OXFORD
INSIGHT
MATHEMATICS GENERAL
12 HSC COURSE 2
JOHN LEY MICHAEL FULLER
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OXFORD
INSIGHT
MATHEMATICS GENERAL
12 HSC Course 2
JOHN LEY MICHAEL FULLER
1 Oxford University Press is a department of the University of Oxford. It furthers the University’s objective of excellence in research, scholarship, and education by publishing worldwide. Oxford is a registered trademark of Oxford University Press in the UK and in certain other countries. Published in Australia by Oxford University Press 253 Normanby Road, South Melbourne, Victoria 3205, Australia © John Ley, Michael Fuller 2014 The moral rights of the author have been asserted First published 2014 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without the prior permission in writing of Oxford University Press, or as expressly permitted by law, by licence, or under terms agreed with the appropriate reprographics rights organisation. Enquiries concerning reproduction outside the scope of the above should be sent to the Rights Department, Oxford University Press, at the address above. You must not circulate this work in any other form and you must impose this same condition on any acquirer. National Library of Australia Cataloguing-in-Publication data Oxford Insight Mathematics : general 12 : HSC course, pathway 2 / John Ley, Michael Fuller. ISBN 978 019 552378 2 (paperback) For secondary school age. Mathematics–Study and teaching (Secondary). Mathematics–Australia–Textbooks. 510.76 Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this work, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL) under the Act. For details of the CAL licence for educational institutions contact: Copyright Agency Limited Level 15, 233 Castlereagh Street Sydney NSW 2000 Telephone: (02) 9394 7600 Facsimile: (02) 9394 7601 Email:
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1 Credit and borrowing
1
Review set 2A ............................................................. 64
1A Flat-rate loans ...................................................... 2
Review set 2B ............................................................. 65
1B Credit cards ........................................................... 6 1C Reducing-balance loans ..................................... 12 1D Comparing loans ................................................. 19 1E Comparison rates................................................ 20 1F Future and present value .................................... 21
Review set 2C ............................................................. 66 Review set 2D ............................................................. 67 Examination question ................................................ 68 3 Further algebraic skills and techniques
69
Review 1 ..................................................................... 23
3A Simplifying algebraic expressions ...................... 70
Review test 1 .............................................................. 23
3B Algebraic fractions .............................................. 72
Review set 1A ............................................................. 25
3C Indices ................................................................. 72
Review set 1B ............................................................. 26
3D Equations............................................................. 76
Review set 1C ............................................................. 27
3E Practical equations ............................................. 81
Review set 1D ............................................................. 28
3F Changing the subject of a formula...................... 84
Examination question ................................................ 29
3G Simultaneous equations ..................................... 89
2 Further applications of area and volume
31
3H Graphical solution of simultaneous equations ............................................................. 90
2A Area of a circle .................................................... 32
3I The substitution method ..................................... 92
2B Simpson’s rule .................................................... 36 2C Surface area ........................................................ 40 2D Surface area of a cylinder ................................... 46 2E Surface area of a sphere ..................................... 48 2F Further volume.................................................... 50 2G Volume of composite solids ................................ 52 2H Accuracy of measurements ................................ 55 Spreadsheet application 2.1......................................... 57 Spreadsheet application 2.2......................................... 58 Spreadsheet application 2.3......................................... 59 Investigation 2.1 .......................................................... 59
CONTENTS
CONTENTS
3J The elimination method ...................................... 95 Investigation 3.1 .......................................................... 99 Review 3 ................................................................... 100 Review test 3 ............................................................ 100 Review set 3A ........................................................... 102 Review set 3B ........................................................... 102 Review set 3C ........................................................... 103 Review set 3D ........................................................... 103 Examination question .............................................. 104 4 Interpreting sets of data
105
Investigation 2.2 .......................................................... 60
4A Grouped data ..................................................... 106
Investigation 2.3 .......................................................... 61
4B Measures of location ......................................... 108
Review 2 ..................................................................... 62
4C Review of measures of spread .......................... 113
Review test 2 .............................................................. 62
4D Relative merits of mean, mode and median ..... 117
Contents
iii
CONTENTS
4E Shape of frequency distributions ...................... 121 4F Displaying data .................................................. 126 4G Analysing data ................................................... 132 Spreadsheet application 4.1....................................... 136 Spreadsheet application 4.2....................................... 136 Spreadsheet application 4.3....................................... 137 Investigation 4.1 ........................................................ 138 Investigation 4.2 ........................................................ 138 Investigation 4.3 ........................................................ 138 Review 4 ................................................................... 139 Review test 4 ............................................................ 139 Review set 4A ........................................................... 141 Review set 4B ........................................................... 142 Review set 4C ........................................................... 143 Review set 4D ........................................................... 143
199
6A Standardised score ........................................... 200 6B Frequency graphs.............................................. 205 6C The normal distribution .................................... 206 6D Properties of the normal distribution ............... 208 Review 6 ................................................................... 213 Review test 6 ............................................................ 213 Review set 6A ........................................................... 214 Review set 6B ........................................................... 215 Review set 6C ........................................................... 216 Review set 6D ........................................................... 217 Examination question .............................................. 218 7 Multistage events and applications of probability 219
Examination question .............................................. 145
7A Listing the sample space .................................. 220
Cumulative review chapters 1–4
146
7C Ordered arrangements of n items .................... 223
5 Applications of trigonometry
149
5A Review of right-angled triangles ...................... 150 5B Angles of elevation and depression .................. 156 5C Bearings ............................................................ 159
iv
6 The normal distribution
7B The fundamental counting theorem ................. 222 7D Ordered selections ............................................ 225 7E Unordered selections ........................................ 226 7F Probability ......................................................... 229 7G Probability trees ................................................ 233 7H Expected frequency of an event ........................ 237
5D Area of a triangle ............................................... 168
7I Comparing theoretical and experimental results ......................................... 239
5E Sine rule ............................................................ 169
7J Expected value .................................................. 241
5F Cosine rule ........................................................ 173
7K Financial expectation ........................................ 243
5G Problems involving trigonometry...................... 177
Spreadsheet application 7.1....................................... 250
5H Radial surveys ................................................... 181
Investigation 7.1 ........................................................ 250
Practical activities 5.1................................................ 186
Investigation 7.2 ........................................................ 250
Practical activities 5.2................................................ 187
Investigation 7.3 ........................................................ 250
Practical activities 5.3................................................ 187
Investigation 7.4 ........................................................ 251
Investigation 5.1 ........................................................ 187
Investigation 7.5 ........................................................ 251
Investigation 5.2 ........................................................ 188
Investigation 7.6 ........................................................ 252
Review 5 ................................................................... 189
Review 7 ................................................................... 253
Review test 5 ............................................................ 189
Review test 7 ............................................................ 253
Review set 5A ........................................................... 192
Review set 7A ........................................................... 255
Review set 5B ........................................................... 193
Review set 7B ........................................................... 256
Review set 5C ........................................................... 195
Review set 7C ........................................................... 257
Review set 5D ........................................................... 196
Review set 7D ........................................................... 258
Examination question .............................................. 198
Examination question .............................................. 260
Insight Mathematics General 12 HSC Course 2
261
8A Future value of an annuity ................................ 262 8B Future value using tables.................................. 263 8C Present value of an annuity .............................. 269 8D Present value using a table............................... 270 8E Loan repayments and costs .............................. 276 8F Interpreting graphs ........................................... 278 Spreadsheet application 8.1....................................... 281 Spreadsheet application 8.2....................................... 281 Graphics calculator 8.1 .............................................. 282 Graphics calculator 8.2 .............................................. 282 Review 8 ................................................................... 283 Review test 8 ............................................................ 283 Review set 8A ........................................................... 285 Review set 8B ........................................................... 285 Review set 8C ........................................................... 286 Review set 8D ........................................................... 287 Examination question .............................................. 288 Cumulative review chapters 5–8
289
9 Modelling linear relationships
293
10E Position on Earth’s surface ............................. 325 10F Distances on Earth’s surface .......................... 328 10G Local time ....................................................... 330 10H Standard time zones ....................................... 334 Investigation 10.1 ...................................................... 338 Investigation 10.2 ...................................................... 339
CONTENTS
8 Annuities and loan repayments
Investigation 10.3 ...................................................... 339 Investigation 10.4 ...................................................... 340 Review 10 ................................................................. 341 Review test 10 .......................................................... 341 Review set 10A ......................................................... 344 Review set 10B ......................................................... 345 Review set 10C ......................................................... 346 Review set 10D ......................................................... 347 Examination question .............................................. 348 11 Sampling and populations
349
11A Distribution of the sample means .................. 350 11B The capture–recapture technique .................. 352 11C Tables of random numbers ............................. 354 11D Generating random numbers by calculator ... 356 11E Generating random numbers by computer .... 356 11F A random sample ............................................ 358
9A Straight lines ..................................................... 294
Investigation 11.1 ...................................................... 362
9B Linear functions ................................................ 297
Review 11 ................................................................. 363
9C Intersecting graphs ........................................... 301
Review test 11 .......................................................... 363
9D Linear variation ................................................. 304
Review set 11A ......................................................... 364
Graphics calculator 9.1 .............................................. 310
Review set 11B ......................................................... 364
Graphics calculator 9.2 .............................................. 310
Review set 11C ......................................................... 365
Review 9 ................................................................... 311
Review set11D .......................................................... 365
Review test 9 ............................................................ 311
Examination question .............................................. 366
Review set 9A ........................................................... 312 Review set 9B ........................................................... 312
12 Modelling non-linear relationships
Review set 9C ........................................................... 313
12A Quadratic functions ......................................... 368
Review set 9D ........................................................... 314 Examination question .............................................. 314 10 Spherical geometry
315
367
12B Other non-linear graphs ................................. 373 12C Direct variation ................................................ 377 12D Inverse variation.............................................. 381 12E Modelling: extension ....................................... 385
10A Arc lengths of circles ...................................... 316
Investigation 12.1 ...................................................... 387
10B Geometry of Earth ........................................... 318
Technology 12.1 ......................................................... 387
10C Measuring latitude .......................................... 320
Review 12 ................................................................. 388
10D Measuring longitude ....................................... 322
Review test 12 .......................................................... 388
Contents
v
CONTENTS
Review set 12A ......................................................... 389 Review set 12B ......................................................... 390 Review set 12C ......................................................... 390 Review set 12D ......................................................... 391 Examination question .............................................. 392 Cumulative review chapters 9–12 13 Mathematics and health
393 395
429
14A Water availability and usage ........................... 430 14B Volume and collection of water ...................... 438 14C Scale ................................................................ 442 14D Estimating area using grid squares ............... 444 14E The polygon method........................................ 445 14F Simpson’s rule for area .................................. 450 14G Volume using Simpson’s rule ......................... 453 14H Cost of electrical energy ................................. 455
13A Scatter diagrams ............................................ 396
14I The BASIX Certificate ...................................... 461
13B Line of best fit ................................................. 397
Investigation 14.1 ...................................................... 464
13C Correlation ...................................................... 402
Investigation 14 2 ...................................................... 465
13D Least-squares line of best fit .......................... 405
Investigation 14.3 ...................................................... 465
13E Regression line by calculator: extension ........ 409
Investigation 14.4 ...................................................... 465
13F Measurement calculations ............................. 410
Investigation 14.5 ...................................................... 466
13G Medication calculations .................................. 411
Investigation 14.6 ...................................................... 466
13H Life expectancy ............................................... 415
Investigation 14.7 ...................................................... 466
13I Interpreting life expectancy data .................... 418
Review 14 ................................................................. 467
Investigation 13.1 ...................................................... 422
Review test 14 .......................................................... 467
Investigation 13.2 ...................................................... 422
Review set 14A ......................................................... 470
Review 13 ................................................................. 423
Review set 14B ......................................................... 471
Review test 13 .......................................................... 423
Review set 14C ......................................................... 473
Review set 13A ......................................................... 424
Review set 14D ......................................................... 475
Review set 13B ......................................................... 425
Examination question .............................................. 477
Review set 13C ......................................................... 426 Review set 13D ......................................................... 427 Examination question .............................................. 428
vi
14 Mathematics and resources
Insight Mathematics General 12 HSC Course 2
Answers ................................................................... 479 Index ....................................................................... 541
CONTENTS
SYLLABUS GRID Chapter
Name
Outcomes
Content summary
1
Credit and borrowing
MG2H-1, MG2H-3, MG2H-9, MG2H-10
FM4
Credit and borrowing money
2
Further applications of area and volume
MGH-4, MG2H-5, MG2H-10
MM4
Applications of area and volume
3
Further algebraic skills and techniques
MG2H-3, MG2H-9, MG2H-10
AM3
Algebraic manipulation
4
Interpreting sets of data
MG2H-1, MG2H-2, MG2H-7, MG2H-9, MG2H-10
DS4
Using various measures to interpret sets of data
CR1–4
Cumulative review chapters 1–4
5
Applications of trigonometry
MG2H-4, MG2H-5, MG2H-10
MM5
Applications of trigonometry
6
The normal distribution
MG2H-1, MG2H-2, MG2H-7, MG2H-9, MG2H-10
DS5
The normal distribution
7
Multistage events and applications of probability
MG2H-1, MG2H-2, MG2H-8, MG2H-9, MG2H-10
PB2
Multistage events and applications of probability
8
Annuities and loan repayments
MG2H-1, MG2H-3, MG2H-6, MG2H-9, MG2H-10
FM5
Annuities and loan repayments
CR5–8
Cumulative review chapters 5–8
9
Modelling linear relationships MG2H-3, MG2H-9, MG2H-10
AM4
Modelling linear relationships
10
Spherical geometry
MG2H-4, MG2H-5, MG2H-10
MM6
Spherical geometry
11
Sampling and populations
MG2H-1, MG2H-2, MG2H-7, MG2H-8, MG2H-9, MG2H-10
DS6
Sampling and populations
12
Modelling non-linear relationships
MG2H–3, MG2H–9, MG2H–10
AM5
Modelling non-linear relationships
CR9–12
Cumulative review chapters 9–12
13
Mathematics and health
MG2H-2, MG2H-3, MG2H-5, MG2H-7, MG2H-9
FSHe1
Body measurements
MG2H-2, MG2H-5
FSHe2
Medication
MG2H–1, MG2H-2, MG2H-3, MG2H-7, MG2H-9, MG2H-10
FSHe3
Life expectancy
MG2H–1, MG2H-2, MG2H-3, MG2H-4, MG2H-5, MG2H-7, MG2H-9, MG2H-10
FSRe1
Water availability and usage
MG2H–1, MG2H-2, MG2H-3, MG2H-4, MG2H-5, MG2H-10
FSRe2
Dams, land and catchment areas
MG2H–1, MG2H-2, MG2H-5, MG2H-9, MG2H-10
FSRe3
Energy and sustainability
14
Mathematics and resources
Contents
vii
OXFORD
INSIGHT
MATHEMATICS GENERALHSC GENERAL 2
OXFORD
INSIGHT
MATHEMATICS GENERAL
12
12
HSC GENERAL 2
JOHN LEY MICHAEL FULLER
STUDENT BOOK Substantially revised to precisely reflect the requirements of the Mathematics General syllabus in New South Wales. Comprehensive exercise sets incorporate worked examples where students actually need them. Carefully graded worked examples and exercises to support individual learning pathways. A wealth of consolidation and practice: reviews, cumulative reviews, exam-style questions and integrated technology.
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OXFORD
INSIGHCST
MATHEMATI L NEGENERARAL 2 GEHSC
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viii
Insight Maths General 12 HSC Course 2
12
ACT IVATION COD E CAR D i dd 1
CODE INSIDE
Credit and borrowing This chapter is about the mathematics involved in borrowing money, the different types of loans available and credit cards. The main mathematical ideas investigated are: ▶ calculating the principal, interest and repayments for flat-rate loans ▶ calculating payments, charges and balances on credit cards ▶ calculating reducing-balance loan repayments, including the use of tables ▶ modelling loans using graphs ▶ calculating future value and present value using the formula.
FINANCIAL MATHEMATICS
Syllabus references: FM4 Outcomes: MG2H-1, MG2H-3, MG2H-9, MG2H-10
1A
Flat-rate loans
Flat-rate loans calculate the interest charges based on a fixed percentage of the original amount borrowed, the principal. This type of calculation of interest is referred to as simple interest. The formula for simple interest is: Prn I = ____ where 100
P = the principal, the amount borrowed r = the interest rate per time period n = the number of time periods
WORKED EXAMPLE 1 Calculate the simple interest payable on these loans. a $5000 at 6.7% p.a. over 4 years b $2300 at 1.56% per month for 19 months c $980 at 0.03% per day for 23 days Solve
a
$5000 × 6.7 × 4 I = ______________ = $1340 100
b
$2300 × 1.56 × 19 I = ________________ = $681.72 100
c
$980 × 0.03 × 23 I = _______________ = $6.76 100
Think/Apply Substitute the values of P, r and n into the formula: Prn I = ____ 100
EXERCISE 1A 1 Calculate the simple interest payable on a loan of: a $6000 at 5.8% p.a. over 3 years c $780 at 0.025% per day for 19 days
b $3200 at 1.1% per month for 13 months
WORKED EXAMPLE 2 Calculate the simple interest on: a $3000 at 15% p.a. over 17 months
FINANCIAL MATHEMATICS
Solve
2
a
17 $3000 × 15 × ___ 12 I = _______________ = $637.50 100 15 $3000 × ___ × 17 12 _______________ = $637.50 Or I = 100
b
21 $1800 × 13 × ____ 365 ________________ I= = $13.46 100 13 $1800 × ____ × 21 365 ________________ = $13.46 Or I = 100
Insight Mathematics General 12 HSC Course 2
b $1800 at 13% p.a. for 21 days Think Use r = 15% p.a.
17 Convert 17 months to ___ years. 12 Or use n = 17. 15 Convert 15% p.a. to ___% per month. 12 Use r = 13% p.a.
21 Convert 21 days to ____ years. 365 Or use n = 21 days. 13 Convert 13% p.a. to ____% per day. 365
Apply Convert r and n to the same time period.
2 Calculate the simple interest on the following investments. a $5600 at 13% p.a. for 16 months b $2900 at 15% p.a. for 23 days c $7890 at 18.6% p.a. for 11 months d $3540 at 12.8% p.a. for 53 days
WORKED EXAMPLE 3 1
Joanne paid $2940 interest on a loan of $6000 over 3_2 years. What was the interest rate? Solve 6000 × r × 3.5 2940 = ______________ = 210r 100 2940 r = _____ = 14 210 Interest rate = 14% p.a.
Think n is expressed in years, so the interest rate will be per annum.
Apply Substitute the known values into Prn I = ____ and solve the resulting 100 equation.
3 Calculate the annual rate of interest when: a $2156 interest is paid on a loan of $4900 over 4 years b $972 interest is paid on a loan of $2400 over 3 years c $576 interest is paid on a loan of $3000 over 16 months d $2204 interest is paid on a loan of $14 500 over 19 months e $17.20 interest is paid on a loan of $1600 over 43 days.
WORKED EXAMPLE 4 Yvonne paid $36.96 interest on a loan of $3200 at 0.035% per day. What was the term of the loan? Solve
Think
3200 × 0.035 × n 36.96 = _______________ 100 = 1.12n 36.96 n = _____ = 33 1.12 The money was borrowed for 33 days.
The interest is expressed as a rate per day, so n will be measured in days.
Apply Substitute the given values into Prn I = ____ and solve the resulting 100 equation.
4 Calculate the term of the loan if: a $19.40 interest is paid on a loan of $3300 at 0.0245% per day b $1950 interest is paid on a loan of $7500 at 6.5% p.a. c $1359.26 interest is paid on a loan of $4900 at 1.46% per month d $5093.20 interest is paid on a loan of $13 600 at 7.49% p.a. e $7821.07 interest is paid on a loan of $16 900 at 12.6% p.a. FINANCIAL MATHEMATICS
WORKED EXAMPLE 5 Calculate the total amount to be repaid on a loan of $8900 at 11% p.a. over 5 years. Solve/Think $8900 × 11 × 5 Interest = ______________ = $4895 100 Total to be repaid = $8900 + $4895 = $13 795
Apply Total to be repaid = amount borrowed + interest
Chapter 1 Credit and borrowing
3
5 Calculate the total amount to be repaid on a loan of: a $4500 at 13% p.a. over 3 years b $5750 at 0.9% per month over 15 months c $7100 at 0.031% per day over 19 days
d $5290 at 14% p.a. over 17 months.
WORKED EXAMPLE 6 Ella borrows $2500 over 3 years to buy a drum kit. If she repaid a total of $3700, what was the interest rate on the loan?
Solve Amount of interest paid = $3700 − $2500 = $1200 2500 × r × 3 1200 = ____________ 100 = 75r 1200 r = _____ = 16 75 Interest rate was 16% p.a.
Think n is in years, so the interest rate will be per year.
Apply Amount of interest = total amount repaid − amount borrowed
6 What was the rate of interest on the following loans if these repayments were made? a $12 740 on $9800 over 4 years b $33 722.50 on $23 500 over 5 years c $14 157 on $11 700 over 14 months d $7064.96 on $7000 over 29 days
WORKED EXAMPLE 7 Dominic borrows $2200 to buy a guitar. The flat interest rate is 9.75% p.a. and he takes the loan over 2 years. a Find the interest on the loan. b Find the total amount to be repaid. c Find the monthly repayment.
FINANCIAL MATHEMATICS
Solve/Think
4
$2200 × 9.75 × 2 = _______________ 100 = $429
a
I
b
Total to be repaid = $2200 + $429 = $2629
c
$2629 Monthly repayment = ______ 24 = $109.54
Insight Mathematics General 12 HSC Course 2
Apply Determine the total amount to be repaid. total to be repaid Monthly repayment = ________________________ number of months of the loan
7 Chad borrows $14 300 to buy a car. The flat interest rate is 12.5% p.a. and he takes the loan over 3 years. Complete the following to find: $14 300 × □ × □ a Interest on the loan = ________________ = $___ 100 b Total to be repaid = $14 300 + $___ = $___ $□ = $___ c Monthly repayment = ___ □ 8 Monica borrows $5800 to buy a bedroom suite. The simple interest rate is 8.6% p.a. and she takes the loan over 4 years. a Find the interest on the loan. b Find the total amount to be repaid. c Find the monthly repayment.
WORKED EXAMPLE 8 Matthew borrows $4900 to buy a new television set. He makes monthly repayments of $207.57 for 3 years. What was the rate of interest charged on the loan? Solve
Think
Total amount repaid = $207.57 × 36 = $7472.52 Interest charged = $7472.52 − $4900 = $2572.52 4900 × r × 3 2572.52 = ____________ 100 2572.52 = 147r 2572.52 r = _______ 147 = 17.5 Interest rate = 17.5% p.a.
Over 3 years, Matthew will make 36 monthly repayments.
Apply Determine the number of monthly repayments. Total amount repaid = monthly repayment × number of repayments Amount of interest paid = amount repaid − amount borrowed Prn Use I = ____ to find r. 100
9 Therese borrows $7350 to buy a second-hand car. She makes monthly repayments of $211.93 for 5 years. Complete the following to find the rate of interest charged on the loan. Total amount repaid = $211.93 × ___ = $___ FINANCIAL MATHEMATICS
Interest charged = $___ − $7350 = $___ $7350 × r × □ ___ = _____________ 100 ___ = ___ r □ r = __ = ___ □ Interest rate = ___% p.a.
10 Justin borrows $2900 to buy a tool kit for his car. If he makes monthly repayments of $168.93 for 2 years, calculate the rate of interest charged on the loan.
Chapter 1 Credit and borrowing
5
1B
Credit cards
Credit cards are issued by banks and financial institutions as a convenient way for consumers to purchase goods and services from vendors. The vendor is paid by the bank and the bank recovers the money from the cardholder. Naturally various fees and interest charges are applied by the banks for the use of this credit facility. The cardholder receives a monthly statement and must make a minimum payment of 2–5% of the balance owing. If you do not pay the account in full by the due date (for example, you only make the minimum payment) then any amount outstanding will be carried over and interest charges will apply. Credit cards usually have a higher rate of interest than most other consumer loans. Different rates of interest apply to purchases and cash advances for most cards. There is often an annual fee for the use of the card and fees may be charged for exceeding the credit limit or making late payments. Many cards offer interest-free days under certain conditions. Consider the monthly statements on the right for a credit card that has up to 55 interest-free days. The minimum payment is the greater of $10 or 3% of the closing balance. The statement period is from the first to the last day of the month. The due date is 55 days from the first day of the statement period. You will not be charged any interest on retail purchases made during a statement period if you pay the closing balance in full by the due date and you have paid the closing balance of the previous statement in full by the due date.
March statement
Note • As the closing balance for each month is paid in full by the due date there are no interest charges for March and April.
Due date: 24 April
• The minimum payment in April is $10 since this is greater than 3% of $223 ($6.69). • The due date for purchases made in March is 24 April, 55 days from 1 March (the start of the statement period). • The number of interest-free days is up to 55. For example, the number of interest-free days for the purchase of furniture in March is 48 days (from 8 March to 24 April inclusive) and for the purchase of hardware is 39 days (from 17 March to 24 April inclusive).
FINANCIAL MATHEMATICS
• The interest-free period does not apply to cash advances (such as ATM withdrawals). These transactions attract interest from the day they appear on your statement.
Date
Details
Mar 1
Opening balance
Mar 8
Furniture
1680
Mar 17
Hardware
67
Insight Mathematics General 12 HSC Course 2
0
Opening balance:
$0
Closing balance:
$1747
Minimum payment due:
$52.41
April statement Date
Details
April 11
Jeans
April 24
Payment
April 30
Sunglasses
Amount ($) 88 −1747 135
Opening balance:
$1747
Closing balance:
$223
Minimum payment due:
$10
Due date: 25 May
May statement Date
Details
May 9
Groceries
May 17
Electrical goods
May 25
Payment
Amount ($) 48 136 −223
Opening balance:
$223
Closing balance:
$184
Minimum payment due:
$10
Due date: 24 June
6
Amount ($)
EXERCISE 1B Questions 1 to 3 refer to the credit card shown in the statements on the previous page.
1 The minimum payment for this credit card is the greater of $10 or 3% of the closing balance. Calculate the minimum payment due on these closing balances. a $96 b $390 c $1245 d $320. 2 For this credit card, what would be the due date for purchases made in: a June? b December?
c February (not a leap year)?
3 How many interest-free days are available for the purchase of: a jeans? b sunglasses? c groceries?
September statement Date
Amount ($)
Details
1 Sep
Opening balance
0
9 Sep
Clothes
80
18 Sep
Make-up
54
Opening balance:
$0
Closing balance:
$___
Minimum payment due:
$___
Due date: 25 October
October statement Date
Amount ($)
Details
5 Oct
Shoes
180
25 Oct
Payment
29 Oct
Television set
−___ 967
Opening balance:
$___
Closing balance:
$___
Minimum payment due:
$___
Due date: _____
November statement Date
Amount ($)
Details
10 Nov
Groceries
48
16 Nov
DVDs
66
___ Nov
Payment
FINANCIAL MATHEMATICS
4 Complete these statements given that the statement period is from the first to the last day of the month, the minimum payment is the greater of $10 or 4% of the closing balance, there are up to 55 interest-free days and the closing balance is paid in full on the due date.
d electrical goods?
−___
Opening balance:
$___
Closing balance:
$___
Minimum payment due:
$___
Due date: _____
Chapter 1 Credit and borrowing
7
WORKED EXAMPLE 1 Calculate the total amount due on an ATM cash withdrawal of $400 using a credit card if the full amount is repaid after 15 days. The annual percentage rate (APR) for cash for the card is 21.5% and there is a fee of 1.5% of the cash advance amount. Solve
Think
Apply
Daily interest rate = APR ÷ 365 = 0.0589% Interest charges $400 × 0.0589 × 15 = _________________ 100 = $3.53 Cash advance fee 1.5 = ____ × $400 = $6 100 Total amount to be repaid = $400 + $3.53 + $6 = $409.53
Calculate the interest Prn using I = ____. 100 Calculate the cash advance fee.
Interest on a cash withdrawal is calculated daily from the date of the transaction whether or not the card has an interest-free period. Daily interest rate = APR ÷ 365 days Interest charges withdrawal × daily interest rate × number of days = _________________________________________ 100 Total amount due = amount of withdrawal + interest charges + cash advance fee
5 Complete the following to calculate the total amount due on an ATM cash withdrawal of $500 using a credit card if the full amount is repaid after 23 days. The annual percentage rate for cash for the card is 22.9% and there is a fee of 1.5% of the cash advance amount. Daily interest rate = ___% ÷ 365 = ___% $□ × □ × □ Interest =____________ = $___ 100 □ Cash advance fee = ____ × $500 = $___ 100 Total amount due = $500 + $___ + $___ = $___
6 Calculate the total amount due on an ATM cash withdrawal of $450 using a credit card if it is repaid after 17 days. The annual percentage rate for cash is 20.9% and there is a fee of 1.5% of the cash advance amount. 7 Calculate the total amount due on an over-the-counter cash withdrawal of $150 using a credit card if it is repaid after 21 days. The annual percentage rate for cash is 19.8% and the cash advance fee is the greater of $2.50 or 1.5% of the cash advance amount.
WORKED EXAMPLE 2 FINANCIAL MATHEMATICS
Calculate the average daily balance for the month of May using the information in this credit card statement.
8
Insight Mathematics General 12 HSC Course 2
May statement Date
Details
Amount ($)
1 May
Opening balance
64
10 May
Water rates
13 May
Purchase
47
20 May
Payment
−100
28 May
Purchase
75
246
WORKED EXAMPLE 2 CONTINUED
Solve (9 × $64 + 3 × $310 + 7 × $357 + 8 × $257 + 4 × $332) Average daily balance = __________________________________________________ = $238.35 31 Daily balance ($)
Number of days
Aggregated balance ($)
64
9
576
310
3
930
357
7
2499
257
8
2056
332
4
1328
31
7389
Total
$7389 Average daily balance = ______ = $238.35 31 Or use the statistics function on your calculator to find the mean of the 31 scores (daily balances). Think
Apply
The average daily balance is the average of the daily balances for each of the 31 days of the month. The balance at the end of each of the 9 days from 1 May to 9 May = $64. The balance at the end of each of the 3 days from 10 May to 12 May = $64 + $246 = $310. The balance at the end of each of the 7 days from 13 May to 19 May = $310 + $47 = $357. The balance at the end of each of the 8 days from 20 May to 27 May = $357 − $100 = $257. The balance at the end of each of the 4 days from 28 May to 31 May = $257 + $75 = $332.
Average daily balance = sum of balances at the end of each day ÷ number of days in the month
Daily balance ($)
Number of days
38
4
187
11
July statement Date
Details
Amount ($)
1 Jul
Opening balance
38
5 Jul
Phone bill
149
16 Jul
Purchase
75
21 Jul
Payment
−80
25 Jul
Purchase
94 FINANCIAL MATHEMATICS
8 Complete the following table and find the average daily balance for the July statement shown.
Aggregated balance ($)
262 182 276 Total
Chapter 1 Credit and borrowing
9
9 Find the average daily balance for December.
December statement Date
Details
Amount ($)
1 Dec
Opening balance
7 Dec
Council rates
14 Dec
Payment
−100
19 Dec
Purchase
43
24 Dec
Payment
−100
10 Find the average daily balance for June.
87 488
June statement Date
Amount ($)
Details
1 Jun
Opening balance
45
3 Jun
Purchase
68
8 Jun
Purchase
45
16 Jun
Purchase
74
24 Jun
Payment
−150
29 Jun
Purchase
82
WORKED EXAMPLE 3 Calculate the interest charges for purchases in April given that the annual percentage rate is 19.9%.
April statement Date
Details
1 Apr
Opening balance
37
7 Apr
Purchase
53
16 Apr
Purchase
29
23 Apr
Payment
−80
Solve Daily balance ($)
Think
Number of days
Aggregated balance ($)
37
6
222
90
9
810
119
7
833
39
8
312
30
2177
Total FINANCIAL MATHEMATICS
$2177 Average daily balance = ______ = $72.57 30 Daily percentage rate = 19.9% ÷ 365 = 0.054 52% $72.57 × 0.054 52 × 30 Monthly interest charges = ____________________ = $1.19 100
Find the average daily balance for the month. Determine the daily percentage rate by dividing the annual percentage rate by the number of days in a year. Calculate the monthly interest charges using Prn I = ____ where n is the 100 number of days in the month and r is the daily rate.
