mathematics (project maths – phase 2) higher level l.17/20_ms 2/48 page 2 of 48 exams deb...

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L.17/20

Pre-Leaving Certificate Examination, 2013

Mathematics (Project Maths – Phase 2)

Higher Level

Marking Scheme

Paper 1 Pg. 2

Paper 2 Pg. 25

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Mathematics (Project Maths – Phase 2)

Higher Level – Paper 1 Marking Scheme (300 marks)

General Instructions

Instructions

There are three sections in this examination paper:

Section A Concepts and Skills 100 marks 4 questions

Section B Contexts and Applications 100 marks 2 questions

Section C Functions and Calculus (old syllabus) 100 marks 2 questions

Answer all eight 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.

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Summary of Marks

Q.1 a 5C (0, 3, 4, 5) + Q.4 a 10B (0, 7, 10) 5B (0, 3, 5) b i 5B (0, 3, 5) b 5C (0, 3, 4, 5) ii 5B (0, 3, 5) c 5B (0, 3, 5) ii 5B (0, 3, 5) d 5C (0, 3, 4, 5) 25 25 Q.5 a 5B (0, 3, 5) Q.2 a 10C* (0, 5, 8, 10) b 10C (0, 5, 8, 10) b 5B* (0, 3, 5) c 5B (0, 3, 5) c 10B (0, 7, 10) d 10C (0, 5, 8, 10) 25 e 10C (0, 5, 8, 10) f 5B (0, 3, 5) Q.3 a 10C (0, 5, 8, 10) g 5B (0, 3, 5) b i 5B (0, 3, 5) 50 ii 10C (0, 5, 8, 10) 25 Q.6 a 10C (0, 5, 8, 10) b 10C (0, 5, 8, 10) c 10C (0, 5, 8, 10) d 10C (0, 5, 8, 10) e 10C (0, 5, 8, 10) 50 200 Q.7 a 10, Att. 3 b i 10, Att. 3 ii 10, Att. 3 c i 10, Att. 3 ii 10, Att. 3 50 Q.8 a 10, Att. 3 b i 10, Att. 3 ii 10, Att. 3 c 5, Att. 2 + 5, Att. 2 + 5, Att. 2 + 5, Att. 2 50 100 300

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Marking Scheme – Sections A and B

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 E

No of categories 2 3 4 5 6

5 mark scale 0, 3, 5 0, 3, 4, 5

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

15 mark scale 0, 11, 15 0, 8, 14,15 0, 5, 10, 14, 15

20 mark scale 0, 14, 18, 20 0, 6, 12, 18, 20

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)

E-scales (six categories) response of no substantial merit (no credit) response with some merit (low partial credit) response almost half-right (lower middle partial credit) response more than half-right (upper middle partial credit) almost correct response (high partial credit) correct response (full credit)

In certain cases, typically involving incorrect rounding or omission of units, 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.

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Section A Concepts and Skills 100 marks

Answer all four questions from this section.

Question 1 (25 marks)

Let z1 1 i and z2 2 2i, where i2 1.

(a) Find z1.z2 and hence plot z1, z2 and z1.z2 on an Argand diagram.

(1 + i)(–2 + 2i) = 1(–2 + 2i) + i(–2 + 2i) = –2 + 2i – 2i + 2i2 = – 2 – 2 = – 4 + 0i Calculating z1, z2 (5C)

Scale 5C (0, 3, 4, 5) Low partial credit: (3 marks) – Attempt at multiplication.

High partial credit: (4 marks) – Not in form a + bi.

Diagram (5B)

Scale 5B (0, 3, 5) Partial credit: (3 marks) – At least two points plotted. – Axes fully filled in.

(b) Express z1, z2 and z1.z2 in polar form. (5C)

z1 = √2(cos 45º + i sin 45º) z2 = √8(cos 135º + i sin 135º) z1. z2 = 4(cos 180º +i sin 180º)

Scale 5C (0, 3, 4, 5) Low partial credit: (3 marks) – One correct solution.

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

Re

z2(–2, 2)

(1, 1)(0, –4)z1z1.z2

Im

1

–1–1

–2

–2

–3

–3

–4

–4

1

2

2

3

3

4

4

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1(c) Using your answers to parts (a) and (b), explain what happens when you multiply two complex numbers. (5B)

Multiply argument

82 = 16 = 4

Add angles 45º +135º = 180°

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Multiplying argument or adding angles.

(d) Use De Moivre’s theorem to evaluate (z1)6, giving your answer(s) in rectangular form. (5C)

(√2(cos 45º + i sin 45º)6 = 8(cos 270º + i sin 270º) = 8(0 – 1) = –8 + 0i

Scale 5C (0, 3, 4, 5) Low partial credit: (3 marks) – 8 or 270º calculated.

High partial credit: (4 marks) – –8, but not in form –8 + 0i.

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Future population size can be described using the exponential equation P(t) Aebt, where A and b are constants. The size of population size P(t) can be determined at various points in time t. The population of a certain village was 1500 in 2000 and 3560 in 2010.

(a) Find the value of a. Find the value of b, correct to three decimal places. (10C*)

P(t) = Aebt P(0) = Ae0 P(0) = A = 1500 [5 marks, low partial credit] P(10) = Ae10t = 1500e10t =3560

1500

356010 te

e10t = 2.3733 lne10b = ln 2.3733 [8 marks, high partial credit] 10b = 0.864 b = 0.0864 b = 0.086

Scale 10C* (0, 5, 8, 9, 10) Low partial credit: (5 marks) – A calculated.

Partial credit: (8 marks) – An equation with b isolated.

High partial credit: (9 marks) – Not to 3 decimal places.

