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L.16/19

Pre-Leaving Certificate Examination, 2018

Mathematics Ordinary Level

Marking Scheme

Paper 1 Pg. 2 Paper 2 Pg. 36

2018.1 L.16/19_MS 2/76 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 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 for the omission of units in 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

2018.1 L.16/19_MS 3/76 of 75 examsDEB

Summary of Marks – 2018 LC Maths (Ordinary Level, Paper 1)

Section A Section B

Q.1 (a) (i) 10C (0, 4, 7, 10) Q.7 (a) (i) 5C (0, 2, 4, 5) (ii) 10C (0, 4, 7, 10) (ii) 5C (0, 2, 4, 5) (b) 5D (0, 2, 3, 4, 5) (iii) 5C (0, 2, 4, 5) 25 (iv) 5C (0, 2, 4, 5) (b) (i) 10C (0, 4, 7, 10) (ii) 10D* (0, 4, 6, 8, 10) (c) (i) 5C* (0, 2, 4, 5) Q.2 (a) (i) 5C (0, 2, 4, 5) (ii) 5D* (0, 2, 3, 4, 5) (ii) 10D (0, 4, 6, 8, 10) 50 (b) (i) 5C (0, 2, 4, 5) (ii) 5B (0, 2, 5) 25 Q.8 (a) (i) 5C (0, 2, 4, 5) (ii) 5C* (0, 2, 4, 5) (b) 10D (0, 4, 6, 8, 10) Q.3 (a) (i) 10C (0, 4, 7, 10) (c) (i) 10C (0, 4, 7, 10) (ii) 5C (0, 2, 4, 5) (ii) 5D (0, 2, 3, 4, 5) (b) 10C (0, 4, 7, 10) (d) (i) 5D (0, 2, 3, 4, 5) 25 (ii) 5C* (0, 2, 4, 5) (iii) 5B (0, 2, 5) 50 Q.4 (a) (i) 2 × 5C (0, 2, 4, 5) (ii) 5D (0, 2, 3, 4, 5) (b) (i) 5C (0, 2, 4, 5) Q.9 (a) (i) 5C* (0, 2, 4, 5) (ii) 5C (0, 2, 4, 5) (ii) 5C (0, 2, 4, 5) 25 (b) (i) 10C (0, 4, 7, 10) (ii) 10D* (0, 4, 6, 8, 10) (iii) 10D (0, 4, 6, 8, 10) (iv) 10D (0, 4, 6, 8, 10) Q.5 (a) 15D* (0, 6, 10, 13, 15) 50 (b) 10D (0, 4, 6, 8, 10) 25 Q.6 (a) 5B (0, 2, 5) (b) (i) 5C (0, 2, 4, 5) (ii) 5D (0, 2, 3, 4, 5) (c) 10D (0, 4, 6, 8, 10) 25

Current Marking Scheme

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.

Copyright

All rights reserved. This marking scheme and corresponding papers(s) are protected by Irish (EU) copyright law. Reproduction and distribution of these materials or any portion thereof without the written permission of the publisher is prohibited except for the immediate use within a classroom.

2018.1 L.16/19_MS 4/76 of 75 examsDEB


Mathematics


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 nine 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.

Section A Concepts and Skills 150 marks

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

Question 1 (25)

1(a) Joe earns a gross wage of €855 for a standard 38-hour working week. He pays income tax, universal social charge (USC) and pay-related social insurance (PRSI) on his gross wage.

(i) Joe pays income tax at the rate of 20% on the first €670 he earns and 40% on the balance. He has weekly tax credits of €63. How much income tax does Joe pay weekly? (10C)

Tax @ 20% = 670 × 20%

= 670 × 100

20

= €134

Tax @ 40% = (855 – 670) × 40%

= 185 × 100

40

= €74

Gross tax = Tax @ 20% + Tax @ 40% = 134 + 74 = €208

Net tax = Gross tax – Tax credits = 208 – 63 = €145

examsDEB

2018.1 L.16/19_MS 5/76 of 75 examsDEB

Question 1 (cont’d.)

1(a) (i) (cont’d.)

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – Some work of merit, e.g. sets up 20% of €670. – Finds tax correctly @ 20% or @ 40% of some relevant figure and stops or continues incorrectly.

High partial credit: (7 marks) – Finds Gross tax per week correctly, but fails to find or finds incorrect Net tax per week.

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

(ii) Joe also pays USC and PRSI on his gross wage. USC amounts to €29⋅10 each week and he pays PRSI on his gross wage. His net weekly take-home pay is €646⋅70. Find the percentage rate at which Joe pays PRSI. (10C)

Take-home pay = Gross pay – Income tax – USC – PRSI = 855 – 145 – 29·10 – PRSI = €646·70

855 – 145 – 29·10 – PRSI = 646·70 PRSI = 855 – 145 – 29·10 – 646·70 = €34·20

% PRSI = 855

2034⋅ ×

1

100

= 0·04 × 100 = 4%

** Accept students’ answers for Income tax paid from part (i) if not oversimplified.

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – Some work of merit, e.g. calculates 855 – 145 – 29·10 or 855 – 646·70 and stops or continues incorrectly.

High partial credit: (7 marks) – Finds PRSI correctly [ans. €34.20], but fails to find % PRSI or finds incorrect % PRSI.

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

2018 LC Maths [OL] – Paper 1

2018.1 L.16/19_MS 6/76 of 75 examsDEB


1(b) Joe is paid “time and a half” for weekday overtime and weekend work. In addition to income tax and PRSI, he pays USC at the rate of 5% on his extra earnings.

Find the minimum number of hours that Joe must work above his standard working week in order to receive a net weekly take-home pay in excess of €800. (5D)

Additional pay = 800 – 646·70 = €153·30

Normal pay per hour = 38

855

= €22·50

Gross overtime pay per hour = 22·50 × 1·5 = €33·75

Total deductions on each overtime hour = 40% Income tax + 5% USC + 4% PRSI = 49% of €33·75

Net overtime pay per hour = 51% of €33·75 = 17·2125

Overtime required for take-home pay to be above €800

= 212517

30153

⋅⋅

= 8·906318... = 9 hours

** Accept students’ answers for % PRSI from part (a)(ii) if not oversimplified.

Scale 5D (0, 2, 3, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. finds additional pay [ans. €153·30], Normal pay per hour [ans. €22·50] or Gross overtime pay per hour [ans. €33·75] and stops.

Mid partial credit: (3 marks) – Finds ‘Total deductions on each overtime hour’ [ans. 49% of €33·75 or €16·5375] and stops or continues incorrectly.

High partial credit: (4 marks) – Finds ‘Net overtime pay per hour’ [ans. 51% of €33·75 or €17·2125], but fails to finish or finishes incorrectly.


2018.1 L.16/19_MS 7/76 of 75 examsDEB

Question 2 (25)

z1 = 3 + 4i is a complex number, where i2 = –1.

2(a) (i) Let z2 = 2iz1. Find z2, in the form a + bi, where a, b ∈ ℝ. (5C)

z2 = 2iz1 = 2i(3 + 4i) = 6i + 8i2 = 6i + 8(–1) = –8 + 6i

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. substitutes correctly z1 into 2iz1 [ans. 2i(3 + 4i)] and stops or continues incorrectly.

High partial credit: (4 marks) – Correct multiplication, i.e. finds 6i + 8i2, but fails to finish or finishes incorrectly.

(ii) Let z3 = i

z

21 . Find z3, in the form a + bi, where a, b ∈ ℝ. (10D)

z3 = i

z

21

= i

i

2

43 +

= i

i

2

43 + ×

i

i

2

2

−−

= 2

2

4

86

i

ii

−−−

= )1(4

)1(86

−−−−− i

= 4

68 i−

= 2

34 i− // 2 –

2

3i

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

correctly z1 into i

z

21 [ans.

i

i

2

43 +]

and stops or continues incorrectly. – Correctly identifies –2i.

Mid partial credit: (6 marks) – Finds

i

i

2

43 + ×

i

i

2

2

−−

, 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

4

86

i

ii

−−−

or similar, but fails

to finish or finishes incorrectly.


2018.1 L.16/19_MS 8/76 of 75 examsDEB


2(b) (i) Plot each of the points z1, z2 and z3 on the given Argand diagram and label each point clearly. (5C)

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.

(ii) Make one observation about the relative positions of the points you plotted on the diagram above. (5B)

Observations: Any 1: – z2 is twice as far from the origin as z1 / modulus of z2

is twice the modulus of z1 / | z2

| = 2| z1 | //

– z1 is twice as far from the origin as z3 / modulus of z1 is twice the modulus of z3 / |

z1 | = 2| z3

| // – z2 = 2iz1 rotates z1 through 90° anti-clockwise //

– z3 = i

z

21 rotates z1 through 90° clockwise

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Some work of merit, e.g. states that three points are in different quadrants, z2 = 2iz1

rotates z1 anti-clockwise, z3 = i

z

21 rotates

z1 clockwise or similar.


Re( )z

Im( )zz2 / �2iz1 / 8 6� i

z iz i3 1/2 / 8 6� �

z1 /3 4� i

�2�4�6�8 2 4 6

2

�2

4

�4

6

�6

ii

zz

2

32/

2/ 1

3 �

2018.1 L.16/19_MS 9/76 of 75 examsDEB

Question 3 (25)

3(a) (i) Solve for x:

4(5 + 2x) – 5 = 5x – 3(1 – 2x), where x ∈ ℝ. (10C)

4(5 + 2x) – 5 = 5x – 3(1 – 2x) 20 + 8x – 5 = 5x – 3 + 6x 8x + 15 = 11x – 3 8x – 11x = –3 – 15 –3x = –18 3x = 18 x = 6

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – Some work of merit, e.g. any correct attempt at simplifying equation, e.g. 4(5 + 2x) – 5 = 20 + 8x – 5, etc. – Simplifies one term correctly, i.e. –3x or 3x, 18 or –18. – Correct answer, but with no work shown.

High partial credit: (7 marks) – Both multiplications done correctly, i.e. simplifies to –3x = –18 or 3x = 18, but fails to finish or finishes incorrectly. – Error in expanding brackets, but finishes correctly.

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

4(5 + 2x) – 5 = 5x – 3(1 – 2x)

@ x = 6 4(5 + 2(6)) – 5 = 5(6) – 3(1 – 2(6)) 4(5 + 12) – 5 = 30 – 3(1 – 12) 4(17) – 5 = 30 – 3(–11) 68 – 5 = 30 + 33 63 = 63

as 63 = 63, x = 6 is a solution (of the equation)

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

High partial credit: (4 marks) – Finds 63 = 63, but no conclusion given (or incorrect conclusion if not equal).