Apply average daily balance × daily interest rate × number of days in month Interest = _________________________________________________________ 100
10
Insight Mathematics General 12 HSC Course 2
Amount ($)
Note: If you have a credit card with interest-free days and you do not pay the full closing balance of your credit card account by the due date, you will be charged interest on the outstanding balance for that statement period as well as for any transactions made since the end of the statement period. If interest is charged on your account it is debited on the last day of the statement period.
11 Complete the following to calculate the interest charges for purchases in May, given that the annual percentage rate is 18.6%. $□ Average daily balance = ___ = $___ 31 Daily percentage rate = ___% ÷ 365 = ___% $□ × □ × 31 Monthly interest charges = ____________ = $___ 100 Daily balance ($)
Number of days
147
May statement Date
Details
Amount ($)
1 May
Opening balance
147
8 May
Purchase
88
20 May
Purchase
133
25 May
Payment
−100
Aggregated balance ($)
7
1029
235 368 268 Total
31
12 Calculate the interest charges for purchases in September given that the annual percentage rate (APR) is 18.8%.
September statement Date
Details
Amount ($)
1 Sep
Opening balance
135
10 Sep
Purchase
59
19 Sep
Purchase
136
24 Sep
Payment
−100
13 Calculate the interest charges for purchases in each of the months in questions 8 to 10 given that the annual percentage rate is 21.6%.
WORKED EXAMPLE 4 How long will it take to pay off a credit card debt of $1000 if the annual interest rate is 18.5% and you only make the minimum payment each month (assuming no other transactions)? How much interest will you pay?
It would take approximately 8 years to repay this debt and you would pay about $924 in interest.
Think There are calculators available via the internet which will do these calculations for you. Try www.moneysmart.gov.au and go to (more) calculators, then credit card calculator (under borrowing and credit).
FINANCIAL MATHEMATICS
Solve
14 Use an internet calculator to find the time it would take to repay various credit card balances if you only make the minimum repayment each month. How much interest would be charged? How much can be saved by increasing your monthly repayment by $5, $10, etc.?
Chapter 1 Credit and borrowing
11
1C
Reducing-balance loans
A home loan is an example of a reducing-balance loan. In this type of loan the interest is calculated on the balance owing at the start of each repayment period. For home loans this is often monthly, but may be fortnightly or weekly. These are often referred to as monthly reducible loans. This section examines various methods of calculating the balance owing and the monthly payments required. Usually the information most frequently needed by borrowers concerns the amount that can be borrowed and the repayment. Most lending institutions have online calculators that can be used to find the monthly repayment, the amount of interest paid and the effects of changing the interest rate and making extra payments.
WORKED EXAMPLE 1 This table of home loan repayments was generated using a spreadsheet. The annual interest rate is 8.5% and the amount shown is the monthly repayment. Home loan table monthly repayments Years
$160 000
$200 000
$240 000
$280 000
$320 000
$360 000
$400 000
5
$3282.64
$4103.30
$4923.96
$5744.62
$6565.30
$7385.96
$8206.60
10
$1983.78
$2479.72
$2975.66
$3471.60
$3967.54
$4463.48
$4959.44
15
$1575.58
$1969.48
$2363.38
$2757.28
$3151.16
$3545.06
$3938.96
20
$1388.52
$1735.64
$2082.78
$2429.90
$2777.04
$3124.16
$3471.28
25
$1288.36
$1630.46
$1932.64
$2254.64
$2576.72
$2898.82
$3260.92
30
$1230.26
$1537.82
$1845.40
$2152.96
$2460.52
$2768.08
$3075.64
Use the table to find the monthly repayment on a loan of $240 000 over: a 30 years b 25 years c 10 years. Solve
a
$1845.40
b
$1932.64
c
$2975.66
Think Look down the $240 000 column and along the row showing the number of years.
Apply Determine the appropriate column and row in the table.
EXERCISE 1C FINANCIAL MATHEMATICS
1 Use the table in Worked example 1 to find the monthly repayment on a loan of $320 000 over: a 10 years b 15 years c 25 years d 30 years. 2 a Find the monthly repayment on $400 000 over 20 years. b Find the monthly repayment on $360 000 over 20 years. c How much larger is the monthly repayment on $400 000 than that on $360 000 over 20 years? 3 From the amounts shown in the table, what is the maximum loan that can be taken if you can afford these monthly repayments? a $2000 b $2300 c $2480 d $2600
12
Insight Mathematics General 12 HSC Course 2
4 Go to the website of a bank and use their online calculator to check the current results for questions 1 to 3.
WORKED EXAMPLE 2 A home loan of $250 000 is taken out at an interest rate of 7.75% p.a., reducible monthly, with a monthly repayment of $1888. Use the table to calculate the amount still owing after 6 months. Home loan table Months (n)
Principal (P)
1
$250 000
P+I
Interest (I)
Balance = P + I − R
2 3 4 5 6 Solve P+I
Balance = P + I − R
$1614.58
$251 614.58
$249 726.58
$249 726.58
$1612.82
$251 339.40
$249 451.40
3
$249 451.40
$1611.04
$251 062.44
$249 174.44
4
$249 174.44
$1609.25
$250 783.69
$248 895.69
5
$248 895.69
41607.45
$250 503.14
$248 615.14
6
$248 615.14
$1605.64
$250 220.78
$248 332.78
Principal (P)
Interest (I)
1
$250 000
2
Think 7.75% Monthly interest rate = ______ = 0.645 83% 12 $250 000 × 0.645 83 × 1 Interest for 1st month I = _____________________ 100 = $1614.58 Hence: P + I = $250 000 + $1614.58 = $251 614.58 P + I − R = $251 614.58 − $1888 = $249 726.58 The first row can now be completed. The balance at the end of the first month becomes the principal for the second month. Hence for the second month:
Apply Calculate the monthly balance by adding the interest to the principal and then subtracting the repayment. This amount then becomes the principal for the next month.
FINANCIAL MATHEMATICS
Months (n)
$249 726.58 × 0.645 83 × 1 I = ________________________ = $1612.81 100 P + I = $249 726.58 + $1612.81 = $251 339.39 P + I − R = $251 339.39 − $1888 = $249 451.39 The balance at the end of the second month becomes the principal for the third month, etc.
Chapter 1 Credit and borrowing
13
Note: In Worked example 2 the amount of interest being paid each month is reducing because the balance is reducing each month. The total amount repaid over 6 months = 6 × $1888 = $11 328. The total interest paid = $1614.58 + $1612.81 + … = $9660.73 and the principal has been reduced by $11 328 − $9660.73 = $1667.27. If this was a flat-rate loan the total amount of interest paid would be 6 × $1614.58 = $9687.48.
5 A home loan of $420 000 is taken out at an interest rate of 7.2% p.a. with a monthly repayment of $3306.83. Complete the table to calculate the amount still owing after 6 months. Home loan table Months (n)
Principal (P)
Interest (I)
P+I
1
$420 000
$2520.00
$422 520.00
Balance = P + I − R $419 213.17
2 3 4 5 6
6 A home loan of $350 000 is taken out at an interest rate of 7.6% p.a. with a monthly repayment of $2471.26. Complete the table to calculate the amount still owing after 6 months. Home loan table Months (n)
Principal (P)
1
$350 000
Interest (I)
P+I
Balance = P + I − R
2 3 4 5 6
WORKED EXAMPLE 3 This table shows the payment per $1000 on a monthly reducible loan. Find the monthly repayment on a loan of $210 000 over: a 20 years at 7.75% p.a. b 25 years at 8.25% p.a.
FINANCIAL MATHEMATICS
Home loan table
14
Term in years
7%
7.25%
7.5%
7.75%
8%
8.25%
8.5%
5
$19.8012
$19.9194
$20.0379
$20.1570
$20.2765
$20.3963
$20.5164
10
$11.6108
$11.7401
$11.8702
$12.0011
$12.1328
$12.2653
$12.3985
15
$8.9883
$9.1286
$9.2701
$9.4128
$9.5566
$9.7014
$9.8474
20
$7.7530
$7.9036
$8.0559
$8.2095
$8.3644
$8.5207
$8.6782
25
$7.0678
$7.2281
$7.3899
$7.5533
$7.7182
$7.8875
$8.0522
30
$6.6530
$6.8218
$6.9921
$7.1641
$7.3377
$7.5127
$7.6891
Insight Mathematics General 12 HSC Course 2
WORKED EXAMPLE 3 CONTINUED
Solve
Think
Apply
a
210 000 Payment = $8.2095 × _______ 1000 = $1724.00
From the table, the payment on $1000 at 7.75% over 20 years is $8.2095.
b
210 000 Payment = $7.8875 × _______ 1000 = $1656.38
Payment on $1000 over 25 years at 8.25% is $7.8875. 210 000 Thousands borrowed = _______= 210. 1000
Determine the payment on $1000 from the appropriate column and row in the table, Multiply this amount by the number of thousands being borrowed.
7 Use the table from Worked example 3 to find the monthly repayment on each of these loans. Amount ($)
Term (years)
Interest rate (%)
a
260 000
25
7.25
b
350 000
30
7
c
415 000
10
8.25
d
396 000
20
7.75
e
527 500
15
8
f
292 600
5
8.5
WORKED EXAMPLE 4 Home loan balance
The graph shows the amount outstanding after n months for a home loan of $240 000 at 8.4% p.a. with a monthly repayment of $2160. a How much is still owing after 75 months? b When is the amount owing equal to $180 000? c When is the loan half paid? d How long does it take to repay the loan in full?
Amount owing $’000
250 200 150 100 50
Solve
50
Think
a
About $195 000
From the graph, when: n = 75 months, A ≈ $195 000
b
About 90 months
From the graph, when: A = $180 000, n ≈ 90 months
c
About 145 months
From the graph, when: 1 A = _2 × $240 000 = $120 000, n ≈ 145
d
About 215 months
From the graph, when: A = $0, n ≈ 215 months
100 150 200 Number of months
250
Apply Locate the appropriate point on the graph.
Chapter 1 Credit and borrowing
FINANCIAL MATHEMATICS
0
15
8 Use the graph from Worked example 4 on the previous page to answer these questions. a How much is owing after 100 months? b How much is owing after 125 months? c When is the amount owing $100 000? d When is the amount owing $70 000? 9 Use this graph of the amount owing on a home loan to answer these questions. a How much was borrowed? b How much is still owing after 100 months? c How much is still owing after 200 months?
Home loan balance 300
Amount owing $’000
250
d How much is still owing after 12_12 years? e f g h
When is the amount owing $200 000? When is the amount owing $50 000? When is the loan half paid? How long does it take to repay the loan in full?
200 150 100 50 0
50
100 150 200 250 Number of months
300
10 The amount $An owing on a home loan after n months is shown in the table below. n
0
60
120
180
240
300
An
450 000
404 720
346 920
261 260
150 135
0
Plot the values in the table and draw a smooth curve through them to draw a graph of amount owing versus time for this loan. Use the graph to answer these questions. a How much was borrowed? b How much is still owing after 100 months? c How much is still owing after 18 years? d When is the amount owing $300 000? e When is the loan half paid? f How long does it take to repay the loan in full?
11 Here is a personal loan table showing the monthly repayments for amounts from $5000 to $15 000, over 1 to 3 years. The loan is monthly reducible with a rate of 10.45% p.a. Personal loan table FINANCIAL MATHEMATICS
Amount borrowed Years
$5000
$7000
$9000
1
$440.63
$616.88
$793.13
2
$231.76
$324.47
3
$162.39
$227.35
Find the monthly repayment on: a $7000 over 3 years c $13 000 over 1 year
16
Insight Mathematics General 12 HSC Course 2
$11 000
$13 000
$15 000
$969.38
$1145.63
$1321.88
$417.18
$509.88
$602.59
$695.29
$292.31
$357.27
$422.23
$487.18
b $11 000 over 2 years d $15 000 over 3 years
12 From the same financial institution the interest rate for loans in excess of 3 years is higher. The table below shows the monthly repayments for personal loans for amounts from $5000 to $15 000, over 4 to 7 years. The loan interest is calculated on the monthly balance. The rate for 4 to 5 years is 10.95% p.a. and for 6 to 7 years is 11.45% p.a. Personal loan table Amount borrowed Years
$5000
$7000
$9000
$11 000
$13 000
$15 000
4
$129.11
$180.75
$232.39
$284.03
$335.68
$387.32
4.5
$117.68
$164.76
$211.83
$258.91
$305.98
$353.05
5
$108.59
$152.02
$195.46
$238.89
$282.33
$325.76
6
$96.33
$134.86
$173.39
$211.92
$250.45
$288.98
6.5
$91.18
$127.65
$164.12
$200.60
$237.07
$273.54
7
$86.80
$121.52
$156.24
$190.96
$225.68
$260.40
Find the monthly repayment on: a $7000 over 4.5 years c $13 000 over 5 years
b $11 000 over 7 years d $15 000 over 6.5 years
13 The table below shows the personal loan repayments for loans over $15 000. The interest rate increases as the time period increases. It is 9.95% p.a. for 1 to 3 years, 10.45% p.a. for 4 to 5 years and 10.95% p.a. for 6 to 7 years. Personal loan table Years
$15 000
$16 000
$17 000
$18 000
$19 000
$20 000
1
$1318.39
$1406.28
$1494.17
$1582.07
$1669.96
$1757.85
1.5
$900.51
$960.54
$1020.58
$1080.61
$1140.65
$1200.68
2
$691.83
$737.95
$784.07
$830.19
$876.32
$922.44
2.5
$566.82
$604.61
$642.40
$680.19
$717.98
$755.76
3
$483.66
$515.90
$548.14
$580.39
$612.63
$644.87
4
$383.69
$409.27
$434.85
$460.43
$486.01
$511.58
4.5
$349.38
$372.67
$395.96
$419.25
$442.54
$465.83
5
$322.04
$343.51
$364.98
$386.44
$407.91
$429.38
6
$285.13
$304.14
$323.14
$342.15
$361.16
$380.17
6.5
$269.64
$287.61
$305.59
$323.56
$341.54
$359.51
7
$256.44
$273.54
$290.63
$307.73
$324.83
$341.92
FINANCIAL MATHEMATICS
Amount borrowed
a Find the monthly repayment on: i $17 000 over 4.5 years ii $19 000 over 3 years iii $20 000 over 5 years iv $16 000 over 2.5 years v $18 000 over 6.5 years b $17 000 is borrowed over 4 years. i Find the total amount repaid over the term of the loan. ii Find the amount of interest paid over the 4 years.
Chapter 1 Credit and borrowing
17
14 Carla and Ian take out a personal loan of $15 000 to furnish their home unit. Interest is 14.4% p.a. monthly reducible. There is a loan establishment fee of $390 and a monthly account-keeping fee (F) of $10 per month. Their monthly repayments (R) are $413. Complete the following table to calculate how much they still owe after 6 months.
Months (n)
Principal (P)
Interest (I)
P+I
P+I+F
P+I+F−R
1
$15 000 + $390 = $15 390
$184.68
$15 574.68
$15 584.68
$15 171.68
2
$15 171.68
Interest (I)
P+I
P+I+F
P+I+F−R
3 4 5 6
15 Harry takes out a personal loan of $18 000 to buy a car. Interest is 15.6% p.a. monthly reducible. There is a loan establishment fee of $260 and a monthly account-keeping fee (F) of $8 per month. His monthly repayments (R) are $434. Complete the following table to calculate how much he still owes after 6 months.
FINANCIAL MATHEMATICS
Months (n) 1
Principal (P) $18 000 + ___ = ___
2 3 4 5 6
18
Insight Mathematics General 12 HSC Course 2
1D
Comparing loans
EXERCISE 1D This table compares four home loans offered by the same bank. Use the table to answer questions 1 to 3. Options
Super
Honeymoon
No Frills
Fixed
Interest rate
7.30%
6.49% 1st year then 7.30%
6.74%
Various see below
Establishment fee
$600
$600
$600
$600
Monthly fee
$8
$8
$8
$8
Additional repayments
Yes
Up to 20%
Yes
No
Redraw fee
Free
No
$300 initially, then $50 per redraw
No
Increase loan
$300
No
$450
No
$200, then free
$200, then free
$95 each
$200, then free
Portability
$300
$300
$300
$300
Repayment holiday
Free
No
$95
No
Interest-only payments
No
Free
No
Free
Parental leave
Free
No
$95
No
Progress draws
The establishment fee is charged to take out the loan. Redraw is a facility to get back extra funds paid off the loan. A loan is portable if it can be transferred to another property. A repayment holiday is a period of time when no repayments are made. Parental leave allows repayments to stop for a period of time, or be altered if the borrowers have a baby.
1 a b c d e f
Which aspects are common to all four loans? Which loan has the lowest interest rate in the first year? Which loan is the best if you require a redraw facility? Which loans enable unlimited additional payments? How much does it cost to increase a No Frills loan? What is the 4-year fixed rate?
Fixed rates Years
Rate p.a.
1
7.59%
2
8.09%
3
7.99%
4
8.40%
5
8.40%
7
8.75%
10
8.75%
FINANCIAL MATHEMATICS
2 Bryan and Daniella are looking for a home loan. They are building their home and therefore require the ability to have progress payments and estimate that there will be three payments. They will not be making extra payments so they do not need a redraw facility. a Examine each of the four loan types and give points for and against each one based on their needs. b Which loan should Daniella and Bryan take? Explain your answer. 3 Compare the Super and No Frills home loans on a loan of $350 000. a Calculate 7.3% and 6.74% of $350 000. How much is the difference? b How much would two redraws cost with each loan? c How much is parental leave with each loan? d Give three advantages of the Super loan over the No Frills loan. e Give three advantages of the No Frills loan over the Super loan.
Chapter 1 Credit and borrowing
19
1E
Comparison rates
A comparison interest rate helps to identify the true cost of a loan by including the fees and charges relating to the loan as well as the interest at the quoted rate. It can be used to compare various loans from different lenders. The comparison rate is calculated according to a standard formula that takes into account: • the amount of the loan • the term of the loan • the frequency of repayments • the quoted interest rate • the fees and charges associated with the loan (with some exceptions). There may be an establishment fee, ongoing account-keeping fees, fees to have a redraw facility etc.
EXERCISE 1E Below is a comparison rate schedule for secured and unsecured personal loans from LBank. Use the schedule to answer the following questions. Personal loans (LBank) Secured personal loan Amount and term
Fixed rate p.a.
Comparison rate p.a.
Unsecured personal loan Fixed rate p.a.
Comparison rate p.a.
$5000 for 2 years
11.95%
16.05%
13.95%
18.10%
$10 000 for 3 years
11.95%
13.30%
13.95%
15.35%
$15 000 for 4 years
10.95%
11.65%
12.95%
13.64%
$20 000 for 4 years
10.95%
11.46%
12.95%
13.43%
$25 000 for 5 years
10.95%
11.25%
12.95%
13.12%
$30 000 for 5 years
9.95%
10.22%
11.95%
12.14%
$70 000 for 5 years
9.95%
10.04%
11.95%
12.07%
$100 000 for 5 years
9.95%
10.03%
11.95%
12.02%
1 Determine the quoted fixed annual interest rate and comparison rate for a secured personal loan of: a $5000 for a term of 2 years b $10 000 for 3 years c $20 000 for 4 years d $25 000 for 5 years e $30 000 for 5 years.
FINANCIAL MATHEMATICS
2 Find the quoted fixed annual interest rate and comparison rate for an unsecured personal loan of: a $5000 for a term of 2 years b $10 000 for 3 years c $70 000 for 5 years d $100 000 for 5 years. 3 Discuss the following with your class. a Are the rates higher for secured or unsecured loans? Why do you think this is so? b Are the comparison rates the same for a given fixed rate? What are the comparison rates for a quoted fixed rate of 12.95%? Is there a trend? c Are the comparison rates the same for the same fixed rate and term? Are the comparison rates the same for a fixed interest rate of 9.95% over 5 years? Is there a trend?
20
Insight Mathematics General 12 HSC Course 2
4 The following is a comparison rate schedule for VBank. Compare these rates with the LBank table at the beginning of this exercise. From which bank is it cheaper to take a secured personal loan of: a $5000 for 2 years b $25 000 for 5 years? Secured personal loan (VBank) Amount and term
Fixed rate p.a.
Comparison rate p.a.
$5000 for 2 years
11.95%
15.64%
$25 000 for 5 years
10.55%
11.45%
1F
Future and present value
WORKED EXAMPLE 1 Calculate the amount to which $1000 will grow if invested for 3 years at 18% p.a. interest compounded monthly. Solve A = P(1 + r)n
(
1.5 = $1000 × 1 + ____ 100 = $1709.14
)
36
Think
Apply
P = $1000 Interest rate per month 18 = ___% = 1.5% 12 n = 3 × 12 = 36 months
Use the compound interest formula. A = P(1 + r)n where: A = the amount P = the principal, the initial amount r = the interest rate per compounding period, as a decimal n = the number of compounding periods.
In finance the principal P, or initial amount, is known as the present value of the investment. The amount to which the principal grows, A, is known as the future value of the investment. Hence the compound interest formula can be written: FV = PV(1 + r)n
where
FV PV r n
is the future value is the present value is the interest rate per compounding period, as a decimal is the number of compounding periods.
WORKED EXAMPLE 2 Solve FV = PV(1 + r)n
(
2.25 = $5000 × 1 + ____ 100 = $5974.16
Think
)
8
PV = $5000 Interest rate per quarter (of a year) 9 = __% = 2.25% 4 n = 2 × 4 = 8 quarters
FINANCIAL MATHEMATICS
Calculate the future value of $5000 if it is invested for 2 years at 9% p.a. interest compounded quarterly. Apply Use the future value formula: FV = PV(1 + r)n
Chapter 1 Credit and borrowing
21
EXERCISE 1F 1 Calculate the future value of: a $2000 invested for 3 years at 6% p.a. interest compounded monthly b $10 000 invested for 5 years at 6% p.a. interest compounded quarterly c $8000 invested for 4 years at 10% p.a. interest compounded yearly d $25 000 invested for 2 years at 8% p.a. interest compounded monthly e $18 500 invested for 3 years at 5.5% p.a. interest compounded half yearly.
WORKED EXAMPLE 3 Calculate the amount that must be invested now at 8.4% p.a. interest compounded monthly in order to have $20 000 in 5 years time. Solve FV PV = _______n (1 + r) $20 000 = ___________ (1 + 0.007)60 = $13 160.18
Think FV = $20 000 8.4 Interest rate per month = ___% 12 = 0.7% Number of months = 5 × 12 = 60
Apply The future value formula can be rearranged to make the present value the FV subject: PV = _______n . (1 + r) This is the single amount that must be invested when the future value is known.
2 Calculate the single amount that must be invested now to get these future values. a $5000 in 3 years time from an investment at 9% p.a. interest compounded monthly b $12 000 in 2 years time from an investment at 8% p.a. interest compounded quarterly c $10 000 in 5 years time from an investment at 6.6% p.a. interest compounded half yearly d $20 000 in 4 years time from an investment at 5.2% p.a. interest compounded yearly. 3 A business owner must repay a loan of $100 000 in 5 years time. What single amount should he invest now at 6% p.a. interest compounded quarterly in order to be able to repay the debt?
FINANCIAL MATHEMATICS
4 Deborah wants to have a 21st birthday party in 4 years time. She estimates that it will cost $1200. How much should Deborah’s parents invest now at 3.6% p.a. interest compounded monthly in order to pay for the party?
22
Insight Mathematics General 12 HSC Course 2
REVIEW 1 CREDIT AND BORRO WING Language and terminology Here is a list of terms used in this chapter. Explain each term in a sentence. average daily balance, comparison rate, credit, flat-rate loan, future value, interest-free days, principal, present value, reducing-balance loan, simple interest, term
Having completed this chapter you should be able to: •
calculate amount of interest, rate of interest and repayments on flat-rate loans
•
calculate credit card payments, interest charges and balances
•
calculate values in a table of loan repayments
•
compare loans in relation to fees and interest rates
•
calculate future value and present value using the compound interest formula.
1 REVIEW TEST Samantha is going on a holiday. She borrows $3500 over 3 years at a flat interest rate of 8% p.a. Use this information to answer questions 1 and 2.
1 The simple interest charged by the lending institution is: A $908.99 B $840 C $280
D $93.33
2 The total to be repaid by Samantha is: A $3500 B $3780
D $4408.99
C $4340
3 The simple interest payable on a loan of $4350 at a simple interest rate of 9.5% p.a. for a period of 2_12 years is closest to: A $413.25 B $103 312.50 C $1033.00 D $1033.13 4 The simple interest payable on a loan of $2380 at a simple interest rate of 7.45% p.a. for a period of 17 months is: A $251.19 B $251.20 C $2481.88 D $101.88 5 $3340 is borrowed for a period of 11 months. If $4022 is repaid, the simple interest rate per annum, correct to 1 decimal place, is: A 22.3% B 20.4% C 18.5% D 17.0% FINANCIAL MATHEMATICS
6 Steve borrows $5600 to buy a car. The simple interest rate is 10.75% and he takes the loan over 3 years. His monthly payment is: A $50.17 B $172.28 C $205.72 D $2468.67 7 The minimum payment on a credit card is the greater of $10 or 4% of the closing balance. The minimum payment on a closing balance of $320 is: A $10 B $4 C $12.80 D $1.28
Chapter 1 Credit and borrowing
23
8 A credit card offers 55 days interest free, from the start of the statement period. The statement period is from the first to the last day of the month. Assuming that all conditions necessary for the interest-free days have been satisfied, the due date for purchases made in May is: A 24 June B 25 June C 24 May D 25 May Use this credit card statement to answer questions 9 and 10.
August statement
9 The average daily balance for August is: A $393.16 B $402.55 C $406.27 D $415.97
Date
10 If the annual percentage rate is 19.6%, the interest charges for August would be: A $6.33 B $6.54 C $6.48 D $6.70
Amount ($)
Details
1 Aug
Opening balance
246
10 Aug
Water rates
395
16 Aug
Purchases
83
21 Aug
Payment
27 Aug
Purchases
−500 67
11 The home loan table below shows the monthly repayments on loans at an annual interest rate of 8.5%. Home loan table monthly repayments Years
160 000
200 000
240 000
280 000
320 000
360 000
400 000
15
1575.58
1969.48
2363.38
2757.28
3151.16
3545.06
3938.96
20
1388.52
1735.64
2082.78
2429.90
2777.04
3124.16
3471.28
25
1288.36
1630.46
1932.64
2254.64
2576.72
2898.82
3260.92
30
1230.26
1537.82
1845.40
2152.96
2460.52
2768.08
3075.64
From the table, the maximum amount that can be borrowed if you can afford monthly repayments of $2000 is:
A $160 000
B $200 000
C $240 000
D $280 000
12 A home loan of $480 000 is taken out at an interest rate of 6.8% p.a., monthly reducible, with a monthly repayment of $3142. Home loan table Months (n)
Principal (P)
Interest (I)
1
480 000
2720
P+I
Balance = P + I − R
482 720
479 578
2
X
3 The value of X in the home loan table above is:
A $476 436
B $479 153.61
C $479 156
D $479 697.13
FINANCIAL MATHEMATICS
13 The table shows the monthly payment per $1000 on a monthly reducible loan. Term in years
7.75%
8%
8.25%
8.5%
5
20.1570
20.2765
20.3963
20.5164
10
12.0011
12.1328
12.2653
12.3985
The monthly repayment on a loan of $15 600 at 8.25% over 10 years is:
A $154.25
24
B $183.98
Insight Mathematics General 12 HSC Course 2
C $195.76
D $191.34
14 The future value of $13 000 invested for 3 years at 8.76% p.a. interest compounded quarterly is: A $13 872.94 B $16 859.48 C $16 724.42 D $35 610.14 15 A business owner must repay a $60 000 loan in full in 4 years time. The single amount, to the nearest dollar, that he should invest now at 5.4% p.a. interest compounded monthly in order to be able to pay the debt is: A $58 932 B $48 617 C $48 368 D $4806 If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question Section
1–6
7–10
11–13
14, 15
A
B
C
F
1A REVIEW SET 1 Angela is going on a holiday. She borrows $4000 over 4 years at a flat interest rate of 9% p.a. a Find the simple interest charged. b Find the total amount to be repaid. 2 Find the simple interest payable on a loan of $7200 at a simple interest rate of 11.5% p.a. for a period of: a 3_12 years b 15 months. 3 $2450 is borrowed over 7 months. If $2789 is repaid, find the simple interest rate per annum correct to 1 decimal place. 4 Vanturo borrows $19 600 to buy a car. The simple interest rate is 9.75% p.a. and he takes the loan over 5 years. a Find the interest on the loan. b Find the total amount to be repaid. c What is Vanturo’s monthly payment? 5 Use the table of home loan monthly repayments in Worked example 1 Section 1C to find the maximum loan that can be taken if you can afford the following monthly repayments. a $1450 b $1900 c $2500 d $2700 6 Use the table in Worked example 3 Section 1C, which shows the payments per $1000 on a monthly reducible loan, to find the monthly payment on a loan of $70 000 over: a 20 years at 7.75% b 25 years at 8.25% 7 The graph in Worked example 4 Section 1C shows the amount outstanding after n months for a home loan of $240 000 at 8.4% p.a. with a monthly repayment of $2160. Use this graph to answer the following questions. a How much is owed after 170 months? b When is the amount owing $170 000? FINANCIAL MATHEMATICS
8 The minimum payment on a credit card is the greater of $10 or 4% of the closing balance. Calculate the minimum payment on these closing balances. a $520 b $240 9 A credit card offers 55 days interest free from the start of the statement period. The statement period is from the first to the last day of the month. Assuming that all conditions necessary for the interest-free days have been satisfied, determine the due date for purchases made in these months. a June b October
Chapter 1 Credit and borrowing
25
Use this credit card statement to answer questions 10 and 11.
September statement
10 Calculate the average daily balance for September.
Date
11 If the annual percentage rate is 19.6%, calculate the interest charges for September.
1 Sep
Opening balance
346
9 Sep
Purchase
195
19 Sep
Purchase
68
24 Sep
Payment
−400
27 Sep
Purchase
53
Details
Amount ($)
12 Calculate the future value of $18 000 invested for 3 years at 7.2% p.a. interest compounded quarterly. 13 A business owner must repay a $160 000 loan in full in 5 years time. What is the single amount that he should invest now at 5.4% p.a. interest compounded monthly in order to be able to pay the debt?
1B REVIEW SET 1 Paul wants to buy a bicycle. He borrows $2300 over 3 years at a flat interest rate of 8.5% p.a. a Find the simple interest charged. b Find the total amount to be repaid. 2 Find the simple interest payable on a loan of $3450 at a simple interest rate of 9.6% p.a. for a period of: 1 a 1__ b 19 months. 2 years 3 Find the simple interest rate if $1840 is repaid on a loan of $1450 over 2_12 years. 4 Monica borrows $1850 to buy a new computer. The simple interest rate is 7.3% p.a. and she takes the loan over 2 years. a Find the interest on the loan. b Find the total to be repaid. c What is Monica’s monthly payment? 5 a Use the table in Worked example 1 Section 1C showing home loan monthly repayments to find the monthly repayment on $280 000 over 20 years. b Find the monthly repayment on $320 000 over 20 years. c How much larger is the monthly repayment on $320 000 than on $280 000 over 20 years? 6 Use the table in Worked example 3 Section 1C, which shows the payments per $1000 on a monthly reducible loan, to find the monthly payment on a loan of $285 000 over: a 15 years at 7.75% b 25 years at 8% FINANCIAL MATHEMATICS
7 The graph in Worked example 4 Section 1C shows the amount outstanding after n months for a home loan of $240 000 at 8.4% p.a. with a monthly repayment of $2160. Use this graph to answer the following questions. a How much is owing after 140 months? b When is the amount owing $150 000? 8 The minimum payment on a credit card is the greater of $12.50 or 5% of the closing balance. Calculate the minimum payment on these closing balances. a $220 b $490
26
Insight Mathematics General 12 HSC Course 2
9 A credit card offers 55 days interest free from the start of the statement period. The statement period is from the first to the last day of the month. Assuming that all conditions necessary for the interest-free days have been satisfied, determine the due date for purchases made in these months. a May b November Use this credit card statement to answer questions 10 and 11.
