Mathematics for the IB Diploma Applications and Interpretation SL Revision Guide [1 ed.] 9781925489897

Citation preview

m

Mathematics @

Applicationstand Interpretation¥SIE

Alanna Hill Andrew!Mollitt Michael Haese

MarkHumphries Ngoc¢iVo

Michael Mampusti

for use with

[BIDiplomaiProgrammee

MATHEMATICS:

APPLICATIONS AND INTERPRETATION

Alanna Hill

B.Sc.

Michael Haese

B.Sc.(Hons.), Ph.D.

Andrew Mollitt

SL REVISION GUIDE

B.Sc., MMT

Mark Humphries Ngoc Vo Michael Mampusti

B.Sc.(Hons.) B.Ma.Sc. B.Ma.Adv.(Hons.), Ph.D.

Haese Mathematics 152 Richmond Road, Marleston,

SA 5033, AUSTRALIA

Telephone: +61 8 8210 4666 Email: [email protected] Web:

www.haesemathematics.com

National Library of Australia Card Number & ISBN

978-1-925489-89-7

© Haese & Harris Publications 2020 Published by Haese Mathematics.

152 Richmond Road, Marleston, First Edition

SA 5033, AUSTRALIA

2020

Trial examinations written by:

Alanna Hill from Sevenoaks School, UK

Andrew Mollitt from United World College of South East Asia, Singapore Michael Haese from Haese Mathematics. Artwork by Brian Houston. Typeset in Australia by Deanne Gallasch and Charlotte Frost. Typeset in Times Roman 10. This book 1s available on Snowflake only. This book has been developed independently connected with, or endorsed by, the IBO.

of the International Baccalaureate

Organization

(IBO).

This book

is in no way

This book is copyright. Except as permitted by the Copyright Act (any fair dealing for the purposes of private study, research, criticism or review), no part of this publication may be reproduced, stored 1n a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the publisher. Enquiries to be made to Haese Mathematics. Copying for educational purposes: Where copies of part or the whole of the book are made under Part VB of the Copyright Act, the law requires that the educational institution or the body that administers 1t has given a remuneration notice to Copyright Agency

Limited (CAL). For information, contact the Copyright Agency Limited.

Acknowledgements: While every attempt has been made to trace and acknowledge copyright, the authors and publishers apologise for any accidental infringement where copyright has proved untraceable. They would be pleased to come to a suitable agreement with the

rightful owner.

Disclaimer: All the internet addresses (URLs) given in this book were valid at the time of publication. While the authors and publisher regret any inconvenience that changes of address may cause readers, no responsibility for any such changes can be accepted by either the authors or the publisher.

FOREWORD The aim of this Guide is to help you prepare for tests and the final examination for the Mathematics: Applications and Interpretation SL course.

* Remember to leave answers correct to three significant figures unless an exact answer 1s more appropriate or a different level of accuracy 1s requested in the question.

This Guide covers all five Topics 1n the Mathematics: Applications and Interpretation SL syllabus. All of the relevant material from the Mathematics: Core Topics SL and the Mathematics: Applications and Interpretation SL textbooks 1s covered.

* Check for key words. If the word “hence” appears, then you must use the result you have just obtained. “Hence, or otherwise” means that you can use any method you like, although 1t 1s likely that the best method uses the previous result.

For each Topic, there 1s a theory summary and a set of skill builder questions.

* [tisimportanttoread the question caretully. Rushing into a question may mean that you miss subtle points. Underlining key words may help.

* The theory summaries highlight the important facts and concepts. They are intended to complement your textbook and International Baccalaureate booklet. When a formula can be found in the formula booklet, it may not berepeated 1n this Guide. * The set of skill builder questions are designed to help consolidate your understanding of each Topic. They should be used to reinforce key ideas, and to identify any arcas of weakness. Within each Topic, the questions are logically ordered according to the chapters of the textbook, so they can be used for test preparation.

Following the coverage of all five Topics, the Guide has ten mixed questions sets, each containing 12 questions. Each set contains questions from every Topic, as well as cross-topic questions. It 1s recommended that you attempt all of the questions 1n a mixed questions set 1in one sitting, as this will give you practice in answering questions from a range of topics in a short time frame. The Guide concludes with four trial examinations, written by IB teachers from around the world. Each trial examination contains two papers: Paper 1, which contains shorter questions, and Paper 2, which contains longer questions. This format 1s consistent with the Mathematics: Applications and Interpretation SL final examination. Full solutions are provided, but 1t 1s recommended that you work through a complete paper before checking the solutions.

We recommend completing each trial examination under exam conditions. You are encouraged to print the formulae summary (seepage S), and have it alongside you as you complete the trial examinations.

* Remember that questions in the examination are often set so that, even if you cannot complete one part, the question can still be picked up in a later part. After completing a trial examination, you should identify areas of weakness. * Return to your notes or textbook and review any material you found challenging. * Ask your teacher or a friend explanation is needed.

for help

* Summarise each Topic. Summaries yourself are the most valuable.

i1f further

that you

make

* If you have had difficulty with a question, try it again later. Do not just assume that you know how to do it once you have read the solution. In addition to the formula booklet, your graphics display calculator is an essential aid. * Make sure you are familiar with the model you will be using. * In trigonometry questions, remember to check that the graphics calculator1s in degrees. *

Important features zooming in or out.

of graphs

may

be

revealed

by

* When using your graphics calculator, it is always important to reflect on the reasonableness of the results. We hope this Guide will help you structure your revision program effectively. Remember that good examination techniques will come from good examination preparation.

* [fyouare having trouble with a question, it 1s often a good strategy to move on to other questions, and return to it later. Time management 1s very important during the examination, and too much time spent on a difficult question may mean that you do not leave yourself sufficient time to complete other questions. * Set out your work clearly with full explanations. A correct answer with no working will not necessarily recerve full marks. * [f you make a mistake, draw a single line through the work you want to replace. Do not cross out work until you have replaced it with something you consider better.

We welcome your feedback:

* Diagrams and graphs should be sufficiently large, well labelled, and clearly drawn.

email:

web:

http://haesemathematics.com [email protected]

4

TABLE OF CONTENTS

TABLE OF CONTENTS FORMULAE SUMMARY TOPIC 1: NUMBER AND ALGEBRA Skill builder questions TOPIC 2: FUNCTIONS Skill builder questions

14 17

TOPIC 3: GEOMETRY AND TRIGONOMETRY Skill builder questions

24 26

TOPIC 4: STATISTICS AND PROBABILITY Skill builder questions

32 38

TOPIC 5: CALCULUS Skill builder questions

52 55

MIXED QUESTIONS Mixed questions set Mixed questions set Mixed questions set Mixed questions set Mixed questions set Mixed questions set Mixed questions set Mixed questions set Mixed questions set Mixed questions set

59

1 2 3 4 3 6 7 8 9 10

59 60 62 64 635 67 69 71 73 75

TRIAL EXAMINATION Paper 1 Paper 2

1

77 77 80

TRIAL EXAMINATION Paper 1 Paper 2

2

83 83 86

TRIAL EXAMINATION Paper| Paper2

3

89 89 92

TRIAL EXAMINATION Paper 1 Paper 2

4

95 95 97

WORKED SOLUTIONS Topic 1 Skill builder questions Topic 2 Skill builder Topic 3 Skill builder Topic 4 Skill builder Topic 5 Skill builder Mixed questions Mixed questions Mixed questions Mixed questions Mixed questions Mixed questions Mixed questions Mixed questions Mixed questions

questions questions questions questions

set set set set set set set set

1 2 3 4 5 6 7 8

100

100 110 129 145 176 191 191 195 199 204 209 213 217 221

Mixed questions set 9 Mixed questions set 10 Trial examination 1 Paper 1 Paper 2 Trial examination 2 Paper 1 Paper 2 Trial examination 3 Paper 1 Paper 2 Trial examination 4 Paper 1 Paper 2

224 230 236 236 239 243 243 247 251 251 254 258 258 261

FORMULAE SUMMARY

FORMULAE

PRINTABLE FORMULAE

SUMMARY

X

SUMMARY

PRIOR LEARNING A = bh, where b is the base, h is the height

A = 2(bh), where b is the base, h is the height

A = 3 (a+ b)h, where a and b are the parallel sides, h is the height A = 7r?,

where r is the radius

(' = 27r,

where r is the radius

V' = lwh,

where [ 1s the length, w 1s the width, & 1s the height

V = mr2h, where r is the radius, h is the height V = Ah,

where A is the area of the cross-section, / is the height

A = 2mrh,

where r is the radius, h is the height

d=+/(x1 —22)?+ (11 — ¥2)? (2171 + 2 2

y1+y2) ;

2

TOPIC 1: NUMBER AND ALGEBRA

_

ARITHMETIC SEQUENCES Up =up + (n—1)d

Sp = ;(21;,1 +(n—1)d)

or

S, =

g(m + Up)

GEOMETRIC SEQUENCES Uy = ugr™ L Sfl, _

COMPOUND

Hlf:'?:; 1)

=

ul(ll_—:n),

” 7& 1

INTEREST

FV = PV x (1+

kn P ) , 100k

where

F'V

1sthe future value

PV

1s the present value

n

1s the number of years

k

1s the number of compounding periods per year

r%

1is the nominal annual rate of interest

EXPONENTS AND LOGARITHMS a* =b


0,

b >0,

a #1

PERCENTAGE ERROR —

|

Va — Vg VE

| x 100%,

where Vg is the exact value and V4 is the approximate value.

5

6

FORMULAE SUMMARY

TOPIC 2: FUNCTIONS STRAIGHT LINES y=ma+cY1

m

—

Yy2 —

or

ar+by+d=0

or

y—y

=m(z—)

QUADRATIC FUNCTIONS f(z) = ax® + bx + ¢ =

axis of symmetryis

z = _%

TOPIC 3: GEOMETRY AND TRIGONOMETRY MEASUREMENT

($1+$2

Y1 -I-yz 31+2:2) 2

v

5 Ah, where A is the area of the base, h is the height

V = 'iTTzh where 7 is the radius, h is the height A = mrl, where r is the radius, [ is the slant height = 11'1"3 where r is the radius A = 47r?, where 7 is the radius

[ = % A=

%

x 2mr, where 6 is the angle measured in degrees, 7 is the radius x mr?, where @ is the angle measured in degrees, r is the radius

TRIGONOMETRY a

b

c

sin A

sinB

sinC

2 =a?+b%—2abcosC

and

TOPIC 4: STATISTICS AND PROBABILITY Interquartile range = Q3 — Q4

Meanof asetofdata

T =

JiTi

n

,

Where

12

cosC = = +2E:1,b

= %absin@’

Z

2,

n = Z fi i=1

=

FORMULAE SUMMARY

PROBABILITY Probability of anevent A

P(A) = :Eé;

P(A) +P(A") =1 P(AuB)=P(A)+P(B) - P(AN B) P(AUB) =P(A) +P(B) for mutually exclusive events P(A|

B)

=

P(AN B)

P(AN B) =P(A)P(B)

forindependent events

Expected value of a discrete random variable X, E(X) =) xzP(X = z)

BINOMIAL

DISTRIBUTION

For X ~ B(n, p): e

Mean E(X)=mnp

e

Variance Var(X) = np(1 — p)

TOPIC 5: CALCULUS DIFFERENTIATION f(x) =2"

=

f'(x) =na™ !

INTEGRATION ol

ndy

o

—

=2 4 C n-+1 (R

T

7

—_

I

Area between a curve y = f(x), where f(x) > 0, and the z-axis = /

b

ydx

TRAPEZOIDAL RULE b /

ydr ~ %h-((yo +Yn) +2(y1

+y2

+ ... + yn_l)),

where

h =

b—a n

7

8

TOPIC 1: NUMBER AND ALGEBRA

TOPIC 1: NUMBER AND ALGEBRA APPROXIMATION AND ESTIMATION Rounding numbers: e

If the digit after the one being rounded is less than 5, round down.

e

Ifthe digit after the one being rounded 1s 5 or more, round up.

Significant figures are counted from the first non-zero digit from the left. For example:

3.413 ~ 3.41

(to 2 decimal places)

0.034 56 ~ 0.0346(to 3 signiticant figures) 236.5 ~ 237

(to the nearest whole number)

You are expected to give answers either exactly or rounded to 3 significant figures unless otherwise specified in the question. A measurement is accurate to :I:% of the smallest division on the scale.

An approximation is a value given to a number which is close to, but not equal to, its true value. An estimation 1s a value which 1s found by judgement or prediction instead of carrying out a more accurate measurement. If the exact value 1s Vg

e

and the approximate value 1s V4

then:

absolute error = | V4 — Vg |

e

percentage error = |Va — Vg | x 100% VE

SCIENTIFIC NOTATION (STANDARD FORM) A number is in scientific notation if it is written in the form

a x 10

where

1 < a < 10 and

k € Z.

SEQUENCES AND SERIES A number sequence 1s a set of numbers defined by a rule. Often, the rule 1s a formula for the general term or nnth term of the sequence.

A sequence which continues forever 1s called an infinite sequence. A sequence which terminates is called a finite sequence.

Arithmetic sequences In an arithmetic sequence, each term differs from the previous one by the same fixed number. Upy1 — Up = d forall

n € ZT,

where d is a constant called the common difference.

For an arithmetic sequence with first term ©; and common difference d, the nth term is w,, = u; + (n — 1)d.

Geometric sequences In a geometric sequence, each term 1s obtained from the previous one by multiplying by the same non-zero constant, called the common ratio 7. Un 41 Upt1 = Ty, SO we can find r = —— forall n € Z7. Un

For a geometric sequence with first term ©; and common

ratio r, the nth term is u,, = u;r"!.

Series A series 1s the sum of the terms of a sequence. For a finite sequence with n terms, the corresponding series 1s S,, = u1 + ug + .... + Uy

Using sigma notation or summation notation we write

wu; + us + ug + .... + u,, as

T

»_ ug. k=1

Fora finite arithmetic series, S,, = ;—l(vl + u,) For a finite geometric series with r # 1, S,, =

or S, = ;—1(21.51 + (n — 1)d). uyp(r™ —1) r—1

TOPIC

1: NUMBER

AND ALGEBRA

9

Compound interest The value of a compound interest investment after n time periods 1s

Un = uo(l +12)" where

wg

and

7

1s the initial value of the investment

1s the interest rate per compounding period.

To find the real value of the investment, we divide by the inflation multiplier each year. You should be able to use the TVM solver on your calculator to solve problems involving compound interest investments and loans.

Depreciation Depreciation is the loss in value of an item over time. The value of an item after n years is u,, = ug(1 — d)" where

and

wug 1S the initial value of the item

d

1s the rate of depreciation per year.

POLYNOMIAL EQUATIONS The highest power of x in a polynomial equation is called its degree. If a polynomial equation has degree n then it may have up to n real solutions. You should be able to use your graphics calculator to solve polynomial equations.

EXPONENTIALS AND LOGARITHMS Laws of exponents

aibdait—lag a" fl_n

. =

a

T

aMm—7n

a

n

—

1 fl_n

and

1 {l__fl’ =

a

n

(i) =i a\"

a”

The logarithm in base 10 of a positive number is the power that 10 must be raised to in order to obtain that number. If 10* = b for b > 0, we say that x is the logarithm of b in base 10, and write x = logb. log 10® =

and

10'°8% = 2 forany

z > 0.

The natural logarithm is the logarithm in base e. The natural logarithm of x is written as Inz or log, x. lne®

=z

and

e

= ¢

forall

= > 0.

10

TOPIC 1: NUMBER AND ALGEBRA

SKILL BUILDER QUESTIONS 1

A circle has diameter 6.24 cm. Find the area of the circle, rounding your answer to: a

2

3 significant figures

b

4 decimal places.

Estimate using a one figure approximation: a

382 x21

b

6.91 x 0.875

¢

383107 =213

3

The speed of a cricket player’s bowl is recorded as 141.6 kmh™", rounded to 1 decimal place. In what range of values does the actual speed s lie?

4

The dimensions of a block of land are measured to be values for the actual area A of the block of land?

5

Terry measures the dimensions of a box as 15 ¢cm by 12 ¢cm by 8 cm, rounded to the nearest centimetre.

6

7

Use Terry’s measurements to estimate the volume of the box.

b

The actual dimensions of the box are 15.3 cm by 11.8 ¢cm by 8.4 cm.

I

Find the actual volume of the box.

il

Find the absolute and percentage error in Terry’s estimate.

The radius of a circle 1s measured as 7 cm, rounded to the nearest centimetre.

a

Use this measurement to estimate the area of the circle.

b

Find the boundary values for the area of the circle.

¢

Hence find the maximum percentage error in the estimate.

Express in exponent form with a prime number base:

b 125 x 5

8

Write without brackets:

9

Simplify:

a (—3m?)* a 404471

b (23)"

(z°+ :1::_2)2

a’bs 42000

b

Qm,_z_f T Mt e

C

12{{3 b—9°

b

0.0000678

¢

526000 000

5 43'3521;’0:

¢ (0.00008)*

Use technology to solve:

b

3z°+ 7z — 3z =2

Solve using technology: a

{2:1: — 3y = 2

{3:1: — Ty = —8

dr + 3y = 5

16

(z*— 2?)(z° + 3)

b

a 8r—2=3x7 15

¢ 22421 427!

Use your calculator to evaluate the following, giving your answer in scientific notation:

a (3.57 x 10%) x (2.38 x 10) 14

¢ 752t x (45t%)°

Write in scientific notation:

a 13

.

Write without negative exponents:

a 12

b (g)

¢ —

Expand the brackets and write in simplest form:

a 11

rounded to the nearest metre. What are the boundary

a

a 64

10

17 m x 22 m,

c

6x + 11y = 12

A triangle is defined by the lines with equations a

Find the coordinates of the triangle’s vertices.

b

Find the area of the triangle.

y =2+ + 2,

:}31+y?1+2232_1_3 32—

4+ y =9,

and = = 4.