2018.1 L.16/19_MS 10/76 of 75 examsDEB


3(b) Solve the inequality:

2(1 + 2x) – 8x ≥ –7, x ∈ ℤ,

and show the solution set on the number line below. (10C)

Solution

2(1 + 2x) – 8x ≥ –7 2 + 4x – 8x ≥ –7 2 – 4x ≥ –7 –4x ≥ –7 – 2 –4x ≥ –9 4x ≤ 9

x ≤

4

9

x ≤ 24

1

x = {2, 1, 0, –1, –2, –3, ...}

Number line

�2�3 2�1 0 1 3

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – Some work of merit, e.g. any correct transposition. – Mishandles inequality sign, but finishes

to x ≥ 4

9 or 2

4

1.

– Correct answer, but with no work shown.

High partial credit: (7 marks) – Finds x ≤ 4

9 or 2

4

1, but no solution set

given or no number line shown.

– Finds x ≤ 4

9 or 2

4

1, but number line

shows x ∈ ℕ or ℝ.


2018.1 L.16/19_MS 11/76 of 75 examsDEB

Question 4 (25)

The diagram shows the graph of the function f (x) = x2 + 2 in the domain –3 ≤ x ≤ 4, x ∈ ℝ.

4(a) (i) On the same diagram, draw the graph of the function g(x) = 2x + 5, x ∈ ℝ. (5C)

g(x) = 2x + 5

@ x = 0 g(0) = 2(0) + 5 = 5 (0, 5) ∈ g

@ x = 1 g(1) = 2(1) + 5 = 7 (1, 7) ∈ g

@ x = 2 g(2) = 2(2) + 5 = 9 (2, 9) ∈ g

@ y = 0 0 = 2x + 5 2x = –5

x = –2

5 // –2

2

1

(–22

1, 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.


g x( )

f x( )

�2�3 2�1 1 3 4

2

12

4

6

8

10

x

y

2018.1 L.16/19_MS 12/76 of 75 examsDEB


4(a) (i) (cont’d.)

Hence, 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 = 3

** 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 3. – Finds correctly y co-ordinate of one point of intersection, i.e. 3 or 11.

High partial credit: (4 marks) – Finds correctly (–1, 3) and (3, 11). – Finds correctly y co-ordinate of both points of intersection, i.e. 3 and 11.

(ii) Verify your answer to part (i) above by using algebra to solve g(x) = f (x). (5D)

g(x) = f (x) 2x + 5 = x2 + 2 x2 – 2x + 2 – 5 = 0 x2 – 2x – 3 = 0

(x – 3)(x + 1) = 0

x – 3 = 0 x = 3

x + 1 = 0 x = –1

Scale 5D (0, 2, 3, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. states relevant ‘–b’ formula. – Equates 2x + 5 = x2 + 2 and stops or continues incorrectly.

Mid partial credit: (3 marks) – Finds correct quadratic equation, i.e. x2 – 2x – 3 = 0 and stops or continues incorrectly.

High partial credit: (4 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.


2018.1 L.16/19_MS 13/76 of 75 examsDEB


4(b) (i) Find f ′(x), the derivative of f (x). Hence, find the value of x at which the tangent to the graph of f (x) is parallel to g(x). (5C)

f (x) = x2 + 2 f ′(x) = 2x

tangent to the graph of f (x) is parallel to g(x) slope of tangent = slope of g(x)

g(x) = 2x + 5 = mx + c mg = 2 mtangent = 2

f ′(x) = mg 2x = 2 x = 1

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. states that ‘slope of tangent equal to f ′(x)’ or similar. – f (x) correctly differentiated [ans. 2x] and stops or continues incorrectly. – Finds correct slope of g(x) [ans. 2] and stops or continues incorrectly.

High partial credit: (4 marks) – Finds f ′(x) and mg / mtangent correctly

and equates f ′(x) = mg / mtangent , but fails


(ii) Hence, find the equation of this tangent. (5C)

f (x) = x2 + 2 f (1) = (1)2 + 2 = 3

point of contact = (1, 3) mtangent = 2

y – y1 = m(x – x1) y – 3 = 2(x – 1) ... award full marks = 2x – 2 2x – y = –3 + 2 2x – y = –1

or

y = mx + c 3 = 2(1) + c 3 = 2 + c c = 3 – 2 = 1 y = 2x + 1 ... award full marks 2x – y = –1

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down correct relevant formula. – Finds correct value of f (1).

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


2018.1 L.16/19_MS 14/76 of 75 examsDEB

Question 5 (25)

5(a) Find the two values of x for which 2x2 + 4x – 5 = 0. Give your answer correct to one decimal place. (15D*)

x = a

acbb

2

42 −±−

x = )2(2

)5)(2(4)4(4 2 −−±−

= 4

40164 +±−

= 4

564 ±−

= 4

...48331474 ⋅±−

x = 4

...48331474 ⋅−−

= 4

...48331411⋅−

= –2⋅870828... ≅ –2⋅87

x = 4

...48331474 ⋅+−

= 4

...4833143⋅

= 0⋅870828... ≅ 0⋅87

Scale 15D* (0, 6, 10, 13, 15) Low partial credit: (6 marks) – Some work of merit, e.g. writes down correct relevant ‘–b’ formula. – a, b and c explicitly identified. – Some correct substitution into ‘–b’ formula and stops or continues incorrectly. – Attempt at factorising quadratic equation.

Mid partial credit: (10 marks) – Substitutes correctly into ‘–b’ formula and stops or continues incorrectly.

– Finds x = 4

244 −±−

and stops or continues.

High partial credit: (13 marks) – Substitutes correctly into ‘–b’ formula,

i.e. x = 4

564 ±−, 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 final answer is incorrectly rounded or not rounded - apply only once to each section (a), (b), (c), etc. of question.


2018.1 L.16/19_MS 15/76 of 75 examsDEB


5(b) When two numbers are added together, the sum is equal to 46. When the smaller number is subtracted from the larger number, the result is equal to 12. By writing two equations to represent this information, or otherwise, find the values of both numbers. (10D)

Two equations:

Let a be the larger number and b be the smaller number

a + b = 46 a – b = 12

Values of both numbers:

a + b = 46 a – b = 12

2a = 58 a = 29

a + b = 46 29 + b = 46 29 + b = 46 – 29 = 17

or a – b = 12 29 – b = 12 –b = 12 – 29 = –17 b = 17

or

a + b = 46 (× 1) a – b = 12 (× –1)

a + b = 46 –a + b = –12 2b = 34 b = 17

a + b = 46 a + 17 = 46 a = 46 – 17 = 29

or a – b = 12 a – 17 = 12 a = 12 + 17 = 29

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down one equation in terms of a and b, i.e. a + b = 46 or a – b = 12.

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.


2018.1 L.16/19_MS 16/76 of 75 examsDEB

Question 6 (25)

The first three patterns in a sequence of patterns of tiles are shown below.

Pattern 1 Pattern 2 Pattern 3

6(a) Draw the next two patterns in the sequence. (5B)

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Some work of merit, e.g. correct number of vertical tiles in both patterns or correct number of horizontal tiles in both patterns (i.e. 6 in Pattern 4 and 7 in Pattern 5). – Draws one correct pattern.

6(b) (i) Based on the patterns shown, complete the table below. (5C)

Pattern number (n)

Number of White Tiles

Number of Grey Tiles

1 2 7

2 6 10

3 12 13

4 20 16

5 30 19

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

High partial credit: (4 marks) – Three, four or five correct entries.


Pattern 4 Pattern 5

2018.1 L.16/19_MS 17/76 of 75 examsDEB


6(b) (cont’d.)

(ii) Show that the number of white tiles in each pattern forms a quadratic sequence. (5D)

Pattern Number (n) 1 2 3 4 5

Number of White Tiles 2 6 12 20 30

Change (1st difference) 4 6 8 10

Change of change (2nd difference)

2 2 2

Conclusion: – as the 2nd difference is constant, the number of white tiles in each pattern forms a quadratic sequence

Scale 5D (0, 2, 3, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down one or two 1st differences and stops.

Mid partial credit: (3 marks) – Finds all 1st differences for the pattern sequence given, i.e. 4, 6, 8, 10, and stops or continues incorrectly.

High partial credit: (4 marks) – Finds 2nd differences for the pattern sequence given, but no conclusion given or incorrect conclusion.


2018.1 L.16/19_MS 18/76 of 75 examsDEB


6(c) Assuming the pattern continues, the number of white tiles in the nth pattern of the sequence is given by the formula Wn = n2 + bn + c, where b, c ∈ ℤ. Find the value of b and the value of c. (10D)

Wn = n2 + bn + c

W1 = (1)2 + b(1) + c 2 = 1 + b + c b + c = 2 – 1 b + c = 1

W2 = (2)2 + b(2) + c 6 = 4 + 2b + c 2b + c = 6 – 4 2b + c = 2

b + c = 1 (× –1) 2b + c = 2 (× 1)

–b – c = –1 2b + c = 2 b = 1

b + c = 1 or 1 + c = 1 c = 1 – 1 = 0

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

= 0

or

b + c = 1 (× 2) 2b + c = 2 (× –1)

2b + 2c = 2 –2b – c = –2 c = 0

b + c = 1 or b + 0 = 1 b = 1

2b + c = 2 2b + 0 = 2 b = 1

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. finds one or both equations in terms of b and c for two patterns. [In the case of W1 and W2, b + c = 1 and/or 2b + c = 2.]

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

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


2018.1 L.16/19_MS 19/76 of 75 examsDEB

Section B Contexts and Applications 150 marks

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

Question 7 (50) Fiona wishes to save up to buy a car. She joins a savings scheme in her local credit union.

She plans to save €250 per month and to increase this amount by €5 every month.

7(a) (i) Complete the table below to show Fiona’s monthly contributions for the period shown. (5C)

Month (n) 1 2 3 4 12 24

Contribution (€) 250 255 260 265 305 365

Monthly Contribution:

Month 2 = 250 + 5 = €255

Month 3 = 255 + 5 = €260

Month 4 = 260 + 5 = €265

Month 12 = 250 + 5(11) = 250 + 55 = €305

Month 24 = 250 + 5(23) = 250 + 115 = €365

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

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


(ii) Show that Fiona’s monthly contributions form an arithmetic sequence. (5C)

Month (n) 1 2 3 4

Contribution (€) 250 255 260 265

Change (1st difference) 5 5 5

Conclusion: Any 1: – as the 1st difference is constant, the monthly contributions

form an arithmetic sequence // – an arithmetic sequence is one in which the difference

between any two successive terms is a constant // etc.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down one or two 1st differences and stops.

High partial credit: (4 marks) – Finds all 1st differences for the sequenceof monthly contributions given, but no conclusion given or incorrect conclusion.


2018.1 L.16/19_MS 20/76 of 75 examsDEB


7(a) (cont’d.)

(iii) Find, in terms of n, a formula that gives Fiona’s monthly contribution in the nth month of the scheme. (5C)

Tn = a + (n – 1)d

a = T1 = 250

d = T2 – T1 = 255 – 250 = 5

Tn = 250 + (n – 1)5 = 250 + 5n – 5 = 5n + 245

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 = 250 + (n – 1)5, but fails to finish or finishes incorrectly.

