7 8

4 5 6 6 7 8 9

6 7 8 8 9 10 11

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – At least six correct outcomes.

High partial credit: (7 marks) – No more than six incorrect outcomes.

5(b) It costs €2·50 to play the game. A score of 6, 7 or 8 wins €3. A score of 9, 10 or 11 wins €10. Calculate the expected value for the game and determine whether it is a fair game or not. (15D*)

From the above table:

P(score of 6, 7 or 8) = 36

16

= 9

4

P(score of 9, 10 or 11) = 36

4

= 9

1

Expected value, E(X) = Σ x.P(x)

= 3 × 9

4 + 10 ×

9

1

= 9

12 +

9

10

= 9

22

= 2·444444... ≅ €2·44

Conclusion: game is unfair – as €2·44 < €2·50 (cost to play game)




5(b) (cont’d.)

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

Scale 15D* (0, 6, 10, 13, 15) Low partial credit: (6 marks) – Some work of merit, e.g. writes down correct relevant formula for expected value. – Finds one correct probability,

i.e. P(score of 6, 7 or 8) = 36

16 /

9

4,

or P(score of 9, 10 or 11) = 36

4 /

9

1.

Mid partial credit: (10 marks) – Finds two correct probabilities and stops or continues incorrectly.

High partial credit: (13 marks) – Finds E(X) = 3 ×

9

4 + 10 ×

9

1, but fails

to evaluate or evaluates incorrectly.

– Finds E(X) = 2·444444... or €2·44, but omits or incorrect conclusion.

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



Question 6 (50)

A restaurant offers a special two-course set menu for lunch as shown below.

Two-Course Set Menu €18 per person

Tomato Soup Chicken Wings

Garlic Bread

*******************

Beef Burger Roast Chicken

Vegetarian Lasagne Salmon

6(a) Assuming that only one selection from each course can be chosen, how many different combinations are possible? (5B)

# Combinations = 3 (starters) × 4 (main courses) = 12

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Some work of merit, e.g. identifies 3 as the number of starter options or 4 as the number of main course options. – Uses incorrect operation to find # Combinations, i.e. 3 + 4 or 7.

6(b) (i) If all combinations are equally likely, what is the probability that a customer selects chicken for both courses? (5C)

P(chicken for both courses)

= 3

1 ×

4

1

= 12

1 or 0·083333...


Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. finds at least one correct probability. – Adds probabilities together instead of multiplying them.


3

1 ×

4

1, but fails to evaluate or

evaluates incorrectly. – Correct answer, but no work shown.




6(b) (cont’d.)

(ii) What is the probability that a customer selects a vegetarian option (no meat or fish) for both courses? (5C)

Vegetarian options = tomato soup, vegetarian lasagne = garlic bread, vegetarian lasagne

# Combinations = 2 × 1 = 2

P(vegetarian option) = 12

2

= 6

1 or 0·166666...

or

P(vegetarian starter) = 3

2

P(vegetarian main course) = 4

1


2 ×

4

1

= 12

2

= 6

1 or 0·166666...


Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down all vegetarian options [method ] or ‘P(vegetarian option) = P(vegetarian starter) × P(vegetarian main course)’ [method ]. – Finds # Combinations = 2 [method ].

– Finds P(vegetarian option) = x

2, x ≠ 12,

x > 2 [method ]. – Finds one correct probability [method ].– Adds probabilities together instead of multiplying them [method ].


3

2 ×

4

1, but fails to evaluate or

evaluates incorrectly [method ]. – Correct answer, but no work shown [method or ].




6(b) (cont’d.)

(iii) Amy goes to the restaurant for lunch on three consecutive days. What is the probability that she selects a vegetarian option for both courses on two of these days? (5C)


1 ... from part (b)(ii)

P(not vegetarian option) = 1 – 6

1

= 6

5

P(vegetarian option, vegetarian option, not vegetarian option (in any order))

=

2

3 ×

6

1 ×

6

1 ×

6

5

= )!23(!2

!3

− ×

216

5

= 112

123

××××

× 216

5

= 3 × 216

5

= 216

15 or

72

5 or 0·069444...