(b) Determine the population size of the village in 2020, correct to three significant figures. (5B*)

P(t) = Aebt P(20) = 1500e0.086(20) = 1500e1.72 = 1500(5.5845) = 8376 = 8380

Scale 5B* (0, 3, 4, 5) Low partial credit: (3 marks) – P(20) fully substituted.

High partial credit: (4 marks) – Not to 3 significant figures.

(c) During what year will the population of the village reach 15 000? (10B)

P(t) = 1500e0.086t = 15000

e0.086t = 1500

15000

e0.086t = 10 [6 marks, partial credit] ln e0.086t = ln 10 0.086t = 2.3025

t = 086.0

3025.2

= 26.77

population will reach 15 000 by 2026

Scale 10B (0, 7, 10) Partial credit: (7 marks) – e0.086t = 10 or similar calculation.

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(a) Solve the simultaneous equations: (10C)

x y z 2x 3y 2z x 2y 10.

2x + 2y = 2z –2x – 3y = –2z –y = 0 x –z = 0 2x – 2z = 0 x = 10 z = 10

Scale 10C (0, 5, 8, 10) Low partial credit: (5 marks) – Reduced to two equations.

High partial credit: (8 marks) – At least one value fully evaluated.

(b) (i) Write the following as a single fraction: (5B)

x

1 +

y2

1 .

xy

x

xy

y

22

2 =

xy

yx

2

2

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

(ii) Hence, or otherwise, show that

(x 2y)

yx 2

11 4,

given that x, y 0 and x, y . (10C)

(x + 2y)

xy

yx

2

2 4 [6 marks, low partial credit]

xy

yxyx

2

44 22 4

22 44 yxyx 8xy

22 48 yxyx 0 [8 marks, high partial credit]

(x – 2y)2 0

Scale 10C (0, 5, 8, 10) Low partial credit: (5 marks) –

xy

yx 2 fully substituted.

High partial credit: (8 marks) – Quadratic equation found.

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(a) Write a polynomial function for the following graph in its simplest form. (10B)

)3)(1)(4)(6( xxxx = 0

)34)(2410( 22 xxxx = 0

72962430401034 223234 xxxxxxxx = 0

7266136 234 xxxx = 0

Scale 10B (0, 7, 10) Partial credit: (7 marks) – )3)(1)(4)(6( xxxx or

)x)(x)(x)(x( 3146 .

�40

�80

40

80

120

160

2�2�4�6�8 4

x

y

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(b) (i) Using the same axis and scales, sketch graphs of the functions f : x ׀ | x 6 | and g : x ׀ | x 2 |. (5B)

Scale 5B (0, 3, 5) Partial credit: (3 marks) – One correct point or line identified.

(ii) Use your graph to solve the inequality | x 6 | | x 2 |. (5B)

x > 2

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Incorrect area shaded, i.e. | x 6 | > | x 2 |.

(iii) Verify your answer algebraically. (5B)

(x – 6)2 < (x + 2)2 x2 – 12x +36 < x2 + 4x + 4 [3 marks, low partial credit] –16x < –32 –x < –2 x > 2 [5 marks, full credit]

Scale 5B (0, 3, 5) Partial credit: (3 marks) – (x – 6)2 < (x + 2)2 or –x < –2.

x–6 < x+2

y f(x)

(g)x

x33 4

4

4 55 6

6

6 7 811 2

2

2 0

33 4

4

4 55 6

6

6 7 811 2

2

2 0

y f(x)

(g)x

x

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Section B Contexts and Applications 100 marks

Answer both Question 5 and Question 6.


A clothing company produces one type of shirt. Market research has found that if the company prices the shirts at €30 each, they will sell 500 units per week. It was also found that if the price was set at €55 each, the company will sell none. The clothing company prices the shirts at €x each, where 30 x 55.

(a) Draw a straight line graph to represent possible sales per week. (5B)

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

(b) Find an expression for sales per week, in terms of x. (10C)

(P, Q): (30, 500) (55, 0)

= 25

500 =

7

100

Q – 0 = 7

100(x – 55) [8 marks, high partial credit]

7Q = –100x + 5500

Q = 7

5500100 x

Q = 20(x – 55) Q = –20x + 1100 Q = 1100 –20x

Scale 10C (0, 5, 8, 10) Low partial credit: (5 marks) – Slope identified. – Point identified.

High partial credit: (8 marks) – Equation fully filled-in.

30 40 50 6010 20 55 Price (�)

Quantity

500

0

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5(c) Write an expression for the value in euro of weekly sales, in terms of x. (5B)

sales × price = (1100 – 20x)x = 1100x – 20x2

Scale 5B (0, 3, 5) Partial credit: (3 marks) – sales × price. – Answer above × x.

(d) Given that the clothing company’s fixed costs are €2000 per week and production costs are €20 for each shirt, find an expression for costs per week. (10C)

Costs = 2000 + 20(1100 – 20x) = 2000 + 22000 –400x = 24000 – 400x

Scale 10C (0, 5, 8, 10) Low partial credit: (5 marks) – 2,000 written. – 20(1100 – 20x) or similar.

High partial credit: (8 marks) – 2000 + 20(1100 – 20x) or similar.

(e) Show that the weekly profit is 20x2 1500x 24 000. (10C)

1100x –20x2 – (24000 – 400x) –20x2 + 1100x + 400x – 24000 –20x2 + 1500x – 24000

Scale 10C (0, 5, 8, 10) Low partial credit: (5 marks) – sales – cost.

High partial credit: (8 marks) – (sales – cost) filled in.

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5(f) A graph of the clothing company’s weekly profits as a function of x is shown below. Use the graph to determine the price that the company should charge in order to maximise profits. (5B)

€37.50

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Maximum value indicated on graph.

