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Project: MathWorks 10output Date: 07/20/11File name: MW11-WorkbookCover.indd

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©2009–10PacificEducationPress

MathWorks11Workbook Solutions

Pacifi c Educational Press

ISBN 978-0-9867141-3-9

This workbook is designed to accompany the MathWorks 11 student resource by providing extra practice problems based upon real-world scenarios.

RELATEDMATHWORKS 11RESOURCESStudent Resource . . . . . . . . . . . . . . . . . . . 978-1-89576-692-9Student Resource CD . . . . . . . . . . . . . . . . 978-0-98671-410-8Teacher Resource . . . . . . . . . . . . . . . . . . . 978-0-98671-415-3Teacher Resource CD . . . . . . . . . . . . . . . . 978-0-98671-416-0

Pacif ic Educational Press University of Brit ish Columbiawww.pacif icedpress.ca

SW-COC-002358M

athWorks11

Workbook

MathWorks 11Workbook

SolutionS

Pacific Educational PressVancouver, Canada

Copyright Pacific Educational Press 2011

ISBN 978-1-926966-05-2

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, electronic, mechanical, photocopying, recording, or otherwise without the prior written permission of the publisher.

WriterKatharine Borgen, Vancouver School Board and University of British Columbia

ConsultantsKatharine Borgen, PhD, Vancouver School Board and University of British Columbia

DesignFive Seventeen

Layout and IllustrationSharlene Eugenio Helen Luk Eva Neeseman Five Seventeen

Cover PhotographMason Morfix/Getty Images

EditingKatrina PetrikJordie Yow

SlopeandRateofChange 5

1.1 Rise Over Run 5

1.2 Grade, Angle of Elevation, and Distance 10

1.3 Rate of Change 15

Chapter Test 21

GraphicalRepresentations 23

2.1 Broken Line Graphs 23

2.2 Bar Graphs 28

2.3 Histograms 31

2.4 Circle Graphs 34

Chapter Test 39

SurfaceArea,Volume,andCapacity 41

3.1 Surface Area of Prisms 41

3.2 Surface Area of Pyramids, Cylinders, Spheres, and Cones 50

3.3 Volume and Capacity of Prisms and Cylinders 57

3.4 Volume and Capacity of Spheres, Cones, and Pyramids 63

Chapter Test 68

1

2

3

MathWorks 10 Workbook Solutions 3

Contents

Contents continued TrigonometryofRight

Triangles 72

4.1 Solving for Angles, Lengths, and Distances 72

4.2 Solving Complex Problems in the Real World 76

Chapter Test 82

ScaleRepresentations 85

5.1 Scale Drawings and Models 85

5.2 Two-Dimensional Representations 90

5.3 Three-Dimensional Representations 94

Chapter Test 98

FinancialServices 100

6.1 Choosing an Account 100

6.2 Simple and Compound Interest 105

6.3 Credit Cards and Store Promotions 110

6.4 Personal Loans, Lines of Credit, and Overdrafts 115

Chapter Test 120

PersonalBudgets 123

7.1 Preparing to Make a Budget 123

7.2 The Budgeting Process 127

7.3 Analyzing a Budget 131

Chapter Test 139

6

5

4

7

MathWorks 10 Workbook Solutions 4

Chapter 1Slope and Rate of Change

1.1 Rise Over Run

Review: wORking with RatiO and slOpe

Build YOuR skills, p. 11

1. Calculatetheratioofconcentratetowater.

concentrate:water=250:750

concentrate:water=(250÷250):(750÷250)

concentrate:water=1:3

2. Calculatetheratioofbluepaintto

yellowpaint.

blue paintyellow paint

blue paintyellow p

= 2 31 7..

aaint

blue paintyellow paint

= ××

=

2 3 101 7 10

2317

.

.

3. a)Calculatetheratioofoatstoraisins.

oats:raisins=6:1

b) Calculatetheratioofalmondstococonut.

almonds:coconut 2: 34

almonds:coconut ):

=

= × ×((2 4 34

4))=almonds:coconut 8:3

c) Calculatethetotalamountofingredients.

total=oats+almonds+raisins+coconut

total=6+2+1+ 34

total=9 34

cups

Calculatetheratioofoatstototal

ingredients.

oats:total 6:9

oats:total 6:

oats:total

=

= +( )34

364

34

== ( )= × ×( )=

6:

oats:total ):

oats:total :

394

6 4 394

4

24

(

339

oats:total ):(

oats:total 8:13

= ÷ ÷

=

( )24 3 39 3

4. Setupaproportiontosolveforx,thenumber

ofcansofpaintneeded.415 20

20 415 20

20

20 415

5 3

=

× = ×

× =

≈

x

x

x

x. L

About5.3Lofoilwillbeneeded.

5. Setupaproportiontosolveforx,thenumber

ofcentimetresusedtorepresent45kmof

actualground.

1025 45

45 1025 45

45

45 1025

18

=

× = ×

× =

=

x

x

x

xcm

Onthemap,45kmwillberepresentedas18cm.

6. Setupaproportiontosolveforx,theamount

ofcornneeded.

31 5 4 5

4 5 31 5 4 5

4 5

4 5 31 5

9

. .

.. .

.

..

=

× = ×

× =

=

x

x

x

xcups

new skills: wORking with slOpe


7. a) m

m

m

=

=

=

riserun

or approx. 0.17

318

16

b) m

m

m

=

=

=

riserun

or 0.25

1 255

14

.

c) m

m

m

m

=

=

= ÷÷

=

riserun

or 1.25

108

10 28 2

54

d) m

m

m

m

=

=

= ÷÷

=

riserun

or approx. 0.44

818

8 218 2

49

e) m

m

m

m

=

=

= ××

=

riserun

or approx

2 43 5

2 4 103 5 10

2435

.

.

.

.

.. 0.69

8. Setupaproportiontosolveforx,therun.

m

x

xx

x

x

x

=

=

× = ×

× =

= ÷

riserun

3190

400

3190

400

3190

400

400 33190

25 333x ≈ m

Thehillcoversahorizontaldistanceof

about25333m.

MathWorks11WorkbookSolutions Chapter1 SlopeandRateofChange 6

9. Solveforx,therunofthestaircase.

m

x

xx

x

x

x

=

=

× = ×

=

≈

riserun

c

0 95 210

0 95 210

2100 95

221

.

.

.

mm

Therunofthestaircaseisabout221cm.

10.Solveforx,theriseofthestreet.

m

x

x

x

=

=

× =

≈

riserun

m

0 5428

28 0 54

15 1

.

.

.

Theriseofthestreetisabout15.1m.

11. m

m

m

m

m

=

=

=

= ÷÷

=

riserun

or appro

3 56 0

3560

35 560 5

712

.

.

xx. 0 58.

12.Solveforx,therunofthestairway.

m

x

xx

x

x

x

=

=

× = ×

× =

=

riserun

0 89 203

0 89 203

0 89 203

2030

.

.

.

..89

228x ≈ cm

Therunofthestairwayisabout228cm.

13.Solveforx,theriseoftheslide.

m

x

x

x

=

=

× =

=

riserun

2.55 m

1710 1 5

1 5 1710

.

.

pRactice YOuR new skills, p. 17

1.

Rise Run

Slope

Asafraction

m =( )riserun

Asadecimal

18m 63mm

m

m

=

= ÷÷

=

186318 963 927

m

m

=

≈

270 29.

21m 49mm

m

m

=

= ÷÷

=

214921 749 737

m

m

=

≈

370 43.

1.2cm 0.6cmm

m

=

=

1 20 62

.

.2

12.4mm 4.6mmm

m

m

m

=

=

= ÷÷

=

12 44 6

12446

124 246 2

6223

..

m

m

=

≈

62232 70.

300ft 900ft m

m

=

=

30090013

m

m

=

≈

130 33.


2.

Rise Run Slope15ft 60ft

m

m

m

=

=

=

riserun156014

12cm

m =

=

=

=

riserun12run

run

run cm

0 375

120 37532

.

.

0.375

m =

=

× =

=

riserunrise

rise

m rise

95 16

16 95

28 8.

16m 95

m =

=

× =

=

riserunrise

rise

in rise

327 42

42 327

192

42in 327

63mm =

=

=

=

riserun63run

run

run m

3 0

633 021

.

.

3.0

19.5ftm =

=

=

=

riserun

runrun

run ft

0 25 19 5

19 50 2578

. .

..

0.25

3. m

m

m

m

=

=

= ÷÷

=

riserun

or approx. 0.4

56120

56 8120 8

715

77

4. Setupaproportiontosolveforx,the

horizontaldistancefromthebaseoftheladder

tothehouse.

41

22

4 22

4 22

224

5 5

=

× = ×

=

=

=

x

xx

x

x

x

x . ft

Thebaseoftheladdershouldbe5.5ftfrom

thehouse.

5. Setupaproportiontosolveforx,theriseof

theditch.

31 5 25

25 31 5

50

.

.

=

× =

=

x

x

xcm

Thedrainageditchwilldrop50cmovera

horizontaldistanceof25m.

6. m

x

x

x

=

=

× =

≈

riserun

m

0 6432

32 0 64

20 5

.

.

.

Thehillwillrise20.5m.

7. Calculatethedifferenceinelevation.

895–752=143m

Converttokilometres.143m=0.143km

Calculatetheslope.

m

m

m

=

=

=

riserun

or approx.

0 1431 2

1431200

0 12

..

.


8. Roof:

m

m

m

m

m

=

=

=

= ÷÷

=

riserun

or 0

3 28

3280

32 1680 16

25

4

.

.

Trusses:

m

m

m

m

m

=

=

=

= ÷÷

=

riserun

or 0.4

1 64

1640

16 840 8

25

.

Theroofandthetrusseshavethesame

slope,0.4.

9. Therunoftheactualhousewillbehalfthe

totalwidth.

10.8m÷2=5.4m

Therunofthedollhousewillbehalfthetotal

width.

1.6m÷2=0.8m

Setupaproportiontosolveforx,theriseof

theroof.

2 45 4 0 8

0 8 2 45 4

0 36

.

. .

. ..

.

=

× =

≈

x

x

xm

Theriseofthedollhouseroofwillbe

about0.36m.


1.2 grade, angle of elevation, and distance

Review: the pYthagORean theORem and the tangent RatiO

tan

tan ..

tan ..

.

θ =

=

= ( )≈ °

−

oppadj

E

E

E

9 46 5

9 46 5

55 3

1

2. a) tan

tan

tan

.

θ

θ

θ

θ

=

=

= ( )≈ °

−

oppadj

1125

1125

23 7

1

m

m

m

=

=

=

riserun

1125

0 44.

b) tan

tan

tan

.

θ

θ

θ

θ

=

=

= ( )≈ °

−

oppadj

2914

2914

64 2

1

m

m

m

=

=

≈

riserun

2914

2 07.


1. a) h a b

h

h

h

2 2 2

2 2 2

2 2

1 9 4 3

1 9 4 3

4 7

= +

= +

= +

≈

. .

. .

. cm

tan

tan ..

tan ..

.

θ =

=

= ( )≈ °

−

oppadj

G

G

G

1 94 3

1 94 3

23 8

1

b) r a b

r

r

r

2 2 2

2 2 2

2 2

7 8 8 2

7 8 8 2

11 3

= +

= +

= +

≈

. .

. .

. m

tan

tan ..

tan ..

.

θ =

=

= ( )≈ °

−

oppadj

S

S

S

8 27 8

8 27 8

46 4

1

c) f a b

f

f

f

2 2 2

2 2 2

2 2

6 5 9 4

6 5 9 4

11 4

= +

= +

= +

≈

. .

. .

. in


3. a)

2 2 2

2 2 2

2 2

3 7

3 7

9 49

7 6

= +

= +

= +

= +

≈

a b

. ft

b) tan

tan

tan

.

θ

θ

θ

θ

=

=

= ( )≈ °

−

riserun

37

37

23 2

1

4. a)

2 2 2

2 2 2

2 2

2 2 5 8

2 2 5 8

4 84 33 64

6

= +

= +

= +

= +

≈

a b

. .

. .

. .

.22 m

b) tan

tan ..

tan ..

.

θ

θ

θ

θ

=

=

= ( )≈ °

−

riserun

2 25 8

2 25 8

20 8

1

5. a) Setupaproportiontosolveforx,therun

ofthewheelchairramp.

2 530

2 4

2 530

2 4

2 530

2 4

2 4 2 530

. .

. .

. .

. .

=

× = ×

× =

= ÷ (

x

xx

x

x

x ))≈x 28 8. m

b)

2 2 2

2 2 2

2 2

28 8 2 4

28 8 2 4

4 84 33 64

= +

= +

= +

= +

≈

a b

. .

. .

. .

228 9. m

new skills: wORking with gRade

6. percent grade riserun

percent grade

= ×

= ×

100

0 95

100.

ppercent grade = 18%

7. a) Converttheruntoinches.

10ft×12in/ft=120in

m

m

m

=

= −

≈ −

riserun

2 5120

0 021

.

.

b) percent grade riserun

percent grade

= ×

= ×

100

0 021 10. 00

2 1percent grade = . %

8. a) percent grade riserun

percent grade

13.

= ×

= ×

100

100m

55 = ×

=

=

m

m

m

100

13 5100

0 135

.

.

b) tan

tan

tan .

tan ( . )

.

θ

θ

θ

θ

θ

=

=

=

=

≈

−

riserun

m

0 135

0 135

7 7

1

°°


c) m =

=

× =

≈

riserun

rise

rise

m rise

0 13515

15 0 135

2 03

.

.

.

new skills: wORking with pitch

9. Apitchof2:5meansaslopeof 25

.

m

h

h

h

=

=

× =

≈

riserun

m

25 4 8

4 8 25

1 92

.

.

.

Thelean-toreaches1.92mupthebuilding.

10. m

m

m

m

m

m

=

=÷

=

=

= ÷÷

=

riserun

3 65 4 2

3 62 7

3627

36 936 9

.( . )

.

.

443

Thepitchoftheroofis4:3.

11. m

x

x

x

x

=

=÷

=

× =

=

riserun

ft

25 7 2

25 3 5

3 5 25

1 4

( )

.

.

.

Theriseoftheroofis1.4ft.

pRactice YOuR new skills, p. 32

1. a) tan

tan

tan

.

tan

θ

θ

=

=

= ( )≈ °

=

−

oppadj

op

A

A

A

3529

3529

50 4

1

ppadj

tan

tan

.

B

B

B

=

= ( )≈ °

−

2935

2935

39 6

1

alteRnative sOlutiOn

90

90 50 4 39 6

2935

° − =

° − ° = °

=

=

A B

m

m

. .

riserun

or approx.. 0.83

b) tan

tan

tan

.

tan

θ

θ

=

=

= ( )≈ °

=

−

oppadj

A

A

A

68102

68102

33 7

1

ooppadj

tan

tan

.

B

B

B

=

= ( )≈ °

−

10268

10268

56 3

1



90

90 33 7 56 3

10268

32

° − =

° − ° = °

=

=

=

A B

m

m

m

. .

riserun

or 1..5c) tan

tan

tan

.

tan

θ

θ

=

=

= ( )≈ °

=

−

oppadj

op

A

A

A

2619

2619

53 8

1

ppadj

tan

tan

.

B

B

B

=

= ( )≈ °

−

1926

1926

36 2

1


90

90 53 8 36 2

1929

° − =

° − ° = °

=

=

A B

m

m

. .

riserun

or approx.. 0.73

d) tan

tan

tan

tan

θ

θ

=

=

= ( )= °

=

−

oppadj

oppa

A

A

A

2216

2216

54

1

ddj

tan

tan

B

B

B

=

= ( )= °

−

1622

1622

36

1

alteRnative sOlutiOn:

90

90 54 36

1622

811

° − =

° − ° = °

=

=

=

A B

m

m

m

riserun

or approxx. 0.73

2. a)

2 2 2

2 2 2

2 2

1 2 7 2

1 2 7 2

1 44 51 84

7

= +

= +

= +

= +

≈

a b

. .

. .

. .

.33m

b) tan

tan

tan ..

tan ..

θ

θ

θ

θ

=

=

=

= −

oppadj

riserun

1 27 2

1 27 2

1 (( )≈ °θ 9 5.

3. a) tan

tan

tan

.

tan

θ

θ

θ

1

1 1

1

1

2

35

0 7

=

=

° =

≈

=

riserun

riser

m

m

m

uun

tan

tan

θ =

° =

=

m

m

m

2

2

2

45

1



percent grade

1

1 1

100

100

= ×

= ×m

ppercent grade

percent grade

percen

1

1

0 7 100

70

= ×

=

.

%

tt grade riserun

percent grade

perce

2

2 2

100

100

= ×

= ×m

nnt grade

percent grade

2

2

1 100

100

= ×

= %

4. a) m

m

m

=

=

=

riserun

or

0 0152 5

3500

0 006

..

.

b) m

x

x

x

=

=

× =

=

riserun

m

3500 12

12 3500

0 072.

Thepipewillhaveadropof0.072mor7.2cm.

c) tan

tan

tan .

tan ( . )

.

θ

θ

θ

θ

θ

=

=

=

=

≈

−

riserun

m

0 006

0 006

0 3

1

44°

5. a) Calculatethechangeinaltitude.

1257–982=275m

2 2 2

2 2 2

2 2 2

2 2

3000 275

3000 275

3000 275

2

= +

= +

− =

− =

a b

a

a

a

9987 m ≈ a


percent grade

= ×

= ×

100

2753000

1100

9 2percent grade = . %

6. a) Theroofhasariseof2ftandarunof5ft.

Thismeansapitchof2:5.

b) m

m

m

=

=

=

riserun

25

0 4.

c) percent grade riserun

percent grade

pe

= ×

= ×

100

25

100

rrcent grade = 40%

7. Calculatetheslopeofeachroof.

m

m

m

m

m

m

1

1

1

2

2

2

4 212

0 35

7 820

0

=

=

=

=

=

=

riserun

riserun

.

.

.

..39

Theroofofthesecondhouseissteeper.

8. Calculatethechangeinelevation.

1132–1070=62m

percent grade riserun

percent grade

= ×

= ×

100

621300

1000

4 8percent grade ≈ . %


1.3 Rate of change

new skills: using cOORdinates tO calculate slOpe

Forline2,usepoints(4,2)and(4,9).

my yx x

m

m

m

22 1

2 1

2

2

2

9 24 4

70

= −−

= −−

=

= undefined

3. a) Iftheslopeis65 ,thismeansthatforarise

of6units,thereisarunof5units.Starting

atA,goup6unitsthentotheright5units.

JoinAtoB.

b) Iftheslopeis 32 ,theriseis–3andtherun

is2.StartatAandgodown3units,thento

theright2units.JoinAtoC.

0 5 100

5

10

A

y

x


1. Forline1,usepoints(5,3)and(10,6).

my yx x

m

m

12 1

2 1

1

1

6 310 5

35

= −−

= −−

= or 0.6

Forline2,usepoints(2,3)and(6,9).

my yx x

m

m

m

22 1

2 1

2

2

2

9 36 2

64

32

= −−

= −−

=

= or 1.5

Line2hasagreaterslope,soitissteeper.

2. Forline1,usepoints(5,6)and(9,6).

my yx x

m

m

m

12 1

2 1

1

1

1

6 69 5

04

0

= −−

= −−

=

=


new skills: wORking with slOpe as a Rate OF change

0 150

600.00

p

Mon

ey o

wed

($)

Weeks

w

Amount Owed by Jenita, per Week

Calculatetheslopeusingtwoofyourdata

points,forexample(600,0)and(5,400).

my yx x

m

m

m

= −−

= −−

= −

= −

2 1

2 1

400 6005 0

2005

40

Theslopeofthelineis–40.Theslope

representstheamountofmoneyJenita

paysperweek.

6. a) CalculateGeorge’searningsbymultiplying

bytherateofcommission,12%(expressed

asadecimal,0.12).

$500.00×0.12=$60.00

$1000.00×0.12=$120.00

$2000×0.12=$240.00

b) George’searningsare0.12timeshissales.

Leterepresenthisearningsandsrepresent

totalsales.

e=0.12s

4. a) Dividethedistancewalkedbythetimein

hours.

3km÷0.5h=6km/h

Reggie’saveragewalkingrateis6km/h.

b) Theindependentvariableistime.Ifyou

weretographthedata,timewouldbeon

thex-axis.

c) Dividethetotaldistancebyhiswalking

rate.

36km÷6km/h=6h

ItwouldtakeReggie6hourstowalk

themarathon.

5. a) Letpbetheamountowedandwbethe

numberofweeksshehaspaid.

p=600–40w

ThisequationshowsthatJenitastarted

offowing$600.00,butpaysback

$40.00perweek.

b) Constructagraphthathastimeinweeks

onthex-axis,andmoneyowedonthe

y-axis.Calculate2or3datapointsby

determininghowmuchmoneyJenitaowes

afterdifferentnumbersofweeks.

At0weeks,Jenitaowed$600.00,soone

pointwillbe(0,600).

After3weeks,Jenitawillhavepaidback

$120.00andwillstillowe$480.00.The

datapointforthiswillbe(3,480).

After5weeks,Jenitawillowe$400.00,so

thedatapointwillbe(5,400).


Theindependentvariableisthetotalsales.

Itwillbeonthex-axisofthegraph.

c)

050

0.00

1000

.00

1500

.00

2000

.00

2500

.000

30.00

60.00

90.00

120.00

150.00

180.00

210.00

240.00

270.00

300.00

e

Earn

ings

($)

Sales ($)

s

George’s Earnings Based on Sales

my yx x

m

m

m

= −−

= −−

=

=

2 1

2 1

120 601000 500

60500

0 12.

Theslopeis0.12or325 .Itrepresentshis

rateofcommission.

d) e s

s

s

s

=

=

=

=

0 12

210 00 0 12

210 000 12

1750 00

.

$ . .

$ ..

$ .

IfGeorgeearnedacommissionof$210.00,

hesold$1750.00worthoffurniture.

7. a) Dividetheamountofwaterinthepoolby

theamountoftime.

2250gal÷45min=50gal/min


2250gal÷0.75h=3000gal/h

Thepoolwasbeingfilledatarateof50gal/

minor3000gal/h.

b) Dividethetotalcapacitybytherate

offilling.

13500÷3000=4.5h


13500÷50=270min

Itwilltake4.5hor270mintofillthepool.

8. a) Dividethetotaldistancetravelledbythe

totaltime.

380km÷5h=76km/h

b) Calculatethedistancehetravelledbetween

theendofthe5thhourandtheendof

the9thhour.

764–380=384km

Dividebythetimetravelledatthe

increasedspeed(4hours).

384km÷4h=96km/h

9. a) Dividethetotalnumberofpages

bytherate.

321pages÷0.6pages/min=535min

Convertthetimeinminutestohours

andminutes.

535min÷60min/h≈8.92h

0.92h×60min/h≈55min


ItwilltakeMaryam535minuteor8hours

55minutestoreadthebook.

b) Multiplytheratebythenumberofminutes

spentreading.

2h=120min

0.6pages/min×120min=72pages

After2hours,shewillhaveread72pages.

pRactise YOuR new skills, p. 50

1. a) my yx x

m

m

m

= −−

= −−

=

=

2 1

2 1

204 04 0

2044

51

b) my yx x

m

m

m

m

m

= −−

= −−

=

=

= ÷÷

2 1

2 1

4 5 316 12

1 54

1540

15 540 5

.

.

== 38

0 375or .

c) my yx x

m

m

= −−

= −−

= −

2 1

2 1

6 325 150

3125

or –0.024

2.

0 1 2 3 4 50

1

2

3

4

5

y

x

3. a) Answerswillvary,dependingonthe

numberofhourschosen.Answersfortimes

of1hto5hareprovided.

Hours Henrik’s earnings Javier’s earnings

1 $12.20 $11.10

2 $24.40 $22.20

3 $36.60 $33.30

4 $48.80 $44.40

5 $61.00 $55.50

b)

0 1 2 3 4 5 6 7 8 9 100

10.0020.0030.0040.0050.0060.0070.0080.0090.00

100.00110.00120.00130.00

e

Earn

ings

($)

Hours worked

h

Earnings per Hours Worked

Henrik

Javier

c) Calculatehowmuchmoneyeachwillmake

after8hours.


Henrik=$12.20/h×8h=$97.60

Javier=$11.10×8h=$88.80

Henrikwillearnmoremoneyafteran

8-hourshift.

d) Therateofearningistheslopeoftheline

onthegraph.

Henrik:

my yx x

m

m

m

= −−

= −−

=

=

2 1

2 1

61 05 0

615

12 20.

Henrikearns$12.20/h.

Javier:

my yx x

m

m

m

= −−

= −−

=

=

2 1

2 1

55 50 05 0

55 505

11 10

.

.

.

Javierearns$11.10/h.

4. WriteanequationshowingJanice’searnings

comparedtosales.

earnings=0.04×sales

Substituteinherearningsandsolveforx,

hersales.earnings sales= ×

=

=

0 04

7560 00 0 04

7560 000 04

.

$ . .

$ ..

x

x

$$ .189000 00 = x

Thesellingpricewas$189000.00.

5. a) Considerthedataasifitwereplottedona

graphofTime(inhours;onthex-axis)vs.

Distancetravelled(inkilometres;onthe

y-axis).Thedatapointswouldbe:(2.25,40)

and(4,75).Solvefortheslope,which

representstherateoftravel.

my yx x

m

m

m

= −−

= −−

=

=

2 1

2 1

75 404 2 25

351 75

20

.

.

Rita’saveragerateoftravelwas20km/h.

b) CalculatehowfarRitawilltravelin2hours

(fromthe4-hourpointtothe6-hourpoint).

2h×20km/h=40km

Addthistothedistanceshehadalready

travelledinthefirst4hours.

75km+40km=115km

After6hours,Ritawillhave

travelled115km.

6. a) Dividetheirtotaldistancesbytheir

walkingtimes.

Sheila:

28.8km÷6h=4.8km/h

Brandy:

32.2km÷7h=4.6km/h

Sheilaisthefasterwalker.


b)

050

0.00

1000

.00

1500

.00

2000

.00

2500

.000

30.00

60.00

90.00

120.00

150.00

180.00

210.00

240.00

270.00

300.00

e

Earn

ings

($)

Sales ($)

s

George’s Earnings Based on Sales

Theslopesofthelinesonthegraphare

equaltotheirwalkingrates(Sheila:4.8

km/h;Brandy:4.6km/h).

c) Dividethetotaldistancebythewalkingrate.

Sheila:

10km÷4.8km/h=2.1h

Brandy:

10km÷4.6km/h=2.2h

ItwilltakeSheila2.1handBrandy2.2hto

walk10km.

7. a) Time elapsed (minutes)

Volume of water remaining (L)

0 480000

30 435000

60 390000

90 345000

120 300000

150 255000

b)

0 30 60 90 120 150 1800

100 000

200000

300 000

400 000

500 000

v

Volu

me

of w

ater

rem

aini

ng (L

)

Time (min)

t

Rate of Water Flow

c) Calculatetheslope.

my yx x

m

m

m

= −−

= −−

= −

= −

2 1

2 1

435000 480 00030 0

4500030

15000

Theslopeistherateatwhichthewater

volumedecreases,–1500L/min.

d) Dividethetotalvolumebytherate.

480000L÷1500L/min=320min

Itwilltake320minforthetanktoempty.


chapteR test, p. 55

1. m

m

m

m

m

=

=

=

= ÷÷

=

riserun

or 0.4

1 84 5

1845

18 945 9

25

.

.

2. m

x

x

x

=

=

× =

≈

riserun

cm

23 250

250 23

167

Therailingwillbeabout167cmhigher.