2y 1 52 — 4

TOPIC 1: NUMBER AND ALGEBRA

17

18

Consider the sequence

20

8, 13, 18, 23, 28, ....

a

Show that the sequence is arithmetic.

¢

Find the 42nd term.

d

Determine whether each number is a member of the sequence:

b

Find the formula for its general term.

i

153

i

4067

Find £ given the consecutive arithmetic terms:

a 19

11

3,k 11

b

2

k+4,

k2+11

¢

k-5,

2k, 2k?

An empty hamster cage has mass 800 g. When 5 hamsters are placed in the cage, the total mass 1s 1400 g. a

Find the average mass of the hamsters in the cage.

b

Hence write an arithmetic sequence for u,,, the approximate total mass when n hamsters are placed in the cage.

Find the general term wu,, of the geometric sequence which has: a

us =324

and

w9 = 78732

Consider the sequence

b

ugs=—-10

and

w19 = —160

2, 23, 6, 61/3, ...

a

Show that the sequence is geometric.

b

Find the formula for its general term.

¢

Find the 10th term.

d

Find the first term which exceeds 1000.

An endangered species of bird has population 217. However, with a successful breeding program it is expected to increase by 42% each year. a

Find the expected population size after:

I b

23

D years

i

10 years.

How long will it take for the population to reach 30 000?

Paige invests €500 in an account that pays 7.2% p.a. compounded monthly.

The amount of money in Paige’s account at the end of each month follows a geometric sequence with common ratio r. a

Find the value of r.

b

Find the value of the account after 3 years.

¢

Given that inflation averages 2% p.a. over the 3 years, find the real value of the investment after 3 years.

24

Stan invests £3500 for 33 months at 8% p.a. interest compounded quarterly. Find its maturing value.

25

How much should I invest now to produce $30 000 in 5 years’ time, if the money can be invested at a fixed rate of 4.8% p.a. interest compounded monthly?

26

A television was purchased for £2000, and depreciates at 30% p.a. for 3 years.

27

a

Find the value of the television at the end of this period.

b

By how much has the television depreciated?

Lauren deposits $10 000 in an account that compounds interest monthly. 4.5 years later, the account has balance $12 000. @

What annual rate of interest did the account pay?

b

How long will it take Lauren to double her deposit?

The first four terms of an arithmetic sequence are

29

30

a

Write down the common difference d.

¢

Find the sum of the first 20 terms.

51, 45, 39, 33.

b

Find the 20th term

wu9y.

b

7+ 125+ 18+235+

d

the integers from 1 to 200 not divisible by 3.

b

Find the 27th term

Find the sum of:

a

114+ 15419423+

.... to 20 terms

¢

1-2+3-445-6+4+7—....

An arithmetic sequence has terms %y

to100 terms

w7y = 1 and

w5

a

Find the first term

¢

Find the sum of the first 27 terms of the series.

= —23.

and common difference d.

wuo~.

...+ 106

12

TOPIC 1: NUMBER AND ALGEBRA

31

The first term of a finite arithmetic series 1s 18 and the sum of the series 1s —210. The common difference 1s —3. Suppose there are n terms in the series.

a 32

33

Consider the arithmetic sequence

b

Find n such that S,, = 140.

Find the sum of: 10+5+2%+1fi+....

to 8 terms

b

2+ 10+ 50+ 250 + .... to 10 terms

b

6—12+24

20

) 3x(=2)2

k=1

Find the sum of the series: a

10+14+184+22+ ...+ 138

—48+96 — .... + 1536

Emma takes out a home loan of $120 000 at 7.2% p.a. interest compounded monthly. The loan is to be repaid over 20 years.

a

Calculate the monthly repayment.

b

Calculate the amount of money still owing on the loan after one year.

¢

d

36

Hence find n.

7, 10, 13, 16, 19, ....

Write down an expression for the sum of the first n terms 5,,.

¢

35

b

a

a

34

Show that 2(39 — 3n) = —210.

i1

Calculate the amount paid in the first year of the loan.

il

By how much has the principal been reduced at the end of the first year?

ili

Explain why the loan does not reduce by the full amount of your first year’s repayment.

Suppose that after 1 year, the interest rate falls to 6.95% p.a. I

Calculate the new monthly repayment.

il

If Emma is able to keep paying the original repayments, how much earlier will the loan be paid off?

Oscar decides to start a new business venture which involves taking out a bank loan. The bank charges an interest rate of

6.55% p.a. compounded quarterly.

His quarterly repayments are $933.62, and must be repaid over 8 years. a

How much did Oscar borrow?

b

How much interest will he pay over the 8 year period?

¢

37

i

Calculate the outstanding balance at the end of the sixth year.

Il

At the end of the 6th year, Oscar pays a lump-sum of $3000 off the loan. Assuming his repayments remain the same, how much sooner will Oscar repay the loan?

Mary takes out a loan of $10 000 to purchase a car. The bank charges an interest rate of 8% p.a. compounded monthly. Mary will repay the loan with quarterly repayments over 5 years.

Calculate the quarterly repayment.

38

b

Find the balance of the loan after 3 years.

¢

After 3 years, Mary decided she would like to have the loan paid off in 1 year’s time. repayment increase to for this to occur?

What must her quarterly

Cassie made an 1nitial investment of €2000 into a savings account, and followed 1t with regular deposits of €500 per quarter. The account pays 1.2% interest per quarter, and inflation is 0.3% per quarter.

39

a

Explain why the real interest rate is approximately 0.9% per quarter.

b

Find the real value of Cassie’s investment after 5 years.

Bill collects $81 000 as his share of a lottery win. He decides to retire from work and buy an annuity to provide $2000 per month, until he gets a pension in four years’ time.

a

What annual interest rate, compounded monthly, is needed for Bill’s plan to work?

b

How much will Bill actually receive each month over the period of the annuity if he receives 77 p.a. interest compounded monthly?

TOPIC 1: NUMBER AND ALGEBRA

40

41

Celia and Mike want to provide each of their two children with $200 per month for the next ten years. Interest on the investment is 5.2% p.a. calculated monthly.

a

How much should be invested now to provide such an annuity?

b

Calculate the total interest earned over the term of the annuity.

Anne places her retirement savings of £400 000 into an annuity fund which returns 4.9% p.a. interest compounded quarterly. She plans to make monthly withdrawals of £3000 from the fund. a

How long will her money last?

b

How much should she withdraw each month if she wants the money to last for 20 years?

42

Simplify:

43

Use your calculator to write the following in the form

a log(10° x 1000°) a

L4

2

b

¢ log(2! x 5t) 10* where x is correct to 4 decimal places:

200

In(ef x e?)

b

¢

IH(E%)

¢

Use your calculator to write the following in the form a

46

b log( oo)

0.02

Simplify:

a 45

13

47

b

In acoustics,

the intensity of sound

L =10 lmg(fi) 0

dB,

¢?Iné

d

e* where k is correct to 4 decimal places:

500

i1s measured

¢ n3

¢ in decibels

where I is the sound intensity, and

(dB).

The

I = 10~1? w/m? 3 X 102

sound

0.023 intensity

level (SIL)

1s given by

is the reference sound intensity.

a

Find, in dB, the SIL of a snare drum with sound intensity

w/m”°.

b

A lawn mower has SIL 85 dB. Find its sound intensity in w/m?, giving your answer in the form k € Z.

a x 10%, 1 < a < 10,

14

TOPIC 2: FUNCTIONS

TOPIC 2: FUNCTIONS PROPERTIES OF LINES The gradient of the line passing through A(x1,y1)

and B(xz2, y2) is m = y-Step _ Y2 — U1 x-step

T2 — T1

The gradient of any horizontal line is zero. The gradient of any vertical line is undefined. The y-intercept of a line is the value of y where the line cuts the y-axis.

The x-intercept of a line is the value of x where the line cuts the z-axis.

PARALLEL AND PERPENDICULAR LINES The gradients of parallel lines are equal. The gradients of perpendicular lines are negative reciprocals.

| gradient m;

1

m; = ——

mo

gradient ms

EQUATION OF A LINE The equation of a line can be presented in: e

gradient-intercept form

e

general form

e

point-gradient form

y = max + ¢ where m 1s the gradient and c 1s the y-intercept.

ax + by = d

y —y; = m(x — xq)

You should be able to find the equation of a line given:

e

its gradient and the coordinates of any point on the line

e

the coordinates of two distinct points on the line.

FUNCTIONS

y = f(x)

A relation between variables x and y is any set of points in the (x, y) plane. A function is a relation in which no two different ordered pairs have the same z-coordinate or first component. For each value of= there is at most one value ofy or f(x). We test for functions using the vertical line test. A graph is a function if no vertical line intersects the graph more than once. For example, the graph of the circle 2 + y? = 1 shows that this relation is not a function.

'

]

AY

//

The domain of a relation 1s the set of values that = can take. To find the domain of a function, remember that we cannot:

e

divide by zero

e

take the square root of a negative number.

The range of a relation is the set of values that y or f(x) can take.

INVERSE FUNCTIONS A function is one-to-one if, for each value of y, there is only one value of . One-to-one functions satisfy the horizontal line test. If a function

f(z)

is one-to-one, it has an inverse function

y = f~1(x) isareflection of y = f(z) in the line y = . The domain of f~! The range of f—!

is equal to the range of f. is equal to the domain of f.

f~!(z).

If f maps x to y, then f~!

maps y back to .

TOPIC 2: FUNCTIONS

15

GRAPHS OF FUNCTIONS The x-intercepts of a function are the values of & for which The y-intercept of a function is the value of y when

y = 0. They are the zeros of the function.

x = 0.

An asymptote is a line that the graph approaches or begins to look like as it tends to infinity in a particular direction. : vertical

.

AY

horizontal e asymptote

Al

[

'

------------------- Pl

asymptote / =Y

L/

|1

To find vertical asymptotes, look for values of « for which the function is undefined. To find horizontal asymptotes, consider the behaviour as * — +oc.

You should be able to use technology to: e

graph a function

e

find the domain and range

e

find axes intercepts

e

find turning points

e

find asymptotes

e

find where tunctions meet.

MODELLING Mathematical models are developed using a modelling cycle: Step 1:

Pose a real-world problem. Make assumptions which simplify the problem without missing key features.

Step 2: Step 3:

Develop a model which represents the problem with mathematics. This may involve a formula or an equation. Test the model by comparing its predictions with known data. If the model is unsatisfactory, return to Step 2.

Step 4: Step 5.

Reflect on your model and apply it to your original problem, interpreting the solution in its real-world context. If appropriate, extend your model to make it more general or accurate as needed. reject

Y Posea

problem

aceept | Ezg?t ?fl

Testthe

Develop | |

|—|

real-world

the mfl?j;

model

a model

Extend

the model

A

You should be able to solve systems of equations using technology to find unknown parameters in models.

QUADRATIC FUNCTIONS A quadratic function has the form

y = ax?® + bz + ¢, a # 0.

The graph 1s a parabola with the following properties: o e

e

Itisconcave upit a > 0 .

)

\/

x = S :

Its vertex has xz-coordinate i

/\

—b

Its axis of symmetryi1s

—b

by substituting = = o

and concave down it a < 0.

—b

:

o

.

i

The y-coordinate of 1ts vertex 1s found

(L

.

o

)

symmet

Into the function.

1l

A

>

If a > 0 the vertex is a minimum turning point.

>

If a < 0 the vertex 1s a maximum turning point.

:

.

:

.

g

-

_ / x-intercept

5 7

vertex/minimum

You should be able to use technology to find points at which: e

a linear function meets a quadratic

4

e

two quadratic functions meet.

§ :

P

Y\ . ¢ R

y-intercept

Y

>

¢ v

16

TOPIC 2: FUNCTIONS

VARIATION MODELS Variation models have the form e

[If n >0

y = ax™,

n € Z, n # 0.

we have direct variation.

The graph passes through the origin

e

[f n —1—-12

b

f(z) =5z — 2°

¢

y=28z>—-2xr—3

b

f(z)=(x+1)°

¢

y=3x°+4x —4

Find the axes intercepts of:

a 15

Hence sketch the graph of y = f(x).

y= 2z —1)(x+3)

Sketch the graph of each function by considering the coefficient of 2 and the axes intercepts.

a

y=2x°—2r—8

b

16

A quadratic function has axis of symmetry

17

The quadratic function a

Find the value of b.

f(z)=—-2z+1)(z—3) x = —1,

¢

y=—=(z—4)°

and one of its x-intercepts 1s 2. Find the other z-intercept.

f(xz) = 22% + bax — 3 has axis of symmetry b

x = 6.

Find the coordinates of the vertex.

TOPIC 2: FUNCTIONS

18

For each of the following quadratics:

I

a

il

Find the axis of symmetry.

ili

Find the coordinates of the vertex, and state whether it is a maximum turning point or a minimum turning point.

iv

Sketch the quadratic.

v

y=—(r—1)(z+3)

b

State the domain and range.

y=2(x+7)(x—2)

Find the equation of the quadratic with graph: a

YA

o

£

19

Find the axes intercepts.

b

2\/5

AY

%

/

Y

20

24

y=2°—4r—-5

and

y =3z

—11

b

y=2°—-2

and

y=-2°4+x+6

y=-222+5r

b

The daily profit made by a local baker selling a

23

Y

and

y=5—2x

Find, to 3 significant figures, the coordinates of the points of intersection of: a

22

9\5?

Find the coordinates of the point(s) of intersection of: a

21

(3,6)

£ “L

18

Copy and complete this table.

y=52°—2

homemade pies is given by

20 | 40 | 60 | 80 | 100

P

100

b

Use the points in a to sketch the graph of P against x.

¢

Find: I

the number of pies that need to be sold to maximise the profit

il

the maximum possible daily profit

il

the number of pies that need to be sold to make a profit of $200

iv

the amount of money the baker loses if no pies are sold.

y=2%—4z+4

P = —0.052% + 92 — 60 dollars.

x | 0]

300

and

340

Jacob’s rainwater tank started leaking. The amount of water in the tank after £ hours is given by W (t) = 1000 — 0.5¢ litres.

a

Find W(0), and interpret your answer.

b

Find ¢ when W (¢) = 700, and explain what this represents.

¢

How long will it take for the tank to empty?

Consider the graph of y = f(x)

alongside.

AY

Decide whether each statement is true or false:

25

26

(4,5)

a

0 1sin the domain of f.

b

0 is in the range of f.

¢

6 is in the range of f.

d

3isin the domain of f.

e

21sin the range of f.

A function f isdefinedas

i

f(—4)

(6,5) (—3,2)

1 b

f(z) = +x +4

a

Find:

i

f(0)

b

Sketch y = f(x).

¢

Hence write down the range of f(x).

for -4 km

a

I

>

Yy

y

(6,2)

_ z) y=f(

e P

-

AY

c

AY

b

LY

a

on the same set of axes. In

Isabelita works by herself

il

Arturo works by himself

lii

Isabelita and Arturo work together?

Discuss any assumptions you have made in @, and whether or not they are reasonable.

An oven technician charges a call-out fee of $50 and an additional $20 per hour for the duration of the appointment.

Sketch the graph of ' against£ for 0

1=1,..,n

T

.

{zi, z2, ...., z,,}

.. +ppn

=1

describes the probability distribution of X

We can also describe the probability distribution of X using a probability mass function P(x) = P(X = x). The expectation of a discrete random variable X is E(X) =

= > x;p;. 1=1

A game where X is the gain to the player is said to be fair if E(X) = 0. The mode is the data value x; whose probability p; is the highest.

THE BINOMIAL DISTRIBUTION In a binomial experiment there are two possible results: success and failure. Suppose there are n independent trials of the same experiment with the probability of success being a constant p for each trial. If X represents the number of successes in the n trials, then X has a binomial distribution, and we write

X ~ B(n, p).

The binomial probability mass function is P(X = z) = (7)p“(1 — p)"~* where © =0, 1, 2, ..., n. You should be able to use your calculator to find: e

P(X = x) using the binomial probability distribution function

e

P(X < x) or P(X > x) using the binomial cumulative distribution function.

If X ~B(n,p),

o

then:

E(X)=p=mnp

THE NORMAL

e

Var(X)=mnp(l—p)

e

0= /Var(X) = /np(l — p)

DISTRIBUTION

If the random variable X has a normal distribution with mean p and variance o , we write X ~ N(u, o2).

The probability density function is f(x) =

f(x)

x—

L

o\ 27

e\ 2

-3(%5*)

for x € R.

is a bell-shaped curve which is symmetric about x = p.

It has the property that:

o

=~ 68% ofall scores lie between

o

~ 95% ofall scores lie between

o

=~ 99.7% of all scores lie between

1 — o and p+ o 1 — 20

and

11 — 30

and

u + 20 p + 30.

You should be able to use your calculator to find normal probabilities for the situations:

e

P(X a)

e

Pla< X

Hy:

>

pug

(one-tailed hypothesis)

>

Hi:

pu < pug

(one-tailed hypothesis)

>

Hi:

1 +# puo

(two-tailed hypothesis, as 11 # pg could mean

p > pg or p < o).

A test statistic 1s a random variable that summarises the information in a sample. The distribution of the test statistic under the assumptions of Hy is called the null distribution. The p-value of a test statistic 1s the probability of a result that 1s as or more “extreme” being observed 1f H

1s true.

The significance level « of a statistical hypothesis test is the largest p-value that would result in rejecting Hy. Any p-value less than or equal to « results in H being rejected. [f a statistical hypothesis test has significance level «, the probability of a Type I error is c. The significance level may be given as a decimal or a percentage.

General procedure Step 1:

Formulate statistical hypotheses.

