Cambridge IGCSE Mathematics Extended Practice Book

21 Ratio, rate and proportion

3 ćJT TQFFEoUJNF HSBQI TIPXT UIF TQFFE PG B DBS JO LN I BHBJOTU UIF UJNF JO NJOVUFT
120
100
80

Speed (km/h) 60
40
20

Tip 0 123456
Time (minutes)
"DDFMFSBUJPO DIBOHF JO
TQFFE Ÿ UJNF UBLFO "T (a) 8IBU JT UIF TQFFE PG UIF DBS BęFS
UIF VOJUT PG TQFFE JO UIJT (i) UXP NJOVUFT
FYBNQMF BSF NFUSFT QFS (ii) TJY NJOVUFT
TFDPOE
UIFO UIF VOJUT PG
BDDFMFSBUJPO BSF NFUSFT QFS (b) 8IFO JT UIF DBS USBWFMMJOH BU LN I
TFDPOE QFS TFDPOE ćJT JT (c) $BMDVMBUF UIF BDDFMFSBUJPO PG UIF DBS JO LN I
XSJUUFO BT m s−2 PS N T (d) 8IBU EJTUBODF EJE UIF DBS DPWFS JO UIF ĕSTU TJY NJOVUFT

4 ćJT TQFFEoUJNF HSBQI TIPXT UIF TQFFE PG B USBJO JO N T BHBJOTU UIF UJNF JO TFDPOET

30

20
Speed
(m/s)

10

0 20 40 60 80 100 120
Time (seconds)

(a) 8IFO XBT UIF USBJO BDDFMFSBUJOH BOE XIBU XBT UIF BDDFMFSBUJPO
(b) 8IFO EJE UIF USBJO TUBSU EFDFMFSBUJOH BOE XIBU XBT UIF EFDFMFSBUJPO
(c) 8IFO UIF USBJO XBT USBWFMMJOH BU DPOTUBOU TQFFE
XIBU XBT UIF TQFFE JO LN I
(d) 8IBU EJTUBODF EJE UIF USBJO USBWFM JO UXP NJOVUFT

Unit 6: Number 143

21 Ratio, rate and proportion

21.5 Proportion
t 1SPQPSUJPO JT B DPOTUBOU SBUJP CFUXFFO UIF DPSSFTQPOEJOH FMFNFOUT PG UXP TFUT
t 8IFO RVBOUJUJFT BSF JO EJSFDU QSPQPSUJPO UIFZ JODSFBTF PS EFDSFBTF BU UIF TBNF SBUF ćF HSBQI PG B EJSFDUMZ

QSPQPSUJPOBUF SFMBUJPOTIJQ JT B TUSBJHIU MJOF QBTTJOH UISPVHI UIF PSJHJO

t 8IFO RVBOUJUJFT BSF JOWFSTFMZ QSPQPSUJPOBM
POF JODSFBTFT BT UIF PUIFS EFDSFBTFT ćF HSBQI PG BO JOWFSTFMZ
QSPQPSUJPOBM SFMBUJPOTIJQ JT B DVSWF

t ćF VOJUBSZ NFUIPE JT VTFGVM GPS TPMWJOH SBUJP BOE QSPQPSUJPO QSPCMFNT ćJT NFUIPE JOWPMWFT ĕOEJOH UIF
WBMVF PG POF VOJU PG XPSL
UJNF
FUD
BOE VTJOH UIBU WBMVF UP ĕOE UIF WBMVF PG B OVNCFS PG VOJUT

Tip Exercise 21.5

*G x BOE y BSF EJSFDUMZ 1 %FUFSNJOF XIFUIFS " BOE # BSF EJSFDUMZ QSPQPSUJPOBM JO FBDI DBTF

QSPQPSUJPOBM
UIFO x JT UIF (a) A 6
y

TBNF GPS WBSJPVT WBMVFT PG B 600

x BOE y

(b) A 5

B

(c) A
B

2 " UFYUCPPL DPTUT

(a) 8IBU JT UIF QSJDF PG TFWFO CPPLT
(b) 8IBU JT UIF QSJDF PG UFO CPPLT

3 'JOE UIF DPTU PG ĕWF JEFOUJDBMMZ QSJDFE JUFNT JG TFWFO JUFNT DPTU

4 *G B N UBMM QPMF DBTUT B N TIBEPX
ĕOE UIF MFOHUI PG UIF TIBEPX DBTU CZ B N UBMM
QPMF BU UIF TBNF UJNF

5 " DBS USBWFMT B EJTUBODF PG LN JO UISFF IPVST BU B DPOTUBOU TQFFE

(a) 8IBU EJTUBODF XJMM JU DPWFS JO POF IPVS BU UIF TBNF TQFFE
(b) )PX GBS XJMM JU USBWFM JO ĕWF IPVST BU UIF TBNF TQFFE
(c) )PX MPOH XJMM JU UBLF UP USBWFM LN BU UIF TBNF TQFFE

6 " USVDL VTFT MJUSFT PG EJFTFM UP USBWFM LJMPNFUSFT

(a) )PX NVDI EJFTFM XJMM JU VTF UP USBWFM LN BU UIF TBNF SBUF
(b) )PX GBS DPVME UIF USVDL USBWFM PO MJUSFT PG EJFTFM BU UIF TBNF SBUF

7 *U UBLFT POF FNQMPZFF UFO EBZT UP DPNQMFUF B QSPKFDU *G BOPUIFS FNQMPZFF KPJOT IJN
JU POMZ
UBLFT ĕWF EBZT 'JWF FNQMPZFFT DBO DPNQMFUF UIF KPC JO UXP EBZT

(a) %FTDSJCF UIJT SFMBUJPOTIJQ
(b) )PX MPOH XPVME JU UBLF UP DPNQMFUF UIF QSPKFDU XJUI

(i) GPVS FNQMPZFFT
(ii) FNQMPZFFT

144 Unit 6: Number

21 Ratio, rate and proportion

8 "ęFS B UTVOBNJ
UFO QFPQMF IBWF FOPVHI GSFTI XBUFS UP MBTU UIFN GPS TJY EBZT BU B TFU SBUF
QFS QFSTPO

(a) )PX MPOH XPVME UIF XBUFS MBTU
JG UIFSF XFSF POMZ ĕWF QFPQMF ESJOLJOH JU BU UIF
TBNF SBUF

(b) "OPUIFS UXP QFPQMF KPJO UIF HSPVQ )PX MPOH XJMM UIF XBUFS MBTU JG JU JT VTFE BU UIF
TBNF SBUF

9 /JDL UPPL GPVS IPVST UP DPNQMFUF B KPVSOFZ BU LN I .BSJF EJE UIF TBNF KPVSOFZ
BU LN I )PX MPOH EJE JU UBLF IFS

10 " QMBOF USBWFMMJOH BU BO BWFSBHF TQFFE PG LN I UBLFT IPVST UP DPNQMFUF B KPVSOFZ
)PX GBTU XPVME JU OFFE UP USBWFM UP DPWFS UIF TBNF EJTUBODF JO UFO IPVST

21.6 Direct and inverse proportion in algebraic terms
t 8IFO UXP RVBOUJUJFT BSF EJSFDUMZ QSPQPSUJPOBM UIFO P kQ
XIFSF k JT B DPOTUBOU
t 8IFO UXP RVBOUJUJFT BSF JOWFSTFMZ QSPQPSUJPOBM UIFO PQ k
XIFSF k JT B DPOTUBOU

Exercise 21.6

Tip 1 (JWFO UIBU a WBSJFT EJSFDUMZ XJUI b BOE UIBU a XIFO b

(a) ĕOE UIF WBMVF PG UIF DPOTUBOU PG QSPQPSUJPOBMJUZ k

ćF TZNCPM ∝ NFBOT (b) ĕOE UIF WBMVF PG a XIFO b
QSPQPSUJPOBM UP
2 ćF UBCMF TIPXT WBMVFT PG m BOE T 4IPX UIBU T JT EJSFDUMZ QSPQPSUJPOBM UP m

m
T

3 F JT EJSFDUMZ QSPQPSUJPOBM UP m BOE F XIFO m
(a) 'JOE UIF WBMVF PG F XIFO m
(b) 'JOE UIF WBMVF PG m XIFO F

4 *O UIF GPMMPXJOH SFMBUJPOTIJQ
xy k 'JOE UIF NJTTJOH WBMVFT

x 6b
y a c

5 y ∝ 1 BOE y XIFO x
x

(a) 'JOE UIF WBMVF PG y XIFO x

(b) 'JOE UIF WBMVF PG x XIFO y

6 y JT EJSFDUMZ QSPQPSUJPOBM UP x
BOE y XIFO x

(a) 8SJUF UIF FRVBUJPO GPS UIJT SFMBUJPOTIJQ
(b) 'JOE y JG x
(c) 'JOE x JG y

Unit 6: Number 145

21 Ratio, rate and proportion

7 y JT JOWFSTFMZ QSPQPSUJPOBM UP x BOE y XIFO x

(a) 8SJUF UIF FRVBUJPO GPS UIJT SFMBUJPOTIJQ

(b) 'JOE y JG x

(c) 'JOE x JG y 1
3

8 a WBSJFT JOWFSTFMZ XJUI b BOE a XIFO b

(a) 'JOE UIF WBMVF PG b XIFO a

(b) 'JOE UIF WBMVF PG a XIFO b

9 ćF SFMBUJPOTIJQ CFUXFFO x BOE y JT HJWFO BT y = 10
x

(a) 'JOE UIF WBMVF PG y XIFO x

(b) $BMDVMBUF x XIFO y

10 " MFOHUI PG XJSF N MPOH JT DVU JOUP B OVNCFS y
PG FRVBM MFOHUIT x N


(a) 4IPX UIBU y JT JOWFSTFMZ QSPQPSUJPOBM UP x
(b) 8SJUF BO FRVBUJPO UP EFTDSJCF UIF SFMBUJPOTIJQ CFUXFFO x BOE y
(c) 'JOE UIF WBMVF PG y XIFO x N