3. tan

tan.

. tan

.

θ =

° =

° =

≈

riserun

m

202 6

2 6 20

0 95

x

x

x

Theriseofthewaterslideisabout0.95m.

4. a) m

x

x

x

x

=

=

=

=

=

riserun

0 375 3 75

0 375 3 75

3 750 375

10

. .

. .

..

fft

Theshortesthorizontaldistancefrom

thebaseofthepatiotothebaseofthe

rampis10ft.

b)

2 2 2

2 2 2

2 2

10 3 75

10 3 75

100 14 0625

1

= +

= +

= +

= +

≈

a b

.

.

.

00 7. ft

Thesurfaceoftherampmustbeatleast

10.7ftlong.

c) tan

tan

tan .

tan ( . )

.

θ

θ

θ

θ

θ

=

=

=

=

≈

−

riserun

m

0 375

0 375

20

1

66°

Theangleofelevationoftheramp

willbe20.6°.

5. a) Apitchof4:12meansaslopeof 412

.This

canbesimplified.

m

m

m

=

= ÷÷

=

412

4 412 4

13

or about 0.33

b) tan

tan

tan

tan

.

θ

θ

θ

θ

θ

=

=

=

= ( )≈ °

−

riserun

m

13

13

18 4

1

Theangleofelevationoftheroofis

about18.4°.


6. percent grade riserun

= ×

= ×

=

×

100

615

100

6100 15

15 61

x

x

000

0 9

=

=

x

x. km

Theverticalchangeis0.9kmor900m.

7. a)

2 2 2

2 2 2

2 2

6 8 125 9

6 8 125 9

126 1

= +

= +

= +

≈

a b

. .

. .

. m

Therailroadtrackwillbeabout

126.1mlong.


percent grade

= ×

=

100

6 8125 9

..

××

≈

100

5 4percent grade . %

8. Apitchof2:5meansaslopeof25 .

m

x

x

x

=

=

× =

=

riserun

m

25 3 5

3 5 25

1 4

.

.

.

Theroofwillrise1.4m.

9. a)

0 1 2 3 4 5 6 7 80

10.00

20.00

30.00

40.00

50.00

60.00

70.00

80.00

90.00

100.00

110.00

120.00

e

Earn

ings

($)

Hours of work

h

Rebecca’s Earnings per Hour

b) Choosetwopointsonthegraph,suchas(1,

15)and(2,30).

my yx x

m

m

m

= −−

= −−

=

=

2 1

2 1

30 152 1

151

15

Theslopeofthegraphis15,which

representsRebecca’searningsperhour.

c) CalculateRebecca’searningsafter8hours.

8h×$15.00/h=$120.00

After8hours,shewillhave

earned$120.00.


Chapter 2Graphical Representations

2.1 Broken Line Graphs

BuiLd Your SkiLLS, p. 60

1. a) Hespentabout$3.00onWednesdayand

nothingonFriday.

b) Hespentslightlylessthan$19.00forlunch

onSunday.Hemayhavegoneoutwith

friendsforlunch.

2. a) ThegraphshowsKatie’sheartrateeach

hourinbeatsperminutefrom8:00am

to9:00pm.

b) Herheartratewaslowestat8:00am,2:00

pm,5:00pmand9:00pm.Itwas68beats

perminute.Katiewasprobablyrelaxedat

thesetimes.

c) Katie’sheartratewashighestat1:00pm.

Shemayhavejustbeeninvolvedinsome

physicalactivity.

3. a) TheWeek1dotisthevalueVincebought

thestockat.Studentscancountthat

therearefiveequalsegmentsbetween0

and3.00.

Calculatethevalueofeachsegment.

3 005

0 60. .=

Thedotisjustbelowthethirdsegment.

0 60 3 1 80. .× =

Thedotisjustbelow$1.80soyoucanestimate

Vinceboughtthestockatavaluenear$1.75.

b) Youcanestimatethatthestockwasworth

$12.15atWeek4.

12.15–1.80=10.40

Hewouldhavemade$10.40perstock.

4.

1 2 3 4 5 6 7 80

5

10

15

20

25

30

35

Canadian Review Newspaper

Num

ber o

f sal

es

Week

n

w

Salesofthemagazineseemtobeslowly

increasing.

5.

0 1 2 3 4 5 6 7 8 9 10 11 120

1

2

3

4

5

6

Jessica’s Weight ChartW

eigh

t (kg

)

Weeks since birth

w

t

Jessicalostweightduringthefirstweek,

andstayedthesamebetweenthe8thand9th,

butotherthanthat,therewasaweightgain

eachweek.

6. a)

1 2 3 4 5 6 7 8 9 10 11 12 13 140

100

200

300

400

500

600

Record of Driving Distances

Dist

ance

(km

)

Day

d

t

b) Answerswillvarybutshouldinclude

thingssuchas:

Thegraphindicatesthedistancehedrove

eachdayandthatthesevaryconsiderably.

Theanomaliesofzeroareprobably

hisdaysoff.

c) Thegraphisanokayrepresentation.

However,sinceitisnotlookingatatrend

overtime,butisdiscretedata,itdoesn’t

givemuchusefulinformation.

7. Answerswillvaryslightlybutshouldsuggest:

Thérèsewasdrivinghercarfrom8:30amuntil

10:30.Itwasparkedfrom10:30to12:30at

whichtimesheputgasinthetank.Shethen

droveforanhour(until1:30)andstopped

againforanhouruntil2:30.Between2:30and

4:30,shedroveagain,butnotsofar,andshe

stoppedforthelasthour.droveabit

8. a) Thecostseemstobedecliningslightly.

b) Drawastraightlinebetweenthepointin

JulyandSeptemberandestimatewhereit

wouldmeetforAugust.Answerswillvary

butshouldbeapproximately$325.00.

c) Answerswillvary,butshouldbebetween

$300.00and$325.00.

9. a)

c

Num

ber o

f cel

l pho

nes

sold

Month

m

Number of Cell Phones Sold to Males and Females in the Last Year

Male

Female

0

50

100

150

200

250

300

350

400

450

500

Dec.Nov.Oct.Sep.

Aug.

Jul.

Jun.

MayApr.Mar.Feb.

Jan.

MathWorks11WorkbookSolutions Chapter2 GraphicalRepresentations 24

b) Thenumberofcellphonesbought

fluctuatedgreatlyfrommonthtomonth,

butoveralltheretendedtobeanincrease.

c) No,thereisnorealrelationshipassome

monthsfemalesboughtmoreandother

monthsitwasmales.

d) Itisnotveryuseful.Astackedbargraphor

evenabargraphwouldbetterindicatethe

numberssold.

10.a)

s

Stoc

k pr

ice

($)

Month

m

Stock Price

Stock A

Stock B

0

0.50

1.00

1.50

2.00

2.50

3.00

Jun.

MayApr.Mar.Feb.

Jan.

Dec.Nov.Oct.Sep.

Aug.

Jul.

b) StockAstartedoffhighandmadeasudden

dropinAugustandSeptember,increasing

againabitoverthenextfewmonths,

thendroppingmoreinMarch.StockB

pricesfluctuatedslightlyovertheyear

butremainedrelativelyevencompared

toStockA.

c) Answerswillvarybutmayinclude:

• SheshouldsellstockAbecauseithas

lostalotanddoesn’tlooklikeitisgoing

toincrease.

• SheshouldsellstockBbecauseithasnot

madeheranymoneyandlikelywon’t.

• SheshouldsellstockBbecauseshe

needstomakeupthemoneysheloston

A,soifshebuysmoreofthemshecan

maybedothat.

11.a) Thegraphrepresentstheamountofmoney

Marciaspentweeklyongroceries.

b) Itappearsthatherexpendituresincreased

greatlyoverthe6weeks.

c) Theverticalaxisstartsat$19.50

ratherthan$0.

d)

1 2 3 4 5 60

5.00

10.00

15.00

20.00

25.00

Marcia’s Weekly Grocery Expenditures

Week

Amou

nt s

pent

($)

s

w

12.a) Thefirstgraphmakesthedecreaselook

morerapidbecausethesamewidth(years)

representstwiceasmanyyears.

b) Thesecondgraphisbetterbecausemore

yearsarerepresented.

c) Estimatingthehalfwaypointbetween

1911and1921shouldgetstudentsa

numbernear61%.

d) Estimatingthe50%pointbetweenstudents

shouldgetapproximately1952.


e) Thegraphindicatesadeclinesostudents

shouldestimateabout17%ofthe

populationwillberuralin2011.

PractiSe Your New SkiLLS, p. 76

1. a) Shewalkedapproximately56000stepsthe

firstweekand81000thelastweek.

b) Shehadthefeweststepsduringweeks5

and9,eachwithabout54000steps.

c) Withtheexceptionofthetwolowweeks,

weeks5and9,therewasageneral

trendofgradualincreaseinnumberof

stepsperweek.

2.

1999

2000

2001

2002

2003

2004

2005

2006

2007

2008

2009

0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

Area of Forest Burned in Canada, 1999–2009

Num

ber o

f hec

tare

s bu

rned

(in

mill

ions

)

Year

n

y

Thereisnogeneraltrendofincreaseor

decreasealthoughifoneyearwashigh,

thenextyeartendedtobelowerforafew

years.Thereseemedtobealotoffiresfrom

2002to2006.

3. Answerswillvarybutshouldincludethe

followingideas:Atthebeginningoftheyear

therewasasteadyriseinprices,butfromJune

toAugusttheydropped.Pricesremainedabout

thesamefromAugusttoSeptember,thenrose

againuntiltheendoftheyear.Thehighest

priceswereinDecember.

4. a) BrandAhasafluctuatingmarketwithno

generaltrend.BrandBhashadasteady

marketwithfewfluctuations.BrandChas

hadsomeofthehighestsales,buthashada

fairlyconsistentdropinsales.

b) Answerswillvarybutshouldinclude

thingssuchas:

• KeepbrandAandBbecauseBhasa

consistentrecordofsales,andAbecause,

althoughitisnotconsistent,itisn’tona

downwardslidelikeC.

• KeepBandCbecauseBhasaconsistent

recordofsalesandCbecause,although

ithashadadropinsalesitseemstobe

steadyingoffwhereasAisinconsistent.

5. a)

0

50

100

150

200

Number of Books in the Library

Num

ber o

f boo

ks

MonthJan. Feb. Mar. Apr. May Jun. Jul. Aug.

Historical fictionLove stories

Science fiction


b) Thereisageneral,slowtrendofincrease

innumberoftitlesavailableindicatingthat

therearemorebooksbeingadded.The

declinewouldlikelymeanthatthelibrarian

gotridofsometitles.


2.2 Bar Graphs


1.

0

10

20

30

40

50

60

OtherOut-sourced to

development provider

Out-sourced to

service provider

In-house IT with help from provider

In-house IT dept.

How software was obtained and installed

Software Installations

Perc

enta

ge o

f tot

al s

oftw

are

inst

alla

tions

2. a) Heightisrecordedhorizontallyratherthan

vertically.Also,hisagegoesfromhighestat

thebottomtolowestatthetop.

b)

0

50

100

150

200

Jamie’s Height

Heig

ht (c

m)

Age (years) 10 11 12 13 14 15 16 17 18

3. Thebarsareofdifferentwidthsrepresenting

differentperiodsoftime.Thebarsarethree

dimensionalgivingafalseperspective.

4. a) Studentsmaychoosethefirstgraph

becauseitisthemorehonestrepresentation

becauseitstartsatzero.

b) Approximately22.5billiontonnes.

c) Approximately27billiontonnes.

d) Themisleadinggraphmaybechosen

becausesomeonemightwantto

deliberatelymisleadtheirviewers.Alsothe

secondgraphdoesallowpeopletoread

thedatamorepreciselywhichinsome

situationsmaybeimportant.

5. Will’s Test Results

0

20

40

60

80

100

654321

Perc

enta

ge

Test number

p

t

Answerswillvary,butshouldinclude:

• Thebrokenlinegraphisbetterbecauseit

iseasiertoreadtrendsinhistestresults.

6. a) Thenumberofticketssoldperdaytends

todecrease.Thisisprobablybecausefans

wantedtobesurethattheygotticketsas


soonaspossible.Thepeoplewhobought

laterweremaybenotaskeenontheband.

b) Thebargraphisthebetterrepresentation

becauseitmoreclearlyindicatesthe

numberofticketssold.Thelinegraph,

ontheotherhand,makesitlooklikea

continuum.

7. a)

0

50.00

100.00

150.00

200.00All households with children

Lowest income households

Eatin

g

Educa

tion

Recrea

tion

and c

ulture

Commun

icatio

n

Transp

ortati

onHea

lth

Heat a

nd

electr

icity

Clothin

g

Food a

nd

drink

Average Weekly Household Expenses, Canada

Aver

age

wee

kly

hous

ehol

d ex

pens

e ($

)

Item

b) Answerswillvarybutmayinclude:The

lowerincomefamiliesspentlessonall

items,butheatandelectricitytheywere

quitecloseandalsoforcommunications.

Healthlooksfairlyclose,butitisactually

half.Thepoorerfamiliesspentmuchless

ontransportation,foodandeatingout,as

wellasrecreationandculture.Theyspent

only1/5oftheamountofallfamilieson

education.

8. a) ThemostvisaswereappliedforinJuneand

theleastinJanuary.Approximately675or

680visaswereappliedforinJulyand125

inJanuary.

b) Thegreatestdiscrepancybetween

applicationsandapprovalswasinJuly.

VisaapplicationsinJuly:535

VisaapprovalsinJuly:310

535–310=225

Theapproximatedifferencebetween

approvalsandapplicationsinJulywas225.

9. a) Theyaregoingdownbecausethey

representnegativenumbers.The

temperaturewasbelowzero.

b) ThereisnogreenbaronSaturdaybecause

itrepresentsthehightemperatureofthe

daywhichwaszero.

c) OnWednesdaytherewasnochangein

temperature.Itstayedat-6°C.

d) Itseemedtowarmupovertheweek.

10.a)

0

100

200

300Girls

Boys

GolfBaseballBasketballHockeyFootball

Favourite Sport to Watch on TV

Num

ber o

f stu

dent

s

Sport


0

200

400

600

GirlsBoys

GolfBaseballBasketballHockeyFootball

Favourite Sport to Watch on TVN

umbe

r of s

tude

nts

Sport

b) Addallthevaluestogethertofindthetotal.

135+121+243+265+101+75+79+

15+18+2=1054

1054tookthesurvey.

c) Hockeyisthemostpopularsporttowatch.

d) Thedoublebargraphindicatesthe

differencesinpreferencesbetweenthe

genders.Thestackedwouldbemoreuseful

ifyouwereinterestedintotals.


1.

0 50 100 150 200

Number of Users of Fitness Equipment per day

Equi

pmen

t

Number of users

Station

ary bi

ke

Tread

mill

Ellipt

ical cr

oss-tra

iner

Staircl

imbe

r

2. a) Themostpopularmoviesare

number4and5.

b) Theleastpopularisnumber8.

3. Moreboyshavebeentakinghomeeconomics

classesbutthenumberofgirlshasstayed

aboutthesame.

4. a) Since2008,thenumberofhousingstarts

hasdroppedconsiderably,butthereisno

particularpredictablepattern.

b)

0

200

400

600

800

1000Multiple

Single

20102009200820072006

Housing Construction Projects Started in Abbotsford, BC, per Year

Num

ber o

f pro

ject

s

Year

Answerswillvarybutshouldbealongthe

linesofoneoftheseanswers:

Thelinegraphbecauseitshowsthe

risesandfalls,andeachcanbelookedat

individually.

Thebargraphbecauseitismorevisualand

showsthecomparisonssidebysideaswell

asacrosstime.


2.3 Histograms


1. a) Approximately125householdsspendless

than$1000.00

b) Between$1000.00and$2000.00is

themostcommonamountspenton

renovations.

c) Wecannotbesurebecausethegraphsays

5000.00+butitwasover$5000.00.

2. a) 3800peoplespendbetween10and15

hoursontheinterneteachweek.

b) Addallhourslessthan15hours.

3800+1700+1200=6700

6700peoplespendlessthan15hoursa

weekontheinternet.

c) 2600peoplespendmorethan30hourson

theinterneteachweek.

d) Youcanaddthetotalfromb)totherestof

thevalues.

6700+4900+4800+3700+

2600=22700

Approximately22700people

weresurveyed.

3.

0

5

10

15

20

25

65+55453525

Ages of Employees

Num

ber o

f peo

ple

Age

4. a)

0

20

40

60

80

100

70+6050403020

Ages of Audience Members at Theatre Presentation

Freq

uenc

y

Ages


b) Addthevaluesofpeoplebelow30.

68+35=103

Thereare103peopleundertheageof30

whoattendedthepresentation.

c) Wecannottellexactlyexceptthatheorshe

wasundertheageof20.


1. a) Thereare8studentswhoreceivedamark

between70%and80%.

b) Addthetotalsofstudentsbelow60%.

2+1+6=9

Thereare9studentswhoscored

below60%.

c) Wecannotbesurebutitwasover90%and

lessthanorequalto100%.

2. a) Thereare2employeeswhoearnover

$100000.00

b) Addthevaluesofemployeeswhoearn

between$30000.00and$50000.00.

8+12=20

Thereare20employeeswhoearnbetween

$30000.00and$50000.00.

3. a)

0

5

10

15

500+400300200100

Fixed Rate Real Estate Housing Sales

Num

ber o

f hou

ses

Selling price (in thousands) ($)

b) Wecannotbesurebutitwas

below$100000.

c) Addthetotalsofhousesthatsoldfor

between$100000.00and$300000.00.

2+7=9

Thereare9housesthatsoldforbetween

$100000.00and$300000.00.


4. a)

0

10

20

30

40

Oct. 20

Nov. 20

Sep. 2

0

Aug. 2

0Ju

l. 20

Jun.

20

Mar. 21

Hurricane Occurrences, by Time of YearPe

rcen

tage

of h

urric

ane

occu

rrenc

es

Dates

b) Hurricanesareunlikelytooccurduring

themonthsfromlateNovemberofoneyear

untilmidMarchofthenext.

c) Ahurricaneismostlikelytooccurfrom

lateAugustuntillateSeptember.


2.4 circle Graphs


1. a) Fromreadingthegraphstudentscansee

that14%ofstudentschosegreen.

Calculate14%ofthetotalnumber

ofstudents:

171 0 14 23 94× =. .

Thereare24studentswhochosegreenas

theirfavouritecolour.

b) Studentscanreadthegraphandseethat

25%ofstudentschoseblue.

Calculate25%ofthetotalnumber

ofstudents:

171 0 25 42 75× =. .

Themostpopularcolourisblue.43

studentschoseitastheirfavouritecolour.

2. a) Frankspendsequalamountsonsavings

andentertainmentat10%each.

b) UsealgebratosolveforFrank’s

totalsavings.

250 0 10

2500 10

2500

=

=

=

.

.

x

x

x

Heearns$2500permonth.

c)

0

10

20

30

40

50

Saving

s

Transp

ortati

on

Enter

tainm

ent

Clothin

gFoo

d

Housin

g

Expenditures

Perc

enta

ge

Item

Answersmayvary,butthecirclegraphis

thebetterdepictionherebecauseitiseasier

toseewhatproportioneachexpenditure

accountsforofthewhole.

3. a) Addpercentagestogether.

15+17+2=34

Thefaucet,showerandbathuse34%of

thewater.

b) Addpercentagestogether.

2+21=23

Thedishwasherandclotheswasheruse

23%ofthewater.


c) Addthetwohighestpercentagestogether.

27+21=48

Thetoiletandtheclotheswasheraddto

abouthalfthewateruseat48%.

4. Findthetotalnumberofstudentssurveyed.

39+44+21=104

Findoutthepercentageeachpetrepresentsof

thetotal.

Dog:

39104

0 375

0 375 37 5

=

=

.

. . %

Cat:

44104

0 423

0 423 42 3

=

=

.

. . %

Rodent:

21104

0 202

0 202 20 2

=

=

.

. . %

Calculatehowmanydegreesofthecircleeach

petrepresents:

Dog:0 375 360 135. × ° = °

Cat:0 423 360 152. × ° ≈ °

Rodent:0 202 360 73. × ° ≈ °

Usethedatacalculatedtodrawthe

followinggraph:

Rodent

Cat

Dog

38%

42%

20%

Type of Pet

5. Findthetotalnumberofstudentssurveyed.

39+44+21+24=128

Findoutthepercentageeachpetrepresentsof

thetotal.

Dog:

39128

0 305

0 305 30 5

=

=

.

. . %

Cat:

44128

0 344

0 344 34 4

=

=

.

. . %

Rodent:

21128

0 164

0 164 16 4

=

=

.

. . %

Nopet:

24128

0 188

0 188 18 8

=

=

.

. . %



petrepresents:

Dog:0 305 360 110. × ° ≈ °

Cat:0 344 360 124. × ° ≈ °

Rodent:0 164 360 59. × ° ≈ °

Nopet:0 188 360 68. × ° ≈ °


followinggraph:

30.4%

34.4%

16.4%

Type of Pet

None

Rodent

Cat

Dog18.8%

6. Findthetotalmoneyfundraised.

$750+$325+$375+$150+$100=$1700

Findoutthepercentageeachfundraising

methodrepresentsofthetotal.

Parentaldonations:

7501700

0 441

0 441 44 1

=

=

.

. . %

Chocolatebarsales:

3251700

0 191

0 191 19 1

=

=

.

. . %

Hotlunches:

3751700

0 221

0 221 22 1

=

=

.

. . %

Concertproceeds:

1501700

0 088

0 088 8 8

=

=

.

. . %

Classroomdonations:

1001700

0 059

0 059 5 9

=

=

.

. . %


fundraisingeffortsrepresents.

Parentaldonations:0 441 360 159. × ° ≈ °

Chocolatebarsales:0 191 360 69. × ° ≈ °

Hotlunches:0 221 360 80. × ° ≈ °

Concertproceeds:0 088 360 32. × ° ≈ °

Classroomdonations:0 059 360 21. × ° ≈ °


followinggraph.

Classroom donations

Concert proceeds

Hot lunches

Chocolate bar sales

Parental donations

School Fundraising for International Disaster Relief

44%

19%

22%

9%

6%

7. a) Findthetotalcontainerssold.

68+45+127+93+76+12=421


Findoutthepercentageeachbrandof

hairsprayrepresentsofthetotal.

A:

68421

0 162

0 162 16 2

=

=

.

. . %

B:

45421

0 107

0 107 10 7

=

=

.

. . %

C:

127421

0 302

0 302 30 2

=

=

.

. . %

D:

93421

0 221

0 221 22 1

=

=

.

. . %

E:

76421

0 181

0 181 18 1

=

=

.

. . %

F:

12421

0 029

0 029 2 9

=

=

.

. . %

Calculatehowmanydegreesofthecircle

eachhairspraybrandrepresents.

A:0 162 360 58. × ° ≈ °

B:0 107 360 39. × ° ≈ °

C:0 302 360 109. × ° ≈ °

D:0 221 360 80. × ° ≈ °

E:0 181 360 65. × ° ≈ °

F:0 029 360 10. × ° ≈ °


followinggraph.

Brand F

Brand E

Brand D

Brand C

Brand B

Brand A

Hairspray Sales in One Month

3%

16%

11%30%

22%18%

b) Theinformationcouldhavebeendisplayed

onabargraphorabrokenlinegraph.

0

50

100

150

Brand FBrand EBrand DBrand CBrand BBrand A


Brand

Num

ber o

f sal

es

0

50

100

150

Brand F

Brand E

Brand D

Brand C

Brand B

Brand A


Brand

Num

ber o

f sal

es



1. a) Addthevaluesinthegraphtoworkout

thetotals.

52+14+6+65=137

b) Dividetheamountofpeoplewholistento

rockbythetotaltogetthepercent.

52137

0 380

0 380 38

=

=

.

. %

Thereare38%ofpeoplewhoprefer

rockmusic.

c) Calculatethepercentageofstudentswho

preferR&B.

14137

0 102= .

Calculatehowmanydegreesofacirclethat

percentagerepresents.

0 102 360 37. × ° ≈ °

R&Brepresents37°ofthecircle.

2. Calculatethenumberofdegreeseach


Travelpackage:0 60 360 216. × ° = °

Tourismoffice:0 15 360 54. × ° = °

Friendrecommendation:0 10 360 36. × ° = °

Subtracttheotherpercentagestoworkout

whatpercentwerewalk-ins.

100–60–15–10=15%

15%=0.15

Walk-ins:0 15 360 54. × ° = °


followinggraph.

Walk-in

Recommended by friendRecomended by tourism office

Travel package

60%

15%

10%

20%

Deciding Factors for Choosing Greg’sTour

3. Findthetotalnumberofbooks.

21+5+32+10+2+3=73

Sciencefiction:

2173

0 288

0 288 28 8

=

=

.

. . %

Biography:

573

0 068

0 068 6 8

=

=

.

. . %

Historicalfiction:

3273

0 438

0 438 43 8

=

=

.

. . %

Poetry:

1073

0 137

0 137 13 7

=

=

.

. . %

Autobiography:

273

0 027

0 027 2 7

=

=

.

. . %

Self-help:

373

0 041

0 041 4 1

=

=

.

. . %


Calculatethenumberofdegreeseach


Sciencefiction:0 288 360 104. × ° ≈ °

Biography:0 068 360 24. × ° ≈ °

Historicalfiction:0 438 360 158. × ° ≈ °

Poetry:0 10 360 36. × ° ≈ °

Autobiography:0 02 360 7. × ° ≈ °

Self-help:0 041 360 15. × ° ≈ °


followinggraph.

cHaPter teSt, p. 119

1. a)

0

5

10

15

20

25

2010

2009

2008

2007

2006

2005

2004

2003

2002

2001

Company Profits, 2001–2010

Year

Prof

it (in

thou

sand

s of

dol

lars

)

p

y

b) Theprofitwashighestin2003at$25000.

c) 2003wasapeakyearforthecompany

wheretheirprofitmarginwentwayup.It

thendroppeddowntowhatithadbeen

theprevioustwoyearsandmadeasteady

increaseover3yearsandseemstohave

leveledoffforthelastthreeyears.

2. a)

0

10

20

30

40

Number sold

Number bought

Brand DBrand CBrand BBrand A

Type

Num

ber o

f mot

ors

Number of Motors Bought and Sold

b) BrandChasthesmallestdifference.Use

subtractiontocalculatethedifference.

15–14=1

BrandChasthesmallestdifference,

whichis1.

c) BrandBhasthelargestdifference.Use

subtractiontocalculatethedifference.

32–22=10

BrandBhasthelargestdifference,

whichis10.

3. a)

0

2000

4000

6000

8000

10 000

Jan. Feb. Mar. Apr. May Jun. Jul. Aug. Sep. Oct. Nov. Dec.

Month

Num

ber o

f vid

eo re

ntal

s

Video Rentals by Month


0

2000

4000

6000

8000

10000

Month

Num

ber o

f vid

eo re

ntal

sVideo Rentals by Month

Jan. Feb

.Mar. Apr.

May Jun.

Jul.

Aug.

Sep.

Oct. Nov. Dec.

n

m

Eitherabrokenlinegraphorabargraph

(verticalorhorizontal)couldbeusedbecauseit

isdiscretedata.

b) Octoberhadthemostrentalsat8978.

c) Rentalstendtobelowerinthesummer

months,butareotherwisestable.


Chapter 3Surface Area, Volume, and Capacity

3.1 Surface Area of Prisms

Build Your SkillS, p. 127

1. a) Itisacircle.