(g) Hence, calculate the number of shirts that will sell per week at this price. (5B)

Q = 1100 – 20(37.5) = 1100 –750 = 350 shirts

Scale 5B (0, 3, 5) Partial credit: (3 marks) – 37.5 put into equation.

Profit (€)

10 20 30 40

–2,000

2,000

4,000

6,000

50 60x

€37.50

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Lisa has won a major prize in a lottery game. When she goes to collect her prize, she is offered one of the following options:

Option A: Receive a payment of €1500, at the beginning of each month for 25 years.

Option B: Receive a single payment lump sum immediately.

Lisa is unsure of which option to take.

Initial exploration:

(a) Lisa feels that if she takes option A, it may give her a regular income in the future. She plans to put the monthly payment in a bank while she decides what to do. The bank is offering a rate of interest which corresponds to an annual equivalent rate (AER) of 3·5%.

Find the rate of interest per month that would, if paid and compounded monthly, correspond to an annual equivalent rate (AER) of 3·5%. (10C)

(1 + i)12 = (1.035)1

(1 + i) = 12

1

)035.1(

1 + i = 1.002871 i = 0.002871

Rate per month = 0.2871% or 0.29%

Scale 10C (0, 5, 8, 10) Low partial credit: (5 marks) – (1 + i)12 given. – (1.035) given.

High partial credit: (8 marks) – (1 + i) = 12

1

)035.1( .

(b) After three months, Lisa decides what she wants to do. She plans to continue saving all of the money as part of a pension for the future. Find the present values of the first three monthly payments lodged in her bank account. (10C)

First payment: 1500(1.002871)3 = 1512.96 Second payment: 1500(1.002871)2 = 1508.63 Third payment: 1500(1.002871)1 = 1504.31

** Accept answer from appropriate value in (a).

Scale 10C (0, 5, 8, 10) Low partial credit: (5 marks) – One correct value implied or calculated.

High partial credit: (8 marks) – Two correct values implied or calculated.

(c) Show that the total value of Lisa’s pension, assuming an annual equivalent rate (AER) of 3·5% over the period of the payments, can be represented by a geometric series. (10C)

1500(1.002871)1 + 1500(1.002871)2 + + 1500(1.002871)300 a = 1500(1.002871) = 1504.31 r = 1.002871 n = 300 (12 × 25) geometric series


Scale 10C (0, 5, 8, 10) Low partial credit: (5 marks) – Geometric series expressed.

High partial credit: (8 marks) – At least two of a, r, n stated.

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6(d) Alternatively, Lisa could have accepted a single payment lump sum (option B). How much would this payment need to be to match the future value of Lisa’s pension plan? (10C)

1

)1(

r

raS

n

n

= 1002871.1

)1002871.1(31.1504 300

= 1002871.1

)1363316.2(31.1504

= 002871.0

)363316.1(31.1504

= 714,332.95


Scale 10C (0, 5, 8, 10) Low partial credit: (5 marks) – Sn with substitution.

High partial credit: (8 marks) – Sn fully substituted.

(e) Lisa was worried that if she received a large sum of money, she would spend it carelessly and then it would be gone. Assuming that she would spend no more than €750 every month, how much better off would Lisa be if she accepted the single payment lump sum (option B) and save the remainder as a pension under the same conditions? (10C)

714,332.95(1.035)25 = 1,688,143.76 750(1.002871) = 752.15

= 1002871.1

)1002871.1(15.752 300

= 1002871.1

)1363316.2(15.752

= 002871.0

)363316.1(15.752

= 357,164.10 714,332.95 – 357,164.10 = 357,168.85


Scale 10C (0, 5, 8, 10) Low partial credit: (5 marks) – New series fully substituted.

High partial credit: (8 marks) – 357,164.10 or Sn fully evaluated.

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Marking Scheme – Section C 1. Penalties of three types are applied to student’s work, as follows:

Blunders - mathematical errors / omissions B (3) Slips - numerical errors S (–1) Misreadings (provided task is not oversimplified) M (–1)

Frequently occurring errors to which these penalties must be applied are listed in this marking scheme. They are labelled B1, B2, B3, ..., S1, S2, S3, ..., M1, M2, M3, ..., etc. Note that these lists are not exhaustive.

2. When awarding attempt marks, e.g. Att. (3), it is essential to note that:

any correct relevant step in a part of a question merits at least the attempt mark for that part if deductions result in a mark which is lower than the attempt mark, then the attempt mark

must be awarded a mark between zero and the attempt mark is never awarded.

3. Worthless work must be awarded zero marks.

4. The phrase “hit or miss” means that partial marks are not awarded - the student receives all of the relevant marks or none.

5. The phrase “and stops” means that no more work of merit is shown by the student.

6. Special notes relating to the marking of a particular part of a question are indicated by a double asterisk. These notes immediately follow the relevant solution.

7. The sample solutions for each question are not intended to be exhaustive - there may be other correct solutions.

8. Unless otherwise indicated, accept the best of two or more attempts - even when attempts have been cancelled.

9. The same error in the same section of a question is penalised once only.

10. Particular cases, verification and answers derived from diagrams (unless requested) qualify for attempt marks at most.

11. A serious blunder, omission or misreading merits the attempt mark at most.

12. Allow comma for decimal point, e.g. €3.50 may be written as €3,50.

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Section C Functions and Calculus (old syllabus) 100 marks

Answer both Question 7 and Question 8.