21.7 Increasing and decreasing amounts by a given ratio

t :PV DBO JODSFBTF PS EFDSFBTF BNPVOUT JO B HJWFO SBUJP
t new amount
old amount = x
new amount = x × old amount 'PS BO JODSFBTF
x y 'PS B EFDSFBTF x y
y y

Exercise 21.7

1 ćF QSJDF PG BO BSUJDMF DPTUJOH JT JODSFBTFE JO UIF SBUJP 8IBU JT UIF OFX QSJDF

2 "O BQBSUNFOU QVSDIBTFE GPS IBT EFDSFBTFE JO WBMVF JO UIF SBUJP
8IBU JT JU XPSUI OPX

3 " DBS QVSDIBTFE GPS IBT EFDSFBTFE JO WBMVF JO UIF SBUJP
(a) 8IBU JT UIF OFX WBMVF PG UIF DBS
(b) )PX NVDI NPOFZ XPVME UIF PXOFS MPTF JG TIF TPME UIF DBS GPS UIF EFDSFBTFE WBMVF

4 *ODSFBTF JO UIF SBUJP

Mixed exercise 1 " USJBOHMF PG QFSJNFUFS NN IBT TJEF MFOHUIT JO UIF SBUJP
(a) 'JOE UIF MFOHUIT PG UIF TJEFT
(b) *T UIF USJBOHMF SJHIU BOHMFE (JWF B SFBTPO GPS ZPVS BOTXFS

2 0O B ĘPPS QMBO PG B TDIPPM
DN SFQSFTFOUT N JO UIF SFBM TDIPPM
8IBU JT UIF TDBMF PG UIF QMBO

146 Unit 6: Number

21 Ratio, rate and proportion

3 " DBS USBWFMT BU BO BWFSBHF TQFFE PG LN I

(a) 8IBU EJTUBODF XJMM UIF DBS USBWFM JO

(i) IPVS (ii) 4 1 IPVST (iii) NJOVUFT
2

(b) )PX MPOH XJMM JU UBLF UIF DBS UP USBWFM

(i) LN (ii) LN (iii) LN

4 ćJT USBWFM HSBQI TIPXT UIF KPVSOFZ PG B QFUSPM UBOLFS EPJOH EFMJWFSJFT

Distance (km)500

400

300

200

100

0 1234567
Time (hours)

(a) 8IBU EJTUBODF EJE UIF UBOLFS USBWFM JO UIF ĕSTU UXP IPVST
(b) 8IFO EJE UIF UBOLFS TUPQ UP NBLF JUT ĕSTU EFMJWFSZ 'PS IPX MPOH XBT JU TUPQQFE
(c) $BMDVMBUF UIF BWFSBHF TQFFE PG UIF UBOLFS CFUXFFO UIF ĕSTU BOE TFDPOE TUPQ

PO UIF SPVUF
(d) 8IBU XBT UIF BWFSBHF TQFFE PG UIF UBOLFS EVSJOH UIF MBTU UXP IPVST PG UIF KPVSOFZ
(e) )PX GBS EJE UIF UBOLFS USBWFM PO UIJT KPVSOFZ
5 ćJT TQFFEoUJNF HSBQI TIPXT UIF TQFFE JO N T GPS B DBS KPVSOFZ

20

Speed
(m/s) 10

0 10 20 30 40 50 60
Time (seconds)

(a) 'PS IPX MPOH XBT UIF DBS BDDFMFSBUJOH
(b) $BMDVMBUF UIF SBUF BU XIJDI JU EFDFMFSBUFE GSPN UP TFDPOET
(c) 8IBU EJTUBODF EJE UIF DBS DPWFS EVSJOH UIF ĕSTU TFDPOET
(d) )PX NBOZ NFUSFT EJE UIF DBS UBLF UP TUPQ PODF JU TUBSUFE EFDFMFSBUJOH

6 /JOF TUVEFOUT DPNQMFUF B UBTL JO UISFF NJOVUFT )PX MPOH XPVME JU UBLF TJY TUVEFOUT UP
DPNQMFUF UIF TBNF UBTL JG UIFZ XPSLFE BU UIF TBNF SBUF

7 " DVCF XJUI TJEFT PG DN IBT B NBTT PG HSBNT 'JOE UIF NBTT PG BOPUIFS DVCF NBEF
PG UIF TBNF NBUFSJBM JG JU IBT TJEFT PG DN

Unit 6: Number 147

21 Ratio, rate and proportion

8 ćF QSFTTVSF P
PG B HJWFO BNPVOU PG HBT JT JOWFSTFMZ QSPQPSUJPOBM UP UIF WPMVNF V
PG
UIF HBT
(a) &YQSFTT UIF SFMBUJPOTIJQ CFUXFFO P BOE V JO B HFOFSBMJTFE FRVBUJPO
(b) *U JT GPVOE UIBU P XIFO V 'JOE UIF WBMVF PG P XIFO V

9 4BMNB IBT B QIPUPHSBQI NN MPOH BOE NN XJEF ćJT JT UPP CJH GPS IFS BMCVN
4IF SFEVDFT JU PO UIF DPNQVUFS TP UIBU JU JT NN MPOH
(a) *O XIBU SBUJP EJE 4BMNB SFEVDF UIF QIPUPHSBQI
(b) $BMDVMBUF UIF OFX XJEUI PG UIF QIPUP DPSSFDU UP UXP EFDJNBM QMBDFT

148 Unit 6: Number

22 More equations, formulae and
functions

22.1 Setting up equations to solve problems
t :PV DBO TFU VQ ZPVS PXO FRVBUJPOT BOE VTF UIFN UP TPMWF QSPCMFNT UIBU IBWF B OVNCFS BT BO BOTXFS
t ćF ĕSTU TUFQ JO TFUUJOH VQ BO FRVBUJPO JT UP XPSL PVU XIBU OFFET UP CF DBMDVMBUFE 3FQSFTFOU UIJT BNPVOU

VTJOH B WBSJBCMF VTVBMMZ x
:PV UIFO DPOTUSVDU BO FRVBUJPO VTJOH UIF JOGPSNBUJPO ZPV BSF HJWFO BOE TPMWF

JU UP ĕOE UIF BOTXFS

Tip Exercise 22.1 A

8IFO ZPV IBWF UP HJWF 1 " SFDUBOHMF JT DN MPOHFS UIBO JU JT XJEF *G UIF SFDUBOHMF JT x DN MPOH
XSJUF EPXO
B GPSNVMB
ZPV TIPVME
FYQSFTT JU JO TJNQMFTU UFSNT (a) UIF XJEUI JO UFSNT PG x)
CZ DPMMFDUJOH MJLF UFSNT (b) B GPSNVMB GPS DBMDVMBUJOH UIF QFSJNFUFS P
PG UIF SFDUBOHMF
(c) B GPSNVMB GPS ĕOEJOH UIF BSFB A
PG UIF SFDUBOHMF JO TJNQMFTU UFSNT
Tip 2 ćSFF OVNCFST BSF SFQSFTFOUFE CZ x
x BOE x

ćFSF JT VTVBMMZ POF XPSE (a) 8SJUF B GPSNVMB GPS ĕOEJOH UIF TVN S
PG UIF UISFF OVNCFST
PS TIPSU TUBUFNFOU JO UIF (b) 8SJUF B GPSNVMB GPS ĕOEJOH UIF NFBO M
PG UIF UISFF OVNCFST
QSPCMFN UIBU NFBOT AFRVBM 3 ćF TNBMMFTU PG UISFF DPOTFDVUJWF OVNCFST JT x
UP &YBNQMFT BSF UPUBM

HJWFT
UIF BOTXFS JT
UIF (a) &YQSFTT UIF PUIFS UXP OVNCFST JO UFSNT PG x
SFTVMU JT
QSPEVDU PG JT
BOE (b) 8SJUF B GPSNVMB GPS ĕOEJOH UIF TVN S
PG UIF OVNCFST
TVN PG JT 4 4BNNZ JT UXP ZFBST PMEFS UIBO .BY 5BZP JT UISFF ZFBST ZPVOHFS UIBO .BY *G .BY JT x ZFBST
PME
XSJUF EPXO

(a) 4BNNZ T BHF JO UFSNT PG x
(b) 5BZP T BHF JO UFSN PG x
(c) B GPSNVMB GPS ĕOEJOH UIF DPNCJOFE BHFT PG UIF UISFF CPZT

Exercise 22.1 B

1 ćFSF BSF NPSF HJSMT UIBO CPZT JO B DMBTT PG TUVEFOUT )PX NBOZ CPZT BSF UIFSF

2 " SFDUBOHMF IBT BO BSFB PG DN BOE UIF CSFBEUI JT DN 'JOE UIF MFOHUI PG UIF SFDUBOHMF

3 ćFSF BSF UFO UJNFT BT NBOZ TJMWFS DBST UIBO SFE DBST JO B QBSLJOH BSFB *G UIFSF TJMWFS BOE
SFE DBST BMUPHFUIFS
ĕOE UIF OVNCFS PG DBST PG FBDI DPMPVS