March statement
10 Calculate the average daily balance for March.
Date
11 If the annual percentage rate is 21.6%, calculate the interest charges for March.
1 Mar
Opening balance
721
9 Mar
Purchase
356
19 Mar
Purchase
49
24 Mar
Payment
−600
27 Mar
Purchase
73
Amount ($)
Details
12 Calculate the future value of $12 000 invested for 5 years at 5.2% p.a. interest compounded monthly. 13 What single amount must be invested now at 6.9% p.a. interest compounded monthly in order to have $25 000 in 3 years time?
1C REVIEW SET 1 Georgia is going on a holiday. She borrows $6000 over 3 years at a flat interest rate of 8% p.a. a Find the simple interest charged. b Find the total to be repaid. 2 Find the simple interest payable on a loan of $4670 at a simple interest rate of 7.9% p.a. for these periods. a 3_12 years b 21 month c 33 days 3 Tim borrows $2800 to go on a cruise. He repays $3430 over a period of 2 years. Calculate the simple interest rate. 4 Natalie borrows $5500 to buy a car. The simple interest rate is 11.5% p.a. and she takes the loan over 4 years. a Find the interest on the loan. b Find the total to be repaid. c What is Natalie’s monthly payment? 5 Use the table in Worked example 1 Section 1C, which shows loan monthly repayments, to find the monthly repayment on a loan of $400 000 over: a 10 years b 30 years. FINANCIAL MATHEMATICS
6 Use the table in Worked example 3 Section 1C, which shows the payments per $1000 on a monthly reducible loan, to find the monthly payment on a loan of $185 000 over: a 20 years at 7.25% b 25 years at 8.5%. 7 The graph in Exercise 1C question 9 shows the amount owing on a home loan of $300 000 against repayment periods. Use the graph to answer the following questions. a How much is owing after 15 years? b When is the amount owing $190 000?
Chapter 1 Credit and borrowing
27
8 The minimum payment on a credit card is the greater of $10 or 3% of the closing balance. Calculate the minimum payment on these closing balances. a $460 b $310 9 A credit card offers 45 days interest free from the start of the statement period. The statement period is from the first to the last day of the month. Assuming that all conditions necessary for the interest-free days have been satisfied, determine the due date for purchases made in these months. a July b December Use this credit card statement to answer questions 10 and 11.
March statement
10 Calculate the average daily balance for March.
Date
11 If the annual percentage rate is 22.4%, calculate the interest charges for March.
1 Mar
Opening balance
526
10 Mar
Purchase
156
21 Mar
Purchase
38
23 Mar
Payment
−200
25 Mar
Purchase
52
Details
Amount ($)
12 Calculate the future value of $50 000 invested for 6 years at 4.9% p.a. interest compounded monthly. 13 What single amount must be invested now at 6.2% p.a. interest compounded quarterly in order to have $30 000 in 5 years time?
1D REVIEW SET 1 Henry wants to buy a sound system. He borrows $1750 over 2 years at a flat interest rate of 7.3% p.a. a Find the simple interest charged. b Find the total to be repaid. 2 Find the simple interest payable on a loan of $7500 at a simple interest rate of 9.9% p.a. for these periods. a 2 _12 years b 19 months c 35 days 3 Madeline takes out a loan to buy a car and she repays $24 800 on a loan of $15 200 taken over a period of 4 years. Calculate the simple interest rate.
FINANCIAL MATHEMATICS
4 Rebecca borrows $8500 to buy a piano. The simple interest rate is 7.5% p.a. and she takes the loan over 4 years. a Find the interest on the loan. b Find the total to be repaid. c What is Rebecca’s monthly payment?
28
Insight Mathematics General 12 HSC Course 2
5 Use the table in Worked example 1 Section 1C, which shows home loan monthly repayments, to find the monthly repayment on a loan of $320 000 over: a 30 years b 25 years. 6 Use the table in Worked example 3 Section 1C, which shows the payments per $1000 on a monthly reducible loan, to find the monthly payment on a loan of $580 000 over: a 30 years at 7.25% b 15 years at 8.25%. 7 The graph in Exercise 1C question 9 shows the amount owing on a home loan of $300 000 against repayment periods. Use the graph to answer the following questions. a How much is owing on the home loan after 90 months? b When is the amount owing $120 000? 8 The minimum payment on a credit card is the greater of $15 or 5% of the closing balance. Calculate the minimum payment on these closing balances. a $880 b $226 9 A credit card offers 45 days interest free, from the start of the statement period. The statement period is from the first to the last day of the month. Assuming that all conditions necessary for the interest-free days have been satisfied, determine the due date for purchases made in these months. a January b April Use this credit card statement to answer questions 10 and 11.
June statement
10 Calculate the average daily balance for June.
Date
11 If the annual percentage rate is 22.1%, calculate the interest charges for June.
1 Jun
Opening balance
589
8 Jun
Purchase
135
15 Jun
Purchase
68
21 Jun
Payment
−450
24 Jun
Purchase
33
Details
Amount ($)
12 Calculate the future value of $25 000 invested for 4 years at 5.9% p.a. interest compounded six-monthly. 13 What single amount must be invested now at 5.5% p.a. interest compounded monthly in order to have $35 000 in 2 years time?
1 EXAMINATION QUESTION (15 MARKS)
b Lilly paid $2275 interest on a flat-rate loan of $14 000 over 2_12 years. What was the interest rate?
FINANCIAL MATHEMATICS
a Kerri borrows $8000 to buy a car. The flat interest rate is 8.6% and she takes the loan over 5 years. i Find the amount of interest charged. (1 mark) ii Find the total amount to be repaid. (1 mark) iii Find the monthly repayment. (1 mark) (1 mark)
Chapter 1 Credit and borrowing
29
c Use this credit card statement to answer parts i and ii. May statement Date
Amount ($)
Details
1 May
Opening balance
174
12 May
Electricity bill
327
17 May
Purchase
85
21 May
Payment
−200
26 May
Purchase
166
i Complete the following table to calculate the average daily balance for the month. Daily balance ($)
Number of days
Aggregated balance ($)
174
11
1914
586
4
2344
386
5
1930
502
6
3012
(3 marks)
501
Total
ii Calculate the interest charges for May if the annual percentage rate is 18.4% p.a. d The table shows the payment per $1000 on a monthly reducible loan. Use the table to find the monthly repayment on a $290 000 loan over 15 years at 7.25% p.a. Term (years)
7%
7.25%
7.5%
5
19.8012
19.9194
20.0379
10
11.6108
11.7401
11.8702
15
8.9883
9.1286
9.2701
20
7.7530
7.9036
8.0559
e A home loan of $320 000 is taken out over 30 years at an interest rate of 8.5% p.a. monthly reducible. The monthly repayment is $2460.52. Complete the next line of the table below to find the amount owing on the loan at the end of the second month. Months (n)
Principal (P)
Interest (I)
P+I
Balance = P + I − R
1
320 000
2266.67
322 266.67
319 806.15
(2 marks) (1 mark)
(2 marks)
FINANCIAL MATHEMATICS
2
30
f Calculate the future value of $10 000 invested at 10% p.a. compounding quarterly for 6 years.
(1 mark)
g A company owner must repay a loan of $120 000 in 5 years time. What single amount of money must he invest now at 9% p.a., compounding monthly, in order to be able to repay the debt?
(2 marks)
Insight Mathematics General 12 HSC Course 2
Further applications of area and volume This chapter deals with area, surface area and volume. The main mathematical ideas investigated are: ▶ calculating areas of plane shapes with curved surfaces ▶ calculating the area of composite figures ▶ using Simpson’s rule to approximate area ▶ calculating surface area and volume of solids ▶ determining errors in measurement.
MEASUREMENT
Syllabus references: MM4 Outcomes: MGH-4, MG2H-5, MG2H-10
2A
Area of a circle
In the Preliminary Mathematics General course the formula for the area of a circle was given. Area of a circle = π × radius squared A = πr 2
r
WORKED EXAMPLE 1 Find the area of each circle, correct to 1 decimal place.
a
b 5 cm
7m
Solve
Think
a
A = πr 2 = π × 52 = 78.539… = 78.5 cm2
A = πr 2
b
A = πr 2 = π × (3.5)2 = 38.484… = 38.5 m2
A = πr 2 Diameter is 7 m, so halve the diameter to get the radius of 3.5 m.
Apply Always use the radius to find the area. If given the diameter, then halve it to find the radius.
EXERCISE 2A 1 Use your calculator to find the area of each circle, giving answers correct to 1 decimal place. a b c d e 9 cm 3 cm
4.7 mm
6.8 cm
7 mm
MEASUREMENT
2 Use your calculator to find the area of each circle, giving answers correct to 1 decimal place. a b c d e 12 cm
32
Insight Mathematics General 12 HSC Course 2
50 mm
16 cm
4.8 mm
10.2 cm
3 Calculate the area of these circles. Evaluate the answers correct to 1 decimal place. a b c d 18 mm
31 cm
15 cm
11 cm
e
f 17 cm
g
h
6 cm
3.9 cm 17.2 mm
i radius = 31 mm l diameter = 9.3 cm
j diameter = 15 cm m radius = 17.4 cm
k diameter = 67.5 cm n radius = 23.8 cm
WORKED EXAMPLE 2 a Determine what fraction of a circle is drawn. b Determine the area to the nearest whole cm2.
6.8 cm
Solve
Think
Apply
1
There are 360° in a circle. The fraction is the angle 90° divided by 360°.
1
The
Calculate the fraction of the circle by dividing the angle by 360°, then multiply by the normal area formula.
a
90 1 ____ =_ 360 4
b
Area = _4 of area of circle = _4 × πr2
symbol means 90°.
1
= _4 × π × (6.8)2 = 36.316… = 36 cm2
4 Determine what fraction of a circle is drawn, then calculate the area correct to the nearest cm2. a b c
120°
11 cm
7.8 cm
15.3 cm
60° 9.3 cm
f 40° 6.2 cm
MEASUREMENT
e
d
8.3 cm
Chapter 2 Further applications of area and volume
33
WORKED EXAMPLE 3 27 cm
Calculate the area enclosed by this composite figure to the nearest cm2.
13 cm
Solve
Think
Apply
There are two semicircles
The ends combine to make a circle with a 6.5 cm radius. Area 1 = πr2 = π × (6.5)2 = 132.732… cm2 = 132.7 cm2 Area 2 = length × breadth = 13 cm × 27 cm = 351 cm2 Total area = Area 1 + Area 2 = 132.7 + 351 = 483.7 = 484 to the nearest cm2
Area 1
1 Area 2
27 cm
1
and one rectangle.
Divide the composite shape into simpler plane shapes. Calculate the areas then add them together.
2 Area 1
Add the areas of the semicircles and the rectangle together.
13 cm
5 Determine the area, correct to the nearest cm2, enclosed by these composite figures. a b c 5 cm
8 cm 12 cm 3 cm 10 cm
d
e
f
7.4 cm
6.3 cm
27.2 cm
16.3 cm
8.6 cm
WORKED EXAMPLE 4 Determine the area of this annulus correct to 1 decimal place. An annulus is the area between two concentric circles (that is, circles with the same centre but different radii).
MEASUREMENT
Solve A = π(R2 − r2) = π(5.52 − 2.52) = 75.398… = 75.4 cm2
34
Insight Mathematics General 12 HSC Course 2
11 cm
5 cm
Think
Apply
The large radius is 11 ÷ 2 = 5.5 cm. The small radius is 5 ÷ 2 = 2.5 cm.
The area of an annulus formula is A = π(R2 − r2).
WORKED EXAMPLE 5 Find the area of this half annulus correct to 1 decimal place.
7 cm 16 cm
Solve 1
A = _2 × π(R2 − r2) 1
= _2 × π(82 − 3.52) = 81.289… = 81.3 cm2
Think
Apply
Halve the formula as only half an annulus is shown. Find the radius of each circle first.
The formula should only be used if both circles have the same centre. Otherwise find each area separately and subtract.
6 Determine the area of the shaded region correct to 1 decimal place. a b 6 cm
c
2 cm 3.1 cm 8.3 cm
12 cm
d
5 cm
e
3 cm
f
5 cm 15 mm
80° 7 cm
3 cm
h
g
m
9 cm
6c
m
8c
10 cm
i r 3.6 cm
8.9 cm
7 The diagram shows a stained glass window in a church. a Calculate the area of the window correct to 2 decimal places. b The cost of replacing the window is $540/m2. Calculate the cost of replacing this window.
MEASUREMENT
14.7 cm
5m
3m
Chapter 2 Further applications of area and volume
35
8 A 3 m wide path is placed around a circular pond that has a diameter of 6 m. Find the area of the path correct to 1 decimal place.
3m 6m
9 The race track shown is to be resurfaced. a Calculate the area to be resurfaced, to the nearest metre. b Calculate the cost of the resurfacing, given a cost of $9.90/m2. Give your answer to the nearest $10.
20 m
15 m
40 m
2.2 m
10 The diagram shows a pool surrounded by a rectangular paved area. a Calculate the paved area (correct to the nearest square metre). b If paving costs $34.90/m2, or part thereof, calculate the cost of paving the area surrounding the pool.
5m
Pool 4.2 m
2B
Simpson’s rule
9m
In the Preliminary Mathematics General course, the traverse survey was used to find the area of fields with straight boundaries. If the property being surveyed has an irregular boundary, such as a river, another rule can be used to find the area. The rule is called Simpson’s rule. This rule gives an estimate of the area. To find the area a number of measurements need to be found. This diagram shows the measurements needed for the calculation. The distance between offsets must be equal.
MEASUREMENT
Simpson’s rule is: h A ≈ __(df + 4dm + dl) where 3 h = the equal distance between offsets df = the first offset dm = the middle offset dl = the last offset
36
Insight Mathematics General 12 HSC Course 2
df
dm dl h
h
WORKED EXAMPLE 1 Use Simpson’s rule to find an approximation for the area of this irregular field. 25 m
42 m
30 m
31 m
30 m
Solve
Think
Apply
h A ≈ __(df + 4dm + dl) 3 30 ≈ ___ × (25 + 4 × 42 + 31) 3 ≈ 2240 m2
h = 30, df = 25, dm = 42, dl = 31
The equal distance between offsets is the value for h. If the offsets are not equally spaced, Simpson’s rule cannot be used.
WORKED EXAMPLE 2
22 m
Find an approximation for the area of this field. 10 m 16 m 10 m 28 m
Solve
Think
Apply
h A ≈ __(df + 4dm + dl) 3 10 ≈ ___ × (22 + 4 × 16 + 28) 3 ≈ 380 m2
h = 10, df = 22, dm = 16, dl = 28.
The equal distance between offsets does not have to be in the horizontal direction.
EXERCISE 2B 1 Use Simpson’s rule to find an approximation for the area of these fields. (Measurements are in metres.)
b 18
23
16 30
c 35
33
30
d
30
25
14
5 18
30
e
36
16
31
f
420
18 420 MEASUREMENT
a
350
25
65 14
25
33
750
800
16 48
16
Chapter 2 Further applications of area and volume
37
g
420
h
420 250
50 40
450
40
WORKED EXAMPLE 3 Use Simpson’s rule twice to find the area of this field.
15 m
15 m
15 m A2
A1 18 m
15 m
12 m
14 m
16 m
20 m
Solve
Think
Apply
h A1 ≈ __(df + 4dm + dl) 3 15 ≈ ___ × (18 + 4 × 14 + 12) 3 ≈ 430 m2 h A2 ≈ __(df + 4dm + dl) 3 15 ___ × (12 + 4 × 16 + 20) ≈ 3 ≈ 480 m2 Total area = 430 + 480 = 910 m2
The distance between offsets is 15 so h = 15. The offsets are df = 18, dm = 14, dl = 12.
When Simpson’s rule is used twice in this way, the middle offset is the dl for the left side and the df for the right side. It is the only offset used for both areas.
The distance between offsets is 15 so h = 15. The offsets are df = 12, dm = 16, dl = 20. The last offset for the first area is now the first offset for the second area.
2 Use Simpson’s rule twice to find the area of each field. All measurements are shown in metres. a b c 41 25 25 25 25 25 58 40
32
36
34
33
16
20
46
16
28
16 10
d
10
10
e 10
MEASUREMENT
15 210
18
45
42
20
20
38
18
30
20
36
Insight Mathematics General 12 HSC Course 2
14
f 18
20
33
16
10
15
36
18
21
130
120 110
47
110
110
110
3 Use two applications of Simpson’s rule to find the area of these sections of land. a b c 12 m 23 m 37 m 33 m 40 m 28 m
35 m
17 m
45 m
45 m
50 m
30 m 52 m
40 m
38 m
61 m
8m 30 m
33 m
26 m 50 m
68 m
28 m
4 The diagram of a dam is shown. a Use the measurements and Simpson’s rule to find the combined area of the dam and land. b Use Simpson’s rule to find the land area. c Find the area of the dam. d The average depth of water is 10 m. Give an estimate for the volume of water in the dam.
Dam 145 m 150 m 60 m
25 m 120 m
5 a Use Simpson’s rule to find the area of this dam. b If the average depth is 18 m, find an estimate for the volume of water in the dam.
120 m
50 m
50 m 22 m
28 m
63 m Dam 72 m 37 m 18 m 50 m
7 The diagram shows the cross-section of a river. a Use Simpson’s rule twice to find an estimate of the area of the cross-section. b If water flows past a particular point at 2 metres per minute, find the volume of water passing that point in 10 minutes. c How much water flows past in an hour?
5m
5m
12 m
8m
8m
8m
5m
11 m
8m
5m
13 m
8m 5m
14 m
12 m
9m
4m
17 m MEASUREMENT
6 The diagram shows the cross-section of a river. a Use Simpson’s rule twice to find an estimate of the area of the cross-section. b If the water flows at 10 metres per minute, find the volume of water passing a particular point on the river over 1 minute. Hint: This is the same as finding the volume with a ‘height’ of 10 m. c How much water flows past in 1 hour?
50 m
INVESTIGATION 2.1
Chapter 2 Further applications of area and volume
39
2C
Surface area BACK
The surface area of any shape refers to the sum of the areas of each face. This box has 6 faces: 2 sides, front and back, and top and bottom.
TOP E
SID
SID
E
BOTTOM FRONT
EXERCISE 2C 1 State the number of faces for each solid. a b
c
e
f
g
WORKED EXAMPLE 1 Calculate the total surface area of this rectangular prism.
Top 56 cm2
MEASUREMENT
Front 24 cm2
40
e Sid 2 m c 2 3
Solve
Think
Apply
Total surface area = (2 × 32) + (2 × 24) + (2 × 56) cm2 = 64 + 48 + 112 cm2 = 224 cm2
The rectangular prism has 6 faces. Area of two sides = (2 × 32 cm2) Area of front + area of back = (2 × 24 cm2) Area of top + area of bottom = (2 × 56 cm2)
Count the number of faces first. Make sure all areas are calculated. If some faces are the same, the calculations are simpler.
Insight Mathematics General 12 HSC Course 2
2 Calculate the surface area of each solid, using the area given. a b Top
c
96 cm2 20 cm2
e Sid 2 m c 0 5
Front 70 cm2
Sloping face 234 cm2
de 2 Si cm Back 0 A = 60 cm2 8 1 Base 170 cm2
d
e
f
2
A A = 48 cm
2
00 =2
2 A= Base 160 cm2
6 cm
m
8c
cm
2
2
m
A = 85 cm2
A=
c 24
18.2 cm
Base 40 cm2
A = 36 cm2
WORKED EXAMPLE 2 Calculate the surface area of each solid.
a
b 4 cm 5 cm 7 cm
4 cm
a
Surface area = 6 × area of one face =6×s×s = 6 × 4 × 4 cm2 = 96 cm2
b
Area of front and back = 2 × area of front = 2 × 7 × 4 cm2 = 56 cm2 Area of sides = 2 × area of one side = 2 × 5 × 4 cm2 = 40 cm2 Area of top and bottom = 2 × area of top =2×7×5 = 70 cm2 ∴ Surface area = 56 + 40 + 70 cm2 = 166 cm2
Think All faces are identical. Multiply the area of one face by 6.
Rectangles
Apply Count all faces. If some faces are the same then the number of calculations are reduced. A cube always has six identical faces. A rectangular prism has three pairs of identical faces.
Rectangles
Rectangles
MEASUREMENT
Solve
Chapter 2 Further applications of area and volume
41
3 Calculate the surface area of each cube. a b
c
5 cm
6 cm
9 cm
d
e
f
3 cm
12 cm
15 cm
4 Calculate the surface area of cubes with sides of the following lengths. a 4.8 cm b 0.7 m d 9.03 cm e 0.31 m
c 17 mm f 14.4 cm
5 Calculate the surface area of these rectangular prisms. a b
c
4 cm 7 cm
3 cm
15 cm
10 cm
25 cm 8 cm
8 cm 4 cm
d
e 6 cm 8 cm
45 mm
12 cm
60 mm
f
17 mm
2m 30 m
130 m
MEASUREMENT
6 Calculate the surface area of the rectangular prisms with the dimensions given in the table. Drawing a diagram can help.
42
Length
Breadth
Height
a
9 cm
12 cm
7 cm
b
11 cm
20 cm
5 cm
c
8m
9m
4m
d
12 mm
15 mm
6 mm
e
15.3 m
6.4 m
4.3 m
Insight Mathematics General 12 HSC Course 2
7 Find the surface area of the solid formed by each net. a b
c 6 cm 5 cm
6 cm
10 cm
4 cm
WORKED EXAMPLE 3 Calculate the surface area of this triangular prism. 4 cm 15 cm 6 cm
Think
A1 = 2 × area of one triangle =2× =2×
1 _ 2 (b × h) 1 _ 2×6×4
Apply
Step 1: Surface area of two triangles (front and back)
cm2
= 24 cm2
All prisms are solid objects that have one pair of identical ends. All other sides are rectangles. The cross-section is the same all along its length. The shape of the ends give the prism its name (for example, triangular prism).
Triangles
A2 = 2 × surface area of two rectangles = 2_______ × l × b ___ 2 b = √4 + 32 = √25 = 5 Thus: A2 = 2 × 15 × 5 cm2 = 150 cm2 A2 = l × b = 15 × 6 cm2 = 90 cm2 Total surface area = 24 + 150 + 90 cm2 = 264 cm2
Step 2: Surface area of two rectangles (identical) Use Pythagoras’ rule to find the breadth of the rectangle.
Step 3: Surface area of one rectangle (base)
8 Calculate the surface area of these triangular prisms. a b
Rectangle
c 10 cm
8 cm
Rectangles
17 cm
MEASUREMENT
Solve
15 cm
12 cm
5 cm
18 cm
4 cm 12 cm
25 cm 16 cm
Chapter 2 Further applications of area and volume
43
d
e
f
41 cm
12 cm 8 cm
5 cm
9 cm
13 cm
4 cm
15 cm
80 cm 13.6 cm
5 cm
6 cm
g
h 17 cm
i
15 cm
9 cm
4.8 cm
8 cm
24.4 cm 7.8 cm
15 cm
20 cm
21.6 cm
24 cm
28 cm
WORKED EXAMPLE 4 Calculate the surface area of the composite prism.
10 cm 8 cm
6 cm 7 cm 12 cm
Solve 1
A1 = _2 b × h 1 = _2 × 12 × 8 cm2
Think A2 = l × b = 12 × 6 cm2
MEASUREMENT
= 48 cm2 = 72 cm2 Composite area = 48 + 72 = 120 cm2 Area front and back = 2 × 120 = 240 cm2 Area of sides = 2 × l × b =2×7×6 = 84 cm2 Area of roof = 2 × l × b = 2 × 10 × 7 = 140 cm2 Area of base = l × b = 12 × 7 = 84 cm2 Total surface area = (front/back) + (sides) + (roof) + (base) = 240 + 84 + 140 + 84 = 548 cm2
44
Insight Mathematics General 12 HSC Course 2
Step 1: For the front and back, find the composite area and multiply by 2. 1
Triangle
2
Rectangle
Step 2: The sides are rectangles. Area is two times l × b. Step 3: The roof is rectangles. Area is two times l × b. Step 4: The base is a rectangle. Area is l × b.
Apply There are many pairs of equal-sized faces. Count the faces to ensure that all areas are calculated. The base can be any planar shape, but all the other sides of a prism are rectangular.
9 Calculate the total surface area of these composite shapes. 17 cm a b
c
5.7 cm
8 cm 8 cm
28 cm
18.2 cm 9 cm
12 cm
15 cm
5 cm
12
d
6.2 cm
4 cm
.9
e
cm
f
38 cm
17.8 cm
17 cm
9.2 cm 12.3 cm
18.1 cm
7 cm
32.9 cm
3 cm
26 cm
15 cm
33 cm 50 cm
6.4 cm
8 cm
29 cm
14 cm
10 A gift box measures 10 cm by 12 cm by 20 cm. a Calculate its total surface area. b Gift wrapping paper costs $3.50 a sheet. If each sheet covers 290 cm2, calculate the total cost of wrapping the gift box. (Remember: You must purchase complete sheets.) 11 a Calculate the wall area of Belinda’s room. Calculate the wall area of Tania’s room. Which room has the greater wall area? Tania’s room
Belinda’s room 2.2 m
5.6 m
2m
3.1 m
6.3 m 2.3 m
b A 4-litre tin of paint covers approximately 10 square metres and costs $34.95. Calculate the number of 4-litre tins of paint needed to paint Tania’s room. c Calculate the cost of purchasing paint for Tania’s room.
12 Which cereal box has the greater surface area?
12 cm 26 cm 22.5 cm
5.5 cm
15.5 cm
13 The interior walls and floor of an in-ground swimming pool are to be repainted. a Calculate the total surface area of the four walls and floor of the pool. b How many cans of paint are needed if one can covers 70 m2? 2.3 m c What is the cost of repainting, if each can costs $82.50?
MEASUREMENT
9 cm
20 m 9m
Chapter 2 Further applications of area and volume
45
2D
Surface area of a cylinder r
In order to find a formula for the surface area of an open cylinder we examine the following diagram showing an open cylinder cut perpendicular to the base. The cylinder, when folded out, forms a rectangle.
The width of the rectangle is equal to the height of the cylinder. The length of the rectangle is equal to the circumference of the circle that formed the top or bottom. The circumference of a circle is given by C = 2πr, therefore the area of the curved surface of the cylinder is A = 2πrh. A closed cylinder has a circle on each end. Cut out it looks like this. A = 2(area of circle) + area of rectangle = 2 × πr2 + 2πr × h = 2πr2 + 2πrh
h
h
2π r r
h
h 2π r r
Cut
The surface area of a closed cylinder is given by A = 2πr2 + 2πrh.
WORKED EXAMPLE 1 Find the outer surface area of these open cylinders, correct to 1 decimal place.
a
b
5 cm Open 12 cm
22 m Open
7m
MEASUREMENT
Solve
46
Think
a
A = 2πrh = 2 × π × 5 × 12 = 376.991… = 377.0 cm2
Use A = 2πrh, r = 5, h = 12.
b
A = 2πrh = 2 × π × 11 × 7 = 483.805… = 483.8 m2
Use A = 2πrh, diameter is 22 m so r = 11, h = 7.
Insight Mathematics General 12 HSC Course 2
Apply As the cylinders are open, only the curved surface area needs to be calculated. Use the formula A = 2πrh.
EXERCISE 2D 1 Find the outer surface area of these open cylinders, correct to 1 decimal place. 5 cm a b c
12 cm
9.2 cm 17 cm
d
14 cm
23 cm
e
4.2 cm
f 16 cm 8.6 cm
9.7 cm 20 cm 2.3 cm
2 Find the outer surface area of open cylinders with these dimensions. Give your answer correct to 1 decimal place. a radius = 4 cm height = 12 cm b diameter = 18 cm height = 13 cm c radius = 3 cm height = 5 cm d diameter = 11.2 cm height = 23.6 cm e diameter = 15.8 cm height = 9.2 cm
WORKED EXAMPLE 2 Find the surface area of these closed cylinders correct to 1 decimal place.
a
4 cm
b
3.6 mm
18 cm
Solve
Think
Apply The cylinders are closed, so the area of the top and bottom circles is added to the curved surface area.
a
A = 2πrh + 2πr2 = 2 × π × 4 × 18 + 2 × π × 42 = 552.920… = 553.0 cm2
Use A = 2πrh + 2πr2 with r = 4 and h = 18.
b
A = 2πrh + 2πr2 = 2 × π × 1.8 × 12.4 + 2 × π × 1.82 = 160.598… = 160.6 mm2
Halve the diameter of 3.6 mm to find the radius: r = 1.8 mm and h = 12.4 mm.
Chapter 2 Further applications of area and volume
MEASUREMENT
12.4 mm
47
3 Find the total surface area of these closed cylinders correct to 1 decimal place. a b c 13 cm
20 cm 25 cm 3 cm
20 cm
8 cm
e
d
f
13.2 cm
17 cm
11.4 m
38 cm 24.6 cm 3.2 m
4 Find the outer surface area of these cylinders that are open at one end only. Open a b
8 cm
2.8 m
Open
15 cm
1.3 m
5 A cylindrical wheat silo is 40 m high and 20 m in diameter. a Find the surface area of the silo including the top (closed at one end only.) b If paint covers 8 m2/L, find the number of 10-litre cans of paint needed to paint the outside of the silo with one coat. c Paint costs $115.20 per 10-litre can. Find the cost of paint for the silo.
2E
Surface area of a sphere
MEASUREMENT
The formula for the surface area of a sphere of radius r is A = 4πr2.
48
Insight Mathematics General 12 HSC Course 2
r
WORKED EXAMPLE 1 Find the surface area of these spheres correct to 1 decimal place.
a
b 3 mm 12.4 cm
Solve
Think
Apply
a
A = 4πr2 = 4 × π × 32 = 113.097 3… = 113.1 mm2
Radius is 3 mm and use A = 4πr2.
b
A = 4πr2 = 4 × π × (6.2)2 = 483.051 286… = 483.1 cm2
Diameter is 12.4 cm so r = 6.2 cm.
The only variable is the radius. If diameter is given then halve it to find radius.
EXERCISE 2E 1 Find the surface area of these spheres correct to 1 decimal place. a b
c
7 cm 15 cm
20 cm
2 Find the outer surface area of these open hemispheres correct to 2 decimal places. a b c 11.49 m 8.42 cm
1.39 m
MEASUREMENT
3 Find the surface area of these closed composite figures correct to 1 decimal place. a b
6.4 m
11.5 cm
8.2 cm 4.3 m
Chapter 2 Further applications of area and volume
49
4 How many spheres of 15 cm diameter can be painted with 1 L of glitter paint if the paint covers 10 m2? 5 Determine the total area of leather in 20 dozen cricket balls each with diameter 7 cm.
INVESTIGATION 2.2 Investigation 2.2 should be completed before commencing Section 2F.
2F
Further volume h
h
h
Triangular-based pyramid
Square-based pyramid
Cone
In Investigation 2.2 we found that the volume of a pyramid or cone is given by: 1
Volume = _3 (area of base × height)
or
1
V = _3 Ah
WORKED EXAMPLE 1 Find the volumes of these solids.
a
b 8 cm
12 cm
6 cm 10 cm
a
Solve
Think
Apply
Volume = _3 × area of base × height
The base is a square side 10 cm and h = 12 cm.
For all pyramids and cones 1 the volume is _3 the area of the base multiplied by the height.
1 1
= _3 × 10 × 10 × 12 cm3 MEASUREMENT
= 400 cm3
b
1
Volume = _3 × area of base × height 1
= _3 × (π × 62) × 8 cm3 = 96π cm3 ≈ 301.6 cm3
50
Insight Mathematics General 12 HSC Course 2
The base is a circle of radius 6 cm and the height is 8 cm.