(iv) Using your formula, or otherwise, find in which month Fiona contributes €400. (5C)

Tn = 5n + 245 = 400

5n + 245 = 400 5n = 400 – 245 = 155

n = 5

155

= 31 // 31st month

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

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down Tn = 400 and stops. – Attempts to find Tn = 400 using list method (terms to at least T10 ).

High partial credit: (4 marks) – Finds T40 correctly using list method. – Equates 5n + 245 = 400 correctly, but fails to finish or finishes incorrectly.


2018.1 L.16/19_MS 21/76 of 75 examsDEB


7(b) (i) Find, in terms of n, a formula that gives the sum of Fiona’s total contributions from the first to the nth month of the scheme. (10C)

Sn = 2

n{2a + (n – 1)d}

a = 250 d = 5 ... part (a)(iii)

Sn = 2

n{2(250) + (n – 1)(5)}

= 2

n{500 + 5n – 5}

= 2

n(5n + 495) //

2

5 2n +

2

495n

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

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

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

i.e. Sn = 2

n{2(250) + (n – 1)(5)}, but fails


(ii) Fiona is guaranteed to receive a fixed return of €14,000 at the end of three years if she does not take any money out of the savings scheme until that time.

Find the percentage return that Fiona will receive on her savings. Give your answer correct to two decimal places. (10D*)

3 years = 3 × 12 = 36 months

Sn = 2

n(5n + 495) //

2

5 2n +

2

495n ... part (b)(i)

Sum of contributions over 3 years, S36

S36 = 2

36(5(36) + 495)

= 18(180 + 495) = 18(675) = €12,150

Guaranteed return = 14,000 – 12,150 = €1,850

% Return = 150,12

850,1 ×

1

100

= 15·226337... ≅ 15·23%


2018.1 L.16/19_MS 22/76 of 75 examsDEB


7(b) (ii) (cont’d.)

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

Scale 10D* (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down ‘3 years = 36 contributions’ or similar. – Some correct substitution into Sn formula.– Substitutes correctly into T36 [ans. 425].

Mid partial credit: (6 marks) – Substitutes correctly into relevant formula,

i.e. S36 = 2

36(5(36) + 495), but fails


High partial credit: (8 marks) – Finds correct value for Guaranteed return [ans. €1,850], but fails to finish or finishes incorrectly.



7(c) Another option is for Fiona to borrow €14,000 from a finance company to buy a car. The loan is to be repaid in equal monthly repayments over the term of the loan. Interest is charged monthly at an annual percentage rate (APR) of 13·5% on the amount borrowed for the entire term of the loan.

(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 corresponds to an APR of 13·5%, correct to two decimal places. (5C*)

(1 + r)12 = 1 + i (1 + r)12 = 1 + 0·135 = 1·135

1 + r = 1·13512

1

= 1·010608... r = 1·010608... – 1 = 0·010608... r = 1·060859...% ≅ 1·06%

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. identifies i = 0·135 or 1 + i = 1 + 0·135 = 1·135. – Divides or multiplies 1·135 by 12. – Some correct substitution into formula and stops or continues.

High partial credit: (4 marks) – Finds 1 + r = 1·13512

1

or 1·010608..., but fails to finish or finishes incorrectly.




2018.1 L.16/19_MS 23/76 of 75 examsDEB


7(c) (cont’d.)

(ii) Fiona decides to borrow the money and repay the loan over 3½ years. Calculate, correct to the nearest euro, the amount of her monthly repayments. (5D*)

3½ years = 3½ × 12 = 42 months

Total amount repayable over the term of loan F = P(1 + i)t = 14,000(1 + 0·0106)42 = 14,000(1·0106)42 = 14,000(1·557149...) = 21,800·094668... ≅ €21,800

Amount of monthly repayments

= 42

800,21

= 519·047619... ≅ €519

Scale 5D* (0, 2, 3, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down ‘3½ years = 42 repayments’ or similar. – Writes down correct relevant formula and stops. – Some correct substitution into relevant formula and stops or continues.

Mid partial credit: (3 marks) – Substitutes correctly into formula, i.e. 14,000(1·0106)42 or 14,000(1·557149...), and stops or continues incorrectly.

High partial credit: (4 marks) – Finds correct value of F [ans. €21,800], but fails to finish or finishes incorrectly.




2018.1 L.16/19_MS 24/76 of 75 examsDEB

Question 8 (50)

A company seeking new investors predicts that the projected value of their investment will follow the function:

v(t) = 5t2 – 180t + 10,000,

where v is the value of the investment, in euro, and t is the time, in months, after the investment is made.

8(a) (i) How much are potential investors asked to invest initially? (5C)

v(t) = 5t2 – 180t + 10,000

@ t = 0 v(0) = 5(0)2 – 180(0) + 10,000 = 0 + 0 + 10,000 = €10,000

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

High partial credit: (4 marks) – Substitutes correctly into v(t) equation, i.e. v(0) = 5(0)2 + 180(0) + 10,000, but fails to finish or finishes incorrectly.


(ii) How long will it take before their investment is profitable? (5C*)

v(t) = 5t2 – 180t + 10,000

investment is profitable when v(t) > 10,000 5t2 – 180t + 10,000 > 10,000 5t2 – 180t > 10,000 – 10,000 > 0 t(5t – 180) > 0

t > 0 ... not applicable

5t – 180 > 0 5t > 180

t > 5

180

> 36 months

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down ‘profitable when v(t) > 10,000’. – Substantially correct approach to solving quadratic equation.

High partial credit: (4 marks) – Solves quadratic equation correctly, i.e. finds t = 0 and 36, but fails to identify correct answer. – Correct answer, but no work shown.

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


2018.1 L.16/19_MS 25/76 of 75 examsDEB


8(b) Use calculus to find how much the investment is worth when it reaches its minimum value. (10D)

Minimum value when dt

dv = 0

v(t) = 5t2 – 180t + 10,000

dt

dv = 10t – 180

= 0

10t – 180 = 0 10t = 180 t = 18 months

Value of investment @ t = 18 v(18) = 5(18)2 – 180(18) + 10,000 = 1,620 – 3,240 + 10,000 = €8,380

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

‘Minimum value when dt

dv = 0’ and stops.

– One term correctly differentiated.

Mid partial credit: (6 marks) – Both terms correctly differentiated and stops or continues incorrectly.

High partial credit: (8 marks) – Equates linear equation correctly, i.e. v ′(t) = 10t – 180 = 0, but fails to find correct value of t. – Finds correct value of t [18 months], but fails to finish or finishes incorrectly. – Correct answer, but no work shown.


8(c) (i) Use the value function, v(t), to complete the table, showing the projected value of the investment over time. (10C)

Time (t ) 6 12 18 24 30 36 42

Value (€) 9,100 8,560 8,380 8,560 9,100 10,000 11,260

v(t) = 5t2 – 180t + 10,000

@ t = 18 v(18) = 5(18)2 – 180(18) + 10,000 = 1,620 – 3,240 + 10,000 = €8,380

@ t = 24 v(24) = 5(24)2 – 180(24) + 10,000 = 2,880 – 4,320 + 10,000 = €8,560

@ t = 30 v(30) = 5(30)2 – 180(30) + 10,000 = 4,500 – 5,400 + 10,000 = €9,100


2018.1 L.16/19_MS 26/76 of 75 examsDEB


8(c) (i) (cont’d.)

@ t = 36 v(36) = 5(36)2 – 180(36) + 10,000 = 6,480 – 6,480 + 10,000 = €10,000

@ t = 42 v(42) = 5(42)2 – 180(42) + 10,000 = 8,820 – 7,560 + 10,000 = €11,260

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

High partial credit: (7 marks) – Three or four correct entries.


(ii) Use the data in the table to draw a graph to represent the projected value of the investment over the period of time indicated. Label your graph clearly. (5D)

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

Scale 5D (0, 2, 3, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. plots correctly one or two points. – Draws correct general shape of graph.

Mid partial credit: (3 marks) – Three or four points plotted correctly with correct general shape of graph.

High partial credit: (4 marks) – Five or six points plotted correctly with correct general shape of graph. – All points plotted correctly, but not joined or joined by straight lines.


1

1

1

1

8000

9000

10000

11000

12000

Time (months)

Va

lue

of

inv

est

me

nt

(�)

6 10 4012 18 24 30 36 42

x

v x( )

w x( )

2018.1 L.16/19_MS 27/76 of 75 examsDEB


8(d) The projected value of another investment opportunity is predicted to follow the function:

w(t) = 70t + 8,000,

where w is the value, in euro, and t is the time, in months, after the investment is made.

(i) On the same axes above, draw a graph to show the projected value of this investment for 0 ≤ t ≤ 42, t ∈ ℝ. Label your graph clearly. (5D)

w(t) = 70t + 8,000

@ t = 0 w(0) = 70(0) + 8,000 = 0 + 8,000 = €8,000 (0, 8,000) ∈ w

@ t = 6 w(6) = 70(6) + 8,000 = 420 + 8,000 = €8,420 (6, 8,420) ∈ w

@ t = 12 w(12) = 70(12) + 8,000 = 840 + 8,000 = €8,840 (12, 8,840) ∈ w

@ t = 18 w(18) = 70(18) + 8,000 = 1,260 + 8,000 = €9,260 (18, 9,260) ∈ w

@ t = 24 w(24) = 70(24) + 8,000 = 1,680 + 8,000 = €9,680 (24, 9,680) ∈ w

@ t = 30 w(30) = 70(30) + 8,000 = 2,100 + 8,000 = €10,100 (30, 10,100) ∈ w

@ t = 36 w(36) = 70(36) + 8,000 = 2,520 + 8,000 = €10,520 (36, 10,520) ∈ w

@ t = 42 w(42) = 70(42) + 8,000 = 2,940 + 8,000 = €10,940 (42, 10,940) ∈ w

Scale 5D (0, 2, 3, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. finds correctly co-ordinates of one point on graph of w(t).– Plots correctly one point on graph of w(t) (no work shown).

Mid partial credit: (3 marks) – Finds correctly one point on graph of w(t) and plots point correctly. – Finds correctly two points on graph of w(t), but fails to plot either point. – Plots correctly two points on graph of w(t) (no work shown).

High partial credit: (4 marks) – Finds and plots two points correctly, but points not joined or not joined by straight line.


2018.1 L.16/19_MS 28/76 of 75 examsDEB


8(d) (cont’d.)

(ii) Use your graphs to estimate the time interval for which the projected value of this investment is higher than that of the other investment. (5C*)

Using graph: Time interval = 40 (±1) – 10 (±1) = 30 (±2) months

** Accept answer based on students’ graph in part (c)(ii) if not oversimplified.

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. marks clearly one or both points of intersection on graph (with or without labels), but no figures. – Finds correctly x co-ordinate of one point of intersection, i.e. 10 or 40. – Finds correctly y co-ordinate of one point of intersection, i.e. 8,700 or 10,800.