4 /BEJSB JT ZFBST ZPVOHFS UIBO IFS GBUIFS /BEJSB T NPUIFS JT UXP ZFBST ZPVOHFS UIBO IFS
GBUIFS 5PHFUIFS /BEJSB
IFS NPUIFS BOE GBUIFS IBWF B DPNCJOFE BHF PG 8PSL PVU UIFJS BHFT

Unit 6: Algebra 149

22 More equations, formulae and functions

5 ćF QFSJNFUFS PG B QBSBMMFMPHSBN JT DN *G UIF MFOHUI JT UISFF UJNFT UIF CSFBEUI
DBMDVMBUF
UIF EJNFOTJPOT PG UIF QBSBMMFMPHSBN

6 ćSFF JUFNT
X
Y BOE Z BSF TVDI UIBU UIF QSJDF PG Y JT D NPSF UIBO UIF QSJDF PG X BOE UIF
QSJDF PG Z JT D DIFBQFS UIBO UIF QSJDF PG X 5XP PG JUFN X QMVT UISFF PG JUFN Y QMVT TJY PG
JUFN Z DPTU 8IBU JT UIF QSJDF PG FBDI JUFN

7 3VTIEJ T GBUIFS JT QSFTFOUMZ ĕWF UJNFT IJT BHF *O UISFF ZFBST UJNF
IF XJMM CF GPVS UJNFT
3VTIEJ T BHF )PX PME JT 3VTIEJ OPX

8 " DPODFSU UJDLFU DPTUT 1FOTJPOFST BOE TUVEFOUT QBZ 'PS POF QFSGPSNBODF
NPSF
EJTDPVOU UJDLFUT UIBO SFHVMBS QSJDFE UJDLFUT XFSF TPME BU UIF EPPS *G XBT DPMMFDUFE
IPX
NBOZ SFHVMBS QSJDFE UJDLFUT XFSF TPME

22.2 Using and transforming formulae
t 5P DIBOHF UIF TVCKFDU PG B GPSNVMB

– FYQBOE UP HFU SJE PG BOZ CSBDLFUT
– VTF JOWFSTF PQFSBUJPOT UP JTPMBUF UIF WBSJBCMF ZPV SFRVJSF

t 8IFO B GPSNVMB DPOUBJOT TRVBSFE UFSNT PS TRVBSF SPPUT
SFNFNCFS UIBU B TRVBSFE OVNCFS IBT CPUI B OFHBUJWF BOE B
QPTJUJWF SPPU

t ćF WBSJBCMF UIBU JT UP CF NBEF UIF TVCKFDU NBZ PDDVS NPSF UIBO PODF JO UIF GPSNVMB *G UIJT JT UIF DBTF
HBUIFS UIF MJLF
UFSNT BOE GBDUPSJTF CFGPSF ZPV FYQSFTT UIF GPSNVMB JO UFSNT PG UIF TVCKFDU WBSJBCMF

REWIND Exercise 22.2

You have met this topic in 1 ćF WBSJBCMF UIBU OFFET UP CF UIF TVCKFDU JT JO CSBDLFUT
chapter 6. This exercise uses the $IBOHF UIF TVCKFDU PG FBDI GPSNVMB UP UIBU WBSJBCMF
same principles to transform
slightly more complicated (a) U T V W (V)
formulae. (V)
(b) V W
U ¦ T (B)

(c) A = C (B)
B (Q)
(Q)
(d) A = B (Q)
C (P)

(e) P Q

(f) P Q

(g) P Q R

(h) Q P

(i) Q PR (P)

(j) Q (P − R) (P)
(Q)
(k) PQ R

2 0IN T MBX JT B GPSNVMB UIBU MJOLT WPMUBHF V

DVSSFOU I
BOE SFTJTUBODF R
(JWFO UIBU V RI

(a) DIBOHF UIF TVCKFDU PG UIF GPSNVMB UP I
(b) ĕOE UIF DVSSFOU JO BNQT
GPS B WPMUBHF PG WPMUT BOE B SFTJTUBODF PG PINT

150 Unit 6: Algebra

22 More equations, formulae and functions

3 ćF BSFB PG B DJSDMF DBO CF GPVOE VTJOH UIF GPSNVMB A ɀr

(a) $IBOHF UIF TVCKFDU PG UIF GPSNVMB UP r
(b) ' JOE UIF MFOHUI PG UIF SBEJVT PG B DJSDMF XJUI BO BSFB PG NN (JWF ZPVS BOTXFS DPSSFDU

UP UISFF TJHOJĕHBOU ĕHVSFT

4 ćF GPSNVMB GPS ĕOEJOH EFHSFFT $FMTJVT GSPN EFHSFFT 'BISFOIFJU JT C 5 (F − 32)
9

(a) .BLF F UIF TVCKFDU PG UIF GPSNVMB

(b) 'JOE UIF UFNQFSBUVSF JO EFHSFFT 'BISFOIFJU XIFO JU JT ¡$

(c) ć F UFNQFSBUVSF JO ,FMWJO DBO CF GPVOE VTJOH UIF GPSNVMB
K C
XIFSF K JT UIF

UFNQFSBUVSF JO ,FMWJO BOE C JT UIF UFNQFSBUVSF JO EFHSFFT $FMTJVT 6TF UIJT JOGPSNBUJPO

UPHFUIFS XJUI UIF QSFWJPVT GPSNVMB
BT OFDFTTBSZ
UP ĕOE UIF ,FMWJO FRVJWBMFOU PG ¡'

22.3 Functions and function notation
t " GVODUJPO JT B SVMF PS TFU PG JOTUSVDUJPOT
GPS DIBOHJOH BO JOQVU WBMVF JOUP BO PVUQVU WBMVF
t ćF TZNCPM G x
JT VTFE UP EFOPUF B GVODUJPO PG x ćJT JT DBMMFE GVODUJPO OPUBUJPO 'PS FYBNQMF G x
x o
t 5P ĕOE UIF WBMVF PG B GVODUJPO
TVDI BT G

ZPV TVCTUJUVUF UIF HJWFO OVNCFS GPS x BOE DBMDVMBUF UIF WBMVF PG UIF

FYQSFTTJPO

t " DPNQPTJUF GVODUJPO JT B DPNCJOBUJPO PG UXP PS NPSF GVODUJPOT ćF PSEFS JO XIJDI UIF GVODUJPOT BSF XSJUUFO JT
JNQPSUBOU HG NFBOT ĕSTU BQQMZ G BOE UIFO BQQMZ H

t ćF OPUBUJPO GPS UIF JOWFSTF PG B GVODUJPO PG G JT Go ćJT JT SFBE BT G JOWFSTF :PV DBO UIJOL PG UIF JOWFSTF PG B GVODUJPO
BT JUT SFWFSTF "QQMZJOH UIF JOWFSTF NFBOT XPSLJOH CBDLXBSET BOE VOEPJOH EPJOH UIF JOWFSTF
PG FBDI PQFSBUJPO

Tip Exercise 22.3

'JOEJOH UIF JOWFSTF PG B 1 G x
x
DBMDVMBUF
GVODUJPO JO QSBDUJDF NFBOT
NBLJOH x UIF TVCKFDU PG UIF (a) G
(b) G o
(c) G
(d) G m

GPSNVMB JOTUFBE PG y
2 (JWFO G x → x
(iii) G

(iii) G a b)
(a) 8SJUF EPXO UIF FYQSFTTJPO GPS G x

(b) $BMDVMBUF

(i) G
(ii) G


(c) 4IPX XIFUIFS G
G
G


(d) 'JOE

(i) G a
(ii) G b


(e) 'JOE a JG G a


3 (JWFO h : x → 5 x


(a) 8SJUF EPXO UIF FYQSFTTJPO GPS I x


(b) 'JOE

(i) I
(ii) I o


4 *G G x
x BOE H x
x o
ĕOE

(a) GH (b) HG

5 *G H x
x BOE I x
x
ĕOE HI


Unit 6: Algebra 151

22 More equations, formulae and functions

6 8SJUF EPXO UIF JOWFSTF PG FBDI PG UIF GPMMPXJOH GVODUJPOT VTJOH UIF DPSSFDU OPUBUJPO

(a) G x
x (b) G x
x o (c) G x
x (d) f (x) = x
(c) GH
o x) −2
x
7 (JWFO G x
x o BOE g(x) = 2
ĕOE

(a) GH x
(b) HG x


(d) HG
o x
(e) Go Ho x
(f) Ho Go x)

Mixed exercise 1 " DFSUBJO OVNCFS JODSFBTFE CZ TJY JT FRVBM UP UXJDF UIF TBNF OVNCFS EFDSFBTFE CZ GPVS

8IBU JT UIF OVNCFS

2 ćF TVN PG UISFF DPOTFDVUJWF OVNCFST JT 'JOE UIF OVNCFST

3 ;PSJOB JT B RVBSUFS PG UIF BHF PG IFS NPUIFS
XIP JT ZFBST PME *O IPX NBOZ ZFBST UJNF XJMM
TIF CF B UIJSE PG IFS NPUIFS T BHF