EXERCISE 2F 1 Find the volume of each solid. a
b
12 cm
c 10 cm
8 cm
8 cm
d
10 cm
3 cm
e
10 cm
f 7 cm
12 cm
8 cm
5 cm 5 cm
6 cm
4 cm
4 cm
2 Find the volume of the following solids correct to 1 decimal place where necessary. a a pyramid with a triangular base 4 cm wide and 5 cm high, and a pyramid height of 6 cm b a cone with base radius 8 cm and height 12 cm c a square-based pyramid of height 12 cm and base edges 6 cm 3 Find the volume of each solid to the nearest whole number. a b
c
50 cm
9 cm 8 cm 75 cm 5 cm
12 cm
6 cm
d
e 40 cm
7 cm
5.2 cm
f 8 cm
20 cm
4.3 cm
30 cm
In the Preliminary Mathematics General course you examined the volumes of prisms, cylinders, pyramids and cones. This section revises formulas for calculating volumes of these solids. The formulas involved are: V = Ah V = lbh V = πr2h
Pyramid
V = _3 Ah
Cone
V = _3 πr2h
1
A is the area of the base MEASUREMENT
Prism Rectangular prism Cylinder
A is the area of the base
1
Chapter 2 Further applications of area and volume
51
4 Find the volume of these solids. a
b
c
0.5 m
21 cm 3m
32 cm 45 cm
1.3 m
e
d
f 5.5 cm
1.2 m
12.5 cm
6.3 m 6.5 cm 8.9 m
0.8 m
5 Find the volume of the following solids. a a 10 m by 8 m by 5 m rectangular prism b a cylinder of base radius 7 cm and height 10 cm
2G
Volume of composite solids
WORKED EXAMPLE 1 Find the volumes of these solids, correct to 1 decimal place.
a
b 3m 28 cm
4m
16 cm
4m
Solve
a
1
V = πr2h + _3 πr2h
1
= π × 22 × 4 + _3 × π × 22 × 3 ≈ 62.83 m3
b
1
V = _2 × volume of cylinder 1
MEASUREMENT
= _2 × πr2h 1
= _2 × π × 82 × 28 ≈ 2814.9 cm3
52
Insight Mathematics General 12 HSC Course 2
Think
Apply
The radius is 2 m. Volume of a cylinder is V = πr2h.
Reduce the composite solids to solids for which you have volume formulas.
1
Volume of a cone is V = _3 πr2h. The radius is 2 m. Volume of a cylinder is V = πr2h and this is half a cylinder.
1 Find the volume of each solid. a
b 5 cm
A = 20 cm2
5 cm 20 cm 8 cm
d
c
10 cm
15 cm 10 cm
15 cm 20 cm
e
f
3m 1m
12 m
3m 5m 1m
10 m
2 Find the volume of these solids correct to 1 decimal place where necessary. a b
c 40 cm
1.4 m 12 cm
2m
14 cm
6m
30 cm
4.2 m
d
e
f 4 cm
5 cm
8 cm
5 cm 5 cm 12 cm 9 cm 7 cm
h
1.2 cm
i 3.1 cm
5.3 cm 8.4 cm 3.7 cm
14 cm
MEASUREMENT
g
12 cm
21 cm 6 cm 9 cm
Chapter 2 Further applications of area and volume
53
j
k
l 5 cm
6 cm
11.2 cm
3 cm
15 cm 9 cm
10 cm
6.5 cm
12 cm
12.5 cm
m
n
6 cm
8 cm
2 cm
5 cm
o
70 cm 15 cm
3 cm 10 cm
30 cm
4 cm
50 cm
9 cm
3 Find the total volume of a piece of steel with length 6 m and cross-section as shown.
1 cm
8 cm
4 A rectangular swimming pool 6 m by 5 m by 2 m deep costs $0.50 per cubic metre per month to maintain. What is the cost of maintaining the pool for a year?
6 cm
5 Concrete costs $150/m3. What will it cost to concrete a driveway 20 m long and 3 m wide to a depth of 12 cm?
MEASUREMENT
6 The diagram represents the entrance of a cave cut into rock. a Use Simpson’s rule to find the area of the entrance of the cave. b If the depth is 40 m, find the volume of rock removed. 18 m 9m
INVESTIGATION 2.3 54
Insight Mathematics General 12 HSC Course 2
12 m
10 m
10 m
2H
Accuracy of measurements
Most measurements involve reading some kind of scale. No matter how many subdivisions the scale has, the object being measured is likely to fall between two of them. Generally, it is possible to decide if a measurement is closer to one marking on the measuring instrument than another. Therefore measuring instruments are accurate to half the smallest division on the scale. As a result, every measurement is an approximation. Most measurements may come close but never match the scale perfectly. A ruler has markings in millimetres and therefore is accurate to the nearest half a millimetre. A given measurement is accurate to ±_12 of the smallest division on the scale. 1
The percentage error is used to show the overall accuracy of a measurement. An error of _2 cm on a measurement of 5 cm is much more significant than the same error on a measurement of 1000 cm.
WORKED EXAMPLE 1 a Find the range of values for a measurement of 36 mm accurate to the nearest mm. b Calculate the percentage error. Solve
a
The measurement is accurate to the nearest 0.5 mm. The range of values is 35.5 mm ⩽ the measurement < 36.5 mm.
b
0.5 ±___ × 100% = ±1.4% 36
Think
Apply
The smallest value, or lower bound, is 35.5 mm and the largest value, or upper bound, must be less than 36.5 mm.
Add and subtract the error to find the range of values.
Divide the error by the measurement and multiply by 100%.
The percentage error gives an indication of the significance of the error.
EXERCISE 2H
MEASUREMENT
1 How accurate are the following measuring instruments? a A ruler marked in centimetres b A scale marked in grams c A measuring cylinder marked in mL d A trundle wheel that clicks every metre e Vernier callipers marked in hundredths of a millimetre f A clinometer marked in degrees g A metre rule marked in graduations of 10 cm h A thermometer marked with a line between every degree i A compass with 4 divisions for each 10 degrees j A watch with markings every 5 minutes
Chapter 2 Further applications of area and volume
55
2 Aaron and Jamal measured the length of the school netball court. Aaron measured it to be 30 m and Jamal measured it to be 29.5 m. The boys told the teacher their answers and the teacher said that they were both correct. Explain the teacher’s answer. 3 Find the range of values for the following measurements. State the upper and lower bounds. a 54 mm b 5m c 23 cm d 3.8 km e 4.3 mm f 18.6 cm g 12.36 cm h 2.71 km i 200 m to the nearest metre j 200 m to the nearest 100 m k 200 m to the nearest 10 m l 20 cm 4 Calculate the percentage error for the measurements in question 3. 5 Five students measured the length of their classroom using the same measuring instrument. Their measurements were 8.3 m, 7.8 m, 8.0 m, 7.9 m, 10.5 m. a One measurement is obviously incorrect. Which is it, and explain one possible way of making an error. b What answer would you give for the length of the classroom? Why? c How accurate are the measurements? d Comment on the possible sizes of the divisions on the scale of the measuring instrument. 6 A rectangle has length 15 cm and width 9 cm. a Calculate the area of the rectangle. b Use the lower bound for each measurement and recalculate the area. c Use the upper bound for each measure and recalculate the area. d What would be a reasonable estimate for the area? Explain your answer. 7 Repeat question 6 for a circle radius 7.3 cm. 8 A rectangular storage box has dimensions 122 cm by 75 cm by 53 cm. a Calculate the volume of the box. b Use the lower bound for each measurement and calculate the volume of the box. c Use the upper bound for each measurement and calculate the volume of the box. d What would be a reasonable answer for the volume of the box? Explain your answer. 9 a A cone has dimensions as shown. Calculate the volume of the cone. b Calculate the upper and lower bounds for the volume of the cone. c Give a reasonable measure for the volume of the cone. Explain your answer.
8.5 cm
MEASUREMENT
SPREADSHEET APPLICATION 2.1 SPREADSHEET APPLICATION 2.2 SPREADSHEET APPLICATION 2.3 56
Insight Mathematics General 12 HSC Course 2
4.9 cm
SPREADSHEET APPLICATION 2.1 The size of containers 200 cm
This exercise may be completed using scale models or a spreadsheet. You are given a square piece of metal of side length 2 m and asked to design a water tank in the shape of an open rectangular prism by cutting squares from the corners and bending the metal. This is shown in the diagram.
200 cm x cm
The formula for the volume is given by V = x(200 − 2x) . 2
x cm
The purpose of this exercise is to produce the container that holds the maximum amount of water. This information has been entered into the spreadsheet below. Formula in cell B4: =$B$1−2*A4 Formula in cell C4: =A4*B4^2 A
B
Square side
200
3
Height (x)
Length (200 − 2x)
Volume (V)
4
5
190
180 500
5
10
180
324 000
6
15
170
433 500
7
20
160
512 000
8
25
150
562 500
9
30
140
588 000
10
35
130
591 500
11
40
120
567 000
12
45
110
544 500
13
50
100
500 000
14
55
90
445 500
15
60
80
384 000
16
65
70
318 500
17
70
60
252 000
18
75
50
187 500
19
80
40
128 000
20
85
30
76 500
21
90
20
36 000
22
95
10
9 500
23
100
0
0
1
C
D
E
F
G
2 800 000 700 000 600 000 500 000 400 000 300 000 200 000 100 000 0
0
20
40
60
80
100
120
From this spreadsheet the prism with maximum volume has height 35 cm and length 130 cm. MEASUREMENT
1 Modify the spreadsheet to find the value(s) for maximum volume more accurately. 2 Experiment with squares of other sizes. 3 Build your own models using cardboard and investigate the dimensions. 4 Write a report outlining your findings. Chapter 2 Further applications of area and volume
57
SPREADSHEET APPLICATION 2.2 Maximum or minimum You are provided with enough material to make a container with surface area of 660 cm2. You are instructed to make the ‘best’ container possible. You must explain why this is the ‘best’ container. The following spreadsheet assumes that the container is a square prism. This may not be the ‘best’ container, but it is a starting point. The information is entered into the spreadsheet as shown below. A
B
C
D
E
1
L (cm)
B (cm)
H (cm)
Vol. (cm3)
SA (cm2)
2
1
1
=(660−2*A2^2)/(4*A2)
=A2*B2*C2
660
=B2+1
=(660−2*A3^2)/(4*A3)
=A3*B3*C3
660
3
=A2+1
The fill down command is used to give the following spreadsheet, and hence a scatter plot graph showing this information. A
B
C
D
E
1
L (cm)
B (cm)
H (cm)
Vol. (cm )
SA (cm )
2
1
1
164.50
164.50
660
3
2
2
81.50
326.00
660
4
3
3
53.50
481.50
660
5
4
4
39.25
628.00
660
1000
6
5
5
30.50
762.50
660
800
7
6
6
24.50
882.00
660
8
7
7
20.07
983.50
660
9
8
8
16.63
1064.00
660
10
9
9
13.83
1120.50
660
11
10
10
11.50
1150.00
660
12
11
11
9.50
1149.50
660
13
12
12
7.75
1116.00
660
14
13
13
6.19
1046.50
660
15
14
14
4.79
938.00
660
16
15
15
3.50
787.50
660
17
16
16
2.31
592.00
660
18
17
17
1.21
348.50
660
19
18
18
0.17
54.00
660
F
G
H
I
2
1200
Volume (cm3)
3
Volume of square prism (fixed surface area)
600 400 200 0
0 2 4 6 8 10 12 14 16 18 20 Length (cm)
MEASUREMENT
From this information the container that would hold the most has a maximum volume of 1150 cm3 with dimensions 10 cm × 10 cm × 11.5 cm. There may be a container with a slightly larger volume if smaller increments are used for the changes in lengths. It may be that maximum volume is the ‘best’ container or there may be some other criteria.
1 Modify the spreadsheet to check if there is a container with a greater volume. (Use 0.1 increments.) 2 Investigate containers with other shapes; for example, rectangular prism, cylinder, etc. 58
Insight Mathematics General 12 HSC Course 2
SPREADSHEET APPLICATION 2.3 The ‘best’ container You are asked to design a container that will hold 1000 mL. It needs to be the ‘best’ container. What shape container will you choose and what will its dimensions be? You must use a spreadsheet to explain your choice. It can be similar to the one in Spreadsheet application 2.2. Here are some examples. r
x cm
h (x + 2) cm
x cm Square prism
Rectangular prism with length 2 cm more than the width
Cylinder
Modify your spreadsheets for other solid shapes, such as a square pyramid, a rectangular prism with length 2.5 cm longer than the width, a rectangular pyramid, a sphere, etc.
INVESTIGATION 2.1 The area of a playground
7.6 m
5 cm
9m
14 cm 10 m
10 m
16.1 cm
8m
9.8 m
9.8 m 8.6 m
14.4 cm
5.1 cm
16 cm
Two groups of students are asked to find the area of their playground. They decide to use Simpson’s rule. The diagram of the playground as drawn by the first group of students is shown on the left below. The diagram of the playground as drawn by the second group is shown on the right below.
1 Calculate the area of the playground using the measurements made by group 1. 2 Calculate the area of the playground using the measurements made by group 2. 3 Why are the answers different? Which group is correct? Explain your answer. 5 Give an approximation for the area of the playground. Explain your answer. 6 How would you obtain a more accurate estimate for the area of the playground? Give at least two different ways.
59
INVESTIGATION 2.2 Volumes of solids and pyramids Equipment needed: cardboard, scissors.
1 Construct and cut out the nets of the open cube and the open square pyramid using the information on the diagrams. The cube and the pyramid have the same base and height. Use sticky tape to make the objects. Fill the pyramid with sand and see how many pyramids of sand it takes to fill the cube. Construct other cubes and square pyramids of your own and test.
30.6 mm
25 mm Open cube
25 mm
Open pyramid with square base
2 Construct and cut out the nets of the open cylinder and open cone using the information on the diagrams. The cylinder and the cone have the same circular base and height. Use sticky tape to make the objects. Fill the cone with sand and see how many cones of sand it takes to fill the cylinder. Construct other cylinders and cones of your own and test. 94 mm 20 mm 216° 15 mm
20 mm
Open cylinder Open cone
60
INVESTIGATION 2.3 The volume of a dam An estimate needs to be found of the volume of water contained in an irregular shaped dam.
1 a Measurements of depth are taken using a boat crossing the dam once. A field diagram is shown. 38 m A′
A 42 m
B′
B 40 m
b The cross-section line AA′ is shown at right. Calculate A the area of the cross-section using Simpson’s rule. 10.5 m c Using the cross-section as a base and distances 38 m and 42 m as heights, calculate an approximate volume for the dam. d Use 1 m3 = 1000 L to find the capacity.
40 m 8m
2 a Another field diagram of the dam is drawn by taking more measurements.
12 m
A′ 7.3 m
C′
C 26 m
A′
A 22 m
D′
D 20 m
B′
B
b Section CC′ is shown at right. Calculate the area of this section using Simpson’s rule. c Section DD′ is shown at right. Calculate the area of this section. d Section BB′ is shown at right. Calculate the area of this section. e Use AreaBB′ × 20 m + AreaDD′ × 22 m + … to find an estimate for the volume of the dam in m3. Find the capacity.
C 5.3 m D 8.3 m
B 9m
35 m
35 m 6.1 m
38 m
38 m 7.1 m
42 m
C′ 4.2 m D′ 8.2 m
42 m 10 m
B′ 9m
Chapter 2 Further applications of area and volume
MEASUREMENT
3 Use your answers from parts 1 and 2 to give a reasonable estimate for the capacity of the dam. Explain your answer. a Describe some limitations of the methods used. b How could the process be improved? Explain. c How accurate should the measurement of capacity be? Explain your answer. 61
REVIEW 2 FURTHER APPLICATIONS OF AREA AND VOLUME Language and terminology 1 When discussing solid shapes, explain what you understand by these terms. a surface area b volume 2 Here is a list of terms used in this chapter. Explain each term in a sentence. annulus, circle, composite figure, cone, cross-section, cylinder, estimate, figure, lower bound, offset, percentage error, plane shape, prism, pyramid, range of values, Simpson’s rule, sphere, upper bound
Having completed this chapter you should be able to: •
find the area of circles and composite figures including an annulus
•
find approximations to areas using Simpson’s rule with one or two applications
•
find the surface area of prisms and open and closed cylinders
•
find the surface area of a sphere
•
find the volume of composite solids
•
interpret the results of measurements and calculations and make judgements about reasonableness and errors.
2 REVIEW TEST 1 The area of this circle is closest to: A 102.1 cm2 C 32.5 cm2
B 51.0 cm2 D 25.5 cm2
2 The area of this part of a circle is closest to: A 1.2 cm2 C 43.1 cm2
B 408.3 cm2 D 433.2 cm2
5.7 cm
38° 11.4 cm
3 The shaded area is closest to: A 16.3 m2 C 9.03 m3
B 67.1 m2 D 11.8 m2
4.3 m
2.1 m
MEASUREMENT
4 The area of the shaded annulus is closest to: A 4 cm2 C 125.6 cm2
B 31.4 cm2 D 125.7 cm2
3 cm
7 cm
62
Insight Mathematics General 12 HSC Course 2
5 Using Simpson’s rule, the area of the field is approximately: A 600 m2 B 700 m2 2 C 800 m D 900 m2
18 m
18 m
26 m
20 m
28 m
6 Using Simpson’s rule twice, the area of this figure is closest to: A 8000 m2 B 6000 m2 2 C 4000 m D 2000 m2
68 m
48 m
33 m 50 m
41 m
7 The surface area of this solid is: A 300 cm2 C 140 cm2
B 21 cm2 D 280 cm2
50 m 49 m
5 cm 6 cm 10 cm
8 A solid with the same cross-section throughout its length is a: A prism B pyramid C cone
D sphere
9 The surface area of this closed cylinder, correct to 1 decimal place, is: A 539.0 m2 B 539.1 m2 2 C 1842.7 m D 1842.8 m2
3.2 cm 15.6 cm
10 The surface area of a sphere with diameter 17.8 cm, correct to 2 decimal places, is: A 995.38 cm2 B 3981.52 cm2 C 3981.53 cm2 D 2952.97 cm2 11 The volume of this figure, correct to the nearest whole number, is: A 575 mm2 B 841 mm2 C 1416 mm2 D 5176 mm2
19 mm
12 The range of values for a measurement of 23.6 cm is: A 23–24 cm B 22.6–24.6 cm C 23.5–23.7 cm D 23.55–23.65 cm
FINANCIAL MATHEMATICS MEASUREMENT
8 mm 13 mm
If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question Section
1–4
5, 6
7, 8
9
10
11
12
A
B
C
D
E
G
H
Chapter 2 Further applications of area and volume
63
2A REVIEW SET 1 Find the area of these figures, correct to 1 decimal place. a b 4.5 cm
c
1.8 m
6.3 cm
2 Find the shaded area correct to 1 decimal place. a
b
5 cm 6 cm
10 cm
12 cm
3 Use Simpson’s rule twice to find the area of this field.
33 m 20 m 18 m
30 m 20 m
6m
33 m
4 Consider this cylinder. Find the surface area if the cylinder is: a open at both ends b closed at both ends c open at one end only. 18.3 cm
5 Find the volume of this cone.
11.7 cm
1.5 m
6 Find the volume of these solids. a
MEASUREMENT
26 cm
1.3 m
b 2.8 cm
106 cm
43 cm
3.3 m
7 Between what values would the area of a circle of radius 3.5 cm lie? 64
Insight Mathematics General 12 HSC Course 2
2B REVIEW SET 1 Find the area of these figures, correct to 1 decimal place. a b 118 mm
c 3
1.
68 cm
m
43°
2 Find the shaded area correct to 1 decimal place. a
b
8.6 cm
5.8 cm 11.5 cm
1.1 cm
3 Use Simpson’s rule twice to find the area of this field. 48 m 40 m
26 m
50 m
50 m 30 m 56 m
4 Consider this cylinder. Find the surface area if the cylinder is: a open at both ends b closed at both ends c open at one end only. 67 cm
5 Calculate the surface area of these solids. a
48 cm
b 5 cm
6.3 m
4 cm
7.8 m 10 cm
15.3 m
68 m
FINANCIAL MATHEMATICS MEASUREMENT
6 Find the volume of these solids. a
6 cm
b 2.1 m 3.8 m 7.5 m
53 m 6.8 m 48 m
7 Find the percentage error for a measurement of 600 m correct to the nearest 100 m.
Chapter 2 Further applications of area and volume
65
2C REVIEW SET 1 Find the area of these figures, correct to 1 decimal place. a b 0.32 m
c
113 mm 6.4 cm
2 Find the shaded area correct to 1 decimal place. a
b
18 mm
40 cm 43 mm
3 Use Simpson’s rule to find the area of this field.
31 m
26 m
35 m
20 m
20 m
4 Consider this cylinder. Find the surface area if the cylinder is: a open at both ends b closed at both ends c open at one end only. 2.4 m
5 Find the surface area of this sphere.
1.3 m
8.4 cm
6 Find the volume of these solids. a 15 cm
8 cm
b
2 cm
MEASUREMENT
4 cm 28 cm
10 cm
7 Find the upper limit for the area of a square of side length 6.4 cm. 66
Insight Mathematics General 12 HSC Course 2
8 cm
2D REVIEW SET 1 Find the area of these figures, correct to 1 decimal place. a b 11.3 cm
c
58.6 cm 4.3 cm
100°
2 Find the shaded area correct to 1 decimal place. a
b 3.5 cm
6 cm
15 cm
3 Use Simpson’s rule twice to find the area of this field.
53 m
51 m
28 m 15 m
15 m
15 m
26 m 15 m
4 Consider this cylinder. Find the surface area if the cylinder is: a open at both ends b closed at both ends c open at one end only. 326 mm
5 Find the surface area of this sphere. 215 mm
6 Find the volume of these solids. 1.4 m a
2.3 m
b
FINANCIAL MATHEMATICS MEASUREMENT
3.2 m
5 cm
8 cm 9 cm
3.5 m
7 Find the range of values for a measurement of 4.3 km.
Chapter 2 Further applications of area and volume
67
2 EXAMINATION QUESTION (15 MARKS) a A fence is measured as 21.6 m. i What is the range of values for this measurement? ii Calculate the percentage error.
(1 mark) (2 marks)
b Calculate the area of this annulus.
(2 marks)
4.7 m 8.2 m
c A cylindrical bucket is open at one end only. Calculate the area of plastic used to make the bucket.
(2 marks)
12 cm 8.1 cm
d Calculate the surface area of this rectangular prism.
(2 marks) 2m 3m 6m
e An irregular field of length 40 m is shown.
22 m
18 m
8m
20 m 10 m
40 m
i Calculate the area using one application of Simpson’s rule. ii Recalculate the area using two application of Simpson’s rule. iii Give a reason why the areas are not the same.
(1 mark) (2 marks) (1 mark)
f A 10 m beam has this uniform cross-section. Calculate the volume in m3.
(2 marks)
50 cm
MEASUREMENT
80 cm
68
Insight Mathematics General 12 HSC Course 2
Further algebraic skills and techniques The main mathematical ideas investigated are: ▶ adding and subtracting like algebraic terms ▶ adding and subtracting algebraic fractions ▶ applying the index laws in algebra ▶ solving linear equations ▶ changing the subject of a formula ▶ solving linear simultaneous equations.
ALGEBRA AND MODELLING
Syllabus references: AM3 Outcomes: MG2H-3, MG2H-9, MG2H-10
3A
Simplifying algebraic expressions
Algebraic expressions can be simplified by collecting like terms. Like terms, as the name suggests, have exactly the same pronumeral part. The numbers may vary. Like terms are the only terms that can be added or subtracted. This section begins with a review of some of the simplifications used in the Preliminary Mathematics General course.
WORKED EXAMPLE 1 Simplify, where possible, by collecting like terms. a 9a − a b 3xy − 5xy Solve
c 4x − x2 + 7x
Think
a
9a − a = 8a
9a and a are like terms.
b
3xy − 5xy = −2xy
3xy and 5xy are like terms.
c
4x − x2 + 7x = 11x − x2
4x and 7x are like terms.
Apply Collect the like terms by carrying the sign in front of each term. x and x2 are not like terms.
EXERCISE 3A 1 Simplify, where possible, by collecting like terms. a 3x + 7x b 6x − 3x d 3p − 2p e x2 − 7x2 g 7ab + 5ba h x + 7 + 2x j p + 3p − 8 k p2 + 2p + 4p2
c f i l
x+x x2 + 3x 8a2 + 14a2 − 6a2 7k + k − 8
WORKED EXAMPLE 2 Simplify the following by collecting like terms. a 4x − 9x + 3 − x b 4 − 2a + 3a + 7
ALGEBRA AND MODELLING
Solve/Think
a
4x − 9x + 3 − x = −5x + 3 − x = −6x + 3
b
4 − 2a + 3a + 7 = a + 11
c
a + 8 − 3a + 7 = −2a + 15
2 Simplify the following by collecting like terms. a x + 2 + 3x + 6 b 4 + 5a + 7 − 2a 2 2 d n − 2n + 3n + 2n e 8 − 7x − 6 + 2 2 2 g a +a+a+a h 5t − 4t + 7 + t
70
Insight Mathematics General 12 HSC Course 2
c a + 8 − 3a + 7 Apply Collect the like terms by carrying the sign in front of each term.
c 2a + 4b + 4a − 2b f x2 + 2x + 3 − 5x i −5a − 3a + 3 − 5
3 Simplify, where possible, by collecting like terms. a −8l − 4 + 2l − 7 b x2 + 2x − 7x − x2 d ab + b − 2ab − 5b e −x − 6 − 2x − 3 g 3a − 2 − a + 2 − a h a2b + a2b − 3a2b + 8a j 4p5 + 5p4 − 2p5 − 6p4 k 8m2 − 5n2 + 7n2 − 4m2
c f i l
4x − 2y − (−x) + y 5t − (−t) + 6 − 3t 5d − 2c + d − 2c + 4 6s2t + 5s2 − 8s2t − 6s2
WORKED EXAMPLE 3 Expand and simplify the following. a 3 + 2(2 − 5x)
b 4 − 3(x − 2)
Solve/Think
a
3 + 2(2 − 5x) = 3 + 4 − 10x = 7 − 10x
b
4 − 3(x − 2) = 4 − 3x + 6 = 10 − 3x
Apply When expanding brackets, multiply every term inside the brackets by the term outside the brackets. Expand and collect like terms.
4 Expand and simplify the following. a 7 + 2(3x +5) b 4 − 3(1 − 2x) c 18 − (5x + 8) d 7x − (3 − 4x) e 5x + 3(2x + 1) f 7 − 5(1 − 3x) g 2x − (3x + 11) h 7 − 2(x − 3) i 5 − 6(3x − 1) j 2(3x − 5) + 7x k 3(x − 2) + 4 l 5(x − 3) − 3x
WORKED EXAMPLE 4 Expand and simplify the following. a 5(x + 2) + 2(5 − x) b x(2 − x) − 4(x − 3) Solve/Think
Apply
5(x + 2) + 2(5 − x) = 5x + 10 + 10 − 2x = 3x + 20
b
x(2 − x) − 4(x − 3) = 2x − x2 − 4x + 12 = −x2 − 2x + 12
5 Expand and simplify these expressions. a 3(x + 4) + 2(x − 1) b d 2(1 − x) − 4(x − 2) e g 2x(x + 2) − 4(4 − x) h j −x − 6 − 7(x + 3) k
Expand and collect like terms.
4(y + 1) + 2(y + 3) d(d + 2) + 3(d − 3) 3x(x + 4) + 5(x + 2) −(3 − x) + 2(4 − x)
c f i l
ALGEBRA AND MODELLING
a
3(p + 1) − 4(p − 5) 2x(x + 1) + 3(x + 2) 6(x − 3) − (5x + 4) 3(5 − 2x) − (4 − x)
Chapter 3 Further algebraic skills and techniques
71
3B
Algebraic fractions
WORKED EXAMPLE 1 Simplify the following. 9t ___ 7t a ___ + 20 15
5k b ___ − 2k 6
Solve
Think
Apply Change the fractions to equivalent fractions with a common denominator.
a
9t ___ 7t 27t 28t ___ + = ___ + ___ 20 15 60 60 55t = ___ 60 11t = ___ 12
Lowest common denominator = 60: 9t ___ 27t 7t 28t ___ = and ___ = ___ 20 60 15 60 55t ÷ 5 ___ 55t _______ 11t ___ = = 60 60 ÷ 5 12
b
5k 5k 12k ___ − 2k = ___ − ____ 6 6 6 7k = −___ 6
Lowest common denominator = 6: 12k 2k = ____ 6
EXERCISE 3B 1 Complete the following to simplify the expressions. 3y ___ 7y __ 7y ___ □ ___ − = − = ___ 8 8 4 8 2 Simplify the following expressions. 2y ___ 3y 11x ___ 8x ___ a ____ b + + 20 3 20 4 11p __ p 4x ___ 3x ___ ____ e 5 + f 10 − 2 4 3y 2p i ___ j 2p + ___ +y 7 5
ALGEBRA AND MODELLING
3C
7a ___ 2a c ___ − 8 3 5a ___ 3a ___ g 7 − 14 2m k m − ___ 3
Indices
INVESTIGATION 3.1 From Investigation 3.1 we see that the laws for the use of indices are: am × an = am + n am ÷ an = am − n (am)n = amn (ab)n = anbn
72
Insight Mathematics General 12 HSC Course 2
14t __ 3t d ___ − 9 5 5z __z h __ + 6
l
3 3k 2k − ___ 8
WORKED EXAMPLE 1 Simplify the following. a p7 × p8
b t 16 ÷ t 4
c (m5)4
Solve
Think
a
p7 × p8 = p15
p7 × p8 = p7 + 8
b
t 16 ÷ t 4 = t 12
t 16 ÷ t 4 = t 16 − 4
c
(m5)4 = m20
(m5)4 = m5 × 4
Apply Use the index laws to simplify.
EXERCISE 3C 1 Simplify the following. a k10 × k6 d m11 × m10
b p12 ÷ p7 e x25 ÷ x5
c (t3)5 f (y7)5
WORKED EXAMPLE 2 y4 Simplify __7 . y Solve
Think
y4 1 __ = y−3 or __3 y7 y
Apply Use the index law for division or divide both the numerator and the denominator by the lowest power of the variable.
Using the index law for division: y4 __ = y4 − 7 = y−3 y7 or y4 ÷ y4 ____ y4 ______ 1 1 __ __ = 7 4 = 7−4 = 3 7 y ÷ y y y y
2 Simplify the following. m3 k7 a ___5 b ___ m k10
t2 c __6
y e __7
x d __4
t
y
x
WORKED EXAMPLE 3 4
Simplify 3m5n3 × _5 m4n7.
4
Think 12
9 10 3m5n3 × _5 m4n7 = __ 5mn
4
4
3m5n3 × _5 m4n7 = 3 × _5 × m5 + 4 × n3 + 7
Apply Multiply the coefficients then the pronumerals.
ALGEBRA AND MODELLING
Solve
3 Simplify the following.
a 3k7 × _14 k3
b 4b4 × 2b3
c 2x7 × 3x2
d p4q3 × p5q2
e 3m5n × 5m2n3
f 4x4y7 × _25 xy
g 5a3b4c2 × 2a2bc3 × 7a4b2c
Chapter 3 Further algebraic skills and techniques
73
WORKED EXAMPLE 4 Simplify the following. 8t9 a ___7 6t
xy b ____5 7 3
xy
Solve
Think
Apply
a
8t9 ___ 4t2 ___ 7 = 3 6t
8t9 __ t9 4 __ ___ 7 = 3 × 7 6t t 4 __ = × t2 3
Simplify the numerical fraction and use the index law to divide the pronumerals.
b
x7y3 __ x6 ____ 5 = 2 y xy
x7y3 __ y3 x7 __ ____ 5 = x × 5 xy y 1 = x6 × __2 y
Separate the pronumerals and use the index law for division.
4 Simplify the following. 12y7 4x5 a ____2 b ____ 10x 3y3 a8b4 f ____ 2 6
m3n7 g ____ 6 5
ab
mn
5x3 c ___ x7
m5 d ____6
k2m h ____ 5 3
i
km
x y e ____ 4 3 10 7
2m 6x3y2 _____ 8xy7
xy
WORKED EXAMPLE 5 Simplify the following. a (2t 5)3
b (a3b7)4
Solve
Think
a
(2t 5)3 = 8t15
(2t 5)3 = 23 × (t 5)3
b
(a3b7)4 = a12b28
(a3b7)4 = (a3)4 × (b7)4
5 Simplify the following. a (2m4)3
d (x3y4)5
Apply Use (ab)n = an × bn.
b (7x5)2
c (3y7)3
e (m4n)6
f (_23 a5b4)4
WORKED EXAMPLE 6
ALGEBRA AND MODELLING
9x4y3 ____ x7y2 Simplify _____ . × 5x2y 3xy2 Solve 3x8y2 x7y2 _____ 9x4y3 ____ _____ 2 × 2 = 5 5x y 3xy
74
Insight Mathematics General 12 HSC Course 2
Think
Apply
9x4y3 ____ 9x11y5 x7y2 _____ ______ 2 × 2 = 5x y 3xy 15x3y3 5 9 x11 y = ___ × ___3 × __3 15 x y 3 = __ × x8 × y2 5
Multiply and divide the coefficients, then multiply and divide the pronumerals using the index laws.