(a) Let f (x) x3 6x k, where k and x . Taking x1 1 as the first approximation of a root of the function f (x) 0 and x2 2 as the

second approximation of this root, use the Newton-Raphson method to find the value of k. (10, Att. 3)

)(xf = kxx 63

)(' xf = 63 2 x

)1(f = k 61 = 5k

)1('f = 63 = 3

1nx = )('

)(

n

nn xf

xfx

2 = 3

51

k

2 = 3

51

k

6 = 3 + k – 5 6 = k – 2 –k = –2 – 6 –k = –8 k = 8

Blunders (–3) B1: Differentiation. B2: f(1). B3: f '(1). B4: Algebraic.

Slips (–1) S1: Numerical.

Attempts (3) A1: )(' xf or f(1) or )1('f evaluated.

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7(b) (i) Differentiate sin x with respect to x from first principles. (10, Att. 3)

)(xf = sin x

)( hxf = sin (x + h)

)()( xfhxf = sin(x+ h) – sin x

sin (A) – sin (B) = 2 cos 2

BAsin

2

BA

= 2cos 2

xhx sin

2

xhx

h

xfhxf )()( =

2

1

2

12

sin2

2cos2

h

hhx

h

xfhxf )()( = cos

2

2 hx

2

2sin

h

h

0limith

h

xfhxf )()( = 0

limith cos

2

2 hx × 0

limith

2

2sin

h

h

= cos x + 1 = cos x

Blunders (–3) B1: Trig formula.

B2: × 2

1.

B3: Limits.

Slips (–1) S1: Numerical. S2: Trig values

Attempts (3) A1: f(x + h).

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7(b)(ii) Given that y )sin(cos2

2sin

xx

x

, find the value of

dx

dy when x 0. (10, Att. 3)

Let u = sin 2x

dx

dy = 2cos 2x

v = 2cos x + 2sin x

dx

dv = –2sin x + 2cos x

dx

dy =

2)sin2cos2(

)cos22sin2)(2(sin)2cos2)(sin2cos2(

xx

xxxxxx

x = 0

2)0sin20cos2(

)0cos20sin2)(0(sin)0cos2)(0sin20cos2(

= 2)2(

)20)(0()2)(02(

= 4

4

= 1

Blunders (–3) B1: Differentiation. B2: Trig values. B3: Substitution.

Slips (–1) S1: Numerical. S2: Trig values.

Attempts (3) A1: Error in differentiation formula. A2: Some correct differentiation.

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7(c) A curve is defined by the equation x2y xy2 6.

(i) Find dx

dy in terms of x and y. (10, Att. 3)

22 xyyx = 6

2xu yv u = x v = y2

xdx

dy2

dx

dy

dx

dv

dx

dy= 1

dx

dyy

dx

dv2

xydx

dyx 22 + 22 y

dx

dyxy = 0

)2( 2 xyxdx

dy = –y2 – 2xy

dx

dy =

xyx

xyy

2

22

2

Blunders (–3) B1: Differentiation. B2: uv formula. B3: Indices. B4: Transposition.

Attempts (3) A1: Correct differentiation. A2: Error in differentiation formula.

(ii) Determine whether the tangents are parallel at the point x = 2. (10, Att. 3)

x = 2 4y + 2y2 = 6 2y2 + 4y – 6 = 0 y2 + 2y – 3 = 0 (y – 1)(y +3) y = +1 y = –3

x = 2 y = 1

dx

dy =

44

41

= 8

5

x = 2 y = –3

dx

dy =

124

129

= 8

3

tangents are not parallel

Blunders (–3) B1: Factors.

B2: dx

dy.


Attempts (3) A1: Quadratic equation.

A2: dx

dy evaluated.

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(a) Find 2

324

x

xx dx. (10, Att. 3)

dxxx 22 31

3

33

13

3

xxx + C

Cx

xx

3

13

Blunders (–3) B1: Integration. B2: Indices. B3: No ‘+ C’.

Attempts (3) A1: Only ‘C’ correct.

Worthless (0) W1: Differentiation instead of integration.

(b) (i) Evaluate

1

2

245 xx

dx

. (10, Att. 3)

5 – 4x – x2 1 + 4 – 4x – x2 (1)2 + (2 – x)2

22 )2()1( x

dx =

1

2

1

1

2sin

x

sin–1 1 – sin–1 0

2

– 0 =

2

Blunders (–3) B1: Integration. B2: Completing the square. B3: In correct order of applying limits.

Slips (–1) S1: Numerical. S2: Trig value.

Attempts (3) A1: Square completed.

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8(b)(ii) Evaluate 3

0

1

12

x

x dx. (10, Att. 3)

Let u = x + 1

dx

du = 1

du = dx 2u = 2x + 2 2u – 1 = 2x + 1

duu

u

12

duu

u2

12

duuu 2/12/12

2

1

2

32 2/12/3 uu

2/12/3

23

4u

u

123

143

xx

123

1442

3

4433

23

44

3

)8(4

3

20 –

3

10 =

3

10

Blunders (–3) B1: Integration. B2: Differentiation. B3: Indices. B4: Limits. B5: Incorrect order of applying limits. B6: Not changing limits


Attempts (3) A1: Differentiation unless required.

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8(c) The diagram shows the curve y x

1,

and the line 4x 8y 33 0.

Calculate the area of the shaded region enclosed by the curve and the line. (5, 5, 5, 5, Att. 2, 2, 2, 2)

4x + 8y – 33 = 0 8y = 33 – 4x

y = 8

433 x

= x

1

8 = 33x – 4x2 4x2 – 33x +8 = 0 = 4x2 – 33x – x +8 = 4x(x – 8) – 1(x – 8) = (4x – 1) (x – 8)

x = 4

1 x = 8 (5, Att. 2)

8

4

1 8

433 x = x

2

1

8

33 dx

= 8

4

1

2

4

1

8

33

xx

= [33 – 16] –

64

1

32

33

= 17 –

64

65

= 64

1023

(5, Att. 2)

dxx

8

4

11

= 84

1ln x

= ln 8 – ln 4

1

= ln 4

8

= ln 32 (5, Att. 2)

Area = 32ln64

1023

(5, Att. 2)

Blunders (–3) B1: Error in area formula. B2: Error in integration. B3: Error or incorrect integration limits. B4: | value |. B5: Fails to substitute limits.