4 $FESJD IBT NPSF UIBO /BUIJ 5PHFUIFS UIFZ IBWF )PX NVDI EPFT FBDI
QFSTPO IBWF

5 4JOEJ
+POBT BOE .P QVU NPOFZ UPHFUIFS UP DPOUSJCVUF UP DIBSJUZ +POBT QVUT JO IBMG UIF
BNPVOU UIBU 4JOEJ QVUT JO BOE .P QVUT JO MFTT UIBO UXJDF UIF BNPVOU UIBU 4JOJE QVUT JO
8PSL PVU IPX NVDI FBDI QFSTPO QVU JO

6 "MUPHFUIFS BEVMUT BOE DIJMESFO UBLF QBSU JO B GVO IJLF "EVMUT QBZ BOE DIJMESFO QBZ
*G XBT DPMMFDUFE
IPX NBOZ DIJMESFO UPPL QBSU

7 .BLF b UIF TVCKFDU PG FBDI GPSNVMB

(a) a = 5(b ) + (b − 3)
3
9

(b) a b o
b)

8 (JWFO f (x) = 2x − 3
ĕOE Go x

5

9 G x
x BOE g(x) = x − 2
ĕOE
5

(a) Go x
(b) Go
(c) UIF WBMVF PG a JG G a


(d) Č x
(e) Ho G


152 Unit 6: Algebra

23 Transformations and matrices

23.1 Simple plane transformations

t " USBOTGPSNBUJPO JT B DIBOHF JO UIF QPTJUJPO PS TJ[F PG B QPJOU PS B TIBQF
t ćFSF BSF GPVS CBTJD USBOTGPSNBUJPOT SFĘFDUJPO
SPUBUJPO
USBOTMBUJPO BOE FOMBSHFNFOU

o ćF PSJHJOBM TIBQF JT DBMMFE UIF PCKFDU 0
BOE UIF USBOTGPSNFE TIBQF JT DBMMFE UIF JNBHF 0′


t " SFĘFDUJPO ĘJQT UIF TIBQF PWFS

o 6OEFS SFĘFDUJPO
FWFSZ QPJOU PO B TIBQF JT SFĘFDUFE JO B NJSSPS MJOF UP QSPEVDF B NJSSPS JNBHF PG UIF PCKFDU
1PJOUT PO UIF PCKFDU BOE DPSSFTQPOEJOH QPJOUT PO UIF JNBHF BSF UIF TBNF EJTUBODF GSPN UIF NJSSPS MJOF
XIFO
ZPV NFBTVSF UIF EJTUBODF QFSQFOEJDVMBS UP UIF NJSSPS MJOF

o 5P EFTDSJCF B SFĘFDUJPO ZPV OFFE UP HJWF UIF FRVBUJPO PG UIF NJSSPS MJOF

t " SPUBUJPO UVSOT UIF TIBQF BSPVOE B QPJOU

o ćF QPJOU BCPVU XIJDI B TIBQF JT SPUBUFE JT DBMMFE UIF DFOUSF PG SPUBUJPO ćF TIBQF NBZ CF SPUBUFE DMPDLXJTF PS
BOUJDMPDLXJTF

o 5P EFTDSJCF B SPUBUJPO ZPV OFFE UP HJWF UIF DFOUSF PG SPUBUJPO BOE UIF BOHMF BOE EJSFDUJPO PG UVSO

t " USBOTMBUJPO JT B TMJEF NPWFNFOU

o 6OEFS USBOTMBUJPO
FWFSZ QPJOU PO UIF PCKFDU NPWFT UIF TBNF EJTUBODF JO UIF TBNF EJSFDUJPO UP GPSN UIF JNBHF

5SBOTMBUJPO JOWPMWFT NPWJOH UIF TIBQF TJEFXBZT BOE PS VQ BOE EPXO ćF USBOTMBUJPO DBO UIFSFGPSF CF

EFTDSJCFE VTJOH B DPMVNO WFDUPS ⎝⎛⎜ x⎞ XIFSF x JT UIF NPWFNFOU UP UIF TJEF BMPOH UIF x BYJT
BOE y JT UIF
y⎠⎟

NPWFNFOU VQ PS EPXO ćF TJHO PG UIF x PS y WBMVF HJWFT ZPV UIF EJSFDUJPO PG UIF USBOTMBUJPO 1PTJUJWF NFBOT UP
UIF SJHIU PS VQ BOE OFHBUJWF NFBOT UP UIF MFę PS EPXO

t "O FOMBSHFNFOU JOWPMWFT DIBOHJOH UIF TJ[F PG BO PCKFDU UP QSPEVDF BO JNBHF UIBU JT TJNJMBS JO TIBQF UP UIF PCKFDU

o The enlargement factor = length length of a side on the image object 8IFO BO PCKFDU JT FOMBSHFE GSPN B
of the corresponding side on the

ĕYFE QPJOU
JU IBT B DFOUSF PG FOMBSHFNFOU ćF DFOUSF PG FOMBSHFNFOU EFUFSNJOFT UIF QPTJUJPO PG UIF JNBHF -JOFT

ESBXO UISPVHI DPSSFTQPOEJOH QPJOUT PO UIF PCKFDU BOE UIF JNBHF XJMM NFFU BU UIF DFOUSF PG FOMBSHFNFOU

t ćF USBOTGPSNBUJPOT BCPWF DBO CF DPNCJOFE

Tip Exercise 23.1 A

3FĘFDUJPO BOE SPUBUJPO :PV XJMM OFFE TRVBSFE QBQFS GPS UIJT FYFSDJTF
DIBOHF UIF QPTJUJPO BOE
PSJFOUBUJPO PG UIF PCKFDU 1 %SBX BOE MBCFM BOZ SFDUBOHMF ABCD
XIJMF USBOTMBUJPO POMZ (a) 3PUBUF UIF SFDUBOHMF DMPDLXJTF ° BCPVU QPJOU D -BCFM UIF JNBHF A′B′C′D′
DIBOHFT UIF QPTJUJPO (b) 3FĘFDU A′B′C′D′ BCPVU B′D′
&OMBSHFNFOU DIBOHFT
UIF TJ[F PG UIF PCKFDU UP
QSPEVDF UIF JNBHF CVU JUT
PSJFOUBUJPO JT OPU DIBOHFE

Unit 6: Shape, space and measures 153

23 Transformations and matrices

2 .BLF B DPQZ PG UIF EJBHSBN CFMPX BOE DBSSZ PVU UIF GPMMPXJOH USBOTGPSNBUJPOT

y
8
A6

4

2

BC
x

–8 –6 –4 –2 0 H2 4 6 8
–2 I

DE –4

K J
–6

G F –8

(a) 5SBOTMBUF ΔABC UISFF VOJUT UP UIF SJHIU BOE GPVS VOJUT VQ -BCFM UIF JNBHF DPSSFDUMZ
(b) 3FĘFDU SFDUBOHMF DEFG BCPVU UIF MJOF y = − -BCFM UIF JNBHF DPSSFDUMZ
(c) (i) 3PUBUF QBSBMMFMPHSBN HIJK ° BOUJDMPDLXJTF BCPVU QPJOU
−


(ii) 5 SBOTMBUF UIF JNBHF H ′I ′J ′K ′ POF VOJU MFę BOE ĕWF VOJUT VQ

3 .BLF B DPQZ PG UIF EJBHSBN CFMPX BOE DBSSZ PVU UIF GPMMPXJOH USBOTGPSNBUJPOT

(a) ΔABC JT USBOTMBUFE VTJOH UIF DPMVNO WFDUPS ⎛ 10⎞ UP GPSN UIF JNBHF A′B ′C ′ %SBX BOE
MBCFM UIF JNBHF ⎝⎜ −9⎠⎟

(b) 2VBESJMBUFSBM PQRS JT SFĘFDUFE JO UIF y BYJT BOE UIFO USBOTMBUFE VTJOH UIF DPMVNO
⎛ 0⎞

WFDUPS ⎝⎜ −6⎠⎟ %SBX UIF SFTVMUBOU JNBHF P′Q′R′S ′

154 Unit 6: Shape, space and measures

23 Transformations and matrices

4 'PS FBDI PG UIF SFĘFDUJPOT TIPXO JO UIF EJBHSBN
HJWF UIF FRVBUJPO PG UIF NJSSPS MJOF

y
10

8A
D D' 6

4 A'
x
2
C' 2 4 6 8 10
–10 –8 –6 –4 –2 0
B
–2
C

–4
B'

–6

–8

–10

5 $PQZ UIF EJBHSBN GPS RVFTUJPO BOE ESBX UIF SFĘFDUJPO PG FBDI TIBQF "o%
JO UIF x BYJT

6 *O FBDI PG UIF GPMMPXJOH
GVMMZ EFTDSJCF BU MFBTU UXP EJČFSFOU USBOTGPSNBUJPOT UIBU NBQ UIF
PCKFDU POUP JUT JNBHF

y
10

8

6

4

2 x
2 4 6 8 10
–10 –8 –6 –4 –2 0
–2

–4

–6

–8

–10

Unit 6: Shape, space and measures 155

23 Transformations and matrices

REWIND Exercise 23.1 B
You dealt with enlargement and
scale factors in chapter 21. 1 'PS FBDI QBJS PG TIBQFT
HJWF UIF DPPSEJOBUFT PG UIF DFOUSF PG FOMBSHFNFOU BOE UIF TDBMF GBDUPS
PG UIF FOMBSHFNFOU
You can find the centre of
enlargement by drawing lines y
through the corresponding vertices
on the two shapes. The lines will 8
meet at the centre of enlargement.
When the image is smaller than 6 B'
the object, the scale factor of the
‘enlargement’ will be a fraction. 4 B
A' A 2 24
x
–8 –6 –4 –2 0 D 68
C' –2
C –4 D'
–6

