higher level marking scheme - m. selkirk confey …...higher level marking scheme paper 1 pg. 2 ......

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J.18/20

Pre-Junior Certificate Examination, 2017

Maths Higher Level

Marking Scheme

Paper 1 Pg. 2

Paper 2 Pg. 38

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Mathematics

Higher Level – Paper 1 Marking Scheme (300 marks)

Structure of the Marking Scheme

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

Scale label A B C D

No. of categories 2 3 4 5

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

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

Marking scales – level descriptors

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

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

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

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

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

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

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

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

examsDEB

DEB 2014 (Paper 1)

Scale label A B CNo of categories 2 3 4

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

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

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

Name/version:

Printed: Whom:

Checked:

To: Ret’d:

Updated: Whom:

Name/version:

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

2017.1 J.18/20_MS 3/72 of 71 examsDEB

Summary of Marks – 2017 JC Maths (Higher Level, Paper 1)

Q.1 (a) 5C (0, 2, 4, 5) Q.7 (a) (i) 5A (0, 5) (b) 5C (0, 2, 4, 5) (ii) 5C (0, 2, 4, 5) (c) 5B (0, 3, 5) (b) 10D* (0, 4, 6, 8, 10) (d) (i) 5A (0, 5) 20 (ii) 5C (0, 2, 4, 5) 25 Q.8 (a) (i) 5C (0, 2, 4, 5) (ii) 5C (0, 2, 4, 5) Q.2 (a) (i) 10D (0, 4, 6, 8, 10) (b) (i) 5C (0, 2, 4, 5) (ii) 5B (0, 3, 5) (ii) 5C (0, 2, 4, 5) (iii) 5B (0, 3, 5) (iii) 5C* (0, 2, 4, 5) (b) (i)

10C (0, 4, 7, 10)

25 (ii) 30 Q.9 (a) 10D (0, 4, 6, 8, 10) (b) (i) 5B (0, 3, 5) Q.3 (a) 5C (0, 2, 4, 5) (ii) 5C (0, 2, 4, 5) (b) (i) 5C (0, 2, 4, 5) (c) 10D (0, 4, 6, 8, 10) (ii) 5C (0, 2, 4, 5) 30 (c) 5B (0, 3, 5) 20 Q.10 (a) 5B (0, 3, 5) (b) 5C (0, 2, 4, 5) Q.4 (a) 5D (0, 2, 3, 4, 5) (c) 10C (0, 4, 7, 10) (b) 5C (0, 2, 4, 5) 20 (c) (i) 5B (0, 3, 5) (ii) 5C (0, 2, 4, 5) 20 Q.11 (a) 5B (0, 3, 5) (b) 5C (0, 2, 4, 5) (c) 5C (0, 2, 4, 5) (d) 5B (0, 3, 5) Q.5 (a) 5C (0, 2, 4, 5) (e) 5C (0, 2, 4, 5) (b) Values 10D (0, 4, 6, 8, 10) (f) 10C (0, 4, 7, 10) Graph 5D (0, 2, 3, 4, 5 ) 35 (c) (i) 5B* (0, 3, 5) (ii) 5B* (0, 3, 5) 30 Q.12 (a) 10C (0, 4, 7, 10) (b) (i) 5C (0, 2, 4, 5) (ii) 5C (0, 2, 4, 5) Q.6 (a) (i) 5C (0, 2, 4, 5) 20 (ii) 5C (0, 2, 4, 5) (b) 10D (0, 4, 6, 8, 10) (c) 5C (0, 2, 4, 5) 25

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

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Mathematics


General Instructions

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

2. Questions do not necessarily carry equal marks.

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

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

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

Q.1 (Suggested maximum time: 10 minutes) (25)

1(a) Write down the fraction that is half-way between 2

1 and

3

2. (5C)

2

1(difference) =

2

1(3

2 –

2

1)

= 2

1(

6

34 −)

= 2

1(6

1)

= 12

1

Half-way fraction = 2

1 +

12

1

= 12

16 +

= 12

7

or

examsDEB

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Q.1 (cont’d.) 1(a) (cont’d.)

Half-way between 2

1 and

3

2

Half-way between 6

3 and

6

4

= 6

3 21

= 627

= 2

7 ×

6

1

= 12

7

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

[ans. 6

1] and stops.

– Finds 6

3 21

or similar and stops or fails

to finish correctly.

High partial credit: (4 marks)

– Finds 12

1 or

627

and stops or fails to finish

correctly.

1(b) Write the following four numbers in order, from the smallest to the biggest.

8

1, 1·3 × 10–1, 0·14, 11·5%. (5C)

8

1 = 0·125

1·3 × 10–1 = 0·13

0·14 = 0·14

11·5% = 0·115

Order: = 11·5% // 0·115

8

1 // 0·125

1·3 × 10–1 // 0·13

0·14

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

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

2017 JC Maths [HL] – Paper 1

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Q.1 (cont’d.)

1(c) Mount Everest is the highest mountain peak on Earth, located in the Mahalangur mountain range in Nepal and Tibet. Its summit is 8800 metres above sea level, correct to the nearest 100 metres.

Complete the inequality below, where h is the height of Mount Everest. (5B)

8,750 m ≤ h < 8,849 m

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

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

1(d) On Budget Day in October 2016, Ireland’s National Debt had fallen to €184,932,672,554.

(i) Express this figure correct to six significant figures. (5A)

€184,932,672,554 ≅ €184,933,000,000

Scale 5A (0, 5) Partial credit: (0 marks) – Hit or miss.

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

(ii) At the peak of the recession in 2012, Ireland’s National Debt was €215,539 million. Using your answer to part (i), find the percentage decrease in our National Debt. Give your answer correct to one decimal place. (5C)

€215,539 million = €215,539,000,000

Decrease in debt = €215,539,000,000 – €184,933,000,000 = €30,606,000,000

% Decrease in debt = 000,000,539,215

000,000,606,30 ×

1

100

= 14·199750...% ≅ 14·2%

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

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

High partial credit: (4 marks) – Correct expression for % Decrease in debt, but fails to finish or finishes incorrectly. – Incorrect decrease, from substantially correct work, but finishes correctly.

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


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Notes:


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2(a) U is the set of natural numbers less than 20. P is the set of divisors of 15. Q is the set of divisors of 18. R is the set of prime numbers less than 20.

(i) Use this information to fill in the Venn diagram below. (10D)

P = {1, 3, 5, 15} Q = {1, 2, 3, 6, 9, 18} R = {2, 3, 5, 7, 11, 13, 17, 19}

U

2

4 8

9

18

6

17 19

137

1014

1216

Q

3

5

1P

R

15

11

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. at least four correct elements of P, Q or R stated. – One to four elements correctly placed.

Middle partial credit: (6 marks) – Five to eight elements correctly placed.

High partial credit: (8 marks) – Nine to twelve elements correctly placed.

(ii) Hence, or otherwise, write down the highest common factor of 15 and 18. (5B)

From Venn diagram Factors of 15 and 18 = (P ∩ Q) = {1, 3} HCF of 15 and 18 = 3

or

Alternative method

15

15 ÷ 3 = 5 5 ÷ 5 = 1

15 = 3 × 5

18

18 ÷ 2 = 9 9 ÷ 3 = 3 3 ÷ 3 = 1

18 = 2 × 3 × 3

HCF of 15 and 18 = 1 × 3 = 3


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Q.2 (cont’d.)

2(a) (ii) (cont’d.)

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

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Some work of merit, e.g. writes down (P ∩ Q) = {1, 3} and stops or continues incorrectly. – Finds correctly 15 and/or 18 as a product of prime numbers, but fails to find or finds incorrect HCF of the two numbers.

(iii) Calculate #(P ∪ Q ∪ R)′. (5B)

P ∪ Q ∪ R = {1, 2, 3, 5, 6, 7, 9, 11, 13, 15, 17, 18, 19} (P ∪ Q ∪ R)′ = {4, 8, 10, 12, 14, 16} #(P ∪ Q ∪ R)′ = 6

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

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Some work of merit, e.g. writes down P ∪ Q ∪ R or (P ∪ Q ∪ R)′ and stops or continues incorrectly.

2(b) U is the universal set. A and B are two subsets of U. #U = 28, #A = 18 and #B = 8. (10C)

(i) Find, with the aid of the Venn diagram, the minimum value of #(A ∪ B)′.

Let #(A ∩ B) = 0 #(A ∪ B) = 18 + 8 = 26 min. #(A ∪ B)′ = 28 – 26 = 2

(ii) Find, with the aid of the Venn diagram, the maximum value of #(A ∪ B)′.

Let #(A ∩ B) = 8 #(B / A) = 0 and #(A / B) = 18 – 8 = 10 #(A ∪ B) = 10 + 8 + 0 = 18 max. #(A ∪ B)′ = 28 – 18 = 10

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – Some work of merit in one part, e.g. identifies #(A ∩ B) = 0 [min. value] or #(A ∩ B) = 8 [max. value]. – Finds #(A ∪ B) = 26 [min. value]. – Finds #(A ∪ B) = 18 [max. value].

High partial credit: (7 marks) – One part fully correct. – Two parts correct, but answers swapped.– Finds correct #(A ∪ B) in both parts, but fails to finish or finishes incorrectly.