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

‘P(not vegetarian option) = 1 – 6

1 or

6

5’.

– Finds 6

1 ×

6

5,

6

1 ×

6

5 or

6

5 ×

6

5 and

evaluates correctly.


6

1 ×

6

1 ×

6

5 or similar, but fails to

evaluate or evaluates incorrectly.

– Finds

2

3 +

6

1 +

6

1 +

6

5 or similar, but

fails to evaluate or evaluates incorrectly.– Correct answer, but no work shown.




6(c) The restaurant wishes to add one more option to the set menu. To which course should this option be added in order to maximise the number of different combinations possible? Justify your answer. (5C)

Extra starter: # New combinations = 4 (starters) × 4 (main courses) = 16

Extra main course: # New combinations = 3 (starters) × 5 (main courses) = 15

as 16 > 15, add extra option to starter to maximise number of combinations


Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down ‘4 (starters) × 4 (main courses)’ or ‘3 (starters) × 5 (main courses)’. – Finds # New combinations (extra starter) = 16 or # New combinations (extra main course) = 15. – Correct answer without supporting calculations.

High partial credit: (4 marks) – Finds # New combinations (extra starter) = 16 and # New combinations (extra main course) = 15, but omits or incorrect conclusion.



Section B Contexts and Applications 150 marks Answer all three questions from this section.

Question 7 (45) ** 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 to each section (a), (b), (c), etc. of the question.

The diagram shows the cylinder P, which has a radius of 3 cm and a height of 10 cm.

7(a) (i) Find the curved surface area of cylinder P. Give your answer in terms of π. (5C*)

Curved surface area = 2πrh = 2π(3)(10) = 60π cm2

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down correct relevant formula for curved surface area and stops. – Some correct substitution into formula and stops.

High partial credit: (4 marks) – Substitutes almost correctly into formula (allow one incorrect or omitted substitution) and finishes correctly. – Substitutes correctly into formula, but fails to finish or finishes incorrectly. – Finds answer in decimal or fraction form.

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


10 cm

3 cm



7(a) (cont’d.)

(ii) Make an accurate scaled drawing below of the net of cylinder P, using the scale 1:2. Record the dimensions of the net on your drawing. Give your answers correct to two decimal places. (10D*)

** Actual size as per examination paper.

Scale 10D* (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. draws a circle or rectangle to any scale and stops. – Finds the length of the circle. – Finds one scaled dimension correctly.

Mid partial credit: (6 marks) – Draws one surface correctly to scale with no dimensions indicated. – Finds all scaled dimensions, but fails to draw or draws incorrect diagram.

High partial credit: (8 marks) – Draws one surface correctly to scale with dimensions indicated. – Draws all surfaces correctly to scale, but no dimensions indicated.

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


10 cm

2 (3) = 18 85 cm��

3 cm

3 cm



7(b) Another cylinder Q is an enlargement of cylinder P. The curved surface area of cylinder Q is 375π cm2.

(i) Show that the scale factor of the enlargement is 2·5. (10C)

scale factor = k CSAQ = k2CSAP

k2 = P

Q

CSA

CSA

= ππ

60

375

= 6·25 or 4

25

k = 256⋅ or 4

25

= 2·5 or 2

5

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down correct relevant formula and stops.

– Finds k = ππ

60

375 or similar, and stops

or continues.

High partial credit: (7 marks) – Substitutes correctly into formula,

i.e. finds k2 = ππ

60

375 or 375π = k2(60π),


(ii) Hence, find the radius and height of the cylinder Q. (5C*)

| rP | = 3

| rQ | = k | rP

| = 2·5(3) = 7·5 cm

| hP | = 10

| hQ | = k | hP

| = 2·5(10) = 25 cm

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down correct relevant formula and stops. – Substitutes correctly into formula to find | rQ

| or | hQ |, but fails to evaluate or

evaluates incorrectly.