–8

2 $PQZ FBDI TIBQF POUP TRVBSFE HSJE QBQFS 6TJOH QPJOU 9 BT B DFOUSF PG FOMBSHFNFOU BOE B
TDBMF GBDUPS PG UXP
ESBX UIF JNBHF PG FBDI TIBQF VOEFS UIF HJWFO FOMBSHFNFOU

a
b
c
d


A

X C D
B X X

X

156 Unit 6: Shape, space and measures

23 Transformations and matrices

Exercise 23.1 C

1 5SJBOHMF ABC JT UP CF SFĘFDUFE JO UIF y BYJT BOE JUT JNBHF ΔA′B′C′ JT UIFO UP CF SFĘFDUFE JO
UIF x BYJT UP GPSN ΔA″ B″ C″

y

8
A

6

4 B
C

2

x
––11 2 4 6 8

(a) %SBX B TFU PG BYFT
FYUFOEJOH UIFN JOUP UIF OFHBUJWF EJSFDUJPO $PQZ ΔABC POUP ZPVS
HSJE BOE ESBX UIF USBOTGPSNBUJPOT EFTDSJCFE

(b) %FTDSJCF UIF TJOHMF USBOTGPSNBUJPO UIBU NBQT ΔABC EJSFDUMZ POUP ΔA″ B″ C″

2 4IBQF " JT UP CF FOMBSHFE CZ B TDBMF GBDUPS PG UXP
VTJOH UIF PSJHJO BT UIF DFOUSF PG

FOMBSHFNFOU
UP HFU TIBQF # 4IBQF # JT UIFO USBOTMBUFE
VTJOH UIF DPMVNO WFDUPS ⎛ −8⎞
UP HFU
⎝⎜ 1 ⎟⎠

TIBQF $

y
8

6

4

2A

x
–1–1 2 4 6 8

(a) %SBX B TFU PG BYFT
FYUFOEJOH UIF x BYJT JOUP UIF OFHBUJWF EJSFDUJPO $PQZ TIBQF " POUP
ZPVS HSJE BOE ESBX UIF UXP USBOTGPSNBUJPOT EFTDSJCFE

(b) 8IBU TJOHMF USBOTGPSNBUJPO XPVME IBWF UIF TBNF SFTVMUT BT UIFTF UXP USBOTGPSNBUJPOT

3 " USBQF[JVN ABCD XJUI JUT WFSUJDFT BU DPPSEJOBUFT A


B


C

BOE D

JT UP
CF SFĘFDUFE JO UIF MJOF x = ćF JNBHF JT UP CF SFĘFDUFE JO UIF MJOF y =

(a) %SBX UIF TIBQF BOE TIPX JUT QPTJUJPO BęFS FBDI SFĘFDUJPO -BCFM UIF ĕOBM JNBHF '′
(b) %FTDSJCF UIF TJOHMF USBOTGPSNBUJPO ZPV DPVME VTF UP USBOTGPSN ABCD UP '′

Unit 6: Shape, space and measures 157

23 Transformations and matrices

23.2 Vectors

t " WFDUPS JT B RVBOUJUZ UIBU IBT CPUI NBHOJUVEF TJ[F
BOE EJSFDUJPO

t 7FDUPST DBO CF SFQSFTFOUFE CZ MJOF TFHNFOUT ćF MFOHUI PG UIF MJOF SFQSFTFOUT UIF NBHOJUVEF PG UIF WFDUPS

BOE UIF BSSPX PO UIF MJOF SFQSFTFOUT UIF EJSFDUJPO PG UIF WFDUPS " WFDUPS SFQSFTFOUFE CZ MJOF TFHNFOU AB TUBSUT

BU A BOE FYUFOET JO UIF EJSFDUJPO PG B

t ćF OPUBUJPO a
b
c PS AB
AB JT VTFE GPS WFDUPST x⎞
t 7FDUPST DBO BMTP CF XSJUUFO BT DPMVNO WFDUPST JO UIF GPSN ⎝⎜⎛ y⎠⎟

" DPMVNO WFDUPS SFQSFTFOUT B USBOTMBUJPO IPX UIF QPJOU BU POF FOE PG UIF WFDUPS NPWFT UP HFU UP UIF PUIFS

FOE PG UIF MJOF
1⎞ SFQSFTFOUT B USBOTMBUJPO POF VOJU JO UIF x EJSFDUJPO BOE UXP VOJUT JO UIF y EJSFDUJPO
ćF DPMVNO WFDUPS ⎛⎝⎜ 2⎠⎟

t 7FDUPST BSF FRVBM JG UIFZ IBWF UIF TBNF NBHOJUVEF BOE UIF TBNF EJSFDUJPO

t ćF OFHBUJWF PG B WFDUPS JT UIF WFDUPS XJUI UIF TBNF NBHOJUVEF CVU UIF PQQPTJUF EJSFDUJPO 4P
UIF

OFHBUJWF PG a JT –a BOE UIF OFHBUJWF PG AB JT AB

t 7FDUPST DBOOPU CF NVMUJQMJFE CZ FBDI PUIFS
CVU UIFZ DBO CF NVMUJQMJFE CZ B TDBMBS B OVNCFS

x⎞
.VMUJQMZJOH BOZ WFDUPS ⎛ y⎠⎟ CZ B TDBMBS k
HJWFT ⎛ kx⎞
⎝⎜ ⎝⎜ ky⎟⎠

t 7FDUPST DBO CF BEEFE BOE TVCUSBDUFE VTJOH UIF AOPTF UP UBJM NFUIPE PS USJBOHMF SVMF
TP

⎛ x1 ⎞ + ⎛ x2 ⎞ = ⎛ x1 + x2 ⎞
⎝⎜ y1 ⎟⎠ ⎝⎜ y2 ⎟⎠ ⎝⎜ y1 + y2 ⎠⎟

5P TVCUSBDU WFDUPST
ZPV OFFE UP SFNFNCFS UIBU TVCUSBDUJOH B WFDUPS JT UIF TBNF BT BEEJOH JUT OFHBUJWF

4P AB − CA = AB + AC ⎛⎞

t :PV DBO VTF 1ZUIBHPSBT UIFPSFN UP ĕOE UIF NBHOJUVEF MFOHUI
PG B WFDUPS *G UIF WFDUPS JT ⎜⎝ y⎠⎟ UIFO JUT NBHOJUVEF JT

x2 + y2 ćF OPUBUJPO ]B] PS ]AB] JT VTFE UP SFQSFTFOU UIF NBHOJUVEF PG UIF WFDUPS

t " WFDUPS UIBU TUBSUT GSPN UIF PSJHJO 0
JT DBMMFE B QPTJUJPO WFDUPS *G OA JT B QPTJUJPO WFDUPS
UIFO UIF DPPSEJOBUFT PG

O NVTU CF

BOE UIF DPPSEJOBUFT PG QPJOU A IBWF UP CF UIF TBNF BT UIF DPNQPOFOUT PG UIF DPMVNO WFDUPS OA

t :PV DBO VTF QPTJUJPO WFDUPST UP ĕOE UIF NBHOJUVEF PG BOZ SFMBUFE WFDUPS CFDBVTF ZPV LOPX UIF DPPSEJOBUFT PG UIF

QPJOUT :PV DBO UIFO VTF 1ZUIBHPSBT UP XPSL PVU UIF MFOHUIT PG UIF TJEFT PG B SFMBUFE SJHIU BOHMFE USJBOHMF

Remember, a scalar is a quantity Exercise 23.2 A 1 6TJOH UIF QPJOUT PO UIF HSJE
FYQSFTT
without direction, basically just a FBDI PG UIF GPMMPXJOH BT B DPMVNO
number or measurement. y WFDUPS
10
(a) (b) BC
8

6 (c) AE (d) BD
AB (e) DB (f) EC

4 (g) CD (h) BE

(i) 8IBU JT UIF SFMBUJPOTIJQ CFUXFFO

2 BE BOE CD

Remember, equal vectors have the (j) 8IBU JT AB + BC

same magnitude and direction; –6 –4 –2 0 2 4 6 8 x (k) 8IBU JT AE − AB
negative vectors have the same 10

magnitude and opposite directions. ED (l) *T BC = ED
–2

158 Unit 6: Shape, space and measures

23 Transformations and matrices

Tip 2 3FQSFTFOU UIF GPMMPXJOH WFDUPST PO TRVBSFE QBQFS

8IFO ZPV ESBX B WFDUPS (a) ⎛ 2⎞ (b) CD = ⎛ −1⎞ (c) ⎛ 4⎞ (d) ⎛ 3⎞
PO TRVBSFE QBQFS ZPV DBO AB = ⎜⎝ 3⎟⎠ ⎝⎜ 4 ⎟⎠ EF = ⎝⎜ 5⎠⎟ GH = ⎝⎜ −4⎟⎠
DIPPTF BOZ TUBSUJOH QPJOU
3 'JOE UIF DPMVNO WFDUPS UIBU EFTDSJCFT UIF USBOTMBUJPO GSPN UIF PCKFDU UP JUT JNBHF JO FBDI PG
Remember, A is the object and A’ UIF GPMMPXJOH FYBNQMFT
is the image.
y
10