U [28]

A = [18 ] B

[ 0 ] [8 ][18 ]

B = [8 ]


U [28]

A = [18 ] B

[ 8 ] [0 ][10 ]

B = [8 ]

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3(a) A supermarket sells a single 32⋅5 g packet of cheese and onion flavoured crisps for €0⋅545, excluding VAT. If the current rate of VAT on crisps is 23%, calculate the price of a packet of crisps including VAT. Give your answer correct to the nearest cent. (5C)

VAT = 0·545 × 100

23

= €0·12535

Price (inc. VAT) = 0·545 + 0·12535 = 0·67035 ≅ €0·67 or 67c

or

Price (inc. VAT) = €0·545 × 100

123

= €0·67035 ≅ €0·67

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit, e.g. writes down VAT = 0·545 × 23% or Price (inc. VAT) = €0·545 × 123%. – Correct formula for VAT, but error(s) calculating value.

High partial credit: (4 marks) – Finds correct VAT and stops. – One error calculating Price (inc. VAT), but finishes correctly.

3(b) The same supermarket is also offering the following multipack deals:

Special Offer 1

6-pack (6 × 25 g) of cheese and onion

flavoured crisps only €1·99

Special Offer 2

14-pack (14 × 25 g) of variety

flavoured crisps only €3·99

(i) Which special offer is better value? Give a reason for your answer. (5C)

Answer – Special offer 2

Reason Special offer 1

Cost per packet = 6

991⋅

= €0·331666... or 33·166666...c

Special offer 2

Cost per packet = 14

993⋅

= €0·285 or 28·5c

** Accept other appropriate answers.

CRISPS


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Q.3 (cont’d.) 3(b) (i) (cont’d.)

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. calculates cost per 25g packet for Special offer 1 or Special offer 2. – Correct answer but no reason given.

High partial credit: (4 marks) – Finds cost per packet for both offers, but no conclusion or incorrect conclusion given.

(ii) Paula says “The price per gram when purchasing a 6-pack of cheese and onion crisps is almost 50% cheaper than the price per gram when purchasing a single packet.” Is she correct? Give a reason for your answer. (5C)

Answer – no

Reason Single packet

Cost per g = 532

670

⋅⋅

= €0·020615... or 2·061538...c

Special offer 1

Cost per g = 25

...3316660⋅

= €0⋅013266... or 1·326666...c

% Difference = ...0206150

...0132660...0206150

⋅⋅−⋅

× 1

100

= 35·648799... = 35·65%

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

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. calculates cost per g for single packet and/or cost per g per packet in special offer 1. – Correct answer but no justification given.

High partial credit: (4 marks) – Finds cost per g for both offers and gives correct conclusion, but fails to find or find incorrect % Difference. – Finds % Difference, but no conclusion or incorrect conclusion given.

(c) Give two limitations of special offers in general. (5B)

Limitations Any 2: – customers are encouraged to buy more than they need // – customers are encouraged to buy products / goods they

don’t need // – customer may want to buy only one variety // – may contribute to ‘impulse buying’ / buying on the

spur of the moment // – may contribute to a person getting into financial

difficulties // etc.


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


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4(a) For each of the following, write down the value of x, where 21 ≤ x ≤ 27 and x ∈ ℕ. (5D)

Description Value of x

A square number 25

A multiple of 7 21

A factor of 66 22

A cube number 27

A prime number 23

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

Middle partial credit: (3 marks) – Two or three correct values of x.

High partial credit: (4 marks) – Four correct values of x.

4(b) 27 + 75 – 12 = k 3 , where k ∈ ℤ. Find the value of k. (5C)

27 + 75 – 12 = 9 3 + 25 3 − 4 3

= 3 3 + 5 3 − 2 3 = 6 3

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down one or more of 27 = 9 3 , 75 = 25 3

or 12 = 4 3 .

– Uses decimal versions to find k 3 .

High partial credit: (4 marks) – Finds 3 3 + 5 3 – 2 3 , but fails to finish or finishes incorrectly.

4(c) (i) Simplify (3 x )6. (5B)

(3 x )6 = (x3

1

)6

= x 3

6

= x2

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Some work of merit, e.g. writes down

(3 x )6 = (x3

1

)6 and stops or continues incorrectly.


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Q.4 (cont’d.)

4(c) (ii) Simplify 5

32

2

)10(3

y

yy ×. (5C)

5

32

2

)10(3

y

yy × =

5

32

2

000,13

y

yy ×

= 5

5

2

000,3

y

y

= 1,500

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down (10y)3 = 1,000y3.

High partial credit: (4 marks) – Finds

5

5

2

000,3

y

y, but fails to finish or

finishes incorrectly.


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A school has a rectangular vegetable garden. The wall of the school building forms one side of the garden and a fence, of length 16 m, is used to form the remaining three sides, as shown. The width of the garden is x m.

5(a) Show that the enclosed area of the garden, A m2, is given by the function A(x) = 16x – 2x2. (5C)

Width = x m Length = 16 – (x + x) = 16 – 2x m

Area = w × l = x(16 – 2x) = 16x – 2x2 m2

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down l + x + x = 16 or similar and stops. Finds l = 16 – 2x and stops or continues incorrectly.

High partial credit: (4 marks) – Finds x(16 – 2x), but fails to finish or finishes incorrectly.

5(b) On the axes below, draw the graph of the function A(x) = 16x – 2x2 for 0 ≤ x ≤ 8, where x ∈ ℝ.

Values (10D)

Table (1st method)

x = 0 1 2 3 4 5 6 7 8 16x 0 16 32 48 64 80 96 112 128 –2x2 0 –2 –8 –18 –32 –50 –72 –98 –128 y = A 0 14 24 30 32 30 24 14 0

or Table (2nd method)

A(x) = 16x –2x2 = y A(0) = 0 0 = 0 A(1) = 16 –2 = 14 A(2) = 32 –8 = 24 A(3) = 48 –18 = 30 A(4) = 64 –32 = 32 A(5) = 80 –50 = 30 A(6) = 96 –72 = 24 A(7) = 112 –98 = 14 A(8) = 128 –128 = 0

x


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Q.5 (cont’d.)

5(b) (cont’d.) or Substitution method

A(x) = 16x − 2x2 A(0) = 16(0) – 2(0)2 = 0

A(1) = 16(1) – 2(1)2 = 14

A(2) = 16(2) – 2(2)2 = 24

A(3) = 16(3) – 2(3)2 = 30

A(4) = 16(4) – 2(4)2 = 32

A(5) = 16(5) – 2(5)2 = 30

A(6) = 16(6) – 2(6)2 = 24

A(7) = 16(7) – 2(7)2 = 14

A(8) = 16(8) – 2(8)2 = 0

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

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

High partial credit: (8 marks) – Six, seven or eight points, (x, y) or (y, x), correctly identified.


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Q.5 (cont’d.)

5(b) (cont’d.)

Graph (5D)

Points (0, 0), (1, 14), (2, 24), (3, 30), (4, 32), (5, 30), (6, 24) , (7, 14) , (8, 0)

** Accept values calculated from previous work (nine co-ordinates needed). ** If no points are worked out, but all points are correctly graphed, award marks

for the graph in both parts.

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

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

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

2 25· 5 5·7

A x( )

Width, (m)x

10 2 3 4 5 6 7 8

Are

ao

fV

eget

ab

leG

ard

en,

(m)

A2

10

0

20

30

40

x

32

26


2017.1 J.18/20_MS 17/72 of 71 examsDEB

Q.5 (cont’d.)

5(c) Use your graph in part (b) to answer the following questions.

(i) Estimate the maximum possible area of the garden. (5B*)

From graph Maximum area = 32 m2

** Accept students’ answers based on fully plotted graph in part (b).

Scale 5B* (0, 3, 5) Partial credit: (3 marks) – Some work of merit, e.g. draws vertical line from 4 m and intersects graph with or without horizontal line from point of intersection to the y-axis. – Answer outside tolerance (±1 grid box), but inside tolerance of ±2 grid boxes. – Gives 4 m as final answer.

* 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.

(ii) Find the range of values of x for which the area of the garden is greater than 26 m2. (5B*)

From graph Area (> 26 m2): 2·25 m ≤ x ≤ 5·75 m

** Accept students’ answers based on fully plotted graph in part (b).

Scale 5B* (0, 3, 5) Low partial credit: (3 marks) – Some work of merit, e.g. draws horizontal line from 26 m2 and intersects graph with or without vertical line(s) from points of intersection to the x-axis. – Answer outside tolerance (±1 grid box), but inside tolerance of ±2 grid boxes. – One correct x-intercept.

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


2017.1 J.18/20_MS 18/72 of 71 examsDEB


6(a) p = 2a + c.

(i) Write a in terms of p and c. (5C)

p = 2a + c 2a = p – c

a = 2

1(p – c) or

2

cp − or

2

p –

2

c.

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

High partial credit: (4 marks) – 2a = p – c, but fails to finish or finishes incorrectly.

(ii) Hence, find the value of a when p = 5 and c = 5

1. (5C)

a = 2

1(p – c)

= 2

1(5 –

5

1)

= 2

1(

5

125 −)

= 10

24

= 5

12 or 2

5

2 or 2·4

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

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


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Q.6 (cont’d.)