A r

A

A

circle

mm

=

= ×

=

π

π

2

2

2

44

6082 1.

Theareaofthecircleis6082.1mm2.

b) Itisatriangle.

A bh

A

A

triangle

m

=

=

=

12

12

5 7 9 3

26 5 2

( . )( . )

.

Theareaofthetriangleis26.5m2.

c) Itisaparallelogram.

A bh

A

A

parallelogram

cm

=

=

=

( . )( . )

.

5 3 14 8

78 4 2

Theareaoftheparellogramis78.4cm2.

d) Itisatriangle.

A bh

A

A

triangle

in

=

=

=

12

12

9 16

72 2

( )( )

Theareaofthetriangleis72in2.

2. a) Itisarectangle.

A lw

A

A

rectangle

m

=

= ×

=

5 2 8 5

44 2 2

. .

.

Theareais44.2m2.

b) Itisaparallelogram.

A bh

A

A

parallelogram

m

=

= ×

=

5 2 8 5

44 2 2

. .

.

Theareais44.2m2.

c) Itisatriangle.A bh

A

A

parallelogram

cm

=

=

=

( . )( . )

.

5 3 14 8

78 4 2

Theareais22.1m2.

d) Itisatriangle.

A bh

A

A

triangle

in

=

=

=

12

12

9 16

72 2

( )( )

Theareais22.1m2.

e) Itisatriangle.

A bh

A

A

triangle

m

=

=

=

12

12

8 5 5 2

22 1 2

( . )( . )

.

Theareais22.1m2.

f) Areasofrectanglesandparallelograms

arethesameifthebaseandheightarethe

sameasthelengthandwidth.Areasof

trianglesarethesameiftheyhavethesame

heightandwidth.

3. a) Dividetheshapeintotwoseparate

rectangles.

A1

A2

5 in

7 in

6.5 in

8 in

Calculatetheareaofthefirstrectangle.A

A

1

12

5 6 5

32 5

= ×

=

.

. in

Calculatetheareaofthesecondrectangle.

A

A

A

2

2

22

5 7 8

12 8

96

= + ×

= ×

=

( )

in

Addthetwoareastogethertogetthetotal.A A A

A

A

= +

= +

=

1 2

2

32 5 96

128 5

.

. in

Thetotalareais128.5in2.

AlTErNATE SoluTioN

Dividetheshapeintotwoseparate

rectangles.

A1

A2

5 in

7 in

6.5 in

8 in

Calculatetheareaofthefirstrectangle.

A

A

A

1

1

12

8 6 5 5

14 5 5

72 5

= + ×

= ×

=

( . )

.

. in

Calculatetheareaofthesecondrectangle.A

A

2

22

8 7

56

= ×

= in


A

A

= +

= +

=

1 2

2

72 5 56

128 5

.

. in

Thetotalareais128.5in2.

b) Dividetheshapeintoarectangleand

atriangle.

Calculatetheareaofthetriangle.

A

A

A

1

1

12

12

4 2 9 6 7 8

12

4 2 1 8

3 8

= −

=

≈

( . )( . . )

( . )( . )

. m

Calculatetheareaoftherectangle.A

A

2

22

7 8 4 2

32 8

= ×

≈

. .

. m

MathWorks11WorkbookSolutions Chapter3 SurfaceArea,Volume,andCapacity 42


A

A

= +

= +

=

1 2

2

3 8 32 8

36 6

. .

. m

Thetotalareais36.6m2.

4. Youranswermaylooklikethis.

8.8m

3.6m

1.8m4.6m

2.6m

A1

A2

A3

A1

A2

A3 3.6m

2.4m

1.8m4.6m

4.2m

2.6m

6.2m

A1

A2

A3

2.4m

1.8m4.6m

2.6m

6.2m

A1

A2

A3

7.0m

3.6m

1.8m4.6m

2.6m

6.2m

Calculatetheareaofthefirstrectangle.A

A

1

12

4 6 3 6

16 6

= ×

≈

. .

. m

Calculatetheareaofthesecondrectangle.

A

A

2

22

2 6 8 8

22 9

= ×

≈

. .

. mCalculatetheareaofthethirdrectangleA

A

3

32

3 6 1 8

6 5

= ×

≈

. .

. m

Addthethreeareastogetthetotal.A A A A

A

A

= + +

= + +

=

1 2 3

2

16 6 22 9 6 5

46

. . .

m

Theareais46m2.

AlTErNATE SoluTioN 1

A

A

A

A

12

22

32

2

28 5

10 9

6 5

45 9

≈

≈

≈

=

.

.

.

.

m

m

m

m

Theareais45.9m2.Thisisdifferentfromthe

originalsolutionduetorounding.


A

A

A

A

12

22

32

2

28 5

6 2

11 2

45 9

≈

≈

≈

=

.

.

.

.

m

m

m

m

Theareais45.9m2.Thisisdifferentfromthe

originalsolutionduetorounding.


A

A

A

A

12

22

32

2

16 6

18 2

11 2

46 0

≈

≈

≈

=

.

.

.

.

m

m

m

m

Theareais46m2.


5. Prism Shape of base Right or oblique Shape of lateral faces Name of prisma) triangular oblique parallelogram obliquetriangularprism

b) rectangular right rectangular rightrectangularprism

c) pentagonal right rectangular rightpentagonalprism

d) trapezoidal oblique parallelogram obliquetrapezoidalprism

e) octagonal right rectangular rightoctagonalprism

f) rectangular oblique parallelogram obliquerectangularprism

6. a)

15in

5in

5in

5in

b)

14cm

2.6cm

3cm3cm

3cm

7. Ralphiscorrect.Manon’snetismissingoneof

thesidepanels.

8. a)

6cm6cm

6cm

6cm

8cm

8cm

10cm

Theformulaforthesurfaceareaisthetotal

surfaceareaofeachfaceaddedtogether.

Youcangroupsomeofthefacesastheyare

identical.

SA A A A

SA

SA

= × + × + ×

= × + × + ×

=

2 2 2

2 8 6 2 6 10 2 8 10

2

1 2 3

( ) ( ) ( )

(448 2 60 2 80

96 120 160

376 2

) ( ) ( )+ +

= + +

=

SA

SA cm

Thesurfaceareais376cm2.


b)

4in5in 5in

5in

3in3in3in

8in



Youcangroupthetriangularfacesasthey

areidentical.

SA A A A A

SA

SA

= × + + +

= × × ×( ) + × + × + ×

= +

2

2 12

4 3 4 8 3 8 5 8

12

1 2 3 4

332 24 40

108 2

+ +

=SA in

Thesurfaceareais108in2.

c)12cm

12cm8cm

15cm

8cm 8cm

9cm

Youneedtocalculatethelengthofthesides

thataretheslantsofthetrapezoid.Youcan

dothisusingthePythagoreantheorem.

x

x

x

x

x

2 2 2

2 2

1 5 8

1 5 8

2 25 64

66 25

8 1

= +

= +

= +

=

=

.

.

.

.

. cm

Calculatetheareaofthetrapezoidalfaces.

A h a b

A

A

A

trapezoid = +

= × + + +

= ×

=

( )

[( . . ) ]

2

8 1 5 1 5 9 92

8 212

884 2cm



Youcangroupsomeofthefacesastheyare

identical.

SA=2×A1+2×A2+A3+A4

SA=2×(8.1×15)+2(84)+9×15+

(1.5+9+1.5)×15)

SA=2(121.5)+2(84)+9×15+

(1.5+9+1.5)×15

SA=243+168+135+180

SA=726cm2

Thetotalsurfaceareais726cm2.

9. Theformulaforthesurfaceareaisthetotal


Studentsshouldrecognizethatthetwoseparate

surfacesonthetopusethesameamountof

areaasthebottomsurfaceandcanbegrouped.

Alsoanumberoftheotherfaceshavethesame

dimensionsandcanbegrouped.


SA=2×A1+4×A2+4×A3

SA=2×(1.5×1.5)+4×(1×1.5)+

4×(2×0.5)

SA=2×2.25+4×1.5+4×1

SA=4.5+6+4

SA=14.5cm2

Thesurfaceareais14.5m2.

10.Solvetheareaoftheirregularfigureby

dividingitinto3regularshapes.A

A

A

1

1

12

20 5 20 5 30 5 5 5

100 100 100

300

= × + × + ×

= + +

=

( – – )

m

Thetotalsurfaceareaisthesurfaceareaofeach

faceaddedtogether.Studentsshouldrecognize

thatthethreeseparatesurfacesonthetopuse

thesameamountofareaasthebottomsurface

andcanbegrouped.Alsoanumberofthe

otherfaceshavethesamedimensionsandcan

begrouped.

SA=2×A1+2×A2+2×A3+2×A4

SA=2×300+2×(20×15)+2×

[(20−5)×15]+2×(30×15)

SA=2×300+2×300+2×225+2×450

SA=600+600+450+900

SA=2550cm2

Theareaofthefigureis2550cm2.

11.a) Thetotalsurfaceareaisthesurfaceareaof

eachfaceaddedtogether.Anumberofthe

faceshavethesamedimensionsandcan

begrouped.

SA=2×A1+2×A2+2×A3

SA=2×(90×70)+2×(70×60)+

2×(60×90)

SA=2×6300+2×4200+2×5400

SA=12600+8400+10800

SA=31800cm2

Thesurfaceareaoftheboxis31800cm2.

b) Usethesurfaceareacalculatedtoworkout

howmuchwoodhewillneed.

w=31800×20%+31800

w=31800×0.20+31800

w=6360+31800

w=38160cm2

Theareaofwoodhewillneedtobuyis

38160cm2.

12.Solvetheareaoftheirregularfigureby

dividingitinto2regularshapes.

A1=6×1.5+(8–6)×1.5

A1=9+3

A1=12ft2

Thetotalsurfaceareaisthesurfaceareaofeach

faceaddedtogether.Anumberofthefaceshave

thesamedimensionsandcanbegrouped.

SA=2×A1+A2+2×A3+A4

SA=2×12+8×1+2×(6×1)+

(6–1.5)×1

SA=24+8+12+4.5

SA=48.5ft2

Thetotalsurfaceareaoftheductis48.5ft2.


PrAcTiSE Your NEw SkillS, P. 167

1. a) 42cm

30cm

30cm

25cm

25cm

30cm 30cm

Thetotalsurfaceisthesurfaceareaofeachface

addedtogether.Anumberofthefaceshavethe

samedimensionsandcanbegrouped.

SA=2×A1+2×A2+2×A3

SA=2×(42×25)+2×(30×25)+

2×(42×30)

SA=2100+1500+2520

SA=6120cm2

Thesurfaceareaoftheprismis6120cm2.

b) 48cm 15cm

25cm

15cm

15cm

20cm25cm 25cm

Thetotalsurfaceisthesurfaceareaof

eachfaceaddedtogether.Anumberofthe

faceshavethesamedimensionsandcan

begrouped.

SA=2×A1+A2+A3+A4

SA=2×(0.5×12×25)+15×48+

20×48+25×48

SA=300+720+960+1200

SA=3180cm2

Thesurfaceareaoftheprismis3180cm2.

2. Thetotalsurfaceisthesurfaceareaofeachface


samedimensionsandcanbegrouped.SA A A

SA

SA

SA

= × + ×

= × × ×( ) + × ×

= +

2 2

2 12

4 1 5 2 12 2 5

6 60

1 2

. ( . )

== 66 2ft

Thetroughwillneed66ft2ofwoodtobuildit.

3. Thetotalsurfaceisthesurfaceareaofeachface


samedimensionsandcanbegrouped.SA A A A

SA

= + × + ×

= × + × × + × ×1 2 32 2

1 2 0 6 2 0 6 0 4 2 1 2 0. . ( . . ) ( . .44

0 72 0 48 0 96

2 2 2

)

. . .

.

SA

SA

= + +

≈ m

Aaronwillneed2.2m2ofglasstomakethe

fishtank.


4. Thesurfaceareaofthegrainelevatorcan

bedividedinto8differentsections:the4

rectangularsectionsoftheside,andthe4

sectionsonthefrontofthebuilding.

6’

12’12’

12’

80’

30’32’

A1

A2

A5

A6

A7 A3

A4A8

Notethatthebackoftheelevatorandthe

othersidenotvisibleareequalinsurfaceareto

thefrontandsidearea,sotheirsurfaceareas

willbemultipliedby2tocalculatethetotal

surfacearea.

Youcanignoretheareaofthebottomofthe

elevator.

Beginbycalculatingtheareasofthe4sections

ofthefrontface.

A bh

A

A

A w

A

A

A

1

1

12

2

2

22

2

12 62

36

12 12

144

=

= ×

=

=

= ×

=

ft

ft

33

3

32

4

4

12

12

12 32 12

264

32

= +

= +

=

=

=

( )

( )( )

a b h

A

A

A w

A

ft

××

=

50

160042A ft

A bh

A

A

A w

A

A

A

1

1

12

2

2

22

2

12 62

36

12 12

144

=

= ×

=

=

= ×

=

ft

ft

33

3

32

4

4

12

12

12 32 12

264

32

= +

= +

=

=

=

( )

( )( )

a b h

A

A

A w

A

ft

××

=

50

160042A ft

Next,calculatetheareasofthe4rectangular

sectionsofthesideface.

Forthefirstrectangle,A5,usethePythagorean

theoremtosolveforthewidth.

a b c

w

w

w

w

2 2 2

2 2 2

2 2

12 2 6

6 6

72

8 5

+ =

÷ + =

+ =

=

≈

( )

. ft

A w

A

A

A w

A

A

5

5

52

6

6

6

30 8 5

254 6

30 12

360

=

= ×

≈

=

= ×

=

.

. ft

ftt2

Forthethirdrectangle,A7,usethePythagorean

theoremtosolveforthewidth.

a b c

w

w

w

w

2 2 2

22 2

2 2

32 122

12

10 12

244

15 6

+ =

−( ) + =

+ =

=

≈. ft


A w

A

A

A w

A

7

7

72

8

8

30 15 6

468 6

30 80 12 12

=

= ×

≈

=

= × − −

.

.

(

ft

−−

=

6

150082

)

A ft

Finally,addalltheareastogetherandmultiply

by2(becausethebackandothersidefacesare

thesameasthefrontandsidefacesshown).

SA=2(A1+A2+A3+A4+A5+A6+A7+A8)

SA=2(36+144+264+1600+254.6+

360+468.6+1500)

SA=9254ft2

SA=3180cm2

ThesurfaceareaofthegrainThesurfacearea

ofthegrainelevatoris9254ft2.


3.2 Surface Area of Pyramids, cylinders, Spheres & cones


1. Dividetheirregularfigureintoarectangleand

asemicircletocalculatetwoareasthatcanbe

addedtogetthetotal.

A lw r

A

A

A

= +

= × + × × ( )= +

=

12

3 5 3 25 12

3 52

11 4 4 8

16

2

2

π

π. . .

. .

..2 2ft

Theareaofthefigureis16.2ft2.

2. Dividetheirregularfigureintoatriangleand

asemicircletocalculatetwoareasthatcanbe

addedtogetthetotal.

A bh r

A

A

= +

= × × + × × ( )= +

12

12

12

6 82

6 8 12

6 82

11 56 1

2

2

π

π. . .

. 88 16

29 7 2

.

.A ≈ cm

Theareaofthefigureis29.7cm2.

3. Tofindtheareaofthefigurecalculatethe

areaoftherectangleandthensubtractthe

semicirclefromit.

A lw r

A

A

A

= −

= × − × × ( )= −

12

310 250 12

902

77 500 3180 9

2

2

π

π

.

≈≈ 74 319 2mm

Theareaoftheshapeis74319mm2.

4. Tofindtheareaofthetankaddtheareaofthe

circularfacestotheareaoftheside.

SA r rh

SA

SA

= +

= × × + × × ×

≈ +

2 2

2 1 5 2 1 5 5

14 1 47 1

2

2

( )

. .

. .

π π

π π

SSA ≈ 61 2 2. m

Thesurfaceareaofthetankis61.2m2.

5. Tofindthesurfaceareaofthepipefindthe

areaoftheside.Theendsofthepipearehollow

sotheycanbeignored.

SA rh

SA

SA

=

= × × ×

≈

2

2 4 52

18 8

265 8 2

π

π . .

. cm

Thesurfaceareaofthepipeis265.8cm2.

6. Tofindthesurfaceareaofthefigureaddthe

areasofeachsurface.Thetoptwosurfaces

canbeaddedtogethertobeequivalenttothe

bottomsurface.

SA=2πr1h1+2πr1h1+2×(πr2)

SA=2×π×3.8×9.6+2×π×(3.8+4.2)×

4.8+2×π×(3.8+4.2)2

SA≈229.2+241.3+402.1

SA≈872.6mm2

Thesurfaceareaofthefigureis872.6mm2.


7. Tocalculatethesurfaceareacalculatethearea

ofthesquarebase.A lw

A

A

1

1

12

28 28

784

=

= ×

= cm

Calculatetheareaofoneofthetriangularsides.

A bh

A

A

2

2

22

12

12

28 21

294

=

=

=

( )( )

cm

Calculatethetotalsurfacearea.

SA A A

SA

SA

= +

= +

=

1 2

2

4

784 4 294

1960

( )

( )

cm

Thesurfaceareaofthepyramidis1960cm2.

8. Calculatetheareaofoneofthetriangles.

A bh

A

A

=

=

=

12

12

18 28

252 2

( )( )

cm


SA A

SA

SA

= ×

= ×

=

5

5 252

1260 2cm

Theexposedareaofthepyramidis1260cm2.

9. Calculatetheareaoneofthetriangles.

A bh

A

A

=

=

≈

12

12

12 6 18 7

117 8 2

( . )( . )

. cm


SA A

SA

SA

= ×

= ×

=

8

8 117 8

942 4 2

.

. cm

Thelateralsurfaceareaofthepyramidis

942.5cm2.

10.Findtheareaofthebaseofthepyramid.

A lw

A

A

1

1

12

12 12

144

=

= ×

= cm

Findtheareaofoneofthetriangles.

A bh

A

A

2

2

22

12

12

12 8

48

=

=

=

( )( )

cm


SA A A

SA

SA

= +

= +

=

1 2

2

4

144 4 48

336

( )

cm

Thetotalsurfaceareaofthepyramidis384in2.

11.Findtheheightofoneofthetrianglesusingthe

Pythagoreantheorem.

h

h

h

h

h

2 22

2 2

16 162

16 8

256 64

192

13 9

= − ( )= −

= −

=

≈ . cm

Findtheareaofoneofthetriangles.

A bh

A

A

=

=

=

12

12

16 13 9

111 2 2

( )( . )

. cm

Calculatethetotalsurfacearea.SA A

SA

SA

=

= ×

=

4

4 111 2

444 8 2

.

. cm

Theareaofthepyramidis444.8cm2.


12.Firstusethesurfaceareayouknowtosolvefor

theslantheight.

SA A A

SA lw bh

h

s

s

= +

= + ( )= × + × × ×( )

1 24

4 12

680 16 16 4 12

16

6800 256 32

424 32

13 25

= +

=

=

h

h

h

s

s

s.

Theslantheightofthepyramidis13.25cm.

UsetheslantheightandthePythagorean

theoremtosolvefortheheightofthepyramid.

h

h

h

h

h

2 22

2 2

13 25 162

13 25 8

175 6 64

111 6

1

= − ( )= −

≈ −

≈

≈

.

.

.

.

00 6. cm

Theheightofthepyramidwas10.6cm.

13.Calculatetheareaofthecircularbase.

A r

A

A

12

12

12

28

2463 0

=

=

≈

π

π( )

. cm

Calculatetheareaofthelateralsurface.

A rs

A

A

2

2

22

28 82

7213 1

=

= × ×

=

π

π

. cm


SA A A

SA

SA

= +

= +

=

1 2

2

2463 7213 1

9676 1

.

. cm

Thetotalareaoftheconeis9676.1cm2.

14.Tocalculatethesurfaceareaoftheconeadd

theareaofthecircularbasetotheareaofthe

lateralsurface.

SA r rs

SA

SA

= +

= ( ) + ( ) ( )

= +

π π

π π

π

2

213 62

13 62

9 8

46 24 6

. . .

. 66 64

112 88

354 6 2

.

.

.

π

πSA

SA

=

≈ cm

Thesurfaceareaoftheconeis354.6cm2.

15.Calculatetheslantheightoftheconeusingthe

Pythagoreantheorem.

s h r

s

s

s

s

2 2 2

2 2 2

2 2

20 16

20 16

656

25 6

= +

= +

= +

=

= . in

Tocalculatethesurfaceareaoftheconeadd


lateralsurface.

SA r rs

SA

SA

SA

= +

= +

= +

=

π π

π π

π π

2

216 16 25 6

256 320

2

( ) ( )( . )

0091 0 2. in

Thesurfaceareaoftheconeis2091.0in2.

16.Calculatetheslantheightoftheconeusingthe

Pythagoreantheorem.

s h r

s

s

s

s

2 2 2

2 2 2

2 2

39 7 45 7

39 7 45 7

3664 6

60

= +

= +

= +

≈

≈

. .

. .

.

..5 mm




lateralsurface.

SA r rs

SA

SA

= +

= +

≈ +

π π

π π

π

2

245 7 45 7 60 5

2088 5 2

( . ) ( . )( . )

. 7764 9

15247 2

. π

SA ≈ mm

Thesurfaceareaoftheconeis15247mm2.

17. Calculatetheslantheightoftheconeusingthe

Pythagoreantheorem.

s h r

s

s

s

s

2 2 2

2 22

2 2

32 282

32 14

1220

34 9

= +

= + ( )= +

≈

≈ . cm



lateralsurface.

SA rs

SA

SA

SA

=

= ( ) ( )

≈

≈

π

π

π

282

34 9

488 6

1535 0 2

.

.

. cm

Thesurfaceareaoftheconeis1535.0cm2.

18.Calculatethesurfaceareaofasphere.

SA r

SA

SA

=

= × ×

=

4

4 1 3

21 2

2

2

2

π

π .

. m

Thesurfaceareaofthesphereis21.2m2.

19.Calculatethesurfaceareaofasphere.

SA r

SA

SA

=

= × × ( )=

4

4 24 82

1932 2

2

2

2

π

π .

. mm

Thesurfaceareaofthesphereis1932.2mm2.

20.Togetthesurfaceareaofahemisphere

calculatehalfthesurfaceareaofasphereand

theareaofthebase.

SA r r

SA

SA

= × +

= × × × + ×

≈ +

12

4

12

4 18 5 18 5

684 5

2 2

2 2

π π

π π

π

. .

. 3342 3

3225 8 2

.

.

π

SA ≈ cm

Thesurfaceareaofthehemisphereis

2150.4cm2.

21.Thesurfaceareaofthefigureisthesurfacearea

ofahemisphere,thelateralsurfaceareaofa

cylinderandalateralsurfaceareaofacone.

Calculatethesurfaceareaofahemisphere.

SA r

SA

SA

12

1

2

12

12

4

12

4 0 82

1 01

= ×

= × × × ( )≈

π

π .

. m

Calculatethesurfaceareaofthelateralcylinder.

SA rh

SA

SA

2

2

22

2

2 0 82

2 2

5 53

=

= × × ×

≈

π

π . .

. m

Calculatethesurfaceareaofthelateralsides

ofthecone.

SA rs

SA

SA

3

3

32

0 82

1 4

1 76

=

= × ×

≈

π

π . .

. m



SA SA SA SA

SA

SA

= + +

= + +

=

1 2 3

2

1 01 5 53 1 76

8 3

. . .

. m

Thetotalsurfaceareaofthefigureis8.3m2.

22.Findthelateralsurfaceareaofthetopcylinder.

SA rh

SA

SA

1

1

12

2

2 2 2

25 1

=

= × × ×

≈

π

π

. ft

Findthelateralsurfaceareaofthe

middlecylinder.

SA rh

SA

SA

2

2

22

2

2 3 2

37 7

=

= × × ×

≈

π

π

. ft

Findthelateralsurfaceareaofthe

bottomcylinder.

SA rh

SA

SA

3

3

32

2

2 4 2

50 3

=

= × × ×

≈

π

π

. ft

Findthesurfaceareaofthecircleon

thebottom.

SA r

SA

SA

42

42

42

4

50 3

=

= ×

≈

π

π

. ft

Thethreeexposedsurfacesonthetopifadded

togetherareequivalenttothesurfaceareaof

thecircleonthebottom.Calculatethetotal

surfacearea.

SA SA SA SA SA

SA

SA

= + + + ×

= + + + ×

=

1 2 3 42

25 1 37 7 50 3 2 50 3. . . .

2213 7 2. ft

Thetotalexposedsurfaceareaofthecylinderis

213.6ft2.

23.Findthesurfaceareaofthehemisphere.

SA r

SA

SA

12

1

2

12

12

4

12

4 182

508 9

= ×

= × × × ( )≈

π

π

. cm

Findthelengthoftheslantofthecone.

s

s

s

s

s

2 22

2 2

12 182

12 9

144 81

225

15

= + ( )= +

= +

=

= cm

Findthelateralsurfaceareaofthecone.

SA rs

SA

SA

2

2

22

182

15

424 1

=

= × ×

≈

π

π

. cm

Addthetotalstogether.

SA SA SA

SA

SA

= +

= +

=

1 2

2

508 9 424 1

933 0

. .

. cm

Thesurfaceareaofthefigureis933.0cm2.

PrAcTicE Your SkillS, p. 167

1. Calculatetheareaofthebottomofthecake

panbycalculatingtheareaofacircle.

SA r

SA

SA

12

1

2

12

92

63 6

=

= × ( )≈

π

π

. in


Calculatetheareaofthesideofthecakepanby

calculatingthelateralsurfaceofacylinder.

SA rh

SA

SA

2

2

22

2

2 92

1 5

42 4

=

= × × ×

≈

π

π .

. in

Calculatethetotalarea.

SA SA SA

SA

SA

= +

= +

=

1 2

2

63 6 42 4

106

. .

in

Hewilluse106in2tomakethecaketin.

2. Findthesurfaceareaofonecupbycalculating

thelateralsurfaceareaofthecone.Convertthe

valuesfrommmtocm.

SA rs

SA

SA

c

c

c

=

= × ×

≈

π

π 32

6

28 3 2. cm

Usethatvaluetofindhowmuchareneededto

produce50cups.

SA

SA

= ×

=

28 3 50

1415 2

.

cm

Tomake50conicalcupsyouwouldneed

1415cm2ofpaper.

3. Calculatetheareaofthefrontandbackofthe

hutbycalculatingtheareaofacircle.Ifyou

ignorethedooratthispointyoucandothisby

calculatingtheareaofonewholecirclebecause

thefrontandbackareeachequaltohalf.

SA r

SA

SA

12

1

2

12

202

314 2

=

= ( )≈

π

π

. ft

Calculatetheareaofthedoor.

SA

SA

2

22

7 8

56

= ×

= ft

Calculatetheareaofthehalfofa

lateralcylinder.

SA rh

SA

SA

3

3

32

12

2

12

2 202

48

1508 0

= ×

= × × × ×

≈

π

π

. ft

Calculatethetotalareabyaddingthe

areaofthefront,backandthesidesand

subtractingthedoor.

SA SA SA SA

SA

SA

= − +

= − +

=

1 2 3

2

314 2 56 1508 0

1766 2

. .

. ft

Therewouldneedtobe1766.2ft2of

corrugatedsteel.

4. Findthesurfaceareaofoneofthetriangles.

A bh

A

A

t

t

t

=

=

=

12

12

3 5 12 4

21 7 2

( . )( . )

. ft

Usetheareacalculatedtosolveforthelateral

surfacearea.