Step 2:

Choose a significance level for the test. This is a threshold for making a decision, like the confidence levels we saw previously.

Step 3.

Use data from a sample to calculate a test statistic.

Step 4.

Calculate a p-value for the test statistic. This is the probability of that test statistic occurring under the assumptions of one of the hypotheses.

Step 5:

Make decisions about the hypotheses.

Step 6:

Interpret the decision in the context of the problem.

The one-sample t-test The t-test is used to test hypotheses about a population mean p when: the population is normally distributed the population variance is unknown. For a t-test of Hy:

1t = pp using a sample of size n with sample mean = and sample standard deviation s:

the test statistic 1s ¢ =

T — o 3

v

the null distribution is 7' ~ ¢,

the p-value calculation depends on H;: >

If Hy:

> po, p-value =P(T

> 1).

>

If Hi:

p < pp, p-value =P(T

< t).

>

If Hyi: p# pg, p-value =2 x P(T > |t]).

The two-sample t-test The two-sample t-test is used to compare the means of two samples from different populations. If the populations have means 141 and po, the null hypothesis has the form: Hy:

{11 = po

or equivalently

Ho: pg — pr2 =0 You should be able to use technology to calculate the test statistic and p-value. In this course you are expected to assume equal variances and hence use the pooled two-sample t-test on your calculator.

TOPIC 4: STATISTICS AND PROBABILITY

37

The x° goodness of fit test The y* goodness of fit test is used to determine whether a probability distribution fits a set of data. Consider a scenario with k categories. Let p; be the population proportion of individuals 1n category 7, where

py +psa—+....+pp = 1.

The hypotheses in a y* goodness of fit test have the form:

Ho: p1 = po1, P2 = pPoz2, ..., and pr = pox H,:

atleast one of p; # po;

where pg; 1s the population proportion of category ¢ under the null hypothesis.

x2,. = >_ (Jobs — fexp)®

The test statistic for a x* goodness of fit test is: where

fops

fe:-:p

1s an observed frequency

fexp 1s an expected frequency. Degrees of freedom (df) refers to the number of values that are “free to vary”. For a y* goodness of fit test, df = number of categories — 1. p-value = probability of observing a value greater than or equal to xZ . y?-distribution i “extreme” values

-

p-value

X?Elh}

For a y? goodness of fit test, we denote the critical value as Xgrit'

A

Since we use the upper tail of the null distribution in calculating the p-value, the critical

region is the set of values > 2. . .

.

2

2

The mnequality xC,.. = X5

.

.

.

.

.

18 called the rejection inequality.

critical region

——

p-value = o

Xgril

XEH]E

The ? test for independence The x? test for independence is used to determine if two variables in a contingency table are independent or not. It is a special case of the y* goodness of fit test. The hypotheses for the v test for independence are

Hy:

the variables are independent

H:

the variables are dependent

The test statistic for the x* test for independence is calculated in a similar way to the y* goodness of fit test. The expected

frequency of each cell in the contingency table is given by fexp =

The p-value and critical value yZ.

row

sum

X

column

otal

sum

for the x test for independence are calculated in the same way as the x* goodness of fit test.

For a contingency table which has r rows and ¢ columns, df = (» — 1)(c — 1).

38

TOPIC 4: STATISTICS AND PROBABILITY

SKILL BUILDER QUESTIONS 1

Gerard wants to estimate the average height of the 500 students at his school. He randomly selects a sample of 10 students, and uses a tape measure to find the height of each student.

Explain why this approach may produce a: 2

coverage error

b

measurement error.

The students at Hoylebury Middle School are to be surveyed on their attitudes on wearing : . school uniform. The numbers of students in each year level are shown. a

b

¢ 3

a

1

What are the advantages of surveying 50 students?

il

What are the disadvantages of surveying all students?

Boys | Girls

How many Year 8 boys will be selected?

iil

How many girls will be selected in total?

135

140

Year 9

130

145

125

130

Year 10 |

A stratified sample system is used to select 50 students. I

Year &

Explain why a stratified sample 1s better than a random sample in this case.

Marie is organising a staft lunch in a large office building. She asks the first 10 people to visit her office for their preferences, and then makes a decision.

&

5

6

a

Explain why this 1s a convenience sample.

b

In what ways will Marie’s sample be biased?

¢

Suggest a more appropriate sampling method that Marie should use.

A ticket inspector checks the tickets of every 20th passenger leaving a train terminal, starting from the 8th passenger. a

Identify the sampling method used. Explain your answer.

b

List the next six passengers to be checked.

¢

Given that 5000 passengers left the terminal that day, find the number of passengers checked.

C(lassify each variable as categorical, discrete, or continuous: a

The number of houses on a particular street.

¢

The brand of laptop someone uses.

b

The number of hours spent travelling on an airplane.

A random sample of people were asked “How many devices have you used to browse the internet in the last month?”. The results are displayed in the column graph.

A frequency

How many people were surveyed? b

Find the mode of the data.

¢

What

percentage

of people

10

browsed

the internet using

1 or

1

o L]

d 7

8

1

o

2 devices?

I

3

2

4

5

P—

6

Describe the distribution of the data.

Each student in a class writes down the total number of children in their families:

1

2

4

3

2

1

1

2

1

5

4

2

2

1

3

2

3

4

1

2

1

2

2

1

a

Explain why the data 1s discrete.

b

Construct a frequency table to organise the data.

¢

Draw a column graph to display the data.

d

Describe the distribution of the data.

e

In what percentage of families are there 3 or more children?

The following marks out of 100 were obtained by students in a Chemistry examination: 9

29

40

33

75

74

64

40

53

95

60

&8

58

51

66

5H3

&7

43

91

58

b7

68

32

65

77

44

T2

24

55

17

a

Construct a tally and frequency table for this data using class intervals 0-9,

b

Draw a column graph of the data.

d

If the number of marks required to pass was 50, what percentage of students passed the examination?

¢

10-19,

...., 90 - 99.

Write down the modal class.

.

devices

TOPIC 4: STATISTICS AND PROBABILITY

9

10

The heights of a s:.:}mple of emperor penguins were measured. The results are given in the table alongside.

39

Height (h cm) | Frequency

a

Explain why height is a continuous variable.

105 < h < 110 110 < h < 115

3 -

b

How many emperor penguins were measured?

115 < b < 120

14

¢

Construct a frequency histogram to display the data.

120 < h < 125

19

d

Describe the distribution of the data.

125 < h < 130

8

e

What is the modal class? Explain what this means.

130 < h

Y 17

18

Find the intervals where

is increasing or decreasing:

a

f(r)=5-3x

b

f(z)=22>—-T7z+6

¢

f(z)= —;

d

f(z)=22°—92°+ T2 +6

£

For each of the following functions, find and classify all stationary points. a

19

f(x)

f(z)=ua2°— 22

b

f(z) =2*—22° +42° — 8

¢

f(z)=2x+-

f(x) = 22° + ax + b has a stationary point at (1, 1). a

Find the values of a and b.

b

Find the position and nature of all stationary points.

b

f(fl:):fl’iz—%

Find the greatest and least values of:

21

22

a

f(::a:):a:3—23:2

¢

f(z)=2a°—622+120—10

—-1

L, passes through

(—2, —1)

y = mx + c.

and (4, 8).

Find the point of intersection of L

and L».

Lin wants to calculate the height of a mobile phone tower. He measures the angle of elevation from point A to the top of the tower C, then moves 60 m closer to point B and takes a second measurement. The information is given in the diagram alongside. a

Calculate the measure of AéB.

b

Determine the height of the tower.

Consider the arithmetic sequence with

a -

¢

4

L has gradient —5 and passes through (0, 4). Find the equation of L1, giving your answer in the form

L

2

Find the signal strength for an object 18 km away.

Find the first term u; ToieaAl dle

A

o

tic

cmccae

suln

0 ale

on

o

Uic

A

us = 18 and

ug = 39.

and common difference d.

L5 il

1Ist

T

10U

o dmcisan

LCTITS

A e o

01

o otz

it

60 m

g

b o

s e

UHIC COITCSPONUIIE

gl g

Find the 12th term of the sequence. D

allticLc

LS

SCrics.

Consider the function f(x) = ax® — bz?. Theline y = x — 6 isatangentto y = f(x) at z = 3. a

Find the constants a and b.

¢

Graph y = f(x)

b

Find the point where the tangent meets y = f(x) again.

and y = z — 6 on the same set of axes.

B

D

MIXED QUESTIONS

Consider this Voronoi diagram for the sites A(—4, 1), B(2, 3), C(0, —1)

D(0, —3). a

and

\ 3|

Identity the site which 1s closest to:

i

(2, 2)

i

OB

A,

(-5, —3)

b

Find the equation of the green edge. axr + by + d = 0.

¢

Find the area of cell C.

A manufacturer produces triangular prism shown.

AY

61

i

(6, —4)

Write your answer

3 in the form

P

3

G

y

B 3D

-

Y wooden

door-stops with the shape of the

a

Calculate the height / correct to 4 significant figures.

~—_~

.

T

b

Determine the area of the triangular end of the prism.

¢

The volume of the door-stop is 60 cm®. Determine its length .

d

Calculate the total surface area of each door-stop. Give your answer correct to 3 significant figures.

h cm

™ \

10 om

The management of a large shopping centre chain sent a survey team to one of its suburban shopping centres. Between 10 am and 3 pm on a very busy Thursday, 100 people in the main mall were asked the following multiple choice question: “At which type of shopping centre do you prefer to shop?”

A

suburban

B

central city

C

equally preferred

a

Give rwo reasons why this survey is likely to contain a coverage error.

b

The results were:

D

neither

E

no opinion

suburban 33%, central city 8%, equally preferred 51%, neither 4%, no opinion 4%

Management concluded that “more than four times as many people prefer suburban shopping to the central city”. Explain why this conclusion is unreasonable.

The current in an electrical circuit ¢ milliseconds after it is switched off is given by I(t) = 40e~ %1% amps. a

What current was flowing in the circuit initially?

b

What current was flowing in the circuit after 100 milliseconds?

¢

Sketch /()

d

How long did it take for the current to fall to 1 amp?

and I =1

on the same set of axes.

Soraya borrowed £7000 to go on a holiday. The bank charged an interest rate of % p.a. compounded monthly. She will repay the loan with monthly repayments of £220 for 3 years. a

Findr.

b

Find the total interest charged on the loan.

¢

Suppose Soraya chose to repay the loan over 4 years instead of 3 years. I

10

Find Soraya’s monthly repayment.

ii

How much extra interest will Soraya pay?

The number of people at a music festival ¢ hours after 12 pm is modelled using the function

N(t) = at® + bt? + ct + d.

There were 2000 people at the festival at 12 pm. The number of people at the festival was increasing at 200 people per hour at 6 pm, and decreasing at 500 people per hour at 8 pm. The festival closed at 10 pm.

a State four conditions for N(¢)

11

and N’(t) you can deduce from the information given.

b

Hence find the function N ().

¢

Predict the maximum number of people at the festival, and the time when this occurred.

A tinned food company examined a sample of its tins of corn and tins of ) o . pineapple for defects. The results are summarised in the table alongside. a

How many tins were included in the sample?

b

Estimate the probability that the next randomly selected tin: I i

is not defective

Il

is defective, given it is a tin of corn.

is a defective tin of pineapple

Defective | Not defective Corn

37

031

Pineapple

24

617

62

MIXED QUESTIONS

12

Monica is a police officer. She wants to investigate whether house break-ins in her city are equally likely to occur on each day of the week.

She compiles the following data for 140 randomly chosen house break-ins over the past year. Day of the week

Monday | Tuesday | Wednesday | Thursday | Friday | Saturday | Sunday

Number of break-ins

15

11

17

18

27

29

23

Monica will perform an appropriate 2 test at a 5% significance level. The critical value of y? for this test is 12.59. Write down the hypotheses that Monica should test. b

Assuming the null hypothesis is true, how many break-ins from Monica’s sample would be expected to occur on each day?

¢ Calculate the test statistic x2,,. . d

Hence determine whether the break-ins in this city are equally likely to occur on each day of the week.

MIXED QUESTIONS SET 3 1

2

3

A 3 m

long ladder 1s resting against the wall of a house. The ladder makes a 20° angle with the wall.

a

Calculate the distance from the foot of the ladder to the base of the wall.

b

The ladder suddenly slips so that its foot moves 1 m further away from the base of the wall. Calculate the angle that the ladder now makes with the wall.

The fluoride concentration of lakes in a particular region was found to be 3 x 10~% g per litre. a

One lake has 5.6 x 10° notation.

b

Another lake contains

litres of water. Find the amount of fluoride in the lake, giving your answer in scientific

4.13 x 107 g of fluoride. Find the volume of the lake.

AY

A graph of the quadratic y = az® + bz + ¢ is shown alongside, including the vertex V

and y-intercept.

9

a

Determine the value of c.

b

Use the axis of symmetry to write an equation involving a and b.

¢

Use the point

d

Find a and b.

V(L,7)

(1, 7) to write another equation involving a and b.


T

MIXED QUESTIONS

7

63

The table shows the amount of petrol remaining in a motorbike’s fuel tank and the number of kilometres travelled. The capacity of the tank is 10 litres.

Remaining fuel (x litres) | 10 | Distance (y km)

8§

6

Plot this data on a scatter diagram.

¢

Interpret the y-intercept of the regression line.

d

The motorbike has travelled 220 km since its tank was refilled. Il

2

1

0 | 90 | 190 | 260 | 330 | 370

a

i

4

b

Find the equation of the regression line for y against .

Use your equation to estimate the amount of fuel left in the tank. Find the average distance travelled per litre over the 220 km.

The probability of Mark waking up early is 0.8 . If he wakes up early, he will pack lunch with probability 0.6 . If he does not wake up early, he will pack lunch with probability 0.15. a

Display the sample space of possible outcomes on a tree diagram.

b

Hence determine the probability that Mark will pack lunch today.

A farmer owns a triangular field ABC.

A

D is the point on [BC] such that [AD] bisects BAC. The farmer divides the field into two parts A, and As by constructing a straight fence [AD] of length 2 m.

10

11

12

a

Use the cosine rule to calculate the length of [BC].

b

Find the total area of the field.

¢

Find, in terms of x, the area of:

d

Hence find z.

I

Ay

65m /30°30° T m

i

A

B

Let D(t) m be the distance between two stunt motorcyclists ¢ seconds after they start riding. motorcyclists were initially 42 m apart, and that D’(t) = 0.8t — 8.

a

Find D(1).

b

Find the distance between the motorcyclists after 2 seconds.

¢

Find the minimum distance between the motorcyclists, and the time at which it occurs.

Melinda bought a car valued for £55000. 9.25% p.a. compounded monthly.

It is known that the

She borrowed the money for the car over 7 years, with interest charged at

a

Calculate her monthly repayments.

b

Calculate the outstanding debt after two and a half years.

¢

The car depreciates at 15% p.a. After 7 years, Melinda sold the car at its depreciated value. Find the total cost of the car to Melinda, taking depreciation and interest into account.

A gardener has been asked to perform maintenance on Globe Park, a circular lawn with radius 30 m. This involves:

e

mowing the interior of the lawn

e

using a line trimmer to tidy the perimeter of the lawn.

From previous experience, the gardener knows that: e

a circular lawn with radius 10 m takes 30 minutes to maintain

e

a circular lawn with radius 20 m takes 110 minutes to maintain.

... (1) ... (2)

The gardener believes the time to maintain a circular lawn with radius » m can be modelled by 7" = ar 4+ b minutes, where a and b are constants. a

Assuming the gardener’s model i1s correct, write two equations connecting a and b.

b

Hence find a and b.

¢

Explain why this model 1s not appropriate for small values of r.

d

Use this model to predict how long it will take to maintain Globe Park.

64

MIXED QUESTIONS

e

Given that it actually took 230 minutes to maintain Globe Park, find the percentage error in the prediction in d.

f

The gardener’s friend suggests a model of the form

7' = pr® + gr, where p and ¢ are constants.

I

Explain why a model of this form is reasonable.

il

Use the information from (1) and (2) to find p and ¢ for this model.

lli

Is this model better at predicting the time taken to maintain Globe Park? Explain your answer.

MIXED QUESTIONS SET 4 1

'

Eliza walked 6.1 km (rounded to 1 decimal place) in 82 minutes (rounded to the nearest minute). Find the /ower bound for Eliza’s average speed, in kmh™".

Consider the graph of y = 3 + 22 — 22,

AY

The point F lies on the curve. Let M be a point close to F with x-coordinate a

What is the y-coordinate of M?

b

Find the gradient of [FM] in terms of A.

¢

Hence find the gradient of the tangent to the curve at F.

2 + h.

The data below are the recent sale prices, in thousands of dollars, of houses in two neighbourhoods. Neighbourhood A:

Neighbourhood B:

275

281

320

265

305

258

310

430

285

290

297

345

195

230

269

300

258

273

325

300

412

370

297

505

340

333

290

428

305

520

360

410

275

320

431

410

a

Is the data discrete or continuous?

b

Use technology to find the five-number summary for each data set.

¢

Display the data in a parallel box plot.

d

Compare and comment on the distributions of each data set.

Bags of rice are sold at a Jakartan wholesale market. The price per bag, P, if b bags are bought 1s shown below:

b (bags)

30

35

40

45

50

P (rupiah) [ 38000 | 36000 | 34000 | 32000 | 30000

a

Determine the function P(b).

¢

Do you think this model can be used to predict the cost of 150 bags of rice? Explain your answer.

The function f can be written in the form

b

FHence predict the total cost of purchasing 60 bags of rice.

f(z) = a(x — p)(z — q) where p > q.

a

Write down the values of p and ¢.

b

Find a.

¢

Write down the equation of the axis of symmetry.