6 Simplify the following. a3b4 × a7b2 a __________ a5b3 (a2b3)4 d ______ a5b2 x3y4 × x4y5 g __________ x2y2 × x7y10
xy ×xy c _________ 2 2 4 2 7 3
m7n8 b ___________ 2 4 3 3
mn ×mn 4x8y10 e ___________ 3 5x y × 2x2y2
5 4
xy ×xy
a5b2 × a2b5 f __________ 3 9 ab
3m n × 2mn h ____________ 3 6 3 5 2
4
4m n × 5m n
7 Complete the following to simplify. a3 a3 1a × a × a a __3 = a□ − □ = a□ b __3 =__________ = ___ 1 a a a×a×a
c a0 = ___
WORKED EXAMPLE 7 Simplify the following.
a 270
b x0
c 5x0
Solve
d (5x)0
Think
a
270 = 1
b
x0 = 1
c
5x0 = 5
5x0 = 5 × x0 =5×1
d
(5x)0 = 1
(5x)0 = 50 × x0 =1×1
Apply Use a0 = 1.
8 Simplify the following. a y0 d 2x0 + 4 g 2x0 + 4y0 j 4m0 + 3n0
b e h k
c f i l
(3y)0 (2x)0 + 4 (2x)0 + (4y)0 (3k2)0 + 2k
3y0 5k0 − 1 6t 0 − (6t)0 7(x0 + 5)
WORKED EXAMPLE 8 Expand and simplify 5x2(3x3 − 4) − 4x2(x3 − 6). Apply
5x2(3x3 − 4) − 4x2(x3 − 6) = 15x5 − 20x2 − 4x5 + 24x2 = 11x + 4x 5
9 Simplify the following. a 3x2(6x − 3) + 2x2(7x + 5) c 2p4(5p3 − p) + 3p4(2p3 − 4p) e 5y2(2y5 + 3y3) − 4y2(3y5 + 2y3)
2
Expand using the index laws and then collect like terms. ALGEBRA AND MODELLING
Solve/Think
b 4x3(2x3 + 7) − x3(3x3 − 4) d 3x5(2x4 − 5x2) − x5(3x4 − 4x2)
Chapter 3 Further algebraic skills and techniques
75
3D
Equations
WORKED EXAMPLE 1 Solve the following equations. x a 6x = 41 b __3 = −4
c x+7=3
Solve
Think
a
x = 6_6
6x = 41 6x ___ 41 ___ = 6 6 5 x = 6_6
b
x = −12
x __ = −4 3 x __ × 3 = −4 × 3 3 x = −12
c
x = −4
x+7=3 x+7−7=3−7 x = −4
d
x=3
x − 4 = −1 x − 4 + 4 = −1 + 4 x=3
5
d x −4 = −1 Apply Add or subtract the same number to or from both sides of the equation. Multiply or divide both sides by the same number.
EXERCISE 3D 1 Solve the following equations.
a 5p = −30
b 7k = 31
x c __7 = 4
p d __6 = −5
e p + 7 = 16
f x + 8 = 16
g x−3=9
h y − 4 = 26
i k − 24 = −25
j x + 11 = 0
k x − 17 = 0
x l __5 = 0
WORKED EXAMPLE 2 Solve the following equations. a 8x − 7 = −31
ALGEBRA AND MODELLING
Solve
76
b 19 − 3x = 7 Think
a
8x − 7 = −31 8x = −24 x = −3
8x − 7 + 7 = −31 + 7 8x = −24 8x 24 ___ = −___ 8 8 x = −3
b
19 − 3x = 7 19 − 3x − 19 = 7 − 19 −3x = −12 x=4
19 − 3x − 19 = 7 − 19 −3x = −12 −3x −12 ____ = ____ −3 −3 x =4
Insight Mathematics General 12 HSC Course 2
Apply Add or subtract the same number to or from both sides of the equation, then divide both sides by the same number.
2 Solve these equations. a 3x − 10 = 2 d 9a + 10 = −8 g 5 − 4x = 21
b 4p + 7 = 3 e 4x − 5 = 9 h 12 − 7x = 15
c 6x − 25 = −5 f 17 − 2x = −3 i 18 = 17 + 5m
WORKED EXAMPLE 3 Solve the following equations. x a __5 + 7 = 3
x 3
b 4 − __ = 2
Solve
Think
a
x __ +7=3 5 x __ = −4 5 x = −20
x __ +7−7=3−7 5 x __ = −4 5 x __ × 5 = −4 × 5 5 x = −20
b
x 4 − __ = 2 3 x −__ = −2 3 x=6
x 4 − __ − 4 3 x −__ 3 x −__ × −3 3 x
3 Solve these equations. m a __ +1=6 3 y d ___ − 3 = −7 10 t g 1 − __ = 5 6
Apply Add or subtract the same number to or from both sides of the equation, then multiply both sides by the same number.
=2−4 = −2 = −2 × −3 =6
k b __2 − 3 = 7
p c __5 + 6 = 2
x 2 q h 2 − __7 = −1
f 9 − __ = −1
e 5 − __ = 2
m 4
2w i ___ −9=6 5
WORKED EXAMPLE 4 2x Solve ___ = 7. 3
2x ___ =7 3 2x = 21 1
x = 10_2
Think 2x ___ ×3=7×3 3 2x = 21 2x ___ 21 ___ = 2 2
Apply Multiply both sides by the same number, then divide both sides by the same number. ALGEBRA AND MODELLING
Solve
1
x = 10_2
4 Solve the following equations. 2x a ___ =5 3 5x d ___ = 3 4
3x b ___ =2 4
2x c ___ = −4 5
3x e ___ = −5
5x f ___ = 14 6
2
Chapter 3 Further algebraic skills and techniques
77
WORKED EXAMPLE 5 2x 5 Solve ___ = __ . 4 3 Solve
Apply
5 2x ___ × 3 = __ × 3 4 3 15 2x = ___ 4 15 __ 2x ___ 1 ___ = × 2 4 2
5 2x __ ___ = 4 3 15 ___ 2x = 4 15
Think
7
_ x = __ 8 or 1 8
15
Multiply both sides of the equation by the same number. then divide both sides by the same number. (Dividing by 2 is the same as 1 multiplying by _2 .)
7
_ x = __ 8 or 1 8
5 Solve the following equations. 3m __ 2x __ 3 1 a ___ b ___ = = 5 4 4 2
2p __ 4 c ___ = 3 7
3k 2 d ___ = −__ 5
5t 5 e __ = −__ 9
3
4
WORKED EXAMPLE 6 Solve the following equations. a 5x + 3 = 3x − 1
b 14 − 3x = 2 − x
Solve
Think
a
5x + 3 = 3x − 1 2x + 3 = −1 2x = −4 x = −2
5x + 3 = 3x − 1 5x + 3 − 3x = 3x − 1 − 3x 2x + 3 = −1
b
14 − 3x 14 − 2x −2x x
14 − 3x = 2 − x 14 − 3x + x = 2 − x + x 14 − 2x = 2
=2−x =2 = −12 =6
6 Solve the following equations. a 6x + 2 = 2x + 14 b 3x + 8 = 13 − 2x e 2x − 10 = x + 4 f 5x − 7 = 13 + 7x
Apply Eliminate the variable on the right-hand (or left-hand) side by adding or subtracting it to or from both sides of the equation. Complete the solution as for previous examples.
c 4x − 5 = 3x − 7 g 3x − 6 = 9 − 2x
d 10 − 2x = 4 − 5x h 4x − 5 = 5x − 2
WORKED EXAMPLE 7 ALGEBRA AND MODELLING
Solve these equations. 4 a __n = 22 Solve
a
78
4 __ n 4 4 ___ 22 n
= 22 = 22n =n 2
= __ 11
Insight Mathematics General 12 HSC Course 2
6 b _____ =3 x+1
n c _____ =3 n+1 Think
4 __ n = 22 4 __ × n = 22 × n n 4 = 22n 22n 4 2 ___ = ____ ∴ n = __ 11 22 22
Apply Multiply both sides of the equation by the same number, then divide both sides by the same number.
WORKED EXAMPLE 7 CONTINUED
Solve
Think
Apply
b
6 _____ =3 x+1 6 = 3(x + 1) = 3x + 3 3 = 3x x =1
6 _____ =3 x+1 6 _____ × (x + 1) = 3 × (x + 1) x+1 6 = 3(x + 1) = 3x + 3 6 − 3 = 3x + 3 − 3 3 = 3x 3 3x ___ = __ 3 3 x =1
Multiply both sides of the equation by the same number, subtract the same number from both sides, then divide both sides by the same number.
c
n _____ n+1 n n −2n
n _____ =3 n+1 n _____ × (n + 1) = 3 × (n + 1) n+1 n = 3(n + 1) n = 3n + 3 n − 3n = 3n + 3 − 3n −2n = 3 3 −2n ___ ____ = −2 −2
Multiply both sides by the same number, subtract the same number from both sides, then divide both sides by the same number.
∴
=3 = 3(n + 1) = 3n + 3 =3 3
n = −_2
3
n = −_2
7 Solve these equations. 3 a __x = 15
8 b __x = 5
7 c __x = −2
4 d _____ =1 x+1
WORKED EXAMPLE 8 w w Solve this equation: __ + __ = −14 5 2
w w __ __ + = −14 5 2 5w + 2w = −140 7w = −140 w = −20
8 Solve the following equations. k k a __3 + __ = 10 2 y __ y __ d 2− =3 6
Think Lowest common denominator = 10 w __ w __ + × 10 = −14 × 10 5 2 w w __ × 10 + __ × 10 = −140 5 2 5w + 2w = −140 7w = −140 −140 w _____ __ = 7 7 w = −20
(
)
z z b __3 + __5 = 4 3x x e ___ − __ = −4 2 5
Apply Multiply both sides of the equation by the lowest common denominator of the fractions.
ALGEBRA AND MODELLING
Solve
2x ___ 3x c ___ =1 + 3 4 3k 4k ___ ___ − f =2 4 3
Chapter 3 Further algebraic skills and techniques
79
WORKED EXAMPLE 9 x Solve __ + 4 = 2x − 11. 3 Solve x __ + 4 = 2x − 11 3 x + 12 = 6x − 33 −5x + 12 = −33 −5x = −45 x= 9
9 Solve the following equations. x a __3 + 2 = x − 4 2x d ___ − 1 = 3x − 15 3
Think
Apply
Lowest common denominator = 3 x (__ + 4) × 3 = (2x − 11) × 3 3 x + 12 = 6x − 33 x + 12 − 6x = 6x − 33 − 6x −5x = −45 x =9
Multiply both sides of the equation by the lowest common denominator of the fractions.
x 2 x x e __5 − 1 = __ + 2 3
c 2x − 5 = ___ + 1
3x 4
b x + 3 = __ + 1
WORKED EXAMPLE 10 Solve these equations. a x2 = 9
b y3 = 7
Solve
a
x = 3 or −3
b
y = √7 = 1.9 (1 decimal place)
3
__
10 Solve the following equations. a y3 = 8 d x2 = 36
Think __
Apply
__
Take the square root of both sides. Note: 3 or −3 is often written as ±3.
x = √9 or x = −√9
__
__
√y3 = √7
Take the cube root of both sides.
b x2 = 100 e y3 = 10
c y3 = 35 f x2 = 7
3
3
WORKED EXAMPLE 11 Solve the following equations. __ a √x = 5
b
ALGEBRA AND MODELLING
Solve (√x )2 = 52
b
x = 64
(3√x )3 = 43
__
d √k = 1 80
__
x = 25
3
Insight Mathematics General 12 HSC Course 2
__
√x
=4
Think
a
11 Solve these equations. __ a √x = 3
3
__
b
3
__
√x
=2
___
e √2p = 10
Apply Square both sides. Cube both sides.
c
__
√m 3
=7
___
f √2w = 5
3E
Practical equations
A formula links two or more pronumerals according to a rule. If one pronumeral is expressed in terms of the others, it is called the subject of the formula. If the pronumeral to be found is the subject of the formula then its value is found directly by substitution. If the pronumeral to be found is not the subject then, following substitution, the resulting equation is solved to find its value.
WORKED EXAMPLE 1 a Find the value of t if v = 117, u = 5, a = 8 and v = u + at. I
b Find the value of N if R = 23, I = 4 and R = __ + 1. N Solve
a
b
v = u + at 117 = 5 + 8t 112 = 8t t = 14 I R = __ + 1 N 4 23 = __ + 1 N 4 22 = __ N 2 4 N = ___ = ___ 22 11
Think In the formula, replace v by 117, u by 5 and a by 8. Solve the resulting equation.
Apply Substitute the given values into the formula and solve the resulting equation.
Replace R by 23 and I by 4. Solve the resulting equation.
EXERCISE 3E 1 a Find the value of t if v = 102, u = 18, a = 7 and v = u + at. b Find the value of a if v = 54, u = 12, t = 14 and v = u + at. I c Find the value of N if R = 3, I = 7 and R = __ + 1. N I d Find the value of N if R = −3, I = 5 and R = __ + 1. N 2 Use the formula d = _12 ct to find the value of:
b t when d = 320 and c = 16
2Rn 3 Use the formula a = _____ to find the value of: n+1 a R when a = 12 and n = 24 b R when a = 10 and n = 36
c n when a = 8 and R = 5
E 4 Use the formula I = _____ to find the value of: R+r a E when I = 8, R = 15, r = 3 b R when I = 2, E = 24, r = 3
c r when I = 10, E = 35, R = 2
1 1 1 5 Find the value of R1 if __ = __ + __ given: R R1 R2 a R = 1.12 and R2 = 2.24
ALGEBRA AND MODELLING
a c when d = 100 and t = 8
b R = 1.5 and R2 = 4.5
Chapter 3 Further algebraic skills and techniques
81
WORKED EXAMPLE 2 __
√
l a Use the formula T = 2π __g to find the value of l when T = 8 and g = 10. ____
√4π
3 3V b Use the formula r = ___ to find V when r = 2.
Solve ___
a
Think ___
√
√
8 l l 8 = 2π ___ ∴ ___ = ___ 10 2π 10 8 2 ___ l ___ = 2π 10 8 2 10 × ___ = l 2π l = 16.2 (1 decimal place)
( ) ( )
___
b
√
3V 3V 2 = ___ ∴ 8 = ___ 4π 4π 32π = 3V 32π ____ =V 3 V = 33.5 (1 decimal place) 3
Apply
In the formula, replace T by 8 and g by 10. Solve the resulting equation.
Substitute the given values and solve the resulting equation.
In the formula, replace r by 2 and solve the resulting equation.
__
√
l 6 Use the formula T = 2π __ g to find the value of l when g = 10 and: a T = 12 b T=7 ____
√
64h 7 Use the formula d = ____ to find the value of h when: 5 a d = 75 b d = 18
c T = 3.9 c d = 112
___
8 Use the formula v = √2gr to find the value of r when g = 10 and: a v = 15 b v = 48 __
m 9 Use the formula P = E√__ r to find the value of:
a m when P = 14, E = 20, r = 8
b r when P = 10, E = 8, m = 6
____
√
3 3V 10 Use the formula r = ___ to find V when: 4π a r=3
b r=5
WORKED EXAMPLE 3 ALGEBRA AND MODELLING
a Use the formula l = ar n − 1 to find the value of r when l = 864, a = 4 and n = 4. b Use the formula b2 = a2(1 − e2) to find the value of e when a = 12 and b = 6. Solve
a
82
l 864 216 r3 r
= ar n − 1 = 4r 4 − 1 = r3 = 216 ____ 3 = √216 = 6
Insight Mathematics General 12 HSC Course 2
Think
Apply
Replace l by 864, a by 4 and n by 4. Solve the resulting equation.
Substitute the given values and solve the resulting equation.
WORKED EXAMPLE 3 CONTINUED
Solve
b
b2 62 62 ___ 122 1 __ 4 3 −__ 4 e2
= a2(1 − e2) = 122(1 − e2)
Think Replace a by 12 and b by 6. Solve the resulting equation.
= 1 − e2 = 1 − e2 = −e2 3
= _4 __ 3 e = ± __ = ±0.866 4
√
11 Use the formula l = arn − 1, r > 0, to find the value of r when: a l = 448, a = 7, n = 4 b l = 225, a = 9, n = 3
c l = 359.375, a = 23, n = 4
12 Use the formula b2 = a2(1 − e2) to find the value of e when: a a = 15, b = 7 b a = 6.5, b = 4.9 13 Use the formula E = _12 mv2 to find the value of v when: a E = 135 and m = 2.8 b E = 428 and m = 6.9 PD 2l 14 Use the formula T = _____ w to find the value of D when P = 20, l = 5, w = 2.5 and T = 360.
15 Use the formula A = PR3 to find R given A = 1191 and P = 1000. 16 Use the formula A = P(1 + r)3 to find r when A = 27 388 and P = 24 000.
WORKED EXAMPLE 4 Use the formula (1 + r)n = 2 to find the value of n when r = 0.05. Give n to the nearest whole number. This formula is used to find the time taken for an investment at a particular interest rate to double.
(1 + r)n = 2 (1.05)n = 2 (1 + 0.05)n = 2 This equation cannot be solved using any of the methods so far. The only way to solve it at this time is to use trial and error. too small Try n = 10 (1.05)10 = 1.63 too big n = 20 (1.05)20 = 2.65 15 close n = 15 (1.05) = 2.08 14 very close n = 14 (1.05) = 1.98 n = 14
17 Use trial and error to solve these equations. a (1.08)n = 2 b (1.13)n = 2.66
Think r = 0.05
c (1.22)n = 3.3
Apply Substitute the value for r and solve the resulting equation by trial and error. ALGEBRA AND MODELLING
Solve
d (1.06)n = 1.6
Chapter 3 Further algebraic skills and techniques
83
18 Washing removes 20% of a deep stain each wash. A formula representing this is (0.8)n = x where n is the number of times the article is washed and x is the decimal fraction of the stain remaining. When 5% = 0.05 of the stain remains, the equation to be solved is (0.8)n = 0.05. Use trial and error to find the number of times the article is to be washed. 19 Repeat question 18 to find the number of times the article has been washed when 10% of the stain remains.
3F
Changing the subject of a formula
In the formula p = rx + d, p is said to be the subject of the formula because it is expressed in terms of the other variables r, x and d. This section uses the same rules as those used for solving equations, to change the subject of a formula.
WORKED EXAMPLE 1 Make x the subject of these formulas. a x+3=y b 3x = p
x c __2 = t
d x−5=r
Solve/Think
a
x+3 =y x+3−3=y−3 x =y−3
b
3x = p p 3x __ ___ = 3 3 p __ x= 3 x __ =t 2 x __ ×2=t×2 2 x = 2t
c
d
Apply Apply the rules for solving equations to rearrange the formula so that x is the subject.
x−5=r x−5+5=r+5 x=r+5
ALGEBRA AND MODELLING
EXERCISE 3F 1 Make x the subject of the following formulas. a x+2=y b x+5=y d 5x = p e 7x = q x x g __3 = t h __5 = r j x−3=p k x−7=m
84
Insight Mathematics General 12 HSC Course 2
c y=x+8 f r = 2x x 8 z=x−1
i m = __ l
WORKED EXAMPLE 2 Make x the subject of the following formulas. a x+a=b b x−k=t
c mx = z
Solve/Think
a
x+a=b x+a−a=b−a x=b−a
b
x−k=t x−k+k=t+k x=t+k
c
mx = z mx __ z ___ m =m z x = __ m
d
x __ w=p x __ w ×w = p × w x = pw
x d __ w=p Apply
Apply the rules for solving equations to rearrange the formula so that x is the subject.
2 Make x the subject of the following formulas. a x+y=z b x+n=m
e ax = c
f dx = e
i p=x+q
j m=x−k
c x−p=q x g __v = w k a = bx
d x−k=n x h __ = t
l
k x t = __ b
WORKED EXAMPLE 3
Solve/Think
w
b zx = __ y Apply
a
x __ b __ a=c b x __ __ a×a=c×a ba ___ ab x = ___ c or c
Apply the rules for solving equations to rearrange the formula so that x is the subject.
b
w zx = __ y w __ zx 1 1 __ × __ = __ z y×z 1 w x = __ yz
1 To divide by z, multiply by its reciprocal __z .
3 Make x the subject of these formulas. p x x k __ a __ b __t = __ m=q m k d d ax = __t e zx = __e
ALGEBRA AND MODELLING
Make x the subject of these equations. x b a __a = __c
x a c __y = __
b m f kx = __ n
Chapter 3 Further algebraic skills and techniques
85
WORKED EXAMPLE 4 Make x the subject of these formulas. a y = 2x + 3 b m = 4 − 3x
c y = mx + b
Solve/Think
a
2x + 3 = y 2x + 3 − 3 = y − 3 y−3 2x _____ ___ = 2 2 y_____ −3 x= 2
b
4 − 3x = m 4 − 3x − 4 = m − 4 −3x ______ m−4 ____ = −3 −3 −(m − 4) x = _________ 3 −m + 4 _______ = 3
c
mx + b = y mx + b − b = y − b y−b mx _____ ___ m = m y−b x = _____ m
4 Make x the subject of these formulas. a y = 3x + 2 b y = 5x + 1 e y = 6x − 5 f y = 3x − 5 i m = 7 + 5x j m = 6 − 5x m y = mx + 4 n y = mx − 3 5 Make y the subject of these equations. a a + by = c b p + qy = r d a − b = cy e x − ky = t g a + by + r = 0 h k = r − py
Apply Apply the rules for solving equations to rearrange the formula so that x is the subject.
c g k o
y = 2x + 5 m = 3 + 2x m = 4 − 3x y = mx + c
d h l p
y = 7x − 4 m = 5 + 3x m = −8 − 6x y = mzx
c xy − p = r f c = 4 − ry i r + px = qy
WORKED EXAMPLE 5 ALGEBRA AND MODELLING
Make y the subject of these equations. b 7 5 a a = __y b __y = __c Solve/Think
a
86
5 a × y = __ y×y ay = 5 ay 5 ___ = __ a a 5 y = __ a
Insight Mathematics General 12 HSC Course 2
m __y c __ y =x Apply Apply the rules for solving equations to rearrange the formula so that y is the subject.
WORKED EXAMPLE 5 CONTINUED
Solve/Think
b
c
7 b __ __ y =c 7 b __ __ y×y =c×y 7y b = ___ c 7y b × c = ___ c ×c bc = 7y 7y bc ___ = ___ 7 7 bc bc ___ = y or y = ___ 7 7
Apply Apply the rules for solving equations to rearrange the formula so that y is the subject.
m __ __y y =x m __ __y y ×y =x×y y2 m = __ x y__2 m×x = x ×x mx = y2 or y2 = mx ∴
___
y = ±√mx
6 Make y the subject of these equations. a x a t = __y b __y = c 5 m __ w __ a e __ f __ n=y y =b y p 5 4 i __y = __ j __y = ______ m m+n
6 7 c __b = __y
x 3 d __y = __z
n m __ k _____ p−q= y
y a+b l __3 = _____ y
y w g __ = __ y 2
y b h __a = __y
WORKED EXAMPLE 6 a Make u the subject of v2 = u2 + 2as. b Make y the subject of x = y3 + a.
a
v2 = u2 + 2as u2 + 2as = v2 u2 = v2 − 2as
Apply Make u2 the subject then take the square root of both sides.
ALGEBRA AND MODELLING
Solve/Think
________
u = ±√v2 − 2as
b
x = y3 + a y3 + a = x y3 = x − a
Make y3 the subject then take the cube root of both sides.
_____
y = 3√x − a
Chapter 3 Further algebraic skills and techniques
87
7 Make x the subject of these equations. a y = x2 + 7 b z = x2 − 2ab x2 − y2 d T 2 = ______ e p = m + x3 k g mx3 + y3 = k h y2 = a2(1 − x2)
c mx2 + y2 = b2 f y = 5 − x3
WORKED EXAMPLE 7 Make x the subject of these equations. _______ a y = √mx − 5
3
______
b y = √kx + b
Solve/Think
a
Apply
_______
Square both sides of the equation to eliminate the square root.
√mx − 5 = y
mx − 5 = y2 mx = y2 + 5 y2 + 5 x = ______ m
b
3
_____
Cube both sides of the equation to eliminate the cube root.
√kx + b = y kx + b = y3 kx = y3 − b y3 − b x = ______ k
8 Make x the subject of these equations. ___
______
_______
a y = √πx
b z = √3x + 11
c p = √ax + b
d z = 3√ax
e y = √2x − 5
3
f y = 3√πx + g
___
______
______
9 Change the subject of the formula to the pronumeral shown in parentheses. a V = 1bh (h) b F = ma PRN c A = _12 bh (b) d I = ____ 100 1 _ e P = 2 h(x + y) (y) f C = 2πr
g E = mc
(c)
i T = a + (n − 1)d
(d)
k E = _12 mv2
(v)
m v2 = u2 + 2as
2
___
√4π
ALGEBRA AND MODELLING
3 3V o r = ___
Q C
q e = IR + __
(r) (h) (r) (a)
(a) (V)
p V = IR − E
(I)
(Q)
r A = πr2
(r)
1 11 a Make C the subject of the formula x = _____ . 2π f C b Evaluate C when x = 0.05 and f = 3.
Insight Mathematics General 12 HSC Course 2
1
(R)
s = ut + _2 at 2 ___ mr n v = ___ π
10 a Make F the subject of the formula C = _59 (F − 32). b Evaluate F when C = 25.
88
h A = πr + 2πrh a j S = _____ 1−r 2
(a)
l
√
(r)
3G
Simultaneous equations
Consider the following tables of values for the equations given. y = 2x + 1
x+y=4
x
−2
−1
0
1
2
x
−2
−1
0
1
2
y
−3
−1
1
3
5
y
6
5
4
3
2
Note: x = −2 and y = −3 is a solution of the equation y = 2x + 1 because, when we substitute these values into the equation, the statement is true: −3 = 2 × −2 + 1 is true. This equation has an infinite number of solutions as x = −1 and y = −1, x = 0 and y = 1, x = 1 and y = 3 are also solutions. Similarly the equation x + y = 4 has an infinite number of solutions as x = −2 and y = 6, x = −1 and y = 5, x = 0 and y = 4 are solutions. Note: x = 1 and y = 3 is a solution of both equations. We say that the equations y = 2x + 1 and x + y = 4 are solved simultaneously by the values x = 1 and y = 3. (As both are linear equations, there is only one solution.) Hence to solve a pair of equations simultaneously, or solve a pair of simultaneous equations, means to find the values of the variables that make both equations true.
WORKED EXAMPLE 1 Complete the following table for each equation and solve y = 3x and x + y = 8 simultaneously. −3
x
−2
−1
0
1
2
3
Solve y = 3x x
−3
−2
−1
0
1
2
3
y
−9
−6
−3
0
3
6
9
−3
−2
−1
0
1
2
3
y
11
10
9
8
7
6
5
Apply
For y = 3x, when: x = −3, y = 3 × −3 = −9 x = −2, y = 3 × −2 = −6 etc.
Complete a table of solutions for each equation and find the values of x and y that are a solution of both.
For x + y = 8, when: x = −3, −3 + y = 8, y = 11 x = −2, −2 + y = 8, y = 10 etc. From the tables, x = 2 and y = 6 satisfy both equations simultaneously.
x+y=8 x
Think
x = 2 and y = 6 is the solution.
ALGEBRA AND MODELLING
y
EXERCISE 3G 1 Complete the tables given below to solve each pair of simultaneous equations. y = 2x x+y=6 x y
−3
−2
−1
0
1
2
3
x
−3
−2
−1
0
1
2
3
y
Chapter 3 Further algebraic skills and techniques
89
2 y = 2x − 1
x+y=8
0
x
1
2
3
x
4
0
1
2
3
4
6
7
8
9
y
y
3 y = 3x − 5
x + y = 23
5
x
6
7
8
x
9
5
y
y
3H
Graphical solution of simultaneous equations
The method in section G can be quite time consuming and difficult if we do not know what values of x to try. We were fortunate that one of the x values chosen in the table produced the same y value for both equations. In this section you will learn a graphical method of solution that avoids this problem by choosing any three values of x.
WORKED EXAMPLE 1 Complete the tables below and solve the simultaneous equations y = 2x + 1 and x + y = 4 graphically. y = 2x + 1 x+y=4 −2
x
0
x
2
Solve y = 2x + 1 x
−2
0
2
y
−3
1
5
x+y=4 x
−2
0
2
y
6
4
2
y 6
y = 2x + 1
ALGEBRA AND MODELLING
4 2 −1
x+y=4 1
2
x
−2 −4
The solution is x = 1, y = 3.
90
0
2
y
y
−2
−2
Insight Mathematics General 12 HSC Course 2
Think
Apply
For y = 2x + 1, when: x = −2, y = −4 + 1 = −3 x = 0, y = 0 + 1 = 1 x = 2, y = 4 + 1 = 5 These solutions of the equation can be written as ordered pairs (−2, −3), (0, 1), (2, 5) and plotted as points on a number plane. Draw the straight line through these points. For x + y = 4, when: x = −2, −2 + y = 4, y = 6 x = 0, 0 + y = 4, y = 4 x = 2, 2 + y = 4, y = 2 These solutions of the equation can be written as ordered pairs (−2, 6), (0, 4), (2, 2) and plotted as points on the same number plane. Draw the straight line through them. The point of intersection is (1, 3). As this point lies on both lines, its coordinates satisfy both equations; that is, x = 1 and y = 3 is the solution of the simultaneous equations.
Draw the graph of each equation by finding any three solutions of each and plotting these as points on a number plane. The coordinates of the point of intersection of the two straight lines give the solution of the simultaneous equations.
EXERCISE 3H 1 Complete the tables given and solve the simultaneous equations graphically. a y=x+2 y = 2x − 1 x
−2
0
x
2
0
2
0
2
4
−1
0
1
y
y
b y = 2x + 2
y = 3x + 1 0
x
2
4
x
y
y
c 2x − y = 2 x
−2
x+y=4
−1
0
1
x
y
y
2 Solve the following simultaneous equations graphically. a x + y = 1 and 3x + 2y = 1 b 3x − y = 2 and x − y = −4 3 The cost ($C) of hiring a car from company A is $50 per day; that is, C = 50d, where d is the number of days for which the car is hired. Company B charges a flat fee of $60 plus $40 per day; that is, C = 60 + 40d. a Complete the following tables and graph the equations. C = 50d C = 60 + 40d d
0
5
10
d
C
0
5
10
C
b Solve the equations simultaneously. c The point of intersection gives the number of days for which the cost is the same under both plans. After how many days is it cheaper to hire from company B?
4 Joanna is a salesperson who earns her income from commissions. Her supervisor offers her a choice of two methods of weekly payment: i a straight commission of 10% of her sales: I = 0.1S where I is income and S is sales ii a retainer of $300 plus 6% of sales: I = 300 + 0.06S. a Complete the following tables and graph each equation. I = 0.1S S($)
0
5000
10 000
5000
10 000
I($) I = 300 + 0.06S S($)
0
I($)
b Find the simultaneous solution of the equations. c What sales would Joanna have to achieve to earn more income using method ii?
91
5 A new car with a petrol engine uses 10.0 L/100 km. If the cost of petrol is $1.50/L then the cost ($C) of fuel to drive the car is given by C = 0.15d, where d is the distance travelled. The same car with a diesel engine costs $2000 more to purchase but only uses 6.0 L/100 km. The cost of diesel fuel is $1.60/L. Compared with the petrol engine, the cost of driving this car is given by C = 2000 + 0.096d.
a Complete the following tables and graph each equation. C = 0.15d d
C = 2000 + 0.096d 0
20 000
C
40 000
d
0
20 000
40 000
C
b Find the simultaneous solution of the equations. c How far do you have to travel before the cost of the two models is the same?