SA A

SA

SA

t= ×

= ×

=

6

6 21 7

130 2 2

.

. ft

Thelateralsurfaceareaofthehexagonal

pyramidis130.2ft2.


5. Calculatethesurfaceareaoftheconebyadding

thebasetothelateralsurfacearea.

SA r rs

SA

SA

SA

= +

= ( ) +

≈ +

≈

π π

π π

2

212 12 15

452 4 565 5

1

( )( )

. .

0017 9. ft

Thesurfaceareaoftheconeis1017.9ft2.

6. Calculatethesurfaceareaofthetennisball.

SA r

SA

SA

=

= × × ( )=

4

4 6 72

141 0

2

2

2

π

π .

. cm

Thesurfaceareaofthetennisballis141.0cm2.


3.3 Volume and capacity of Prisms and cylinders


1. a) Usetheformulaforvolume.

V A h

V lwh

V

V

base= ×

=

= × ×

≈

15 7 18 8 12 5

3690 3

. . .

cm

Thevolumeoftheprismis3690cm3.

Since1Lequals1000cm3,dividethe

volumeby1000.

3690 1000 3 69÷ = . L

Theprismhasacapacityof3.69L.

b) Usetheformulaforvolume.

V

V

= × ×

≈

2 75 2 75 4 5

34 0 3

. . .

. m

Thevolumeis34m3.

Sinceweonlyknowtheconversionofcm3

toLwemustconvertm3tocm3.

Thereare100centimetresinametre.So1

m3isequalto(100cm)3,whichequals

1000000cm3.Somultiplytheamount

inm3by1000000cm3/m3.Then

since1Lequal1000cm3,dividethe

volumeby1000.

34 1 000 000 34 000 000

34 000 000 1000 3400

3× =

÷ =

cm

00 L

Thecapacityofprismis34000L.

c) Converttheunitsfrominchesto

centimetres.1 5 2 54 1 5

3 81

3 75 2 54

. ( . . )

.

. ( .

in cm

1.5 in cm

in

= ×

=

= ××

≈

= ×

3 75

3 9 53

2 25 2 54 2 25

. )

.

. ( . . )

cm

.75 in cm

in cm

22.25 in cm≈ 5 72.

Usetheformulaforvolume.V

V

= × ×

≈

3 81 9 53 5 72

208 3

. . .

cm

Since1mLequals1cm3thecapacity

is208mL.

AlTErNATE SoluTioN

Ifyouknowtheconversionsyoucouldalso

dothecalculationsinimperial.

Usetheformulaforvolume.

V

V

= × ×

≈

1 5 3 75 2 25

12 7 3

. . .

. in

Since1USgalis231in3,dividethevolume

by231.

12 7 231 0 055. .÷ ≈ US gal

Thecapacityis0.055USgal.

2. Usethevolumeformulatosolvefortheheight.

V A h

V lwh

h

h

base= ×

=

= × ×

≈

142 5 2 7 8

3 5

. .

. m

Theheightoftheprismis3.5m.


3. Wecanusethevolumeformulatosolveforthe

missingheight.

l w h l w h

h

h

1 1 1 2 2 2

2

2

18 12 32 14 20

6912 280

24 7

=

× × = × ×

=

. cm ≈≈ h2

Theheightofthesecondprismmust

be24.7cm.

4. Calculatethetotalvolumeofthehole.

V lwh

V

V

=

= × ×

=

35 25 12

10500 3m

Todeterminehowmanytripsmustbemade

dividethetotalvolumebythevolumethe

trailercancarry.t

t

= ÷

=

10500 15

700

Theexcavationwilltake700trips.

5. a) 9.4cm

16.8cm

6.6cm

7.5cm

V1

V2

4.2cm

b)

V1

V2

7.4cm

16.8cm6.6cm

11.9cm

4.2cm

6. Dividethefigureintotworegularsections.

30 m

8 m9 m 9 m

8 m

12 m

11 m

V1

V2

Calculatethevolumeofthetopsectionof

theshape.

V lwh

V

V

V

1

1

1

13

30 9 9 12 11

12 12 11

1584

=

= − − × ×

= × ×

=

( )

m

Calculatethevolumeofthebottomsectionof

theshape.

V lwh

V

V

2

2

23

30 12 8

2880

=

= × ×

= m

Calculatethetotalvolume.

V V V

V

V

= +

= +

=

1 2

3

1584 2880

4464 m

Thevolumeofthefigureis4464m3.

AlTErNATE SoluTioN

Dividethefigureintothreeregularsections.

30 m

8 m9 m 9 m

8 m

12 m

11 mV1 V2

V3


Calculatethevolumeoftheleftsectionof

theshape.

V lwh

V

V

1

1

13

9 12 8

864

=

= × ×

= m

Calculatethevolumeofthemiddlesectionof

theshape.

V lwh

V

V

V

2

2

2

2

30 9 9 12 11 8

12 12 19

2736

=

= − − × × +

= × ×

=

( ) ( )

m33

Theleftandrightsectionshavethesame

dimensionssoatthispointcalculatethe

totalvolume.

V V V

V

V

= +

= +

=

2

2 864 2736

4464

1 2

3

( )

( )

m

Thevolumeofthefigureis4464m3.

7. Dividethefigureintotwosections.

9.4 cm

w

16.9 cm

7.3 cm

8.5 cm

V1

V2

Calculatewasanexpressionoftheleftsection

oftheshape.

V lwh

V w

V w

1

1

1

16 9 9 4

158 86

=

= × ×

=

. .

.

Calculatewasanexpressionoftheright

sectionoftheshape.

V lwh

V w

V w

2

2

2

7 3 8 5 9 4

130 67

=

= × × +

=

. ( . . )

.

Since1Lequals1000cm3,multiplythe

capacityby1000togetthevolume.

1 8 1000 1800

1800

3

3

. × =

=

cm

cmV

Usethevaluescalculatedtosolveforw.

V V V

w w

w

w

= +

= +

=

≈

1 2

1800 158 86 130 67

1800 289 53

6 2

. .

.

. cm

Thedepthoftheprismis6.2cm.

AlTErNATE SoluTioN

Dividethefigureintotwosections.

V19.4 cm

w

16.9 cm

7.3 cm

8.5 cm

V2

Calculatewasanexpressionofthebottom

sectionoftheshape.

V lwh

V w

V w

1

1

1

16 9 7 3 9 4

227 48

=

= + × ×

=

( . . ) .

.


Calculatewasanexpressionofthetopsection

oftheshape.

V lwh

V w

V w

2

2

2

7 3 8 5

62 05

=

= × ×

=

. .

.

Asaboveconvertthecapacityofthecontainer

toanexpressionofvolumetoget1800cm3.

Usethevaluescalculatedtosolveforw.

V V V

w w

w

w

= +

= +

=

≈

1 2

1800 227 48 62 05

1800 289 53

6 2

. .

.

. cm

Thedepthoftheprismis6.2cm.

8. Calculatethevolumeofthecylinder.

V A h

V r h

V

V

base= ×

=

= × ( ) ×

≈

π

π

2

2

3

152

36

6362 cm

Thevolumeofthecylinderis6362cm3.

Since1Lequals1000cm3,dividethevolume

by1000togetthecapacity.

6362 1000 6 4÷ ≈ . L

Thecapacityofthecylinderis

approximately6.4L.

9. Calculatethevolumeofthetopcylinder.

V r h

V

V

t

t

t

=

= × ( ) ×

≈

π

π

2

2

3

102

20

1570 8. cm

Calculatethevolumeofthemiddlecylinder.

V r h

V

V

m

m

m

=

= × ( ) ×

≈

π

π

2

2

3

202

20

6283 2. cm

Calculatethevolumeofthebottomcylinder.

V r h

V

V

b

b

b

=

= × ( ) ×

≈

π

π

2

2

3

402

20

25132 7. cm


V V V V

V

V

t m b= + +

= + +

≈

1570 8 6283 2 25132 7

32987 3

. . .

cm

Thevolumeofthecylindersis32987cm3.

Since1Lequals1000cm3,dividethevolume

by1000togetthecapacity.

32987 1000 33 0÷ ≈ . L

Thecapacityofthecylindersis

approximately33L.

10.Convertthecapacityofthecanintoan

expressionofvolume.

V

V

= ×

=

3 24 1000

3240 3

.

cm

Usetheformulaforvolumetosolveforthe

heightofthecan.

V A h

V r h

h

h

base= ×

=

= × ( ) ×

≈

π

π

2

2

3240 15 562

3240 190 16

17

.

.

..0 cm ≈ h

Theheightofthecanis17.0cm.



1. a) Calculatethevolumeofthefigure.

V A h

V lwh

V

V

base= ×

=

= × ×

≈

6 8 4 2 3 9

111 4 3

. . .

. cm

Thevolumeis111.4cm3.

b) Calculatethevolumeofthefigure.

V A h

V r h

V

V

base= ×

=

= × ×

≈

π

π

2

2

3

2 1 2 5

34 6

. .

. cm

Thevolumeis34.6cm3.

c) Calculatethevolumeofthecylinder.

V r h

V

V

c

c

c

=

= × ×

≈

π

π

2

2

3

0 8 5 2

10 5

. .

. m

Calculatethevolumeoftherectangularprism.

V lwh

V

V

r

r

r

=

= × ×

≈

4 5 4 5 3 5

70 9 3

. . .

. m

Calculatethetotalvolume.V V V

V

V

c r= +

= +

=

10 5 70 9

81 4 3

. .

. m

Thevolumeofthefigureis81.4m3.

2. Itshouldbeapparentthatarectangularprism

isbiggertoprovethisfactitisnecessaryto

calculatevolumeofeach.Calculatethevolume

oftherectangularprism.

V lwh

V

V

r

r

r

=

= × ×

=

16 16 25

6400 3cm

Calculatethevolumeofthecylinder.

V r h

V

V

c

c

c

=

= × ( ) ×

≈

π

π

2

2

3

162

25

5026 5. cm

Therectangularprismisbigger.

3. Thesiloisshapedlikeacylindersousethe

formulaforthevolumeofacylinder.

V r h

V

V

=

= × ( ) ×

≈

π

π

2

2

3

242

70

31 667 ft

Thesilocanhold31667ft3ofgrain.

4. a) Calculatethevolume.Convertallvalues

tometres.

V lwh

V

V

=

= × ×

≈

6 4 5 0 15

4 1 3

. .

. m

Thevolumeofconcreteneededis4.1m3.

b) Usetheconversionrateprovidedto

calculatetheweight.

W

W

= ×

=

2400 4 1

9840

.

kg

Thepatiowillweigh9840kg.

5. Convertthecapacityofthewatertovolume,

1Lequals1000cm3.

V

V

= ×

=

15 1000

15 000 3cm


Usetheformulaforvolumetosolvefor

thedepth.

V lwh

h

h

h

=

= × ×

=

=

15 30 20

15 600

25

000

000

cm

Thedepthofwaterinthetankis25cm.

6. Convertthecapacityofthetanktovolume.

V

V

= ×

=

20 000 1000

20 3000 000 cm

Convertthevaluetom3.Thereare100

centimetresinametre.So1m3isequalto

(100cm)3,whichequals1000000cm3.

Converttometresbydividingthevalueby

1000000.

20 1 20 3000 000 000 000 m÷ =

Usetheformulaforvolumetosolveforthe

heightofthetank.

V r h

h

h

h

=

= × ( ) ×

≈

≈

π

π

2

2

20 2 42

20 4 52

4 4

.

.

. m

Theheightofthetankisapproximately4.4m.

AlTErNATE SoluTioN

Insteadofconvertingthevolumefromcm3

tom3youcouldhaveinsteadconvertedthe

diameterfrommetrestocentimetres.Usethe

formulaforvolumetosolvefortheheight

ofthetank.

V r h

h

h

=

= × ( ) ×

≈

π

π

2

2

20 2402

20 45 239

4

000 000

000 000

442 cm ≈ h

Theheightofthetankisapproximately

442cmor4.4m.


3.4 Volume and capacity of Spheres, cones, & Pyramids


1. a) Calculatethevolumeofthesphere.

V r

V

V

=

= ×

=

43

43

8 5

2572

3

3

3

π

π .

cm

Thevolumeofthesphereis2572cm3.

b) Calculatethevolumeofthesphere.

V r

V

V

=

= × ( )≈

43

43

782

248

3

3

3

π

π

475 cm

Thevolumeofthesphereis248475cm3.

2. a) Calculatethevolumeoftheoutersphere.

V r

V

V

O

O

O

=

= ×

≈

43

43

76

1

3

3

3

π

π

838 778 cm

Calculatethevolumeoftheinnersphere.

V r

V

V

I

I

I

=

= ×

≈

43

43

46

407

3

3

3

π

π

720 cm

Subtractthevolumeoftheinnersphere

fromthevolumeoftheoutersphere.

V V V

V

V

O I= −

= −

=

1 407

1 3

838 778 720

431 058 cm

Thevolumeofthespacebetweenthespheresis

1431058cm3.

b)Convertthevolumetocapacity,1L

equals1000cm3.

1 1000 1431431 058 L÷ ≈

Thecapacityofthespacebetweenthetwo

spheresisapproximately1431L.

3. a) Usetheformulaforcalculatingthevolume

ofapyramid.

V lwh

V

V

=

= × × ×

=

13

13

15 17 19

1615 3in

Thevolumeofthepyramidis1615in3.

b) Usetheformulaforcalculatingthevolume

ofapyramid.

V lwh

V

V

=

= × × ×

=

13

13

5 4 6

40 3ft

Thevolumeofthepyramidis40ft3.

c) Calculatethevolumeofthetoppyramid.

V lwh

V

V

1

1

13

13

13

9 8 4 3 6 2

87 1

=

= × × ×

≈

. . .

. cm

Calculatethevolumeofthe

bottompyramid.


V lwh

V

V

2

2

23

13

13

9 8 4 3 4 8

67 4

=

= × × ×

≈

. . .

. cm


V V V

V

V

= +

= +

=

1 2

3

87 1 67 4

154 5

. .

. cm

Thevolumeofthetwopyramidsis

154.5cm3.

AlTErNATE SoluTioN

Studentsmayhavenoticedthatsincethetwo

pyramidsshareabaseyoucouldalsohave

calculatedthevolumeasifthetwopyramids

werejustonebigpyramid.

V lw h h

V

V

= +

= × × × +

= × ×

13

13

9 8 4 3 6 2 4 8

13

9 8 4 3

1 2( )

. . ( . . )

. . ××

≈

11

154 5 3V . cm

Thevolumeofthetwopyramidsis154.5cm3.

4. Converttheunitsfromfeettometres.

20 20 0 305

6 10

9 9 0 305

9

ft m

20 ft m

ft m

= ×

=

= ×

( . )

.

( . )

ft m

ft m

22 ft m

ft

≈

= ×

=

=

2 75

22 22 0 305

6 71

15

.

( . )

.

(( . )

.

15 0 305

15 4 58

×

≈

m

ft m

Calculatethevolumeofthepyramid.

V lwh

V

V

1

1

1

13

13

6 71 6 10 4 58 2 75

13

6 71 6

=

= × × × −

= × ×

. . ( . . )

. .. .

.

10 1 83

25 013

×

=V m


V lwh

V

V

2

2

23

6 71 6 10 2 75

112 6

=

= × ×

=

. . .

. m

Calculatethetotalvolume.V V V

V

V

= +

= +

=

1 2

3

25 0 112 6

137 6

. .

. m

Thevolumeofthefigureis137.6m3.

Convertfromvolumetocapacity.

137 6 1000 137. × = 600 L

Thecapacityofthefigureis137600L.

AlTErNATE SoluTioN

Ifyouknowtheconversionratesfrom

imperialvolumetocapacityyoucouldhave

alsocalculatedthesolutionasfollowsusing

imperialmeasures.


V lwh

V

V

V

1

1

1

1

13

13

22 20 15 9

13

22 20 6

880

=

= × × × −

= × × ×

=

( )

ftt3


Calculatethevolumeoftherectangularprism.V lwh

V

V

2

2

23

22 20 9

3960

=

= × ×

= ft


V V V

V

V

= +

= +

=

1 2

3

880 3960

4840 ft

Thevolumeofthefigureis4840ft3.

Convertfromvolumetocapacity,1ft3equals

7.48USgallons.

4840 7 48 36 203× ≈. US gal

Thecapacityofthefigureis36203USgallons.

5. Calculatethevolumeofthepyramid.

V lwh

V

V

p

p

p

=

= × × ×

=

13

13

28 16 12

1792 3in


V lwh

V

V

r

r

r

=

= × ×

=

28 16 12

5376 3in

Calculatethedifferenceinthetwovolumes.

x V V

x

x

r p= −

= −

=

5376 1792

3584 3in

Thedifferenceinthevolumesis3584in3.

6. a) Calculatetheheightofthepyramidusing

thePythagoreantheorem.

h

h

h

h

2 22

2 2

25 352

25 17 5

318 75

17 9

= − ( )= −

=

≈

.

.

. mm


V lwh

V

V

=

= × × ×

≈

13

13

35 35 17 9

7309 3

.

mm

Thevolumeofthepyramidis7309mm3.

b) Calculatetheheightofthepyramidusing


h

h

h

h

2 22

2 2

19 232

19 11 5

228 75

15 1

= − ( )= −

=

≈

.

.

. in


V lwh

V

V

=

= × × ×

≈

13

13

23 23 15 1

2663 3

.

in


c) Calculatetheheightofthepyramidusing


h

h

h

h

2 22

2 2

2 12 92

24 4 5

555 75

23 6

= ×( ) − ( )= −

=

≈

.

.

. in



V lwh

V

V

=

= × × ×

≈

13

13

9 9 23 6

637 3

.

in


d) Calculatetheheightofthepyramidusing


h

h

h

h

2 22

2 2

16 142

16 7

207

14 4

= − ( )= −

=

≈ . mm


V lwh

V

V

=

= × × ×

≈

13

13

14 12 14 4

806 3

.

mm

Thevolumeofthepyramidis806mm3.

7. a) Calculatethevolumeofthecone.

V A h

V r h

V

V

base= × ×

=

= × ×

≈

13

13

13

5 14 5

379 6

2

2

3

π

π .

. in

Thevolumeoftheconeis379.6in3.

b) Calculatethevolumeofthefirstcone.

V r h

V

V

12

12

13

13

13

6 18

678 6

=

= × ×

≈

π

π

. cm

Calculatethevolumeofthesecondcone.

V r h

V

V

22

22

23

13

13

6 24

904 8

=

= × ×

≈

π

π

. cm


V V V

V

V

= +

= +

≈

1 2

3

678 6 904 8

1583

. .

cm

Thetotalvolumeofthefigureis

approximately1583cm3.

c) Calculatethevolumeofthecone.

V r h

V

V

12

12

13

13

13

3 6

56 5

=

= × ×

≈

π

π

. m

Calculatethevolumeofthecylinder.

V A h

V r h

V

V

base2

22

22

23

3 5

141 4

= ×

=

= × ×

≈

π

π

. m


V V V

V

V

= +

= +

≈

1 2

3

56 5 141 4

198

. .

m

Thetotalareaofthefigureis198m3.

8. Usetheformulaforthevolumeofaconeto

solvefortheheight.

V r h

h

h

h

=

= × ×

≈

≈

13

4071 5 13

12

4071 5 150 8

27

2

2

π

π.

. .

mm

Theheightoftheconeis27mm.


9. UsethePythagoreantheoremtocalculatethe

heightofthecone.

h

h

h

h

2 2 2

2 2

15 8

15 8

161

12 7

= −

= −

=

≈ . cm

Calculatethevolumeofthecone.

V r h

V

V

=

= × ×

≈

13

13

8 12 7

851

2

2

3

π

π .

cm

Thevolumeoftheconeis851cm3.


1. a) Calculatethevolumeofthesphere.

V r

V

V

sphere =

= ×

≈

43

43

6 3

1047

3

3

3

π

π .

mm

Thevolumeofthesphereis1047mm3.

b) Calculatethevolumeofthecone.

V r h

V

V

cone =

= × ×

≈

13

13

6 3 12 4

515

2

2

3

π

π . .

mm

Thevolumeoftheconeis515mm3.

c) Calculatethevolumeofthepyramid.

V lwh

V

V

pyramid =

= × × ×

≈

13

13

3 6 5 2 4 9

30 6 3

. . .

. mm

Thevolumeofthepyramidis30.6mm3.

d)Calculatethevolumeofthehemisphere.

V r

V

V

13

13

13

12

43

12

43

4 3

166 5

= ×

= × ×

≈

π

π .

. mm


V r h

V

V

22

22

23

13

13

4 3 5 4

104 6

=

= × ×

≈

π

π . .

. mm


V V V

V

V

= +

= +

≈

1 2

3

166 5 104 6

271

. .

mm

Thevolumeofthefigureis271mm3.

2. Calculatethevolumeofthewatertower.

V r

V

V

=

= × ( )≈

43

43

31 22

15

3

3

3

π

π .

902 ft

Convertthevolumeintocapacity.

15 902 7.48 118 947 US gal.× ≈

Thecapacityofthesphericalwatertoweris

118947USgal.

3. a) Calculatethevolumeofthepyramid.

V lwh

V

V

p

p

p

=

= × × ×

=

13

13

14 14 45

2940 3cm



V r h

V

V

c

c

c

= ×

= × × ( ) ×

≈

13

13

142

45

2309

2

2

3

π

π

cm

Calculatethedifferenceinvolumes.

x

x

= −

=

2940 2309

631 3cm

Thedifferenceinvolumesis631cm3.

b) Thedifferenceincapacityisthedifference

involumeconvertedtocapacity.One

centimetrecubedisequaltoonemillilitre.

Thedifferenceincapacityis631mL.

4. Calculatethevolumeofthepileofgrain.

V r h

V

V

=

= × ( ) ×

≈

13

13

4 52

1 2

6 4

2

2

3

π

π . .

. m

Thereare6.4m3ofgraininthepile.

5. Calculatethevolumeofthefirstsphere.

V r

V

V

13

13

13

43

43

3 4

164 6

=

= ×

≈

π

π .

. cm

Calculatethevolumeofthesecondsphere.

V r

V

V

23

23

23

43

43

2 8

92 0

=

= ×

≈

π

π .

. cm

Calculatethevolumeofthethirdsphere.

V r

V

V

33

33

33

43

43

4 6

407 7

=

= ×

≈

π

π .

. cm


V V V V

V

V

= + +

= + +

≈

1 2 3

3

164 6 92 0 407 7

664

. . .

cm

Thevolumeofwaterdisplacedis664cm3.

chAPTEr TEST, p. 194

1. a) Calculatetheslantheightofthetriangle

usingthePythagoreantheorem.

h

h

h

h

2 22

2 2

8 62

8 3

55

7 4

= − ( )= −

=

≈ . cm

Calculatetheareaofoneofthetriangles.

A bh

A

A

1

1

12

12

12

6 7 4

22 2

=

= × ×

=

.

. cm

Calculatetheareaofthebase.

A lw

A

A

2

2

22

6 6

36

=

= ×

= cm


SA A A

SA

SA

= +

= × +

=

4

4 22 2 36

124 8

1 2

2

.

. cm

Thetotalsurfaceareais124.8cm2.


b) Calculatethesurfaceareaofeachface.

A

A

A

A

1

12

2

22

15 8

120

8 8

64

= ×

=

= ×

=

cm

cm


SA A A

SA

SA

= +

= × + ×

=

4 2

4 120 2 64

608

1 2

2cm

Thetotalsurfaceareais608cm2.

c) Calculatethesurfaceareaofthe

triangularface.

A bh

A

A

1

1

12

12

12

5 8

20

=

= × ×

= cm

Calculatetheslantheightofthetriangle

usingthePythagoreantheorem.

s

s

s

s

2 2 2

2 2

8 5

8 5

89

9 4

= +

= +

=

≈ . cm

Calculatetheareaofeachrectangularsurface.

A

A

A

A

A

A

2

22

3

32

4

4

5 18

90

8 18

144

9 4 18

16

= ×

=

= ×

=

= ×

≈

cm

cm

.

99 2cm


SA A A A A

SA

SA

= + + +

= × + + +

=

2

2 20 90 144 169

443

1 2 3 4

2cm

Thesurfaceareaoftriangularprism

is443cm2.

2. Calculatetheareaofeachfaceofthebox.

A

A

A

A

A

A

1

12

2

22

3

32

3 2

6

2 7 5

15

3 7 5

22 5

= ×

=

= ×

=

= ×

=

m

m

m

.

.

.


SA A A A

SA

SA

SA

= + +

= × + × + ×

= + +

=

2 2 2

2 6 2 15 2 22 5

12 30 45

8

1 2 3

.

77 2m

Thesurfaceareaoftheboxis87m2.

Calculatethevolumeofthebox.

V lwh

V

V

=

= × ×

=

3 2 7 5

45 3

.

m

Thevolumeoftheboxis45m3.

3. a) Calculatetheareaofcircularbase.

SA r

SA

SA

12

12

12

4 5

63 6

=

= ×

≈

π

π .

. cm

Calculatetheareaofthelateralside

ofthecan.

SA rh

SA

SA

2

2

22

2

2 4 5 5

141 4

=

= × ×

=

π

π .

. cm

SA SA SA

SA

SA

= +

= × +

=

2

2 63 6 141 4

268 6

1 2

2

. .

. cm


Thesurfaceareaofthecanis268.6cm2.

b) Calculatethevolumeofthecan.

V A h

V r h

V

V

base= ×

=

= × ×

≈

π

π

2

2

3

4 5 5

318 1

.

. cm

Thevolumeofthecanis318.1cm3.

c) Usethevolumeofthecantocalculatethe

capacityofthecan.Eachcentimetrecubed

isequalto1mLsothecapacityofthecan

is318.1mL.

4. a) Calculatethesurfaceareaoftheball.

SA r

SA

SA

=

= ×

=

4

4 56

39 408

2

2

2

π

π

mm

Thesurfaceareaoftheballis39408mm2.

b) Calculatethesurfaceareaoftheball.

V r

V r

V

V

=

=

= ×

≈

43

43

43

56

735 619

3

3

3

3

π

π

π

mm

Thevolumeofthesphereis735619mm3.

5. Calculatetheslantheightthepyramid.

s

s

s

s

2 22

2 2

9 112

9 5 5

111 25

10 5

= + ( )= +

=

≈

.

.

. cm

Calculatetheareaofthetriangularfacesof

thepyramid.

A bh

A

A

1

1

12

12

12

11 10 5

57 8

=

= × ×

≈

.

. cm

Calculatetheareaofthelateralfacesofthe

topprism.

A

A

2

22

11 16

176

= ×

= cm

Calculatetheareaofthelargefrontfaceon

thebottom.

A

A

3

32

25 14

350

= ×

= cm

Calculatetheareaofthesidefaceon

thebottom.A

A

4

42

11 14

154

= ×

= cm

Calculatetheareaofthebottomofthefigure.A

A

5

52

25 11

275

= ×

= cm

Calculatetheareaofthespaceontopof

thelargerrectanglethatisfilledbythe

smallerfigure.

A

A

6

62

11 11

121

= ×

= cm


SA=4A1+4A2+2A3+2A4+2A5–2A6

SA=4×57.8+4×176+2×350+2×154+

2×275–121

SA=231.2+704+700+308+550–121

SA=2372.2cm2

Thesurfaceareaofthefigureis2372.2cm2.



V lwh

V

V

1

1

13

13

13

11 11 9

363

=

= × × ×

= cm

Calculatethevolumeofthetop

rectangularprism.