AY \


T

90 \/ Y

Trains A and B are 10 km apart, and are approaching the same train station. Train A 1s 8 km from the train station on the bearing 071°. Train B is on the bearing 296° from the train station.

a

Display this information on a diagram.

b

Find the bearing of train B from train A.

¢

Train B is travelling at an average speed of 7 ms™'. Find, to the nearest second, the time it will take for train B to reach the train station.

A conical funnel is 80 cm wide and 50 ¢m high. a

Estimate the capacity of the funnel in mL. Write your answer in the form where

b

1 < a < 10

and

|

2 (km) (s

MIXED QUESTIONS

10

67

Find the equation of the sine function shown in the graph.

4l

P

>

\/

T

Yy

11

a

Use technology to find all solutions to z* — 22 — 7z 4+ 5 = 0.

b

Hence find the area of the shaded triangle.

AY

y=x°—22°—Tx+5 T

o

w1

:I.'

Y

12

The distance travelled by two similar toy cars after rolling down a slope was measured 40 times each. The measurements were rounded to the nearest tenth of a metre. Redcar

56

64

42

53

6.1

4.5

Blue car

54

46

39

6.2

58

45

5H4

6.1

Median distance

4.8 m

4.5 4.1 3.9

5.6 53 54

5.7 42 5.7

48 62 48

39 74 54

56 54 5.7

6.1 58 6.1

5.9 4.5 64

Shortest distance Longest distance Q, Lower quartile Q5 Upper quartile

3.2m 6.7 m 4.1 m 5.4m

Complete this table of cumulative frequencies for the red car data.

b

Draw the cumulative frequency graph for the distance travelled by the red car.

¢

Use the graph to find the following statistics for the red car:

Number of rolls

Distance 3.0

40

s

(m (m)

frequency

d a, is one-to-one.

2]

96

TRIAL EXAMINATION 4

6

|[Maximum mark: 4]

The Voronoi diagram with points A(—1, 3), B(2, 6), C(5, 3), D(—2, —2), and E(4, —2) locations of recycling centres in a city.

of

below represents the

Y

4 Aeo

-C

2 """‘""---..._,_____“

-

"""""---..._,_____“

—25

—9

2

1

D) a'

D#

—T

‘E

Charlotte lives at (4, 1). Which recycling centre should she visit? b

7

A new recycling centre is to be built as far away as possible from the existing recycling centres. Assuming it must be located inside the region defined by the existing recycling centres, where should the new recycling centre be placed?

[1] 3]

[Maximum mark: 6] A savings scheme is offering an interest rate of 2.5% per annum, compounded half-yearly. Astria wants to save £10 000. She works out that she can save £500 a year, which she will deposit on the 1st of January each year.

8

a

How much will she have saved after 5 years?

13]

b

How many years will it take for Astria to save the full amount? Give your answer as an integer.

13]

[Maximum

mark: 4]

The life expectancy of a person in the United States of America, ¢ years after 1900, is given by

L =10.54+13.91n(t 4+ 10) years,

9

t > 0.

a

Estimate the life expectancy in the United States of America in 1975.

2]

b

In what year was the life expectancy 60 years?

2]

[Maximum mark: 5] The volume, V cm?, of a tin of radius

cm and with fixed surface area, is given by the formula

V = 250r — 7. dl

10

)

dV

Flfld

[1]

E

.

..

dV

e

2]

b

Find the positive value of » when

0.

¢

Find the value of V' which corresponds to this value of r.

2]

[Maximum mark: 7]

The number of words in the first 20 sentences of Chapter 1 of The Hunger Games by Suzanne Collins are as follows: 12,

11

17,

12,

4, 7,

7, 9,

18,

13,

18,

7, 5,

11,

14,

14,

3, 5, 25,

9,

10

a

Calculate the mean for the data.

12

b

Calculate the IQR for the data.

12

¢

What is the largest data value that is not an outlier?

3.

[Maximum mark: 7 AN

W

r

l

Sarah will cycle 2000 miles over a number of days for charity. She cycles 12 miles on day 1, and increases this distance by 10% each day. a

How many days will 1t take for her to complete this challenge?

b

What is the greatest number of miles that she will complete in a single day?

[4] 3]

TRIAL EXAMINATION 4

12

Q7

[Maximum mark: 7] A theory claims that, when sweet peas with red flowers and sweet peas with white flowers are crossed, the next generation of sweet peas have red, white, and pink flowers in the proportions Ll > o> and % respectively.

The outcomes 1n an actual experiment are as follows: tlowers.

24 with red flowers, 34 with white flowers, and 62 with pink

A x? goodness of fit test at a 1% significance level was conducted on the data.

13

a

State the null and alternative hypotheses.

2]

b

State the number of degrees of freedom.

1]

¢

Find the p-value.

2]

d

What is the conclusion of the test? Give reasons for your answer.

2]

|[Maximum mark: 8] The diagram below shows a tower, PT, with height 30 m and points A and B which lie on the same horizontal plane as the base of the tower. The angle of elevation to the top of the tower is 15° from A, and 13 from B. The angle BTA i1s 60° as shown. P

30m

60°

Calculate the distance AB.

14

|[Maximum mark: 4]

A renowned butcher is blindfolded and asked to taste and arrange eight cuts of meat in order of price. The correct order is A, B, C, D, E, F, G, H while the order chosen by the butcher was A, (B, D), C, G, (E, F, H). The brackets indicate cuts of meat which the butcher assigned the same price. a

True rank

Butcher s rank |

b

2]

Copy and complete this table of rankings:

1 |

2

1

Determine the value of 75 as a measure of the correlation between the butcher’s opinion and the correct order.

PAPER 2

2]

CALCULATOR, 90 MINUTES

|[Maximum mark:

17]

Lucy was assigned a project on 3D shapes and decided to model her crayon as a cylinder and a cone. She drew the diagram below and added measurements to 1 decimal place. =

9.2cm

i

D.Qlcm -

7.8cm

>

a

C(Calculate the height of the cone.

1]

b

Find the total volume of the crayon.

4]

¢

Calculate the slant height of the cone.

2]

08

TRIAL EXAMINATION 4

A label 1s put around the crayon so that only the cone and base of the crayon is visible. d

6]

What percentage of the crayon’s surface area is being covered by the label?

After a few weeks of use the top of the crayon looks like this:

E

x 100%0

497 — 42.257 | | 42.25m x 1007 ~ 16.0% If the exact area Vg

was

56.257 ¢cm“,2 the percentage error _= [ Va—-Vg| % E

x 100%0

497 — 56.257|

B 56.25m ~ 12.9% the maximum percentage error in the estimate ~ 16.0%.

g

x 1007

WORKED SOLUTIONS

7

a

o2

64—832

—

b

e

125 x5

B

=5

3

>::5

I

fl

3

37—

o

26

101

9?}1

(97)" (32)211 32m

—

_

8

a

,

(—3m3?*=(-3)*x(m?? —

b

W) 2

2\ °

215

= (my,) 25

81??112

C

34?1

321'11—4?1

.

Ts’t x

e

(4st?)’ = Ts*t x 4°s°t”

mfiylfl

=7

32

9

a 4%4+41=141

BK

= (11) .

6435t10

— 4485°¢10

b (23) 7 = ()77

_ 5

X

¢ 22+2' 42 =442+1

2

-

1

= fiuj 2 ~ 2

112

_ 16 —

10

a

(2 +2

%)

=@

+22° xa

=z*+24+

121

2+ (27 7)°

b

" *

(z* — 2%)(2® + 3)

= 2* x 2

4+ 32* + (—2?) x 2% + (—2?) x 3

=z’ + 3z* — 2% — 322 11

a

23

—3

2

ab " =a

i

X 3

2m_2fl3

b

e

—

3

—

.n2

XM

="

o -

13

a

49000

b

L? T

n*(79)

78 X 1

o

— 12 x a

—3

X

1

=3

—12x — x b° &

_1255

&

a’

omT

0.0000678

¢

526000000

— 4.2 x 10000

— 6.78 x 0.000 001

— 5.26 x 100000 000

— 4.2 x 10%

— 6.78 x 107°

— 5.26 x 10°

Using technology:

a (3.57 x 10°) x (2.38 x 10%) = 8.4966 x 10° ' | e ¢ 14

5

=2 xm 7% x

b?

12

12(1_3

o AO1X107T 3.45 x 108

3451015

(0.000 (JS)""l — 4.096 x 10~17

a

)

8r — 2 = 372

—3z“+8x

BatDedforn ( 1000. Using a graphics calculator with Y1 = 2 x v/3 A (X — 1), we view a table of values: atfilegforn] (dc)Feal ¥1=2%(43) ~(x-1) 11 12

486 841.77

14

2525.3

13

LR

1458

FORMULA] P13 IGRITIP [ EDIT

(GPH-CON)[GPH-PL

The first term to exceed 1000 1s u13 = 1458.

22

There 1s a fixed percentage increase each year, so the population forms a geometric sequence with uy = 217 and r=1.42.

the population after n years is u,, = 217 x (1.42)".

a

i wus =217 x (1.42)°

il wo =217 x (1.42)"

~ 1252.86

~ 7233.41

The expected population size after 5 years is approximately 1250 birds.

b

The expected population size after 10 years 1s approximately 7230 birds.

We need to find when 217 x (1.42)™ = 30 000.

Using technology, n ~ 14.1, so it will take approximately 14.1 years for the population

to reach 30 000.

HethDedlom]) (G7c)Fes) Eq:217x1.42*=30000 L pi=saaog Dot

Rgt=30000 REPEAT

23

a

Ifthe interest rate per annum is 7.2%, then the interest rate per month r=1+1

=1+ 0.006 — 1.006

b

The interest is calculated monthly, so n = 3 x 12 = 36 time periods. Uz

—

Up

X

TSG

— 500 x (1.006)%° ~ 620.15

The value of the account after 3 years 1s €620.15.

¢ real value x (1.02)° = €620.15 [

val

real value =

€620.15

1.02)3

= €584.38 24

.

:

:

:

i

33

:

There is 1 time period every 3 months, so n = e

11 time periods.

: : : . . . 8% Each time period the investment increases by 1 = 7

.

:

0 2%.

wuy; = ug x (1 + )"

the amount after 33 months is

= 3500 x (1.02)'" ~ 4351.81

{2% = 0.02}

The maturing value of the account is £4351.81. 25

The initial investment uq is unknown. There are » =5

x 12 = 60

time periods.

Each time period the investment increases by i = Now

wugy = ug X (1 + i)fio

30000 = ug x (1.004)%° ELO

{0.4% = 0.004}

30 000

___'Eiiffiiiiafi_ffi#fzz;{il()'lél

[ need to invest $23 610.14 now.

4.8%

i

0.4%.

i = T'IQQ% = 0.6% = 0.006.

WORKED SOLUTIONS

26

a

us=ugx (1—d)° = 2000 x (0.7)° — 686

b

105

Depreciation = £2000 — £686

{30% = 0.3}

= £1314

So, after 3 years the value of the television is £686.

27

a N =45x12=54, ClY =12

PV =-10000,

PMT =0,

FV =12000,

P/Y =12,

PV

Compound 1%

=

—-10000,

PMT

=

T

0,

FV

=

20000,

P/Y

=

12,

Interest

=54

0 /0

=

2V O

Hornd)

Compound

n

=

=4.058437598

PMT=0 FV =12000 P/Y=12

The account paid about 4.06% interest per annum. /% = 4.06, G Y=a12

[Norm1]

n

1% ~ 4.06

b

E

4

A AHORTZN

Interest

=1

=205.2174663

N =~ 205.2

28

a

It will take 206 months or 17 years and 2 months for Lauren to double her deposit.

P/Y=12 Cn J(T% JCPV

The commmon difference

= g(ul + up)

b

Unp = up + (n—1)d

d = —6.

C

uoo = w1 + 19d

_ %(51

=21+ 19 x -6 =51 —114

10

= —63 29

a

The series is arithmetic with

Now

vy = 11,

d = 4, and

n = 20.

S, = 2 (2u1 + (n — 1)d) So0= 22(2x 11 + 19 x 4) = 10(22 + 76) = 980

b

7+125+

184235+

....4+ 106

The series 1s arithmetic with

«w; = 7, d = 5.5,

and

w,, = 106.

First we need to find n. Now

u,, = 106

up + (n—1)d = 106

‘

74+ 5.5(m —1) = 106 5.5(n — 1) = 99

= 22 x 113

n =19

Cc 1-24+3—-4+5—-6+4+7— 1+3+5+7+.... and

n

Using Sn = 3 (1 +un) . Si9 = (7 +106)

n—1=18

— 1073.5

... to1l00 terms can be expressed as two separate arithmetic series: where

—2—4—6—-—8— ... where

Using S,, = 2(21@1 + (n—1)d),

uy =1, uv1 = -2,

d =2, d=

n =50 -2,

n =50

the sum of the first series = 22 (2(1) + 49(2))

= 25(2 + 98) — 2500 and the sum of the second series = 2}(2(—2) + 49(—2))

— 95(—4 — 98) = —2550

the sum of both series = 2500 + (—2550)

— 50

So,

1—24+3—-44+5—-6+7—....

to 100 terms 1s —50.

_ 63)

% —192

_ _1920

(PMTCFV

0

$

106

WORKED

d

SOLUTIONS

The integers from 1 to 200 which are not divisible by 3 are

1, 2,4, 5, 7, 8, ...., 200.

The sum of these integers can be expressed as two separate arithmetic series A and B: and

SApA=14+4+7+...+196 4+ 199

where

ui =1, d =3,

u,, = 199

Sp=2+5+8+....+197+ 200

where

uy =2, d =3,

u,, = 200

Now for S,

u, =u;1+(n—1)d

andfor

S,

(n—1)d

199 =143(n—1)

.

200=243(n—1)

198 =3(n — 1)

c.

198 =3(n—1)

66 =n —1

S,

n =67

Using

u, =u1+

66=n—1

c.on=067

S, = 2 (w1 +un),

Sa=F(14199)=6700

and

Sp= F(2+200) = 6767

The total sum = S4 + Sp = 6700 + 6767 = 13467

30

a

wur=1 w1y

=

c.oup+6d=1 +

up

..

—23

=

14d

{using u,, =uy + (n—1)d} —23

Using technology to solve these equations simultaneously, we find that vy = 19 and d = —3.

HetiDegfom] [@)Fed Sk R ¢

2.1

a“x?%fcn

1mmmE

ST

-23 b

Up = u1 + (n — 1)d

C

ugy =19+ 26(—3)

{usinga}

= —959

e

19

REPEAT

Dy — %(*ul + Up)

oo

Sor =219+

(—59))

{from b}

= 2L x (—40) = —540

31

a

Sy — 2(21@1 + (n—1)d)

—210 = 2(2 x 18 — 3(n — 1)) .

(36 —3n+3) =-210 5(39 = 3n) = —210

b

From a,

n 5(39 — 3n) = —210

fl(39

—

3?‘1)

—

=] Deg Real Xgflffifnflitaw

—420

C

3

= [Math||Dea)[Norm !x=d@

0}

The sequence is arithmetic with w; = 7 and d = 3.

Now

S5, = g(Zul + (n — 1)d) Sn

=

%(2(7)

== 3(?1 . 1))

S, = 2(14 +3n — 3) Sn = 5 (11 4 3n) b

S, = 140

5(11

T

Sfl“)

—

140

e fornl) (I7c)Fe)

| {LlSlIlg

aX?2 +t;}{+c=E a}

C

3

11

(@) Fead

]

ax}ffi +bX+c=0

EFED

HE[-II.BE}

n(11 + 3n) = 280 11n + 3n®

= 280

~280

3n? + 11n — 280 = 0 Using technology,

n =8

{n > 0}

REPEAT

S

WORKED SOLUTIONS

33

a

The series 1s geometric with n = 8.

5, =

Now

u;

=

10,

r =

%,

and

b

The series is geometric n = 10.

r—

10((3)® — 1)

o8 =

.

;— 1

vy

=

2,

r

=

5,

and

— 1)

S, = ui(r

Now

o 1

with

107

r— 1

_ 205" - 1)

P10= T

= 19.921 875

= &G04 Bl

20

€ > 3x (=22 =3x(-22+3x(=2)*+...+3x (-2)? The series is geometric with u; = 3 x (—2)° = —24, r = —2, and n = 20.

Now §, = — ("= r—1

=24((-2)%" - 1)

S20 =

o

= 8 388600 34

a

10+

14 + 18 + 22 + .... + 138

w1 = 10,

Now

1s arithmetic with

b

6—124+24 —48+ 96 — .... + 1536

d =4

up =06, 1 = —2

u; + (n—1)d = 138 10 4+ 4(n — 1) = 138 4(n—1) =128 n—1=

Now u "~ = 1536 s 6x (=2)"1 =1536 -, n=9 {using technology} .

32

n — 33

So, the sum is

So, the sum 1s

ou (L=

%(ul + ug3) = %(10 + 138) 9449

a

N

=20x12

=240,

C/Y =12 L

I%

=

7.2,

PV

=

120000,

FV

=0,

P/Y

=

12,

C/Y

=12,

I%

19

=

()

Compound

Interest

7.2,

PV

=

P/ Y= 120000,

PMT

=

—-944.82,

P/Y

=

12,

T Forad

CumEnunH

Interest

Ti

o

e 2772

After 1 year, ? $117 211.33 is still owing on the loan. i

1

by S120000

FV ~117211.33 ¢

T

Yo 75

The monthly repayment is $944.82. N

6(1—(—2)9)9

— 1026

PMT ~ —944.82

b

Tn

_ 6 9 _ E(l — (—2) ) =2 X 513

— 33(148) 35

is geometric with

P/Y=12

I_IITIITIIWIITI

Amountpaid = $944.82 x 12 — $11337.84

i

$120000 — $117211.33 = $2788.67

lli

The load does not decrease by the full amount of the monthly repayment as the payment is used to pay interest as W

d

|

Aall 11

A

as

t

10

vraadry

1reauce

‘l"lfl. Le

N=19x12=228, P/Y

_

12!.