High partial credit: (4 marks) – Substitutes correctly into formula to find | rQ

| and | hQ |, but fails to evaluate or

evaluates incorrectly either length.

– Finds correct | rQ | or | hQ

|.

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




7(c) (i) Find the volume of cylinder Q in terms of π. (5C*)

VQ = πr2h = π(7·5)2(25) = 1,406·25π cm3

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down correct relevant formula for volume of a cylinder and stops. – Some correct substitution into formula and stops.

High partial credit: (4 marks) – Substitutes almost correctly into formula (allow one incorrect or omitted substitution) and finishes correctly. – Substitutes correctly into formula, but fails to finish or finishes incorrectly. – Finds answer in decimal or fraction form.

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

(ii) A sphere has the same volume as the cylinder Q. Find the radius of the sphere, correct to two decimal places. (10C*)

VSphere = 3

4πr3

= VQ = 1,406·25π

3

4πr3 = 1,406·25π

r3 = 4

3(1,406·25)

= 1,054·6875

r = 3 6875054,1 ⋅

= 10·179066... ≅ 10·18 cm

Scale 10C* (0, 4, 7, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down correct relevant formula for the volume of a sphere and stops.


4πr3 = 1,406·25π or similar, but

fails to finish or finishes incorrectly.

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



Question 8 (35)

The horizontal surface of a wall shelf is in the shape of part of a sector of a circle, as shown. The radius of the circle is 104 cm.

The back edge of the shelf, which attaches to the wall, is a chord of the circle. The length of this back edge is 160 cm.

8(a) Find the measure of , the angle subtended at the centre of the circle by the arc of the sector. Give your answer, in degrees, correct to one decimal place. (10C*)

Cosine Rule:

a2 = b2 + c2 – 2bc cos A 1602 = 1042 + 1042 – 2(104)(104)cos

cos | ∠ | = )104)(104(2

160104104 222 −+

= 632,21

600,25816,10816,10 −+

= 632,21

968,3−

= –0·183431... | ∠ | = cos–1(–0·183431...) = 100·569725... ≅ 100·6°

Scale 10C* (0, 4, 7, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down correct formula for cosine rule and stops.– Some correct substitution into relevant formula for cosine rule.

High partial credit: (7 marks) – Correct substitution into formula for cosine rule, but fails to finish or finishes incorrectly. – Substitutes almost correctly into formula for cosine rule (allow one incorrect or omitted substitution) and finishes correctly.– Finds cos , but fails to find or finds incorrect value for | ∠ |. – Incorrect calculator mode - apply once only [Radian: ans. = 1·755272...; Gradian: ans. = 111·744139...]. – Finds correct answer, but no work shown.



104cm

160 cm

Shelf

�



8(a) (cont’d.)

or

Trigonometric method:

= 2

θ

sin | ∠ | = |Hyp|

|Opp|

| Opp | = 2

160

= 80 cm

sin | ∠ | = 104

80

= 0·769230... | ∠ | = sin–1(0·769230...) = 50·284862...

| ∠ | = 2(50·284862...) = 100·569725... ≅ 100·6°

Scale 10C* (0, 4, 7, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down correct relevant trigonometric ratio (sin) and stops.

– Finds = 2

θ and | Opp | =

2

160or 80.

– Some correct substitution into correct trigonometric ratio.

High partial credit: (7 marks) – Correct substitution into trigonometric ratio and correctly manipulated, i.e. | ∠ | = sin–1(0·769230...) or 50·284862..., but fails to finish or finishes incorrectly. – Finds correct answer, but no work shown.



80 cm

�104 cm



8(b) Find the area of the horizontal surface of the wall shelf, as shown by the shaded region in the diagram above. Give your answer correct to two significant figures. (10D*)

Area of sector = 360

θπr2

= 360

6100⋅ π(104)2

= 9,495·373038...