V lwh

V

V

2

2

23

11 11 16

1936

=

= × ×

= cm

Calculatethevolumeofthebottom

rectangularprism.

V lwh

V

V

3

3

33

25 11 14

3850

=

= × ×

= cm


V V V V

V

V

= + +

= + +

=

1 2 3

3

363 1936 3850

6149 cm

Thevolumeofthefigureis6149cm3.

6. Calculatethevolumeofthepile.

V r h

V

V

=

= × ( ) ×

≈

13

13

6 82

2 8

33 9

2

2

3

π

π . .

. m

Thevolumeofthepileis33.9m3.

7. Calculatetheareaofonefaceofthepyramid.

A bh

A

A

1

1

12

12

12

18 12

108

=

= × ×

= m


SA A

SA

SA

=

= ×

=

4

4 108

432

1

2m

Calculatethecostoftheroofingmaterial.

432 5 75 2484× =. $

Thecostoftheroofingmaterialis$2484.00.


Chapter 4Trigonometry of Right Triangles

4.1 Solving for Angles, Lengths, and Distances

BuiLD Your SkiLLS, p. 200

1. a) tan ..

tan .

tan ( . )

.

θ

θ

θ

θ

=

=

=

= °

−

7 912 3

0 6423

0 6423

32 7

1

Thetrigonometricratioequals0.6423and

θis32.7°.

b) sin ..

sin .

sin ( . )

.

θ

θ

θ

θ

=

=

=

= °

−

1 82 3

0 7826

0 7826

51 5

1


θis51.5°.

c) cos ..

cos .

cos ( . )

.

θ

θ

θ

θ

=

=

=

= °

−

5 57 75

0 7097

0 7097

44 8

1


θis44.8°.

2. FindthevalueofhusingthePythagoreantheorem.

c a b

h

h

h

2 2 2

2 2 2

2 2 2

2

19 8 10 6

19 8 10 6

19 8 10

= +

= +

− =

= −

. .

. .

. .66

392 04 112 36

279 68

16 7

2

h

h

h

= −

=

=

. .

.

. cm

Thevalueofhis16.7cm.

3. Usethetanratiotosolveforthevalueofthe

sharedside.

tan

tan

.

3519

35 19

13 3

° =

= ° ×

≈

y

y

y cm

UsethePythagoreantheoremtosolveforx.

x

x

x

x

x

2 2 2

2 2

33 13 3

33 13 3

1089 176 89

1265 89

= +

= +

= +

=

=

.

.

.

.

335 6. cm

Thevalueofxis35.6cm.

4. Usethetanratiotosolvefortheotheranglein

thetriangle.

tan

tan

α

α

α

=

= )(≈ °

−

811

811

36

1

Usetheanglecalculatedtosolveforθ.

θ=180°–90°–36°

θ=54°

Thevalueofθis54°.

5. Usethecosratiotosolveforx.

cos .

.cos

.

28 6 3

6 328

7 1

° =

= °=

x

x

x m

Usethevalueofxandthesinratioto

solvefory.

sin.

sin. .

sin.

. si

280 8

280 8 7 1

287 9

7 9

° = +

° = +

° =

= ×

yx

y

y

y nn

.

28

3 7

°

=y m

Thevalueofxis7.1mandthevalueof

yis3.7m.

6. Usethetanratiotosolvefortheangleofoffset.

tan ..

tan ..

.

θ

θ

θ

=

= )(= °

−

12 614 9

12 614 9

40 2

1

Usetheangleofoffsetandthesinratiotofind

thelengthofthetravelpipe.

sin . .

.sin .

.

40 6 12 6

12 640 6

19 4

° =

= °=

t

t

t cm

Theangleofoffsetis40.2°andthelengthof

thetravelpipeis19.4cm.

ALTErNATE SoLuTioN

UsethePythagoreantheoremtofindthelength

ofthetravelpipe.

t

t

t

t

2 2 2

2 2

12 6 14 9

12 6 14 9

158 76 222 01

380

= +

= +

= +

=

. .

. .

. .

..

.

77

19 5t = cm

7. Usethetanratiotosolveforthedepthof

theravine.

tan

tan

.

52120

120 52

153 6

° =

= × °

=

d

d

d m

Thedepthoftheravineis153.6m.

8. Usethetanratiotosolveforthewidthof

theriver.

tan

tan

.

28500

500 28

265 9

° =

= × °

=

AB

AB

AB m

Thewidthoftheriveris265.9m.

9. Usethesinratiotosolvefortheleftside.

sin . .

.sin .

.

68 1 1 75

1 7568 1

1 89

° =

=°

=

x

x

x m

MathWorks11WorkbookSolutions Chapter4 TrigonometryofRightTriangles 73

Usethesinratiotosolvefortherightside.

sin . .

.sin .

.

49 4 1 75

1 7549 4

2 30

° =

= °=

y

y

y m

Usethevaluescalculatedtofindtheperimeter.

ℓ=5+2.75+1.89+2.30

ℓ=11.94m

Paulinewillneedtobuild11.94moffence.

10.Findtheangleoftherightsideofthetriangle.

θ=131.6°–90°

θ=41.6°

Usetheanglecalculatedandthesinratioto

findtheheightofthetriangle.

sin .

.

41 63

2 00

° =

=

h

h m

Calculatetheareaofthetriangle.

a

a

t

t

=

=

12

2 8

8 2

( )( )

m

Calculatetheareaofthesquare.a

a

s

s

= ×

=

8 8

64 2m

Calculatethetotalarea.a

a

= +

=

8 64

72 2m

Soo-Jinwillneedtobuy72m2.

PrAcTiSE Your NEw SkiLLS, p. 210

1. a) Usethetanratiotosolveforx.

tan.

.

629 8

18 4

° =

=

x

x m

b) Usethecosratiotosolveforx.

cos.

.

7825 9

5 4

° =

=

x

x m

c) UsethePythagoreantheoremtosolveforx.

x

x

x

x

x

2 2 2

2 2

8 5 19 5

8 5 19 5

72 25 380 25

452 5

= +

= +

= +

=

. .

. .

. .

.

== 21 3. cm

d) Usethetanratiotosolveforx.

tan ..

tan ..

.

x

x

x

=

= ( )= °

−

1 519 5

1 519 5

4 4

1

e) Usethesinratiotosolveforx.

sin ..

sin ..

.

x

x

x

=

= ( )= °

−

96 898 1

96 898 1

80 7

1

f) Usethesinratiotosolveforx.

sin .

.sin

.

48 14 3

14 348

19 2

° =

= °=

x

x

x cm

g) Usethecosratiotosolveforx.

cos .

.cos

.

66 3 5

3 566

8 6

° =

= °=

x

x

x m

h) Usethecosratiotosolveforx.

cos ..

cos ..

.

x

x

x

=

= ( )= °

−

3 15 8

3 15 8

57 7

1


2. Usethecosratiotofindthelengthof

theladder.

cos .

.cos

.

65 1 8

1 865

4 3

° =

= °=

m

Thelengthoftheladderis4.3m.

3. Usethetanratiotofindtheheightofthetree.

tan.

. tan

.

3812 4

12 4 38

9 7

° =

= × °

=

t

t

t m

Theheightofthetreeis9.7m.

4. Usethesinratiotofindtheangleofdepression.

sin

sin

.

θ

θ

θ

=

= ( )= °

−

200350

200350

34 8

1

Usethecosratiotofindthe

horizontaldistance.

cos .

cos .

.

34 8350

350 34 8

287 4

° =

= × °

=

h

h

h m

Theangleofdepressionis34.8°andthe

horizontaldistancetravelledis287.4m.

5. Weknowthattheroofoftheplayhouseis

anisoscelestrianglebecauseeachlengthof

theroofisequivalent.Whenyoudividean

isoscelestriangleintotwoeventriangleyou

knowthateachwillhavethesamedimensions.

Thebaseofthewholetriangleis4ftandeach

legsofthesmallertriangleswillbeequalto

halfofthebaseofthewholetriangle.

4 2 2÷ =

Solvefortheangleofthebaseofthe

smallertriangle.

127°–90°=37°

Usethevaluescalculatedandthecosratioto

solvefora.

cos

cos

.

37 2

237

2 5

° =

= °=

a

a

a ft

Usethetanratiotosolveforb.

tan

tan

.

372

2 37

1 5

° =

= × °

=

b

b

b ft

Usetheknowledgeofthepropertiesoftriangles

tosolveforc.

c=180°–90–37°

c=53°

Usethevaluewecalculatedforbtosolveford.

d=5–1.5

d=3.5ft


4.2 Solving complex Problems in the real world

BuiLD Your SkiLLS, p. 215

1. Usethetanratioforthebaseofthelarger

triangle.

tan .

.tan

.

37 4 9

4 937

6 5

° =

= °=

y

y

y m

Usethevalueyoucalculatedandthecosratio

tosolveforx.

cos.

. cos

.

376 5

6 5 37

5 2

° =

= × °

=

x

x

x m

Thevalueofxis5.2m.

2. a) UsethePythagoreantheoremtosolveforx.

x

x

x

x

x

2 2 2

2 2

9 4 8 6

9 4 8 6

88 36 73 96

162 32

1

= +

= +

= +

=

=

. .

. .

. .

.

22 7. cm

Usethevaluecalculatedforxandthecos

ratiotosolvefory.

cos .

.cos

.

48 12 7

12 748

19 0

° =

= °=

y

y

y cm

b) Usethetanratiotosolveforbothpartsofx.

tan.

. tan

.

tan.

326 7

6 7 32

4 19

486 7

1

1

1

2

2

° =

= × °

=

° =

=

x

x

x

x

x 66 7 48

7 44

4 19 7 44

11 6

2

1 2

. tan

.

. .

.

× °

=

= +

= +

=

x

x x x

x

x m

Usethecosratiotosolvefory.

cos .

.cos

.

48 6 7

6 748

10 0

° =

= °=

y

y

y m

3. Usethesinratiotosolveforh.

sin.

. sin

.

5321 4

21 4 53

17 1

° =

= × °

=

h

h

h cm

Solveforthehypotenuseofthelowertriangle

usingthesinratio.

sin .

.sin

.

53 8 6

8 653

10 8

1

1

1

° =

= °=

H

H

H


Usethevaluecalculatedtofindthehypotenuse

oftheuppertriangle.

H2=21.4–10.8

H2=10.6cm

Usethepropertiesoftrianglestocalculatethe

valueofthetopangleinthetoptriangle.

θ=180°–90°–53°

θ=37°

Usetheanglecalculated,thelengthofthe

hypotenuseandthecosratiotosolveforx.

cos.

. cos

.

3710 6

10 6 37

8 5

° =

= × °

=

x

x

x cm

Usingthesinratiosolvefory.

sin.

. sin

.

3710 6

10 6 37

6 4

° =

= × °

=

y

y

y cm

ALTErNATE SoLuTioN

Alternatelyyoucouldcalculatetheotherangle

intheuppertriangle.Startbycalculatingthe

otherangleinthelowertrianglebyusingthe

propertiesoftriangles.α

α

= 180° − ° − °

= °

90 53

37

Usethepropertiesofstraightlinestocalculate

thelowerangleoftheuppertriangle.β

β

= ° − ° − °

= °

180 90 37

53

Usethevalueofthelowerangleandthesin

ratiotocalculatethevalueofx.

sin.

. sin

.

5310 6

10 6 53

8 5

° =

= × °

=

x

x

x cm

Usethevalueofthelowerangleandthecos

ratiotocalculatethevalueofy.

cos.

. cos

.

5310 6

10 6 53

6 4

° =

= × °

=

x

x

x cm

4. Usethetanratiotocalculatethehorizontal

distancebetweenbuildings.

tan

tan

.

25 200

20025

428 9

° =

= °=

x

x

x m

Usethetanratioandthehorizontaldistanceto

calculatethedifferenceinheightsofB1andB2.

tan.

. tan

.

40428 9

428 9 40

359 9

° =

= × °

=

h

h

h m

CalculatetheheightofB2.

HeightofB2=359.9+200

HeightofB2≈560m

TheheightofB2isapproximately560m.

5. a) Findtheheightoftheladderatitsshortest.

sin

sin

.

6518

18 65

16 3

1

1

1

° =

= × °

=

h

h

h ft

Findtheheightoftheladderatitslongest.

sin

sin

.

6532

32 65

29 0

2

2

2

° =

= × °

=

h

h

h ft

Findthedifferencesinthetwoheights.

x=29.0–16.3

x=12.7ft


Theladderreaches12.7ftfurtherupthe

buildingwhenitisfullyextended.

b) Findthedistanceoftheladderat

itsshortest.

cos

cos

.

6518

18 65

7 6

1

1

1

° =

= × °

=

d

d

d ft

Findthedistanceoftheladderat

itslongest.

cos

cos

.

6532

32 65

13 5

2

2

2

° =

= × °

=

d

d

d ft

Findthedifferencesinthetwodistances.

x=13.5–7.6

x=5.9ft

Thebaseoftheladdermustbe5.9ft

furtherfromthebuildingwhenitisfully

extended.

6. Therearetwosituations.

180m

32° 50°

180m

32° 50°

SolveforthedistanceZolaisfromthecell

phonetower.

tan

tan

32 180

18032

288

° =

= °≈

z

z

z m

SolveforthedistanceNaeemisfromthecell

phonetower.

tan

tan

50 180

18050

151

° =

= °≈

n

n

n m

Calculatetheirdistanceiftheyareonthesame

sideofthetower.

d1=288–151

d1=137m

Iftheyareonthesamesideofthetowerthey

are137mapart.

Calculatetheirdistanceiftheyareonopposite

sideofthetower.

d2=288+151

d2=439m

Iftheyareontheoppositesideofthetower

theyare439mapart.

7. a) Solveforthefirstsegmentusingthe

tanratio.

tan .

.tan

.

25 14 9

14 925

32 0

1

1

1

° =

= °=

x

x

x m

Thesecondsegmentis8m.


Solveforthethirdsegmentusingthe

tanratio.

tan .

.tan

.

47 26 8

26 847

25 0

3

3

3

° =

= °=

x

x

x m

Addeachhorizontaldistancetogetherto

findthetotal.

d=32.0+8+25.0

d=65.0m

Thetotalhorizontaldistancecoveredby

thisportionofthetrackis65.0m.

b) Thedistancetravelledbythecaroverthe

firstsegmentoftrackisthehypotenuse.

Usethesinratiotocalculatethe

hypotenuse.

sin .

.sin

.

25 14 9

14 925

35 3

1

1

1

° =

= °=

h

h

h m

Thesecondsegmentoftrack

coveredis8m.

Thedistancetravelledbythecaroverthe

thirdsegmentoftrackisthehypotenuse.

Usethesinratiotocalculatethe

hypotenuse.

sin .

.sin

.

47 26 8

26 847

39 3

3

3

3

° =

= °=

h

h

h m

Addeachdistanceoftrackthecartravels.

d=35.3+8+39.3

d=82.6m

Thecartravels82.6malonghisportionof

thetrack.

8. a) UsethePythagoreantheoremtocalculate

thestraight-linedistance.

d

d

d

d

d

2 2 2

2 2

185 100

185 100

34225 10000

44225

21

= +

= +

= +

=

≈ 00 3. km

b) Usethetanratioandthestraight-line

distancecalculatedtocalculatetheangleof

elevation.

tan.

tan.

.

θ

θ

θ

=

= ( )= °

−

7210 3

7210 3

1 9

1

9. Usethesinratiotocalculatethehorizontal

distancefromSylvietothetree,whichisthe

sameasthehypotenuseofthetrianglebetween

Mathieu,Sylvieandthetree.

sin

sin

.

73 89

8973

93 1

° =

= °=

d

d

d m

Usethetanratioandthedistancecalculatedto

calculatetheheightofthenest.

tan.

. tan

.

3593 1

93 1 35

65 2

° =

= × °

=

h

h

h m

Theheightofthenestis65.2m.


PrAcTiSE Your NEw SkiLLS, p. 223

1. a) Usethecosratiotosolvethevalueofx.

cos.

. cos

.

328 4

8 4 32

7 1

° =

= × °

=

x

x

x cm

Usethesinratiotosolvefory.

sin .

.sin

.

54 6 8

6 854

8 4

° =

= °=

y

y

y cm

b) Usethepropertyofrightanglesto

solveforθ.

θ=90°–57°

θ=33°

Usethecosratiotosolveforx.

cos .

.cos

57 21 8

21 857

40

° =

= °=

x

x

x cm

Usethecosratiotosolvefory.

cos

cos

.

33 40

4033

47 7

° =

= °=

y

y

y cm

c) Usethetanratiotosolveforx.

tan

tan

.

15 21

2115

78 4

° =

= °=

x

x

x ft

Usethesinratiotosolvefory.

sin

sin

.

15 21

2115

81 1

° =

= °=

y

y

y ft

Usethetanratiotosolveforθ.

tan ..

tan ..

θ

θ

θ

=

= ( )= °

−

70 678 4

70 678 4

42

1

2. Calculatethebaseofthetriangleofwhichxis

thehypotenuse.

2 2 4× = ft

UsethePythagoreantheoremtocalculatethe

valueofx.

x

x

x

x

x

2 2 2

2 2

4 7 5

4 7 5

16 56 25

72 25

8 5

= +

= +

= +

=

=

.

.

.

.

. ft

Calculatethebaseofthetriangleofwhichyis

thehypotenuse.

2 4 8× = ft

UsethePythagoreantheoremtocalculatethe

valueofy.

x

x

x

x

x

2 2 2

2 2

7 5 8

7 5 8

56 25 64

120 25

11 0

= +

= +

= +

=

≈

.

.

.

.

. ft


UsethetanratiotosolveforG.

tan .

tan .

.

G

G

G

=

= ( )≈ °

−

7 58

7 58

43 2

1

3. a) Calculatethedifferenceinheightsofthe

buildings.

147–109=38ft

Usethedifferenceinheightandthetan

ratiotocalculatethedistancebetweenthe

buildings.

tan

tan

.

41 38

3841

43 7

° =

= °=

x

x

x m

Thebuildingsare43.7mapart.

b) Usethetanratiotocalculatetheangleof

depression.

tan.

tan.

.

θ

θ

θ

=

= ( )= °

−

10943 7

10943 7

68 2

1

Theangleofdepressionis68.2°.

4. a) Usethetanratiotocalculatetheheight

ofthetree.

tan

tan

.

20100

100 20

36 4

° =

= × °

=

h

h

h m

Theheightofthetreeis36.4m.

b) Calculatethedistanceheisfromthetree

usingthetanratio.

tan .

.tan

.

36 36 4

36 436

50 1

° =

= °=

x

x

x m

Subtractthisvaluefromhisoriginal

distancetocalculatehowfarhepaddled.

d=100–50.1

d=49.9m

Hepaddled49.9mclosertothetree.

5. Usethetanratiotocalculatethedistancefrom

Point1tothebaseofthecliff.

tan

tan

38 200

20038

256

° =

= °≈

x

x

x m

Usethedistancecalculatedandthetanratioto

determinetheheightofthecliff.

tan

tan

32256

256 32

160

° =

= × °

≈

h

h

h m

Theheightofthecliffis160m.


chAPTEr TEST, p. 226

1. a) Usethesinratiotosolveforx.

sin.

. sin

.

473 8

3 8 47

2 8

° =

= × °

=

x

x

x m

b) UsethesinratiotosolveforX.

sin ..

sin ..

X

X

X

=

= ( )= °

−

4 37 5

4 37 5

35

1

c) UsethetanratiotosolveforX.

tan ..

tan ..

.

X

X

X

=

= ( )= °

−

3 81 1

3 81 1

73 9

1

d) Usethecosratiotosolveforx.

cos .

.cos

.

54 23 4

23 454

39 8

° =

= °=

x

x

x in

2. a) Usethecosratiotosolveforx.

cos .

.cos

.

42 7 5

7 542

10 1

° =

= °=

x

x

x cm

UsethetanratiotosolveforY.

tan ..

tan ..

.

Y

Y

Y

=

= ( )= °

−

10 19 2

10 19 2

47 7

1

b) UsethetanratiotosolveforX.

tan ..

tan ..

.

X

X

X

=

= ( )= °

−

4 23 1

4 23 1

53 6

1

Usethetanratiotosolvefory.

tan ..

. tan .

.

53 66 2

6 2 53 6

8 4

° =

= × °

=

y

y

y cm

UsethePythagoreantheoremtosolveforz.

z

z

z

z

z

2 2 2

2 2

8 4 6 2

8 4 6 2

70 56 38 44

109

10 4

= +

= +

= +

=

=

. .

. .

. .

. cm

c) UsethePythagoreantheoremtosolveforx.

x

x

x

x

x

2 2 2

2 2

4 4 3 9

4 4 3 9

19 36 15 21

34 57

5

= +

= +

= +

=

=

. .

. .

. .

.

.99 cm

UsethetanratiotosolveforY.

tan ..

tan ..

.

Y

Y

Y

=

= ( )= °

−

4 43 9

4 43 9

48 4

1

UsethePythagoreantheoremtosolveforz.

z

z

z

z

z

2 2 2

2 2

4 4 2 9

4 4 2 9

19 36 8 41

10 95

3 3

= −

= −

= −

=

=

. .

. .

. .

.

. cm


3. Dividethepatiointoarectangleandatriangle.

Calculatetheareaoftherectangle.a

a

r

r

= ×

=

8 19

152 2m

Tocalculatetheareaofthetriangleyouwill

needtoknowthebaseandtheheightofthe

triangle.Calculatethebaseofthetriangle.

b=19–6.5

b=12.5m

Theangleonthebottomofthetrianglecan

bedeterminedbysubtractingthevalueofa

straightanglefromthereflexangle.θ

θ

= ° − °

= °

205 180

25

Usethatangleandthesinratiotocalculatethe

heightofthetriangle.

sin.

. sin

.

258 3

8 3 25

3 5

° =

= × °

=

h

h

h m

Usethebaseandheightofthetriangleto

calculateit’sarea.

a bh

a

a

t

t

t

=

=

≈

12

12

12 5 3 5

21 9 2

( . )( . )

. m

Calculatethetotalarea.

a a a

a

a

t r= +

= +

=

152 21 9

173 9 2

.

. m

TheareaofPaolo’spatiois173.9m2.

4. Tosolvestudentswillneedtocalculatethe

lengthofthetopofthekiteandthelength

ofthebottomofthekiteandaddtheresults

together.

Usethecosratiotosolveforthelengthofthe

topofthekite.

cos

cos

.

742 50

50 37

39 9

1

1

1

°( ) =

= × °

=

cm

Usethecosratiotosolveforthelengthofthe

bottomofthekite.

cos

cos

.

802 62

62 40

47 5

2

2

2

°( ) =

= × °

=

cm

Addthetwolengthstogetherforthetotal.

= +

= +

=

1 2

39 9 47 5

87 4

. .

. cm

Thelengthofthekiteis87.4cm.

5. Usetheangleofdepressiontothetopofthe

buildingtocalculatethedifferenceinheightof

thebuilding.

tan

tan

7131

31 71

90

° =

= × °

=

x

x

x m

Subtractthedifferenceinheightsofthe

buildingsfromtheheightofthebuilding

Chung-hoisstandingon.

h=135–90

h=45m

Theheightoftheotherbuildingis45m.


6. a) Usethetanratiotocalculatetheheightof

thenest.

tan

tan

.

4275

75 42

67 5

° =

= × °

=

h

h

h m

Thenestis67.5mhigh.

b) FindhowfarBernhardtisfromthetree.

d

d

= +

=

75 30

105 m

Usethetanratiotofindtheangleof

elevation.

tan .

tan .

.

θ

θ

θ

=

= ( )= °

−

67 5105

67 5105

32 7

1

Theangleofelevationheseesthenest

atis32.7°.

7. a) UsethePythagoreantheoremtofindthe

straight-linedistance.

d

d

d

d

d

2 2 2

2 2

75 35

75 35

5625 1225

6850

82 8

= +

= +

= +

=

= . km

Thestraight-linedistancetotheairport

is82.8km.

b) Usethetanratiotosolvefortheangleof

elevation.

θ

θ

= ( )= °

−tan.

.

1 682 8

4 1

Theangleofelevationis4.1°.


Chapter 5Scale Representations

5.1 Scale Drawings and Models

BuilD Your SkillS, p. 233

1. a) Solveforx.

38

= x168

168 × 38

= x168

× 168

168 × 38

= x

63 = x

b) Solveforx.

x13

= 791

13 × x13

= 791

× 13

x = 791

× 13

x = 1

c) Solveforx.

x7

= 30105

7 × x7

= 30105

× 7

x = 30105

× 7

x = 2

d) Solveforx.

408x

= 49

x × 408x

= 49

× x

408 = 49

x

94

× 408 = 49

x × 94

94

× 408 = x

918 = x

e) Solveforx.

90198

= 5x

x × 90198

= 5x

× x

90198

x = 5

19890

× 90198

x = 5 × 19890

x = 5 × 19890

x = 11

MathWorks11WorkbookSolutions Chapter5 ScaleRepresentations 85

f) Solveforx.

34289

= 2x

x × 34289

= 2x

× x

34289

x = 2

28934

× 34289

x = 2 × 28934

x = 2 × 28934

x = 17

2. a) Solveforx.

56

= 12x

x × 56

= 12x

× x

56

x = 12

65

× 56

x = 12 × 65

x = 12 × 65

x = 14.4

b) Solvefork.

715

= 9k

k × 715

= 9k

× k

715

k = 9

157

× 715

k = 9 × 157

k = 9 × 157

k ≈ 19.3

c) Solveform.

1.24.9

= m7.3

7.3 × 1.24.9

= m7.3

× 7.3

7.3 × 1.24.9

= m

1.8 = m

d) Solveforp.

p85

= 7639

85 × p85

= 7639

× 85

p = 7639

× 85

p = 165.6

3. SetuparatioforTomandMary’sagegiventhe

informationprovided.

34

= 15M

M × 34

= 15M

× M

34

M = 15

43

× 34

M = 15 × 43

M = 20years

Maryis20yearsold

4. Setuparatioshowingtherelationshipbetween

thedistancesandthetimetravelled.

355d

= 78.5

d × 355d

= 78.5

× d

355 = 78.5

d

8.57

× 355 = 78.5

d × 8.57

8.57

× 355 = d

431km ≈ d


Georginawillhavetravelled431kmin

8.5hours.

5. Convertbothdistancestothesameunits.

1850km=185000000cm

Usethistodeterminethescalestatement.

scalestatement=map:original

scalestatement=3.7:185000000

scalestatement=1:5000000

6. a) Thescalestatementis200:1.

b) Determinethescalefactor.

scalefactor = modeloriginal

scalefactor = 2001

scalefactor = 200

Usethescalefactortocalculatethewidth

oftheoriginal.


200 = 2x

x × 200 = 2x

× x

200x = 2

x = 0.01cm

Theactualwidthofthehairis0.01cm.

7. Converttheheightstobethesameunits.

1.93m=193cm

Determinethescalestatement.

scalestatement=card:original


scalestatement=1:35.7

Thescaleusedtoprintthecardwas1:35.7.

8. Usetherelationshipabouttheheightofthe

manandtheheightsintheimagetosetupa

ratiotosolvefortheheightoftheflagpole.

2.3cm1.78m

= 7.6cmf

f × 2.31.78

= 7.6f

× f

2.31.78

f = 7.6

1.782.3

× 2.31.78

f = 7.6 × 1.782.3

f ≈ 5.9m

Theheightoftheflagpoleis5.9m.

9. a) Thepictureisapprox.4.75cmlong.

Convertthesizeofthebelugainto

centimetres.