6(b) Solve the equation x2 – 6x + 4 = 0. Give each answer in surd form. (10D)

x2 – 6x + 4 = 0 x2 – 6x + 9 – 5 = 0 (x – 3)2 – 5 = 0 (x – 3)2 = 5 x – 3 = ± 5

x – 3 = 5

x = 3 + 5

or x – 3 = – 5

x = 3 – 5

or

x = a

acbb

2

42 −±−

x = )1(2

)4)(1(4)6()6( 2 −−±−−

= 2

16366 −±

= 2

206 ±

= 2

546 ±

= 2

526 ±

= 3 ± 5

x = 3 + 5 or x = 3 – 5

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down ‘–b’ formula and identifies a, b and c. – Finds (x – 3)2 – 5 = 0 and stops.

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

i.e. x = )1(2

)4)(1(4)6()6( 2 −−±−−

and stops. – Both factors correct (not looking for ‘= 0’).

High partial credit: (8 marks) – Each of the following steps correct: correct formula and correctly identifies a, b and c, fully correct substitution, works out solution to general surd form [ans. 3 ± 5 ].

– Finds only one value of x [ans. 3 + 5

or 3 – 5 ], but fails to find or finds incorrectly other value of x. – Finds x – 3 = ± 5 , but fails to finish or finishes incorrectly.


2017.1 J.18/20_MS 20/72 of 71 examsDEB

Q.6 (cont’d.)

6(c) Write the following as a single fraction in its simplest form.

2

32 −x –

5

74 −x. (5C)

2

32 −x –

5

74 −x =

)5)(2(

)74(2)32(5 −−− xx

= 10

1481510 +−− xx

= 10

12 −x

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. identifies 10 as the denominator with some (incorrect) numerator. – Correct numerator, i.e. multiplies 5(2x – 3) – 2(4x – 7), but no denominator.

High partial credit: (4 marks) – Finds )5)(2(

)74(2)32(5 −−− xx or similar,

but fails to finish or finishes incorrectly.


2017.1 J.18/20_MS 21/72 of 71 examsDEB

Notes:


2017.1 J.18/20_MS 22/72 of 71 examsDEB


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

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

Aoife is a sales representative for a pharmaceutical company. She earns a basic pay of €2750 per month plus commission of 1⋅5% on her total sales.

For the month of December, Aoife’s gross pay, including commission, was €4,700.

7(a) (i) Calculate Aoife’s commission for the month of December. (5A)

Aoife’s commission = 4,700 – 2,750 = €1,950

Scale 5A (0, 5) Partial credit: (0 marks) – Hit or miss.


(ii) Hence, calculate Aoife’s total sales for the month of December. (5C)

1⋅5% of sales = 1,950

1% of sales = 51

950,1

⋅

= 1,300 Total sales (100%) = 1,300 × 100 = €130,000

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. indicates division by 1·5.

High partial credit: (4 marks) – Finds €1,300 or €1,300 × 100, but fails to finish or finishes incorrectly. – Correct answer with no supporting work or correct answer with invalid supporting work.



2017.1 J.18/20_MS 23/72 of 71 examsDEB

Q.7 (cont’d.)

The standard rate of tax is 20% and the higher rate of tax is 40%. The standard rate cut-off point is €33,800 per annum.

7(b) Aoife pays €1026⋅34 in income tax for December. Calculate Aoife’s tax credits for the month. (10D*)

Tax @ 20% = 12

800,33 × 20%

= 2,816·67 × 100

20 or 2,816·67 × 0·20

= 563·333333... ≅ €563·33

Tax @ 40% = (4,700 – 2,816·67) × 40%

= 1,883·33 × 100

40 or 1,883·33 × 0·40

= €753·332 ≅ €753·33

Gross tax = Tax @ 20% + Tax @ 40% = 563·33 + 753·33 = €1,316·66

Net tax = Gross tax – Tax credit €1026·34 = 1,316·66 – Tax credit Tax credits = 1,316·66 – 1026·34 = €290·32

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

SRCOP [ans. 12

800,33 or 2,816·67]

and stops. – Finds tax @ 20% or @ 40% correctly of some relevant figure, but fails to continue correctly.

Middle partial credit: (6 marks) – Finds tax @ 20% and @ 40% correctly, but fails to find or finds incorrect figure for Gross tax.

High partial credit: (8 marks) – Finds Gross tax correctly, but fails to find or finds incorrect figure for monthly Tax credits.

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



2017.1 J.18/20_MS 24/72 of 71 examsDEB


8(a) The nth term of a sequence is given by the formula

Tn = 5n2.

(i) Write down the first three terms of the sequence. (5C)

Tn = 5n2

T1 = 5(1)2 = 5

T2 = 5(2)2 = 20

T3 = 5(3)2 = 45

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. some correct substitution into relevant formula. – One correct term.

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

(ii) Is 1000 a term in this sequence? Justify your answer. (5C)

Answer – no

Justification – Tn = 5n2 5n2 = 1,000 n2 = 200 n = 200 = 14·142135... ∉ ℕ

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Correct answer, but no justification given.– Some work of merit, e.g. equates 5n2 = 1,000 or n2 = 200 and stops.

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


2017.1 J.18/20_MS 25/72 of 71 examsDEB

Q.8 (cont’d.)

8(b) Stephen plays golf.

The table below shows the average distance, in metres, he regularly hits the ball with various golf clubs.

Golf Club Putter 9 Iron 8 Iron 7 Iron 6 Iron 5 Iron

Average Distance (m) 75 80 85 90 95 100

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

Answer – linear

Justification – as the first differences are constant, the average distance forms a linear sequence (see table)

Club Avg. D (m) 1st Diff.

P 75

5 9 80

5 8 85

5 7 90

5 6 95

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Correct answer but no justification given.– Some work of merit, e.g. finds one first difference.

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


2017.1 J.18/20_MS 26/72 of 71 examsDEB

Q.8 (cont’d.) 8(b) (cont’d.)

(ii) On the grid below, draw a graph showing the pattern of average distances for the different golf clubs. (5C)

Scale 5D (0, 2, 4, 5) Low partial credit: (2 marks) – One or two points correctly plotted, labelled or not.

High partial credit: (4 marks) – Between three and five points correctly plotted, labelled or not. – All points correctly plotted, but not joined together with line.

(iii) Using your graph, or otherwise, estimate the average distance Stephen would expect to achieve with a 3 Iron golf club. (5C*)

From graph Distance (3 Iron) = 110 m

** Accept students’ answers based on fully plotted graph in part (b)(ii).

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. extends graph of points plotted as far as 3 Iron (with or without x-axis labelled). Draws vertical line from 3 Iron position, but does not extend graph to intersect.

High partial credit: (4 marks) – Extends line correctly as far as 3 Iron and draws vertical line from 3 Iron with or without horizontal line from point of intersection to the y-axis. – Answer outside tolerance (±1 grid box), but inside tolerance of ±2 grid boxes. – Correct answer with no supporting work.


Golf Club

P 9 8 34567

Av

era

ge

Dis

tan

ce(m

)

10

0

20

30

40

50

60

70

80

90

100

110

120

x


2017.1 J.18/20_MS 27/72 of 71 examsDEB


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

–21 ≤ 3 – 6x < 12, x ∈ ℤ. (10D)

Solution

–21 ≤ 3 – 6x 6x ≤ 21 + 3 6x ≤ 24 x ≤ 4

and 3 – 6x < 12 –6x < 12 – 3 < 9

x > 6

9

−

> –2

3

–2

3 < x ≤ 4

or

–21 ≤ 3 – 6x < 12 ... subtract 3 –21 – 3 ≤ 3 – 6x – 3 < 12 – 3 –24 ≤ –6x < 9

–24 ≤ –6x < 9 ... divide by –6

6

24

−−

≥ 6

6

−− x

> 6

9

−

4 ≥ x > –2

3

–2

3 < x ≤ 4

Number line

0 1 2 3 4 5�1�2�3

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

Middle partial credit: (6 marks) – Finds x > –2

3 or x ≤ 4 (accept answer

with or without inequality sign).

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


2017.1 J.18/20_MS 28/72 of 71 examsDEB

Q.9 (cont’d.)

9(b) Factorise fully each of the following expressions.

(i) 27x2 – 45x (5B)

27x2 – 45x = 9x(3x – 5)

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

(ii) 6xy – 3sy – 4tx + 2st (5C)

6xy – 3sy – 4tx + 2st = 3y(2x – s) – 2t(2x – s) = (3y – 2t)(2x – s)

or

6xy – 3sy – 4tx + 2st = 6xy – 4tx – 3sy + 2st = 2x(3y – 2t) – s(3y – 2t) = (2x – s)(3y – 2t)

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

High partial credit: (4 marks) – Finds one correct factorisation - both inside and outside bracket, e.g. 3y(2x – s).– Finds two correct factorisations, but with sign errors.


2017.1 J.18/20_MS 29/72 of 71 examsDEB

Q.9 (cont’d.)

9(c) A 99 Flake ice cream, more commonly known as a 99 or ninety-nine, is an ice cream cone in which a Flake bar is inserted.