8 B'

C6 B
4
E'
C' 2 x

–10 –8 –6 –4 –2 0 2 4 6 8 10
E –2 D

–4 A'

A D'
–6

–8
–10

Exercise 23.2 B

1 (JWFO UIBU a = ⎝⎛⎜ 1⎞
b = ⎝⎛⎜ −2⎞ BOE c = ⎝⎜⎛ 0⎞
ĕOE
3⎠⎟ 4 ⎟⎠ −3⎟⎠

(a) b (b) a (c) − c (d) a + b (e) b + c

(f) a + b + c (g) a + b (h) b ¦ c (i) − a ¦ b (j) a ¦ b + c

2 ćF EJBHSBN TIPXT UIF QBUUFSO PO B ĘPPS UJMF ćF UJMFT BSF TRVBSFT EJWJEFE JOUP GPVS
DPOHSVFOU USJBOHMFT CZ UIF JOUFSTFDUJOH EJBHPOBMT PG FBDI TRVBSF

A B D

aG ć F WFDUPST = a
OC = b BOE OG = c BSF TIPXO 6TF UIJT
c JOGPSNBUJPO UP XSJUF FBDI PG UIF GPMMPXJOH JO UFSNT PG a
b BOE c

F (a) DE (b) AD (c) AG (d) OB

(e) OE (f) CD (g) BF + FD (h) DE + EF
(i) 4BF 3EF (j) 1 OC + 3GB
2

O bC E

Unit 6: Shape, space and measures 159

23 Transformations and matrices

3 %SBX BOZ QPTJUJPO WFDUPS OA B PO B TFU PG BYFT BOE UIFO JOEJDBUF

(a) –a (b) a (c) 1 a (d) o a (e) − 1 a
2 2

Tip 4 ćF GPMMPXJOH WFDUPST BSF ESBXO PO B DN TRVBSFE HSJE $BMDVMBUF UIF NBHOJUVEF PG FBDI

*G ZPV DBOOPU SFNFNCFS WFDUPS JO DFOUJNFUSFT (JWF ZPVS BOTXFST DPSSFDU UP UXP EFDJNBM QMBDFT
UIF SVMF GPS ĕOEJOH UIF
NBHOJUVEF PG B WFDUPS
(a) MD = ⎛ 4⎞ (b) PQ = ⎛ −2⎞
UIFO ESBX UIF WFDUPS PO B ⎝⎜ 5⎠⎟ ⎜⎝ 7 ⎠⎟
HSJE BT UIF IZQPUFOVTF PG B
SJHIU BOHMFE USJBOHMF :PV ⎛ 9⎞ ⎛ −13⎞
XJMM RVJDLMZ TFF IPX UP (c) XY = ⎜⎝ −12⎠⎟ (d) ST = ⎝⎜ −12⎠⎟
XPSL PVU JUT MFOHUI VTJOH
1ZUIBHPSBT UIFPSFN 5 1PJOU O JT UIF PSJHJO 1PJOU X JT


QPJOU Y JT
o
BOE QPJOU Z JT o
o
'JOE UIF WBMVF
PG

(a) ]OX] (b) ]OY ]
(c) ]OZ] (d) ]XY ]

6 (JWFO UIF QPTJUJPO WFDUPST OA = ⎛ −6⎞
OB = ⎛ −2⎞ BOE OC = ⎛ 5⎞
⎝⎜ 2 ⎠⎟ ⎝⎜ −4⎠⎟ ⎝⎜ 1⎠⎟

(a) HJWF UIF DPPSEJOBUFT PG QPJOUT A
B BOE C

(b) FYQSFTT WFDUPST AB
BC BOE CA BT DPMVNO WFDUPST

7 *O 6XYZ
x BOE YZ y M JT UIF NJEQPJOU PG XY &YQSFTT XZ
ZX BOE MZ JO UFSNT PG
x BOE y Y

x y
M

Tip XZ

8IFO UXP WFDUPST BSF ⎛1⎞ ⎛ 3⎞ ⎛ −2⎞
FRVBM UIFZ IBWF UIF TBNF 8 a ⎝⎜ 2⎠⎟
b ⎜⎝ 0⎠⎟ BOE c ⎜⎝ 5 ⎠⎟
NBHOJUVEF BOE EJSFDUJPO

IFODF UIF MJOFT BTTPDJBUFE (a) 8SJUF B DPMVNO WFDUPS GPS x
y BOE z
JG
XJUI UIFN BSF FRVBM BOE
QBSBMMFM (i) x a b c
7FDUPST XJUI UIF TBNF (ii) y a o b o c
EJSFDUJPO BSF QBSBMMFM
FWFO (iii) z a b o c
JG UIFZ EP OPU IBWF UIF
TBNF NBHOJUVEF (b) $BMDVMBUF BOE HJWF ZPVS BOTXFST UP UISFF TJHOJĕDBOU ĕHVSFT

Tip (i) ]x] (ii) ]y] (iii) ] z]

%SBX B EJBHSBN UP 9 *O UIJT USJBOHMF
X BOE Y BSF UIF NJEQPJOUT PG AB BOE AC SFTQFDUJWFMZ D JT UIF NJEQPJOU PG
SFQSFTFOU UIJT TJUVBUJPO XY
AX a BOE AY b
BOE UIFO BQQMZ UIF WFDUPS B
DBMDVMBUJPOT
X
a

AD

b
Y

C

160 Unit 6: Shape, space and measures

23 Transformations and matrices

(a) &YQSFTT UIFTF WFDUPST JO UFSNT PG a BOE b
HJWJOH ZPVS BOTXFST JO UIFJS TJNQMFTU GPSN

(i) XY (ii) AD (iii) BC

(b) 4IPX UIBU XY ]] BC BOE JT IBMG JUT MFOHUI

10 $BSMPT KPHHFE GPS NJOVUFT BU B TUFBEZ QBDF PG LN I JO B OPSUI FBTUFSMZ EJSFDUJPO )F
UIFO XBMLFE XFTUXBSET VOUJM IF XBT EVF OPSUI PG IJT TUBSUJOH QPJOU
XIFSF IF TUPQQFE GPS
MVODI *G $BSMPT KPHHFE IPNF USBWFMMJOH EVF TPVUI BU UIF TBNF QBDF BT IF TFU PČ
IPX NBOZ
NJOVUFT EJE UIF MBTU MFH PG IJT KPVSOFZ UBLF IJN (JWF ZPVS BOTXFS UP POF EFDJNBM QMBDF


23.3 Further transformations

t " TIFBS JT B USBOTGPSNBUJPO UIBU LFFQT POF MJOF PG B TIBQF ĕYFE JOWBSJBOU
BOE NPWFT BMM PUIFS QPJOUT PO UIF TIBQF

QBSBMMFM UP UIF JOWBSJBOU MJOF

o ćF EJTUBODF UIBU UIF QPJOUT NPWF JT QSPQPSUJPOBM UP UIF EJTUBODF PG UIF QPJOUT GSPN UIF JOWBSJBOU MJOF
UIF
DPOTUBOU PG QSPQPSUJPOBMJUZ JT DBMMFE UIF TIFBS GBDUPS

o 5P EFTDSJCF B TIFBS ZPV OFFE UP HJWF UIF TIFBS GBDUPS BOE TUBUF XIJDI MJOF JT JOWBSJBOU

t " POF XBZ TUSFUDI JT B USBOTGPSNBUJPO UIBU FOMBSHFT UIF TIBQF JO POF EJSFDUJPO
QFSQFOEJDVMBS UP BO JOWBSJBOU MJOF

o ćF EJTUBODF UIBU UIF QPJOUT NPWF GSPN UIF JOWBSJBOU MJOF JT QSPQPSUJPOBM UP UIFJS PSJHJOBM EJTUBODF GSPN UIF
JOWBSJBOU MJOF
UIF DPOTUBOU PG QSPQPSUJPOBMJUZ JT DBMMFE UIF TDBMF GBDUPS

o 5P EFTDSJCF B POF XBZ TUSFUDI ZPV OFFE UP HJWF UIF TDBMF GBDUPS BOE TUBUF XIJDI MJOF JT JOWBSJBOU

t " UXP XBZ TUSFUDI JT B DPNCJOBUJPO PG UXP POF XBZ TUSFUDIFT BU ¡ UP FBDI PUIFS

o ćF POMZ JOWBSJBOU QPJOU PO UIF PCKFDU JT UIF QPJOU XIFSF UIF UXP QFSQFOEJDVMBS JOWBSJBOU MJOFT DSPTT UIJOL PG UIJT
BT UIF PSJHJO PG UIF TUSFUDI o POF TUSFUDI JT UIFO JO y EJSFDUJPO BOE UIF PUIFS JT JO x EJSFDUJPO


o " UXP XBZ TUSFUDI BČFDUT CPUI UIF x BOE y DPPSEJOBUFT PG FBDI QPJOU CZ UIF SFTQFDUJWF TDBMF GBDUPST

Exercise 23.3

1 %SBX UIF JNBHF PG FBDI TIBQF BęFS UIF HJWFO USBOTGPSNBUJPO

(a) 4IFBS GBDUPS
x JT JOWBSJBOU (b) 4IFBS GBDUPS
y JT JOWBSJBOU

y y
10 10
9 9
8 8
7 7
6 6
5 5
4 4
3 3
2 2
1 1

x x
–1–10 1 2 3 4 5 6 7 8 9 10 –1–10 1 2 3 4 5 6 7 8 9 10

Unit 6: Shape, space and measures 161

23 Transformations and matrices

Tip (c) 4IFBS GBDUPS
y BYJT JT JOWBSJBOU (d) 4USFUDI
TDBMF GBDUPS
y BYJT JT JOWBSJBOU

t :PV DBO UIJOL PG B POF y y
10 10
XBZ TUSFUDI BT B TIFBS 9 9
QFSQFOEJDVMBS UP UIF 8 8
JOWBSJBOU MJOF 7 7
6 6
t " TDBMF GBDUPS k M 5 5
4 4
SFEVDFT UIF TJ[F PG UIF 3 3
TIBQF JO B TUSFUDI BOE 2 2
UIF JNBHF XJMM CF DMPTFS 1 1
UP SBUIFS UIBO GVSUIFS
GSPN
UIF JOWBSJBOU MJOF x x
–1–10 1 2 3 4 5 6 7 8 9 10 –1–10 1 2 3 4 5 6 7 8 9 10