4.2m=420cm


Thescalestatementis=picture:original


scalestatement=1:84

Thescalestatementforthepictureis1:84.

whichis47:420

b) Usethescalestatementtosetuparatioto

determinethelengthoftheactualalligator.

184

= 5.9a

a × 184

= 5.9a

× a

a84

= 5.9

a ≈ 496cmor5.0m

Thelengthofthealligatoris5.0m.


c) Convertthesizeoftheostrichinto

centimetres.

1.9m=190cm

Usethescalestatementtosetuparatio

todeterminetheheightoftheostrichin

thepicture.

184

= o190

190 × 184

= o190

× 190

19084

= o

2.3cm ≈ o

Theheightoftheostrichinthe

pictureis2.3cm

10.a) Usethescalestatementtodeterminethe

sizeofamite.

12.51

= 3.8m

m × 12.5 = 3.8m

× m

12.5m = 3.8

12.5m ÷ 12.5 = 3.8 ÷ 12.5

m ≈ 0.30cm

Theactualsizeofthemiteis

0.30cmor3mm.

b) Usethescalestatementtodeterminethe

heightofthedrawingofthecat.

12.51

= c30

30 × 12.5 = c30

× 30

375cm = c

Theheightofthecatwouldbe375cmor

3.75mifdrawnatthatscale.

c) Itisunlikelythatthesamescalewould

beusedasthesizesareverydifferent.For

thecat,youwouldprobablyuseascaleof

1:c.Thatis,youwouldrepresentitsmaller

thanitactuallyis.Theonlytimeyoumight

usethesamescaleforthemisif,forsome

reason,youneededtocomparehowsmall

themitewouldbeonthecat.

Practice Your New SkillS, p. 241

1. Converttheobjecttocentimetressothescale

canbedetermined.

7.8m=780cm

Determinethescalefactor.

scalefactor =pictureoriginal

scalefactor = 1.5780

scalefactor = 1520

Thescalefactorusedis

1520

.

2. Usetheschooltosolveforthelengthofthe

shoreline.

3cm100km

= xcm2719km

2719 × 3100

= x2719

× 2719

2719 × 3100

= x

81.6cm ≈ x

Thelengthoftheshorelineonthemapis.

3. a) Measurethemasterbedroomtoget

approximately1.69in.


Convertthemeasurefromfeettocentimetres.

12.5 × 12 = 150in

Calculatethescaleusingthemeasurement.

scalestatement=1.69in:150in

scalestatement=1:88.8

Thescaleusedtocreatethefloorplanis1:88.8

or13:1200.

b) Measurethefamilyroomtodetermine

theapproximatedimensions,whichare

1.375inby2.125in.Usethescale

statementtocalculatethedimensions.

1.375 × 88.8 = 122.1inor10.2ft

2.125 × 88.8 = 188.7inor15.7ft

Thedimensionsofthefamilyroomare

122.1in×188.7inor10.2ft×15.7ft.

c) Measurethebedroomtodetermine

theapproximatedimensions,which

are1.313inby1.125in.Usethescale

statementtocalculatethedimensions.

1.313 × 88.8 = 116.6inor9.7ft

1.125 × 88.8 = 99.9inor8.3ft

Thedimensionsofthesmallerbedroomare

116.6in×99.9inor9.7ft×8.3ft.

4. a) Convertthemeasureinmetresto

centimetres.

298m=29800cm




Thescaleusedtobuildthemodelwas

2504.4b.Calculatetheheightoftheantenna.

355 − 298 = 57m

Convertthemeasureinmetresto

centimetres.

57m:5700cm

Usethescalecalculatedtodeterminehow

longthemodelantennabe.

5700 ÷ 2504 ≈ 2.3cm

Theheightoftheantennainthemodelwill

be2.3cm.

5. Usethescaletocalculatetheactualdimensions

ofthebookcase.

7.8 × 30 = 234cm

5.4 × 30 = 162cm

1 × 30 = 30cm

Thedimensionsofthebookcaseis234cm×

162cm×30cm.

6. CalculatetheactualheightoftheTDCanada

TrustBuilding.

16.6 × 10 = 166m

CalculatetheactualheightoftheNexen

Building.

15.0 × 10 = 150m

Calculatethedifferenceinheights.

166–150=16m

Theactualdifferenceinheightsofthetwo

buildingsis16m.



1.

Front

Top

Side

2. a)

Front

b) No,theremightbesomeblockswecannot

see,hiddenbythetower.

3.1in

FrontView

7in

3in

4in

8in

24in

SideView

2in

10in

10in

4. Scale1:40

1m

102cm

1m

5cm

SideView

10cm

110cm

0.5m20cm 0.5m

10cm

104cm

FrontView

two-Dimensional representations5.2


5.

SideComponent

ShelfComponent

KickPlateComponent

a ×2 B ×4 c ×1

150cm

30cm

90cm

30cm

90cm

6cm

6. a) Measurethepatterntodeterminethe

scale.Whenmeasuringyoushouldget

approximately2in.Convertthemeasurein

feettoinches.

1foot=12inches

Determinethescale.

2in:12in

1:6

Thescaleusedtodrawthediagramis1:6.

b) Shecanmakeaquiltwhichwillhave

squarepatches,butitwillbesmallerthan

sheplannedbecausesheforgotaboutthe

seamallowance.

c) Calculatethelengthandwidthofeach

blockbyaddingtheseamallowanceon

eitherend.

12 + 3

8+ 3

8= 12 3

4in

Shemustcuteachblocktobe

12 3

4in × 12 3

4in

7. Asamplegraphmightlooklikethis.

Scale1:20

a

B ×480cm

150cm

90cm6cm

PractiSe Your New SkillS, p. 252

1.

TopView

FrontViewSideView

2. a)

TopView

FrontViewSideView


b) No,therecouldbeadditionalblocks

orgapsthatwecannotseefromthe

currentdiagram.

3.

TopView

FrontView

Scale1:6

4. a) Thewidthofthebuildingsinthediagram

isapproximately2cmandtheheightis

approximately4cm.Converttheactual

widthandheighttocentimetres.

7m=700cm

14m=1400cm

Determinethescaleusingthewidth.

scalestatement = 2:700


Thoughnotabsolutelynecessaryyoucan

confirmyourcalculationsbyrecalculating

thescaleusingtheheight.



Thescaleusedtodrawthediagramis1:350.

b) Calculatethedimensionsofthe

modelhouses.

700 ÷ 100 = 7cm

1400 ÷ 100 = 14cm

Themodelswillbe7cmwideby14cm

tallhigh.

c) No,wedonotknowwhatthebackpartsof

thehouseslooklike.

5. a) Shewillneedtoprovideafrontandatop

view.Shedoesnotneedtoprovideside

viewsbecauseallthesidesarethesame.

Shedoesnotneedtoprovideatopview,

becausetheroofcanbeseenonthefront

view,sinceallthesidesarethesame.

b)

Scalestatementswillvarybasedonthe

measurements.


6. Scale1:20

0.7m

0.9m

Front

0.55m

0.9m

Lid

0.3m

0.9m

Top

0.9m

0.9m

Back

0.8m

0.9m

Base

0.7m 0.9m

0.9m0.55m

0.8m

Side×2

7. Scale1:5

a ×2

22cm

5cm

B ×2

62cm

5cm

c ×10

40cm

2cm


5.3 three-Dimensional representations


1. Usethemeasurementprovidedtodetermine

thescale.Thereare9dotsthatequal18feet.

9dots:18ft

1:2

Thescaleis1dot:2feet.

Countallthedesiredlengths.

x = 1

y = 5

z = 8

h = 4

Usingthescalecalculatethelengths.

x = 1 × 2

x = 2ft

y = 5 × 2

y = 10ft

z = 8 × 2

z = 16ft

h = 4 × 2

h = 8ft

2. Scale1:20

3. Scale1:60


4. A

B

Thetwowillnotlookexactlyalikebecauseit

dependsonwhereyouplaceyourvanishing

pointandhowdeepyouchoosetomakeyour

box.Thus,evenifyouchoosetomakethem

thesamedepth,theywilllookslightlydifferent

becauseoftheangletothevanishingpoint.

5. V

6.

7.

8.


9.


1. Seediagramonthenextpage.

2.

3. a)

b)

c) Theperspectiveandtheanglestothe

vanishingpointaredifferent.

4.

B ×3 20cm

14cm

Scale1:5

a ×232cm

20cm

5. a) Scale1:5

10cm

8cm

a ×4 10cm

10cm

B ×4

6cm

6cm

c ×2

b)


5.3Three-DimensionalRepresentations,PractiseYourNewSkills,Question1.

Scale1:1


chaPter teSt, p. 273

1. a)

1128

Usethescaletodeterminethescalefactor.


scalefactor = 1128

Thescalefactorofthemodelis

1128

.

b) Usethescaletodeterminethewingspanof

theactualairplane.

30 × 128 = 3840cmor38.4m

Thewingspanoftheactualairplaneis

3840cmor38.4m.

2. Convertthekilometrestocentimetres.

350 × 100000 = 35000000cm

Simplifytodeterminethescale.

7cm:35000000cm

7:35000000

1:5000000

7:35000000Thesimplestexpressionofthe

scaleis1:5000000.

3. Setuparatiotosolvefortheheightof

theflagpole.

4.81.3

= h190

190 × 4.81.3

= h190

× 190

190 × 4.81.3

= h

702cm ≈ h

Theheightoftheflagpoleis702cm.

alterNate SolutioN

Itispossibletogetthesamesolutionifyouset

theratiosupdifferently.

h4.8

= 1901.3

4.8 × h4.8

= 1901.3

× 4.8

h = 1901.3

× 4.8

h = 702cm

Theheightoftheflagpoleis702cm.

702cmorapprox7m

4. a) Usethemeasurementsprovidedto

determinethescalestatement.

scalestatement=7cm:105cm

scalestatement=1:15

Thescaleofthedollhouseis1:15.

b)1

15Usethescalestatementtodetermine

thescalefactor.

scalefactor = dollhouseoriginal

scalefactor = 115

ThescalefactorCharlotteisusingis1

15 .

c) Multiplytheactualheightofthewindowto

determinetheheightofthewindowinthe

dollhouse.

h = 126 × 115

h = 8.4cm

Theheightofthewindowis8.4cm.


5.

Side Front

Top

6.

7.

8. a) Scale1:10

45cm

25cm

a ×1

27cm

7cm

B ×4

Scalestatementswilldifferdependingon

thediagrams.

b)


Chapter 6Financial Services

6.1 Choosing an Account


1. a) TheValueAccountallows10self-service

transactions.Additionalself-service

transactionscost$0.50each,andallteller-

assistedtransactionscost$1.00each.

CalculateFrédéric’stotalnumberofself-

servicetransactions.

2+4+5+10=21

Tenofthosetransactionswillbefree.He

willbecharged$0.50eachfortheother11

transactions.

11×$0.50=$5.50

Hemaintainstherequiredminimum

monthlybalance,sohisaccountfeeis

waived.Hewillbechargedonly$5.50.

b) TheSelf-serviceAccountallows25free

self-servicetransactions.Frédéricmakes

fewertransactionsthanthis,sohewillnot

havetopayanytransactionfees.Healso

maintainstherequiredminimummonthly

balance,sohewillnotbechargedthe

accountfee.

Hewillnothavetopayservicecharges.

2. a) TheBonusSavingsAccountallowsonly2

debittransactions.Allothertransactions

cost$1.25each.TwoofThom’stransactions

willbefree,andall27otherswillbe

chargedat$1.25each.

27×$1.25=$33.75

Thereisnoaccountfee,soThomwillowe

$33.75inservicecharges.

b) TheFull-serviceAccountallows40free

transactions,soThomwillnothavetopay

anytransactionfees.However,hedoesnot

maintaintheminimummonthlybalance,

sohewillhavetopaytheaccountfee

of$24.50.

3. TheValueAccountallows10self-service

transactions.Additionalself-service

transactionscost$0.50each,andallteller-

assistedtransactionscost$1.00each.

LiPing’s8self-servicetransactionsarefree.

Her2teller-assistedtransactionswillcost

$1.00each.Shedoesnotmaintaintherequired

minimummonthlybalance,soshewillhave

topaythe$3.90accountfee.LiPingwill

thereforeoweatotalof$5.90infees.

4. Transaction Description Withdrawal Deposit Balance$1798.53

Directdeposit Paycheque $1432.51 $3231.04

ATM Cash $200.00 $3031.04

Bankcard Groceries $63.95 $2967.09

Bankcard Clothes $75.32 $2891.77

Bankcard Movie $24.50 $2867.27

Teller Hydrobill $89.56 $2777.71

Directdeposit Paycheque $1432.50 $4210.21

ATM Cash $100.00 $4110.21

Auto-withdrawal Loanpayment $375.86 $3734.35

Bankcard Groceries $154.32 $3580.03

Bankcard Gas $56.23 $3523.80

Bankcard Dinner $25.38 $3498.42

Auto-withdrawal Rent $575.00 $2923.42

Bankcard Books $123.45 $2799.97

Auto-withdrawal Monthlybankingfee $24.50 $2775.47

Mitramaintainstheminimummonthlybalanceof$2000.00,soshewillnothavetopaytheaccount

fee.Herendofmonthbalanceis$2775.47.


ATM Cash $50.00 $532.98–$50.00=$482.98

Bankcard Movie/snack $18.54 $482.98–$18.54=$464.44

Auto-deposit Paycheque $238.21 $464.44+$238.21=$702.65

Auto-payment Loan $385.21 $702.65–$385.21=$317.44

Bankcard Groceries $115.87 $317.44–$115.87=$201.57

ATM Cash $100.00 $201.57–$100.00=$101.57

Auto-withdrawal Accountfee $3.90 $101.57–$3.90=$97.67

Marek’saccountallows10freeself-servicetransactions,sohedoesnothavetopayanytransaction

fees.However,hedoesnotmaintaintherequiredminimummonthlybalance,sohedoeshavetopay

theaccountfeeof$3.90.Hisfinalbalancewillbe$97.67.

MathWorks11WorkbookSolutions Chapter6 FinancialServices 101


ATM Cash $40.00 $785.53–$40.00=$745.53

ATM Cash $100.00 $745.53–$100.00=$645.53

Bankcard Groceries $45.98+$1.25servicecharge

$645.53–$45.98–$1.25=$598.30

Bankcard Movie $12.95+$1.25servicecharge

$598.30–$12.95–$1.25=$584.10

ATM Babysitting $1.25servicecharge $30.00 $584.10+$30.00–$1.25=$612.85

Bankcard Snacks $9.59+$1.25servicecharge

$612.85–$9.59–$1.25=$602.01

Directdeposit Paycheque $1.25servicecharge $187.37 $602.01+$187.37–$1.25=$788.13

Auto-withdrawal Telephone $58.82+$1.25servicecharge

$788.13–$58.82–$1.25=$728.06

ATM Cash $100.00+$1.25servicecharge

$728.06–$100.00–$1.25=$626.81

Auto-withdrawal Accountfee – – $626.81

WiththeBonusSavingsAccount,onlyNoah’sfirsttwotransactionsarefree.Eachadditional

transactioncosts$1.25.However,thereisnomonthlyaccountfee.Noah’sfinalbalanceis$626.81.

PrACtiSe Your New SkillS, p. 287

1. CalculateGeorgina’sfees.

($0.50×8)+($2.00×3)=$10.00

2. Rolfe’sSelf-serviceAccountallows25freeself-

servicetransactions,sohewillnothavetopay

anytransactionfees.

CalculateRolfe’sminimummonthly

balance.Sinceyoudon’tknowtheorderof

histransactions,assumehemadeallthe

withdrawalsbeforedepositinganymoney.

$2675.32–(6×$100.00)=$2075.32

Therequiredminimummonthlybalanceis

$1500.00,andRolfedefinitelyhadthatmuch

inhisaccountatalltimessohewillnothaveto

paythemonthlyaccountfee.

Rolfewillnothavetopayanyfeesthismonth.

3. IfPatricemaintainshiscurrent

spendinghabits,hewillprobablyhave

aminimummonthlybalanceofatleast

$2000.00.Hemakesabout15self-service

transactionsamonth.

UseatabletoshowPatrice’sfeeswitheachtype

ofaccount.(Seetableonnextpage)

PatriceshouldgetaFull-serviceAccountora

Self-serviceAccount.TheFull-serviceAccount

giveshimthemostfreetransactionssoitwould

givehimthemostflexibility,buthewouldneed

tocontinuetomaintainaminimummonthly

balanceof$2000.00.


Value Account Self-service Account

Full-service Account

Bonus Savings Account

Monthlyfee Waived Waived Waived Nofee

Transactionfees 10free

5×$0.50=$2.50

25free 40free 2free

13×$1.25=$16.25

Totalcost $2.50 Free Free $16.25

4. a) Jeanette’sValueAccountallows10freeself-servicetransactions.Eachadditionalself-service

transactionwillcost$0.50andteller-assistedtransactionswillcost$1.00.

Transaction Description Withdrawal Deposit Balance$4986.54

ATM Cash $250.00 $4986.54–$250.00=$4736.54

Bankcard Dinner $25.32 $4736.54–$25.32=$4711.22


Directdeposit Paycheque $524.66 $4565.29+$524.66=$5089.95

Bankcard Movie $12.98 $5089.95–$12.98=$5076.97

ATM Cash $100.00 $5076.97–$100.00=$4976.97

Bankcard Gas $48.96 $4976.97–$48.96=$4928.01

Teller Utilities$123.23+$1.00servicecharge $4928.01–$123.23–$1.00=$4803.78

Auto-withdrawal Rent $550.00 $4803.78–$550.00=$4253.78

Directdeposit Paycheque $524.65 $4253.78 –$524.65=$4778.43


ATM Cash$100.00+$0.50servicecharge $4613.19–$100.00–$0.50=$4512.69

Teller Phone$47.25+$1.00servicecharge $4512.69–$47.25–$1.00=$4464.44

Bankcard Misc.$12.32+$0.50servicecharge $4464.44–$12.32–$0.50=$4451.62

Bankcard Dinner$15.88+$0.50servicecharge $4451.62–$15.88–$0.50=$4435.24

ATM Cash$200.00+$0.50servicecharge $4435.24–$200.00–$0.50=$4234.74

Bankcard Prescription$32.54+$0.50servicecharge $4234.74–$32.54–$0.50=$4201.70

b) No,sheshouldswitchtoaFull-serviceAccountbecauseshekeepsenoughmoneyinheraccount

forthefeetobewaivedandshewouldhavemoretransactionsincludedforfreewithheraccount.


5. UseatabletocalculateMariya’sbalance.HerValueAccountallows10freeself-servicetransactions.

Eachadditionalself-servicetransactionwillcost$0.50andteller-assistedtransactionswillcost$1.00.

(NotethatMariyalikelydidnotcompletethetransactionsinthisorder,becausetheyresultina

significantnegativebalancemid-month,butthisdoesnotmatterforcalculatingthefinalbalanceat

theendofthemonth.)


ATM Cash $200.00 $432.98–$200.00=$232.98

ATM Cash $200.00 $232.98–$200.00=$32.98

ATM Cash $200.00 $32.98–$200.00=–$167.02

TellerPhonebillpayment

$68.21+$1.00servicecharge –$167.02–$68.21–$1.00=–$236.23

Bankcard Groceries $103.56 –$236.23–$103.56=–$339.79

Bankcard Movie $12.87 –$339.79–$12.87=–$352.66

Bankcard Meal $15.89 –$352.66–$15.89=–$368.55

Bankcard Meal $22.48 –$368.55–$22.48=–$391.03

Directdeposit Paycheque $658.42 –391.03+$658.42=$267.39

OnlineCreditcardbillpayment $243.56 $267.39–$243.56=$23.83

Auto-withdrawal Accountfee $3.90 $23.83–$3.90=$19.93

Mariyadidnotmaintaintheminimummonthlybalanceof$1000.00,soshedoeshavetopaythe

$3.90accountfee.Herclosingbalanceis$19.93.


6.2 Simple and Compound interest


1. a) I=Prt

I=($1000.00)(0.025)(1)

I=$25.00

b) I = Prt

I =($1000.00)(0.05)(1)

I =$50.00

c) I = Prt

I =($1000.00)(0.025)(2)

I =$50.00

d) I = Prt

I =($2000.00)(0.025)(1)

I =$50.00

e) Whentheprincipalandthetermstaythe

samebuttheratedoubles—forexample,

fromquestiona)toquestionb)above—the

amountofinterestearneddoubles.

f) Whentheprincipalandtheratestaythe

samebutthetermdoubles—forexample,

fromquestiona)toquestionc)above—the

amountofinterestearneddoubles.

2. I = Prt

I =($600.00)(0.0375)(5)

I =$112.50

A = P+I

A =$600.00+$112.50

A =$712.50

3. I = Prt

I =($1000.00)(0.0450)(10)

I =$450.00

A = P+ I

A =$1000.00+$450.00

A =$1450.00

4.

A = P 1 + rn( )nt

A = ($5000.00) 1 + 0.031( )1× 2

A = ($5000.00)(1.03)2

A = $5304.50

5. Option1:$4000.00investedat3.50%per

annum,compoundedannually,for3years

A = P 1 + rn( )nt

A = ($4000.00) 1 + 0.0351( )1× 3

A = ($4000.00)(1.035)3

A ≈ $4434.87

Option2:$4000.00investedat3.50%simple

interestfor3years

I = Prt

I =($4000.00)(0.0350)(3)

I =$420.00

A = P+ I

A =$4000.00+$420.00

A =$4420.00


difference=Option1–Option2

difference=$4434.87–$4420.00

difference=$14.87

6.

A = P 1 + rn( )nt

A = ($8000.00) 1 + 0.0251( )1× 5

A = ($8000.00)(1.025)5

A ≈ $9051.27

I =A – P

I =$9051.27–$8000.00

I =$1051.27

7. a)

A = P 1 + rn( )nt

A = ($4000.00) 1 + 0.041( )1× 8

A = ($4000.00)(1.04)8

A ≈ $5474.28

b)

A = P 1 + rn( )nt

A = ($4000.00) 1 + 0.042( )2× 8

A = ($4000.00)(1.02)16

A ≈ $5491.14

c)

A = P 1 + rn( )

nt

A = ($4000.00) 1 + 0.044( )

4 × 8

A = ($4000.00)(1.01)32

A ≈ $5499.76

d)

A = P 1 + rn( )nt

A = ($4000.00) 1 + 0.0412( )12× 8

A = ($4000.00)(1.00333)32

A ≈ $5505.58

8. Compoundedannually:

A = P 1 + rn( )nt

A = ($10 000.00) 1 + 0.03751( )1×1

A = ($10 000.00)(1.0375)

A = $10 375.00

Compoundeddaily:

A = P 1 + rn( )nt

A = ($10 000.00) 1 + 0.0375365( )365×1

A ≈ $10 382.10

difference=daily–annually

difference=$10382.10–$10375.00

difference=$7.10

9. Theinvestmentiscompoundedquarterly,so

thetermofeachinterestperiodisone-quarter

ofayear,or0.25.

Interest period

Investment value at beginning of period

Interest earned (I = Prt)

Investment value at end of period

1 $3000.00$3000.00×0.0325×0.25≈$24.38

$3000.00+$24.38=$3024.38

2 $3024.38$3024.38×0.0325×0.25≈$24.57

$3024.38+$24.57=$3048.95

3 $3048.95$3048.95×0.0325×0.25≈$24.77

$3048.95+$24.77=$3073.72

4 $3073.72$3073.72×0.0325×0.25≈$24.97

$3073.72+$24.97=$3098.69

5 $3098.69$3098.69×0.0325×0.25≈$25.18

$3098.69+$25.18=$3123.87

6 $3123.87$3123.87×0.0325×0.25≈$25.38

$3123.87+$25.38=$3149.25

7 $3149.25$3149.25×0.0325×0.25≈$25.59

$3149.25+$25.59=$3174.84

8 $3174.84$3174.84×0.0325×0.25≈$25.80

$3174.84+$25.80=$3200.64


Studentscanchecktheirfinalanswerbyusing

thecompoundinterestformula.

A = P 1 + rn( )nt

A = ($3000.00) 1 + 0.03254( )4 × 2

A ≈ $3200.64

10.a) Yearstodoubleinvestment=72÷(interest

rateasapercent)

Yearstodoubleinvestment=72÷4

Yearstodoubleinvestment=18

Itwilltakeabout18yearsforthe

investmenttodoubleinvalue.

b) Yearstodoubleinvestment=72÷(interest

rateasapercent)

Yearstodoubleinvestment=72÷2.45

Yearstodoubleinvestment=29.4

Itwilltakeabout29.4yearsforthe


c) Yearstodoubleinvestment=72÷(interest

rateasapercent)

Yearstodoubleinvestment=72÷1.95

Yearstodoubleinvestment=36.9




1. a) I = Prt

I =($400.00)(0.0125)(8)

I =$40.00

A = P+ I

A =$400.00+$40.00

A =$440.00

b) I = Prt

I =($750.00)(0.0275)(5)

I =$103.13

A = P+ I

A =$750.00+$103.13

A =$853.13

c) I = Prt

I =($1000.00)(0.0450)(10)

I =$450.00

A = P+ I

A =$1000.00+$450.00

A =$1450.00

d) I = Prt

I =($1200.00)(0.0395)(9)

I =$426.60

A = P+ I

A =$1200.00+$426.60

A =$1626.60

2. a)

A = P 1 + rn( )nt

A = ($400.00) 1 + 0.012512( )12× 8

A ≈ $442.05

b)

A = P 1 + rn( )nt

A = ($750.00) 1 + 0.027512( )12× 5

A ≈ $860.42

c)

A = P 1 + rn( )nt

A = ($1000.00) 1 + 0.045012( )12×10

A ≈ $1566.99


d)

A = P 1 + rn( )nt

A = ($1200.00) 1 + 0.039512( )12× 9

A ≈ $1711.27

3. Theinvestmentiscompoundedsemi-annually,

sothetermforeachinterestperiodis0.5years.

Interest period

Investment value at beginning of period

Interest earned (I = Prt)

Investment value at end of period

1 $1000.00

$1000.00×0.0385×0.5=$19.25

$1000.00+$19.25=$1019.25

2 $1019.25

$1019.25×0.0385×0.5=$19.62

$1019.25+$19.62=$1038.87

3 $1038.87

$1038.87×0.0385×0.5=$20.00

$1038.87+$20.00=$1058.87

4 $1058.87

$1058.87×0.0385×0.5=$20.38

$1058.87+$20.38=$1079.25

Studentscanchecktheiranswerusingthe

compoundinterestformula.

A = P 1 + rn( )nt

A = ($1000.00) 1 + 0.03852( )2× 2

A ≈ $1079.25

4. a)

A = P 1 + rn( )nt

A = ($4000.00) 1 + 0.0300365( )365× 3

A ≈ $4376.68

b)

A = P 1 + rn( )nt

A = ($4000.00) 1 + 0.0300365( )365×10

A ≈ $5399.37

5. a) Yearstodoubleinvalue=72÷(interest

rateasapercent)

Yearstodoubleinvalue=72÷2.80

Yearstodoubleinvalue≈26

Itwilltakeabout26yearsforthe


b)

A = P 1 + rn( )nt

A = ($1000.00) 1 + 0.02801( )1× 26

A ≈ $2050.32

6. a)

A = P 1 + rn( )nt

A = ($1000.00) 1 + 0.025012( )12× 5

A ≈ $1133.00

b) Yearstodoubleinvalue=72÷(interest

rateasapercent)

Yearstodoubleinvalue=72÷2.5

Yearstodoubleinvalue=28.8



7. Chooseaprincipal,suchas$1000.00,andtest

bothinvestmentoptions.