Mary buys 2 plain ice cream cones and 3 ninety-nines costing €10⋅20. Joe buys 3 plain ice cream cones and 4 ninety-nines costing €14⋅20. Calculate the cost of a flake. (10D) Mary: 2x + 3y = 10·20 Joe.: 3x + 4y = 14·20

2x + 3y = 10·20 (×3) 3x + 4y = 14·20 (×–2)

6x + 9y = 30·60 –6x – 8y = –28·40 y = 2·20

2x + 3y = 10·20 2x + 3(2⋅20) = 10·20 2x + 6⋅6 = 10·20 2x = 10·20 – 6·60 = 3·60 x = 1·80

or 3x + 4y = 14·20 3x + 4(2·20) = 14·20 3x + 8·80 = 14·20 3x = 14·20 – 8·80 = 5·40 x = 1·80

or

2x + 3y = 10·20 (×4) 3x + 4y = 14·20 (×–3)

8x + 12y = 40·80 –9x – 12y = –42·60 –x = –1·80 x = 1·80

2x + 3y = 10·20 2(1⋅80) + 3y = 10·20 3⋅60 + 3y = 10·20 3y = 10·20 – 3·60 = 6·60 y = 2·20

or 3x + 4y = 14·20 3(1·80) + 4y = 14·20 5·40 + 4y = 14·20 4y = 14·20 – 5·40 = 8·80 y = 2·20

Cost of flake = y – x = 2·20 – 1·80 = €0·40 or 40c

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Any work of merit, e.g. multiplying any equation by appropriate constant to facilitate cancellation of x or y term. – Finds one or both expressions correctly, i.e. 2x + 3y = 10·20 and/or 3x + 4y = 14·20. – Finds either variable (x or y) correctly by trial and error, but fails to verify in both equations or verifies incorrectly.

Middle partial credit: (6 marks) – Finds first variable (x or y) correctly, but fails to find second variable or finds incorrectly. – Finds both variables (x and y) correctly with no work shown. – Finds both variables (x and y) by trial and error, but does not verify into both equations.

High partial credit: (8 marks) – Finds both variables (x and y) correctly, but fails to find or finds incorrect cost of flake.



2017.1 J.18/20_MS 30/72 of 71 examsDEB


A triangle has three sides as follows

3x + 2, 2x – 5, x + 1

where x ∈ ℝ.

10(a) Calculate the length of the perimeter of the triangle in terms of x. Give your answer in its simplest form. (5B)

Perimeter = (3x + 2) + (2x – 5) + (x + 1) = 3x + 2x + x + 2 – 5 + 1 = 6x – 2

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Some work of merit, e.g. adds two sides correctly. – Finds correct one component.

10(b) Show that the triangle is isosceles when x is equal to 6. (5C)

Side 1 3x + 2 = 3(6) + 2 = 18 + 2 = 20

Side 2 2x – 5 = 2(6) – 5 = 12 – 5 = 7

Side 3 x + 1 = 6 + 1 = 7

triangle is isosceles as side 2 / 2x – 5 and side 3 / x + 1 are equal in length (7)

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. finds length of one or two sides (different lengths) of triangle when x = 6.

High partial credit: (4 marks) – Finds correct length of two relevant sides of triangle, but no conclusion or incorrect conclusion given.


2017.1 J.18/20_MS 31/72 of 71 examsDEB

Q.10 (cont’d.)

10(c) Mark says that x equal to 6 is the only value of x such that the triangle is isosceles. Is he correct? Justify your answer. (10C)

Answer – yes

Justification – Consider: Side 1 = Side 2 3x + 2 = 2x – 5 x = –7

3x + 2 = 3(–7) + 2 = –21 + 2 = –19 as side < 0, it can’t be a triangle

or 2x – 5 = 2(–7) – 5 = –19 as side < 0, it can’t be a triangle

or x + 1 = –7 + 1 = –6 as side < 0, it can’t be a triangle

– Consider: Side 1 = Side 3 3x + 2 = x + 1 2x = –1

x = –2

1

2x – 5 = 2(–2

1) – 5

= –1 – 5 = –6 as side < 0, it can’t be a triangle

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – Some work of merit, e.g. equates correctly Side 1 = Side 2 or Side 1 = Side 3 and finds correct x value. – Correct answer but no justification given.

High partial credit: (7 marks) – Equates correctly Side 1 = Side 2 or Side 1 = Side 3 and find correct x values and finishes correctly. – Equates correctly Side 1 = Side 2 and Side 1 = Side 3 and finds correct x values, but fails to finish or finishes incorrectly.


2017.1 J.18/20_MS 32/72 of 71 examsDEB


Two gyms, A and B, offer special rates to students for the month of July. The grid below shows the graphs of the total cost of membership for the number of days used, for each gym, during July.

11(a) Which gym has a joining fee? Justify your answer with reference to the graph above. (5B)

Answer – gym A

Justification Any 1: – consider the equation of a line y = mx + c that represents

the cost of gym A - as c = 20, therefore €20 is charged whether a student uses the gym or not //

– as the equation of the line representing gym A cuts the y-axis at 20, this indicates a cost of €20 before going to the gym or not //

– as the equation of the line representing gym A goes through (0, 20), this means that cost is €20 when the number of days used is zero // etc.

** Accept other appropriate answers and interpretations.

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


Number of Days

5 10 2015 25

To

tal

Co

st(�

)

10

0

20

30

40

50

60

70

80

90

100

x

B

A

6 666...·

33 333...·

2017.1 J.18/20_MS 33/72 of 71 examsDEB

Q.11 (cont’d.)

11(b) How might a student decide which gym is better value for money? Justify your answer. (5C)

Answer – determine the point of intersection (6·666..., 33·333...) and consider the case above / below this point

Justification Any 1: – below this point (6·666 visits), gym B is better value

for money // – above this point (6·666 visits), gym A is better value

for money

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. finds the point of intersection of the two graphs. – Correct answer but no explanation or incorrect explanation given.

High partial credit: (4 marks) – Valid explanation, but no conclusion stated [e.g. above the point of intersection one gym will be more expensive than the other, etc.] – Correct answer, but incomplete or unsatisfactory explanation given.

11(c) Find the slope (rate of change) of the graph for gym A. Explain what this value means. Refer to both the cost and the number of days the gym is used in July. (5C)

Slope of gym A

(0, 20), (25, 70) (x1, y1) (x2, y2)

mA = 12

12

xx

yy

−−

= 025

2070

−−

= 25

50

= 2

or

mA = run

rise

= 25

50

= 2

Explanation – after paying an initial fee of €20 for the month, the cost will be €2 per day to attend the gym

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

High partial credit: (4 marks) – Finds correct slope, but incomplete or unsatisfactory explanation given. – Correct explanation, but fails to find or finds incorrect slope.


2017.1 J.18/20_MS 34/72 of 71 examsDEB

Q.11 (cont’d.)

11(d) Write down a formula to represent the total cost for the number of days gym A is used in July. State clearly the meaning of any letters you use in your formula. (5B)

mA = 2

Cost of gym A c = 20 + 2x

where c = cost x = number of days gym used

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Correct expression for cost, but no explanation of terms used.

11(e) Write down a formula to represent the total cost for the number of days gym B is used in July. State clearly the meaning of any letters you use in your formula. (5C)

Slope of gym B:

(0, 0), (10, 50) (x1, y1) (x2, y2)

mB = 12

12

xx

yy

−−

= 010

050

−−

= 10

50

= 5

or

mB = run

rise

= 10

50

= 5

Cost of gym B c = 5x

where c = cost x = number of days gym used

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

High partial credit: (4 marks) – Finds correct slope, but no expression or incorrect expression for cost. – Correct expression for cost, but no explanation of terms used.


2017.1 J.18/20_MS 35/72 of 71 examsDEB

Q.11 (cont’d.)

11(f) Use your formulae from parts (d) and (e) to justify your answer in part (b). (10C)

gym A: c = 20 + 2x gym B: c = 5x

5x = 20 + 2x 5x – 2x = 20 3x = 20

x = 3

20

= 6·666... or 2x – c = –20 (×1) 5x – c = 0 (×–1)

2x – c = –20 –5x + c = 0 –3x = –20

x = 3

20

= 6·666... c = 5x

= 5(3

20)

= 3

100

= 33·333...

or c = 20 + 2x

= 20 + 2(3

20)

= 3

60 +

3

40

= 3

100

= 33·333...

point of intersection (6·666..., 33·333...)

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – Some work of merit, e.g. equates 5x = 20 + 2x and stops or continues incorrectly [Method ]. – Sets up two simultaneous equations and multiplies equation(s) by appropriate constant(s) to facilitate the cancellation of x or c term [Method ].

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


2017.1 J.18/20_MS 36/72 of 71 examsDEB


12(a) Six functions are defined as follows: (10C)

e(x) = 4 – 3x – x2 f (x) = 4 g(x) = (x – 2)2

h(x) = x2 + x – 6 j(x) = 2x k(x) = 54 x

The table below shows the sketches of all six functions. Write the name of the function into the box underneath the correct sketch for each of the

functions.

y

x

x

y

x

y

f (x) j(x) h(x)

x

y

y

x

y

x

e(x) k(x) g(x)

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – One or two functions correctly identified.

High partial credit: (7 marks) – Three or four functions correctly identified.


2017.1 J.18/20_MS 37/72 of 71 examsDEB

Q.12 (cont’d.)

12(b) Let f be the function f : x 2 →׀x2 + 9, where x ∈ ℝ.

(i) Find the values of x for which f(x) = 9x. (5C)

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

x =

2

3

or x – 3 = 0 x = 3

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. equates 2x2 + 9 = 9x or 2x2 – 9x + 9 = 0 and stops or continues incorrectly.