Slips (–1) S1: Numerical slip in calculations (max. of –3).

Attempts (2, 2, 2, 2) A1: Some attempt at integration.

Worthless (0) W1: Differentiation instead of integration except

where other work merits attempt mark.

x

y

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.


Project Maths (Phase 2)


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 2 questions

Students must answer all eight questions, as follows:

In Section A, answer:

Questions 1 to 5 and either Question 6A or Question 6B.

In Section B, answer Questions 7 and 8.

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.

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Summary of Marks

Q.1 a i 5B (0, 3, 5) Q.6A a 5B (0, 3, 5) i 5B (0, 3, 5) b 5B (0, 3, 5) + b i 5B (0, 3, 5) 5B (0, 3, 5) ii 5B (0, 3, 5) 10C (0, 4, 8, 10) iii 5B (0, 3, 5) 25 25 OR Q.2 a 5B (0, 3, 5) Q.6B a 15C (0, 8, 12, 15) b 10C (0, 4, 8, 10) b 10B (0, 5, 10) c 5A (0, 5) 25 d 5B (0, 3, 5) 25 Q.7 a 15C (0, 8, 12, 15) b 15C (0, 8, 12, 15) Q.3 a 5B (0, 3, 5) c 15C (0, 8, 12, 15) b 10C (0, 4, 8, 10) d i 10C (0, 4, 8, 10) c 10C (0, 4, 8, 10) ii 5B (0, 3, 5) 25 iii 5B (0, 3, 5) e 10C (0, 4, 8, 10) Q.4 a 10C (0, 4, 8, 10) 75 b 5C (0, 3, 4, 5) c 10C (0, 4, 8, 10) Q.8 a 15C (0, 8, 12, 15) 25 b 10C (0, 4, 8, 10) + 10C (0, 4, 8, 10) + Q.5 a 5B (0, 3, 5) 10B* (0, 5, 10) b 10C (0, 4, 8, 10) c 10C (0, 4, 8, 10) c 10C (0, 4, 8, 10) d i 10C (0, 4, 8, 10) 25 ii 10C (0, 4, 8, 10) 75 300

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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 E

No of categories 2 3 4 5 6

5 mark scale 0, 3, 5 0, 3, 4, 5

10 mark scale 0, 5, 10 0, 4, 8, 10

15 mark scale 0, 8, 12, 15

20 mark scale 0, 7, 18, 20 0, 7, 10, 18, 20

25 mark scale 0, 15, 20, 22, 25 0, 5, 10, 15, 20, 25

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)

E-scales (six categories) response of no substantial merit (no credit) response with some merit (low partial credit) response almost half-right (lower middle partial credit) response more than half-right (upper middle partial credit) almost correct response (high partial credit) correct response (full credit)

In certain cases, typically involving incorrect rounding or omission of units, 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.

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Project Maths (Phase 2)


Section A Concepts and Skills 150 marks

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

1. The Venn diagram shows the probability of two events A and B occurring. (a) (i) Find the value of P(A B). (5B)

= 1 – (0.4 + 0.1 + 0.3) = 1 – (0.8) = 0.2

Scale 5B (0, 3, 5) Partial credit: (3 marks) – 0.4 + 0.1 + 0.3 = 0.8 or states total = 1.

(ii) Verify that P(A B) P(A) P(B) P(A B) and hence, state what this tells us about A and B. (5B)

= 0.6 = 0.3 + 0.5 – 0.2 = 0.6 = 0.6 = events mutually exclusive

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Shows equal, but no conclusion.

examsDEB

A

S

B

0�4

0�1 0�3

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1(b) A fair coin is tossed four times giving either ‘heads’ or ‘tails’ each time. (i) Complete the sample space below to show all the possible outcomes. (5B)

H H H H

H H H T

H H H H H T H H T H H H T T H H

H H H T H T H T T H H T T T H T

H H T T H T T T T H T T T T T T

H H T H H T T H T H T H T T T H

Scale 5B (0, 3, 5) Partial credit: (3 marks) – At least 3 more combinations. (ii) E is the probability that the first three results are ‘heads’ and F is the probability that the

fourth result is ‘heads’. Find P(E) and P(F).. (5B)

P(E) = 16

2 =

8

1

P(F) = 16

8 =

2

1

Scale 5B (0, 3, 5) Partial credit: (3 marks) – P(E) correct only.

– P(F) correct only.

(iii) Show that P(E and F) P(E) P(F) and hence, state what this tells us about E and F. (5B)

16

1 =

8

1 ×

2

1 =

16

1

= events independent

Scale 5B (0, 3, 5) Partial credit: (3 marks) – 16

1 or

8

1 ×

2

1, but no conclusion.

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2. One of the highlights of the Olympic Games is the marathon. It is a long-distance running event with an official distance of 42·195 kilometres that is usually run as a road race. It is one of the original events of the modern Olympic Games revived in 1896 although it did not feature in the original Games in ancient Greece.

A sports researcher, wants to investigate the connection, if any, between athletes’ performances in the marathon and the athletes’ ages. He takes a random sample of 15 athletes who completed the race in the 2012 London Olympic Games and compares the race times of these athletes to each of their ages. The results are shown in the table below.