C/Y

fififififififififi

1

lJllllLallJ{l.l

—

I% =695,

19

PV

=117211.33,

FV =0,

%% :?i$gll

PMT ~ —927.42

P

The new monthly repayment is $927.42.

i /% =6.95 P/Y

=12,

PV =117211.33,

C/Y

— 12

Enmgg%gd

PMT = —944.82,

33

4156112

|:1E§_|| P/Y= P

FV =0,

N =~ 219.5

It will take 220 months to pay off the rest of the loan with the original repayments. the loan will be paid off 8 months earlier.

Interest

Cflmdest PV

—117211

PMT=-944.

e

82

33

e

108

36

WORKED SOLUTIONS

a N =8x4 =232 C/Y =4 PV

I%=655

PMT =-93362,

FV =0,

P/Y = 4,

~23110.12

[Hormd]

Compound

Interest

n

=32

PV

=23110.1243

1%

R

=6.55

P/Y=4

Oscar borrowed $23110.12.

b

E

JCPV_J(PMT JCEV_

(Cn_T%

4

(0D

Interest = $933.62 x 8 x 4 — $23110.12 — $6765.72

¢

i N=6x4=24 I% =655 P/Y =4, C/Y =4

PV = 2311012,

PMT = —933.62,

6947.33

=24

Interest

[% =6.585 PV =23110.12 PMT=-933.62 FV =-6947.324187 P/Y=4 (Cn_ JT% PV _J[(PMTJ[EV_

So, the outstanding balance is $6947.33 . /% = 6.55, PV = P/Y =4, C/Y =4

[Norm1)

Compound n

FV =~ —6947.33

i

E

— 3000,

PMT

=

-933.62,

FV

=

0,

E [ornd Compound n

Interest

Ry

J )TN 2

=4.417244824

[% =6.55 PV =3947.33 PMT=-933.62 FV =0 P/Y=4 J (n JT% JCPV J(PMT ] EV_JETEED

N =~ 4.417

It will take another 5 quarters to pay off the rest of the loan.

time saved = 8 x 4 — (6 x 4 + 5) =32 —29 — 3 quarters

37

a

N=5x4=20, PMT

I%=8,

PV =10000,

FV =0,

P/Y =4,

C/Y =12

0

Compound n

~ —612.36

[% PV

The quarterly repayment is $612.36 .

FV

~

I% =28,

PV =10000,

PMT = —612.36,

P/Y = 4,

PMT

% =8,

PV

Compound

=4483.17,

FV

=0,

P/Y =4,

C/Y =12

=12

J[PMTJ[FV]

Interest

R

Norm1

Compound

n

~ —1177.77

1% PV

J J(PMT I FV_JEENEEY

JCPV

.

=4

Interest

+End

=8 =4483.17

SR

WL TN

P/Y=4

Cn

a

)PV [Norm1)

P/Y=4 Cn 1%

The quarterly repayment must increase to $1177.77.

38

J[(T%

J

[% =8 PV =10000 PMT=-612. 36 FV =-4483.168913

—4483.17

N=1x4=4,

Interest

=

=8 =10000

cl

n

The balance of the loan is $4483.17 .

¢

=20

P/Y=4

[(Cn_

b N=3x4=12, ClY =12

Hormi

JT%

JCPV_(PMTICEV]

4

real interest rate multiplier x 1.003 = 1.012 1.012

real interest rate multiplier = =i

~ 1.008 97

real interest rate ~ 0.897% ~ 0.9% b

N=5x4=20,

C/Y =4

I%=12

PV

=

-2000,

PMT

=

FV ~12413.68

After 5 years, Cassie will have €12 413.68 in her savings account.

Now

real value x (1.003)°** = €12413.68 €12413.68 real value = (1.003)20 ~ €11691.81

—-500,

P/Y

= 4,

B Form1 Compound n =20 [% =1.2

Interest

PV =-2000 PMT=-500 FV =12413.67839 P/Y=4

Ln

J 1%

J{ PV

JLPMT |l FV

+End]

4

JELLqw

WORKED

39

N =4x12 =48, ClY =12

PV = —81000,

PMT = 2000,

FV =0,

P/Y

= 12,

& ormd) Compound Interest n =48 If LR GERELE!

I% ~ 8.60

PMT 2000 FV =0 P/Y=12

Bill needs to receive 8.60% p.a. compounded monthly. N =48, PMT

I% =7

PV =-81000,

FV =0,

[

P/Y

=12,

C/Y

=12

I%

=

JCT%

JCPV

[Formi)

Compound n =48

=~ 1939.64

N = 10 x 12 = 120, P/Y =12, C/Y =12

n

[=]

J[PMT ] FV

5.2,

PMT

=

200 x 2 =

400,

FV

=

0,

E formd) Compound n =120 1% =5.2 PV

PV ~ 37367.19

109 =0

|

=2

Interest

1% =7 PV =-81000 PMT=1939.645818 FV =0 PIY 12 | [ T% JC PV _J[PMTJFV

Bill would receive $1939.64 per month.

40

SOLUTIONS

Interest

| [~+End|

=-37367.18576

PMT=400 FV =0 PIY 12 I L 1% Hfi?IflWWflF?IJMHF

So $37 367.19 should be invested now to provide such an annuity. Total interest = $200 x 2 x 12 x 10 — $37 367.19 = $48 000 — $39 367.19 — $10632.81

41

I% =49,

PV

C/Y =4

=-400000,

PMT

= 3000,

FV

=0,

P/Y =12,

|

Compound

n =192.5335227 1% =4.9 PV =-400000 PMT=3000 FV =0 P/Y=12 [__7FEFVTWTHWTTTT7]WEF

N =~ 192.5 Anne’s money will last for 193 months, or 16 years and 1 month. N =20 x C/lY =4 PMT

12 = 240,

I%

= 4.9,

PV

= —400000,

FV

=0,

P/Y

O

lflg(log

X

b

(103)6)

= log(10? x 10°%)

log(ig;)

b

—

log(2" x 5%)

= log((2 x 5)")

:

— log(10")

7)

= 1

102.3010

~

—= k44

21n6

d

6_11]3

::(Ehlb)

]H(EI_WH)

'

lfl(ek+flj

;3-10—{L6990

c

ln(eim)

b

¢ 0.02 =10"800

200 = 1008200

100'3010

In(e® x e*)

10

2)

] I% "fi?HFfllflF?IIWEfl

o

2 = 109872

Li

C

10™

log (10"

=

=9+ 3b ~

[n

= log(

)

::log(109+3b)

43

1

— 6

-

b

iy — 81“ a7

500

_3 1

46

When

I =3 x 1072,

¢ 0.023 =00

= Eln 500 ~ 0-2146

s 58901

L=

lfllflg(:};li

—2

~ o 31723 o

)

~ 105 dB

I

When L = 85, 85 = 10log( - ) using technology,

[ ~ 3.16 X 10~4

i

Eq:85=10xlog

mxfi

Lf?é:gélgggqqfigxfifln

Rgt=85 REPEAT

(Elnd)—l

=31

2

= 36 45

Interest

[End

=~ 2613.40

log(10? x 1000°) —

[Horm)

Compound

Anne should withdraw $2613.40 each month. 42

Interest

.

110

WORKED

SOLUTIONS

TOPIC 2 SKILL BUILDER QUESTIONS 1

a

y—6=—-2(z—(-5))

The equation of the line is

y—6=—-2(x+5) y—6=—2x—10

2

b

The equation of the line is y =

a

Thelineis parallelto

.

ool

y =

—2r—4

T + 9.

22 —y = —3

or y = 2x + 3 which has gradient 2.

the line has gradient 2 and passes through the equation of the line is

(5, 3).

y—3=2(x—5) y—3

=2z — 10 y=2x—7

b

The line is perpendicular to y = —4x + 3, which has gradient —4. ',

the line has gradient i and passes through

',

the equation of the line is

(—1, 5).

y—5=z(x—(-1)) y—5=}l($+1)

1o+

Y — O

1

1T Ty

Y

3

Substituting * = —1 .

and y = —6

.

oy1—y1

The gradient is b

oz

1

21

into the equation gives 10—-4

7(—1) — (—6) = k k=—-7+6=—1

_fi_

(3

3 =3

The equation of the line is y = 3z + c.

The line passes through (—3,4), so

¢

4 =3(—-3)+c¢

When

=0,

When

y =0,

c=13 *. 5

a

the equationis

b

MY

e

the gradient is —4.

o\l

D\\

For 7z + 4y = 14:

When

z =0,

\‘_i

\ -

13

\

X

4y = 14 Y=

.

479

\ 7\

2 \ 7z + 4y = 14

3

\

So, the y-intercept is %

\\y =—4x+8 o\

Yy 6

32+ 13=0 _

B the z-intercept is.13 — 0}

~ 0.8409

The weight of the sample on any given day is reduced to about 84.09% of this weight the following day.

C

W(t) =100 x (%)

|

W (6) = 100 x (%)

{from b}

\6

~ 35.4

The weight of the sample after 6 days is about 35.4 mg. i

d

Whel‘l

W(t)

=

—

60

60,

o,

100

t=~2.95

X

i};'l;EgHE]Ifi?_c:;-lc:}n]rdinatas

Q_é

\\

{using technology}

It will take about 2.95 days for the weight to fall to 60 mg.

o PN

3

2

-1

JJJJJJ 1

X=2.947882377

i

When

W(t)

=30,

{using technology} It will take about 6.95 days for the weight to fall to 30 mg.

y = e’ — 1 1s a vertical translation of y = e* by 1 unit downwards.

b

The y-intercept 1s 0, and the horizontal asymptote is y = —1. AY

¢

y — e3

4

5

B

I;ITSEEC?[

¥Y=60

YIi=100%23 ?2=3nm{

5

\k

gl 1 ¢ % 4 & & 7 & 3 INTSECT

X=6.947882377

a

3

[E}:E]:fi‘ls_n{::}.;rdinates

30 = 100 x 2_&

t =~ 6.95

58

2

isahorizontal stretch of y = ¢* with scale factor 3.

The y-intercept is 1, and the horizontal asymptote 1s y = 0.

y = 2e® factor 2.

is a vertical stretch of y =

¥Y=30

¢® with scale

The y-intercept 1s 2, and the horizontal asymptote is y = 0.

WORKED

{

6

The domain of f(z)

i

----------------------- e

is {z | x € R}.

The range of f(z) is {y |y < 6}.

f(z)=6—e""""

/5 P

=

I

127

SOLUTIONS

>T

As r — —o0,

y= f(x) —> —oo.

As x — 00,

y= f(x)— 6.

From the graph in a:

v 60

61

a

For y= f(x) = sin4ax: theamplitudeis

e

the principal axisis y =0

e

the period 1s

a

| Il i

f(z) =k

has one solution if £ < 6.

ii

f(z) =k

hasno solutions if k > 6.

For y = f(z) = —2sin§ — 1:

e

|1| =1

360

I

_ 90°.

e

the amplitudeis

|—2| =2

e

the principal axisis y = —1

e

the periodis —00"

7ope.

2

amplitude=1

I

amplitude = 3

principal axisis y = 2

Il

principal axisis y =0

period =

360°

= 360°

iv

il

AY

period =

360°

= 180°

v

Yy=sInzx 1 2

y=3cos2x

:

>

180°..

m

Y=COST

i

amplitude=|—-6| =6

Il

principal axisis y = —1

Il

principal axisis y = 10

i

period =

360°

aus

i

k2| =

amplitude =1

¢

= 720°

.

Il

period=

iv

161AY

360°

= 120° v=10 y =10 —— 6sin6sin 3z3z

121 8__

o

T

4--

-

3

Y

62

a

AY f(x)=acos2zx+d (120°, 3.5)

(45°, 2) “‘tmfsz’

‘“m__##’jfig

Y

f(45°%) = 2

acos90° +d = 2 d =2

f(120°)

= 3.5

cooacos240° +d = 3.5 —%a+2:3.5 —%(L —

1.5

a=

—3

_____ 60°

y=sinzx

-:"'""L-i-:____:______j____.-—l

120°

180°

240°

300°

360°

-

*

128

WORKED SOLUTIONS

b

A

T

(90°.7) f(r)=asinz+d

(210°,1)

el

~—

>

Y

£(90°) =7

£(210°) =1

asin90° +d =17

cooasin210°+d =1

a+d=7

..(1)

oo—za+d=1

..(2)

fath)Deg)Horml] [d7c] Real

n H?ECE&J

flnjx+4fln1f=4zg a

1[ 2

|

c

1 -0.5

SOLVE [[M4RI9|CLEAR]

1 1

7

EDIT

REPEAT

Solving (1) and (2) simultaneously using technology gives @ =4 63

a

High tide is 4.7 metres, and low tide is 2.4 metres below it.

and d = 3.

A H (m)

b

low tide 1s 4.7 — 2.4 = 2.3 metres.

So,

amplitude = L ; 20 F Bk

H=1.2 cas(%t}“ + 3.5

=kl

period = 12.3 hours

0

=2 =123

p = 120

and the principal axis1s

— 24

0

{ (hours)

H = i ;L 2

d= 3.5 H=1.2

64

a

cm(%t)g

+ 3.9

For T'(t) = 5sin(15t)° + 24: e

the amplitudeis

|5 =5

e

the period is %’5” = 24 hours

e

the principal axisis 1" = 24.

T(t) =5sin(15t)° + 24

:

) b

1

2pmis2hours after 12 noon.

I

5

12

:

13

:

21

t

(h

! (hours)

9 pmis 9 hours after 12 noon.

T(2) = 5sin(15 x 2)° + 24

T(9) = 5sin(15 x 9)° + 24

= 5sin 30° + 24

= 5sin 135° + 24

=5(3) +24 = 26.5 at 2 pm, the temperature inside Pam’s caravan 1s 26.5°C. ¢

:

The minimum temperature inside Pam’s caravan is 24 — 5 = 19°C,

(1

~ 27.5 at 9 pm, the temperature inside Pam’s caravan 1s about 27.5°C.

which occurs when

18 hours after 12 noon is 6 am the following day. So, the minimum temperature inside Pam’s caravan occurs at 6 am the following day.

¢ = 18.

WORKED SOLUTIONS

TOPIC 3 SKILL BUILDER QUESTIONS 1

a

Perimeter = 2 X line segment length 4 inner arc length + outer arc length

— 2 x 14 + 200 x 28 360 x 27 x 14 + 259 360 x 27

~ 219 cm b

Area

—

outer sector area

—

inner sector area

9 260 360>
0}

r =~ 14.9

the radius of the beach ball 1s approximately 14.9 cm.

T x6+2x6x5

129

WORKED

SOLUTIONS

Let the pipe have length [ m.

|

le—— Im _,|

\"'f[}l

m

Now

volume of concrete = volume of whole cylinder — volume of hollow section

3=m(1)° x1—7(0.9)% x I

i l2m

5. 3=r7l—0.8lnl

7y

nol=—— ~503

3

3 = 0.1971

.

the pipe is approximately 5.03 m long.

10 cm

Perimeter = 27 + arc length 40 = 2(10) arc length = 20

0 cm

b

Now,

arclength = %

+ arc length

X 277

% 21(10) © 20 = -2 360 O

20 = =

=22

T

area =
0}

11cm

F

11cm

F

Let BEC be a. ftan o =

;121

o = tan~! (\/32_1) ~ 15.1° The angle 1s about 15.1°. d

The projection of [AM] onto the base plane is [DM].

.

the required angle is AMD.

Let DM be x cm.

Using Pythagoras in ADCM,

2° = 11* + 5° z? = 146

r =146

Let AMD be o tana =

im

a =_ tan

—1

(\/fi4 fi)

The angle 1s about 18.3°.

s

~ 18.3

o

{z > 0}

WORKED SOLUTIONS 23

a

The projection of [CD] onto the base plane is [CO].

Z4D(0,0,6)

G

. the required angle 1s OCD. tan 0 =g

0 = tan"'(2) ~ 40.6° The angle 1s about 40.6°.

The projection of [OF] onto the base plane 1s [OB].

the required angle is BOF.

Now

FB = 6 units

and

OB = /(5 —0)2+ (7—0)2+ (0 — 0)2

= V25 +49

v

..

/C(0,7,0)

— /74 units

B(5,7,0)

6 1 74

tanf =

0 = tan_l(

6 ) ~ 34.9° VT4

The angle is about 34.9°.

ZAD(0,0,6)

The projection of [AG] onto the base plane is [AC].

the required angle is GAC.

Now

GC = 6 units

and

AC = +/(0—5)2+(7—0)2+ (0 —0)2

+ 49 = V25

v

.

/C(0,7,0)

— v/ 74 units .

— 6

0 = tan (%)

~ 34.9°

The angle is about 34.9°. 24

a

M is the midpoint of [AD] which is (

O+ (—10)

S

44+4

6+6

) or (—5,4,6).

Let CMD be 6. Now

DM = 5 units

and

CD = +/(—10 — (=10))2 + (4 — 0)2 + (6 — 0)2

= /02 4 42 4 62 =

/D2 units VD2

tanf =

:

0 = tan~! (¥22) ~ 55.3° CMD =~ 55.3° C

G(0,7,6)

I

The

required

angle

is DON,

where

N

has coordinates

(—10, 4, 0). Now

and

DN

= 6 units

NO = /(=10 —0)2 + (4 — 0)2 + (0 — 0)?

— V/(~10)? + 42+ 07

tan

— v/ 116 units 6 = 71To 6 _= tan

—1

6 (v,m)

The angle 1s about 29.1°.