High partial credit: (4 marks) – Both points of intersection on graph with figures indicated, but fails to find or finds incorrect time interval. – Finds correctly y co-ordinate of both points of intersection, i.e. 8,700 and 10,800. – Finds correct answer for time interval, but no work shown on graph. – Final answer outside of tolerance, but work shown on student’s graph.

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

(iii) Which of the above investment opportunities would you recommend? Give a reason for your answer. (5B)

First investment / v(t)

– better return from this investment in the long term / after first 3 years // etc.

** Accept any other appropriate reason.

Second investment / w(t) Any 1: – return on this investment is always positive // – if you have to withdraw from the investment at any time,

you will not have lost any money // – smaller initial investment required // etc.

** Accept any other appropriate reason.

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Answer incomplete, but with some merit.

* No marks for recommendation if reason omitted.


2018.1 L.16/19_MS 29/76 of 75 examsDEB

Question 9 (50)

A club proposes to erect two floodlighting towers at opposite corners of its rectangular playing field of length 80 m and width 60 m. A power cable is required between the diagonal corners, A and B.

The cost of digging a trench to lay the power cable is €20 per m, while the cost of boring underground to install the cable is €40 per m.

The club does not wish to dig a trench across the playing field.

9(a) One option is to dig a trench for 20 m from A along the edge of the playing field and then to bore underground from this point diagonally across the field to B, as shown.

(i) Use the Theorem of Pythagoras to find y, the distance required to be bored underground. Give your answer correct to one decimal place. (5C*)

Theorem of Pythagoras:

| Hyp |2 = | Opp |2 + | Adj |2 y2 = (80 – 20)2 + (60)2 = (60)2 + (60)2 = 3,600 + 3,600 = 7,200

y = 200,7

= 84·852813... ≅ 84·9 m

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 y2 = 7,200 or y = 200,7 , but fails


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

(ii) Hence, find the total cost of this option. (5C)

Total cost of option = (20 m × €20) + (84·9 m × €40) = 400 + 3,396 = €3,796

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down correct formula for Pythagoras’ theorem.– Finds cost of one part correctly, i.e. 20 m × €20 or 84·9 m × €40.

High partial credit: (4 marks) – Finds (20 m × €20) + (84·9 m × €40), but fails to finish or finishes incorrectly.



B

A60 m

20 m

80

m

y

60 m

80 – 20 = 60 m

2018.1 L.16/19_MS 30/76 of 75 examsDEB


9(b) To investigate other options, a general formula for y (distance to be bored underground) is derived for different values of x (length of trench to be dug).

(i) Show that the general formula for y is 22 60)80( +− x . (10C)

| Hyp |2 = | Opp |2 + | Adj |2 y2 = (80 – x)2 + (60)2

y = 22 60)80( +− x

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down correct formula for Pythagoras’ theorem.– Finds 80 – x and stops or continues incorrectly. – Some correct substitution into formula for Pythagoras’ theorem.

High partial credit: (7 marks) – Finds y2 = (80 – x)2 + (60)2, but fails to finish or finishes incorrectly.

(ii) Find the value of y for each value of x given in the table below. Give your answers correct to one decimal place. (10D*)

x (metres) y (metres) Total Cost (€)

10 92⋅2 3,888

20 84·9

30 78·1

40 72·1

50 67·1

60 63·2

70 60·8

80 60·0

y = 22 60)80( +− x

@ x = 20

y = 22 60)2080( +−

= 600,3600,3 +

= 200,7

= 84·852813... ≅ 84·9 m

@ x = 30

y = 22 60)3080( +−

= 600,3500,2 +

= 100,6

= 78·102496... ≅ 78·1 m

B

A

x

60 m

80

m

y


60 m

80 – x

2018.1 L.16/19_MS 31/76 of 75 examsDEB



@ x = 40

y = 22 60)4080( +−

= 600,3600,1 +

= 200,5

= 72·111025... ≅ 72·1 m

@ x = 50

y = 22 60)5080( +−

= 600,3900 +

= 500,4

= 67·082039... ≅ 67·1 m

@ x = 60

y = 22 60)6080( +−

= 600,3400 +

= 000,4

= 63·245553... ≅ 63·2 m

@ x = 70

y = 22 60)7080( +−

= 600,3100 +

= 700,3

= 60·827625... ≅ 60·8 m

@ x = 80

y = 22 60)8080( +−

= 600,30 +

= 600,3

= 60 m

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

Mid partial credit: (6 marks) – Three or four correct entries.

High partial credit: (8 marks) – Five or six correct entries.

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


2018.1 L.16/19_MS 32/76 of 75 examsDEB


9(b) (cont’d.)

(iii) Write a formula in x and y for the total cost of installing the power cable. (10D)

Cost = 20x + 40y

Hence, complete the table above to show the total cost for each option given.

x (metres) y (metres) Total Cost (€)

10 92⋅2 3,888

20 84·9 3,796

30 78·1 3,724

40 72·1 3,484

50 67·1 3,684

60 63·2 3,728

70 60·8 3,832

80 60·0 4,000

Cost = 20x + 40y

@ x = 20 Cost = 20(20) + 40(84·9) = 400 + 3,396 = €3,796

@ x = 30 Cost = 20(30) + 40(78·1) = 600 + 3,124 = €3,724

@ x = 40 Cost = 20(40) + 40(72·1) = 600 + 2,884 = €3,484

@ x = 50 Cost = 20(50) + 40(67·1) = 1,000 + 2,684 = €3,684

@ x = 60 Cost = 20(60) + 40(63·2) = 1,200 + 2,528 = €3,728

@ x = 70 Cost = 20(70) + 40(60·8) = 1,400 + 2,432 = €3,832

@ x = 80 Cost = 20(80) + 40(60·0) = 1,600 + 2,400 = €4,000


2018.1 L.16/19_MS 33/76 of 75 examsDEB


9(b) (iii) (cont’d.)

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down cost of one part correctly, i.e. 20x or 40y.– One or two correct entries, but no evidence of formula.

Mid partial credit: (6 marks) – Correct formula for cost, i.e. 20x + 40y, but fails to evaluate any for value. – Three or four correct entries, but no evidence of formula.

High partial credit: (8 marks) – Correct formula for cost, i.e. 20x + 40y, but not all values evaluated (at least two).– Five or six correct entries, but no evidence of formula.


(iv) Use the formula from part (i) to find the shortest route between A and B. Hence, state whether this is the cheapest option. Justify your answer by calculation. (10D)

y = 22 60)80( +− x

Shortest route when x = 0

y = 22 60)080( +−

= 22 6080 +

= 600,3400,6 +

= 000,10

= 100 m

Cost = 20x + 40y Cost = 20(0) + 40(100·0) = 0 + 4,000 = €4,000

Conclusion: – the shortest route is not the cheapest option

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down formula for y and states ‘Shortest route when x = 0’ or similar. – Some correct substitution into formula for y and stops or continues incorrectly.

Mid partial credit: (6 marks) – Finds y = 000,10 or 100 and stops

or continues incorrectly.

High partial credit: (8 marks) – Finds cost of this option [ans. €4,000], but no conclusion given or incorrect conclusion.


2018.1 L.16/19_MS 34/76 of 75 examsDEB

Notes:


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Notes:


2018.1 L.16/19_MS 36/76 of 75 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, 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

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 for the omission of units in 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 DNo 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

2018.1 L.16/19_MS 37/76 of 75 examsDEB

Summary of Marks – 2018 LC Maths (Ordinary Level, Paper 2)

Section A Section B

Q.1 (a) (i) 5C (0, 2, 4, 5) Q.7 (a) (i) 5C (0, 2, 4, 5) (ii) 5C (0, 2, 4, 5) (ii) 10D (0, 4, 6, 8, 10) (iii) 5C (0, 2, 4, 5) (b) 10D* (0, 4, 6, 8, 10) (b) (i) 5C (0, 2, 4, 5) (c) (i) 5C (0, 2, 4, 5) (ii) 5C (0, 2, 4, 5) (ii) 5D (0, 2, 3, 4, 5) 25 (iii)

5B (0, 2, 5)

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

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

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

Current Marking Scheme

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

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All rights reserved. This marking scheme and corresponding papers(s) are protected by Irish (EU) copyright law. Reproduction and distribution of these materials or any portion thereof without the written permission of the publisher is prohibited except for the immediate use within a classroom.

2018.1 L.16/19_MS 38/76 of 75 examsDEB


Mathematics


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 nine 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.

Section A Concepts and Skills 150 marks

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

Question 1 (25)

A survey of 136 students was carried out. They were asked whether they were studying French (F) or Spanish (S). Of those surveyed, 87 study French, 85 study Spanish and 54 study both languages.

1(a) (i) Represent this information on the Venn Diagram. (5C)

[ ]

[ ]136

= 18

��

[ ]

[]

87 54

= 33

� ][

[]

85 54

= 31

� ][

SF

54

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down ‘87 – 54’ or ‘85 – 54’ in the correct position on the Venn diagram. – One correct/consistent element calculated or inserted on the Venn diagram.

High partial credit: (4 marks) – Two correct/consistent elements calculated or inserted on the Venn diagram.

examsDEB

2018.1 L.16/19_MS 39/76 of 75 examsDEB


1(a) (ii) A student is chosen at random from those surveyed. Find the probability that the student studies neither language. (5C)

P(student studies neither language)

= 136

18

= 68

9 or 0·132352...

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

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Either #(E) or #(S) correct (answer must be shown in fraction format).

High partial credit: (4 marks) – Finds

136

18, but fails to evaluate or

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

(iii) Two students are chosen at random from those surveyed who study at least one of these languages. Find the probability that both students study one language only. Give your answer correct to two decimal places. (5C)

# students who study at least one language = 33 + 54 + 31 = 118

P(1st student selected studies 1 language only)

= 118

3133 +

= 118

64 //

59

32 // 0·542372...

P(2nd student selected studies 1 language only)

= 1118

13133

−−+

= 117

63 //

13

7 // 0·538461...

P(both students selected study 1 language only)

= 118

64 ×

117

63

= 806,13

032,4 //

903,6

016,2 // 0·292046...


Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. any #(E) or #(S) correct (answer must be shown in fraction format). – Finds one correct probability. – Finds two correct probabilities, but adds values together instead of multiplying.

High partial credit: (4 marks) – Finds 118

64 ×

117

63 (or consistent), but fails

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


2018.1 L.16/19_MS 40/76 of 75 examsDEB


1(b) When the responses of 14 additional students are included in the survey, the probability

that a student chosen at random studies both French and Spanish increases to 5

2.

(i) How many of these additional students study both French and Spanish? (5C)

# students surveyed = 136 + 14 = 150

# students who study both French and Spanish

= 5

2 × 150

= 60 students

Number of new students who study both French and Spanish = 60 – 54 = 6

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down new #(S) [ans. 136 + 14 or 150] and stops.

– Finds 5

2 × any figure.