Race Time (minutes)

Age of Athlete (years)

129 37

131 35

132 33

133 34

134 32

135 32

135 33

136 32

137 30

138 28

139 27

140 27

141 25

143 26

145 24

(a) Explain the following terms in relation to the above information: (5B)

(i) Sample;

– a subset of the population

(ii) Population.

– all the possible subjects or cases of interest

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

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2(b) Create a suitable graphical representation to illustrate the data. (10C)

Scale 10C (0, 4, 8, 10) Low partial credit: (4 marks) – Scatter plot.

High partial credit: (8 marks) – Axes not labelled properly.

(c) What kind of relationship, if any, does the observed data suggest exists between athletes’ performances and the athletes’ ages in the marathon? (5A)

– older runners are faster

Scale 5A (0, 5) – Hit / Miss.

(d) Can you make the same hypothesis for all athletes that compete in the marathon? (5B)

– no

Explain your answer.

– as sample may be too small for the population

Scale 5B (0, 3, 5) Partial credit (3 marks) – Correct answer, but reason not given.

Athlete’s Age (years)

RaceTime(minutes)

20 25 30 35 40

120

130

140

150

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3. l is the line x 3y 11 0 and m is the line x 2y 1 0.

(a) Write down the slope of l and the slope of m. (5B)

Slope of l = –3

1

Slope of m = 2

1

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Only one correct slope

– 3

1 and

2

1 given.

(b) Find the measure of the acute angle, , between l and m. (10C)

Tan = ± 21

211

1 mm

mm

Tan = ± )

2

1()

3

1(1

2

1

3

1

Tan = ±

6

56

5

= ± 1

= 45º = 135º but 45º as is acute

Scale 10C (0, 4, 8, 10) Low partial credit: (4 marks) – Correct formula with some substitution.

High partial credit: (8 marks) – Two solutions given.

(c) The line n cuts the x-axis at the same point as l. The line n also makes the same acute angle, , with l.

Find the equation of n. (10C)

– the line n cuts the x-axis at the same point as l

= x + 3y –11 = 0 = x –11 = 0 = (11, 0) pt = to m slope –2 = y – 0 = 2(x – 11) = y – 0 = –2x – 22 = 2x + y –22 = 0

Scale 10C (0, 4, 8, 10) Low partial credit: (4 marks) – Correct point or slope.

High partial credit: (8 marks) – Correct point and slope.

x

l m

x

y

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4. c1 is a circle with centre (3, 2). The lines l1 : 2x y 3 0 and l2: x 2y 6 0

are tangents to c1, as shown in the diagram.

(a) Find the radius of c1 and hence, write down the equation of c1. (10C)

= (3, 2), 2x + y – 3 = 0

= r = 22 )1()2(

3)2(1)3(2

= 5

326

= 5

5 = 5

= (x – 2)2 + (y – 2)2 = ( 5 ) = (x – 3)2 + (y – 2)2 = 5

Scale 10C (0, 4, 8, 10) Low partial credit: (4 marks) – 5

5 or distance with substitution.

High partial credit: (8 marks) – Equation of circle formula with substitution.

(b) c2, c3 and c4 are images of c1 under different transformations. Describe fully the transformation in each case. (5C)

c2 axial symmetry in the line l2 / rotation of 270º c3 central symmetry in the pt (0, 3) / rotation at 180º c4 axial symmetry in the line l1 / rotation of 90º

Scale 5C (0, 3, 4 ,5) Low partial credit: (3 marks) – One correct substitution.

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

(c) Find the equations of the two circles that touch c1, c2, c3 and c4 . (10C)

= centre (0,3) = (0,3) to (3,2)

= 22 )1()3(

= 10

= 10 – 5

= 10 + 5

inner circle (x – 0)2 + (y – 3)2 = ( 10 – 5 )2

outer circle (x – 0)2 + (y – 3)2 = ( 10 + 5 )2

Scale 10C (0, 4, 8, 10) Low partial credit: (4 marks) – 10 found.

High partial credit: (8 marks) – 10 + 5 and 10 – 5 found.

x

yl1

c1

c2

c3

c4

l2

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5. (a) Show that sin 2x 2sin x cos x. (5B)

sin (A + B) = sin A cos B + cos A sin B sin (x + x) = sin x cos x + cos x sin x = 2 sin x cos x

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Sin (2x) = sin (x + x).

– Sin(A + B) = sin A cos B + cos A sin B.

(b) Solve the equation sin 2x sin x 0 in the domain 0 ≤ x ≤ 2, x . (10C)

= 2 sin x cos x – sin x = 0 = sin x (2cos x – 1) = 0 = sin x = 0 2 cos x = 1

cos x = 2

1

= x = 0º, 180º, 360º 60º, 300º

= x = 0º, 60º, 180º, 300º, 360º

Scale 10C (0, 4, 8, 10) Low partial credit: (4 marks) – Equation in terms of x only.

High partial credit: (8 marks) – Sin x = 0 and cos x = 2

1.

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5(c) Verify your solution from part (b) by sketching the graphs of the functions f : x sin 2x and g : x sin x in the domain 0 ≤ x ≤ 2, x .

Indicate clearly which is f and which is g. (10C)

Scale 10C (0, 4, 8, 10) Low partial credit: (4 marks) – One correct function graphed.

High partial credit: (8 marks) – Two correct graphs, but answers not shown.

0� ��

x

y

2

�

2

3�

x

y

�1

1

g x( ) f x( )

3

�

2

� �

2

3� ��

3

5�

0

0

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6. Answer either 6A or 6B.