~

~ 29.1

o

137

138

WORKED SOLUTIONS

ii The required angle is MEP, where P has coordinates (—5,4,0). Now

MP

and

= 6 units

PE = /(=10 —(=5))2 + (8 —4)2 + (0 — 0)2

= V/(=5)2 + 42 + 02

= tan

_

0 =

V41 units

6

\/T

6 Y A 0 __= tan~"tan—L1f (A7) ~ 43.1

0

The angle is about 43.1°. 25

a

A

Area = %bc sin A

Jm

= 5 X 2 X 3 X sin 82°

&

B

~ 2.97 m*

&

b

A

ABC = 180° — 62° — 39°

{angles in a triangle}

= 79° area — %ac sin B

= 5 X 7 X5 X sin79° ~ 17.2 cm? b

The smallest angle in triangle ABC is opposite the shortest side. BCA is the smallest angle. By the cosine rule: cos BCA

=

2ol a5

2 x 12 x 10

cos BCA = %

= %

BCA = cos! (2) ~ 41.4° €

Area = %ab sin C

~ 5 x 10 X 12 x sin41.4° ~ 39.7 cm? a

A

Using the cosine rule,

AC?

=612

+ 722 — 2 x 61 X 72 x cos 43°

~ 2480.8 AC = 49.808

~49.8cm 43° C

b

61 cm

Let

AEB .

sing

Using

B

— ¢ :

the sine rule

sin @

’

72 sin ¢

—

sin 43°

AC 72 sin 43°

2

27

49.808

~ (0.9859

0 ~ 80.4°

and so

ACB ~ 80.4°

{as AC > 0}

WORKED SOLUTIONS 28

a

Area =

139

rm

%absin@

— % X 4 % 4 % sin0 4 = 8sin b

4m

: 1 811:19—2

6 = 150° b

{since 0 is obtuse}

Using the cosine rule,

x* =4° +4° — 2 x 4 x 4 x cos 150° 33:\/42+42—2>]

a

39

22

O

5

.

b

|&

LinearReg(ax+b) a

b

°

o

. O

25

O

20

=3.45284197

=-9.7520299

r =0.8188878 r2=0.67057723 MSe=29.3021549 y=ax+b [COPY)([DRAW

°

30

40

o

55y 45 40 35

40

o

50

So,

o

r =~ 0.819.

o

E\/\ ¢

25

Rk

~

32 4b 27

12

38 | 18 |

B

List1 | List 2 | List 3 | List 4

SUB 1 2 3

27

16

14

12

10

8

18

The data suggests that there 1s a moderate, positive, linear correlation between the students’ language scores and their mathematics scores. So, “Those who do well in languages also do well in mathematics.” 1s a moderately reasonable statement.

Monthly rainfall (mm)

34

14

Crop yield (tonnes) d |BE

10

15

20

29

30

21

29

31

30

28

A moderate, positive relationship may exist between crop yield and rainfall.

[RadHorm1] [d/c][Real

LinearReg a

(ax+b)

=0.06571428

b =15.86 r =0.79505891 rz=0.63211867 MSe=20.3714285

v=ax+b

So,

¢

r~

(COPYJ[DRAW]

0.795.

The relationship between rainfall and crop yield does not appear to be linear and so Pearson’s correlation coefficient may not be appropriate for this data.

404 crop vield (tonnes) 30 20 10 rainfall ( mm-z

10

15

20

29

30

156

WORKED SOLUTIONS

35 | Exchange rate (pesos/USD) | 2.85 | 2.95 | 2.90 | 2.75 | 2.65 | 2.80 | 3.05 | 2.98 | 2.95

Interest rate (%) a

7.8

4 interest

rate

740 | 7.50 | 7.55 | 7.25 | 7.25 | 7.35 | 7.65 | 7.75 | 7.60

h%)

7.6

7.4

°

A ¢

2.6

°

B

r

(ax+b)

=1.2903368 =3.76734262 =0.9151128

r2=0,83743144 MSe=5.8189x503 y=ax+b

(COPY][DRAW]

?

2.7

a b

o

®

Red Fornd

LinearReg

o

°

¢

>

b

o

ge rate (pesos/USD

2.8

2.9

3

43.1

’

There is a strong, positive, linear relationship between the exchange rate and interest rate.

36 | Median weight (x g) | 88 | 97 | 105 | 110 | 125 | 140 | 145 | 150

Numberinbag (y) | 28 | 26 | 26 |

a

23 |

21 |

19 |

18 |

16

20y y 40 ~31.

[T

e

e

) mea n poin. t ] (120, 22.125)

e

AYD6[ eore et 20

o

o

10

x (g2] =y

b

|

When

50

70

x =100,

100

150

y =~ 26.

il When

If the median weight 1s 100 grams, there are about 26 potatoes in a bag.

¢

=70,

y =~ 31.

If the median weight 1s 70 grams, there are about 31 potatoes in a bag.

The estimate in b i is an interpolation, and the estimate in b ii is an extrapolation. So, the estimate in b i is likely to be more reliable.

60, income ($ x 10 00) 50

I

30

o

[ o

40

20

37

©

20 10

()

age (years) 0

10

20

30

40

o

a

There 1s a strong, positive correlation between the age of an individual and their annual income.

b

No, the relationship is more likely dependent on the amount of professional experience or qualifications an individual has.

¢

604 income ($ x 1000) ~

50

4%6-...-.---..--.. ..... TR ik i

30

20 =

d

When

0

e

| &7 1 8

4

' B

T m e -;----.-;;.-.

,.G"/.?,

%

/

/

Sy(27, 40). mean point age (years)

0

10

20

30

40

-

= = 30, y =~ 45.

The annual income of someone who is 30 years old is approximately $45 000. This is an interpolation, so the estimate s reliable.

WORKED

38

Sample

A | B

Volume (xcm?) | 3 |

Mass(yg) SUB

1 | List

2 | List

1

3

40

s

4

7

50

2

B

3

4

E

F

G |

H

I

J

K

L

7

16

8

5

12

9

6

10

11

3 | List

=

4

[dic](Real

Y

95

“

el

.

Lo

o

n

.

o

GRAPH] CALC J TEST | INTR / DIST JHI=NE

a

360y 300

'

b

O

240

180

o

120

0

|&

LinearReg(ax+b)

a =19.1171662 b =-17.86376 r =0.98019806 rz=0,96078823 F=EEEE22B,081743

*

o

0

2

.

©

.

¢

[COPY)[DREW]

*

&

60

157

| 40|95 | 50 | 160 | 285 | 130 | 65 | 210 | 155 | 90 | 170 | 190

[d/c] Real

List

6 |

D

SOLUTIONS

So,

o

s

r =~ 0.980.

3

r(cm®) 4

O

8

10

12

14

16

-

18

¢

There appears to be a strong, positive correlation between the volume of a sample of silver and its mass.

d

The data point (7, 160) which corresponds to sample D appears to be an outlier. We therefore agree with the jeweller

e

that there 1s a fake sample. 1

|8

i

LinearReg(ax+b) a =19.4604316 b =-24.676258 r =0.9987246 rz=_0,99745083 MSe=16.3069544 yv=ax+b [COPY][DRAW]

When

=17,

y ~ 19.5(7) — 24.7 ~ 112

So, a sample of silver with volume 7 cm? would weigh approximately 112 g.

Using technology, the regression line 1s y ~ 19.5x — 24.7.

39 | Studytime(xh) | 7 | 6 | 3 Result (y %) a

LinearReg

|16 | 15 | 11 | 18 | 32 | 20

56 | 42 [ 25 | 80 | 65 | 60 | 85 | 96 | 90 Using technology, the least squares regression line 1s y ~ 2.43x + 32.0.

(ax+b)

a =2.43264433 b =31.9579472 r =0.91326692 rz=0,83405646 MSe=104.88158 y=ax+b

(COPY J[DRAW]

So, there is a strong, positive correlation between the number of hours that a student studies and their examination result. ¢

Yes, this 1s a causal relationship as spending more time studying for the examination 1s likely to cause a better result.

d

When

y =70,

70~ 2.43xz+ 32.0 38 ~ 2.43x r~

15.6

So, Tony studied for approximately 15.6 hours. e

The y-intercept of the line of best fit ~ 32.0. This indicates that 1f a student did not spend any time studying, they

would obtain a result of 32% on average.

The gradient of the line of best fit ~ 2.43 . This indicates that for every additional hour of study, the result obtained increases by an average of 2.43%.

158

WORKED

40

SOLUTIONS

Time (t days) | O |

1

2

3

Height (hmm) | 5 | 5.7 | 5.7 a

cofem) @R

b

So,

¢

|6

|8

7 | 8]

|83

9

|9

9.3

risvery close to 1 which indicates a very strong correlation between the variables.

b =4.91454545 r =0.99265002 r2=0 98535408 MSe=0.03648484

y=ax+b

o

6.2 | 68 | 7.1

nserelesiel,

[0

4

The sign of 1s positive which indicates that the variables are positively correlated. : : : ; . :

An increase 1n one variable results in an increase in the other.

[COPY](DRAW]

r ~ 0.993.

h~ 04879t +4.9145 I

When

t =14,

h ~0.4879(14) + 4.9145

i

When

A =20,

~ 11.7

20 = 0.4879t + 4.9145

15.0855 ~ 0.4879¢

after 14 days, the grass i1s about 11.7 mm high.

t = 30.9 .

the grass reaches a height of 20 mm after about 30.9 days.

B1 | Age(xyears) | 28 | 40 | 21 | 38 | 30 | 26 | 18 | 32 | 25 | 29 | 20 | 24

Time(ymin) | 20 | 32 | 15 | 40 | 26 | 25 | 19 [ 28 | 21 | 25 | 16 | 22 a

|B

FEIEm @O LinearReg(ax+b) a =0.92969136 b =-1.5606535 r =0.89822155 r2=0.80680195 MSe=10.4504043

y=ax+b

[COPY](DRAW]

So, r ~ (0.898. There 1s a strong, positive correlation between the age of contestants and the time taken to complete the task.

b

i

From the screenshot in a, the linear regression line is y =~ 0.930x — 1.56.

il

The gradient ~ 0.930. complete the task.

This means that an increase of one year in age will add about 0.93 minutes to the time to

42 | Length(mm) | 95 | 83 | 91 | 82 |

Weight (g) | 5.4 | 45| a

6

50 | 4.1 | 3.7

Aweight(g)

3

o

9

|26

|45 | 3.1

|47

b

5 1

o ©

©

°

60

77

|48 | 3.6 | 4.2

a =0.07291841 b =-1.5723113 r =0.96187039 r2=0.92519465 MSe=0.05184722

0

R %o o

37|51

72 |

LinearReg(ax+b)

y=ax+b

’

1 0 LA, d

75 [ 62 | 79 | 63 | 81 | 69 | 94 | 88 |

(COPYJ[DRAW]

So, 7 ~ 0.962.

70

_

|

length (mmj 80

90

_

5

100

¢

There 1s very strong, positive, linear correlation between : length and weight.

From the screenshot in b, the equation of the least squares regression line is y ~ 0.0729x — 1.57.

e

1

When

2z =110 mm, y ~ 0.0729 x 110 — 1.57

i

When

2z =70 mm, y ~ 0.0729 x 70 — 1.57

~ 6.45 g

f

The prediction in e 1i is more likely to be reliable, as it is an interpolation.

~ 3.53 g

WORKED SOLUTIONS

43

a

AY

159

As x increases, y generally decreases. o

°

o

So as the rank of x increases, the rank of y generally decreases.

o o

o o

o o

o

-

€I

A rank of y 8

o

Looking at the first few points, we see that B is the correct rank scatter diagram. o

O

o o

4

o :-_

)

0

b

o

0O

2

4

6

T = 10

&8

rank of x

The rank scatter diagram has a strong, negative linear correlation, so the correct value of Spearman’s rank correlation coefficient is g ~ —0.960 (€).

k4 | Number of matches played (x) | 11 | Highest score () a

AY

:

I

o°

0

2

8

92 | 65 | 71 | 82 [ 21 | 7|55 o

30 ] 60

5 | 10 | 16 |

|8 [ 79|

18 | 60 [ 51

y=ax+b

2018 12

16

20

1, ~ 0.829.

-

¢ | Number of matches played () | 11 | 5 | 10

d

|16 |

2

4

4

7

|12

10|

2.5 | 4.5 | 4.5

Highest score ()

92

165 | 71 [ 82|

21 [ 7|55

|8

|79 ]

18 | 60 | 51

rank ofvy

121

71

3

|11

9

2

e

LN g 50 955s b —0.43485915

(1|

15|

6 | 8

(10

|20

8 | 20|

91

8

[11

2 [ 1|

rank ofx

()R

[1]

&5

MSe=1.93644366 s ~ 0.930.

35 -39

10

40 - 44

46

45 - 49

43

o0+ Total

15 114

4

The scatter diagram in @ shows a non-linear trend.

y=ax+

45 | Time (min) | Frequency

6

Using d, there appears to be a strong, positive, non-linear correlation between number of matches played and highest score.

Lo :B j 3%223232

So,

4

r =0.82905743 r2=0.68733622 MSe=273.382779

So,

8

4 |

Redfornl) ([{TRea)

{

4

2 |

ST b =26.6945945

b

10

oL 0

20115

a

46 P(40 to 44 minutes) ~ 75 ~ (0.404

b

P(at least 50 minutes) ~ 2 ~ (0.132

¢

P(between 35 and 49 minutes inclusive) & at +14164+ s ~ 0.868

160

WORKED SOLUTIONS

46 | Division | 2017 | 2018 | 2019

a

1

4

5

D

2

6

7

8

3

[

12

14

4

18

10

14

5

20

17

16

Total

61

51

D7 4 «+— number of division 1 players in 2017 tournament

P(player in the 2017 tournament played in division 1) ~ 61

die

P(exactly one 4) = =

23

=

gy 3

P(AUB)=P(A)+P(B)—-P(AN B)

o 078 =0.37+ 0.41 — P(A N B) P(ANB)=0 b

51

Since P(AN B) =10, Aand B are mutually exclusive events.

A and B are mutually exclusive Now

P(ANB) =0

P(AUB)=P(A)+P(B)—-P(ANDB) = P(A) + P(B)

0.55 = P(A) + 0.3 P(A) = 0.25

52

P(AUB)=1-P((AUB)") _ 1

Now

=1 583

©. P(ANB) = 24 Box 0X X

a

1

A

B

Box 0X Y

P

%

2

3

B

y

R

5

)

3

1 R

Ri:::

i

3

B

s

R

B< 2

2

Box {;’. Z

;TR 2 2

b

P(AUB)=P(A)+P(B)—-P(ANB) .11 _ 23 | 5

3

B

-

R

R{i::

P(exactly 2 red balls are drawn)

— P(RRB) + P(RBR) + P(BRR)

=g_ 1,3xBx , 3+1 ixdxd il =5 T 20T 30 23 ~ 60

®0® | | ®@ ® X

i

o

P(blue balls are drawn from boxes X and 7))

— P(BBB) + P(BRB)

—1xix3+ixin | 1 ~ 10T 15 Ll G

2 3

161

162

WORKED SOLUTIONS

iii

P(at most one blue ball is drawn) = P(no blue balls are drawn) + P(exactly one blue ball is drawn)

P(RRR) + [P(BRR) + P(RBR) + P(RRB)]

3.2 1 1 .21 .3_.3_1,3_92_2 4> die 1

WORKED SOLUTIONS 67

Thereare

a

11 +8 + 4 + 2 = 25 students. 11 +38

P(H)=_

o

b

11+ 2

N = P(7T")

_ 19 d

=P(HUT)

) P(T|H)= 11+ 8 _ 8

a

(2)

U

19

114844

68

ol

25

P(plays at least one sport)

— 23 25

3

o

— 13

- 25 ¢

25

a

P(:z:)—(m_g)z,

r=20,1,2

b

P(X =2)=P(2)

—

P(2

P(1)=2,

P(0)=2,

P@)=a D=7 ()_51‘ Since P(x) is a probability mass function,

= 3%

n

{from a}

> P(x;) =1 1=

2424

o

"1

‘

a=

4T

%a =

g36

a= ¢

Since P(X =2) = % a

%

is the greatest probability, the mode of the distribution is 2. 4

P1=§=?=Efi0.0816

4 ppp= 0oty a4, 9 1B T 4 T 49 4 49 Ta9 a9 PLTP2 =3 Since

69

p1 + p2 + p3 = 1 = 0.5,

the median 1s 2.

a

X =223,4

b

Let B represent a blue ticket, and R represent a red ticket. The possible selections that can be made are:

1st ticket

2nd ticket

3rd ticket

-

R

i < 21

B

2

BR

RB (X=2)

BBR

RRRB

RRB (X=3)

l (X=4)

4th ticket

0

3__~B 33 R

1

So,

3 R< 1

3

_+B

5

R< :

_~B R

the probability distribution of X is

P(X=2)=2x35+2x=2=2

P(X=3)=5x3x3+5x§xX5=1p

.

PX=2)

PX=4)=5x{x3X5=1 —

¢

__ 3

2

1

2 1

It is most likely that 2 tickets are drawn, so the mode is 2.

d

E(X)=) ‘

|2]3]4 |2

]3|

x;p;

=2=25 70

P(z)=P(X =z)= 4 (z+6), =123 a

165

P(1)=5(1+6) =

P@)=L1(2+6)=2 (2) 214( ) 2;1 =1 i

3

bEmy;;%Hm)

-

:1(2—1)—#2(%)—%3(;—4)

{from a}

166

WORKED

SOLUTIONS

Cups oftea |

71

0

1

2

3

4

5

Probability | 0.1 | 0.07 | 0.16 | 0.37 | 0.21 | 0.09 a

Russell is most likely to drink 3 cups of tea in one day, so the mode is 3 cups.

b

Expected number of cupsoftea=0x 0.14+1x0.074+2x0.16+3 = 2.79 cups

x 0.37+4 x 0.21 + 5 x 0.09

. on average, Russell drinks 2.79 cups of tea per day. 72

a

Let X be the return from each game. 1 3 8 E(X) =40(13) +20(53) +5(13)

[

Winnings

40 + 60 4 40

—

Blue | Red | Yellow

$40 | $20

7 Probability

>

51

$5

53

i8

140

b

The expected return per game is $11.67 . It costs $15 to play.