Area of triangle = 2

1ab sin C

= 2

1(104)(104)(sin100·6°)

= 5,408(0·999955...) = 5,407·759815...

Area of the horizontal surface of the shelf (shaded region) = 9,495·373038... – 5,407·759815... = 4,087·613223... ≅ 4,100 cm2

Scale 10D* (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down correct relevant formula for the area of a sector or area of a triangle and stops.– Some correct substitution into relevant formula for area of sector or triangle. – Finds correct answer for area of sector or triangle, but no work shown.

Mid partial credit: (6 marks) – Finds area of sector = 360

6100⋅ π(104)2 or

area of triangle = 2

1(104)(104)(sin100·6°)

or similar, and stops or continues. – Finds correct area of sector or triangle and stops. – Finds correct answer for area of sector and triangle, but no work shown.

High partial credit: (8 marks) – Finds correct area of sector and triangle, but fails to finish or finishes incorrectly. – Finds correct area for sector or triangle and substitutes almost correctly into other formula (allow one incorrect or omitted substitution) and finishes correctly. – Finds correct answer, but no work shown.

* Deduct 1 mark off correct answer only if final answer(s) not rounded or incorrectly rounded or for the omission of or incorrect use of units (‘cm2’) - apply only once to each section (a), (b), (c), etc. of question.




8(c) A metal trim, of width 6·5 cm, is applied to the front edge of the shelf along the arc of the sector. Find the area of metal trim required. Give your answer correct to the nearest whole number. (10C*)

Length of arc = 360

θ × 2πr

= 360

6100⋅ × 2π(104)

= 182·603327...

Area of trim = 6·5(182·603327...) = 1,186·9216255... ≅ 1,187 cm2

Scale 10C* (0, 4, 7, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down correct relevant formula for the length of an arc and stops. – Some correct substitution into relevant formula for length of arc.

High partial credit: (7 marks) – Correct substitution into formula for

length of arc, i.e. 360

6100⋅ × 2π(104), but

but fails to finish or finishes incorrectly. – Substitutes almost correctly into formula for length of arc (allow one incorrect or omitted substitution) and finishes correctly.– Finds 182·603327..., but fails to find or finds incorrect area of trim. – Finds correct answer, but no work shown.

* Deduct 1 mark off correct answer only if incorrectly rounded or not rounded or for the omission of or incorrect use of units (‘cm2’) - apply only once to each section (a), (b), (c), etc. of question.

8(d) Find the distance from the front edge of the shelf to the wall. Give your answer correct to one decimal place. (5C*)

Theorem of Pythagoras:

| Opp |2 + | Adj |2 = | Hyp |2

| Opp | = 2

601

= 80 | Hyp | = r = 104

| h |2 = 1042 – 802 = 10,816 – 6,400 = 4,416

| h | = 416,4

= 66·452990... ≅ 66·5 cm

| d | = 104 – | h | = 104 – 66·5 = 37·5 cm


80 cm

�

h104 cm

d



8(d) (cont’d.)

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

High partial credit: (4 marks) – Finds | h | = 416,4 , 66·452990... or 66·5,


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

Trigonometric method:

= 2

θ

| ∠ | = 2

6100⋅

= 50·3°

cos | ∠ | = |Hyp|

|Adj|

| h | = 104 × cos 50·3° = 104 × 0·638767... = 66·431853... ≅ 66·4 cm

or

tan | ∠ | = |Adj|

|Opp|

| h | = °⋅350tan

80 ... | Opp | =

2

601 = 80

= ...2045051

80

⋅

= 66·417279... ≅ 66·4 cm

| d | = 104 – | h | = 104 – 66·4 = 37·6 cm

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down correct relevant trigonometric ratio (cos or tan) and stops.

– Finds | ∠ | = 2

6100⋅ or 50·3° and stops.

– Finds cos 50·3° = 0·638767... or tan 50·3° = 1·204505... and stops. – Some correct substitution into correct trigonometric ratio.