6A. (a) Explain the difference between an axiom and a theorem. (5B)

– an axiom is a statement assumed to be true // – a theorem is a statement which has been

proved, deducted from axioms and logical argument

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

(b) ABC is a triangle. Prove that, if a line l is parallel to BC and cuts [ AB ] in the ratio s : t, then it also cuts [ AC ]

in the same ratio. (5B, 5B, 10C)

Given Triangle ABC l \\ BC

To prove

||

||

DB

AD =

||

||

EC

AE =

t

s

Diagram

Proof We prove only the commensurable case. Let l cut [AB] in D in the ratio m : n with natural numbers m, n. Thus there are points. D0 = A, D1, D2,…,Dm –1, Dm = D, Dm +1,..., Dm + n–1, Dm +n = B

Equally spaced along [AB], i.e. the segments: [D0D1], [D1D2],….[DiDi + 1],…[Dm +n – 1Dm +n]

have equal length.

Draw lines D1E1, D2E2, …parallel to BC with E1, E2, … on [AC].

Then all the segments: [AE1], [E1E2], [E2,E3], …, [Em + n – 1C] have the same length [Theorem 11]

and Em = E is the point where l cuts [AC]. [Axiom of Parallels]

Hence E divides [AC] in the ratio m : n

Diagram: (5B)

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Triangle with l drawn.

Given/To prove: (5B)

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Given correct or to prove correct.

Proof: (10C)

Scale 10C (0, 4, 8, 10) Low partial credit: (4 marks) – One correct statement.

High partial credit: (8 marks) – One missing step or steps in incorrect order.

A

Dm+n =B

Dm =D

D1

C = Em+n

E = Em

E1

l

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OR 6B ACDE is a parallelogram.

The points A, D and E lie on the circle which cuts [ AC ] at B.

(a) Prove that BCD is an isosceles triangle. (15C)

| 1| = | 2 | = parallelogram ACDE | 1| + | 4 | = 180 opposite angles of cyclic quadrilateral | 3| + | 4 | = 180 straight line | 1| + | 4 | = | 3| + | 4 | | 1| = | 3| | 2| = | 3| BDC is isosceles

Scale 15C (0, 8, 12, 15) Low partial credit: (8 marks) – | 1| = | 2 |.

– | 1| + | 4 | = 180.

High partial credit: (12 marks) – | 1| = | 3 |.

(b) Prove that | BDE | | AED |. (10C)

| 3| = | 5 | = alternate angles | 3| = | 1 | = proved above | 1| = | 5| | AED| = | BDE |

Scale 10C (0, 4, 8, 10) Low partial credit: (4 marks) – | 3| = | 5 | or | 3| = | 1 |.

High partial credit: (8 marks) – | 3| = | 5 | and | 3| = | 1 |.

A

D

E

CB

1

4

5

23

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Section B Contexts and Applications 150 marks

Answer Question 7 and Question 8 from this section. (75 marks each)

7 A confectionary manufacturer is launching a new range of chocolates. The disc-shaped chocolates, each of radius 1·5 cm and height 0·75 cm, are packed in two layers in a rectangular box. The configuration of chocolates in each layer is shown below.

(a) Calculate the internal volume of the box. (15C)

8 1.5 = 12 cm 12 1.5 18 cm 2 0.75 1.5 cm 12 18 1.5 324 cm3

Scale 15C (0, 8, 12, 15) Low partial credit: (8 marks) – At least two dimensions calculated.

– Volume = l × b × h given.

High partial credit: (12 marks) – Volume of one layer only.

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A production researcher suggests that if the configuration of the chocolates in each layer of the box was altered, the packaging costs of the chocolates could be reduced. The alternative design of each layer is shown below.

(b) Using the right-angled triangle on the alternative design, calculate p and hence, find the internal volume of this box. (15C)

p2 + 32 = 62 p2 = 36 – 9 p2 = 27 p = 27

= 5.1561 = 10.3923 13 1.5 = 19.5 cm 2 0.75 = 1.5 cm 2 5.1561 = 10.3923 10.3973 19.5 1.5 = 303.9749 = 304 cm3

Scale 15C (0, 8, 12, 15) Low partial credit: (8 marks) – At least two dimensions calculated.

– p calculated.

High partial credit: (12 marks) – Volume of one layer only.

(c) The original box costs €0·80 to produce. Calculate the ratio of the volumes of the two boxes and hence, find the potential savings of using the alternative design if the cost of producing each box is directly proportional to the volume of the box and the company projects sales of 150 000 boxes annually. (15C)

= 324

20 0.80 150,000

= 7,407.41

or

1st box - 324 = 0.80 150,000 120,000.00 2nd box - 304 = 0.7506 150,000 112,592.59

= 7,407.41

Scale 15C (0, 8, 12, 15) Low partial credit: (8 marks) – 120,000 written. – 112,592.59 written. – 0.7506 cent per new box written.

High partial credit: (12 marks) – 120,000 and 112,592.59 written.

p

p6

3

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7(d) There are three different types of chocolates in the box. Half the chocolates are milk chocolates while the ratio of dark to white chocolates is 2 : 1.

(i) What is the probability that a chocolate chosen at random from the box will be a white chocolate? (10C)

24 milk = 48

8 =

6

1

16 dark 8 white

Scale 10C (0, 4, 8, 10) Low partial credit: (4 marks) – 24 / 16 / 8 calculated.

High partial credit: (8 marks) – Correct favourable outcomes.

– Correct total outcomes (probability 0 ≤ p ≤ 1).

(ii) If three chocolates are chosen at random from the box, find the probability that each of the chocolates chosen is made from the same type of chocolate. (5B)

= 48

24

47

23

46

22 +

48

16

47

15

46

14 +

48

8

47

7

46

6

= 776103

14412

,

, +

776103

3603

,

, +

776103

336

,

= 776103

84015

,

,

= 0811

165

,

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Correct favourable outcomes.