So, the expected gain ~ $11.67 — $15 ~ —$3.33 It is not advisable to play this game many times as the player can expect to lose $3.33 on average per game. ¢

Let £ be the number of extra red tickets added to the bag.

P

Winnings .

Probability |\

Dluc

Red | Yellow

$40

$20

$5

1

k+3

8

=5 | 5% |

For the game to be fair, the expected return must equal the cost of each game. 1

k+3

8

E(X) = 40(12+ A) + 20(12+ h) + 5(12+k) = 15 40

20k + 60

2k

124k

40

124k

:

{the cost of the game is $15}

i

20k + 140

2R

20k + 140 = 15(12 + k) 20k + 140 = 180 + 15k bk = 40 k=28 So, 8 extra red tickets should be added to the bag to make the game fair. 73

a

I

1 1 1

b

i

1

2 3

4

n =

1 3

6

= 2 1

4

=3 1

n=4

Let X bethe number of times we get a number greater than 3.

X ~B(1,3) and P(X =) = (4 (3)"(})*

so, (X =9)= ((2)"(2) —4(3)°(2)

AN

~ (0.366

il

ol

Let X be the number of times we get a number greater than 5.

X ~B(4,3) and P(X =2) = ()(D)" ()" So,

P(X>2)=1-P(X

0.6296482639

BRG]

SRR

P(X =16) ~ 0.218 d

[

|g

BinomialCD(10,15,20,> 0.3697883224

SRR

P(X > 16) =~ 0.630

P(10 < X < 15) = 0.370

If more than 8 residents support the construction, then 8 or fewer residents oppose the construction. P(X

L]

< 8) = 0.000102

' Bpd | Bed | InvB |

75

a

Let X be the number of defective items.

n==6,so

X=0,1,2,3,4,5 0r6,

and p = 5% = 0.05

X ~ B(6,0.05)

Using technology,

P(X > 2)=P(X > 3)

Gl

~ 0.002 23

.

)

the manufacturer will have to pay a refund on about

0.002 23 x 100% =~ 0.223%

of boxes. b

2. 22984375::1.:*13

' Bpd | Bed | InvB |

LetY be the number of boxes refunded.

n=10,s0

Y =0,1,2,3,...,0r 10, and p ~ 0.00223

{from a}

i)

i

BinnmialPD(l

b

Y ~ B(10, 0.00223)

0.02185641054

Using technology, P(Y = 1) = 0.0219 So, the probability that Patrick will get a refund for exactly 1 box is about 0.0219. 76

_Bpd | Bed [InvB|

I.ct I be the event of a faulty chip.

P(F) = 0.03 and P(F') = 0.97

e

[f X is the number which are faulty then X ~ B(500, 0.03).

BinomialCD(5,10,500,>

-

0.1140330949

1% is 5 chips and 2% is 10 chips. Using technology, P(5< X < 10)= 0.114

Probability | a

1

3 | 4

| k|

|

5

Since this is a probability distribution,

T

>

P(x;) =1

pthtits= 1

1

1

_

=

Score

-

Caf—

77

Bpd | Bed | InvB.

10,0.00>

168

WORKED SOLUTIONS

b

Let X be the number of 2s rolled.

n = 2400, so X =0,1,2,3,..,0r2400, and p= 5

{from a}

X ~ B(2400, )1 So,

u=mnp -

and

1

i

o = \/np(1 —p)

= /2400 x §x 2

= 300

_

/1600 3

_ w0V V3 V3 78

~40V/3 3

Y ~ B(30, 1) a

u=np

_ 30 —B

6

and

o = /np(1l — p)

X3

b

= N/SO X &+) X 2)

et DegMorn) (d/c] R

BinomialPD| 20,30, D

3.382767959x1508

_ j ’

~ 2.19

P(Y = 20) ~ 3.38 x 108 “

But

6+2y/%

BinomialCD(11,30,30,"

~10.38

-

0.02561625534

P(Y > p+20) =P(Y > 11) ~ 0.0256 79

a

{using technology}

BT

The variable is likely to be normally distributed as the amount of sleep will be generally centred around the mean, but will vary due to factors

b

The variable is not likely to be normally distributed as it is a discrete variable. The number of lollies may vary from bag to bag, so the distribution may appear symmetric.

¢

The variable is not likely to be normally distributed as it is more likely that there would be more people younger than the mean age than there are older. The distribution may be positively skewed.

80

C

L

such as diet, exercise, and so on.

A

A and B both have the same mean, and € has a greater mean.

B has a greater spread, and hence a larger standard deviation than A.

B

a

=4,

o =1

corresponds to €

b

1 =2,

0 =2

corresponds to A

¢

=2,

o =3

corresponds to B

WORKED SOLUTIONS

—9

82

83

—1

23

7

11

a

Approximately 68% of the population lies between 25 and 35.

b

Approximately 95% of the population lies between 20 and 40.

¢

Approximately 99.7% of the population lies between 15 and 45.

X

a P(X 1050) ~ 0.0388 about 3.88% of containers overflow.

1020 1050

X (mL)

169

170

WORKED SOLUTIONS

¢

LetY be the number of containers which overtlow.

n="75s0Y=01,2,3,..,0or75

and p ~ 0.0388

{from b}

Y ~ B(75, 0.0388)

Using technology,

P(Y < 3) =~ 0.668

2

(JZc)ea) BinomialCD(3,75,0.03p 0.6678189784 ]

N REERRNNG:

85

Let the volume of a randomly selected drink be X mL.

So, X ~ N(254, (2.3)). d

=]

Deslforml] (dic)Real

Normal Data

=

C.D :Variable

Lower Upper o 1L

[DeglForml] [d/c)Real

Normal C.D p =0.5

P -9O%1w099° : 254 2.3 1254

Z:Low=-3.913x10°9 Z:Up =0

:,

"None JHNSII

204

P(X 254+ 2

x23)=P(X

254

> 258.6)

258.6

X (mL)

~ (0.0228

we expect about 0.0228 x 80 =~ 2 drinks to have volume at least two standard deviations above the mean. d

1|1

|B

Normal Data Lower Upper o

@) Rea)

C.D :Variable : 250 :9x1099 2.3

B

Do)

@R

Normal C.D P =0.95899408

Z:Low=-1.7391304 Z:Up =3.913x%x19°

:254

:

’ 250

P(X > 250) = 0.959

:

: 254

X (mL)

the operator’s guarantee that at least 95% of drinks will have volume at least 250 mL is valid.

ii

Suppose X ~ N(254, (2.5)%). &

Deg)orml] (d7c]Real

Normal

Data

Lower Upper C

L

C.D

:Variable : 250 :9%109°

12, o : 204

E

Degorm] (dic]Real

Normal

P

C.D

=0.9452007

Z:Low=-1.6 Z:Up =3.6x10°°%

| None |MEB]I

P(X > 250) ~ 0.945 about 94.5% of drinks will have volume at least 250 mL. So, the operator’s guarantee 1s no longer valid.

250

254

X (mL)

WORKED SOLUTIONS

86

a

Letthe volume of sauce in a randomly selected bottle be X mL. X

~ N(SUO.,

171

HfiElen) ok

NormCD(-9x1099,495,2.>

(25)?)

.

0.02275013195

P(X < 495) =~ 0.0228 IR

b

Let Y be the number of bottles which require extra sauce. Y ~ B(200, 0.0228)

et g erm) (/)R

BinomialCD(8,200, 200>

{from a}

[

2990 E4 £HSR00

P(Y > 8) = 0.0892 SRR

87

a

Letthe completion time of a randomly selected run be X seconds.

X ~ N(45, 47) i P(X < 40) ~0.106 ii

0 Hetelform] [fea]

P(two consecutive runs under 40 seconds) = [P(X < 40)]

9

NormCD (-9x1099,40,4, 4>

eah

SRR

e

0.01116187468

[

~ 0.0112 b |

ER)Re)

Normal Data

B

C.D :Variable

Lower Upper c

Cefom) R

Normal C.D p =0.29016878

144 + 47 14 + 45

Z:Low=-0.25 z:Up =0.5

] LIST

Pl 44 45 47

P(44 < X 4) ~ 0.0228

=

X

(kg)

about 2.28% of babies have birth weights in excess of4 kg. il

|e

Normal Data Lower Upper o L

(=3

C.D

Normal

:Variable ‘9 4 :0.3 :3.4

C.D

P =0.88603864 Z:Low=-1.3333333 Z:Up =2

[(None |INELD

:

3

P(3< X 27) ~ 0.401 we expect about

200 x 0.401 ~ 80

runners to have completed the course in more than 27 minutes.

b

8

Fedfomd

Inverse Data Tail Area

4]

save

k 26

Rad Fornd)

Normal :Variable :Left :0.4

Inverse Normal xInv=24.,9866116

- 4

Res:None

v

X (minutes)

P(X 0. Nowif

1

.

v X

(e

o

i

WORKED SOLUTIONS

12

179

y=2a"+2z+1 dy 5 2 + 2 dw—?’x

Now when z = -1,

*.

y=(-1)>+2(-1)+1

= —1-2+1 — 9

the point of contactis

(—1, —2)

', the tangent has equation whichis

and

= 3(—1)% + 2

— 342

and the gradient of the tangent is 5.

y =5(z+ 1) — 2 vy = 5z + 3.

Using technology, the tangent meets the curve again at (2, 13).

[EXE]:Show coordinates

voberav 5

#/

a

ax2

10

£

2

When

=12,

//

—

INTSECT

Y=13

{froma}

y:%_22+1

—2—-4+1

:§_2$

= —1

Now the gradient of the tangent at x = 2 1s —5.

o7 —2(2) =5

i

A

— ({1

—a

-

a 2 4 2 y=-—x°+1=-—x*+1

yzg—qfi+1:ay4—x?+1 dy =

P

»d"éc,:#f#

Eéfi

13

10

So, the point of contact is (2, —1).

i

The equation of tangentis

—a

y = —5(x — 2) + (—1) y=—or+10—1

y = —or +9Y

—a

- =1 a—=4

14

a

f(x) =2° — 22

f'(z) = 322 — 2 Now f(1)=1°—-2(1) = —1 and ff(1)=3(1)*-2=1 ..

the point of contact is (1, —1) and the gradient

Now f(2)=%—$=0and f/(2)= 3+ 12 =3

of the normal 1s —1.

the equation of the normalis

.

whichi1s

y = —(x —1) —1

.

the point of contact is (2, 0) and the gradient of the normal is — 2.3

y = —ux.

the equation of the normal is

whichi1s

15

a

y = —%(:1: —2)4+0

y = —%::r: - %.

f(z)=ax’+bx i

When

=2,

2+ 3y=—4 3y = —6

y = —2

i

f'(z)=2ax+0b Now the normal has equation

Jy=-4—=zx

£(2) = -2

a(2)? + b(2) = -2

4a + 20 = -2

x + 3y = —4

y=—3— 3%

-. the normal has gradient — at © = 2. .

the tangent has gradient 3 at x = 2

" N w2023 2a(2) +b=3 da +b=3

180

WORKED SOLUTIONS

b

Solving

the

da + 2b = —2

system of equations

simultaneously gives

[athlDeg/Bornl]

an X+bn

o

a = 2, b = —5.

E

[dic)Real

Y=CE

an X+bn

|0

a

c

(SOLVE]JEIA3(CLEAR][ EDIT )

¢

([athfegMornl] [(dic)Real Y=Cn

x

2

[REPEAT)

From aii, the normal L has gradient —%.

.

the tangent L has gradient — %

Now from b, f(z) = 22% —5x

and f'(z) = 4z — 5. 4r —5=—=

the tangent has gradient —% when

.

122 — 15 = —1 122 = 14

3

14 7

T=173 =%

() =2) -5 . 16

a

=-%

Qhas coordinates i

f(x)

(&, —%).

isincreasing for % po b

[Deg Hormd)

SUB

List1 | List2 | List 3 | List 4

1R

20.8

3

22.1

2/

4|

23.5|

B

DegHorn)

2—Sample

21.2

24.1|

22.3|

1)

2)

20.7

1) 22.4 ||Freq(2) EPINEDEP > | | (List ) Var |

YIS

tTest

2-Sample

:

t

i

tTest

Y2 =2.4270426

P

=0.01239673

df

: :1

Using technology, the test statistic ¢ ~ 2.43 ¢

[Deg) (Morm)

=20

x1 L1x2

=22.7 =21.76

¥

and p-value ~ 0.0124.

Since p-value < 0.1, we have enough evidence to reject Hy on a 10% level of significance. .

we conclude that the company’s claim 1s valid.

MIXED QUESTIONS SET 5 1

a b

f(1)=1"-2(1)=-1

y

g(1) = V1- =1

f(z)=2*—2x

¢ is one-to-one, so it is invertible. f 1s not one-to-one, so it 1s not invertible.

C

g

»

(ffl)

_ 1 g(x) ~ 7

s

-

r = g(4) 2

a

OTP=90°

/

\/

=>

{radius-tangent}

AOPT is right angled at T. cos

v = %

= cms_l(%) ~~ 72.5° b

Areaof AOPT = % X 9 x 30 X sin «

Area of sector OQT = 32%0 X T2

— 135 sin@ cm?

_= a 359 X T X 9 2

_= dam g cm 2 So, shaded area = area of AOPT — area of sector OQT — 135 sine «v Pln

— 135 Sil‘l(COS_l(%)) - i—g cos_l(%)

{using a}

~ 77.5 cm?

3

a

Francesca adds $0.50 in the first week, $1 the next, $1.50 the next, adding an additional $0.50 each subsequent week.

in the nth week, Francesca adds 0.50n dollars to her money box. Now the last week before her 11th birthday is the 51st week. in the last week before her 11th birthday, Francesca added $0.50 x 51 = $25.50 to her money box.

b

Let P(n) dollars be the amount Pierre had added to his money box after n weeks, and F'(n) dollars be the amount Francesca had added to her money box after n weeks.

Pierre adds $10 each week, so after n weeks he has added 10n dollars.

So,

P(n)=10n P(8) =10 x 8 =80

After 8 weeks Pierre had added $80 to his money box.

210

WORKED

SOLUTIONS

From a, Francesca adds 0.50n dollars in the nth week, so after n weeks she has added

0.50+1+1.50+

.... + 0.50n

Now

1.50 4+ .... + 0.50n

0.50+1+

dollars.

1s an arithmetic series with ©v; = 0.5 and

d = 0.5.

NS

(2x0.54+(n—1)x0.5)

|3

oS

0.50 +1+ 1.50 + ... + 0.50n = = (2u; + (n — 1)d)

( 1+

0.5n — 0.5)

51 + 0.25n2

o

O

NI

(0.5 4+ 0.5n)

So,

F(n)=0.25n+ 0.25n°

F(8) =0.25 x 8 +0.25 x 8% = 18 After 8 weeks, Francesca added $18 to her money box. ¢

There are 52 weeks in 1 year.

Now

P(52) =10 x 52 = 520

and .

F(52) = 0.25 x 52 + 0.25 x 52% = 689

after 1 year, Pierre had $520 + $100 = $620 money box.

in his money box, and Francesca had $687 + $100 = $789

in her

So, Francesca had more money in her money box after 1 year.

4

a

Bis

(6,6,0)

and Cis

(0,6, 0).

b

Volume = 3 X area of base x height

:%x6>=—-13 qg=—13 b

i i

f(0)=3-4"=2 Asz—o00,

-

47— 0

the y-intercept is 2. andso

B(—2,—13

y —3

(

)T

y = 3 1s the horizontal asymptote.

2

d

Therangeis

a

The line passes through 90 — 120

2—-0

{y |y < 3}. —30

= —

(0, 120)

and

(2, 90), so the gradient is 120

= —15.

2

This means that the amount of money left in the subscription account decreases by $15 each month. :

:

:

i.

.

The y-intercept is 120. This means that the initial balance was $120.

90

60 20

0

\H r (months) O

1

2

b

The gradient is —15 and the y-intercept is 120, so the equation of the line is y = —15x + 120.

¢

The account runs out of money when

vy =0

—152 + 120 = 0 15z = 120 r =8 The account will run out of money after 8 months.

3

by (S)

a

y=-+3=az""+3 The normal to y = % + 3 at the point where

.

x = 2 has gradient 2.

the tangentto y = Z + 3 at the point where xIr

So, when

x = 2,

= 2 has gradient —%.

j—i:—% a

1

a

1

227 9 4 (l

2

=

2

When 2 =2, y=%+3=4 So, the point of contact is (2, 4). .

the equation of the normal is

y = 2(x — 2) + 4 =2r—4+4 = 21

3

4

5

6

7

>

214

WORKED SOLUTIONS

We use technology to find where the normal meets the curve again:

[EXE]:Show

coordinates

The normal meets the curve again at (—0.5, —1). 4

a

There are twelve equal angles at the centre of the dodecagon. AQOB

b

360°

=

5

Areaof AAOB

— 30

= % x 6 x 6 X sin 30° — 9cm?