3I
The substitution method
In this section you will solve simultaneous equations using an algebraic method. The substitution method involves substituting one equation into the other and solving for the remaining variable. Check: It is easy to check if the solution is correct by direct substitution into both equations.
WORKED EXAMPLE 1 Solve the following equations simultaneously, using the substitution method. a y = 2x − 11 and x + y = 19 b 2x − y = 9 and x = 2y − 3
ALGEBRA AND MODELLING
a
92
Solve
Think
x + y = 19 x + (2x − 11) = 19 3x − 11 = 19 3x = 30 x = 10 y = 2 × 10 − 11 =9 or 10 + y = 19 y =9 Solution is x = 10 and y = 9.
As y is the subject of the first equation, substitute y = 2x − 11 into x + y = 19. That is, in x + y = 19 replace y by 2x − 11. Collect like terms and solve for x. Substitute x = 10 into y = 2x − 11 (or x + y = 19) to find y. If x = 10 and y = 9 then 9 = 2 × 10 − 11 is true and 10 + 9 = 19 is true.
Insight Mathematics General 12 HSC Course 2
Apply Substitute one equation into the other and solve for the remaining variable. Substitute the value of this variable into either of the equations to find the value of the other variable.
WORKED EXAMPLE 1 CONTINUED
b
Solve
Think
Apply
2x − y = 9 2(2y − 3) − y = 9 4y − 6 − y = 9 3y − 6 = 9 3y = 15 y=5 x=2×5−3=7 or 2x − 5 = 9 2x = 14 x=7 Solution is x = 7 and y = 5.
As x is the subject of the second equation, substitute x = 2y − 3 into 2x − y = 9. Expand, collect like terms and solve for y. Substitute y = 5 into x = 2y − 3 (or 2x − y = 9) to find x. If x = 7 and y = 5 then 2 × 7 − 5 = 9 is true and 7 = 2 × 5 − 3 is true.
Substitute one equation into the other and solve for the remaining variable. Substitute the value of this variable into either of the equations to find the value of the other variable.
EXERCISE 3I 1 Complete the following to solve the simultaneous equations y = 3x − 1 and x + 2y = 5, using the substitution method. x + 2y = 5 x + 2(______) = 5 x + ___x − ___ = 5 ___x − 2 = ___ ___x = ___ ∴ x = ___ Substitute x = ___ into y = 3x − 1. y = 3 × ___ − 1 = ___ The solution is x = ___ and y = ___. Check: Check that your solution is correct by direct substitution into both equations. 2 Use the substitution method to solve the following pairs of simultaneous equations. Check your solution. a y = 3x + 2 b 2x + y = 12 c 3x − 2y = 4 2x + y = 17 x = 3y − 1 y=x+1 d x = 2y − 3 e y = 3x − 1 f 5x − y = 16 x + 3y = 17 x − 3y = −9 x=4−y
WORKED EXAMPLE 2 Solve
Think
Apply
3x − 1 = x − 7 2x − 1 = −7 2x = −6 x = −3 If x = −3, y = 3 × −3 − 1 = −10 The solution is x = −3 and y = −10.
Substitute y = 3x − 1 into the second equation; that is, replace y by 3x − 1 and solve the resulting equation. Substitute x = −3 into y = 3x − 1 (or y = x − 7) to find y.
Substitute one equation into the other and solve for the remaining variable. Substitute this variable into either equation to find the value of the other variable.
Chapter 3 Further algebraic skills and techniques
ALGEBRA AND MODELLING
Solve y = 3x − 1 and y = x − 7 simultaneously using the substitution method.
93
3 Solve the following pairs of simultaneous equations. Check your solution. a y = 2x − 3 b y = 4x + 3 c y = 2x − 7 y=x+5 y=x−3 y = 4x + 8
WORKED EXAMPLE 3 Solve the following equations simultaneously using the substitution method. 1 x+y=1 2x + 3y = −1
2
Solve
Think
From equation 1 , y = 1 − x.
Label the equations 1 and 2 .
Substitute into equation 2 . 2x + 3(1 − x) = −1 2x + 3 − 3x = −1 −x + 3 = −1 −x = −4 x =4 If x = 4, y = 1 − 4 = −3. The solution is x = 4 and y = −3. Check: In 1 : 4 + −3 = 1
Make y the subject of equation 1 . Substitute y = 1 − x into equation 2 . Solve for x. Substitute this value of x into one of the equations to find y.
Apply Make y (or x) the subject of one of the equations and substitute this into the other equation to find the value of the remaining variable.
In 2 : 2 × 4 + 3 × −3 = −1
4 Complete the following to solve these equations simultaneously using the substitution method. 1 x+y=3 2x + y = 1
2
From equation 1 , y = ___ − ___. Substitute into equation 2 .
ALGEBRA AND MODELLING
2x + ___ − ___ = ___ ___ + 3 = ___ x = ___ The solution is x = ___ and y = ___. Or from equation 1 , x = ___ − ___.
Or from equation 2 , y = ___ − ___.
Substitute into equation 2 . 2(___ − ___) + y = 1 ___ − ___ + y = 1 ___ − ___ = 1 −y = ___ The solution is y = ___ and x = ___.
Substitute into equation 1 . ___ + ___ − ___ = 3 ___ − x = 3 ___ = ___ The solution is x = ___ and y = ___.
5 Solve the following equations simultaneously using the substitution method. a 2x + y = 16 b 3x + 2y =10 c x−y=3 3x + y = 21 x + 3y = 1 2x − y = 8 d 3x − 2y = 5 e 5x + 2y = 9 2x + y = 1 2x − 3y = −4
94
Insight Mathematics General 12 HSC Course 2
3J
The elimination method
Sometimes y (or x) cannot easily be expressed as the subject of one of a pair of equations to be solved simultaneously, or the resulting equation, after substitution, is complicated to solve (such as question 5 part e in Exercise 3I). In such cases it is more appropriate to use the elimination method.
WORKED EXAMPLE 1 2x + 3y = 5
1
x − 2y = 7
2
For the equations 1 and 2 above, find:
b 3× e 4×
2
2 1 −2× 2
Solve
Think
a
4x + 6y = 10
Multiply both sides of equation 1 by 2. 2 × (2x + 3y) = 2 × 5
b
3x − 6y = 21
Multiply both sides of equation 2 by 3. 3 × (x − 2y) = 3 × 7
c
3x + y = 12
2x + 3y = 5
1
x − 2y = 7
2
1 + 2
d
x + 5y = −2
2x + 3y = 5
1
x − 2y = 7
2
6x + 16y = 6
7x = 31
f
Multiply both sides of the equation by the stated number and add or subtract like terms as appropriate. Note that if we multiply both sides of an equation by the same number, the new equation formed has the same solutions as the original. For example 2x + 3y = 5 and 4x + 6y = 10 have the same solutions. (Check that x = 4 and y = −1, x = −2 and y = 3 are solutions of both.)
x + 5y = −2
4× 1
8x + 12y = 20
2× 2
2x − 4y = 14
4× 1 −2× 2
6x + 16y = 6
2× 1
4x + 6y = 10
3× 2
3x − 6y = 21
2× 1 +3× 2
7x = 31
2
Apply
3x + y = 12
1 − 2
e
c 1 + 2 f 2× 1 +3×
ALGEBRA AND MODELLING
a 2× 1 d 1 −
EXERCISE 3J 1 3x − y = 2 2x + 3y = 1
1 2
For the equations 1 and 2 above, find:
a 4× d
1
1 − 2
b 5×
2
c
1 + 2
e 5×
1 −2× 2
f 2×
1 −3× 2
Chapter 3 Further algebraic skills and techniques
95
2 4x + 2y = 3 3x − 5y = 4
1 2
For the equations 1 and 2 above, find: 1 + 2
a
d 3×
b
1 +6× 2
3 7x − 3y = 5
1
2x + 4y = 1
2
1 − 2
e 3×
c
1 −4× 2
2 − 1
f 5×
1 +2× 2
For the equations 1 and 2 above, find:
a 2×
1 −7× 2
b 4×
1 +3× 2
WORKED EXAMPLE 2 Solve these simultaneous equations using the elimination method. x + 2y = 1 3x − 2y = 5 Solve 1
x + 2y = 1
2
3x − 2y = 5
1 + 2
Think Label the equations 1 and 2 .
4x = 6 1
x = 1_2
1 If x = 1_2 , substitute x into 1 . 1
1_2 + 2y = 1
As the coefficients of y are opposites, add the equations. This eliminates y. Solve the resulting equation in x. Substitute this value of x into equation 1 (or 2 ) to find y.
2y = −_12 y = −_14
1
Solution is x = 1_2 and y = −_14 .
4 Complete to solve these simultaneous equations using the elimination method. 1 x − 2y = 7 2 −x + 4y = −13 1 + 2 ___ = −6 ___ = ___ ___ = −3 into 1 . ___ − (−6) = ___ x = ___ Solution is x = ___, y = ___.
ALGEBRA AND MODELLING
Substitute
5 Solve simultaneously and check your solution by direct substitution. a 3x + y = 4 b 3x + y = 5 x−y=8 −3x + 2y = 1
c 2x − 3y = −1 4x + 3y = 7
96
Insight Mathematics General 12 HSC Course 2
Apply If the coefficients of y (or x) are opposites, add the equations to eliminate y (or x).
WORKED EXAMPLE 3 Solve the following equations simultaneously. 3x − 2y = 5 x − 2y = 7 Solve 1
3x − 2y = 5
2
x − 2y = 7
Think
1 − 2
2x = −2 x = −1 If x = −1, −3 − 2y = 5 −2y = 8 y = −4 Solution is x = −1 and y = −4.
The coefficients of y are the same, so subtract the equations to eliminate y. Solve the resulting equation to find x. Substitute this value of x into 1 (or 2 ) to find y.
Apply If the coefficients of y (or x) are the same, subtract the equations to eliminate y (or x).
6 Solve these equations simultaneously and check your solution by direct substitution. a 5x − 3y = 4 b 2x − y = 3 c 3x + 5y = 10 d 7x − 3y = 18 2x − 3y = 13 2x + 5y = −3 x + 5y = 4 7x + 5y = −2
WORKED EXAMPLE 4 Solve these equations simultaneously. 1 2x + y = 6 3x − 2y = −5
2
Solve 2× 1 2
4x + 2y = 12 3x − 2y = −5
2× 1 + 2
7x = 7 x=1 If x = 1, 2+y=6 y=4 Solution is x = 1 and y = 4.
Think
Apply
Multiply equation 1 by 2 to make the coefficients of y opposites.
Make the coefficients of y (or x) opposites by multiplying both sides of one of the equations by the appropriate number. Adding then eliminates y (or x).
Find 2 × 1 + 2 (this eliminates y). Solve for x. Substitute this value of x into 1 (or 2 ) to find y.
7 Complete to solve the following equations simultaneously. 1 2x + 3y = 11 ___ × 2 1 1 + ___ × 2
5x − y = 19
ALGEBRA AND MODELLING
2
___ − 3y = ___ 2x + 3y = 11
___x x If x = ___, ___ + 3y 3y y Solution is x = ___, y = ___.
= ___ = ___ = ___ = ___ = ___
Chapter 3 Further algebraic skills and techniques
97
8 Solve the following equations simultaneously. a 4x + y = 7 b 3x − 2y = 13 x − 3y = −8 2x + y = 11 d x + 3y = 11 −2x + 4y = −2 (Hint: Eliminate x.)
c 2x + 3y = −7 3x − y = −5
e −x + 2y = −7 4x − 3y = 13
WORKED EXAMPLE 5 Solve the following equations simultaneously. 1 2x + 3y = 7 5x + y = −2
2
Solve 3× 2 1
Think
15x + 3y = −6 2x + 3y = 7
3× 2 − 1
13x = −13 x = −1 If x = −1, −2 + 3y = 7 3y = 9 y=3 Solution is x = −1 and y = 3.
Apply
Multiply equation 2 by 3 to make the coefficients of y the same. Find 3 × 2 − 1 (this eliminates y). Solve for x. Substitute this value of x into 1 (or 2 to find y.
9 Solve the following equations simultaneously. a 3x + 2y = 0 b 4x − 3y = −1 2x + y = 1 3x − y = −7
Make the coefficients of y (or x) the same by multiplying both sides of one of the equations by the appropriate number. Subtracting then eliminates y (or x).
c 2x + y = −4 3x + 4y = −6
d 5x − 3y = 11
e 2x + 5y = 1
x − 4y = −8 (Hint: Eliminate x.)
x − 3y = −5
WORKED EXAMPLE 6 Solve the following equations simultaneously. 1 5x − 3y = 14 4x + 2y = −2
2
Solve 2× 1
10x − 6y = 28
3× 2
12x + 6y = −6
22x = 22 x=1 If x = 1, 4 + 2y = −2 2y = −6 y = −3 Solution is x = 1 and y = −3.
ALGEBRA AND MODELLING
2× 1 +3× 2
98
Insight Mathematics General 12 HSC Course 2
Think
Apply
Multiply equation 1 by 2 and multiply equation 2 by 3. This makes the coefficients of y opposites. Find 2 × 1 + 3 × 2 (this eliminates y). Solve for x. Substitute this value of x into
Make the coefficients of y (or x) opposites by multiplying the equations by the appropriate numbers. Adding then eliminates y (or x).
equation 1 (or 2 ) to find y. Note: To make the coefficients of x opposites we could multiply equation 1 by 4 and equation 2 by −5 (or multiply equation 1 by −4 and equation 2 by 5). Adding then eliminates x.
10 Solve the following equations simultaneously. a 3x − 4y = 18 b 3x + 5y = −7 2x + 3y = −5 4x − 2y = 8
c 7x − 3y = −4 3x + 2y = −5
d 5x + 4y = 7 4x − 3y = 18
WORKED EXAMPLE 7 Solve the following equations simultaneously. 1 4x + 5y = 17 5x + 3y = 5
2
Solve
Think
3× 1
12x + 15y = 51
5× 2
25x + 15y = 25
3 × 1 − 5 × 2 −13x = 26 x = −2 If x = −2, −8 + 5y = 17 5y = 25 y=5 Solution is x = −2 and y = 5.
Multiply equation 1 by 3 and equation 2 by 5. Find 3 × 1 − 5 × 2 . This eliminates y. Note: To make the coefficients of x the same we could multiply equation 1 by 5 and equation 2 by 4. Subtracting then eliminates x.
11 Solve the following equations simultaneously. a 3x + 4y = 13 b 4x − 2y = 12 2x + 3y = 10 5x − 3y = 16
c 2x + 5y = −3 7x + 3y = 4
12 Solve the following equations simultaneously, by any method. a 6x + 3y = 18 b 3x − y = −5 c y = 2x + 7 x − 2y = −7 3x + 2y = 1 x − y = −5
e z = 3w + 1 2w − 3z = 4
f 5a − 7b = −5
g p = 3q + 2
2a + 6b = −2
p = 5q − 1
Apply Make the coefficients of y (or x) the same by multiplying the equations by the appropriate numbers. Subtracting then eliminates y (or x).
d 2x − 3y = −13 3x − 2y = −12
d 4m − 3n = 15 3m − 2n = 11
h 3t + 5s = 11 2t − 3s = −18
INVESTIGATION 3.1 Indices Complete the following.
ALGEBRA AND MODELLING
1 y3 × y4 = y × y × y × ___ × ___ × ___ × ___ = y□ k×k×k×k×k×k×k×k 2 k8 ÷ k5 = __________________________ k×□×□×□×□ = k□
3 (p2)3 = p2 × p2 × p2 = p□ 4 (km)3 = km × km × km = k × k × k × ___ × ___ × ___ = k□m□
Chapter 3 Further algebraic skills and techniques
99
REVIEW 3 FURTHER ALGEBRAIC SKILLS AND TECHNIQUES Language and terminology Here is a list of terms used in this chapter. Explain each term in a sentence. algebraic fraction, equation, expand, expression, formula, index, index law, indices, like terms, pronumeral, simultaneous equations, subject of the formula, substitute, value, variable
Having completed this chapter you should be able to: • add and subtract like terms • add and subtract algebraic fractions • apply the index laws in algebraic form • solve linear equations • solve equations following substitution of values • change the subject of a formula • solve linear simultaneous equations.
3 REVIEW TEST 1 8x − x = A 8
B 8x
C 7x
D 7
2 5x − 7x + 4 − 3x = A −5x
B −5x + 4
C 5x + 4
D −x
3 −5d − 3c + 2d − 4c + 6 = A 3d − 7c + 6 B 6
C −3d + 7c + 6
D −3d − 7c + 6
4 3 − 4(1 − 2x) = A −1 − 2x
B − 1 + 8x
C −1 + 2x
D 1 − 8x
5 2x(x − 3) + x(x − 7) = A x2 − 13x
B x2 + x
C 3x2 + 13x
D 3x2 − 13x
t 4t 6 __ + __ = 2 5 5t A __ 7
13t B ___ 20
13t C ___ 10
4t2 D ___ 7
1 B __3 k
C k2
1 D __2 k
B xy16
y2 C __ x
D xy3
C 12a3 − 5a2
D 12a3 − a2
ALGEBRA AND MODELLING
k3 7 __6 = k A k3 x4y3 × x2y7 8 _________ = x7y5 y5 A __ x
9 2a2(3a − 1) − 3a2(1 − 2a) = A −5a2 B −a2
100
Insight Mathematics General 12 HSC Course 2
10 3y0 + 2 = A 5
B 3
C 9
D 2
10 C −__ 7
7 D −__ 10
15 C −__ 4
D
C −12
D −24
11 The solution of the equation 6 − 7x = −4 is x = A
10 __ 7
B
7 __ 10
x 12 The solution of the equation _____ = 5 is x = x+3 3 A −_34 B _4
15 __ 4
w w 13 The solution of the equation __ + __ = −10 is w = 3 2
B −8_13
A −25
1 1 1 14 Using the formula __ = __ + __, the value of R1 when R = 0.5 and R2 = 0.7 is: R R1 R2 A 0.2 B −0.2 C _47
D 1_34
15 When the formula V = IR − E is rearranged to make R the subject, then R = V+E V−I E A ______ B _____ C ___ I E VI
E−V D ______ I
___
mr 16 When r is made the subject of v = ___ , then r = 2π 2 2πv 2πv A ____ B ____ m m
√
2πv2 C ____ m2
4π2v2 D _____ m2
17 The simultaneous equations y = _32 x − 3 and y = _14 x + 2 are shown on this graph. The solution is: A x = −8, y = 0 B x = 0, y = −3 C x = 3, y = 4 D x = 4, y = 3
y 4
y = 14 x + 2
2 −1
1 −2
y=
2 3 x 2
3
4
x
−3
−4
18 When y = 2x − 1 is substituted into x − 2y = 8, the resulting x-value is: A −6 B −3_13 C −2
D −7
19 Which pair of simultaneous equations has the solution x = −5, y = 3? A y=x+8 B y=x+8 C x+y=2 4x − 3y = 29 3x − 2y = −21 5x + y = −22
D y−x=8 4x − 3y = 26
3x − 2y = −5
ALGEBRA AND MODELLING
20 Consider the solution of these simultaneous equations. 1 2x + 5y = 3 2
If equation 1 is multiplied by 2, equation 2 is multiplied by 5 and the resulting equations added, then the next line in the solution is: A −11x = 31 B 11x = 31 C 19x = −19 D 19x + 20y = −19 If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question Section
1–5
6
7–10
11–13
14
15, 16
17
18, 19
19, 20
A
B
C
D
E
F
H
I
J
Chapter 3 Further algebraic skills and techniques
101
3A REVIEW SET 1 Simplify the following. a 7ab − 5ba d x2 + 2x + 2x2 + x 2 Simplify the following. a 5m3n2 × 7m4n 4p6q5 _____ 6p2q8 d _____ × p11q 3p3q3
b 7k + k e 2(3x − 5) + 8x
c 5p − 3p + 7 − p f 2(4x − 1) + 3(2x + 1)
2k7m4 b _____ 8k4m7
c (a2b3)5
e 3y 0 + 5
f 5a3(2a2 + 7) − 2a3(3a2 − 4)
3 Solve the following equations. x a 8x = 57 b __ = −2 5
c x − 5 = −3
e 4x − 5 = 3
f 5 − 3x = −4
g 15 − 4x = 3 + x
__z + __z = 1 5 2 m x2 = 16
3 j __x = 2 __ n 3√y = 5
5 k _____ = −3 x+2
i
d x+8=5 x h __ + 1 = 3x − 9 2 2x l _____ = 3 x−3
4 Use the formula E = _12 mv2 to find the value of v when E = 390 and m = 1.95. 5 Make x the subject of the following formulas. a 3x = t b y = mx − 7
___
7 c a = __x
mx d V = √___ π
6 Graphically solve the simultaneous equations y = _12 x + 1 and y = _34 x. 7 Solve the simultaneous equations: y = −3x − 4 and x + 2y = 2.
3B REVIEW SET 1 Simplify the following. a 4x − x d 4y + x2 − 2x2 + y 2 Simplify the following. a 4k2m5 × _14 km2 e 2x0 + (4y)0 3 Solve these equations. a 7p = 64 ALGEBRA AND MODELLING
e 3x − 7 = 5 i
__
√w
=6
b 12x2 − 8x2 e 7 − 4(2x + 3)
c −4x − 5x − 7x + y f 2(2 − 3x) + 4(x − 1)
4w3 b _____8 c (2p3)4 12w f 4p2(3p − 5q2) − 2p2(7p + 3q2)
15p7q8 d ____________ 3p4q2 × 5p3q4
3x b ___ = 5 2 f 4 − 7x = −2 4 j __x = 3
c x + 12 = 7
d 8 − x = 15
g 12 − 7x = 3x + 1 7 k _____ = 4 x−1
h y3 = 8 3x l _____ = 7 x−2
4 Use the formula T = ar n − 1 to find the value of r when T = 189, a = 7 and n = 4.
102
5 Make x the subject of the following formulas. a x+y=m b n = k − 3x
a x c __x = __t
6 Solve the following simultaneous equations. a y = 2x − 3 and y = 6 − x
b 2x + 3y = 0 and 3x − y = −11
Insight Mathematics General 12 HSC Course 2
___
d z = √πx
3C REVIEW SET 1 Simplify the following. a 8p − p d 5 − x2 + 3x2 + 2 2 Simplify the following. 2v2w a _____ 5v3w2 6t 7v3 5t 3v4 ____ d _____ 4 7 × 10t v tv5
b 6x + 7x e 4 − 2(x − 3)
c 2y − 5x − 5y − 2x f 4(x + 1) − 5(2x + 1)
m3n4 b _____ m7n10
c (3yz3)2
e (5x)0 − 2
f 3x2(4x3 − 2) − x2(3x3 + 9)
3 Solve the following equations. 4x a 5x = 16 b ___ = −2 3
c 4 − x = −9
e 4x − 11 = 8
f 3 − 5x = −9
g 8 − 5x = 7 − 2x
3z __ 5z __ − =8 2 6 m y2 = 3
3 j −__x = 2
5 k _____ =1 x−5
i
_____
x d __ + 5 = 3 2 y h __ − 1 = y + 5 3 2x _____ l =3 x−3
n √t + 1 = 4 __
√
l 4 Use the formula T = 2π __ g to find the value of l when T = 4 and g = 10.
5 Make x the subject of the each formula. x a __y = t b y = k − mx
Q c e = IR + __ x
d V = _43 πx3
6 The cost ($C) of hiring a car from company A is given by C = 60d, where d is the number of days the car is hired. For company B, the cost of hire is C = 100 + 40d. a Draw the graphs of cost versus number of days for each company, on the same number plane. b Find the solution of the simultaneous equations C = 60d and C = 100 + 40d. c After how many days is the cost the same for both companies? 7 Solve x − 3y = 11 and 5x − 2y = 16 simultaneously.
1 Simplify the following. a 3x − 2x d a2 − 5a2b2 + 2a2b2 − a2 2 Simplify the following. 2x4y5 _____ 5x2y3 a _____ × 5 2 6 2 4a b 3a7b9 _____ d _____ 3 8 × 6a b 4a10b5
b 4x2 + 8x + x2 e 8 + 3(3x − 2)
c 3x − 4y + 4x − y f 2(p + 1) − 3(p − 2)
w b ___ w10
c (10u5v7)4
e 4p0 + (3q)0
f 5y2(2y4 − 5) − 3y2(3y4 − 4)
3 Solve the following equations. y 1 a 2x = 29 b __ = 1_3 5 e 5 − 3k = −1 f 5v + 3 = 9 − 7v 3 3k i _____ = 2 j _____ = 2 x−4 4−k
c m − 7 = −7 z g 2z − 3 = 5 − __ 2 k t2 = 1
ALGEBRA AND MODELLING
3D REVIEW SET
d 5t + 2 = −3 5p p h ___ − __ = _34 3 6 __ 3 l √y = 3
Chapter 3 Further algebraic skills and techniques
103
2Rn 4 Use the formula a = _____ to find the value of R when a = 95 and n = 19. n+1
5 Rearrange the formula to make the pronumeral in the parenthesis the subject of the formula. __ m a A = P + Prn (n) b T = 2π __ (m) g w p+Q c __ = ______ (w) d S = 2πrh + πr2 (h) w 4
√
6 The cost ($C) of electricity to run a home is $200 per month. This may be written as C = 200m, where m is the number of months. For the same house, the cost of installing a solar system is $4000, but the monthly cost of electricity decreases to $150. The total cost of electricity using a solar system is then C = 4000 + 150m. a Complete the following tables and graph each equation. C = 200m C = 4000 + 150m m
0
50
m
100
0
50
100
C
C
b Find the simultaneous solution of the equations. c After installing the solar system, how long would it take to start saving money? 7 Solve the following equations simultaneously. 2x + 5y = −6 3x − 2y = 29
3 EXAMINATION QUESTION (15 MARKS) a Simplify the following. i 3(5x − 2y) −2(4x − y) t 3t ii __ − __ 4 5 iii 3w0 + 2 8p6q3 × 3p2q5 iv ____________ 12p7q10 b Solve the following equations. k i __ − 2 = 2k + 3 3 4 2 ii _____ = __ x+1 5
(2 marks) (1 mark) (1 mark) (2 marks)
(2 marks) ___
√
ALGEBRA AND MODELLING
3V c Make V the subject of the formula r = ___ . πh d Solve these equations simultaneously. 3x + 4y = −8 2x − 3y = −11
104
Insight Mathematics General 12 HSC Course 2
(2 marks) (2 marks)
(3 marks)
Interpreting sets of data This chapter deals with the use of data displays, measures of location and measures of spread to summarise and interpret one or more sets of data. The main mathematical ideas investigated are: ▶ using data displays ▶ using measures of central tendency: mean, mode, median ▶ understanding the relative merits of the mean, mode and median ▶ finding measures of dispersion: range, interquartile range, standard deviation ▶ comparing displays and summary statistics for two sets of data ▶ working with grouped and ungrouped data.
DATA AND STATISTICS
Syllabus references: DS4 Outcomes: MG2H-1, MG2H-2, MG2H-7, MG2H-9, MG2H-10
4A
Grouped data
In the Preliminary Mathematics General course, scores were organised into frequency distribution tables using groups. This section expands this to include grouped data histograms. It is important to remember the following information about grouped data: • Grouping data makes large numbers of data easier to analyse, but loses detail as individual scores are not available. • Groups must be the same size without overlap or gaps.
WORKED EXAMPLE 1 The following are the marks scored by 48 students in an examination. 71 50 32 55 75 51 35 61 69 58 46 57 55 86 45 67 57 61 21 44 52 28 60 19 12 51 40 49 75 52 49 70 76 62 41 35 63 63 54 50 41 93 63 37 5 50 78 64 a Organise these 48 scores into a grouped frequency distribution using class sizes of: i 10 ii 20 b Draw a frequency histogram for each of the frequency tables. Solve
a
i
Class
Tally
0–9
!
1
10–19
!!
2
20–29
!!
2
30–39
!!!!
4
40–49
!!!! !!!
8
50–59
!!!! !!!! !!!
13
60–69
!!!! !!!!
10
70–79
!!!! !
6
80–89
!
1
90–99
!
1 Total
DATA AND STATISTICS
ii
Class
Tally
48 Frequency
0–19
!!!
3
20–39
!!!! !
6
40–59
!!!! !!!! !!!! !!!! !
21
60–79
!!!! !!!! !!!! !
16
80–99
!!
2 Total
106
Frequency
Insight Mathematics General 12 HSC Course 2
48
Think
Apply
Ensure that the scores are placed in the correct groups. Use the tally column so that all scores are used.
The groups allow the scores to be analysed effectively. It is important to make sure the groups are small enough to discriminate but that there are not so many groups that it is not practical.
WORKED EXAMPLE 1 CONTINUED
Solve
i
Student scores 14 12 10 8 6 4 2 0
Apply
Make sure the columns are joined to each other and the group is shown in the horizontal axis.
Histograms are column graphs with no spaces between the columns. They give an indication of the shape of the distribution.
0– 10 9 –1 20 9 –2 30 9 –3 40 9 –4 50 9 –5 60 9 –6 70 9 –7 80 9 –8 90 9 –9 9
Frequency
b
Think
Class
ii Frequency
Student scores 24 20 16 12 8 4 0
0–19 20–39 40–59 60–79 80–99 Class
EXERCISE 4A 1 The marks scored by 40 students in a test are shown. 30 53 37 34 6 59 62 26 47 63 70 46 49 36 25 44 54 62 46 41 24 49 38 34 22 76 46 55 56 64 49 30 55 22 59 42 32 40 58 37 a Organise this data into a frequency distribution table, using the classes: i 0–4, 5–9, 10–14, 15–19, … ii 0–9, 10–19, … iii 0–19, 20–39, … iv 0–49, 50–99 b Draw a histogram for each frequency distribution table in part a. c Which of the four groups gives the best picture of the data? Explain the advantages or disadvantages of each. d Why would the groupings 0–4, 5–14, 15–29, … be unsuitable?
Chapter 4 Interpreting sets of data
DATA AND STATISTICS
2 A group of 55 students went ten-pin bowling and their scores are shown. 113 81 116 139 144 65 124 119 117 148 50 54 149 71 111 124 79 114 101 121 141 113 95 90 99 149 53 128 145 125 68 76 79 66 66 76 67 118 66 82 98 95 131 121 69 93 75 91 126 92 98 98 66 94 71 a Organise this data into a frequency distribution table, using the classes: i 50–54, 55–59, 60–64, 75–79, … ii 50–59, 60–69, … iii 50–69, 70–89, … iv 50–99, 100–149 b Draw a histogram for each frequency distribution table in part a. c Which of the four groups gives the best picture of the data? Explain the advantages or disadvantages of each. 107
Marks scored 22 20 18 16 14 12 10 8 6 4 2 0
20
–3 40 9 –5 60 9 –7 80 9 10 –99 0 12 –11 0– 9 14 139 0 16 –15 0– 9 18 17 0 9 10 –19 0– 9 21 9
Frequency
3 The following histogram shows grouped data. a Use this histogram to draw another histogram with a group size that is twice as large; that is, 20–59, etc. b Is it possible to draw a histogram with groups of 20–29, 30–39, etc? Explain.
4B
Group
Measures of location
Measures of location, sometimes called measures of central tendency, are single numbers used to represent or summarise a set of data. The commonly used measures of central tendency are the mean, the mode and the median. This section revises the calculation of mean, mode and median for ungrouped data and then introduces the calculations for grouped data. • The mean is the sum of all scores divided by the number of scores. • The mode is the score that occurs most often. • The median is the middle score when the scores have been arranged in ascending order.
WORKED EXAMPLE 1
DATA AND STATISTICS
For the scores 5, 5, 6, 6, 7, 7, 7, 9 find the: a mean b mode
108
c median
Solve
Think
Apply
a
Mean 5+5+6+6+7+7+7+9 = ___________________________ 8 52 ___ = 6.5 = 8
Add all the scores and divide by the total number of scores.
The mean can be calculated using the statistics function of your calculator.
b
Mode = 7
The mode is the score with the highest frequency.
The mode is the most common score.
c
The 4th score is 6 and the 5th score is 7, hence 6+7 Median = _____ = 6.5 2
The median divides the data into two parts, so there are equal numbers of scores below and above it. The number of n+1 scores is n = 8 and _____ = 4.5. 2 This indicates that the median is midway between the 4th and 5th scores. This is shown diagrammatically by crossing off equal numbers of scores from each end: 5 5 6 6 7 7 7 9
If n is odd there is only one middle score. If n is even, then the median is the average of the two middle scores.