Option1:investmentatarateof1.90%per

annum,compoundedannually

A = P 1 + rn( )nt

A = ($1000.00) 1 + 0.01901( )1× 5

A ≈ $1098.68

Option2:investmentatarateof1.75%per

annum,compoundedmonthly

A = P 1 + rn( )nt

A = ($1000.00) 1 + 0.017512( )12× 5

A ≈ $1091.37


Theinvestmentat1.90%perannum

compoundedannuallyisabetterinvestment.

8. a)

A = P 1 + rn( )nt

A = ($5000.00) 1 + 0.02601( )1×10

A = ($5000.00)(1.0260)10

A ≈ $6463.14

b)

A = P 1 + rn( )nt

A = ($5000.00) 1 + 0.02604( )4 ×10

A ≈ $6479.20

c)

A = P 1 + rn( )nt

A = ($5000.00) 1 + 0.026012( )12×10

A ≈ $6482.83

d)

A = P 1 + rn( )nt

A = ($5000.00) 1 + 0.0260365( )365×10

A ≈ $6484.59


6.3 Credit Cards and Store Promotions


1. a) I = Prt

I =($2076.54)(0.1950)(15÷365)

I =$16.64

b) I = Prt

I =($1007.48)(0.2150)(38÷365)

I =$22.55

c) Thetermofthisinvestmentis18months.

Intheinterestcalculation,divide18

monthsbythenumberofmonthsperyear.

I = Prt

I =($2019.64)(0.1850)(18÷12)

I =$560.27

2. CalculatethenumberofdaysforwhichMarcia

willbechargedinterest.Shewillbecharged

interestforMarch10–31andApril1–2.

March10–31=22days(includingMarch10)

April1–2=2days

Totaldays=24

I = Prt

I =($568.93)(0.2400)(24÷365)

I =$8.98

A = P+ I

A =$568.93+$8.98

A =$577.91

Marciawillowe$577.91.

3. a) Adduphertotalpurchases.

$124.32+$187.54+$32.42+$154.21+

$54.24+$654.32=$1207.05

Harleywillowe$1207.05.

b) Calculate5%ofHarley’sunpaidbalance.

$1207.05×0.05≈$60.35

Thisismorethan$10.00,soherminimum

paymentwillbe$60.35.

c) Calculateherbalanceaftershemakesthe

minimumpayment.

$1207.05–$60.35=$1146.70

Calculatehowmanydaysshewillbe

chargedinterest.

November28–30=3days

December1–28=28days

Total=31days

I = Prt

I =($1146.70)(0.1850)(31÷365)

I =$18.02

A = P+ I

A =$1146.70+$18.02

A =$1164.72

4. a) Calculatethenumberofdays.

December10–21=12days(including

December10)


b) I = Prt

I =($550.00)(0.2490)(12÷365)

I ≈$4.50

A = P+ I

A =$550.00+$4.50

A =$554.50

Javierwillowe$554.50.

c) Calculatethenumberofdays.

December10–31=22days(including

December10)

January1–10=10days

Total=32days

I = Prt

I =($550.00)(0.2490)(32÷365)

I ≈$12.01

A = P+ I

A =$550.00+$12.01

A =$562.01

5. Calculatehertotalpurchases.

$28.95+$45.39+$106.15=$180.49

Calculate5%.

$180.49×0.05≈$9.02

Thisislessthan$10.00,soherminimum

paymentis$10.00.

Calculateherunpaidbalance.

$180.49–$10.00=$170.49

Calculatethenumberofdaysshewillbe

chargedinterestonthisbalance.

October30–31=2days

November1–29=29days

Total=31days

Calculateherinterestchargesonthe

unpaidbalance.

I = Prt

I =($170.49)(0.2195)(31÷365)

I ≈$3.18

Next,calculatehowmanydaysshewillbe

chargedinterestontheNovember12purchase.

November12–29=18days(including

November12)

Calculatetheinterestonthispurchase.

I = Prt

I =($119.65)(0.2195)(18÷365)

I ≈$1.30

Addtheunpaidbalance,thenewpurchase,and

thetwointerestcharges.

$170.49+$119.65+$3.18+$1.30=$294.62

6. Calculatethetotalcostofthepaymentplan.

25×$75.00=$1800.00

Calculatethedifferencebetweenthecashprice

andthepaymentplanprice.

$1800.00–$1675.89=$124.11

Calculatetheinterestrate.

I t

r

r

=

= × ×

=

Pr

$ . $ .

$ . $ .

$ .

124 11 1675 89 2

124 11 3351 78

124 1113351 78

0 0370

$ .

.

=

≈

r

r


Multiplyby100toconverttoapercent.

0.0370×100=3.70%

Solwillpayanannualinterestrateof3.70%.

7. Calculatethecostofeachoption.

Option1:

10%downpayment=0.10×$689.98

10%downpayment≈$69.00

Monthlypayments:

6×$115.00=$690.00

Totalcost:

$69.00+$690.00=$759.00

Option2:

24×$35.00=$840.00

Option3:

I = Prt

I =($689.98)(0.2095)(20÷365)

I ≈$7.92

A = P+ I

A =$689.98+$7.92

A =$697.90

Option3isthebestdeal.

8. Calculatethetotalcostofthepaymentplan.

60×$450.00=$27000.00

Calculatethedifferencebetweenthecashprice

andthepaymentplanprice.

$27000.00–$24789.00=$2211.00

Calculatetheannualinterestrate.Thetermis

60months,or5years.

I = Prt

$2211.00 = $24 789.00 × r × 5

$2211.00 = $123 945.00r

$2211.00$123 945.00

= r

0.0178 ≈ r

Multiplyby100toconverttoapercent.

0.0178×100=1.78%

Jacquieispayinganannualinterest

rateof1.78%.


1. a) I = Prt

I =($2987.69)(0.2150)(45÷365)

I ≈$79.19

b) I = Prt

I =($1539.99)(0.2095)(6÷12)

I ≈$161.31

2. CalculateSimona’sminimumpayment.

$1630.45×0.05≈$81.52


paymentis$81.52.

Calculateherbalanceaftershemakesthe

minimumpayment.

$1630.45–$81.52=$1548.93

Calculatehowmanydayssheis

chargedinterest.


June19–30=12days(includingJune19)

July1–18=18days

Total=30days

Calculatetheinterestdue.

I = Prt

I =($1548.93)(0.1950)(30÷365)

I ≈$24.83

A = P+ I

A =$1548.93+$24.83

A =$1573.76

3. a) CalculateVlad’sminimumpayment.

$398.51×0.05≈$19.93

Thisismorethan$10.00,sohisminimum

paymentis$19.93.

Calculatehisbalanceafterhemadethe

minimumpayment.

$398.51–$19.93=$378.58

Calculatethenumberofdayshepaid

interestonthisamount(June13–July12).


July1–12=12days

Total=30days

Calculatetheinterestonthe

unpaidbalance.

I = Prt

I =($378.58)(0.1850)(30÷365)

I ≈$5.76

Calculatethenumberofdaysheischarged

interestontheJune14purchase.


July1–12=12days

Total=29days

Calculatetheinterestonthenewpurchase.

I = Prt

I =($575.54)(0.1850)(29÷365)

I ≈$8.46

Addtheunpaidbalance,thenewpurchase,

andthetwointerestcharges.

$378.58+$575.54+$5.76+

$8.46=$968.34

Vlad’snewbalancewillbe$968.34.

b) Calculate5%ofthebalance.

$968.34×0.05≈$48.42

Thisismorethan$10.00,sohisminimum


4. a) Calculatethetotalcostofthelivingroom

setonthepaymentplan.

$435.00×6=$2610.00

Calculatethedifferencebetweenthecash

priceandthepaymentplanprice.

$2610.00–$2543.90=$66.10

Hewillpay$66.10ininterest.

b) Calculatetheinterest.

I = Prt

I =($2543.90)(0.2275)(30÷365)

I ≈$47.57

Calculatethetotalcost.


A = P+ I

A =$2543.90+$47.57

A =$2591.47

Thelivingroomsetwouldcost$2591.47.

5. Calculatethetotalcostofeachoption,andthe

differencebetweenthepaymentplancostand

thecashprice.

Option1:

$220.00×4=$880.00

$880.00–$859.40=$20.60

Option2:

$150.00×6=$900.00

$900.00–$859.40=$40.60

Calculatetheinterestrateofeachoption.

Option1:

I = Prt

$20.60 = $859.40 × r × 412( )

$20.60 = $286.47r

$20.60$286.47

= r

0.0719 ≈ r

Multiplyby100toconverttheratetoapercent.

0.0719×100=7.19%

Option2:

I = Prt

$40.60 = $859.40 × r × 612( )

$40.60 = $429.70r

$40.60$429.70

= r

0.0945 ≈ r


0.0945×100=9.45%

Consideringinterestrateonly,Option1isa

betterbuy.


6.4 Personal loans, lines of Credit, and overdrafts


1. CalculatehowmuchinterestBaroupaid.

$275.00–$250.00=$25.00

Usethesimpleinterestformula.

I = Prt

$25.00 = $250.00 × r × (15 ÷ 365)

$25.00 = $10.27r

$25.00$10.27

= r

2.43 ≈ r


2.43×100=243%

Theannualinterestrateis243%.

2. a) Calculatetheamountofinterestpaid.

$415.00–$400.00=$15.00

Calculatetheannualinterestrate.

I = Prt

$15.00 = $400.00 × r × (10 ÷ 365)

$15.00 = $10.96r

$15.00$10.96

= r

1.37 ≈ r

Multiplyby100toconverttherateto

apercent.

1.37×100=137%


b) Calculatethedailyinterestrate.

I = Prt

$15.00 = $400.00 × r × 10

$15.00 = $4000.00r

$15.00$4000.00

= r

0.00375 = r


apercent.

0.00375×100=0.375%

Thedailyinterestrateis0.375%.

3.

I = Prt

I = ($200.00)(3.95)(7 ÷ 365)

I ≈ $15.15

Shepaid$15.15ininterest.

4. I = Prt

I =($500.00)(0.0112)(25)

I =$140.00

A = P+ I

A =$500.00+$140.00

A =$640.00

Arletahastorepay$640.00.


5. CalculatehowmuchinterestHelenpaid.

$781.50–$750.00=$31.50

I = Prt

$31.50 = ($750.00)(.0105)t

$31.50 = $7.875t

$31.50$7.875

= t

4 = t

Helenhadthemoneyfor4days.

6. a) I = Prt

I =($1000.00)(0.0050)(60)

I =$300.00

A = P+ I

A =$1000.00+$300.00

A =$1300.00

Hanswillhavetorepay$1300.00.

b)

I = Prt

$300.00 = ($1000.00)(r)(60 ÷ 365)

$300.00 = $164.38r

$300.00$164.38

= r

1.825 ≈ r


apercent.

1.825×100=182.5%

Theannualinterestrateis182.5%.

7. a) UsingthePersonalLoanPayment

Calculatortable,lookup9.00%interest

intheleft-handcolumn.Inthatrow,look

attheentryunderthe2-yearterm:it

is$45.68.

Dividetheamountoftheloan($3000.00)by

$1000.00andmultiplyby$45.68.

($3000.00÷$1000.00)×$45.68=$137.04

Themonthlypaymentwillbe$137.04.

Calculatethetotalamountpaid.

2years×12months/year×$137.04/month

=$3288.96

Calculatethedifferencebetweentheprincipal

andthetotalamountpaid.

$3288.96–$3000.00=$288.96

Thefinancechargeis$288.96.

b) UsingthePersonalLoanPayment




is$30.99.

Dividetheamountoftheloan($2125.00)

by$1000.00andmultiplyby$30.99.

($2125.00÷$1000.00)×$30.99≈$65.85




=$2370.60

Calculatethedifferencebetweenthe

principalandthetotalamountpaid.

$2370.60–$2125.00=$245.60



c) UsingthePersonalLoanPayment




is$22.92.



($11500.00÷$1000.00)×$22.92=$263.58




=$12651.84



$12651.84–$11500.00=$1151.84


8. a) CalculatehowmuchAdelewillhave

toborrow.

$2900.00–$1100.00=$1800.00





is$44.55.



($1800.00÷$1000.00)×$44.55=$80.19


c) Multiplythemonthlypaymentbythe

amortizationperiod.


=$1924.56

d) Addthecostoftheloantotheamountof

herdownpayment.

$1100.00+$1924.56=$3024.56

Thecarwillcost$3024.56.


1. a) CalculatehowmuchinterestSheypaid.

$950.00–$850.00=$100.00

Thetermoftheloanis12days.

I = Prt

$100.00 = ($850.00)(r)(12)

$100.00 = $10 200.00r

$100.00$10 200.00

= r

0.0098 ≈ r


apercent.

0.0098×100=0.98%


b) Tocalculatetheannualinterestrate,the

termneedstobeconvertedtoyears.The

termwillbe12dividedby365.

I = Prt

$100.00 = ($850.00)(r)(12 ÷ 365)

$100.00 = $27.95r

$100.00$27.95

= r

3.58 ≈ r



apercent.

3.58×100=358%


2. Theinterestrateisgivenasadailyrate,sothe

termoftheinvestmentisalsoindays.

I = Prt

I =($250.00)(0.0117)(18)

I =$52.65

A = P+ I

A =$250.00+$52.65

A =$302.65

Carmenwillhavetorepay$302.65.





is$31.34.



($2500.00÷$1000.00)×$31.34=$78.35




=$2820.60



$2820.60–$2500.00=$320.60






is$19.45.



($10000.00÷$1000.00)×

$19.45=$194.50




=$11670.00



$11670.00–$10000.00=$1670.00


c) UsingthePersonalLoanPayment




is$43.31.



($1500.00÷$1000.00)×$43.31≈$64.97




=$1559.28


Calculatethedifferencebetweentheprincipal

andthetotalamountpaid.

$1559.28–$1500.00=$59.28






is$44.77.



($5000.00÷$1000.00)×$44.77≈$223.85


b) Calculatethetotalamountpaid.


=$5372.40

c) Calculatethedifferencebetweenthe


$5372.40–$5000.00=$372.40


5. a) CalculatehowmuchmoneyManonwould

havetoborrowfromthebank.

$3499.99–$1000.00=$2499.99

UsingthePersonalLoanPayment




is$44.55.



($2499.99÷$1000.00)×$44.55≈$111.37


b) Calculatethecostofeachoption.

Option1:


=$2672.88

Addthedownpaymentplusthecost

oftheloan.

$1000.00+$2672.88=$3672.88

Option1costs$3672.88.

Option2:

$50.00+($325.00×12)=$3950.00


Option3:

I = Prt

I =($2499.99)(0.0112)(30)

I ≈$840.00

A = P+ I

A =$2499.99+$840.00

A =$3339.99

Addthedownpaymentplusthecost

oftheloan.

$1000.00+$3339.99=$4339.99


SheshouldchooseOption1.


ChAPter teSt, p. 325

1. Duringthemonth,Salmamade10self-servicewithdrawalsand3deposits,soshewillnotbecharged

anytransactionfees.


ATM Cash $200.00 $2879.54–$200.00=$2679.54



Bankcard Gas $42.54 $2964.63–$42.54=$2922.09

ATM Cheque—reimbursement $175.64 $2922.09+$175.64=$3097.73

Bankcard Dinner $32.42 $3097.73–$32.42=$3065.31

ATM Cash $100.00 $3065.31–$100.00=$2965.31

Auto-withdrawal Hydro $112.21 $2965.31–$112.21=$2853.10


Auto-withdrawal Rent $645.00 $3310.31–$645.00=$2665.31

Bankcard Carrepairs $276.97 $2665.31–$276.97=$2388.34

Bankcard Movie $28.12 $2388.34–$28.12=$2360.22

ATM Cash $200.00 $2360.22–$200.00=$2160.22

Salmamaintainsaminimumbalanceof$2000.00,soshedoesnothavetopaythe$6.00accountfee.

2. a) I = Prt

I =($5000.00)(0.0250)(10)

I =$1250.00

b) A = P+ I

A =$5000.00+$1250.00

A =$6250.00

3.

I = Prt

$82.50 = P × 0.0110 × 5

$82.50 = 0.0550P

$82.500.0550

= P

$1500.00 = P

Theprincipalwas$1500.00.

4.

A = P 1 + rn( )nt

A = ($5000.00) 1 + 0.02501( )1×10

A = ($5000.00)(1.0250)10

A ≈ $6400.42

5.

A = P 1 + rn( )nt

A = ($1000.00) 1 + 0.030012( )12×10

A ≈ $1349.35

I =A –P

I =$1349.35–$1000.00

I =$349.35

Theinvestmentearned$349.35ininterest.


6. a) I = Prt

I =($1629.53)(0.1950)(21÷365)

I ≈$18.28

b) I = Prt

I =($2639.99)(0.2295)(30÷365)

I ≈$49.80

7. a) CalculateMonique’sminimumpayment.

$739.65×0.05≈$36.98


paymentis$36.98.

Calculatehernewunpaidbalance.

$739.65–$36.98=$702.67

Calculatehowmanydaysshewillpay

interestonthisamount.

March13–31=19days

(includingMarch13)

April1–12=12days

Total=31days

Calculatetheinterestonthe

unpaidbalance.

I =Prt

I=($702.67)(0.2085)(31÷365)

I≈$12.44

Calculatethenumberofdaysinterestis

chargedontheMarch16purchase.

March16–31=16days

(includingMarch16)

April1–12=12days

Total=28days

Calculatetheinterestonthenewpurchase.

I = Prt

I =($179.39)(0.2085)(28÷365)

I ≈$2.87

Addtheunpaidbalance,thenewpurchase,

andthetwointerestamounts.

$702.67+$179.39+$12.44+

$2.87=$897.37

ThenewbalanceonherApril12statement

willbe$897.37.

b) Calculate5%ofthenewbalance.

$897.37×0.05≈$44.87



8. a) Calculatetheamountofinterestpaid.

$540.00–$500.00=$40.00

Calculatethedailyinterestrate,using7

daysastheterm.

I = Prt

$40.00 = ($500.00)(r)(7)

$40.00 = $3500.00r

$40.00$3500.00

= r

0.0114 ≈ r


apercent.

0.0114×100=1.14%


b) Calculatetheannualinterestrate,using7

dividedby365astheterm.


I = Prt

$40.00 = ($500.00)(r)(7 ÷ 365)

$40.00 = $9.589r

$40.00$9.589

= r

4.17 ≈ r


9. a) CalculatehowmuchmoneyNathaniel

needstoborrow.

$2055.99–$400.00=$1655.99

UsingthePersonalLoanPayment



attheentryunderthe2-yearterm:itis

$44.89.



($1655.99÷$1000.00)×$44.89≈$74.34


b) Calculatethecostofthebicycleonthe

paymentplan.

12×$180.00=$2160.00

Calculatetheinterestpaidbysubtracting

thecashpricefromthepaymentplanprice.

$2160.00–$2055.99=$104.01

Calculatetheannualinterestrate.

I = Prt

$104.01 = ($2055.99)(r)(1)

$104.01 = $2055.99r

$104.01$2055.99

= r

0.0506 ≈ r


0.0506×100=5.06%

Theannualinterestrateis5.06%.

c) Option1:

Multiplythemonthlypaymentbytheterm.


=$1784.16

Addthedownpayment.

$1784.16+$400.00=$2184.16

ThetotalcostofOption1is$2184.16.

ThetotalcostofOption2wascalculatedin

b)above.Itis$2160.00.

NathanielshouldchooseOption2because

itischeaper.


Chapter 7Personal Budgets

7.1 Preparing to Make a Budget


1. Answerswillvary,particularlyforthereason

behindtheclassification.Possiblesolutions

areprovided.

Item Classification ReasonSemi-monthlypaycheque

Regular Thispaychequewillhappentwiceamonth.

Birthdaygift

Variable Youdon’tknowwhatyou’llgetforyourbirthdayanditonlyhappensonceayear.

Tips Variable Youdon’tknowhowmanytipsyou’llgetinashift.

Interestfrominvestment

Regular Yourinvestmentswillhaveregularinterestpayments.

Taxrefund Variable Youdon’tknowhowmuchtaxyou’llpayorgetrefundedeachyearuntilyoudoyourtaxes.

2. Answerswillvary,particularlyforthereason

behindtheclassification.Possiblesolutions

areprovided.

Item Classification ReasonRent Recurring Youwillhavetopaythe

samerenteverymonth.

Newshoes Variable Youdon’tneedtobuynewshoesoftenenoughtoplanonthis.

Loanpayments

Recurring Youwillhaveanagreementwithwhoeverissuedtheloanonarepaymentschedule.

Carrepairs Unexpected Youcan’tplanwhenyourcarwillbreakdown.

Groceries Variable Youdon’tknowhowmuchyou’llspendongrocerieseachtrip,butyoudohavetodoitregularly.

Mealatrestaurant

Variable Youdon’tknowhowmuchyou’llspendeverytimeyougofordinner,butyoucanbudgetforit.

Replacingbrokencellphone

Unexpected Youcan’tplanwhenyoubreakyourcellphone.

Carinsurance

Recurring Youwillhaveanagreedrateofcarinsurancewithyourinsurancebroker.

3. Item Income ExpensePaycheque:$250.98 Regular

John'sbirthday:$54.25 Variable

Commission:$75.00 Variable

Babysitting:$40.00 Variable

Paycheque:$123.42 Regular

Loanpayment:$125.00 Regular

Donationtoearthquakerelieffund:$25.00

Variable

Carinsurance:$115.32 Regular

Savings:$50.00 Variable

Rent:$450.00 Regular

Cellphonebill:$45.00 Regular

Mealatrestaurant:$56.76 Variable

4. CalculateTonia’stotalincome.

4×$450.00=$1800.00

Calculatehertotalexpenses,includingthe

carrepair.

$775.00+$225.39+$74.00+$66.79+$47.59

+$84.00+$250.00+$243.25=$1766.02

Calculatethedifferencebetweenherincome

andexpenses.

$1800.00–$1766.02=$33.98

Toniawillsave$33.98thismonth.

5. CalculateFranklin’stotalexpenses.

$1250.46+$250.00+$135.76+$50.00+

$245.00+$250.00+$100.00=$2281.22

Calculatethedifferencebetweenhisincome

andexpenses.

$2456.85–$2281.22=$175.63

Franklinwillsave$175.63.Multiplyby12to

findouthowmuchhewillsaveinoneyear.

$175.63×12=$2107.56

Inoneyear,hewillsave$2107.56.

Calculatewhatpercentageofhisincomehis

savingsrepresent.

$ .$ .

. %175 632456 85

100 7 1× ≈

Franklin’ssavingsrepresentabout7.1%of

hisincome.

6. CalculateMarion’stotalincome.

$1037.72×2=$2075.44

Calculatehertotalexpenses.

$825.00+$110.00+$100.00+$150.00+

$25.00+$225.00+$275.00+$100.00+

$25.00+$50.00=$1885.00

Calculatethedifferencebetweenherincome

andhisexpenses.

$2075.44–$1885.00=$190.44

Shesaves$190.44inonemonth.Calculatehow

muchshewillhavesavedafter3months.

$190.44×3=$571.32

No,shewillnotbeabletopaycashforthe

$1399.99TV.Shewillhave$571.32saved.

7. CalculatehowmuchAidansavespermonth.

$1588.25–$1275.00=$313.25

Dividethecostofthecomputerbytheamount

hesavespermonth.

$1798.98÷$313.25≈5.7months

ItwilltakeAidanabout6monthstosavefor

thenewcomputer.

MathWorks11WorkbookSolutions Chapter7 PersonalBudgets 124

8. CalculateKaren’smonthlyexpenses.

$775.00+$175.00+$342.00+$123.00+

$42.00+$225.00+$90.00+$75.00+$50.00

=$1897.00

CalculatehowmuchKarensavesinonemonth.

$2379.00–$1897.00=$482.00

Shesaves$4820.00permonth.Calculatehow

muchmoneyKarenwillsavefromJanuaryto

thebeginningofJuly(ortheendofJune,so

6months).

$482.00×6=$2892.00

Karenwillnothaveenoughmoneysaved

forhertrip.

$3000.00–$2892.00=$108.00

Shewillbe$108.00shortofhergoal.


1. Item Income ExpensePaycheque Regular

Rent Regular

Carinsurance Regular

Gas Variable

Clothing Variable

Cellphonebill Regular

Mother’sDaygiftformom Variable

Groceries Variable

PresentforJoan Variable

Entertainment Variable

Mowinglawns Variable

Utilities Variable

2. a) CalculateHannah’stotalincome.

$400.00+$140.00+$150.00+$150.00+

$400.00=$1240.00


$395.00+$85.00+$185.00+$75.00

+$170.00+$50.00+$25.00+$40.00

=$1025.00

Calculatehowmuchshewillsave

eachmonth.

$1240.00–$1025.00=$215.00

Hannahwillsave$215.00permonth.

b) Calculatewhatpercentageofherincome

hersavingsrepresent.

$ .$ .

%215 001240 00

100 17× ≈

Hannah’ssavingsrepresentabout17%of

herincome.

c) Calculatehowmuchshewillsave

in6months.

$215.00×6=$1290.00

Hannahwillhaveenoughmoneysavedfor

thebicycle.Shewillhave$1290.00saved.

3. CalculatehowmuchmoneyPerseushassaved

inthepast8months.

8×$125.00=$1000.00

Calculatehowmuchmoremoneyheneedsto

savetobeabletoaffordthetrip.

$1900.00–$1000.00=$900.00

Dividetheremainingcostofthetripby6

tofindouthowmuchPerseushastosave

permonth.

$900.00÷6≈$150.00


Hehasbeensavings$125.00amonth,sohe

willneedtosaveanadditional$25.00amonth

tobeabletoaffordthetripin6months’time.

4. a) Income Expenses Babysitting $40.00 Cellphone $40.00

Babysitting $40.00 Entertainment $105.00

Paycheque $375.00 Food $195.00

Paycheque $375.00 Loanpayment $65.00

Tutoring $50.00 Rent $425.00

Tutoring $100.00 Renter’sinsurance

$25.00

Tutoring $50.00 Transportation $80.00

b) CalculateShonda’stotalincome.

$40.00+$40.00+$375.00+$375.00+

$50.00+$100.00+$50.00=$1030.00


$40.00+$105.00+$195.00+$65.00+

$425.00+$25.00+$80.00=$935.00

Shona’stotalincomeis$1030.00andher

expensesare$935.00.

c) Calculatethedifferencebetweenher

incomeandexpenses.

$1030.00–$935.00=$95.00

Shecansave$95.00inonemonth.

Multiplytofindouthowmuchshecansave

inoneyear.

$95.00×12=$1140.00

Inoneyear,Shondacansave$1140.00,

assumingshecontinuestoearn

approximatelythesameincomeandspend

thesameamountonhervariousexpenses.

d) Calculatewhatpercentage$95.00isofher

monthlyincome($1030.00).

$ .$ .

%95 001030 00

100 9× ≈

Shonda’ssavingsrepresentabout9%of

herincome.


7.2 the Budgeting Process


1. a) Answerswillvary.Apossiblebudgetis

provided.

craig aNd StéfaNie’S MoNthlY BudgetIncome ExpensesCraig’smonthlypay

$2800.00 Mortgagepayment

$1850.00

Stéfanie’smonthlypay

$2770.00 Utilities $175.00

Catcare $50.00

Clothing $150.00

Food $400.00

Gas $150.00

Carinsurance $112.00

Gifts $150.00

Entertainment $120.00

Loanpayment $150.00

Homeinsurance

$122.00

Tripfund $250.00

Cellphones $135.00

Miscellaneous $475.00

Savings $1281.00

Total income $5570.00 Total expenses $5570.00

b) Answerswillvary,dependingonthe

budgetcreated.Inthebudgetprovided

above,CraigandStéfaniesave$1281.00

permonth.Multiplyby12tocalculatehow

muchtheywillsaveinoneyear.