High partial credit: (4 marks) – Finds new #(E) [ans. 5

2 × 150 or 60], but

fails to finish or finishes incorrectly. – Correct answer, but no work shown.

(ii) Find the highest possible number of these additional students who study Spanish only. (5C)

# new students surveyed = 14 # new students who study both French and Spanish = 6

Highest possible # new students who study Spanish only = 14 – 6 = 8


Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down ‘14 – # students who study both French and Spanish’ or similar. – Finds 14 – any figure.

High partial credit: (4 marks) – Finds 14 – 6, but fails to contextualise relevant figures. – Correct answer, but no work shown.


2018.1 L.16/19_MS 41/76 of 75 examsDEB

Question 2 (50)

The circle c has a radius of 5 cm and touches both the x-axis and the y-axis, as shown.

2(a) (i) Write down the co-ordinates of the centre of c. Hence, find the equation of c. (5C)

Co-ordinates of centre of c

radius of c = 5 centre of c = (5, 5)

Equation of c

(x – h)2 + (y – k)2 = r2 (x – 5)2 + (y – 5)2 = 52 (x – 5)2 + (y – 5)2 = 25

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down correct relevant circle formula and stops. – Finds correct co-ordinates of centre of c and stops or continues incorrectly.

High partial credit: (4 marks) – Finds correct co-ordinates of centre of c and some correct substitution into relevant circle formula, but fails to finish or finishes incorrectly.

(ii) Verify, using algebra, that the point P(9, 8) is on c. (5C)

Equation of c: (x − 5)2 + (y − 5)2 = 25

P(9, 8): (9 − 5)2 + (8 – 5)2 = 25 (4)2 + (3)2 = 25 16 + 9 = 25 25 = 25

as LHS = RHS, P(9, 8) is on c

or

Centre of circle (5, 5) Distance from (5, 5) to (9, 8)

= 212

212 )()( yyxx −+−

= 22 )58()59( −+−

= 22 )3()4( +

= 916 +

= 25 = 5 = radius of c

P(9, 8) is on c


x

c

y

5

5(5, 5)

2018.1 L.16/19_MS 42/76 of 75 examsDEB

Question 2 (cont’d.) 2(a) (ii) (cont’d.)

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. identifies correct relevant formula. – Substitutes (9, 8) into relevant formula and stops or continues incorrectly.

High partial credit: (4 marks) – Finds 25 = 25 [Method ], but incorrect or no conclusion given. – Finds 5 [Method ], but incorrect or no conclusion given.

2(b) (i) Find the equation of t1, the tangent to c at P. (10D)

Slope of radius at P(9, 8)

Centre (5, 5), P (9, 8) (x1, y1) (x2, y2)

Slope = 12

12

xx

yy

−−

= 59

58

−−

= 4

3

or

Slope = Run

Rise

= 59

58

−−

= 4

3

Slope of tangent at P(9, 8) = ⊥ slope of radius at P(9, 8)

= –3

4

Equation of t1

m = –3

4, P(9, 8)

y – y1 = m(x – x1)

y – (8) = –3

4(x – 9) ... award full marks

3(y – 8) = –4(x – 9) 3y – 24 = –4x + 36 4x + 3y = 60

or

y = mx + c

8 = –3

4(9) + c

8 = –4(3) + c c = 8 + 12 = 20

y = –3

4x + 20

3y = –4x + 20(3) 4x + 3y = 60


x

cc

y

t1

5

5(5, 5)

P(9, 8)

2018.1 L.16/19_MS 43/76 of 75 examsDEB


2(b) (i) (cont’d.)

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down ‘slope of t1 equal to ⊥ slope of radius at P(9, 8)’ or similar. – Identifies correct relevant formula, i.e. slope formula or equation of a line. – Finds correct slope of radius at P(9, 8)

[ans. 4

3] and stops or continues incorrectly.

– Some correct work with P, e.g. correct substitution into relevant line formula (no slope or incorrect slope used).

Mid partial credit: (6 marks) – Finds correct slope of tangent at P(9, 8),

[ans. –3

4] and stops or continues

incorrectly.

High partial credit: (8 marks) – Finds correct slope of tangent at P(9, 8),

[ans. –3

4] with some correct substitution

into relevant 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.

(ii) t2 is another tangent to c at the point Q and t2 is parallel to t1. Find the co-ordinates of Q. (5C)

P(9, 8) → (5, 5) 9 → 5: ↓ 4 8 → 5: ↓ 3

centre (5, 5) → Q(5 – 4, 5 – 3) → Q(1, 2)


Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down ‘slope of t2 equal to slope of t1’ or similar.– Work with a translation which indicates an understanding of the concept. – Any work of merit using algebraic method.

High partial credit: (4 marks) – An incorrect translation applied correctly.– A correct graphical solution. – Finds correct value for either x or y co-ordinate of Q.


x

cy

t1

t2(5, 5)

P(9, 8)

Q

2018.1 L.16/19_MS 44/76 of 75 examsDEB

Question 3 (50)

A company produces plastic balls for children’s ball ponds. Each ball is in the shape of a sphere with a diameter of 6 cm.

3(a) (i) Find the radius of each ball. (5B*)

rball = 2

diameter

= 2

6

= 3 cm

Scale 5B* (0, 2, 5) Partial credit: (2 marks) – Some work of merit, e.g. writes down ‘rball = diameter / 2’ or similar.

– Finds rball = 2

6 or similar, but fails to

simplify or simplifies incorrectly.

* 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.

(ii) Hence, find the volume of each ball, correct to one decimal place. (5C*)

Vball = 3

4 πr3

= 3

4 π(3)3

= 36π = 36(3·141592...) = 113·097335... ≅ 113·1 cm3

** Accept other answers based on π = 3·14 [113·04 ans. 113·0 cm3]

or π = 7

22 [113·142857... ans. 113·1 cm3].

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

High partial credit: (4 marks) – Correct substitution into relevant volume formula, but fails to finish or finishes incorrectly. – One error in substitution into relevant volume formula, but finishes correctly. – Final answer expressed in terms of π.

* Deduct 1 mark off correct answer only if final answer is not rounded or incorrectly rounded or for the omission of or incorrect use of units (‘cm3’) - apply only once to each section (a), (b), (c), etc. of question.


2018.1 L.16/19_MS 45/76 of 75 examsDEB


3(b) The balls are packed in rectangular cases which can hold 240 balls. The width of the case can fit 6 balls and the length can fit 8 balls.

(i) Find the volume of the smallest rectangular case that the balls will fit into. (10D*)

Length, l ≡ 8 balls Length, l = 8 × diameters = 8 × 6 = 48 cm

Width, w ≡ 6 balls Width, w = 6 × diameters = 6 × 6 = 36 cm

Layer = 8 × 6 = 48 balls

Layers required = 48

240

= 5

Height, h ≡ 5 balls Height, h = 5 × diameters = 5 × 6 = 30 cm

Volume of case = l × w × h = 48 × 36 × 30 = 51,840 cm3

Scale 10D* (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down ‘length ≡ 8 balls’, ‘width ≡ 6 balls’ or similar and stops. – Finds l = 48 or w = 36 and stops or continues incorrectly.

– Indicates h ≡ 68

240

×.

Mid partial credit: (6 marks) – Finds l = 48 and w = 36 and stops or continues incorrectly.

High partial credit: (8 marks) – Finds l = 48, w = 36 and h = 30, but fails to find volume or finds incorrect volume.



2018.1 L.16/19_MS 46/76 of 75 examsDEB


3(b) (cont’d.)

(ii) What percentage of the volume of the rectangular case is not taken up by the balls? Give your answer correct to two decimal places. (5D*)

Volume of balls in rectangular case = 240 × 113·1 = 27,144 cm3

Volume of rectangular case = 51,840 cm3 Volume of unoccupied space in rectangular case = 51,840 – 27,144 = 24,696 cm3

% Volume unoccupied

= 840,51

696,24 ×

1

100

= 47·638888... ≅ 47·64%

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

Scale 5D* (0, 2, 3, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down relevant % formula. – Finds correct volume of balls in rectangular case [ans. 27,144 cm3] and stops or continues incorrectly.

Mid partial credit: (3 marks) – Finds correct volume of unoccupied space in rectangular case [ans. 24,696 cm3] and stops or continues incorrectly.

High partial credit: (4 marks) – Finds correct volume of unoccupied space in rectangular case with some correct substitution into % formula, but fails to finish or finishes incorrectly.




2018.1 L.16/19_MS 47/76 of 75 examsDEB

Question 4 (50)

4(a) (i) Construct a triangle ABC, where | AB | = 10 cm, | AC | = 9 cm and | ∠CAB | = 30°. The point A is given to you. Show all construction lines clearly. (10C)

A

BA

10 cm

9 cm

C

30�

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

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – One side correctly constructed, i.e. point B correctly marked on line given or side AC correctly constructed (30°) with or without point C marked. – Arc from point A drawn correctly to find third point (C) not on line given. – Freehand triangle drawn with dimensions.

High partial credit: (7 marks) – Point B correctly marked on line given and side AC correctly constructed (30°) or arc from point A drawn correctly to find third point (C). – Second and third points constructed correctly, but not all sides drawn. – Triangle constructed correctly, but measurements not exact. – Correct triangle drawn, but with no construction lines shown.

(ii) What is the largest angle in the triangle ABC? Give a reason for your answer. (5B)

Largest angle: – | ∠BCA | / | ∠ABC |

Reason: – | ∠BCA | / | ∠ABC | is opposite the largest side (10 cm) of the triangle ABC

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Correct angle identified but no reason or incorrect reason given. – Correct reason but angle not identified.


2018.1 L.16/19_MS 48/76 of 75 examsDEB


4(b) (i) On the same diagram above, construct the circumcentre and the circumcircle of the triangle ABC, using a compass and straight edge only. Show all construction lines clearly. (10D*)

BA

10 cm 501

cm

�

9 cm

C

30�

(ii) Measure the length of the radius of the circle constructed in part (b)(i).

Length of radius = 5·01 cm

** Accept answers within 5 < r ≤ 5·5 cm.

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

Scale 10D* (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. attempt to bisect any side of triangle and stops. – Indicates and labels midpoint of any side (no constructions shown) and stops. – Finds correctly the incentre or centroid.

Mid partial credit: (6 marks) – Further work of merit such as perpendicular bisector of any side constructed correctly.

High partial credit: (8 marks) – Circumcentre established correctly, but no circle drawn. – Both mediators and circumcentre drawn correctly, but no evidence of construction.– Both circumcentre and circumcircle drawn correctly, but radius-length not measured or outside tolerance.

* 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.


2018.1 L.16/19_MS 49/76 of 75 examsDEB

Question 5 (25)

The diagram shows the parallelogram ABCD with [ AB ] extended to E.