– Correct total outcomes (probability 0 ≤ p ≤ 1).

(iii) If three chocolates are chosen at random from the box, find the probability that each of the chocolates chosen is made from a different type of chocolate. (5B)

= 48

24

47

16

46

8 6

= 776103

0723

,

, 6

= 0811

192

,

Scale 5B (0, 3, 5) Partial credit: (3 marks) – Correct favourable outcomes.

– Correct total outcomes × 6 omitted (probability 0 ≤ p ≤ 1).

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7(e) The confectionary company claims that the minimum weight of chocolates in each box is 350 g. The company’s quality control department has determined that the weights of chocolates in each box are normally distributed with mean equal to the specified weight and standard deviation of 8 g. 50 boxes are selected at random and it is discovered that the mean weight of chocolates in these boxes is 348 g.

Use a hypothesis test at the 5% level of significance to decide whether there is sufficient evidence to validate the confectionary company’s claim.

Be sure to state the null hypothesis clearly, and to state the conclusion clearly. (10C)

Ho – the null hypothesis, that the chocolates are within the standard range

Z = errorStandard

350348

Standard error = 80

8 1.13137

13137.1

350348 = 1.7677 1.96

The null hypotheses are correct and they are within the standard range

Scale 10C (0, 4, 8, 10) Low partial credit: (4 marks) – Hypothesis stated correctly.

High partial credit: (8 marks) – Standard error.

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8 Mary has a rectangular hallway measuring 6 m by 1·25 m. She plans on tiling the floor with regular pentagon-shaped

tiles of side 6 cm using a mosaic pattern. She has decided to use a tiling contractor and he has given her a scaled drawing of the proposed work.

(a) A tile on the scaled drawing is reproduced on the square grid as shown. Given the centre of enlargement, O, construct, using a compass and straight edge only,

a full-size tile. (15C)

Scale 15C (0, 8, 12, 15) Low partial credit: (8 marks) – At least one point extended.

High partial credit: (12 marks) – At least three points extended.

O

2 cm

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8(b) The tiling contractor has asked Mary to order the tiles she wants to use. She needs to find out the area that each tile covers in order to calculate the number of tiles needed to complete the work. However, the area of each tile is not shown on her drawing.

She knows that the internal angles of a regular pentagon are 108 and it can be divided up into three isosceles triangles as shown.

Find the area of each tile. Give your answer correct to one decimal place. (10C, 10C, 10B*)

= 2

1ab sin C

= 2

1(6)(6) sin 108

= 2

1(6)(6) sin 0.951056

= 17.1190

Scale 10C (0, 4, 8, 10) Low partial credit: (4 marks) – Formula with some substitution.

High partial credit: (8 marks) – Formula fully substituted.

= x2 = 62 + 62 – 2(6) (6) cos 108 = x2 = 36 + 36 – 72 (–0.3090) = x2 = 94.24922 = x = 9.7082

= 2

1 ab sin C

= 2

1(9.7082)(6) sin 72

= 2

1(9.7082)(6)(0.951056)

= 27.6991

Scale 10C (0, 4, 8, 10) Low partial credit: (4 marks) – Cosine rule with substitution.

High partial credit: (8 marks) – Area formula with substitution.

= 17.1190 = 27.6991 = + 17.1191

= 61.9371 = 61.9 cm2

Scale 10B* (0, 5, 9, 10) Low partial credit: (5 marks) – Three triangles added together.

High partial credit: (9 marks) – Not rounded off to one decimal place.

108�

6 cm

108°6 6

72°

9.7

9.7

108°6 6

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8(c) Mary asked her friend, Sean, to check her calculations. He uses an alternative method to find the area of each tile.

On the scaled drawing, Sean sketches a triangle ABC, where C is the centre of the pentagon. He calculates the area of the tile to be:

)54)(sin2)(71(

2

159

Explain what each of the following numbers represent: (10C)

1.7:

Length of | AC | or | BC |

5:

5 triangles

9:

32 scale factor squared

54:

Angle | CAB | or | CBA |

Scale 10C (0, 4, 8, 10) Low partial credit: (4 marks) – At least one correct answer.

High partial credit: (8 marks) – At least three correct answers.

(d) Mary’s hallway measures 6 m by 1·25 m. The tiling contractor has asked her to purchase 10% more tiles than are required to allow for cutting and wastage.

(i) Assuming the area of the triangular cut-offs required are negligible, how many boxes of tiles does Mary need to order if they are supplied in boxes of 60? (10C)

600 125 = 7500 61.9 = 1211.6 tiles = 60 = 20.19 1.1 = 22.2 = 23 boxes

or

61.9 1.1 = 68.09 75000 68.09 1101 1.1 1332.79 60 22.2 = 23 boxes

Scale 10C (0, 4, 8, 10) Low partial credit: (4 marks) – Area of hall evaluated as 7500 cm2.

High partial credit: (8 marks) – 1211.6 tiles needed.

– 1332.79 tiles needed.

B

C

A

2 cm

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8(d)(ii) When Mary goes to order the tiles, she is informed that this particular size of tile has been discontinued. However a similar regular pentagon-shaped tile of side 5 cm supplied in boxes of 72 is available. Calculate the number of boxes of tiles she now needs to order. (10C)

1332.76 tiles 25

36

72

171919.

26.65 = 27 boxes

Scale 10C (0, 4, 8, 10) Low partial credit: (4 marks) –

6

5or

5

6.

High partial credit: (8 marks) – ×

25

36 or ÷

36

25 given.

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Notes:

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