$1281.00×12=$15372.00

Basedonthebudgetprovided,theywould

save$15372.00inoneyear.

2. a) Answerswillvary.Apossiblebudget

isprovided.

MiNh’S MoNthlY BudgetIncome ExpensesRegular $1665.00 Rent $640.00

Tips $300.00 Utilities,phone,cable,internet

$115.00

Food $300.00

Transportation $110.00


Other $500.00

Savings $90.00



budgetcreated.Inthebudgetabove,Minh

saves$90.00amonth.Calculatehowmuch

hewillsavein6months.

$90.00×6=$540.00

Calculatethedifferencebetweenthecostof

theskisandMinh’ssavings.

$1059.99–$540.00=$519.99

Basedonthisbudget,Minhwillnotbeable

toaffordtheskis.Hewillbe$519.99short.



isprovided.

georgiNa’S MoNthlY BudgetIncome Expenses Monthlypay $3050.00 Mortgage

payment$1025.00

Carinsurance $122.00

Houserepairs $70.00

Housetaxes $350.00

Gifts $45.00

Charity $30.00

Gas $125.00

Loanpayment $220.00



Homeinsurance $125.00

Clothing $130.00

Carmaintenance

$50.00

Food $375.00

Skiingseason’spass

$55.00

Savings $8.00



budgetcreated.Basedonthisbudget,

Georginawillsaveonly$8.00amonth.

Multiplyby12tofindouthowmuchshe

willsaveinoneyear.

$8.00×12=$96.00

Shewillsaveonly$96.00inoneyear.

c) Answerswillvary,dependingonthe


Georginasaves$8.00permonthand

earns$3050.00.

$ .$ .

. %8 003050 00

100 0 3× ≈

Georginasavesonly0.3%ofherincome.

d) Answerswillvary.Optionsinclude:

• decreasingherspendingon

entertainment,clothing,or

miscellaneousexpenses;

• notbuyingaseason’spasstotheskihill;

• sellinghercar;

• findingalessexpensiveplacetolive.

4. a) CalculatehowmuchmoneyJanaeearnsin

oneweekinwages.

40×$10.75=$430.00

Calculatehowmuchshewouldearninone

month,includingtips.

($430.00×4)+$500.00=$2220.00

Calculate80%ofherincome.

$2220.00×0.8=$1776.00

Calculatehowmuchshesavespermonth.

$2220.00–$1776.00=$444.00

b) CalculateJanae’smonthlyfoodexpenses.

$75.00×4=$300.00

Calculatewhatpercentageofhermonthly

incomethisrepresents.Janae’smonthly

incomeis$2220.00.

$ .$ .

. %300 002220 00

100 13 5× ≈

Janaespendsabout13.5%ofher

incomeonfood.

5. a) Calculate12%of$2700.00.

$2700.00×0.12=$324.00

Justinwillsave$324.00amonth.


b) Divide$1500.00bytheamounthesaves

permonth.

$1500.00÷$324.00≈4.6

ItwilltakeJustinabout5monthsto

save$1500.00.

6. CalculatehowmuchmoneyVeejaysaves

permonth.

$1725.00×0.55=$948.75

Calculatehowmuchheneedstosavefor

tuitionplusbooks.

$6000.00+$550.00=$6550.00

Divide$6550.00bytheamounthesaves

permonth.

$6550.00÷$948.75≈6.9

ItwilltakeVeejayabout7monthstosaveup

forcollegetuitionandbooks.



isprovided.

Pierre’S MoNthlY BudgetIncome Expenses Semi-monthlypay

$1475.00 Mortgagepayment

$1450.00

Semi-monthlypay

$1475.00 Utilities/phone $150.00

Groceries $325.00

Loanrepayment

$250.00

Charity $40.00


Clothing $150.00



Savings $135.00



budgetcreated.Basedonthebudget

providedabove,Pierrewillsave$135.00a

month.Calculatehowmuchhewillsave

inoneyear.

$135.00×12=$1620.00

Pierrewillsave$1620.00inoneyear.


isprovided.

chaNtal’S MoNthlY BudgetIncome Expenses Semi-monthlypay

$1110.00 Rent $675.00

Semi-monthlypay

$110.00 Utilities,phone,cable,internet

$150.00

Tips $135.00 Food $400.00



Other $375.00

Savings $305.00




Chantalsaves$305.00permonth.

Calculatewhatpercentageofherincome

thisrepresents.

$ .$ .

%305 002355 00

100 13× ≈

Basedonthisbudget,Chantalsaves13%of

herincome.



Chantalsaves$305.00permonth.Divide

thecostofthecomputerbythisamount.

$1000.00÷$305.00≈3.3


ItwilltakeChantalabout3.3monthsto

saveforhercomputer.Becauseshewillnot

haveenoughmoneysavedatthebeginning

ofthethirdmonth,youshouldroundthe

answerup.Itwilltakeher4monthstosave

forthecomputer.

3. a) CalculateMarcel’smonthlyincomefrom

regularpay,assumingthereare4weeks

inamonth.

(40h/wk×$14.50/h×4wk)=$2320.00

Calculatehismonthlyincomefrom

overtimepay.

6h×(1.5×$14.50/h)=$130.50

Calculatehistotalmonthlyincome.

$2320.00+$130.50=$2450.50

Calculatehistotalexpenses,excludingrent.

$125.00+$250.00+$165.00+$300.00+

$350.00+$400.00=$1590.00

Calculatehowmuchhecanspendonrent.

$2450.50–$1590.00=$860.50

Marcelcanspendamaximumof$860.50

permonthonrent.

b) Dividethecostofthetripbytheamounthe

savespermonth.

$2700.00÷$300.00=9months

ItwilltakeMarcel9monthstosaveupfor

thetriptoEcuador.


7.3 analyzing a Budget


1. a) CalculatehowmuchmoneySukhsaves

permonth.

$2865.00×0.15=$429.75

DividetheamountSukhneedstosave

($6000.00)bytheamounthesaves

permonth.

$6000.00÷$429.75≈14

ItwilltakeSukh14monthstosaveenough

moneyforthecollegecourse.

b) CalculatehowmuchmoneySukhwillhave

tosavepermonthbydividingtheamount

needed($6000.00)by12months.

$6000.00÷12=$500.00

Calculatewhatpercentage$500.00isofhis

income($2865.00).

$ .$ .

. %500 002865 00

100 17 5× ≈

Sukhwillneedtosaveabout17.5%ofhis

incomeinordertobeabletoaffordthe

coursein12months.

2. a) CalculateJuliet’sincome.

$ . %

$ .

$ . .

$

275 00 100 12

275 00 12100

275 00 0 12

275

x

x

x

× =

=

=

.. .

$ ..

$ .

00 0 12

275 000 12

2291 67

=

=

≈

x

x

x

Calculate8%ofherincome.

x

x

x

x

$ .%

$ .

$ ..

$

2291 67100 8

2291 678

100

2291 670 08

× =

=

=

= 22291 67 0 08

183 33

. .

$ .

×

≈x

Juliet’snewmonthlyentertainmentbudget

is$183.33.

b) $ . %

$ .

$ . .

$

275 00 100 12

275 00 12100

275 00 0 12

275

x

x

x

× =

=

=

.. .

$ ..

$ .

00 0 12

275 000 12

2291 67

=

=

≈

x

x

x

Juliet’stotalmonthlyincomeis$2291.67.


3. a) Answerswillvary.Forexample,Petra

coulddecreaseherspendingbyatotalof

$100.00onacombinationofexpenses,

suchasfood,miscellaneous,andclothing.

Shecoulddecreaseherfoodexpensesto

$275.00,miscellaneousto$75.00,and

clothingto$75.00.Theseexpensesareall

variable,andPetracanchoosetobuyless

expensiveitemswithinthesecategories.


suggestionsmadeina).Basedonthe

suggestionsabove,thenewbudgetwould

beasfollows.

Petra’S MoNthlY BudgetIncome Expenses Regular $1400.00 Housing $550.00

Utilities,phone,cable,internet

$175.00

Food $275.00



Clothing $75.00


Charitabledonations

$25.00

Savings $100.00


Toconstructacirclegraphoftheexpenses,

firstcalculatewhatpercentageofPetra’sincome

eachcategoryrepresents.

Housing:($550.00÷$1400.00)×100≈39%

Utilities,etc.:($175.00÷$1400.00)×100≈13%

Food:($275.00÷$1400.00)×100≈20%

Transportation:($75.00÷$1400.00)×100≈5%

Entertainment:($50.00÷$1400.00)×100≈4%

Clothing:($75.00÷$1400.00)×100≈5%

Miscellaneous:($75.00÷$1400.00)×100≈5%

Charitabledonations:($25.00÷$1400.00)

×100≈2%

Savings:($100.00÷$1400.00)×100≈7%

Next,calculatehowmanydegreeseach

percentageofexpensesrepresents.

Housing:0.39×360°≈140°

Utilities,etc.:0.13×360°≈47°

Food:0.20×360°=72°

Transportation:0.05×360°=18°

Entertainment:0.04×360°≈14°

Clothing:0.05×360°=18°

Miscellaneous:0.05×360°=18°

Charitabledonations:0.02×360°≈7°

Savings:0.07×360°≈25°

(Notethatthesectionsaddupto359°rather

than360°;thisisduetorounding.Toconstruct

thegraph,youcanincreasethesizeofoneof

thesectionsby1°.)

Drawacircleanduseaprotractortomeasure

thedegreesforeachcategoryofexpense.Petra’s Monthly Expenses

39%

7%

5%

13%

4%

20%

5%5%

2%Rent

Utilities, phone, cable, internet

Food

Transportation

Entertainment

Clothing

Miscellaneous

Charitable donations

Savings


4. a) Calculatewhatpercentageofhisincome

Sammyspendsoneachcategory.

Housing:($400.00÷$1500.00)×100≈27%

Food:($225.00÷$1500.00)×100=15%

Transportation:($350.00÷$1500.00)

×100≈23%

Entertainment:($250.00÷$1500.00)

×100≈17%


×100≈2%

Savings:($50.00÷$1500.00)×100≈3%

Other:($200.00÷$1500.00)×100≈13%

Sammy’sspendingfallswithintheguidelines

forhousing,food,andcharitabledonations

(butforthislastcategory,justbarely).Heis

currentlyspendingtoomuchontransportation,

entertainment,andother.Heiscurrentlynot

puttingenoughmoneyintosavings.

b) MultiplySammy’smonthlysavingsby

12months.

$50.00×12=$600.00

Sammywillsave$600.00inoneyear.

c) Answerswillvary.Apossiblebudget

isprovided.

SaMMY’S MoNthlY BudgetIncome ExpensesPay $1500.00 Housing $400.00

Food $275.00



Charitabledonations

$125.00

Savings $150.00

Other $200.00



firstcalculatewhatpercentageofSammy’s

incomeeachcategoryrepresents.

Housing:($400.00÷$1500.00)×100≈27%

Food:($275.00÷$1500.00)×100≈18%


×100≈13%


×100=10%


×100≈8%

Savings:($150.00÷$1500.00)×100=10%

Other:($200.00÷$1500.00)×100≈13%

Notethatthepercentagesaddupto99%

ratherthan100%;thisisduetorounding.To

constructthecirclegraph,youcanincreasethe

sizeofoneofthesectionsby1%(forexample,

housingto28%).



Housing:0.28×360°≈101°

Food:0.18×360°≈65°


Transportation:0.13×360°≈47°

Entertainment:0.10×360°=36°


Savings:0.10×360°=36°

Other:0.13×360°≈47°

Notethatthesectionsaddupto361°rather


thegraph,youcandecreasethesizeofone

ofthesectionsby1°(forexample,housingto

100°,sinceyouroundedthepercentageupin

theprevioussection).


thedegreesforeachcategoryofexpense.

Sammy’s Monthly Expenses

$400.00

$275.00

$200.00

$200.00

$150.00

$150.00

$125.00 Housing

Food

Transportation

Entertainment


Savings

Other


isprovided.

willa’S MoNthlY BudgetIncome Expenses Regular $3530.00 Housing $1250.00

Utilities $235.00


Debtrepayment $420.00

Savings $375.00

Food $450.00

Recreation $215.00

Healthandpersonalcare

$300.00



budgetcreatedina).Thefollowinggraphis

providedforthebudgetgivenabove.


firstcalculatewhatpercentageofWilla’s


Housing:($1250.00÷$3530.00)×100≈35%

Utilities:($235.00÷$3530.00)×100≈7%


×100≈8%

Debtrepayment:($420.00÷$3530.00)

×100≈12%

Savings:($375.00÷$3530.00)×100≈11%

Food:($450.00÷$3530.00)×100≈13%

Recreation:($215.00÷$3530.00)×100≈6%

Healthandpersonalcare:($300.00÷

$3530.00)×100≈8%



Housing:0.35×360°=126°

Utilities:0.07×360°≈25°


Debtrepayment:0.12×360°≈43°

Savings:0.11×360°≈40°

Food:0.13×360°≈47°

Recreation:0.06×360°≈22°

Healthandpersonalcare:0.08×360°≈29°



than360°;thisisduetorounding.To

constructthegraph,youcandecreasethe

sizeofoneofthesectionsby1°(forexample,

recreationto21°).



Willa’s Monthly Expenses

$1250.00

$450.00$420.00

$375.00

$300.00

$285.00

$235.00$215.00

Health and personal

Recreation

Food

Savings

Debt repayment

Transportation

Utilities

Housing


1. a) CalculatehowmuchmoneyManjeetsaves

inoneweek.

$550.00×0.14=$77.00

Multiplyby52weeks.

$77.00×52=$4004.00

Manjeetwillsave$4004.00inoneyear.

b) Calculatea5%increasetoManjeet’s

weeklysalary.

$550.00×0.05=$27.50

Calculatehisnewweeklysalary.

$550.00÷$27.50=$577.50

Calculate14%ofhisnewsalary.

$577.50×0.14=$80.85

Multiplyby52weeks.

$80.85×52=$4204.20

Inoneyear,Manjeetwillsave$4204.20.

2. a) CalculateLena’sincome.

$ . %

$ .

$ . .

$

3150 00 100 7

3150 00 7100

3150 00 0 07

31

x

x

x

× =

=

=

550 00 0 07

3150 000 07

45000 00

. .

$ ..

$ .

=

=

=

x

x

x

Lena’sannualincomeis$45000.00.

Calculate10%.

$45000.00×0.10=$4500.00

Inoneyear,Lenawillbeableto

save$4500.00.

$ . %

$ .

$ . .

$

3150 00 100 7

3150 00 7100

3150 00 0 07

31

x

x

x

× =

=

=

550 00 0 07

3150 000 07

45000 00

. .

$ ..

$ .

=

=

=

x

x

x

Lena’sannualincomeis$45000.00.

3. a) AddSean’sincome.

$1300.00+$1300.00+$150.00

=$2750.00


Addhisexpenses.

$1200.00+$175.00+$150.00+$325.00+

$150.00+$125.00+$175.00+$100.00+

$50.00+$100.00+$200.00=$2750.00

Sean’sincomeisequaltohisexpenses,so

hisbudgetisbalanced.

b) Toconstructacirclegraphoftheexpenses,

firstcalculatewhatpercentageofSean’s


Housing:($1200.00÷$2750.00)×100≈44%

Utilities:($175.00÷$2750.00)×100≈6%

Phone,cable,internet:($150.00÷$2750.00)

×100≈5%

Food:($325.00÷$2750.00)×100≈12%


×100≈5%


×100≈5%

Clothing:($175.00÷$2750.00)×100≈6%

Miscellaneous:($100.00÷$2750.00)

×100≈4%


×100≈2%

Medical/healthandotheremergencies:

($100.00÷$2750.00)×100≈4%

Savings:($200.00÷$2750.00)×100≈7%



Housing:0.44×360°≈158°

Utilities:0.06×360°≈22°

Phone,cable,internet:0.05×360°=18°

Food:0.12×360°≈43°


Entertainment:0.05×360°=18°

Clothing:0.06×360°≈22°

Miscellaneous:0.04×360°≈14°


Medical/healthandotheremergencies:

0.04×360°≈14°

Savings:0.07×360°≈25°



thegraph,youcanincreasethesizeofone

ofthesectionsby1°(forexample,housing

to159°).



Sean’s Monthly Expenses

$1200.00

$325.00$200.00

$175.00

$175.00

$150.00

$150.00$125.00

$100.00$100.00 $50.00

Savings Medical/health and other emergencies


Miscellaneous

Clothing Entertainment

Transportation

Food

Phone, cable, internet

Utilities

Housing


c) Fromthecalculationsinb),Seanis

spendingabout44%ofhisincomeon

spending.Thisismorethansuggestedby

theguidelines.

Calculate35%ofSean’sincome.

$2750.00×0.35=$962.50

Heshouldbespendingnomorethan

$962.50onhousing.

d) Answerswillvary.Calculatehowmuch

Seanwillneedtosavepermonth.

$2750.00×0.15=$412.50

$412.50–$200.00=$212.50

Seanneedstoadjusthisbudgettosave

$212.50morethanheiscurrentlysaving

permonth.

Possiblesuggestionsinclude:

• decreasinghisspendingon

entertainment,clothing,miscellaneous,

andcharitabledonations.Hecould

decreasehisexpensestothefollowing

amounts:entertainment,$75.00;

clothing,$75.00,miscellaneous,$75.00;

charitabledonations,$12.50.

• findingalessexpensivemodeof

transportation;or

• gettingalessexpensivephone/cable/

internetpackage.

4. a) Toconstructacirclegraphoftheexpenses,

firstcalculatewhatpercentageofGabriella’s


Housing:($840.00÷$1990.00)×100≈42%

Food:($175.00÷$1990.00)×100≈9%


×100≈6%


×100≈9%

Clothing:($175.00÷$1990.00)×100≈9%

Miscellaneous:($150.00÷$1990.00)

×100≈8%

Loanpayments:($275.00÷$1990.00)

×100≈14%

Savings:($75.00÷$1990.00)×100≈4%

Notethatthepercentagesaddupto101%;

thisisduetorounding.Forthenextstep,you

candecreaseoneofthecategoriesby1%(for

example,housingto41%).



Housing:0.41×360°≈148°

Food:0.09×360°≈32°



Clothing:0.09×360°≈32°

Miscellaneous:0.08×360°≈29°

Loanpayments:0.14×360°≈50°

Savings:0.04×360°≈14°




thegraph,youcanincreasethesizeofone

ofthesectionsby1°(forexample,housing

to149°).



Gabriella’s Montly Expenses

$840.00

$175.00

$175.00

$175.00

$150.00

$125.00$75.00

$275.00

Savings

Loan payments

Miscellaneous

Clothing

Entertainment

Transportation

Food

Housing

b) CompareGabriella’sspendingtothe

guidelines.

Item Gabriella’s spending (percentage)

Guidelines (percentage)

Housing 42% 30–35%

Food 9% 8–15%

Transportation 6% 8–12%

Debtrepayment 14% 0–12%

Entertainment 9% 8–12%

Personal 9%(clothing)+8%(misc.)=17%

8–12%

Savings 4% atleast10%

Gabriellaisspendingtoomuchonhousing,

debtrepayment,andpersonalexpenses.

Sheisspendingbelowtheguidelineson

transportation,andisnotputtingenough

moneyintosavings.

Sheiswithintheguidelinesforfoodand

entertainment.

CalculatehowmuchmoneyGabriella

shouldbespendingineachcategory.

Housing:

$1990.00×0.35=$696.50

Comparetohercurrentspending.

$840.00–$696.50=$143.50

Sheisspendingabout$145.00toomuch

onhousing.

Transportation:

$1990.00×0.12=$238.80

$238.80–$125.00=$113.80

Gabriellacouldspendanadditional

$113.80ontransportationandstillfall

withintheguidelines.

Debt repayment:

$1990.00×0.12=$238.80

$275.00–$238.80=$36.20

Gabriellaisspendingabout$36.00

morethanrecommendedtorepayher

loan.However,thisexpenseislikelynot

negotiablesoshewillhavetocontinue

payingthissameamount.

Personal expenses (clothing and

miscellaneous):

$1990.00×0.12=$238.80

($175.00+$150.00)–$238.80=$86.20

Gabriellaisspendingabout$90.00too

muchonpersonalexpenses.Shecould

decreaseherspendingonclothingto

$100.00andonmiscellaneousto$135.00.


Savings:

$1990.00×0.10=$199.00

$199.00–$75.00=$124.00

Gabriellaissavingabout$125.00lessper

monththanrecommended.

c) DividethetotalsavingsgoalbyGabriella’s

currentsavingspermonth($75.00).

$1000.00÷$75.00≈13.3months

ItwilltakeGabriellaabout13.3monthsto

save$1000.00.Becauseshewillnothave

enoughmoneysavedatthebeginningof

the13thmonth,youshouldroundthe

answerup.Itwilltakeher14monthsto

save$1000.00.

Calculatehowmuchshewouldhavetosave

permonthtosave$1000.00in6months.

$1000.00÷6≈$166.67

Calculatethedifferencefromwhatsheis

currentlysaving.

$166.67–$75.00=$91.67

Ifshewantedtosave$1000.00in6

months,shewouldneedtosave$91.67

morepermonth.Hersavingswouldstillbe

lowerthantheguidelinessuggest.

chaPter teSt, p. 367

1. Item Income ExpenseBiweeklypaycheque:$735.00 Regular

Monthlybuspass:$75.00 Recurring

Gas:$42.95 Variable

Cellphonebill:$39.96 Recurring

Newjacket:$69.95 Variable

Tips:$52.00 Variable

Debtrepayment:$75.00 Recurring

Emergencyvetbillforcat:$250.00

Unexpected

Movieticketandsnack:$20.00 Variable

Dinneratrestaurant:$27.00 Variable

Charitabledonation:$25.00 Variable

Commission:$100.00 Variable


isprovided.

taMara’S MoNthlY BudgetIncome ExpensesSemi-monthly $975.00 Rent $575.00

Semi-monthly $975.00 Utilities,phone,cable,internet

$155.00

Groceries $275.00

Debtrepayment $100.00


Personal $465.00

Savings $55.00



budgetcreated.Basedonthebudgetabove,

Tamarasaves$55.00permonth.

$55.00/month×12months=$660.00

Ifshefollowsthisbudget,Tamarawillsave

$660.00inoneyear.


budgetcreated.Calculatewhatpercentage

$55.00representsof$1950.00.


$ .$ .

%55 001950 00

100 3× ≈

Basedonthisbudget,Tamarasavesabout

3%ofherincome.

3. a) CalculatehowmuchMasaruearnsperweek.

(35h×$11.75/h)+$125.00=$536.25

Calculatewhatpercentageofhisincome

Masaruputsintosavings.

100–85=15%

CalculatehowmuchmoneyMasaruputs

intosavingsinoneweek.

$536.25×0.15≈$80.44

Multiplytofindhowmuchhesaves

inoneyear.

$80.44/week×52weeks=$4182.88

Inoneyear,Masarusavesabout$4182.88.

b) CalculateMasaru’sannualincome.

$536.25/week×52weeks=$27885.00

Calculatewhatpercentage$7200.00

representsofhisincome.

$ .$ .

. %7200 0027 885 00

100 25 8× ≈

Masaruspendsabout26%ofhis

incomeonrent.

4. a)

MadeleiNe’S MoNthlY iNcoMe & exPeNSeSIncome ExpensesSemi-monthly $885.00 Rent $595.00

Semi-monthly $885.00 Utilities,phone,cable,andinternet

$275.00

Tips $250.00 Food $295.00



Other $350.00

Savings $180.00


b) Calculatewhatpercentageofherincome

eachcategoryrepresents.

Rent:($595.00÷$2020.00)×100≈29.5%

Utilities,phone,cable,andinternet:

($275.00÷$2020.00)×100≈13.6%

Food:($295.00÷$2020.00)×100≈14.6%

Transportation:($100.00÷$2020.00)×

100≈5.0%

Entertainment:($225.00÷$2020.00)×

100≈11.1%

Other:($350.00÷$2020.00)×100≈17.3%

Savings:($180.0÷$2020.00)×100≈8.9%

c) Calculatea5%increaseto

Madeleine’sincome.

$2020.00×0.05=$101.00

$2020.00+$101.00=$2121.00

Calculate8.9%of$2121.00.

$2121.00×0.089≈$188.77

Madeleinewillsave$188.77permonth.


5. a) Calculatewhatpercentageofhisincome

Noahspendsoneachcategory.

Housing:($795.00÷$2400.00)×100≈33%

Utilities:($150.00÷$2400.00)×100≈6%

Food:($500.00÷$2400.00)×100≈21%


×100≈16%


×100≈6%

Clothing:($100.00÷$2400.00)×100≈4%

Savings:($30.00÷$2400.00)×100≈1%

Other:($300.00÷$2400.00)×100≈13%

Comparethesepercentagestotheguidelines:

• Housing:withintheguidelines

• Utilities:withintheguidelines

• Food:higherthanrecommended

• Transportation:higherthanrecommended

• Entertainment:withintheguidelines

• Clothing:withintheguidelines

• Savings:lowerthanrecommended

• Other:higherthanrecommended

Forfood,transportation,savings,and

other,calculatehowmuchNoahshould

bespending.

Food:

$2400.00×0.15=$360.00

$500.00–$360.00=$140.00

Tostaywithintheguidelines,Noahwould

needtospend$140.00lessonfood.

Transportation:

$2400.00×0.15=$360.00

$375.00–$360.00=$15.00

Hewouldneedtospend$15.00lesson

transportation.

Other:

$2400.00×0.10=$240.00

$300.00–$240.00=$60.00

Tostaywithintheguidelines,Noahwould

havetospend$60.00lessonsavings.

Savings:

$2400.00×0.05=$120.00

$120.00–$30.00=$90.00

Noahwouldneedtosaveatleast$90.00

morepermonth.

b) $2400.00×0.05=$120.00

Heshouldputatleast$120.00intosavings

permonth.

c) Answerswillvary.Apossiblebudget

isprovided.

Noah’S MoNthlY BudgetIncome ExpensesSemi-monthly $1200.00 Rent $795.00

Semi-monthly $1200.00 Utilities $150.00

Food $360.00



Clothing $100.00

Savings $245.00

Other $240.00




firstcalculatewhatpercentageofNoah’s


Housing:($795.00÷$2400.00)×100≈33%

Utilities:($150.00÷$2400.00)×100≈6%

Food:($360.00÷$2400.00)×100=15%


×100=15%


×100≈6%

Clothing:($100.00÷$2400.00)×100≈4%

Savings:($245.00÷$2400.00)×100≈10%

Other:($240.00÷$2400.00)×100=10%

Notethatthepercentagesaddupto99%;this

isduetorounding.Forthenextstep,you

canincreaseoneofthecategoriesby1%(for

example,housingto34%).



Housing:0.34×360°≈122°

Utilities:0.06×360°≈22°

Food:0.15×360°=54°



Clothing:0.04×360°≈14°

Savings:0.10×360°=36°

Other:0.10×360°=36°



Noah’s Monthly Expenses

$795.00

$150.00

$360.00$360.00

$150.00

$100.00

$245.00

$240.00Other

Savings

Clothing

Entertainment

Transportation

Food

Utilities

Rent


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