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

(ii) Hence, or otherwise, find the values of t for which f (t

1) = t

9. (5C)

Let x = t

1

t

1 =

2

3

t =

3

2

or t

1 = 3

t = 3

1

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

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

x = t

1 or equivalent.

High partial credit: (4 marks) – Correct method and finishes correctly, but one minor error. – Finds only one value of t.


2017.1 J.18/20_MS 38/72 of 71 examsDEB


Mathematics


Structure of the Marking Scheme

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

Scale label A B C D

No. of categories 2 3 4 5

5 mark scale 0, 5 0, 3, 5 0, 2, 4, 5 0, 2, 3, 4, 5 10 mark scale 0, 4, 7, 10 0, 4, 6, 8, 10 15 mark scale 0, 4, 8, 12, 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 (middle partial credit) almost correct response (high partial credit) correct response (full credit)

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

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

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

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

examsDEB

Name/version:

Printed: Whom:

Checked:

To: Ret’d:

Updated: Whom:

Name/version:

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

2017.1 J.18/20_MS 39/72 of 71 examsDEB

Summary of Marks – 2017 JC Maths (Higher Level, Paper 2)

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

10C (0, 4, 7, 10)

(ii) Q.9 15D (0, 4, 8, 12, 15) (iii) 15 (c) 5C (0, 2, 4, 5) 25 Q.10 (a) 5C* (0, 2, 4, 5) (b) 5C* (0, 2, 4, 5) Q.4 (a) 10C (0, 4, 7, 10) (c) 5C* (0, 2, 4, 5) (b) 5B (0, 3, 5) (d) 5C* (0, 2, 4, 5) (c) 5C* (0, 2, 4, 5) (e) 5C* (0, 2, 4, 5) 20 25 Q.5 (a) 5B (0, 3, 5) Q.11 (a) 5C (0, 2, 4, 5) (b) (i) 5B (0, 3, 5) (b) 5C (0, 2, 4, 5) (ii) 5C* (0, 2, 4, 5) (c) 5B (0, 3, 5) (c) (i) 10D* (0, 4, 6, 8, 10) (d) 5C (0, 2, 4, 5) (ii) 5C* (0, 2, 4, 5) 20 30 Q.6 (a) 5A* (0, 5) (b) 5B* (0, 3, 5) (c) 5C* (0, 2, 4, 5) (d) 5A (0, 5) (e) 5C (0, 2, 4, 5) (f) 10C* (0, 4, 7, 10) (g) 5D (0, 2, 3, 4, 5) (h) 5B (0, 3, 5) 45

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.

2017.1 J.18/20_MS 40/72 of 71 examsDEB


Mathematics


General Instructions

1. There are 11 questions on this examination paper. Answer all questions.

2. Questions do not necessarily carry equal marks.

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

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

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



The pie chart opposite shows the sources of energy production in the EU in 2016.

1(a) What type of data is displayed? Put a tick () in the correct box below. (5B)

Numerical Discrete

Numerical Continuous

Categorical Nominal

Categorical Ordinal

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Ticks ‘Categorical Ordinal’ box.

examsDEB

x25%

10%

2x20%

Renewables

Nuclear

Gas

Coal

Oil

2017.1 J.18/20_MS 41/72 of 71 examsDEB

Q.1 (cont’d.)

1(b) What percentage of energy was produced from coal in 2016? (5C)

% energy produced from coal

25 + 10 + x + 20 + 2x = 100 3x + 55 = 100 3x = 100 – 55 = 45 x = 15%

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down any three of 25, x, 10, 20 or 2x. – Finds 25 + 10 + x + 20 + 2x = 100 and stops or continues incorrectly.

High partial credit: (4 marks) – Finds 3x + 55 = 100, but fails to finish or finishes incorrectly.


1(c) Calculate the measure of the angle in the pie chart that represents the percentage of energy produced from nuclear in 2016. (5B*)

x = 15 2x = 30 | angle | = 360° × 30%

= 360° × 100

30

= 108°

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Some work of merit, e.g. writes down 2x = 30. – Writes down | angle | = 360° × 30%

or 360° × 100

30, but fails to finish

or finishes incorrectly.

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

1(d) 75 tonnes of energy was produced from renewables in 2016. How much energy was produced from oil in 2016? (5B*)

25% ≡ 75 tonnes

1% ≡ 25

75

≡ 3 tonnes

10% ≡ 3 × 10% ≡ 30 tonnes

Scale 5B* (0, 3, 5) Partial credit: (3 marks) – Some work of merit, e.g. equates 25% to 75 tonnes and stops. – Finds 1% ≡ 3 tonnes, but fails to finish or finishes incorrectly.

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


2017.1 J.18/20_MS 42/72 of 71 examsDEB


The Great Pyramid of Giza (also known as the Pyramid of Khufu) on the banks of the River Nile in Egypt is the oldest of the Seven Wonders of the Ancient World.

The pyramid consists of four equilateral triangular sides sitting on a square base.

2(a) The base of the pyramid has an area of 52,900 m2. Find the length of one side of its base. (5C*)

Area of square base = l × w = l2

l2 = 52,900

l = 900,52

= 230 m

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down l × w or l2 and stops. – Finds l2 = 52,900, but fails to finish or finishes incorrectly.

High partial credit: (4 marks) – Finds l = 900,52 , but fails to finish



2(b) When it was built, the triangular faces of the pyramid were covered in polished limestone.

(i) Find the height of one of the triangular faces. Give your answer correct to the nearest metre. (5C*)

Consider the perpendicular height of a single triangular face: Using Pythagoras’ theorem

| side |2 = | 2

1base |2 + | h |2

(230)2 = | 2

1(230) |2 + | h |2

| h |2 = 2302 – 1152 = 52,900 – 13,225 = 39,675

| h | = 675,39

= 199·185842... ≅ 199 m

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, writes down correct formula for Pythagoras’ theorem and stops.– Some correct substitution into formula for Pythagoras’ theorem (not stated).

High partial credit: (4 marks) – Fully correct substitution into formula for Pythagoras’ theorem with some manipulation, e.g. | h |2 = 2302 – 1152, 52,900 – 13,225 or 39,675, but fails to finish or finishes incorrectly.


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


2017.1 J.18/20_MS 43/72 of 71 examsDEB

Q.2 (cont’d.) 2(b) (ii) Find the surface area of the polished limestone that would have been required to cover

the pyramid. (5C*)

Surface area = 4 × Area of triangular side

= 4 × 2

1| base | × | ⊥height |

= 4 × 2

1(230) × 199

= 4 × 115 × 199 = 91,540 m2

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

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

High partial credit: (4 marks) – Fully correct substitution into correct area formula, but fails to finish or finishes incorrectly. – Finds correct area of one side of pyramid [ans. 22,885 m2] and stops.

* 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.

2(c) There are two smaller pyramids, Khafre and Menkaure, near the Great Pyramid. The triangular faces of all three pyramids are similar, as shown in the diagram below.

Great Pyramid (Khufu) Khafre Menkaure

(i) Find the percentage decrease in the length of a side of Khafre, whose sides measure 216 m compared to that of the Great Pyramid (Khufu). Give your answer correct to one decimal place. (5B*)

Decrease in length of side = 230 − 216 = 14 m

% Decrease = 230

14 ×

1

100

= 6·086956... ≅ 6·1%

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

Scale 5B* (0, 3, 5) Partial credit: (3 marks) – Some work of merit, e.g. writes down correct formula % Decrease and stops. – Finds correct decrease in length of side.




2017.1 J.18/20_MS 44/72 of 71 examsDEB

Q.2 (cont’d.)

2(c) (ii) The percentage decrease in the length of a side of Menkaure compared to that of Khafre is 49·5%. Calculate the length of the perimeter of the base of Menkaure. Give your answer correct to the nearest metre. (5C*)

Decrease in length of side = 49·5% of 216

= 216 × 100

549⋅

= 106·92 m

Length of side of Menkaure = 216 – 106·92 = 109·08 m

Perimeter of the base of Menkaure = 109·08 × 4 = 436·32 ≅ 436 m

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down 49·5% of 216 or equivalent and stops. – Finds correct decrease in length of side.

High partial credit: (4 marks) – Finds correct length of side, but fails to finish or finishes incorrectly.



(iii) Does the surface area of the faces of the pyramids decrease at the same rate? Give a reason for your answer. (5B)

Answer – no

Reason Any 1: – area is two dimensional - each dimension decreases

by a scale factor, k - therefore area decreases by a factor of k2 //

– area is two dimensional - each dimension decreases at calculated rate, therefore area decreases at increased rate // etc.


Scale 5B (0, 3, 5) Partial credit: (3 marks) – Some work of merit, e.g. reference to scale factor, k or k2 and stops. – Correct answer, but no reason given.


2017.1 J.18/20_MS 45/72 of 71 examsDEB

Notes:


2017.1 J.18/20_MS 46/72 of 71 examsDEB


3(a) The co-ordinate diagram shows the lines l, m, n, o and p. The table shows the equation of each line. Write the letters l, m, n, o and p into the table to match each line to its equation. (10C)

x

yl m

n

op

Equation Line

x – y + 3 = 0 m

x = –2y p

y – 5 = 0 n

y = 2x + 5 l

2x + y = 9 o

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

High partial credit: (7 marks) – Two or three lines correctly labelled.