2 %FTDSJCF UIF TJOHMF USBOTGPSNBUJPO UIBU NBQT FBDI PCKFDU POUP JUT JNBHF
(a) (b)

(c) (d)

162 Unit 6: Shape, space and measures

23 Transformations and matrices

23.4 Matrices and matrix transformation

t " NBUSJY JT BO BSSBZ PG OVNCFST ćF OVNCFST JO UIF NBUSJY BSF DBMMFE UIF FMFNFOUT PG UIF NBUSJY
t ćF PSEFS PG NBUSJY JT UIF OVNCFS PG SPXT NVMUJQMJFE CZ UIF OVNCFS PG DPMVNOT 4P B NBUSJY XJUI UISFF SPXT BOE

GPVS DPMVNOT IBT BO PSEFS PG ¨

t .BUSJDFT PG UIF TBNF PSEFS DBO CF BEEFE BOE TVCUSBDUFE 5P EP UIJT
BEE PS TVCUSBDU UIF DPSSFTQPOEJOH FMFNFOUT PG

UIF UXP NBUSJDFT UP ĕOE UIF TPMVUJPO XIJDI XJMM CF B NBUSJY PG UIF TBNF PSEFS BT UIF PSJHJOBMT


t .BUSJDFT DBO CF NVMUJQMJFE CZ TDBMBST OVNCFST
.VMUJQMZ FBDI FMFNFOU PG UIF NBUSJY CZ UIF TDBMBS :PVS BOTXFS XJMM

CF B NBUSJY PG UIF TBNF PSEFS BT UIF PSJHJOBM

t :PV DBO NVMUJQMZ UXP NBUSJDFT A ¨ B
POMZ JG UIF OVNCFS PG DPMVNOT JO A JT FRVBM UP UIF OVNCFS PG SPXT JO B ćF

FMFNFOUT PG B SPX JO A BSF NVMUJQMJFE CZ UIF DPSSFTQPOEJOH FMFNFOUT PG B DPMVNO JO B BOE UIF QSPEVDUT BEEFE o UIF
SFTVMU HPFT JOUP B QPTJUJPO FRVJWBMFOU UP UIF SPX OVNCFS PG A BOE UIF DPMVNO OVNCFS PG B ćJT JT SFQFBUFE GPS BMM
DPNCJOBUJPOT PG SPXT GSPN A XJUI DPMVNOT GSPN B

⎛ 1 0⎞

t ćF TRVBSF ¨ NBUSJY I ⎝⎜ 0 1⎠⎟ JT DBMMFE UIF identity NBUSJY 8IFO ZPV NVMUJQMZ B ¨ NBUSJY A CZ * PS * CZ B

¨ NBUSJY A UIFO AI A BOE IA A

⎛ 0 0⎞

t ćF TRVBSF ¨ NBUSJY O ⎜⎝ 0 0⎠⎟ JT DBMMFE UIF zero NBUSJY GPS ¨ NBUSJDFT .VMUJQMJDBUJPO CZ UIF [FSP NBUSJY JT

FRVBM UP UIF [FSP NBUSJY FBDI FMFNFOU JT NVMUJQMJFE CZ
HJWJOH BMM SPVOE


t 'PS BOZ ¨ NBUSJY A ⎝⎛⎜ a b⎞ UIF DBMDVMBUJPO ad o bc
JT DBMMFE UIF determinant PG UIF NBUSJY ćF TZNCPM ]A]
c d⎠⎟

EFOPUFT UIF EFUFSNJOBOU PG A

t :PV DBOOPU EJWJEF NBUSJDFT
CVU ZPV DBO ĕOE UIF JOWFSTF PG B OPO TJOHVMBS NBUSJY +VTU BT B OVNCFS ¨ JUT JOWFSTF


B NBUSJY ¨ JUT JOWFSTF I UIF identity NBUSJY
ćF JOWFSTF PG NBUSJY A JT EFOPUFE CZ A–1
TP A ¨ A–1 I XIFSF UIF
JOWFSTF FYJTUT


t ⎛ ⎞
A–1 (ad 1 bc ) ⎛ d b⎞ 1
'PS A ⎜⎝ c d⎠⎟ − ⎜⎝ −c
a ⎠⎟ ćF NVMUJQMJFS (ad − bc) JT UIF SFDJQSPDBM PG UIF EFUFSNJOBOU PG UIF NBUSJY

BOE UIF FMFNFOUT PG UIF NBUSJY JO UIF NVMUJQMJDBUJPO BSF GPVOE CZ TXBQQJOH FMFNFOUT a BOE d BOE DIBOHJOH UIF TJHO

PG FMFNFOUT b BOE c PG UIF PSJHJOBM NBUSJY A

t 8IFO ad bc
UIFO (ad 1 bc) = 1 #FDBVTF 10 JT VOEFĕOFE ZPV DBOOPU EJWJEF CZ

UIF NBUSJY IBT OP JOWFSTF BOE JU
− 0

JT DBMMFE B TJOHVMBS NBUSJY

Exercise 23.4

⎛ −1 2 3 ⎞ ⎛1 4 2⎞
1 (JWFO A ⎝⎜ 4 3 −3⎠⎟ BOE B ⎝⎜ 3 1 0⎠⎟
ĕOE

(a) A B (b) A o B (c) B o A

2 D ⎝⎜⎛ 1 4⎞
5 8⎟⎠

(a) 'JOE UIF EFUFSNJOBOU PG D
(b) 'JOE D–1

Unit 6: Shape, space and measures 163

23 Transformations and matrices

⎛3 0⎞ BOE B ⎛⎝⎜ 1 3⎞
3 (JWFO A ⎜⎝ 1 2⎠⎟ −4 2⎠⎟

(a) 4IPX CZ DBMDVMBUJPO UIBU AB ȶ BA

(b) 'JOE

(i) ]A] (ii) ]AB]

4 (JWFO A ⎛ 1 3 0⎞ BOE B ⎛6 3 −3⎞
ĕOE
⎜⎝ 4 2 −1⎠⎟ ⎜⎝ 7 2 5 ⎠⎟

(a) A B (b) 3A (c) o B

5 (JWFO UIBU C ⎛ 2 x⎞ BOE ]C]
⎝⎜ 1 4⎠⎟

(a) 'JOE UIF WBMVF PG x
(b) 'JOE C–1

23.5 Matrices and transformations

t $PMVNO WFDUPST DBO CF VTFE UP SFQSFTFOU USBOTMBUJPOT PO B HSJE BOE ¨ NBUSJDFT DBO SFQSFTFOU SFĘFDUJPO
SPUBUJPO


FOMBSHFNFOU
TUSFUDI BOE TIFBS USBOTGPSNBUJPOT

t 5P ĕOE UIF JNBHF PG BOZ QPJOU ZPV NVMUJQMZ UIF NBUSJY CZ UIF QPTJUJPO WFDUPS PG UIF QPJOU 3FNFNCFS UIF QPTJUJPO

WFDUPS PG B QPJOU ⎝⎛⎜ ⎞ HJWFT ZPV UIF DPPSEJOBUFT x
y
PG UIF QPJOU

y⎟⎠

t 5P ĕOE XIBU ¨ NBUSJY SFQSFTFOUT B HJWFO USBOTGPSNBUJPO

o TFMFDU UIF QPJOUT

BOE



o ĕOE UIF JNBHFT PG UIFTF QPJOUT VOEFS UIF HJWFO USBOTGPSNBUJPO

o XSJUF EPXO UIF QPTJUJPO WFDUPST PG UIFTF JNBHFT

o UIF QPTJUJPO WFDUPS PG UIF JNBHF PG UIF ĕSTU QPJOU CFDPNFT UIF ĕSTU DPMVNO BOE UIF QPTJUJPO WFDUPS PG UIF JNBHF PG
UIF TFDPOE QPJOU JT UIF TFDPOE DPMVNO PG UIF SFRVJSFE NBUSJY

t 'PS DPNCJOFE USBOTGPSNBUJPOT SFQSFTFOUFE CZ UXP NBUSJDFT USBOTGPSNBUJPO CZ A BOE UIFO CZ B

UIF NBUSJY QSPEVDU

BA EFTDSJCFT UIF DPNCJOFE USBOTGPSNBUJPO /PUF UIBU UIF PSEFS PG NVMUJQMJDBUJPO JT WFSZ JNQPSUBOU JO NBUSJDFT
CFDBVTF AB ȶ BA