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

| h | =104 × 0·638767... or = ...2045051

80

⋅,

but fails to finish or finishes incorrectly. – Finds correct | h | and stops. – Finds correct answer, but no work shown.

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


80 cm

�

h104 cm

d


Question 9 (70)

The heights and arm spans, in cm, of a random sample of students in a school were recorded. The data collected is shown below.

Height Arm Span Height Arm Span

140 143 169 171 140 149 170 162 153 155 173 168 156 163 177 166 160 150 177 177 160 160 178 178 161 159 181 174 162 160 183 170 162 162 187 181 165 157 188 178 165 165 189 173 165 167 191 192

9(a) (i) What type of data was collected? Explain your answer. (5C)

Type of data – numerical continuous

Explanation – numerical: both height and arm span data can be represented by numbers

– continuous: data can take any value between any two whole numbers // 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.

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

(ii) Explain the term random sample. (5B)

Random sample Any 1: – a subset of a population that is representative

of the population // – where every student in the school has an equal chance

of being selected // – an unbiased representation of all students in the

school // etc.


Scale 5B (0, 2, 5) Partial credit: (2 marks) – Insufficient or incomplete explanation, but with some merit, e.g. writes down ‘every student in the school’ or ‘equal chance of being selected’, etc.




9(a) (cont’d.)

(iii) Complete the axes of the scatter graph below and plot the data collected. (10D)

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Correct scales with at least four points correctly plotted.

Mid partial credit: (6 marks) – Correct scales with between five and twelve points correctly plotted.

High partial credit: (8 marks) – Correct scales with more than twelve points correctly plotted. – All points plotted but scales incorrect.

(iv) Describe the correlation, if any, in the context of the data. (5B)

Correlation – weak positive correlation

Context – evidence is weak to conclude that the height and arm span of students are interconnected // etc.


Scale 5B (0, 2, 5) Partial credit: (2 marks) – Correct answer, but explanation given insufficient or incorrect. – Incorrect or no answer given, but sufficient explanation including mention of ‘weak’ or ‘positive’ in answer.


Arm Span (cm)

Hei

gh

t(c

m)

140 150 160 170 180 190 200

130

140

150

160

170

180

190

200



9(b) The mean height of a larger group of 15-year-old girls in school is 163 cm with a standard deviation of 13 cm. Assuming the girls’ heights are normally distributed, use the Empirical Rule to find the percentage of girls who are less than 176 cm in height. (10C)

x = 163

176 = 163 + 13 = x + σ

Empirical Rule states that 68% of data falls within one standard deviation of the mean

68% of the distribution lies in the range x – σ and x + σ

x + σ = 163 + 13 = 176

16% of the distribution lies in the range above x + σ

% of girls less than 176 cm in height = 68 + 16 = 84%

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – Any work of merit, e.g. states that 68% of distribution lies between x – σ and x + σ. – Some correct substitution into x + σ, but fails to finish or finishes incorrectly.

High partial credit: (7 marks) – Finds x + σ = 176, but fails to finish or finishes incorrectly. – Correct answer with no work shown.





9(c) The data of a sample group from this student population was used to create the information shown in the table below.

Height (cm) 145-150 150-155 155-160 160-165 165-170 170-175 175-180 180-185

Number of girls 72 120 178 240 167 90 82 51

(i) Use mid-interval values of the information in the table to estimate the mean height of the girls in this sample. (10C*)

Mean = 51829016724017812072

)515182()825177(...)1205152()725147(

+++++++×⋅+×⋅++×⋅+×⋅

= 000,1

5307,9555,14...300,18620,10 ⋅++++

= 000,1

315,163

= 163·315 cm


correct formula for mean, f

xf

, and stops.

– Writes

sampleinGirls

sampleingirlsallofHeights

and stops or continues. – Finds correct Σ f and stops (ans. 1,000).