5(a) Use the measurements shown on the diagram to find the values of x, y and z. Give a reason for your answer in each case. (10D*)

x:

Answer: – x° = 34°

Reason: – the sum of the three angles in a triangle is equal to 180°

x° = 180° – 44° – 102° = 180° – 146° = 34°

y:

Answer: – y° = 44°

Reason: – as [ AD ] is parallel to [ BC ], ∠DAC and ∠BCA are alternate angles

y° = | ∠BCA | = | ∠DAC | as | ∠DAC | = 44° y° = 44°

z:

Answer: – z° = 78°

Reason: – exterior angle of a triangle is equal to the sum of the two opposite interior angles

z° = | ∠CAB | + | ∠BCA | | ∠CAB | = | ∠ACD | // x° = 34° | ∠BCA | = y° = 44° z° = 34° + 44° = 78°

** Accept students’ answers on the diagram.

Scale 10D* (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down any correct relevant point or geometric statement for x, y or z. – Finds one correct value (x, y or z), but no reason or incorrect reason given.

Mid partial credit: (6 marks) – Finds one correct value (x, y or z) with valid reason given. – Finds two correct values, but no reasons or incorrect reasons given. – Finds two correct values, with one valid reason given.

High partial credit: (8 marks) – Finds two correct values, both with valid reasons given. – Finds all values correct, with one or two valid reasons given.

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


B

D

EA

C7 1 cm�

10 cm

x°y°

z°

102°

44°

2018.1 L.16/19_MS 50/76 of 75 examsDEB


5(b) (i) Find the area of the triangle ABC. Give your answer correct to one decimal place. (10C*)

Area of Δ ABC = 2

1| AC |.| AB | sin | ∠CAB |

| AB | = | DC | = 7·1 cm

| ∠CAB | = | ∠ACD | // x° = 34°

or

| ∠CAB | = | ∠DAB | – | ∠DAC | = | ∠CBE | // z° – 44° = 78° – 44° = 34°

Area of Δ ABC = 2

1(10)(7·1)(sin 34°)

= 2

1(10)(7·1)(0⋅559192...)

= 19·851348... = 19·9 cm2

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

Scale 10C* (0, 4, 7, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down correct relevant formula for the area of a triangle and stops. – Some correct substitution into relevant formula for area of triangle (not given). – Finds correct value for | AB | or | ∠CAB | and stops or continues incorrectly.

High partial credit: (7 marks) – Finds area of Δ ABC = 2

1(10)(7·1)(sin 34°)

or similar, but fails to finish or finishes incorrectly. – Calculates correctly area of triangle with one incorrect substitution. – Incorrect calculator mode - apply once only [Radian: ans. = 18·782435... cm2; Gradian: ans. = 18·070970... cm2]. – Finds correct or consistent answer, but no work shown.



2018.1 L.16/19_MS 51/76 of 75 examsDEB


5(b) (cont’d.)

(ii) Hence, or otherwise, find the area of the parallelogram ABCD. (5C*)

Area of parallelogram ABCD = 2 × area of Δ ABC = 2 × 19·9 = 39·8 cm2


Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down ‘area of parallelogram ABCD equal to 2 × area of Δ ABC’ or similar and stops.

– Finds area of parallelogram = 2

919 ⋅.

High partial credit: (4 marks) – Finds area of parallelogram = 2 × 19·9, or similar/consistent, but fails to finish or finishes incorrectly.



2018.1 L.16/19_MS 52/76 of 75 examsDEB

Question 6 (25) 6(a) A seed company claims that 70% of its seeds will germinate. A research team decided to

investigate this claim and tested all the seeds contained in 6 randomly selected packets. The table below summarises the results of the 150 seeds in each packet.

Packet 1 2 3 4 5 6

Number of seeds that germinated 81 85 78 92 81 83

Using a calculator, or otherwise, calculate the mean and the standard deviation of the number of seeds in each packet that germinated, correct to the nearest whole number. (5C*, 5C*)

Mean:

x = n

x

= 6

838192788581 +++++

= 6

500

= 83·333333... ≅ 83 ... calculator work

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down correct relevant formula for mean. – Finds correct numerator (ans. Σx = 500) or denominator (n = 6) and stops or continues incorrectly.

High partial credit: (4 marks) – Finds x = 6

500, but fails to finish

or finishes incorrectly.


Standard deviation:

= n

xx − 2)(

= 6

...)8378()8385()8381( 222 +−+−+−

= 6

)0()2()9()5()2()2( 222222 +−++−++−

= 6

04812544 +++++

= 6

118

= 4·434711... ≅ 4 ... calculator work

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

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



2018.1 L.16/19_MS 53/76 of 75 examsDEB


6(b) After testing a large number of seed packets, the research team established that the number of seeds in each packet that will germinate is normally distributed with the same mean and standard deviation as those of the sample investigated in part (a) above.

Use the Empirical Rule to find an interval that contains the number of seeds that will germinate in approximately 95% of all packets. (5C)

Empirical Rule states that 95% of data falls within two standard deviations of the mean

x – 2σ = 83 – 2(4) = 83 – 8 = 75

x + 2σ = 83 + 2(4) = 83 + 8 = 91

Interval = [75, 91]

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

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down that 95% of distribution lies between x – 2σ and x + 2σ or similar. – Some correct substitution into x – 2σ or x + 2σ, but fails to finish or finishes incorrectly.

High partial credit: (4 marks) – Finds either x – 2σ or x + 2σ [ans. 75 or 81]. – Correct interval, but with no work shown.

6(c) (i) Find the margin of error, at 95% confidence, for the sample investigated in part (a). Write your answer as a percentage, correct to two decimal places. (5C*)

Size of sample, n = 6 × 150 = 900

Margin of error = n

1

= 900

1

= 0·033333... ≅ 3·33%

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. – Finds n = 900 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.



2018.1 L.16/19_MS 54/76 of 75 examsDEB


6(c) (cont’d.)

(ii) Use your answer to part (c)(i) above to create a 95% confidence interval for the expected rate of germination. Is there sufficient evidence to reject the company’s claim, at the 5% level of significance? (5C)

Margin of error:

Margin = n

1

= 900

1

= 0·033333... ≅ 3·33% ... answer from (b)(i)

Total number of seeds in sample that germinated = 81 + 85 + 78 + 92 + 81 + 83 = 500

Proportion of seeds in sample that germinated, p̂

= 900

500

= 9

5

= 0·555555... ≅ 55·55%

Confidence interval:

@ 95% confidence

p̂ – n

1 < p < p̂ +

n

1

p̂ – n

1 = 55·55 – 3·33

= 52·22%

p̂ + n

1 = 55·55 + 3·33

= 58·88%

55·22% < p < 58·88%

Conclusion:

– as 70% is outside this confidence interval [55·22% < p < 58·88%], we reject the company’s claim (that 70% of its seeds will germinate) at the 5% level of significance

** 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 relevant formula for confidence interval,

p̂ – n

1 < p < p̂ + n

1 , and stops.

– Finds correct value for p̂ [ans. 55.55%].

High partial credit: (4 marks) – Finds 55·55 – 3·33 < p < 55·55 + 3·33, but fails to finish or finishes incorrectly. – Finds correct answer, but no conclusion given or conclusion incomplete.



2018.1 L.16/19_MS 55/76 of 75 examsDEB

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

Question 7 (50)

7(a) A random sample of 120 students from a school with a total student population of 600 were surveyed about their social media preferences.

(i) Explain ‘random sample’ and suggest how it might have been achieved. (5C)

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

population // – a sample in which every member of the population

has an equal chance of being selected // – an unbiased representation of all students in the school // – where every student in the school has an equal chance

of being selected // etc.

How it might have been achieved:

Any 1: – assign every student a number from 1 to 600 and draw

numbers from a hat // – use a computer program, e.g. Excel - assign a number

to each student and use a random number generator // – select every 5th student (120 / 600) alphabetically from

the school register // etc.

** Accept other appropriate material.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Insufficient or incomplete explanation of either part, but with some merit, e.g. writes down ‘every student in the school’ or ‘equal chance of being selected’, etc. for explanation of ‘random sample’.

High partial credit: (4 marks) – Correct explanation of ‘random sample’, but insufficient or no description of how it may be achieved. – Accurate description of how ‘random sample’ achieved, but insufficient or incomplete explanation.


2018.1 L.16/19_MS 56/76 of 75 examsDEB


7(a) (cont’d.)

The pie chart below shows the preferred social media outlets of the students surveyed.

24°

36°

Facebook

WhatsApp

60°240°

Instagram

Snapchat

(ii) Complete the table below to show the preferred social media outlets of the students surveyed. (10D)

Angle measured

Number of students

Snapchat 240° 80

Instagram 60° 20

WhatsApp 36° 12

Facebook 24° 8

1 student ≡ 120

360°

= 3°

Snapchat ≡ 240°

= °

°3

240

= 80 students

or ≡ 240°

= °°

360

240 × 120

= 80 students

Instagram ≡ 60°

= °°

3

60

= 20 students

or ≡ 60°

= °°

360

60 × 120

= 20 students

WhatsApp ≡ 36°

= °°

3

36

= 12 students

or ≡ 36°

= °°

360

36 × 120

= 12 students

Facebook ≡ 24°

= °°

3

24

= 8 students

≡ 24°

= °°

360

24 × 120

= 8 students


2018.1 L.16/19_MS 57/76 of 75 examsDEB


7(a) (ii) (cont’d.)

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

1 student ≡ 120

360° or similar.

– Finds 3° and stops or continues incorrectly.– One or two correct answers (angles measured and/or number of students which angle represents).

Mid partial credit: (6 marks) – Three, four or five correct answers.

High partial credit: (8 marks) – Six or seven correct answers.

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

7(b) A random sample of 20 students from transition year were asked about the lengths of time they spent studying and using social media in the lead-up to the summer exams.

The data below shows the number of hours each student spent studying per week.

4 5 5 8 10 7 12 15 13 12 9 7 6 10 11 6 12 15 14 9

Find the mean, mode and median of the data. (10D*)

Mean:

x = n

x

= 20

91415126...108554 ++++++++++

= 20

190

= 9·5 hours ... calculator work

Mode:

4, 5, 5, 8, 10, 7, 12, 15, 13, 12, 9, 7, 6, 10, 11, 6, 12, 15, 14, 9

Reorganising the data (lowest to highest):

4, 5, 5, 6, 6, 7, 7, 8, 9, 9, 10, 10, 11, 12, 12, 12, 13, 14, 15, 15

Mode = 12

Median:

4, 5, 5, 8, 10, 7, 12, 15, 13, 12, 9, 7, 6, 10, 11, 6, 12, 15, 14, 9

Reorganising the data (lowest to highest):

4, 5, 5, 6, 6, 7, 7, 8, 9, 9, 10, 10, 11, 12, 12, 12, 13, 14, 15, 15

Median = 2

109 +

= 2

19

= 9·5 hours


2018.1 L.16/19_MS 58/76 of 75 examsDEB


7(b) (cont’d.)