3(b) Complete the following sentences. Write one of the letters l, m, n, o and p in each box. (10C)

(i) Lines p and o have negative slopes.

(ii) Lines l and p are perpendicular to each other.

(iii) Lines l and n intersect at the point (0, 5).

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

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


2017.1 J.18/20_MS 47/72 of 71 examsDEB

Q.3 (cont’d.)

3(c) The lines m, n and o intersect at the same point. Find the co-ordinates of this point of intersection. (5C)

n: y – 5 = 0 y = 5

m: x – y + 3 = 0 n ∩ m: x – (5) + 3 = 0 x – 2 = 0 x = 2

n ∩ m = (2, 5) n ∩ m ∩ o = (2, 5)

or

n: y – 5 = 0 y = 5

o: 2x + y = 9 n ∩ o: 2x + (5) = 9 2x = 9 – 5 = 4 x = 2

n ∩ o = (2, 5) n ∩ m ∩ o = (2, 5)

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

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Any work of merit, e.g. multiplying equation(s) by appropriate constant to facilitate cancellation of x or y term. – Finds either variable (x or y) correctly by trial and error, but fails to verify in both equations or verifies incorrectly. – Finds first variable (y) correctly and stops.

High partial credit: (4 marks) – Finds first variable (y) correctly and substitutes value into second equation correctly, e.g. m: x – (5) + 3 = 0 or o: 2x + (5) = 9, but fails to find or finds incorrect value of second variable (x). – Finds both variables (x and y) correctly, but no work shown. – Finds both variables (x and y) by graphical means. – Finds both variables (x and y) by trial and error, but does not verify in all three equations.


2017.1 J.18/20_MS 48/72 of 71 examsDEB



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

| AB | = 6 cm | ∠CAB | = 90° | BC | = 8 cm. (10C)

A

C

B

8cm

6 cm

41�

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

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

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

4(b) On your diagram, measure the angle ∠ABC. Give your answer correct to the nearest degree. (5B)

From diagram | ∠ABC | = 41° (± 2°)

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

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

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


2017.1 J.18/20_MS 49/72 of 71 examsDEB

Q.4 (cont’d.)

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

cos | ∠ABC| = |Hyp|

|Adj|

= ||

||

BC

AB

= 8

6

= 0·75

| ∠ABC | = cos–1(0·75) = 41·409622... ≅ 41°

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

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down correct trigonometric ratio (cos). – Some correct substitution into correct trigonometric ratio (cos). – Finds cos | ∠ABC| and stops [ans. 0·75]. – Using answer from part (b), finds correct value of cos 41° [ans. 0·754709...]

High partial credit: (4 marks) – Correct substitution into trigonometric ratio and correctly manipulated, e.g. | ∠ABC | = cos–1(0·75), but fails to finish or finishes incorrectly.

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


2017.1 J.18/20_MS 50/72 of 71 examsDEB



Landslides occur when the stability of a slope of ground is compromised. This can happen after there has been heavy rainfall.

The sketch shows the incline (slope) of a section of ground on the side of a mountain prior to a landslide.

5(a) Find the incline (slope) of the land in the sketch between the points A and B. (5B)

Incline (slope) = –|Run|

|Rise|

= –10

14

= –1·4

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Some work of merit, e.g. writes down

slope = –|Run|

|Rise| or

|Run|

|Rise| and stops.

– Gives 10

14 or 1·4 as final answer.

5(b) The diagram below shows a sketch of the incline after a landslide. | ∠EDC | = 36°.

36º

Ground line

C

DE23 m


Ground line

A

B10 m

14 m

2017.1 J.18/20_MS 51/72 of 71 examsDEB

Q.5 (cont’d.)

5(b) (i) Find the incline (slope) of the ground shown in the sketch between the points C and D. (5B)

Incline (slope) = –tan | ∠EDC | = –tan 36° = –0·726542...

** Accept students’ answers that are appropriately round.

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Some work of merit, e.g. writes down tan | ∠EDC | or tan 36° and stops.

(ii) Hence, find | CE |, giving your answer correct to the nearest metre. (5C*)

tan | ∠EDC | = |Adj|

|Opp|

= ||

||

DE

CE

= 0·726542... ... answer from part (i)

0·726542... = 23

||CE

| CE | = 23 × 0·726542... = 16·710478... ≅ 17 m

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

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

– Finds 0·726542... = 23

||CE and stops.

High partial credit: (4 marks) – Finds | CE | = 23 × 0·726542..., but fails to finish or finishes incorrectly.




2017.1 J.18/20_MS 52/72 of 71 examsDEB

Q.5 (cont’d.)

5(c) The National Roads Authority has decided to construct a concrete retaining wall, 80 m long, to protect a section of motorway from future landslides.

The cross-section of the retaining wall, of height h, is shown.

(i) Find h, giving your answer correct to the nearest metre. (10D*)

cos | ∠A | = |Hyp|

|Adj|

= 163

5051

⋅⋅−⋅

= 163

1

⋅

= 0·316455... | ∠A | = cos–1(0·316455...) = 71·551284... ≅ 71·6°

tan | ∠A | = |Adj|

|Opp|

tan 71·551284...° = 1

h

| h | = 1 × tan 71·551284...° = 2·997599... ≅ 3 m

Scale 10D* (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. writes down correct trigonometric ratio (cos or tan). – Some correct substitution into correct trigonometric ratio (cos or tan).

– Finds cos | ∠A | = 163

1

⋅ or 0·316455...

and stops.

Mid partial credit: (6 marks) – Finds | ∠A | = cos–1

163

1

⋅ or

cos–1 0·316455... and stops or continues incorrectly.

High partial credit: (8 marks) – Finds | ∠A | = 71·551284...° correctly (accept rounded value), but fails to finish or finishes incorrectly.

– Finds tan 71·551284...° = 1

h or h, but

fails to finish or finishes incorrectly. – Finds | h | = 1 × tan 71·551284...° or similar, but fails to finish or finishes incorrectly.




0 m·5

1 m·5

hm

3·1

6m

A

2017.1 J.18/20_MS 53/72 of 71 examsDEB

Q.5 (cont’d.)

5(c) (cont’d.)

(ii) Given that the wall is 80 m long, find the volume of concrete required to construct the wall. (5C*)

Area of cross section of retaining wall = Area of rectangle + area of triangle

= l × w + 2

1| base | × | ⊥height |

= 3 × 0·5 + 2

1| 1·5 – 0·5 | × | 3 |

= 2

3 +

2

1 × 1 × 3

= 2

3 +

2

3

= 2

6

= 3 m2

Volume of concrete required = Area of cross section × 80 = 3 × 80 = 240 m3

** 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 correct area formula from Tables. – Some correct substitution into correct area formula (not stated).

High partial credit: (4 marks) – Finds correct area of cross section, but fails to finish or finishes incorrectly.

* Deduct 1 mark off correct answer only for the omission of or incorrect use of units (‘m3’) - apply only once per each section (a), (b), (c), etc. of question.


2017.1 J.18/20_MS 54/72 of 71 examsDEB


Transition year students conducted a survey for their “Green Schools” project. They handed out a questionnaire to a sample group of students in their school, which asked how they travel to school and how long it takes. The length of time (in minutes) taken to travel to school by the students surveyed is displayed in the back-to-back stem-and-leaf plot shown.

Junior School Senior School 8 6 0 3 5 7 8 9

8 6 5 4 3 2 1 1 4 5 5 5 7 9 8 6 4 3 1 2 2 3 4 5 8 9 3 3 4 6 7 4 0 5 9

Key: 2 2 means 22 minutes

6(a) What is the mode of the Senior School data? (5A*)

Mode = 15 minutes

Scale 5A* (0, 5) – Hit or miss.

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

6(b) Find the median time taken to travel to school by Junior School students. (5B*)

Median

6, 8, 12, 13, 14, 15, 16, 18, 21, 23, 24, 26, 28

Median = 16 minutes

Scale 5B* (0, 3, 5) Partial credit: (3 marks) – Indication of ‘middle’ time.

* Accept correct answers with or without work shown. * Deduct 1 mark off correct answer only for the omission of or incorrect use of

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


2017.1 J.18/20_MS 55/72 of 71 examsDEB

Q.6 (cont’d.)

6(c) Find the interquartile range of the Junior School data. (5C*)

Graphical means

6, 8, 12, 13, 14, 15, 16, 18, 21, 23, 24, 26, 28

Lower Quartile, Q1 = 2

1312 +

= 12·5 minutes

Upper Quartile, Q3 = 2

2423 +

= 23·5 minutes

Interquartile range = Q3 – Q1 = 23·5 – 12·5 = 11 minutes ... Answer 1

or

Numerical means

6, 8, 12, 13, 14, 15, 16, 18, 21, 23, 24, 26, 28

n = number of results in dataset Lower quartile, Q1

k = 4

n

= 4

13

= 3·25 → 4 Q1 = 4th value in list = 13

Upper quartile, Q3

k = 4

3n

= 4

)13(3

= 9·75 → 10 Q3 = 10th value in list = 23

Interquartile range = Q3 – Q1 = 23 – 10 = 10 minutes ... Answer 2

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down definition of interquartile range and stops.– Some relevant work towards finding upper and/or lower quartile by either method.

High partial credit: (4 marks) – Finds correct value of upper or lower quartile by either method, but fails to finish or finishes incorrectly. – Finds correct value of interquartile range, but no work shown.