¢

Areaof dodecagon = 12 x 9

= 108 cm” ab ~61

804 cof fee expenditure ($) 6\6“ oo

ool oo

=ES-

S

E

o

/

5

20 0

°

P(32,56)

s hours worked

10

0

30

20

>

50

35 40

¢

From the graph, if James works a 35 hour week, he spends about $61.

d

There is a strong positive linear relationship between the length of time James works and the amount he spends on coffee. Since the prediction in € was an interpolation on strongly correlated data, it 1s a reliable estimate. Weight (w g)

2

< 310

Ol O

= S

¢

The data 1s slightly negatively skewed.

d

The modal class is the interval

e

Mean

=

A frequency

S~

Qo Do

w < 325

w < 335

Weight of cereal boxes

W

b

w < 320

o Qo

x KD S Ot O v NN NN

N

W

310 < w < 315

w < 330

b

Ul O

300

< w

Frequency

R

a

/

6

e

N

5

320 < w < 325

A,

305

310

315

320

325

330

335

.

weight(g)

because it has the highest frequency.

312 + 320 + ... + 324

7696

24

24

~ 320.67

The mean of the sample is reasonably close to the average weight that the manufacturer claims. 7

a

Height of cylinder = total height of silo — height of cone — height of hemisphere

_

Sramle

= 3.5 —0.8 — -

m

e

el

4

N

R

)

= 1.8 m 3.0m --------------------

WORKED SOLUTIONS

b

¢

Volume of cone = %wrzh

=1 x 7 x0.9%x 0.8 ~ 0.679 m” 2 Volume of cylinder = 7r=h = : 2

Let the slant height of the cone be s m.

2~ (05 + (09)

s =+/(0.8)2+(0.9)2 ~ 1204 m

09718

1.

~ 4.58 m

{s >0}

Volume of hemisphere = % (577°)

_ 1,4

Surface area of cone = 7rs

N2

~m x 0.9 x 1.204

15 ; g

9

3

~ 1.5 m

~ 3.40 m?

-,

total volume =~ 0.679 + 4.58 + 1.53

= 3 ~6.79m

Surface area of cylinder = 27rh

m"

=2x7mx09x%x1.8

Now

6.79 m® = 6.79 kL

~ 10.2 m*

So, the capacity of the silo 1s about 6.79 kL.

Surface area of hemisphere = 277~

=2 x 7 x (0.9)

~ 5.09 m? total surface area ~ 3.40 + 10.2 4+ 5.09 m? ~ 18.7 m* So, about 18.7 m? of sheet metal was used to make

the silo. 8

P(.’L)

a

~>

xI

E

1000000

P(1000000) ~ T

b

Percentage error =~ |Va—-VEg] T X 100%

~ 72400

~

We estimate there are about 72400 prime numbers

less than 1000 000.

| 72400 — 78 498 | « 100%

;

~1.8%

558

¢ P(z)~ 2000 when é ~ 2000. Using technology, = ~ 19 785.

el Eq:——=2000 n x x=19785.39905 Lft=2000 Rgt=2000

So, there are about 2000 prime numbers less than or equal to 19 785. 9

The distance travelled d 1s directly proportional to the square of the time taken .

d = kt* where k is a constant.

When d =19.6m, t =25, so

19.6 = k(2)° 19.6

k= Vin

4.9

So, d = 4.9¢t°.

a When t =3s,

d=4.9(3)

= 44.1m

b

When d =100m,

100 = 4.9t

M2

.

I O

4.9 100

O

215

216

WORKED SOLUTIONS

QR 2 — =2z“ 2 |+ 8 Q2

{Pythagoras}

©. QR=+/22+64

{as QR >0}

Also QS = PS — PQ =— B

R

8 km

P

g

\flfkm

I

1"y

11 km

So, the length of pipeline under the sea is v x2 + 64 km,

and the length of pipeline overland is (11 — ) km. the cost C'(x) = 5V x? + 64 + 3(11 — )

= 5V

million dollars

x? + 64 + 33 — 3z million dollars.

TGT C' (million dollars)

¢

74 % 70

\

11

68. /( 11, ,68.0)

(6,65)

0

The weight of the radioactive substance at the end of each year forms a geometric sequence with common ratio r = 0.965.

So, percentage decrease = (1 — ) x 100%

= 0.035 x 100% = 3.5% b

W(300) =5 x (0.965)" ~ (0.000114

~1.14 x 1074 The weight of the substance after 300 years is about

¢

Weneedtosolve

1.14 x 10~* grams.

W(t) =1

[EXE]:Show coordinates

5 x (0.965)" =1 Using technology,

t ~ 45.2

it will take about 45.2 years for the weight of the substance to fall below 1 g.

(2, 4) is closest to Q, so we estimate a lead concentration of 47 ppm at (2, 4). (—2, —3)

is closest to R, so we estimate a lead concentration of 62 ppm at (—2, —3).

(—4, —1)

is equally closest to P and R, so we estimate a lead concentration of

28 4 62

0%

S lies in the original cell R, so we construct the perpendicular bisector of [RS] within this cell.

_ 45 ppm at (—4, —1).

y (km)A

We then construct the perpendicular bisector of [PS] within the original cell P.

P

Finally, we remove the segment of the perpendicular bisector of [PR] which now lies within cell S.

(2,4)

1

__#',,——**"“*4“imh

(—4, —1)

E

©

+5

62455 _ 58.5 ppm at (—2, —3). 28+632+55 ~ 48.3 ppm

S?

R

Y

at (—4, —1).

y=2>4+ax®>+bxr+3

The function passes through (1, 8), so

(1)? +a(1)® + b(1) + 3 = 8 l+a+b+3 =28 a+b+4=8

a+b=4

Now

At

dy L

..(1)

= 322 + 2ax + b.

(1, 8) the tangent has equation y = 2x + 6 which has gradient 2.

3(1)% +2a(1) + b =2 3+ 2a+b=2

20 +b=-1

...(2)

We solve equations (1) and (2) simultaneously using technology.

a=—5and

b=9

S|

T~

is now equally closest to P, R, and S, so we estimate a lead

concentration of

(km)

T—

is now equally closest to R and S, so we estimate a lead

concentration of

a

Q

is still closest to Q, so its estimate is unchanged.

(—2, —3)

9

_15

O

Weg Forn]) (/)sl BaAtbai=ts & 1

d

o

1

1

SOLVE|(NARIICLEAR) EDIT

Math)DeglBornl]

an X+bn 4

[REPEA[

Y=Cn

]

[dic]Real

WORKED SOLUTIONS b

From a,

y:x3—5$2+9$+3

229

and

Wdx 302 10p+9 Now when z = —1,

y=(-1)° =5(=1)*+9(-1)+3

and

j—i =3(-1)* —10(—=1) +9

=—1—-95-943 So the point of contactis

(—1, —12)

The equation of the normal is

and the gradient of the normal is — %

y = —o5(z — (—1)) — 12 e

g

Y= 7237 10

a

=3+ 1049

265

%3

We need to find how long it will take for the future value to fall to €0. 1% =4.6,

C/Y =12

PV =-500000,

PMT

= 3000,

FV =0,

P/Y =12,

Compound

1

N ~ 266

Raman will be able to withdraw €3000 for 266 months, and then less in the 267th month. b

N=4x12=48, ClY

I% =4.6,

PV = -500000,

PMT = 3000,

P/Y =12,

=12

N=15x12=180, C/lY =12

i

a

=48

IWHTI

Interest

=4.6

I% = 4.6,

PV = —443028.05,

FV =0,

P/Y =12,

Florn) Compound n, =180 V. =(

g

e

Interest

TR

B

A= 1

|

C

By the cosine rule in ABAP: -~

2

cos BAp —

-7

2 X7

=

2 _

i

—4 42

x10

C

-

P

a

~ 18.2°

By the cosine rule in AAPC:

7em, 41 .

A5

AD — A =60 £N° — BAP CAP

~ 41.8°

— cos_l(%fi)

CP? ~ 10% + 7% — 2(10)(7) cos 41.8° CP =~ /102 + 72 — 2(10)(7) cos 41.8°

~C

cm

{angles in an equilateral triangle}

EGOD—L%QG

133

BAP

BAC = 60° .;

cos BAP = 540

12

Compound

e

Raman can afford to withdraw €3411.82 each month.

b

|P L "TII% —

E?Yi?33028.0463

FALT & 3411.52

11

=0

N

After 4 years, the balance 1s €443 028.05 . ¢

FV

1%

_

PP PRLEY

i3 808" n

FV ~ 443 028.05

Interest

.. CP=~6.68cm

Painting | Sketching | Sculpting | Sum Male

30

30

15

30

Female

20

15

25

60

Sum

20

20

40

140

The expected number of male sculptors is XD

140

x2., we reject the null hypothesis that gender and choice of art speciality are independent. We therefore conclude at the 1% significance level that gender and choice of art specialty are dependent.

MIXED QUESTIONS SET 10 1

a

Area = area of two semi-circles + area of rectangle

_T_

= 7% + 100 x 2r

"4

— 7 x 12.52 + 100 x 25 — 156.257 + 2500 m° b

Using m~ 3,

Lawn

v >

area~ 156.25 x 3 + 2500

¢

~ 2969 m”

o

.

Percentage error

_ Va- Vel

100%

VE

2969 — (156.257 + 2500) | -

156.257 + 2500

~ 0.73% 2

a

x 100%

{2 significant figures}

Total number of Year 7 students = 30 4+ 27 = 57

Let A represent a student selected from class A and B represent a student selected from class B.

or BB)

P(same class) = P(AA

=30« 204 57

‘—-v-—"'

27, 2

1st student

=-

56

i .

2nd student

29 56

A

27 =06

B

=5

A

5

B O

A
0.05, we do not have enough evidence to reject H accept that the cards are distributed as claimed.

at a 5% significance level, so we

R1Al

Total |8 marks|

9

a

a=1

Al

) k= —2

Al

WORKED SOLUTIONS

b

0

Area =

= / = /

—3 0 —3

0

—3

253

f(x)dx

—2(x —1)(x + 3) dx

(M1)

(—22% — 42 + 6) du

= [-22° — 222 + 62]_,

(M1)

=0— (=3(=3)" = 2(=3)" 4+ 6(-3)) = —(18 — 18 — 18)

— 18 units”

Al Total |5 marks]

10

a

AN

ABC = 360° — 230° — (180 — 110)° 110°

AN

AQ\

M1

= 60°

Al

B 230°

C

b

After 2 minutes, Alanis (600 — 3 x 120) = 240 m from B.

\%8

In the same time, Belinda cycles 8 x 120 = 960 m.

M1

Alan

240 m

6027

-

Let the distance between Alan and Belinda be D m.

B

Using the cosine rule,

D? = 240% + 9607 — 2 x 240 x 960 x cos 60° 060 m

M1

= 748 800 D =~ 865

Al

After 2 minutes, the distance between Alan and Belinda 1s about

865 m.

Belinda

Total |6 marks]

11

a

N=3 PV = —-36995 PMT

=0

F'V = 21600 P/Y =1 @/

yL=]l

1% ~ —16.420

MI1Al Al

The annual rate of depreciation was about 16.4%.

b

N =10 1% ~ —16.420 PV = —36995 PMT

=0

P/Y =1 C/Y =1 FV

~ 6154.30

M1AL1 Al

After 10 years, we would expect the forklift to be worth $6154.30 . Total [6 marks]

254

WORKED

12

Hy:

SOLUTIONS

An adult’s age group 1s independent of their opinion about their government’s handling of the COVID-19 pandemic.

Using technology, p-value ~ 0.495.

Al A2

p-value > 0.1, so we do not have enough evidence to reject the null hypothesis, and therefore should accept it.

R1A1

We conclude that at a 107% level of significance, age group and opinion are independent. Total [5 marks]|

13

Using technology, volume ~ 0.0160 x (height)?-°" where r ~ 1.00.

Al Al

Since 7 ~ 1 for the model in a, and the power 3.00 1s consistent with volume being proportional to the cube of a length, it is reasonable to believe that the models are mathematically similar.

R1Al

Using the model from a, if height = 5.17 m then volume ~ 0.016 x 5.17°

~ 2.21 m*

M1

This is about 5.3% more than the actual volume of David, so we conclude the replicas are not to scale.

R1

Total |6 marks]|

14

k=1-0.21-0.36—-0.16 — 0.04 — 0.01 = (.22

Al

Expected number of errors = 25;]:1:1 D;

i=

=

0x021+1x0364+2x0.22+3x0.164+4x0.04+5

x0.01

M1

= 1.49 errors per 100 words.

Al Total |3 marks]

PAPER 2

1

a X ~N(5.38,0.62%)

M1

P(X 6) ~ 0.1587 ~ 0.159

Al

il

Christina would expect ~ 50 x 0.1587 ~ 8 cherries to have mass greater than 6 g.

ili

Let Y be the number of cherries in Christina’s sample with mass greater than 6 g.

MI1Al

Y ~ B(50, 0.1587)

\%8

Using technology, P(Y > 4) =~ 0.967 i

0.9

Al

A sugar content (g)

L

A2

0.8 o

0.7

P

0.6

i

%

° o N6

iil

o

i5

5

53

54

56

55

mass (g)

6

52

From the scatter diagram, the data appears non-linear. We therefore favour Spearman’s rank correlation coefficient rather than applying Pearson’s correlation coefficient directly. Mass (g) rank of mass Sugar content (g)

rank of sugar content |

R1

5.17 1 5.84 | 6.01 | 5.74 | 4.88 | 5.41 | 5.62 | 4.78 | 4.89 | 5.20 4

9

10

8

2

6

7

1

3

D

Al

0.66 | 0.83 1 0.92 |1 0.75 |1 0.59 | 0.71 | 0.75 | 0.60 | 0.58 | 0.65

5

9

10 |

7.9

2

6

7.5

3

1

4

Al

WORKED SOLUTIONS

iv

Using technology,

rs =~ 0.936

v

There i1s a strong positive correlation between the variables.

255

A2 Al Total [16 marks]|

a

The data begins with high tide at ¢ = 0. The cosine model 1s more appropriate, since it starts at its maximum value.

b

.

Amplitude =

maximum

R1A1

— minimum

:

. 33-09 S 2 o 2

Al Al

a~1.2

¢

Mean water depth =

maximum

33409

~ 2.1

-+ minimum

2

.

Al

d~ 2.1

d

Al

Period = time between high tides ~ 12.4 hours 360

-

& 12.4 360

b~ o~ e

Al

29.0

Al

h(t) =~ 1.2co0s(29.0t)° + 2.1

M1

h(14) ~ 1.2 c0s(29.0 x 14) + 2.1 ~ 2.93

..

f

Al

the depth of the water is about 2.93 m.

h(t) =14

when

1.2c0s(29.0t)°+2.1=14

1.2 ¢08(29.0t)° = —0.7 0.7

cos(29.0t)° = = For

7

(M1)

0 3 hours screen time) = 4464+84+..+7T+74+9 32 95

9+7+7T+9

¢

P(Year 11 N < 1 hour screen time) = o

d

P(> 3hours | Year 12) =

9 11+14+9

_ 9 34

e

Hy:

Year group and the amount of screen time are independent

f

6

g

p-value = 0.770

h

0.770 > 0.1,

(0.76958....)

so accept H

Year group and the amount of screen time are independent.

M1

Al A2 (M1)AL1 Al Al Al A2 R1 Al Total [14 marks]

3

a

35°

b

FBC = 180° — 35° = 145°

Al

BFC = 180° — 145° — 29° = §° In ABCEF,

FC

7.0

sin 145° FC

sin 6° —

(AT1)(A1) M1Al

7.0 Sjin 145° sin 6°

— 41.154 .... ~41.2m

¢

InAACE,

.

sin29”

o

FA

= ———

FA = 19.952 ....

The height of the flagpole 1s about 20.0 m.

Al (M1)A1 Al

WORKED

d

SOLUTIONS

In AACF,

tan 29° = AC

=

19.952 ...

(MT)

AC 19.952.... tan 29°

(A1)

= 30.994 ... ‘.

(M1) Al

total distance = 35.994 .... x 4 ~ 144 m

Total |13 marks]

i

2

Yy =+voxr —1

3

2.9

3

3.3912 | 3.7417

4.5

3.9 4.0620

4.3589

4.6368

4.8990

2.9

6

5.1478

2.3802

1 2

h:

/

A2

(A1) 6

vVoxr — 1dx

J2

X

262

%(%) (3 + 5.3852) 4+ 2(3.3912 + 3.7417 + 4.0620 + 4.3589 + 4.6368 + 4.8990 + 5.1478)]

~ 17.215

(M1)A1 Al

6 Vor

—1ldx

2

= 17.222637 ....

A2

~ 17.223

117.215 — 17.223] x 100% 17.223

(M1)A1 Al

~ 0.0464% Area = /

6 2

vV ox — 1 dx — area of triangle

~17.— 22 5 x43x 3

(A1) MI1Al

~ 11.223 m?

Al

Total [15 marks]

X ~N(51.6, 7.5%) Al : “bell curve” shape Al : mean correctly labelled Al : one standard deviation labelled

i T N 1w=>51.6

A (8)

(M1)

P(X < 49) ~ 0.364 -~

36.4% of eggs weighed less than 49 g.

Al

(M1)

P(X > k) =0.2787 k=~ 56.0

Al

Let Y be the number of large eggs selected. Y ~ B(40, 0.2787)

M1

E(Y) =40 x 0.2787 = 11.148 eggs

(M1)

2 of 40 = 16 P(Y = 16)

(accept 11 or 11.1)

(A1) Al

=~ 0.0328

(M1)ALT

T =5H3.1 s~ 13.0 I

Hy:

A3 pog =516

Hy: po > 51.6

Al

WORKED SOLUTIONS

i h

263

p-value =~ 0.267 t ~ 0.630

A2

0.267 > 0.1, so we accept Hy

R1

The eggs are not significantly heavier than last year, so the farmer is not correct.

Al Total [21 marks]