Insight Mathematics General 12 HSC Course 2
EXERCISE 4B 1 For the following scores find the: i mean ii mode iii median. a 7, 7, 8, 8, 9, 10, 10, 10, 11 b 0, 1, 1, 1, 2, 2, 3, 3 c 23, 24, 25, 25, 25, 26, 27, 27, 27, 29, 30 d 103, 104, 105, 105, 106, 106, 106, 107, 108, 110, 111, 112
WORKED EXAMPLE 2 For the scores in the table, find the: a mean b mode
c median.
Solve
a
Frequency (f )
7
4
8
2
9
7
10
6
11
1
Think
Score (x)
Frequency (f )
f×x
7
4
28
8
2
16
9
7
63
10
6
60
11
1
11
20
178
Total
Score (x)
Apply
Add an f x column to find the mean. Calculate the values in the f x column by multiplying the score by the frequency.
Check this answer using the statistics function on your calculator.
The largest value in the frequency column belongs to the score that is the mode.
Look for the score with the highest frequency; that score is the mode.
n+1 In this case _____ = 10.5. 2 Hence the median is the number midway between the 10th and 11th scores. Add a cumulative frequency column. Look for the number in the cumulative frequency column that is equal to or greater than 11.
A cumulative frequency column makes finding the median easier. Find the values in the cumulative frequency column by adding the frequency of each score to the frequency of all the scores less than that score.
b
Mode = 9
c
Cumulative frequency
Score
Frequency
7
4
4
8
2
6
9
7
13
10
6
19
11
1
20
The 10th and 11th scores are both 9. 9+9 Therefore, median = _____ = 9. 2
Chapter 4 Interpreting sets of data
DATA AND STATISTICS
178 Mean = ____ = 8.9 20
109
2 For the following scores, find the: i mean
a
c
ii mode
iii median. b
Score (x)
Frequency (f )
2
13
1
6
3
14
3
7
5
15
8
8
6
16
9
9
4
17
4
Score (x)
Frequency (f )
Score (x)
Frequency (f )
37
15
114
11
38
29
115
16
39
36
116
23
40
13
117
10
41
7
118
12
119
3
Score (x)
Frequency (f )
5
d
WORKED EXAMPLE 3 For the grouped data in the table, find the: a mean b mode
c median.
Solve
DATA AND STATISTICS
a
Class
Class centre (x)
Frequency (f )
f×x
0–3
1.5
2
3.0
4–7
5.5
3
16.5
8–11
9.5
4
38.0
12–15
13.5
9
121.5
16–19
17.5
2
35.0
20
214.0
Total
Class
Frequency (f )
0–3
2
4–7
3
8–11
4
12–15
9
16–19
2
Think
Apply
Find the class centres by averaging the boundary values of the classes: 0+3 _____ = 1.5 2 4+7 _____ = 5.5, etc. 2
For grouped data the class centre is used instead of the score. It gives an approximate value for the mean.
The largest frequency is 9, so the modal class is 12–15.
The mode cannot be found so the modal class is used.
214 Mean = ____ = 10.7 20
b
110
The modal class = 12–15
Insight Mathematics General 12 HSC Course 2
WORKED EXAMPLE 3 CONTINUED
Solve
c
Class centre
Cum. frequency
0–3
1.5
2
4–7
5.5
5
8–11
9.5
9
12–15
13.5
18
16–19
17.5
20
Cumulative frequency
Class
20 18 16 14 12 10 8 6 4 2 0
3
7
19
11 15 Score
From the ogive, median ≈ 12.
Think
Apply
Add a cumulative frequency column to the table. The median is the average of the 10th and 11th scores. From the cumulative frequency column, the 10th and 11th scores are in the 12–15 class, sometimes called the median class. It is not possible to find the median for grouped data from a table. However, it is possible to estimate the median from a cumulative frequency polygon (or ogive). The middle of 20 scores is the average of the 10th and 11th scores. Draw a line across to the ogive and a line down to estimate the median.
The ogive or cumulative frequency polygon is used to estimate the median in grouped data. Ogives for grouped data were demonstrated in Section 7M of the prelimary course. Plot the class endpoint on the horizontal axis and the cumulative frequency on the vertical axis.
3 For the following grouped data, find the: i mean ii modal class
a
Class
Frequency
0–4
b
iii median. c
Class
Frequency
3
0–9
9
6–10
7
10–19
11
3
11–15
12
20–29
15
15–19
2
16–20
5
30–39
5
20–24
1
21–25
3
Class
Frequency
5
1–5
5–9
8
10–14
WORKED EXAMPLE 4
Solve
Stem 3 4 5 6
Leaf 0 2 2 3 8 1 3 6 2 4 4 4 8 8 9 1 5 5 6 7
Think
Apply The stem-and leaf plot shows all scores. Make sure the stem and leaf are combined.
a
Mean = 50.4
Enter each value into a calculator in statistics mode. Enter 30, 32, … not 0, 2, 2. Press the keys for mean, x.
b
Mode = 54
54 occurs 3 times.
c
54 + 54 Median = _______ = 54 2
The median is the average of the 10th and 11th scores.
Chapter 4 Interpreting sets of data
DATA AND STATISTICS
For the data shown in the stem-and leaf plot, find the: a mean b mode c median.
111
4 For the following data, find the: i mean a Stem Leaf 2 3 4 4 8 3 0 1 1 1 5 7 4 2 4 8 9 5 3 3 5 6 6 8
c
Stem 12 13 14 15 16
Leaf 5 8 9 0 0 3 1 2 4 1 3 3 3 4 4
ii mode
iii median. b Stem Leaf 5 6 7 8
2 3 0 1
5 5 6 4 5 7 8 8 8 0 1 2 2 2 2 3 4 7 7 9 2
d Stem Leaf 6 4 3 5
6 6 3 7
7 9 8 5 5 8 8
9 10 11 12 13
3 4 0 2 0
3 4 1 2 0
7 4 3 3 0
4 9 9 3 3 4 5 5 6 1
5 Use the data from the stem-and-leaf plot in Worked example 4. a Complete a frequency distribution table using groups 30–39, 40–49, etc. b Calculate the mean from the grouped data. Compare this with the mean of 50.4 from the stem-and-leaf plot. c Add a cumulative frequency column. d Draw the ogive and estimate the median. Compare this with the exact median value of 54. e Why are some answers different? 6 Repeat question 5 using groups of 30–34, 35–39, etc. Explain any differences in the results. 7 Use the data from the stem-and-leaf plots in question 4. i Construct grouped frequency distribution tables with group size of 10. ii Estimate the mean and compare with the actual mean. iii Construct an ogive and estimate the median. Compare this with the exact median obtained for the stem-and-leaf plot. 8 a The mean mark in a test for the 24 students in Class 12M1 was 70 out of 100. What was the total of the marks scored by this class? b The mean on the same test for Class 12M2, which has 17 students, was 56. What was the total of the marks scored by Class 12M2? c What was the combined mean mark of both classes?
DATA AND STATISTICS
9 The distance from Broken Hill to Tibooburra is 360 km. A motorist travels from Broken Hill to Tibooburra at an average (mean) speed of 90 km/h and returns at an average speed of 60 km/h. What is the average speed of the motorist for the whole journey? 10 a b c d e
Show that the median of the scores 2, 5, 5, 5, 8, 11, 14, 17, 44 lies between the mode and the mean. Show that the mean of the scores 1, 3, 3, 3, 5, 5, 6, 6, 7, 7 lies between the mode and the median. Find a set of scores for which the mode lies between the median and the mean. Find a set of scores for which the mean lies between the median and the mode. Find a set of scores that has the same mean, mode and median.
INVESTIGATION 4.1 112
Insight Mathematics General 12 HSC Course 2
4C
Review of measures of spread
The following sets of marks both have the same mean of 10. Set A: 8, 9, 10, 11, 12 Set B: 0, 1, 10, 19, 20 In set A, the scores are within 2 marks of the mean. In set B, the scores are spread up to 10 marks away from the mean. Hence, as well as having a central representative value for a data set, it is also useful to have a measure of the spread of the data. Such measures of spread are called measures of dispersion. The measures of dispersion we have previously investigated in this course are the range, interquartile range and standard deviation. • The range is the difference between the highest score and the lowest score. • The interquartile range (IQR) is the difference between the upper quartile and the lower quartile. • The standard deviation is a measure the of ‘typical’ distance of all the scores from the mean.
WORKED EXAMPLE 1 Find the range of the following scores. a 3, 3, 4, 5, 5, 6, 6, 6, 8 Solve
b 3, 3, 4, 5, 5, 6, 6, 6, 80 Think
a
Range = 8 − 3 =5
Highest score = 8 Lowest score = 3
b
Range = 80 − 3 = 77
Highest score= 80 Lowest score = 3
Apply The major disadvantage of the range is that it depends only on the extreme values of the data, which may be outliers, such as the 80 in part b.
WORKED EXAMPLE 2
Solve
a
3 3 4 5 5 6 6 6 8 ↑ ↑ ↑ Q2 Q3 Q1 Q1 = 3.5, Q2 = 5, Q3 = 6 IQR = Q3 − Q1 = 6 − 3.5 = 2.5
b
3 3 4 5 5 6 6 6 80 ↑ ↑ ↑ Q2 Q3 Q1 Q1 = 3.5, Q2 = 5, Q3 = 6 IQR = Q3 − Q1 = 6 − 3.5 = 2.5
Think
Apply
There are 9 scores, hence the median (Q2) is the 5th score. The lower quartile (Q1) is the middle score of the lower subset 3 3 4 5. The upper quartile (Q3) is the middle score of the upper subset 6 6 6 8.
After arranging the data in numerical order, the quartiles are the scores that divide the data into four parts, each with the same number of scores. The results are the same for both data sets, as the interquartile range is unaffected by outliers.
Chapter 4 Interpreting sets of data
DATA AND STATISTICS
Find the interquartile range (IQR) of the following scores. a 3, 3, 4, 5, 5, 6, 6, 6, 8 b 3, 3, 4, 5, 5, 6, 6, 6, 80
113
For large data sets: • 25% of the scores are less than Q1 and 75% are greater than it. • 50% of the scores are less than Q2 and 50% are greater than it. • 75% of the scores are less than Q3 and 25% are greater than it. Hence the interquartile range gives a measure of the spread of the middle 50% of the scores. The weakness of the interquartile range as a measure of dispersion is that its value depends on only two scores, the upper and lower quartiles.
Standard deviation The standard deviation is a measure of spread whose value depends on all the scores. It is determined by finding the distances of all the scores from the mean and then calculating an average of these distances. The standard deviation can be calculated using an algebraic formula, but it is simpler and quicker to find values using a scientific calculator.
WORKED EXAMPLE 3 Find the population standard deviation (σn) for the following data set. 7, 8, 9, 9, 11, 12, 13, 13, 13, 14 Solve The calculator displays the value 2.34 (2 decimal places). σn = 2.34
Think Put the calculator into statistics mode and enter the scores. Press the keys for σn for the population standard deviation.
Apply The memory must be cleared before entering new data.
Just as there are different types of averages, there are also two types of standard deviation. It can be shown that, when σn is found for a sample taken from a large population, its value tends to underestimate the standard deviation for the whole population. For this reason the algebraic formula for standard deviation is altered slightly and this value, denoted by σn − 1, is used when finding the standard deviation for a sample. If the data in Worked example 3 was a sample from a larger population, then press the keys for σn − 1. The calculator displays the value 2.47 (2 decimal places). This is the sample standard deviation. In this course only the population standard deviation will be used.
DATA AND STATISTICS
EXERCISE 4C 1 For each set of scores, find the: i range ii interquartile range iii population standard deviation (1 decimal place). a 0, 0, 2, 3, 3, 4, 4, 5, 6 b 8, 8, 8, 9, 9, 11, 12, 12, 13, 14 c 17, 18, 18, 19, 19, 20, 21, 21, 21, 23, 24 d 7, 7, 8, 9, 9, 9, 10, 10, 12, 13, 14, 14 e 1, 2, 3, 4, 5 f 7, 7, 7, 7, 7
114
Insight Mathematics General 12 HSC Course 2
WORKED EXAMPLE 4 For the scores in this frequency table, find the: a range b interquartile range c population standard deviation.
Solve
b
5
63
8
64
11
65
9
66
7 Apply
Highest score = 66 Lowest score = 62
Score
Frequency
Cum. freq.
62
5
5
63
8
13
64
11
24
65
9
33
66
7
40
σn = 1.27 (2 decimal places)
Add a cumulative frequency column to the table. For this set of scores n = 40, n+1 hence _____ = 20.5. 2 The median is the average of the 20th and 21st scores 64 + 64 = _______ = 64. 2 There will be 20 scores below the median and 20 scores above it.
Put your calculator into statistics mode. Enter the scores.
2 For the scores in these frequency tables, find the: i range ii interquartile range
a
62
Think
Range = 66 − 62 = 4
10th score + 11th score Q1 = ____________________ 2 63 + 63 = _______ = 63 2 30th score + 31st score ____________________ Q3 = 2 65 + 65 _______ = 65 = 2 IQR = Q3 − Q1 = 65 − 63 = 2
c
Frequency
For the interquartile range find the median first. It is Q2. Find the upper and lower quartiles by applying the method used for the median to each of the two halves of the data.
Check the calculator instructions to enter multiple data values.
iii population standard deviation (1 decimal place). b
Score
Frequency
4
46
4
18
8
47
3
19
13
48
6
20
4
49
6
21
1
50
5
51
4
52
4
53
3
Score
Frequency
17
DATA AND STATISTICS
a
Score
Chapter 4 Interpreting sets of data
115
WORKED EXAMPLE 5 Use an ogive to calculate the interquartile range for this grouped data.
Cumulative frequency
Solve Class
Frequency
Cum. freq.
0–9
6
6
10–19
9
15
20–29
15
30
30–39
13
43
40–49
7
50
52 48 44 40 36 32 28 24 20 16 12 8 4 0
10
QL 20 M 30 QU 40 Class
Class
Frequency
0–9
6
10–19
9
20–29
15
30–39
13
40–49
7
Think
Apply
Add a cumulative frequency column to the table. Draw the ogive. There are 50 scores. For the lower quartile QL, draw a line across at 12.5 to find the value. For the upper quartile QU, draw a line across at 37.5 to find value.
The median and quartiles for grouped data are estimates. Add a cumulative frequency column and draw the ogive. Estimate from the graph by dividing the total number of scores into quarters. Note: Q1 is also called the lower quartile and Q3 is also called the upper quartile.
50
QU ≈ 35 Median ≈ 25.5 QL ≈ 16 IQR = 35 − 16 = 19
3
i Add a cumulative frequency column to these grouped frequency distribution tables and draw an ogive. ii Estimate the interquartile range.
DATA AND STATISTICS
a
116
b
Class
Frequency
8
0–4
10–19
12
20–29
c
Class
Frequency
10
100–109
5
5–9
12
110–119
12
17
10–14
20
120–129
20
30–39
20
15–19
11
130–139
18
40–49
15
20–24
7
140–149
15
50–59
8
Class
Frequency
0–9
Insight Mathematics General 12 HSC Course 2
4 a Use the class centres to find the population standard deviation for the grouped data given. b Is it possible to find the range and interquartile range from the table? c Draw the ogive for this data and use it to estimate the interquartile range.
Class
Frequency
0–4
5
5–9
6
10–14
8
15–19
7
20–24
4
5 The weights (in kg) of 17-year-old soccer players were 69, 64, 58, 77, 85, 90, 72, 61, 73, 59. Find the standard deviation (σn) of these weights. 6 a Find the standard deviation (σn) for these data sets. i 5, 6, 6, 7, 8, 9 ii 5, 6, 6, 7, 8, 90 b Discuss the effect of outliers on the standard deviation. 7 Determine whether the following are true or false. a The greater the standard deviation the greater the spread of the scores. b The standard deviation can be larger than the range. c If the standard deviation is zero then all the scores are of equal value.
INVESTIGATION 4.2 From Investigation 4.2 you should discover that: If a constant k is added to each of a set of scores then: • the mean is increased by an amount k • the standard deviation remains unchanged. If each of a set of scores is multiplied by a constant k then: • the new mean is k times the original mean • the new standard deviation is k times the original standard deviation.
8 A data set has a mean of 20 and standard deviation of 5. Determine the new mean and standard deviation when: a 4 is added to each score b 10 is added to each score c each score is multiplied by 2 d each score is multiplied by 5 e 3 is subtracted from each score f each score is divided by 5.
Relative merits of mean, mode and median
DATA AND STATISTICS
4D
Recall that the mean, mode and median are measures of central tendency; they are single numbers that are central or typical of all the data under consideration. All these measures are often referred to as ‘averages’, but this can be misleading as there are some sets of data for which one or more are inappropriate. Which measure is the most appropriate in a given situation? There is no general rule; rather, it depends on the nature of the data and the relative merits of each measure.
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WORKED EXAMPLE 1 Find the mean, mode and median for each set of data and comment on the relative merits of each measure. a 5, 5, 6, 6, 6, 7, 80 b 4, 4, 4, 6, 7, 7, 8, 9, 10 c 2, 3, 3, 4, 7, 7, 7, 7, 7 d 2, 4, 5, 7, 8, 9 Solve
Think/Apply
a
Mean = 16.4 Mode = 6 Median = 6
The mode and median are both central and typical for this data. However, 6 of the 7 scores are smaller than the mean, so it is not a central value and it would be inappropriate to use it as a measure of central tendency. The mean has the disadvantage that it is affected by relatively large, or small, atypical values, called outliers (such as the 80 in this data set). These are often caused by errors of measurement or data entry and, if so, should be discarded when calculating the mean. If we ignore the 80, the mean of the remaining scores ≈5.8. Otherwise, in these cases it is more appropriate to use the median or mode, which are not affected by the value of the outliers. Note: It is important to be certain that an outlier is an error before discarding it. A computer decision to reject outlying ozone layer measurements in the late 1980s delayed the discovery of the real decline in the ozone layer (the ozone hole) for some years.
b
Mean ≈ 6.6 Mode = 4 Median = 7
The mode occurs at one end and is therefore not a central value. The mean and median are both good measures of central tendency. Note: There are situations when the mode is important even if it is not central. For example, if the numbers represent shoe sizes then the size sold most often would be important to the shop owner.
c
Mean ≈ 5.2 Mode = 7 Median = 7
For this data neither the median nor the mode is a central value. The mean would be a more appropriate measure of central tendency. Note: If the numbers represented waiting times at a bank, then the mode would be a very important measure.
d
Mean ≈ 5.8 Median = 6
For this data no score occurs more often than any other, thus there is no mode. The mean and median are satisfactory measures of central tendency.
When selecting which of the three measures of central tendency to use as a representative figure for a set of data, consider the following characteristics of each measure: Mean • It is the most commonly used, is easy to understand and is easy to calculate. • Its value depends on the value of every score in the data set. • Its main disadvantage is that its value is easily distorted by very large, or very small, atypical scores (outliers). In this course outliers are either less than QL − 1.5 × 1QR or greater than QU + 1.5 × 1QR. DATA AND STATISTICS
• It has the best sampling stability. If we choose random samples of the same size from a large population, then the mean is the measure that varies the least from sample to sample. • It is not suitable for categorical data. Mode • It is the most typical value within a set of data. • Its value is easy to determine. • It is not affected by outliers.
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Insight Mathematics General 12 HSC Course 2
• • • •
It is the only measure suitable for categorical data. There may be no mode or more than one mode in a data set. It may not be a central value. When comparing random samples of the same size taken from a large population, the mode is the measure that varies the most.
Median • It is easy to understand: there are equal numbers of scores above it and below it. • It is not affected by outliers since its value depends on the number of scores above and below it, not the values of these scores. • It is not suitable for categorical data. • For some unusual patterns of scores it may not be a central value. For example, 2, 3, 3, 4, 7, 7, 7, 7, 7. • When comparing samples, it varies more than the mean but much less than the mode.
EXERCISE 4D 1 Find the mean, mode and median for the following sets of data and comment on the relative merits of each measure. a 17, 19, 20, 22, 23, 24 b 2, 3, 4, 5, 6, 6, 6, 100 c 8, 10, 12, 13, 14, 14, 14, 14, 14, 14 d 9, 9, 10, 11, 11, 12, 13, 13, 15
DATA AND STATISTICS
2 A small printing firm has 30 employees with the following annual salaries: General manager $160 000 Accountant $100 000 3 supervisors $60 000 10 tradesmen $40 000 15 production workers $28 000 a Calculate the mean salary for the employees of this firm. b How many employees earn less than the mean salary? c What is the median salary? d Find the modal salary. e Which ‘average’ would you use in a wage case if you were the: i general manager? ii union representative? f Which measure of central tendency is the most appropriate to represent the salaries of the employees? Give reasons for your answer. 3 The weights, in kilograms, of the players in a women’s hockey team are listed: 55, 56, 63, 48, 58, 56, 48, 54, 55, 62, 64, 157 a Are there any outliers in this data set? Discuss possible reasons for this score. b Calculate the mean weight of the team with and without the outlier. c Find the median weight of the team. d What is the modal weight? e Which measure of central tendency is the most appropriate to represent the weights of the players?
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4 The number of injuries sustained in workplace accidents in a factory over a period of 12 months are listed: 2, 0, 1, 0, 5, 0, 0, 3, 0, 1, 2, 2 a For the number of injuries, find the: i mean ii mode iii median b Which of these ‘averages’ is the most appropriate for this data? Give reasons for your answer.
5 The numbers of cars of each colour sold by a new car dealership are shown in the table. For this data, find, where possible, the: a mean b mode c median. 6 The times, to the nearest minute, spent in the waiting room of a hospital by 15 patients, chosen at random, were: 18, 19, 19, 20, 20, 21, 21, 28, 29, 30, 30, 30, 30, 30, 30. a Find the mean, mode and median of these waiting times. b Discuss the relative merits of these three averages for this data.
Colour
Number of cars
White
17
Grey
14
Red
12
Blue
7
Green
4
Other
5
7 Maria scored the following marks in a series of topic tests: 7, 2, 7, 3, 7, 2, 6, 1 a Find the mean, mode and median for this set of marks. b Which of these ‘averages’ would Maria prefer to tell her parents? Explain your answer. 8 The prices of units sold by a real estate agent in a particular suburb in one month were: $250 000, $320 000, $490 000, $265 000, $280 000. a The real estate agent claimed that, for this month, the ‘average’ price of units in the suburb was $321 000. Which ‘average’ has the estate agent used? b Is this an appropriate ‘average’ to use? c Give reasons for your answer, with reference to the other measures of central tendency.
DATA AND STATISTICS
9 If we choose random samples of the same size from a large population, then: a the measure that varies the least from sample to sample is the: A mean B mode C median b the measure that varies the most is the: A mean B mode C median 10 The measure that is affected the most by outliers is the: A mean B mode
C median
11 Which measure of central tendency depends on the value of every score in the data set? 12 Which is the only measure suitable for categorical data?
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Insight Mathematics General 12 HSC Course 2
4E
Shape of frequency distributions
The shape of frequency curves may be described in terms of smoothness, symmetry and number of modes. Graph A is a smooth curve, Graph B is not smooth. Graph A
Graph B f
f
Graph C is unimodal (has one mode) and graph D is bimodal (has two modes). Graph C
Graph D f
f
Note: A curve with two distinct humps, even if the frequency at each hump is not the same, is called bimodal. If a curve has three or more modes it is sometimes described as being multimodal. Graph E is symmetrical, Graph F is asymmetrical. Graphs which are not symmetrical are said to be skewed. Graph E
f
If the longer tail of the graph is to the left, then the distribution is negatively skewed. This would occur, for example, if we graphed the distribution of the results of a very easy test. Most of the students would score high marks and only a few would score low marks.
f
If the longer tail is to the right, then the distribution is positively skewed. This would occur for the distribution of the results of a very hard test.
f
Graph F
DATA AND STATISTICS
f
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Other shapes that occur regularly enough to be of importance are shown. This curve is described as bell-shaped. It occurs for many naturally occurring characteristics, such as the number of tomatoes on a plant, the number of peas in a pod, the heights of a particular age group of females. More will be learnt about this curve later in this chapter.
This is called a J-shaped distribution because of its similarity to the shape of this letter. As the value of the variable increases so does the frequency of that variable.
f
For a reverse J-shaped distribution, the value of the variable decreases as the frequency of occurrence increases.
f
f
A U-shaped distribution is U-shaped.
A uniform distribution has no mode.
f
f
WORKED EXAMPLE 1
Number of children
Number of matches
DATA AND STATISTICS
e
Points scored
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Insight Mathematics General 12 HSC Course 2
Number of fatalities
Number of peas
d
c
Number of families
b Number of pods
a
Speed
Number of students
Describe the shape, smoothness, symmetry and modality of the following curves.
Score on test
WORKED EXAMPLE 1 CONTINUED
Solve
Think/Apply
a
The curve is bell-shaped, smooth, symmetrical and has one mode (number peas in pod/number of pods).
b
The curve is positively skewed, smooth, not symmetrical and has one mode (number of children/number of families).
Symmetry is the first feature examined, followed by the number of ‘bumps’ or modes.
c
The curve is smooth, symmetrical and bimodal (score on test/ number of students).
d
The curve is not smooth or symmetrical and is multimodal (number of points scored/number of matches).
e
The curve is J-shaped, smooth, not symmetrical; the mode occurs at end of data (speed/number of fatalities).
EXERCISE 4E
Number of days
F
Number of tomatoes
Number of students
I Number of plants
Number of people at concert
Age
Rainfall
H
Time
Number of employees
E
Number of students
Age
Blood pressure
Shoe size
G
C
DATA AND STATISTICS
Team score
D
smooth, reverse J-shape, mode at end of data smooth, unimodal, positively skewed smooth, symmetrical, bell-shaped, unimodal smooth, J-shaped, mode at end of data
Number of people
B
Number of matches
A
b d f h
Number of deaths
1 Match each description with one of the curves shown. a smooth, symmetrical, U-shaped c uniform distribution, no mode e smooth, unimodal, negatively skewed g smooth, symmetrical, bimodal i not smooth, asymmetrical, multimodal
Mark
Chapter 4 Interpreting sets of data
123
WORKED EXAMPLE 2 Comment on the nature of the data graphed in Worked example 1. Solve
Think
Apply
a
The number of peas in a pod is evenly spread on both sides of the mode.
This is a symmetrical distribution.
b
More families have small numbers of children and fewer have large families.
Positive skew means more families at the left side of the graph.
c
Bimodal distributions often occur when two distinct populations are represented on the same graph. The graph shows that, as expected, a large number of Engineering students scored high marks on this test and a large number of Commerce students did poorly.
This is symmetrical with two modes.
d
There is no pattern of high, medium or low number of points scored in the matches recorded.
No real pattern.
e
For vehicles involved in accidents, as the speed of the vehicles increases so does the number of fatalities.
This is a type of exponential graph.
Smooth symmetrical graphs give detail about populations. In a later chapter we will examine symmetrical, unimodal distributions called normal distributions.
2 Comment on the nature of the data graphed in question 1.
WORKED EXAMPLE 3 For the data given in the stem-and-leaf plot: a Determine whether the distribution of scores is symmetrical, positively skewed or negatively skewed. b Are there any outliers? If so, what are they? c Are there any clusters? If so, where are they located?
Solve
a
The frequency curve is negatively skewed.
Think
DATA AND STATISTICS
3 Stem
0
1
9 8
4 2 1
7 6 5 5 3 1 1 0
2
3
4
9 8 7 6 4 3 2
5 4
5
6
b
The score 3 is an outlier.
The score 3 is much smaller than the other scores.
c
This set of scores is clustered in the 40s and 50s.
The rows with the most numbers can form a cluster of scores.
Insight Mathematics General 12 HSC Course 2
Leaf 3 8 1 0 2 4
9 2 4 1 1 3 5 5 6 7 3 3 5 7 8 9 5 Apply
Turn the stem-and-leaf plot on its side and draw a curve to fit the height of each column. Leaf
124
Stem 0 1 2 3 4 5 6
Outliers and clusters show properties of distributions.
3 Consider the data given in the stem-and-leaf plots below. i Determine whether the distribution of scores is symmetrical, positively skewed or negatively skewed. ii Are there any outliers? If so, what are they? iii Are there any clusters? If so, where are they located?
a
Stem 1 2 3 4 5
Leaf 2 3 6 8 8 2 5 6 7 7 8 9 4 4
b Stem Leaf
c Stem Leaf
0 1 2 3 4 5 6
3 4 5 6 7 8
9
5 3 2 0 2 1 4
7 2 4 6 1 1 5 8 8 9 3 3 6 5
9 2 2 0 0 1
5 2 3 1 3 4 4 7 8 0 2 4 5 5 7 9 9 3
WORKED EXAMPLE 4 Consider the data in the stem-and-leaf plot shown. a Calculate the median, QL, QU and IQR. b Calculate the limits for outliers in this data. c Is 50 an outlier? Explain your answer.
Solve
Stem 5 6 7 8 9 10 11
Leaf 0 9 1 1 0 3
9 2 8 1 4 6 6 6 7 8 0 1 3 4 5 8 6
Think
Apply The interquartile range is calculated as usual. The formulas used in this course for outliers are QL − 1.5 × IQR and QU + 1.5 × IQR. The scores can be crossed off from either end to find the median, QL and QU.
a
Median = 96 QL = 88 QU = 103 IQR = 103 − 88 = 15
There are 23 scores so the median is the 12th score. There are 11 scores on either side of the median. QL is the 6th score and QU is the 18th score.
b
QL − 1.5 × IQR = 88 − 1.5 × 15 = 65.5 QU + 1.5 × IQR = 103 + 1.5 × 15 = 125.5
Use the formula for outliers.
c
Yes
The score of 50 is below 65.5. There are no very high outliers.
DATA AND STATISTICS
4 Consider the data in the stem-and-leaf plots in question 3. i Calculate QL, QU and IQR. ii Calculate the values for outliers using QL − 1.5 × IQR and QU + 1.5 × IQR. iii Are there any outliers? Explain your answer.
Chapter 4 Interpreting sets of data
125
4F
Displaying data
Data can be displayed in graphs and tables to give an overall picture of trends and other properties of the data. A multiple display is one in which more than one set of data is graphed on the display so that the data sets can be compared. In this section we will use multiple displays to describe and interpret the relationship between data sets.
EXERCISE 4F Company employment Number of employees
1 Examine this multiple column graph (sometimes called a multiple bar chart) to answer the following questions. a How many company employees were there in 2010? b How many male employees were there in 2011? c How many female employees were there in 2008? d How many more males than females were employed in 2009? e What trends are evident from this display?
200 150
Males Females
100 50 0
2008 2009 2010 2011 2012 2013 Year
2 The horizontal bar graph below shows the apple production for a Tasmanian orchard in 2011 and 2012. a For which varieties was there an increase in production from 2011 to 2012? b For which variety was the increase greatest? c For which variety was the decrease greatest? d Estimate whether the total production in 2012 was greater than or less than in 2011. Tasmanian apple production Red Delicious Golden Delicious Democrat Granny Smith Red Fuji Sturner Crofton
2011 2012
Jonathon Others 0
400
1200 800 Number of cartons
1600
DATA AND STATISTICS
3 a The number of marriages and divorces in 2011 (to nearest 500) for the given states is shown in the table. i Use a double column graph to display this information for each state. ii For each state, express the ratio of number of divorces to number of marriages in the form n : 1 (n to 3 decimal places). iii Which state has the highest number of divorces per marriage? Which has the least? 126
Insight Mathematics General 12 HSC Course 2
State
Marriage
Divorce
NSW
42 000
14 000
Vic
29 000
12 500
Qld
26 000
11 500
SA
8 000
3 500
WA
12 500
5 000
b The number of births and deaths for 2011 is shown for these states in the table. i Display this information using a double horizontal bar chart. ii For each state, express the ratio of the number of births to the number of deaths in the form n : 1. iii Compare and comment on this information.