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


2017.1 J.18/20_MS 56/72 of 71 examsDEB

Q.6 (cont’d.)

6(d) What time in the stem-and-leaf-plot could best be described as an outlier? (5A)

Outlier = 59 minutes

Scale 5A (0, 5) – Hit or miss.

6(e) Fill in the grouped frequency table below using the data from the stem-and-leaf plot. (5C)

Time (minutes) 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60

Frequency 7 13 11 4 1 1

Note: 20 − 30 means at least 20 minutes but less than 30 minutes, etc.

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.

6(f) Use mid-interval values of the data in the grouped frequency table to estimate the mean time taken to travel to school by the students in the sample. Give your answer correct to the nearest minute. (10C*)

Mean = 11411137

)155()145()435()1125()1315()75(

+++++×+×+×+×+×+×

= 37

554514027519535 +++++

= 37

745

= 20·135135... ≅ 20 minutes

Scale 10C* (0, 4, 7, 10) Low partial credit: (4 marks) – Some work of merit, e.g. shows indication of division by 37. – Finds one correct mid-interval value.

High partial credit: (7 marks) – Finds correct numerator, i.e. 745 or (5 × 7) + (15 × 13) + (25 × 11) + (35 × 4) + (45 × 1) + (55 × 1). – Finds incorrect mean using consistent incorrect mid-interval values.

– Finds 37

745, but fails to finish or finishes

incorrectly.


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


2017.1 J.18/20_MS 57/72 of 71 examsDEB

Q.6 (cont’d.)

6(g) Display the data from the grouped frequency table on a histogram. (5D)

Scale 5D (0, 2, 3, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. draws correctly labelled axes (with no bars inserted). – One or two bars correct (each of consistent width).

Mid partial credit: (3 marks) – Three or four bars correct (each of consistent width). – Bar chart correctly drawn, but both axes not numerated or labelled.

High partial credit: (4 marks) – Five bars correct (each of consistent width). – All six bars correct (each of consistent width), but both axes not labelled. – Bar chart correctly drawn.

6(h) The students felt that the wording of Q1 in the survey could have been improved, because some students didn’t take the survey seriously when filling in their answers as shown.

Green Schools Project Survey

Q1: How did you travel to school this morning? by Jet Pack

Q2: How long did it take you to get to school this morning? 25 minutes

Rewrite Q1 of the survey in a more suitable form. (5B)

Any 1: – How do you travel to school? Walk Bus Cycle Car Other //

– How did you travel to school this morning? (Tick one) Walk Bus Cycle Car Other // etc.


Scale 5B (0, 3, 5) Partial credit: (3 marks) – Question written in an unsuitable form, but with deficiencies.


2

12

4

14

6

8

0

10

Time (minutes)

Fre

qu

ency

0 10 20 30 40 50 60

2017.1 J.18/20_MS 58/72 of 71 examsDEB


7(a) An electric appliance store sells three different brands of television, all in various screen sizes and features, as shown.

Televisions Brand Features Screen Size Zony Smart TV 24 inches

Hamsung Full HD 32 inches Fillips 40 inches

49 inches 55 inches 65 inches

(i) Calculate how many different televisions can be purchased from this store. (5B)

# options = 3 × 2 × 6 = 36

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Some work of merit, e.g. mention of the ‘Fundamental Principle of Counting’. – Writes a list of possible options. – Finds 3 × 2 × 6, but fails to finish or finishes incorrectly. – Uses addition instead of multiplication, i.e. # options = 3 + 2 + 6 = 11.

(ii) A customer purchased a television from the store. Assuming all brands and types of televisions are equally likely to be purchased, find the probability that this person purchased a Zony 32 inch smart TV. (5B)

# Zony, 32 inch, Smart TV = 1 # All TV types = 36

P(Zony, 32 inch, Smart TV) = 36

1 or 0·027777...

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

(iii) A customer is chosen at random from those who bought a Fillips TV. Find the probability that this person purchased a smart TV. (5B)

# All Fillips TV = 1 × 2 × 6 = 12 # Fillips, Smart TV = 1 × 6 = 6

P(Smart TV) = 12

6

= 2

1 or 0·5

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Some work of merit, e.g. identifies 6 or 12 and stops or continues.


2017.1 J.18/20_MS 59/72 of 71 examsDEB

Q.7 (cont’d.)

7(a) (iv) Television screen sizes are measured according to the length of their diagonal. A television on special offer has a screen 24 inches in height and 32 inches in width.

Write down the screen size of the television. Give a reason for your answer. (5C*)

Answer – 40 inches

Reason – according to Pythagoras’ theorem: d2 = 322 + 242 = 1,024 + 576 = 1,600 d = 600,1

= 40 inches

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

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

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


Diagonal

32 inches

24 inches

2017.1 J.18/20_MS 60/72 of 71 examsDEB

Q.7 (cont’d.) 7(b) The English-language edition of Scrabble contains 100 tiles, 98 of which have a single letter

printed on them. There are 42 vowels and 56 consonants. To make a game more difficult, the two blank tiles are removed. The board game starts with each player picking tiles from a bag containing the remaining 98 tiles. When a tile is picked from the bag, it is not replaced.

(i) The first player picks two tiles from the bag. Complete the tree diagram below to show all possible outcomes and fill in the boxes in the diagram which represent the probability that vowels or consonants will be picked. (10C)

Letters

42

41

56

42

55

56

98

97

97

97

97

98 Consonant

Consonant

Consonant

Vowel

Vowel

Vowel

Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) – Between one and four outcomes or probabilities correct.

High partial credit: (7 marks) – Between five and eight outcomes or probabilities.

(ii) Find the probability that the tiles selected are either both consonants or both vowels. (5C)

P(consonant, consonant) = 98

56 ×

97

55

= 506,9

080,3

= 679

220

P(vowel, vowel) = 98

42 ×

97

41

= 506,9

722,1

= 679

123 or 0·324005...

P(consonant, consonant or vowel, vowel)

= 679

220 +

679

123

= 679

343 or

97

49 or 0·505154...


2017.1 J.18/20_MS 61/72 of 71 examsDEB

Q.7 (cont’d.)

7(b) (ii) (cont’d.)

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

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

98

56 and

97

55 or

98

42 and

97

41, but does not

attempt to multiply them. – Uses addition and finishes correctly

[ans. 98

56 +

97

55 =

679

773 or equivalent].

Mid partial credit: (3 marks) – Finds one probability correctly,

i.e. P(c, c) = 506,9

080,3 or

679

220

or P(v, v) = 506,9

722,1 or

679

123, but fails

to finish or finishes incorrectly.

High partial credit: (4 marks) – Finds both probability correctly, i.e. P(c, c) and P(v, v), but fails to finish or finishes incorrectly.

(iii) Find the probability that at least one of the tiles selected is a vowel. (5C)

P(at least 1 vowel) = 1 – P(consonant, consonant)

= 1 – 679

220

= 679

459 or 0·675994...

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. identifies 679

220

and stops or continues.

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

679

220, but fails to finish



2017.1 J.18/20_MS 62/72 of 71 examsDEB



8(a) ABCD is a parallelogram with the side AB extended to E to form an external angle.

(i) Find the measure of each of the angles marked x, y and z. Give a reason for your answer in each case. (10D)

Angle: Reason:

x° = 113° x° + 29° + 40° = 180° ... the sum of the three angles in a triangle is equal to 180° x° = 180° – (29° + 54°) = 180° – 83° = 97°

y° = 29° – alternate angles as [ AB ] is parallel to [ DC ]

z° = 126° z° + 54° = 180° ... opposite angles in z° = 180° – 54° a parallelogram are equal = 126° and EB is a straight line / straight angle = 180°

or

z° = x° + y° ... corresponding angles as = 97° + 29° [ AD ] is parallel to [ BC ] = 126°

Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) – Some work of merit, e.g. identifies one correct angle or reason.

Mid partial credit: (6 marks) – Finds two correct angles, but no reason(s) or incorrect reason(s) given. – Gives two correct reasons, but angle(s) not evaluated or evaluated incorrectly.

High partial credit: (8 marks) – Finds two correct angles and correct reasons given. – Finds three correct angles, but no reason(s) or incorrect reason(s) given.

(ii) What type of triangle does DBC represent? Give a reason for your answer. (5B)

Answer – scalene (triangle)

Reason – all three internal angles are unequal and hence the triangle has three unequal sided // etc.

** Accept other appropriate material.

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


AE

CD

xy

29 54

zB

° °

°°°

2017.1 J.18/20_MS 63/72 of 71 examsDEB

Q.8 (cont’d.)

8(a) (iii) State whether triangles DBC and ABD are congruent. Give a reason for your answer. (5C)

Answer – triangles DBC and ABD are congruent

Reason – SSS: | DB | = | DB |, | AD | = | BC | , | AB | = | DC |, //

or

– SAS: | DB | = | DB |, | ∠ABD | = | ∠CDB |, | AB | = | DC | or | AD | = | BC |, | ∠BDA | = | ∠DBC |, | DB | = | DB | or | AB | = | DC |, | ∠DAB | = | ∠BCD |, | AD | = | BC | //

or

– ASA: | ∠ABD | = | ∠CDB |, | DB | = | DB |, | ∠BDA | = | ∠DBC | or | ∠DAB | = | ∠BCD |, | AD | = | BC |, | ∠BDA | = | ∠DBC | or | ∠DAB | = | ∠BCD |, | AB | = | DC |, | ∠ABD | = | ∠CDB | // etc.