Scale 10D* (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down correct relevant formula / description for mean, mode or median. – Substantial relevant work towards finding one value (mean, mode or median). – Finds one correct value (mean, mode or median).

Mid partial credit: (6 marks) – Finds one correct value (mean, mode or median) and substantial relevant work towards finding second value.

High partial credit: (8 marks) – Finds two correct values (mean, mode or median).

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

The standard deviation of the times students spent studying in the sample is 3·4 hours.

7(c) The data below shows the lengths of time the same students spent using social media per week. Each student’s data corresponds to the same position in the previous data set, e.g. the first student surveyed spent 4 hours studying and 21 hours using social media.

21 22 18 10 8 4 2 3 1 5 11 16 18 4 3 5 1 2 1 8

The table below shows the summary statistics for the social media data.

Social media data

Mean 8·2

Mode 1

Median 5

Standard deviation 6·9

(i) Interpret two differences between the statistics data for social media usage and the statistics data for study in part (b). (5C)

Differences:

– on average, students spent less time using social media than studying, mean: 8·2 hours versus 9·5 hours / mode: 1 hour versus 12 hours / median: 5 hours versus 9·5 hours

– the amounts of time spent using social media are more spread out than those for studying / wider range of time spent on using social media than on studying, standard deviation: 6·9 hours versus 3·4 hours

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

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Incomplete or partially incorrect interpretation, but with some merit.

High partial credit: (4 marks) – One correct interpretation with reference to appropriate measure of variability.


2018.1 L.16/19_MS 59/76 of 75 examsDEB


7(c) (cont’d.)

(ii) Complete the scatter graph below to show the relationship between the lengths of time that students spent studying and using social media. (5D)

Time using social media 21 22 18 10 8 4 2 3 1 5 11 16 18 4 3 5 1 2 1 8

Time spent studying 4 5 5 8 10 7 12 15 13 12 9 7 6 10 11 6 12 15 14 9

Scale 5D (0, 2, 3, 4, 5) Low partial credit: (2 marks) – At least four points correctly plotted (with or without axes labelled correctly).– Axes correctly labelled, but no points plotted or plotted incorrectly.

Mid partial credit: (3 marks) – Between five and fifteen points correctly plotted (with or without axes labelled correctly).

High partial credit: (4 marks) – More than fifteen points correctly plotted (with or without axes labelled correctly).– All points correctly plotted but axes not labelled or labelled incorrectly.


Time spent using social media (hours)

Tim

esp

en

tst

ud

yin

g(h

ou

rs)

124 16 20 248

2

12

4

14

6

16

8

10

2018.1 L.16/19_MS 60/76 of 75 examsDEB


7(c) (cont’d.)

(iii) The correlation coefficient between the two sets of data is one of the numbers below. Write the letter corresponding to the correct answer in the box. (5B)

A 0·2 B 0·86

D C –0·2 D –0·86

(iv) What can you conclude from the scatter plot and the correlation coefficient?

Conclusion

– there is a strong negative correlation between the lengths of time that students spent studying and using social media

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Correct correlation coefficient, but incorrect or no conclusion given. – Incorrect or no answer given for coefficient given, but sufficient explanation including mention of ‘strong’ or ‘negative’ in answer.

7(d) (i) Complete the following grouped frequency table to illustrate the data collected about the lengths of time that students spent studying. (5D)

Time spent studying 0 – 5 5 – 10 10 – 15 15 – 20

Number of students 1 9 8 2

[Note: 5 – 10 means 5 hours or more but less than 10 hours, etc.]

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

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

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


2018.1 L.16/19_MS 61/76 of 75 examsDEB


7(d) (cont’d.)

(ii) Use mid-interval values of the data in the table to estimate the mean time that students in this sample spent studying. (5C*)

Mean = 2891

)2517()8512()957()152(

+++×⋅+×⋅+×⋅+×⋅

= 20

3510056752 ++⋅+⋅

= 20

205

= 10·25 hours

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

correct formula for mean, f

xf

, and stops.

– Writes

sampleinStudents

sampleinstudyingspentTimes

and stops or continues incorrectly. – Finds correct Σ f and stops (ans. 20).

High partial credit: (4 marks) – Finds correct Σ f x , but fails to finish or finishes incorrectly [ans. 205].

– 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 (‘hours’) - apply only once to each section (a), (b), (c), etc. of question.


2018.1 L.16/19_MS 62/76 of 75 examsDEB

Question 8 (50) Nadia wishes to map her daily run in a local park on the co-ordinate diagram below. She plots

the starting point of her run at the point O, the origin on the diagram.

8(a) (i) Nadia jogs along a straight pathway, which is represented by the line k, from O to the point A(5, 12). Plot and label this point on the diagram above. (5B)

Scale 5B (0, 2, 5) Partial credit: (2 marks) – Point plotted correctly, but not labelled. – Point plotted and labelled as A, but not in the correct position. – Point (12, 5) plotted and labelled as A, (5, 12) or (12, 5).

(ii) Find the equation of the line k. (5C)

Formula:

mk , slope of line k = 12

12

xx

yy

−−

O(0, 0), A (5, 12) (x1, y1) (x2, y2)

mk = 05

012

−−

= 5

12

Equation of line k y – y1 = m(x – x1)

mk = 5

12, O(0, 0)

y – 0 = 5

12(x – 0)

5y = 12x 12x – 5y = 0


2

12

�2

3

3

4

14

6

16

8

18

10

20

2O�2

x

y

4�4 6�6 8�8 10�10 12�12�14

A(5, 12)

(0, 14 5)�

( 12 1, 0)� �

( 5, 17)�

2018.1 L.16/19_MS 63/76 of 75 examsDEB


8(a) (ii) (cont’d.)

or

mk = 5

12, A(5, 12)

y – 12 = 5

12(x – 5)

5(y – 12) = 12(x – 5) 5y – 60 = 12x – 60 12x – 5y = 0

From graph:

Slope of line k = Run

Rise

= 5

12

Equation of line k y = mx + c

mk = 5

12, O(0, 0)

0 = 5

12(0) + c

c = 0

y = 5

12x + 0

5y = 12x 12x – 5y = 0

or

mk = 5

12, A(5, 12)

12 = 5

12(5) + c

12(5) = 12(5) + c c = 60 – 60 = 0

y = 5

12x + 0

5y = 12x 12x – 5y = 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 slope or the equation of a line and stops.

– Finds correct slope for line k [ans. 5

12]

and stops or continues incorrectly. – Some correct work with O or A, e.g. some correct substitution into relevant line formula (no slope or incorrect slope used).

High partial credit: (4 marks) Finds correct slope for line k and 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.


2018.1 L.16/19_MS 64/76 of 75 examsDEB


8(a) (cont’d.)

(iii) Nadia then jogs partially around a circular pond, c. The point (0, 14·5) is the centre of c and A is a point on c.

Show c on the diagram above and hence, find the radius and circumference of c. Give your answers correct to one decimal place. (10D*)

Plot c on co-ordinate diagram:

Circle c, centre (0, 14·5) which passes through A(5, 12)

Radius of c:

centre (0, 14·5), A(5, 12)

Equation of circle: (x – h)2 + (y – k)2 = r2 (5 – 0)2 + (12 – 14·5)2 = r2 r2 = (5)2 + (–2·5)2 r2 = 25 + 6·25 = 31·25 r = 5231⋅ = 5·590169... r ≅ 5·6 units

or

r = distance from (0, 14·5) to (5, 12)

= 212

212 )()( yyxx −+−

= 22 )51412()05( ⋅−+−

= 22 )52()5( ⋅−+

= 25625 ⋅+

= 2531⋅ = 5·590169... r ≅ 5·6 units

Circumference of c:

Circumference = 2πr = 2(3·141592...)(5·6) = 35·185837... = 35·2 units

** Accept other answers based on π = 3·14 [35·168 ans. 35·2 units]

or π = 7

22 [ans. 35·2 units].

Scale 10D* (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down correct relevant formula for equation of a circle, distance between two points or length of circumference and stops. – Circle c plotted correctly (labelled or not).– Some correct substitution into relevant formula to find radius and stops or continues incorrectly.

Mid partial credit: (6 marks) – Circle c plotted correctly (labelled or not) with some correct substitution into relevant formula to find radius and stops or continues incorrectly.

High partial credit: (8 marks) – Circle c plotted correctly (labelled or not) and finds radius correctly, but fails to find or finds incorrect value for circumference.

* 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 (‘units’) - apply only once to each section (a), (b), (c), etc. of question.


2018.1 L.16/19_MS 65/76 of 75 examsDEB


8(b) Nadia jogs halfway around the circular pond before she takes another straight pathway, represented by the line l, which is parallel to k.

(i) Find the co-ordinates of the point of contact of circle c and line l. (5C)

A(5, 12) → (0, 14·5) 5 → 0: ↓ 5 12 → 14·5: ↑ 2·5

(0, 14·5) → (0 – 5, 14·5 + 2·5) → (–5, 17)

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit which indicates an understanding of the concept. – Any work of merit using algebraic method.

High partial credit: (4 marks) – An incorrect translation applied correctly.– A correct graphical solution. – Finds correct value for either x or y co-ordinate of point of contact.

(ii) Draw line l on the diagram and hence, find the equation of line l. (10D)

Plot line l on co-ordinate diagram:

Point (–5, 17) ml , slope of l, which is parallel to mk

Equation of line l:

mk = 5

12

ml = mk ... parallel lines

= 5

12

Equation of line l

ml = 5

12, (–5, 17)

y – y1 = m(x – x1)

y – 17 = 5

12(x – (–5))

5(y – 17) = 12(x + 5) 5y – 85 = 12x + 60 12x – 5y = –85 – 60 12x – 5y = –145

or

y = mx + c

17 = 5

12(–5) + c

17 = –12 + c c = 17 + 12 = 29

y = 5

12x + 29

5y = 12x + 29(5) 12x – 5y = –145


2018.1 L.16/19_MS 66/76 of 75 examsDEB



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

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down ‘slope of l equal to slope of k’ or similar.– Identifies correct relevant formula for the equation of a line.

– Line l plotted correctly (labelled or not).

– Finds ml = mk = 5

12 (or consistent value).

– Some correct work with (–5, 17), e.g. correct substitution into relevant line formula (no slope or incorrect slope used).

Mid partial credit: (6 marks) – Line l plotted correctly (labelled or not) with some correct substitution into relevant formula for the equation of a line and stops or continues incorrectly.

High partial credit: (8 marks) – Line l plotted correctly (labelled or not) with fully correct substitution into relevant formula for the equation of a line, but fails to finish or finishes incorrectly. – Line l plotted correctly (labelled or not) and substitutes almost correctly into equation of a line formula (allow one incorrect or omitted substitution) and finishes correctly.

(iii) Nadia continues along line l until it crosses the x-axis and, from this point, she returns along the x-axis to her starting point.