High partial credit: (7 marks) – Finds correct Σ f x , but fails to finish or finishes incorrectly.

– Finds f

xf

(with only one error) and

finishes correctly/consistently.

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




9(c) (cont’d.)

(ii) Find the margin of error, at 95% confidence, for this sample. Write your answer as a percentage, correct to one decimal place. (5C*)

Margin of Error = n

1

Number of girls in sample = 72 + 120 + 178 + 240 + 167 + 90 + 82 + 51 = 1,000

For n = 1,000

Margin of error = 000,1

1

= 0·031622... = 3·2%

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down correct formula for margin of error and stops. – Writes down n = 1,000 and stops.

High partial credit: (4 marks) – Substitutes correctly into relevant formula, but fails to finish or finishes incorrectly. – Finds value correctly, but fails to give final answer as a percentage.

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

(iii) Estimate the percentage of girls in the sample who are less than 176 cm in height. (5C)

Within the 175-180 cm interval Number of girls < 176 cm

= 175180

175176

−−

(82)

= 5

1(82)

= 16·4 ≅ 16

Overall number of girls < 176 cm = 72 + 120 + 178 + 240 + 167 + 90 + 16 = 883

% of girls in sample < 176 cm

= 000,1

883 ×

1

100

= 88·3%

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down 72 + 120 + 178 + 240 + 167 + 90 + x or 867 + x and stops. – Finds ‘Number of girls < 176 cm’ within 175-180 cm interval.

High partial credit: (4 marks) – Finds correct answer for ‘Overall number of girls < 176 cm’ [ans. 883], but fails to finish or finishes incorrectly. – Correct answer with no work shown.





9(c) (cont’d.)

(iv) Create a 95% confidence interval for the percentage of girls in the sample who are less than 176 cm in height. (5C)

Margin of error = n

1

= 000,1

1

= 0·031622... = 3·2% ... answer from (ii) = 0·032

% of girls in sample < 176 cm = 88·3% ... answer from (iii) p̂ = 0·883

Confidence interval

p̂ – n

1 < p < p̂ +

n

1

p̂ – n

1 = 0·883 – 0·032

= 0·851

p̂ + n

1 = 0·883 + 0·032

= 0·915

0·851 < p < 0·915 85·1% < p < 91·5%

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

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

p̂ – n

1 < p < p̂ + n

1 , and stops.

– Finds correct margin of error and/or p̂.

High partial credit: (4 marks) – Finds 0·883 – 0·032 < p < 0·883 + 0·032, but fails to finish or finishes incorrectly. – Finds one side only and finishes correctly, i.e. 85·1 < p or p < 91·5.





9(d) A student claims that the proportion of girls in the sample group who are less than 176 cm in height is the same as in the student population of 15-year-old girls.

Use a hypothesis test at the 5% level of significance to decide whether there is sufficient evidence to accept this student’s claim. State clearly the null hypothesis and the alternative hypothesis. Interpret your conclusion in the context of the question. (10D)

From part (b): % of 15-year-old girls in the student population less than 176 cm in height = 84%

H0: 84% of girls in the sample group are less than 176 cm in height H1: 84% or more of girls in the sample group are less than 176 cm in height

From part (c)(iv): Confidence interval: 85·1% < p < 91·5%

Conclusion: – as 84% is outside this confidence interval, 85·1% < p < 91·5%,

we reject the null hypothesis H0 and accept the alternative hypothesis H1

Interpretation: – the percentage of students in the sample who are less than 176 cm in height

is not the same as in the student population of 15-year-old girls

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

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Any work of merit, e.g. states null hypothesis and stops. – Writes down both null and alternative hypotheses and stops.

Mid partial credit: (6 marks) – Writes down correct confidence interval and stops or continues incorrectly, e.g. accepts the null hypothesis.

High partial credit: (8 marks) – Finds correct conclusion, but fails to contextualise answer, i.e. writes down ‘reject the null hypothesis’ and stops.



Notes:



Notes:



Notes:



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