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

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

8(b) P, Q, R and S are points on a circle with centre O. | ∠ROP | = 115°, as shown. (i) Find | ∠RSP | . (5B*)

| ∠RSP | = 2

1| ∠ROP |

= 2

1(115°)

= 57·5°

Scale 5B* (0, 3, 5) Partial credit: (3 marks) – Some work of merit, e.g. writes down that the angle at the centre is equal to twice the angle at the circumference. – Writes down | ∠ROP | = 2 × | ∠RSP | or similar. – Finds | ∠RSP | = 2 × 115° = 230°.

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


P

QR

O

115

S

°

2017.1 J.18/20_MS 64/72 of 71 examsDEB

Q.8 (cont’d.)

8(b) (cont’d.)

(ii) Find | ∠PQR |. (5B*)

| ∠PQR | = 2

1[360° – | ∠ROP |]

= 2

1(360° – 115°)

= 2

1(245°)

= 122·5°

or

| ∠PQR | + | ∠RSP | = 180°

| ∠PQR | = 180° – | ∠RSP | = 180° – 57·5° = 122·5°

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

Scale 5B* (0, 3, 5) Low partial credit: (3 marks) – Some work of merit, e.g. writes down that in a cyclic quadrilateral opposite angles sum to 180°. – Finds 245° and stops [method ]. – Finds | ∠PQR | = 180° – 115° = 65° [method ].

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


2017.1 J.18/20_MS 65/72 of 71 examsDEB


Prove that the angle at the centre of a circle standing on a given arc is twice the angle at any point of the circle standing on the same arc. (15D)

Given Circle, with centre O,

containing points A, B and C.

To prove | ∠COB | = 2| ∠CAB |

Construction Join A to O and extend to D.

Proof

O

CB

A

D

| ∠DOB | = | ∠OAB | + | ∠OBA | ... exterior angle equal to the sum of 2 opposite interior angles

| OA | = | OB | ... both radius of the circle | ∠OAB | = | ∠OBA | ... base angles of isosceles triangle are equal

∴ | ∠DOB | = | ∠OAB | + | ∠OAB | = 2| ∠OAB |

similarly | ∠COD | = 2| ∠CAO |

∴ | ∠COB | = | ∠DOB | + | ∠COD | = 2| ∠OAB | + 2| ∠CAO | = 2(|∠OAB | + | ∠CAO |) = 2| ∠CAB |

∗∗ Some steps may be indicated on the diagram.

Scale 15D (0, 4, 8, 12, 15) Low partial credit: (4 marks) – Some work of merit, e.g. draws correct diagram.

Middle partial credit: (8 marks) – Diagram, Given, To Prove and Construction only; or More than one step missing in proof.

High partial credit: (12 marks) – One step missing in proof; or fully correct but with no reason given; or gets as far as step or equivalent.


2017.1 J.18/20_MS 66/72 of 71 examsDEB



The diagram below is a scale drawing of a dome which is in the shape of a hemisphere. The figure of a man, who is 1⋅8 m tall, standing beside the dome, allows the scale of the drawing to be estimated.

10(a) Estimate the radius of the dome. (5C*)

Scale

From the diagram: Height of man = 2 grid units 2 grid units = 1·8 m

1 grid unit = 2

81⋅

= 0·9 m

Radius

From the diagram: Height of dome = 11 grid units = 11 × 0·9 m = 9·9 m

Radius = 9·9 m

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. equates height of man ≡ 2 grid units or height of dome ≡ 11 grid units. – Writes down height of dome ≡ radius of hemisphere. – Finds correct scale.

High partial credit: (4 marks) – Uses correct scale to determine height / radius of dome, but error(s) in finding radius. – Finds correct diameter of dome [ans. 19·8m].

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


2017.1 J.18/20_MS 67/72 of 71 examsDEB

Q.10 (cont’d.)

10(b) Using your answer to part (a), find the curved surface area of the dome. Give your answer in m2, correct to one decimal place. (5C*) Curved surface area = 2πr2 = 2π(9·9)2 = 196·02π = 615·814991... ≅ 615·8 m2

** Accept students’ answer 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 correct formula for curved surface area of a hemisphere from Tables. – Some correct substitution into relevant formula (not stated) and stops or continues.

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


* No deduction applied for the omission of or incorrect use of units (‘m2’) as units are mentioned in the question.

10(c) The volume of the dome is 2,063·14 m3. Find the radius of the dome. Give your answer in metres, correct to two decimal places. (5C*)

Volume of hemisphere = 3

2 πr3

= 2,063·14 m3

3

2 πr3 = 2,063·14

r3 = π×

×⋅2

3412063

= 985·076787...

r = 3 ...076787985⋅ = 9·950006... ≅ 9·95 m

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

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


* No deduction applied for the omission of or incorrect use of units (‘m2’) as units are mentioned in the question.


2017.1 J.18/20_MS 68/72 of 71 examsDEB

Q.10 (cont’d.)

10(d) Find the percentage change between the actual radius and the estimated radius, correct to one decimal place. (5C*)

Estimated radius = 9·9 m Actual radius = 9·95 m Difference = 9·95 – 9·9 = 0·05 m

% Difference = 959

050

⋅⋅

× 1

100

= 0·005025... × 100 = 0·5025... ≅ 0·5%

Scale 5C* (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. finds correct difference [ans. 0·05 m], but does not attempt to find percentage.

High partial credit: (4 marks) – Finds 959

050

⋅⋅

or 959

050

⋅⋅

× 1

100, but fails to

evaluate or evaluates incorrectly.



10(e) Find the area of the floor of the dome, in terms of π. (5C*)

Area of floor = πr2 = π(9·95)2 = 99·0025π m2

** Accept students’ answer from part (c) 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 area of circle. – Some correct substitution into relevant formula (not stated) and stops or continues.

High partial credit: (4 marks) – Fully correct substitution into relevant formulae, but fails to finish or finishes incorrectly. – Final answer not in terms of π.

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


2017.1 J.18/20_MS 69/72 of 71 examsDEB


The equation of the line l is 3x – y – 1 = 0.

11(a) Find the co-ordinates of the points where l cuts the axes. (5C) l cuts the x-axis

y = 0 3x – (0) – 1 = 0 3x = 1

x =

3

1

co-ordinates = (3

1, 0)

l cuts the y-axis

x = 0 3(0) – y – 1 = 0 –y = 1 y = –1 co-ordinates = (0, –1)

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

– Finds x = 3

1 or y = –1.

– Writes down y = mx + c or y = 3x –1.

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

1, 0) or (0, –1) and stops.

– Finds both x = 3

1 and y = –1.

– Correct points in incorrect boxes.

11(b) Find the length of the line segment between the two points found in part (a). (5C)

(3

1, 0), (0, –1)

length of the line segment = 212

212 )()( yyxx −+−

= 22 )3

10()01( −+−−

= 9

11 +

= 9

10

= 3

10 units

Using Pythagoras’ theorem

| Length |2 = (–1)2 + (3

1)2

= 1 + 9

1

= 9

10

| Length | = 3

10 units


2017.1 J.18/20_MS 70/72 of 71 examsDEB

Q.11 (cont’d.)

11(b) (cont’d.)

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down correct formula for distance of a line segment or Pythagoras’ theorem.

– Some correct substitution into relevant formula (not stated) and stops or continues.

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

– Gives final answer as 9

10 or 1·111...

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

11(c) Find the slope of the line l. (5B)

l: 3x – y – 1 = 0 y = 3x – 1 y = mx + c ml = 3

or

(3

1, 0), (0, −1)

(x1, y1) (x2, y2)

ml = 12

12

xx

yy

−−

=

3

10

01

−

−−

=

311

−

−

= 3

or

ml = run

rise

=

311

= 3

Scale 5B (0, 3, 5) Low partial credit: (3 marks) – Some work of merit, e.g. writes down

correct formula for slope (12

12

xx

yy

−−

or run

rise)

or equation of line (y = mx + c).

– Some indication of –b

a.

– Writes l in the form y = 3x –1 and stops.– Points correctly plotted.


2017.1 J.18/20_MS 71/72 of 71 examsDEB

Q.11 (cont’d.)

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

ml = 3

mk = ⊥ml

= –3

1

mk = –3

1, point (–3, 2)

y – y1 = m(x – x1)

y – 2 = –3

1(x – (–3))

3y – 6 = –(x + 3) 3y – 6 = –x – 3 x + 3y – 6 + 3 = 0 x + 3y – 3 = 0

Scale 5C (0, 2, 4, 5) Low partial credit: (2 marks) – Some work of merit, e.g. writes down correct formula for equation of a line. – Finds correct slope for k and stops. – Some correct substitution into relevant formula (not stated) and stops or continues.

High partial credit: (4 marks) – Finds correct slope for k and continues - one incorrect substitution into line formula, but finishes correctly. – Finds correct slope for k, but both x and y reversed in substitution, but finishes correctly. – Fully correct substitution into line formula, but fails to finish or finishes incorrectly. – Final answer not in the correct form.


2017.1 J.18/20_MS 73/72 Page 73 of 71 examsDEB

higher level marking scheme - m. selkirk confey …...higher level marking scheme paper 1 pg. 2 ......

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