Use algebra to find the co-ordinates of the point of intersection of line l and the x-axis. (5C)

Equation of line l: 12x – 5y = –145

Line l and x-axis intersect @ y = 0 12x – 5(0) = –145 12x = –145

x = 12

145−

= –12·083333... x ≅ –12·1 // –12·08 // –12·083 // –12·0833

** Accept students’ answers from part (b)(ii) for line l if not oversimplified.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down ‘intersection of line l and x-axis when y = 0’ or similar. – Identifies y = 0 and stops or continues incorrectly. – Some correct work with y = 0, e.g. some correct substitution into the equation of line l.

High partial credit: (4 marks) – Fully correct substitution of y = 0 into the equation of line l, but fails to finish or finishes incorrectly. – Uses x = 0 and substitutes correctly into line l and finishes correctly [ans. 22].


2018.1 L.16/19_MS 67/76 of 75 examsDEB


8(c) Given that the co-ordinate diagram is drawn using the scale of 1 cm = 200 m, find the total distance of Nadia’s run. Give your answer correct to the nearest metre. (10D*)

Individual distances:

Distance from O(0, 0) to A(5, 12)

d1 = 212

212 )()( yyxx −+−

= 22 )012()05( −+−

= 22 )12()5( +

= 14425 +

= 169 = 13 units

Distance from A(5, 9) to (–5, 17)

d2 = 2

1(circumference of circle c)

= 2

1(35·2)

= 17·6 units

Distance from (–5, 17) to (–12·1, 0)

d3 = 22 )170())5(112( −+−−⋅−

= 22 )17()17( +⋅−

= 2894150 +⋅

= 41339⋅ = 18·423083... d3 ≅ 18·4 // 18·42 // 18·423 // 18·4231 units

Distance from (–12·1, 0) to O(0, 0)

d4 = 212

212 )()( yyxx −+−

= 22 )00()0112( −+−⋅−

= 2)112( ⋅−

= 12·1 units

Total distance of Nadia’s run:

Total distances on co-ordinate diagram = d1 + d2 + d3 + d4 = 13 + 17·6 + 18·4 + 12·1 = 61·1 units

2 units = 1 cm on co-ordinate diagram 1 unit = 0·5 cm 61·1 units = 30·55 cm

Scale 1 cm = 200 m Total distance of Nadia’s run = 30·55 × 200 = 6,110 m


2018.1 L.16/19_MS 68/76 of 75 examsDEB


8(c) (cont’d.)

** Accept students’ answers from previous parts if not oversimplified.

Scale 10D* (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. indicates on co-ordinate diagram total distance of Nadia’s run. – Identifies correct relevant formula for the distance between two points. – Some correct substitution into distance formula and stops or continues incorrectly.– Finds one correct distance.

Mid partial credit: (6 marks) – Finds two or three correct distances and stops or continues incorrectly.

High partial credit: (8 marks) – Finds all four distances correctly, but fails to finish or finishes incorrectly. – Finds two or three correct distances and finishes (applies scale) correctly.

* 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.

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


2018.1 L.16/19_MS 69/76 of 75 examsDEB

Question 9 (50)

An architect is designing the roof of a new house. The width of the house is 6 m and the length is 15 m. The outline of one design option is shown in the diagram below. The length of one surface of the roof is 4·5 m and the pitch of that roof surface is 30°.

BA

C

4 5 m�

6 m

y

30°

9(a) (i) Find | BC |, the length of the other surface of the roof, correct to two decimal places. (10C*)

Cosine Rule:

a2 = b2 + c2 – 2bc cos A | BC |2 = | AC |2 + | AB |2 – 2| AC |.| AB | cos | ∠CAB | = (4·5)2 + (6)2 – 2(4·5)(6)cos 30° = 20·25 + 36 – 54(0·866025...) = 56·25 – 46·765371... = 9·484628... | BC | = ...4846289⋅ = 3·079712... ≅ 3·08 m

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 | BC |2, but fails to find or finds incorrect value for | BC |. – Incorrect calculator mode - apply once only [Radian: ans. = 6·922457... m; Gradian: ans. = 2·852305... m]. – Finds correct answer, but no 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 use of units (‘m’) - apply only once to each section (a), (b), (c), etc. of question.


2018.1 L.16/19_MS 70/76 of 75 examsDEB


9(a) (cont’d.)

(ii) Find | ∠ABC |, the pitch of the other roof surface. Give your answer correct to the nearest degree. (10C*)

Sine Rule:

A

a

sin =

B

b

sin =

C

c

sin

||sin

||

ABC

AC

∠ =

||sin

||

CAB

BC

∠

||sin

54

ABC∠⋅

= °

⋅30sin

083

sin | ∠ABC | = 083

)30(sin54

⋅°⋅

= 083

)50(54

⋅⋅⋅

= 083

252

⋅⋅

= 0·730519... | ∠ABC | = sin–1(0·730519...) = 46·929961... ≅ 47°


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

High partial credit: (7 marks) – Correct substitution into formula for sine rule, but fails to finish or finishes incorrectly. – Substitutes almost correctly into formula for sine rule (allow one incorrect or omitted substitution) and finishes correctly.– Finds sin | ∠ABC |, but fails to find or finds incorrect value for | ∠ABC |. – Incorrect calculator mode - apply once only [Radian: ans = invalid; Gradian: ans. = 46·168744...°]. – Finds correct answer, but no 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 use of units (‘°’) - apply only once to each section (a), (b), (c), etc. of question.


2018.1 L.16/19_MS 71/76 of 75 examsDEB


9(a) (cont’d.)

(iii) Find y, the height of the roof. (5C*)

Trigonometry:

sin | ∠ | = |Hyp|

|Opp|

sin | ∠CAB | = || AC

y

y = | AC | (sin | ∠CAB |) = 4·5(sin 30°) = 4·5(0·5) = 2·25 m

or

sin | ∠ABC | = ||BC

y

y = | BC | (sin | ∠ABC |) = 3·08(sin 47°) = 3·08(0·731353...) = 2·252569... ≅ 2·25 m

** 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 correct relevant trigonometric ratio (sin) and stops. – Some correct substitution into correct trigonometric ratio. – Finds sin 30° = 0·5 or sin 47° = 0·731353... and stops or continues incorrectly.

High partial credit: (4 marks) – Correct substitution into trigonometric ratio and correctly manipulated, i.e. y = 4·5(0·5) or 3·08(0·731353...), but fails to finish or finishes incorrectly.

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


2018.1 L.16/19_MS 72/76 of 75 examsDEB


9(a) (cont’d.)

(iv) Find the total area of the two roof surfaces of the new house. (5C*)

Total area of roof surfaces = (| AC | × 15) + (| BC | × 15) = (4·5 × 15) + (3·08 × 15) = 67·5 + 46·2 = 113·7 m2

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

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down formula for area of one or both roof surfaces, i.e. | AC | × 15, | BC | × 15, (| AC | × 15) + (| BC | × 15) or similar. – Some correct substitution into formula for one or both roof surfaces.

High partial credit: (4 marks) – Fully correct substitution into formula, i.e. (4·5 × 15) + (3·08 × 15) (or consistent), but fails to finish or finishes incorrectly.

* Deduct 1 mark off correct answer only for the omission of or incorrect use of units (‘m2’) - apply only once to each section (a), (b), (c), etc. of question.

9(b) A condition of planning permission is that the height of the roof does not exceed 2·25 m.

(i) If the pitch of the two roof surfaces is the same, find the maximum pitch of the roof surfaces in order to comply with planning permission. Give your answer in degrees, correct to two decimal places. (10C*)

Pitch of two roof surfaces is equal | ∠CAD | = | ∠CBD | ΔCAB is isosceles

| AC | = | BC | | CD | is common side of both ΔCAD and ΔCBD ΔCAD and ΔCBD are congruent (SAS) | AD | = | DB |

= 2

|| AB

= 2

6

= 3 m

Trigonometry:

Consider ΔCAD

tan | ∠CAD | = ||

||

AD

DC

= 3

252⋅

| ∠CAD | = tan–1

3

252⋅

= tan–1(0·75) = 36·869897... = 36·87°


3 3 A D B

C

2018.1 L.16/19_MS 73/76 of 75 examsDEB


9(b) (i) (cont’d.)

Scale 10C* (0, 4, 7, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down ‘ΔCAD and ΔCBD are congruent’ or ‘ΔCAD and ΔCBD ≡ SAS’ or similar. – Identifies any condition of congruent triangles, i.e. | ∠CAD | = | ∠CBD |, | AC | = | BC | or | CD | = common side. – Finds correct value of | AD | [ans. 3 m]. – Identifies correct relevant trigonometric ratio (tan) and stops. – Some correct substitution into correct trigonometric ratio.

High partial credit: (7 marks) – Correct substitution into trigonometric ratio and correctly manipulated,

i.e. | ∠CAD | = tan–1

3

252⋅ or tan–1(0·75),

but fails to finish or finishes incorrectly. – Finds correct answer, but no 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 use of units (‘°’) - apply only once to each section (a), (b), (c), etc. of question.

(ii) Find the total area of the roof surfaces of the house using this design option. (5C*)

Theorem of Pythagoras:

Consider ΔADC | Opp |2 + | Adj |2 = | Hyp |2 | CD |2 + | AD |2 = | AC |2 | AC |2 = (2·25)2 + (3)2 = 5·0625 + 9 = 14·0625 | AC | = 3·75 m

Total area of roof surfaces = (3·75 × 15) × 2 = 112·5 m2

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

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down correct formula for Pythagoras’ theorem.– Identifies area of one or both roof surfaces, i.e. | AC | × 15, (| AC | × 15) × 2 or similar.– Some correct substitution into formula for Pythagoras’ theorem.

High partial credit: (4 marks) – Finds correct value for | AC | [ans. 3·75 m], but fails to find or finds incorrect total area of roof surfaces.

* Deduct 1 mark off correct answer only for the omission of or incorrect use of units (‘m2’) - apply only once to each section (a), (b), (c), etc. of question.


3 3A D B

C

2·25

2018.1 L.16/19_MS 74/76 of 75 examsDEB


9(b) (cont’d.)

(iii) What does your answer to part (b)(ii) show about the total area of the roof surfaces of this design option compared to the option in part (a)? Explain why this is the case. (5C)

Answer: Any 1: – although the height of the roof structure is the same

in both designs, the total area of the roof surface in this design is less than in the first design //

– although the pitch of both roofs in this design is greater than 30° (first design), the total area of the roof surfaces is less than in the first design // etc.

Explanation: – as the pitch of both roof surfaces is the same, the apex / peak of the roof is equidistant from either end of the base of the roof, minimising the total area of the roof surfaces

** Accept students’ answers based on previous work if not oversimplified.

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Answer incomplete, but with some merit - no corresponding explanation given. – Explanation incomplete, but with some merit - no reason given.

High partial credit: (4 marks) – Answer complete, but no explanation or incomplete explanation given. – Explanation complete, but no answer or incomplete answer given.


2018.1 L.16/19_MS 75/76 of 75 examsDEB

Notes:


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