4 The subdivided vertical column graph shows the number of exploratory space missions sent to the Moon and nearby planets up to 1997. These have been broken down into four types: manned, landers, orbiter/probes and flyby missions. a How many manned missions have been sent to: i the Moon? ii Venus? b How many landers have been sent to: i Mars? ii Venus? c For all types of exploratory space missions, which heavenly body had the: i greatest number of missions? ii least number of missions? d Which planet has had the most flyby missions? 5 The radar chart shows the average monthly rainfall for Sydney and Melbourne. a What is the driest month of the year in each city? b What is the average rainfall in January for: i Sydney? ii Melbourne? c What is the difference between the average June rainfall of Sydney and Melbourne? d What is the difference between the average rainfall of the two cities in their wettest month? e In how many months is the rainfall in Sydney greater than in Melbourne? f Estimate the difference between the total annual rainfall of Sydney and Melbourne.
State
Births
Deaths
NSW
99 000
44 500
Vic
71 500
32 500
Qld
63 500
23 000
SA
20 000
12 000
WA
32 500
11 000
Number of missions
Space missions 52 48 44 40 36 32 28 24 20 16 12 8 4 0
Manned Lander Orbit/probe Flyby
Moon
Mars Venus Target body
Mercury
Monthly rainfall (mm) Jan Dec Nov
Oct
Sydney Melbourne
Feb
120 100 80 60 40 20 0
Mar
Apr
Sep
May Aug
Jun Jul
Month
J
F
M
A
M
J
J
A
S
O
N
D
Sydney
12
12
13
12
12
12
10
10
11
12
11
12
Melbourne
8
7
9
11
14
14
15
15
14
14
12
10
a b c d
DATA AND STATISTICS
6 The average number of rainy days each month for Sydney and Melbourne are given in the following table.
Display this data on a radar chart. In which months does Melbourne have more rainy days than Sydney? What is the total number of rainy days in each city? Based on the information given in questions 5 and 6, which is the wetter city, Sydney or Melbourne?
Chapter 4 Interpreting sets of data
127
WORKED EXAMPLE 1 Draw an area chart to compare the rainfall (in millimetres) in different regions of NSW, over the four seasons, using the information in the table. Northern
North-East
Southern
Western
Summer
300
280
300
140
Autumn
290
280
310
210
Winter
280
220
380
200
Spring
300
270
330
220
Solve NSW seasonal rainfall Western Southern North-East Northern
1200
Rainfall (mm)
1000 800 600 400
Think
Apply
Plot each value on top of the previous value. Plot 300 then add 280 on top to 580 then add 300 to 880 and finally add 140 to 1020. Continue for the other seasons.
This can be done by plotting points and joining with straight lines, as shown, or more simply using the stacked area graph of the Chart option on a spreadsheet package such as Excel.
200 0 Summer
Autumn
Winter
Spring
7 The table shows the number of employees of a manufacturing company for the years 2006 to 2010.
128
Year
Factory
Clerical
Management
2006
45
12
8
2007
40
10
7
2008
38
9
7
2009
30
8
6
2010
28
7
6
a Draw an area chart to compare the number of types of employees for these years. b What was the total number of employees in: i 2006? ii 2010? c What was the change in the total number of employees from 2006 to 2010? d Calculate the percentage change in the total number of employees from 2006 to 2010. e For which type of employee was there the greatest change in number of employees from 2006 to 2010? f Calculate the percentage change, from 2006 to 2010, in the number of: i factory workers ii clerical workers iii management staff. g For which type of employee was there the greatest percentage change in the number of employees? h What percentage of the employees in 2006 were: i working in the factory? ii clerical workers? iii in management positions? i Repeat part h for the year 2010. j What is the change in the percentage composition of the three types of employees from 2006 to 2010?
Spring Winter Autumn Summer
Rainfall (mm)
8 This area chart compares rainfall across the regions of NSW in a particular year. a Estimate the rainfall for the Northern region in: NSW regional rainfall i summer ii autumn 1200 iii winter iv spring. 1000 b Estimate the rainfall for the North-East region in: i summer ii autumn 800 iii winter iv spring. c How much rain fell in the: 600 i Southern region in autumn? 400 ii Western region in spring? d Which region had the: 200 i wettest summer? ii driest summer? 0 iii wettest winter? iv driest winter? Northern
North-East Southern Region
Western
9 a The area graph below shows the sales, expenses, profit before tax and profit after tax of a chain of retail stores for the years 2007 to 2010. In 2007 what were the: i sales? ii expenses? iii profit before tax? iv profit after tax? b Determine these amounts: i expenses in 2008 ii profit before tax in 2009 iii profit after tax in 2010 c In which year(s) did: i sales decrease? ii expenses decrease? iii profit after tax increase? iv profit after tax decrease? Profit and costs
Amount ($ millions)
80
DATA AND STATISTICS
Profit after tax Profit before tax Expenses Sales
100
60 40 20 0 2007
2008
2009
2010
Year
Chapter 4 Interpreting sets of data
129
WORKED EXAMPLE 2 a From the graph find the number of people aged under 20 years
Age distribution for city of 200 000 people Males
who are: i males ii females. b What percentage of the population is under 20 years of age? c What percentage of the population aged 70 years and over are: i males? ii females?
Age 80+ 70–79 60–69 50–59 40–49 30–39 20–29 10–19 0–9
Females
20 16 12 8 4 0 0 4 8 12 16 20 Population (’000)
Solve
a
i
17 000 + 13 000 = 30 000
ii
16 000 + 12 000 = 28 000
b
c
Think
i ii
Apply
Add the numbers in the 0–9 and 10–19 groups.
30 000 + 28 000 = 58 000 58 000 _______ × 100% = 29% 200 000
Calculate the total as a percentage of 200 000.
8000 _______ × 100% = 4% 200 000
Add 70–79 and 80+ then calculate this number as a percentage of the 200 000 total.
11 000 _______ × 100% = 5.5% 200 000
The graph is a special type of horizontal histogram used to display population statistics, called a population pyramid. It consists of two histograms, one for males and one for females, placed back to back. The vertical axis shows age groups and the horizontal axis shows the numbers, or percentages, of each age group in the total population.
10 The population pyramid below gives the age distribution of a town of 7820 people. Males
DATA AND STATISTICS
350
300
250
200
150
Age distribution for town of 7820 people Age 80+ 75–79 70–74 65–69 60–64 55–59 50–54 45–49 40–44 35–39 30–34 25–29 20–24 15–19 10–14 5–9 0–4 100
50
0 0 Population
50
100
Females
150
200
250
a How many males are there in these age groups? i 0–4 ii 35–39 iii 60–64 iv 20–29 b How many females are there in these age groups? i 5–9 ii 30–34 iii 65–69 iv 30–39 c Calculate the percentage (of the total population) of males in these age groups. i 10–14 ii 80+ iii 55–64 130
Insight Mathematics General 12 HSC Course 2
300
350
d Calculate the percentage (of the total population) of females in these age groups. i 15–19 ii 40–44 iii 0–14 e Calculate the percentage of the population that is: i less than 20 years of age ii 65 years or older iii 50 years or older but less than 70. f In which age group is the number of females double the number of males? g Is the birth rate higher for males or females? h What percentage of the 80+ age group are: i males? ii females? i What does the answer for part h indicate about the life expectancy of males and females? 11 The results of a survey of the number of males and females who smoke tobacco is shown in the table. Males
Females
Total
Smokers
4027
4426
8453
Non-smokers
8321
7462
15 783
12 348
11 888
24 236
Total
a b c d e f
What percentage of the people surveyed are smokers? What percentage are male? What percentage of the females surveyed are smokers? What percentage of smokers are female? Discuss the difference between your answers for part c and part d. Under what circumstances would these percentages be the same?
12 The following data concerning blood pressure was collected from a large sample of the population. Low
Normal
High
Total
Number of men
1350
8564
3648
13 562
Number of women
1290
7593
2861
11 744
2640
16 157
6509
25 306
a What percentage of the people in this sample: i are male? ii are female? iv have normal blood pressure? v have high blood pressure? b What percentage of the people with low blood pressure are men? c What percentage of the people with normal blood pressure are men? d What percentage of the people with high blood pressure are women? e What percentage of men have high blood pressure? f What percentage of women have low blood pressure?
iii have low blood pressure?
13 The table shows the place of birth of immigrants to Australia from English-speaking countries. a What percentage of Canadian immigrants are male? Country Males b What percentage of male immigrants are Canadian? Canada 11 600 c What percentage of Irish immigrants are female? d What percentage of female immigrants are Irish? Ireland 27 000 e What percentage of immigrants came from the UK? New Zealand 139 900 f From which countries were there more female than South Africa 24 300 male immigrants? United Kingdom 560 700 g For which country was there the greatest difference between the numbers of male and female immigrants? USA 26 300
Females 12 500
DATA AND STATISTICS
Total
25 400 135 900 25 300 557 900 24 300
Chapter 4 Interpreting sets of data
131
h Use the following formula to answer the questions. number of females − number of males Percentage sex difference = ________________________________ × 100% number of females + number of males i Which country has the greatest percentage sex difference? ii Which country has the smallest percentage sex difference? iii What is the significance of the negative answers?
14 The data in the table gives the proportion of home ownership in each of Australia’s five largest cities. a In which city is the percentage of the population who own their home: i greatest? ii smallest? b In which city is the number of people who own their home: i greatest? ii smallest? c Calculate the average (mean) number of people in these five cities who own their home.
Population (millions)
Population who own their home (%)
Sydney
3.714
69.26
Melbourne
3.189
74.96
Brisbane
1.422
72.74
Perth
1.221
73.55
Adelaide
1.071
70.98
City
SPREADSHEET APPLICATION 4.1
4G
Analysing data
Measures of location and measures of spread can be used to analyse and compare sets of data.
EXERCISE 4G 1 Carrie and Laura played 9 holes of golf. The player with the lowest score wins the hole. Usually the player with the lowest final score wins. Their scores are listed below. Hole
1
2
3
4
5
6
7
8
9
Total
Carrie
5
8
6
3
5
8
4
7
8
54
Laura
4
6
5
3
4
8
4
6
15
55
DATA AND STATISTICS
a For each player, find the: i mean ii mode b How many holes did each player win? c For each set of scores, compare the: i range ii interquartile range d Use the data to present an argument that Laura is the better player. e Use the data to present an argument that Carrie is the better player.
iii median.
iii standard deviation.
2 The marks scored by Julie and Alison in the 8 topic tests they did in Mathematics over the year are shown. Test
132
1
2
3
4
5
6
7
8
Julie
72
71
81
74
72
65
78
44
Alison
71
70
76
73
78
62
76
69
Insight Mathematics General 12 HSC Course 2
a Find the total marks scored by each student. b For the marks scored by each student, find the: i mean ii mode iii median. c In how many tests did Julie score higher marks than Alison? d For of each student’s marks, compare and comment on the: i range ii interquartile range. e Calculate the standard deviation of each student’s scores. f Which student’s performances were more consistent? g The Mathematics Prize was to be given to the better of these two students. Use the data to present a case for each student to receive the award.
3 The histograms represent the distribution of the marks of class 12D in English, Mathematics and Science.
5
6
7 8 Mark
9
10 8 6 4 2 0
Frequency
Frequency
Frequency
10 8 6 4 2 0
Science
Mathematics
English
5
6
7 8 Mark
9
10 8 6 4 2 0
5
6
7 8 Mark
9
a For each subject find the: i mean ii mode iii median iv range v standard deviation. b How many students scored the top mark in: i English? ii Mathematics? iii Science? c How many students scored the bottom mark in: i English? ii Mathematics? iii Science? d In which subject is the spread of scores: i greatest? ii least? e In which subject did the class perform the best? Give reasons for your answer. 4 The histograms below have the same scale on the frequency axis and the same scores on the horizontal axis. Data set 2
4 3 2 1 0
Frequency
Frequency
Data set 1
5
6
7 8 Score
9
4 3 2 1 0
5
6
7 8 Score
9
Chapter 4 Interpreting sets of data
DATA AND STATISTICS
a Determine whether the following statements are true or false. i The means of the distributions are equal. ii The medians of the distributions are equal. iii The modes of the distributions are equal. iv The range of data set 1 is greater than the range of data set 2. v The standard deviation of data set 1 is greater than the standard deviation of data set 2? b i Write out all the scores in data set 1. ii Find the median and the upper and lower quartiles. iii Repeat for data set 2. c Draw box-and-whisker plots on the same axis for the two data sets. What do the plots show? 133
5 The back-to-back stem-and leaf-plot compares the marks gained out Class A Stem Class B of 100 by students in two classes, A and B, on a term test. 8 2 5 8 9 a Are the following statements are true or false? 5 3 2 1 6 0 3 4 6 i The median mark in class A is 72. 8 5 3 1 0 7 0 2 6 6 7 ii The median mark in class B is 76. 2 1 0 8 1 4 8 iii The top mark in both classes in 98. 9 0 9 2 8 iv The lowest mark in both classes is 58. v The range of marks is the same for the two classes. b Draw box plots on the same axis to represent the data. c Count the actual scores to find the percentage of scores between the upper and lower quartiles for each class. 6 The height (nearest cm) of 17-year-old boys and Boys Stem Girls girls are shown in the stem-and-leaf plot. 15 4 a What is the height of the shortest: 15* 5 6 8 9 i girl? ii boy? 4 3 0 0 16 2 3 3 3 4 b What is the range of heights for: 9 9 8 8 8 8 7 7 7 6 5 16* 5 5 5 5 6 6 6 8 8 8 9 9 i girls? ii boys? 4 4 3 2 2 1 1 1 0 17 0 0 0 0 0 1 1 2 c How many students less than 160 cm tall are: 9 7 7 5 5 17* 5 6 7 i girls? ii boys? 1 0 18 d How many students more than 170 cm tall are: 5 18* i girls? ii boys? e Find the modal height for each sex. f Find the median height for each sex. g Which of the measures, mode or median, is more appropriate to use as a measure of location (central tendency) for this data? h Display this data in box plots on the same axis. i Calculate the actual percentage of scores between the lower and upper quartiles. 7 The results of two classes on a History test are given. Class A: 11, 16, 6, 7, 15, 12, 17, 20, 11, 12, 15, 10, 4, 17, 5, 13, 15, 15, 12 Class B: 4, 11, 14, 13, 13, 17, 19, 17, 12, 12, 11, 18, 9, 12, 8, 19, 19, 11, 6, 1 a Construct a back-to-back stem-and-leaf plot for this data using stems 0, 0*, 1, 1*, 2. b How many students in each class scored less than 5? c In which class did more students score a mark greater than 14? d For each set of scores, find the: i mean ii mode iii median. e Which class had the higher range of marks? f Calculate the standard deviation for each set of marks. g Which class performed better? Justify your answer.
DATA AND STATISTICS
8 a This box-and-whisker plot shows the times taken (in seconds) by male and female competitors to complete a triathlon. i What was the fastest time in the event? ii What was the difference in time between the slowest male and slowest female competitor?
Triathlon Females
Males
6500
7000
7500
8000
iii How long did it take the first 50% of the male competitors to finish? iv How long did it take the top 75% of the males to finish? 134
Insight Mathematics General 12 HSC Course 2
8500 9000 Time (s)
9500 10 000 10 500
v How long did it take the first 25% of the females to finish?
vi Which group had the greatest range of times? vii Which group had the smallest interquartile range of times?
viii Which group had the fastest overall times? b The box-and-whisker-plots shown give the breakdown for each leg of the triathlon. Analyse each leg of the triathlon in terms of the fastest times, spread of scores and symmetry of the distribution. Write a short report. Swimming leg of triathlon Females
Males
1250
1500 1750 Time (s)
2000
2250
Cycling leg of triathlon
Running leg of triathlon
Females
Females
Males
Males
3000
3500
4000 4500 Time (s)
5000
5550
1750
9 The table gives information about the average daily maximum temperatures for January in Sydney and Melbourne. Draw box-and-whisker plots on the same scale and compare the temperatures for the two cities. Write a short report on this comparison.
2000
2250
2500 2750 Time (s)
3000
3250
3500
Sydney
Melbourne
Minimum value
25.0
22.4
Lower quartile
25.5
24.0
Median
26.4
25.2
Upper quartile
26.9
27.2
Maximum value
28.9
29.5
Karuah
Buladelah
Mean
59
60
Median
58
58
Lower quartile
52
50
Interquartile range
18
9
Highest speed
85
105
INVESTIGATION 4.3
Lowest speed
35
40
SPREADSHEET APPLICATION 4.2
SPREADSHEET APPLICATION 4.3
10 The speed of the first 100 cars travelling through the towns of Karuah and Buladelah, from 9 am to 10 am on Saturday, were measured and the results shown in the table. a Draw box-and-whisker plots to display this data. b From this data, in which town should the police target speeding, between 9 am and 10 am on a Saturday morning? Justify your answer.
Chapter 4 Interpreting sets of data
DATA AND STATISTICS
1000
135
SPREADSHEET APPLICATION 4.1 Graphing data Follow the steps below to draw a multiple column graph for the information in this spreadsheet. Step 1: Start a new spreadsheet and type in the information in the table. Step 2: Highlight the table. Step 3: Click on the Insert menu and select ‘Column graph’.
Step 4: Choose ‘Multiple column graph’ from the Chart sub-types. Step 5: Investigate the other types of graphs available on the spreadsheet. Step 6: Investigate what happens when you adjust the scales, labels and other features.
1
A
B
C
D
Student
English
Maths
Science
2
Reiko
75
84
68
3
James
65
76
84
4
Wendy
89
65
60
5
Hassan
54
78
75
6
Vincent
62
90
88
Student
100 80 60
English Maths Science
40 20 0
Reiko
James Wendy Hassan Vincent
Step 7: Using the data in questions 3, 6 and 7 of Exercise 4F, compare different representations of the same data and comment on the appropriateness and effectiveness of the displays.
SPREADSHEET APPLICATION 4.2 Statistics Spreadsheets have inbuilt functions to find various statistics for a set of data. The spreadsheet below contains the marks out of 50 obtained on some tests. The marks in cells B2 down to B6 can be summed using the sum function in the formula =SUM(B2:B6). The mean mark can be found using the formula =AVERAGE(B2:B6) or =AVG(B2:B6). The standard deviation can be found using =STDEV(B2:B6).
DATA AND STATISTICS
1
136
A
B
Test
Mark
A
B
1
Test
Mark
2
1
33
2
1
33
3
=A2+1
23
3
2
23
4
=A3+1
45
4
3
45
5
=A4+1
18
5
4
18
6
=A5+1
44
6
5
44
7
Total
=SUM(B2:B6)
7
Total
163
8
Mean
=AVERAGE(B2:B6)
8
Mean
32.6
9
Standard deviation
=STDEV(B2:B6)
9
Standard deviation
12.14
Insight Mathematics General 12 HSC Course 2
1 Enter some marks and find the total, mean and standard deviation. 2 Try these formulas: =COUNT(B2:B6) =MAX(B2:B6) Explain what these formulas do.
=MIN(B2:B6)
3 Investigate the FORMULA function of the spreadsheet.
SPREADSHEET APPLICATION 4.3 Pivot tables A pivot table report is an interactive table that combines and compares large amounts of data. You can rotate its rows and columns to see different summaries of the source data, you can display the details for areas of interest and you can calculate different summaries, such as averages.
• Enter the data from the table into a spreadsheet. • Select ‘Table’ from the Insert menu. • Enter where the data is located and that the table has headers. Insert that information into another spreadsheet. • Enter ‘Distance’ as a column label and ‘Method’ as a row label. • Enter ‘Time’ into the values box, select ‘Value field settings’ then select ‘Average’. Average of time taken
Method
Time taken
100
Jog
22
100
Walk
28
100
Run
14
200
Run
31
200
Walk
50
200
Jog
39
Jog
Run
Walk
100
Jog
24
100
19.8
12.3
28.7
100
Jog
20
200
36
27.3
55.3
100
Walk
33
100
Run
11
200
Run
25
200
Walk
63
200
Jog
40
100
Jog
18
100
Walk
25
100
Run
12
200
Run
26
200
Walk
53
200
Jog
29
100
Jog
15
• Go to pivot table tools: Design, select ‘Grand totals’ and select ‘None’. Select Pivot chart to draw the graph. 60 Average time in seconds
Distance
Training times
50 40
Jog Run Walk
30 20 10 0
100
Distance
200
DATA AND STATISTICS
What to do
Experiment with changing the row and column labels and other aspects of the pivot table.
Chapter 4 Interpreting sets of data
137
INVESTIGATION 4.1 Modelling 1 For the scores 4, 5 and 9, find the: a mean
b median.
2 a Add 3 to each of the scores in question 1, then find the mean and median of these scores. b Is the new mean equal to 3 more than the original mean? c Is the new median equal to 3 more than the original median? 3 a Multiply each of the scores in question 1 by 3, then find the mean and median of this set of scores. b Is the new mean equal to 3 times the original mean? c Is the new median equal to 3 times the original median? 4 a Square each of the each of the scores in question 1, then find the mean and median of these scores. b Is the mean of the squares equal to the square of the mean? c Is the median of the squares equal to the square of the median?
INVESTIGATION 4.2 Comparing two sets of data 1 For the scores 2, 2, 3, 5, 5, 6, 9 and 10, find the: a mean
b population standard deviation (σn).
2 Add 3 to each score in question 1 and recalculate the mean and standard deviation. 3 Add 5 to each score in question 1 and recalculate the mean and standard deviation. 4 Multiply each score by 3 in question 1 and recalculate the mean and standard deviation. 5 Multiply each score by 5 in question 1 and recalculate the mean and standard deviation.
INVESTIGATION 4.3 Comparing two sets of data
DATA AND STATISTICS
Use summary statistics and statistical displays to compare and comment on related pairs of data. • Home scores and away scores for national sporting competitions • The ages of Oscar-winning male actors and female actors • The blood pressure of males versus females • The heights (or weights) of males versus females • The waiting times of customers at a fast-food outlet on two different days of the week What to do
1 Collect data. Determine the means, modes, medians and spreads of the scores. 2 Display the data using multiple displays, including comparative box-and-whisker plots. 3 Comment on similarities and differences between the data sets. 138
Insight Mathematics General 12 HSC Course 2
REVIEW 4 INTERPRETING SETS OF DATA Language and terminology Here is a list of terms used in this chapter. Explain each term in a sentence. area chart, bell-shaped, bimodal, interquartile range, measure of dispersion, measure of location, measure of spread, measures of central tendency, mean, median, mode, multiple display, multimodal, population pyramid, range, population standard deviation, sample standard deviation, skewed, standard deviation, unimodal
Having completed this chapter you should be able to: • determine the mean, mode and median for both grouped and ungrouped data • recognise when it is appropriate to use these measures and location • determine, where possible, the range, interquartile range, population and standard deviation for both grouped and ungrouped data • use multiple displays to describe and interpret the relationships between two sets of data • display and interpret two sets of data on a radar chart • construct and interpret an area chart to compare different sets of data over time • interpret data presented in two-way table form • compare summary statistics from two sets of data • display and interpret data in back-to-back stem-and-leaf plots • display and interpret data in two box-and-whisker plots drawn on the same scale.
4 REVIEW TEST 1 Which statement about grouping data is correct? A Groups do not have to be the same size. B Grouping makes large numbers of data easier to analyse. C Grouping does not lose detail. D Groups must overlap each other. 2 The median of the data 16, 21, 18, 15, 19, 17, 24, 15, 32, 13 is: A 17.5 B 18 C 18.5
D 19
3 The mean of the data 16, 21, 18, 15, 19, 17, 24, 15, 32, 13 is: A 17.5 B 18 C 18.5
D 19
FINANCIAL DATA AND STATISTICS MATHEMATICS
4 In a Science test, a class of 24 boys had a mean of 56 and a class of 16 girls had a mean of 66. The combined mean of the two classes is: A 56 B 60 C 61 D 62 5 Which statement is true when 4 is added to each score in a data set? A The mean increases by 4 and the standard deviation increases by 4. B The mean increases by 4 and the standard deviation stays the same. C The mean stays the same and the standard deviation increases by 4. D The mean stays the same and the standard deviation stays the same. 6 The interquartile range of the scores 5, 5, 6, 7, 7, 7, 8, 8, 9 is: A 2 B 2.5 C 3
D 4
Chapter 4 Interpreting sets of data
139
7 The standard deviation of the scores in the table is: A 17.19 B 6.31 C 6.43 D 4
Class
Frequency
5–9
3
10–14
7
15–19
8
20–24
4
25–29
5
8 Consider the data set, 5, 6, 7, 7, 11, 12, 13. If the value 13 is changed to 3, which of the following is unchanged in value? A mean B median C range D standard deviation 9 Which is the only measure of central tendency that can be used with categorical data? A mean B mode C median D range 10 The graph shown is: A smooth, unimodal, positively skewed B smooth, unimodal, negatively skewed C smooth, symmetrical, bimodal D smooth, symmetrical, U-shaped
f
11 The graph in question 10 could represent: A the class marks for an easy test C a uniform distribution of class marks
B the class marks for a hard test D no recognisable pattern of marks.
12 Which of the following statements about the data in the graph is not true? A Expenses exceeded sales in only one year. B Expenses have increased every year. C Sales have increased every year. D The greatest profit was made in 1994.
Sales and expenses
Amount ($’000)
50 40 30
Sales Expenses
20 10 0
1992 1993 1994 1995 1996 1997 1998 1999 Year
DATA AND STATISTICS
13 A random sample of people were surveyed and the results are shown in the table. Which of the following statements is not true? Born in Australia
Born overseas
Totals
Male
216
24
240
Female
220
40
260
436
64
500
Total
A B C D 140
The percentage of the sample who are females is 52%. The percentage of females who were born overseas is 8%. The percentage of people born overseas who are female is 62.5%. The percentage of males who were born in Australia is 90%.
Insight Mathematics General 12 HSC Course 2
14 Which of the following statements is true for the data sets shown? A The range of data set 2 is greater than the range of data set 1. B The median for data set 2 is greater than the median for data set 1. C The interquartile range is greater for data set 2. D The highest value occurs in data set 2.
Data set 1
Data set 2
100
200
150
15 Which of the following statements is not true? A The median of class B is greater than the median of class A. B The mean of class B is greater than the mean of class A. C The mode of class B is greater than the mode of class A. D The range of class B is greater than the range of class A.
250
300
Class A 8 5 2 2 0 9 7 6 3 1 8 7 4 3 2 9 8 6 5 9
350
300
350
Class B 2 5 3 3 6 9 9 0 1 4 8 8 8 1 1 3 7 0 1 1
4 5 6 7 8
If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question
1
2–4
5–7
8, 9
10, 11
12, 13
14, 15
Section
A
B
C
D
E
F
G
4A REVIEW SET 1 For the scores 5, 5, 6, 7, 7, 8, 8, 8, 10 find the: a mean b mode
c median
2 Find the range and interquartile range of the scores 5, 5, 6, 7, 7, 8, 8, 8, 10. 3 The prices of apartments sold in one month by a real estate agent are listed. $250 000, $280 000, $240 000, $300 000, $290 000, $550 000 a Calculate the: i mean price ii median price. b Which of these measures (mean or median) is the more appropriate average to use for these prices? Give reasons for your answer.
Had accident
No accident
Total
Drinkers
27
19
46
Non-drinkers
10
44
54
37
63
100
Total
FINANCIAL DATA AND STATISTICS MATHEMATICS
4 A survey of drivers by an insurance company yielded the following results.
a What percentage of those surveyed: i were drinkers? ii were not drinkers? iii were involved in an accident? iv were not involved in an accident? b What percentage of drinkers were involved in an accident? c What percentage of drivers who were involved in an accident were drinkers?
Chapter 4 Interpreting sets of data
141
5 The maximum temperature recorded each month for two towns is shown in the table below. Month
J
F
M
A
M
J
J
A
S
O
N
D
Town A
31
30
28
26
23
18
17
19
24
27
28
30
Town B
36
34
30
26
22
17
13
20
25
27
32
35
a For the temperatures for each town, find the: i mean ii mode iii median. b For the temperatures for each town, find the: i range ii interquartile range iii standard deviation. c Graph the maximum temperature for each town on the same radar chart. d Write a short description comparing the temperatures in the two towns.
4B REVIEW SET 1 For the data in the table find the: a mean b mode
c median.
2 a Draw a cumulative frequency graph for the data in the table. b Use your graph to find the: i median ii interquartile range.
Score
Frequency
12
5
13
4
14
7
15
11
16
9
Class
Frequency
1–5
3
6–10
5
11–15
12
16–20
6
21–25
4
3 For the data 22, 23, 23, 25, 26, 26, 27, 29, 31, 31 find the population standard deviation. 4 What percentage of a large set of scores: a is less than the lower quartile? b is greater than the lower quartile? c lies between the lower quartile and the upper quartile? d is less than the upper quartile? e is greater than the upper quartile?
DATA AND STATISTICS
5 Two groups of students are given pairs of shoes to wear to school. The students in the first group (X) are given rubber-soled shoes and those in the second group (Y) are given shoes with soles made of a new synthetic material. The thickness (mm) of the soles of the shoes after 6 months is shown below: Group X: 5, 3, 4, 5, 4, 3, 5, 6, 2, 6, 4, 4, 5, 3, 5, 6, 7, 6, 6, 3, 4, 5, 4, 5 Group Y: 3, 4, 5, 6, 5, 6, 4, 8, 5, 3, 4, 7, 6, 5, 2, 8, 5, 6, 5, 6, 8, 7, 3, 4 a For each group of results, find the: i mean ii mode iii median. b For each group of results determine the: i range ii standard deviation. c Is the new synthetic sole an improvement on the original? Justify your answer.
142
Insight Mathematics General 12 HSC Course 2
4C REVIEW SET 1 For the data in the table, determine the: a mean b modal class.
2 For the data in question 1, find the population standard deviation.
Class
Frequency
7–9
7
10–12
15
13–15
24
16–18
19
19–21
16
3 a Drawn an ogive for the data in this frequency table. Score
0
1
2
3
4
5
6
Frequency
11
8
15
21
28
10
7
b From the ogive, estimate the value of the: i lower quartile ii upper quartile
iii median.
4 A netballer had an average of 16.5 goals after 14 matches. If she scored 21 goals and 24 goals in the next two matches, find her new average. 5 The table shows the mass at birth of some guinea ppigs g and their mass when theyy were 2 weeks old. Mass at birth (g)
Mass at 2 weeks (g)
A
75
210
B
70
200
C
80
200
D
70
220
E
74
215
F
60
200
G
55
206
H
83
230
Using a spreadsheet, or otherwise, draw a multiple graph to display this information. What was the mean mass at birth? What was the mean mass after 2 weeks? What was the mean increase in mass over the 2 weeks?
FINANCIAL DATA AND STATISTICS MATHEMATICS
a b c d
Guinea pig
4D REVIEW SET 1 The mean and standard deviation for a set of data are 12 and 5 respectively. Write down the new mean and standard deviation if: a 3 is added to each score b 5 is subtracted from each score c each score is multiplied by 2 d each score is divided by 2.
Chapter 4 Interpreting sets of data
143
2 The number of goals scored by two basketball players in 20 games are shown below. Adrian
Zoltan
Number of goals
4
5
6
7
8
9
10
Frequency
1
1
2
4
7
4
1
Number of goals
5
6
7
8
9
10
Frequency
3
2
4
5
4
2
a For each player calculate the: i mean ii mode iii median. b For each player, find the: i range ii standard deviation. c Who is the better player? Give reasons for your answers. 3 The results of a History test given to two classes is shown below. Class A: 12, 17, 8, 9, 15, 13, 18, 20, 11, 13, 16, 10, 4, 19, 7, 14, 16, 15, 12 Class B: 3, 8, 14, 12, 12, 17, 19, 17, 13, 11, 10, 14, 9, 12, 6, 18, 18, 11, 5, 1
a Construct a back-to-back stem-and-leaf plot for this data with stems divided into two groups; that is, using stems 0, 0*, 1, 1*, 2. b What is the range for each class? c Find for each class the: i mean mark ii median mark. d Calculate the standard deviation of the marks for each class. 4 A farmer investigated the yield, to the nearest kilogram, of the fruit trees in two different orchards. The results are recorded below. Yield (kg)
10–14
15–19
20–24
25–29
30–34
35–39
40–44
Number of trees (Orchard A)
6
19
23
28
22
17
5
Number of trees (Orchard B)
8
25
30
19
15
13
10
a For each orchard estimate the: i mean yield per tree ii standard deviation. b For each orchard draw cumulative frequency polygons and estimate the: i median yield ii interquartile range of the yields. c Which orchard has more low-yielding trees (