Exercise 23.5

Tip 1 'PS CPUI QBJST PG TIBQFT

:PV EPO U OFFE UP LOPX (i) EFTDSJCF GVMMZ UIF USBOTGPSNBUJPO UIBU NBQT TIBQF " POUP TIBQF #
UIF NBUSJDFT UIBU SFQSFTFOU
EJČFSFOU USBOTGPSNBUJPOT (ii) ĕOE UIF ¨ NBUSJY UIBU SFQSFTFOUT FBDI USBOTGPSNBUJPO
GSPN NFNPSZ CVU ZPV
NVTU CF BCMF UP XPSL (a) y (b) y
UIFN PVU BOE VTF UIFN UP
EFTDSJCF USBOTGPSNBUJPOT 10 10

9 9

8 8

7 7

6 6

5 5

4 4

3 3

A2 B 2

1x 1A B x

–4 –3 –2 –1–01 1 2 3 4 5 6 –1–10 1 2 3 4 5 6 7 8 9 10

164 Unit 6: Shape, space and measures

23 Transformations and matrices

2 " TRVBSF XJUI WFSUJDFT BU P


Q


R

BOE S

JT USBOTGPSNFE CZ UIF

NBUSJY ⎛ 1 0⎞
⎜⎝ 0 4⎠⎟

(a) 'JOE UIF DPPSEJOBUFT PG WFSUFY 3h VOEFS UIJT USBOTGPSNBUJPO

(b) %FTDSJCF UIF USBOTGPSNBUJPO JO XPSET ⎛ 3 0⎞

3 ćF USBOTGPSNBUJPO 5 JT EFĕOFE CZ UIF NBUSJY A XIFSF A ⎝⎜ 0 3⎠⎟

(a) %FTDSJCF UIF USBOTGPSNBUJPO 5 JO XPSET ⎛ −1 0⎞

(b) U "I FT FNDPBOUSEJY U SUBIOBTUG SPFSQNSFBTUFJPOOUT
5UI F
DJTP SNFQCSJOFTFFEO UUSFBEO CTGZP USINF BNUJPBUOS JBYA B XIFSF B ⎝⎜ 0 1⎠⎟ 'JOE

4 'JOE UIF NBUSJY UIBU SFQSFTFOUT UIF DPNCJOFE USBOTGPSNBUJPO SFĘFDUJPO JO UIF x BYJT GPMMPXFE
CZ BO FOMBSHFNFOU PG TDBMF GBDUPS BCPVU UIF PSJHJO

Mixed exercise 1 10 y

8

B 6A

H4
2

E x
–10 –8 –6 –4 –2 0 2 4 6 8 10

–2 C

–4
–6 G
F –8 D

–10

(a) %FTDSJCF B TJOHMF USBOTGPSNBUJPO UIBU NBQT USJBOHMF " POUP

(i) USJBOHMF # (ii) USJBOHMF $ (iii) USJBOHMF %

(b) %FTDSJCF UIF QBJS PG USBOTGPSNBUJPOT UIBU ZPV DPVME VTF UP NBQ USJBOHMF " POUP

(i) USJBOHMF & (ii) USJBOHMF ' (iii) USJBOHMF ( (iv) USJBOHMF )

2 4BMMZ USBOTMBUFE QBSBMMFMPHSBN DEFG BMPOH UIF DPMVNO WFDUPS ⎛ −4⎞ BOE UIFO SPUBUFE JU °
⎝⎜ 5 ⎟⎠
y
8 DMPDLXJTF BCPVU UIF PSJHJO UP HFU UIF JNBHF D′E′F′G′ TIPXO PO UIF HSJE
6
4 D' (a) %SBX B EJBHSBN BOE SFWFSTF UIF USBOTGPSNBUJPOT 4BMMZ QFSGPSNFE PO UIF TIBQF UP TIPX
2 UIF PSJHJOBM QPTJUJPO PG DEFG

(b) &OMBSHF DEFG CZ B TDBMF GBDUPS PG VTJOH UIF PSJHJO BT UIF DFOUSF PG FOMBSHFNFOU -BCFM
ZPVS FOMBSHFNFOU D″ E″ F″ G″

G'

E' F' x

2468

Unit 6: Shape, space and measures 165

23 Transformations and matrices

3 4RVBSF ABCD JT TIPXO PO UIF HSJE

y
10

8 B
6
A4

2C

–10 –8 –6 –4 –2 D 246 x
–2 8 10

–4

–6

–8

–10

0OUP B DPQZ PG UIF EJBHSBN
ESBX UIF GPMMPXJOH USBOTGPSNBUJPOT BOE
JO FBDI DBTF
HJWF UIF
DPPSEJOBUFT PG UIF OFX QPTJUJPO PG WFSUFY B

(a) 3FĘFDU ABCD BCPVU UIF MJOF x = −

(b) 3PUBUF ABCD ° DMPDLXJTF BCPVU UIF PSJHJO

(c) 5SBOTMBUF ABCD BMPOH UIF DPMVNO WFDUPS ⎝⎛⎜ −3⎞
2 ⎠⎟

(d) &OMBSHF ABCD CZ B TDBMF GBDUPS PG VTJOH UIF PSJHJO BT UIF DFOUSF PG FOMBSHFNFOU

4 (JWFO UIBU a = ⎛ 3⎞
b = ⎜⎝⎛ 2⎞ BOE c = ⎛⎝⎜ −2⎞

⎜⎝ 6⎠⎟ −4⎟⎠ −4⎠⎟

(a) ĕOE

(i) a (ii) b + c (iii) a ¦ b (iv) a + b

(b) %SBX GPVS TFQBSBUF WFDUPS EJBHSBNT PO TRVBSFE QBQFS UP SFQSFTFOU

(i) a
a (ii) b
c
b + c (iii) a
b
a o b (iv) a
b
a + b

A 5 ABCDEF JT B SFHVMBS IFYBHPO XJUI DFOUSF O FA x BOE AB y
xy
FB (a) &YQSFTT UIF GPMMPXJOH WFDUPST JO UFSNT PG x BOE y

C (i) ED (ii) DE (iii) FB
E
(iv) EF (v) FD
D
(b) * G UIF DPPSEJOBUFT PG QPJOU E BSF

BOE UIF DPPSEJOBUFT PG QPJOU B BSF


$BMDVMBUF ]EB] DPSSFDU UP UISFF TJHOJĕDBOU ĕHVSFT

166 Unit 6: Shape, space and measures

23 Transformations and matrices

6 %FTDSJCF UIF GPMMPXJOH USBOTGPSNBUJPOT (b) y
(a) B′

B

A x
C

(c) (d)

7 %SBX UIF JNBHF PG ABCD VOEFS B TUSFUDI PG TDBMF GBDUPS 1 XJUI UIF x BYJT JOWBSJBOU
2

Unit 6: Shape, space and measures 167

23 Transformations and matrices

8 %SBX UIF JNBHF PG ABCD VOEFS B TIFBS PG GBDUPS XJUI y JOWBSJBOU

y

10

9

8

7

6

5

4

3
2A

1D B x

–1–10 1 2C 3 4 5 6 7 8 9 10

9 P ⎛ 1 2⎞ BOE Q ⎛ 1 3⎞
⎝⎜ 2 1⎟⎠ ⎝⎜ 2 2⎠⎟ $BMDVMBUF

(a) P Q
(b) P o Q
(c) PQ
(d) QP
(e) ]P]
(f) P–I

⎛ 2 5 a ⎞ ⎛ a⎞ ⎛ 45⎞

10 (JWFO ⎜ 3 4 −61⎟⎟⎠⎟ ⎜ b3⎟⎟⎠⎟ = ⎜ 4c7⎟⎠⎟⎟
ĕOE UIF WBMVFT PG a
b BOE c
⎜⎝⎜ −2 4 ⎝⎜⎜ ⎝⎜⎜

11 $PQZ UIF EJBHSBN POUP B TFU PG BYFT

y
10

9

8

7

6
5A

4

3

2

1 x
CB

–1–10 1 2 3 4 5 6 7 8 9 10

(a) %SBX UIF JNBHF PG ABC BęFS SFĘFDUJPO JO UIF x BYJT
(b) 8SJUF EPXO UIF ¨ NBUSJY A UIBU SFQSFTFOUT UIJT USBOTGPSNBUJPO
(c) &OMBSHF UIF JNBHF A'B'C' CZ B TDBMF GBDUPS PG XJUI DFOUSF PG FOMBSHFNFOU BU


(d) 8SJUF EPXO UIF ¨ NBUSJY B UIBU SFQSFTFOUT UIJT USBOTGPSNBUJPO
(e) 'JOE UIF NBUSJY PG UIF DPNCJOFE USBOTGPSNBUJPO A GPMMPXFE CZ B

168 Unit 6: Shape, space and measures

24 Probability using tree
diagrams

24.1 Using tree diagrams to show outcomes
t " QSPCBCJMJUZ USFF TIPXT BMM UIF QPTTJCMF PVUDPNFT GPS TJNQMF DPNCJOFE FWFOUT
t &BDI MJOF TFHNFOU PS CSBODI SFQSFTFOUT POF PVUDPNF ćF FOE PG FBDI CSBODI TFHNFOU JT MBCFMMFE XJUI UIF PVUDPNF

BOE UIF QSPCBCJMJUZ PG FBDI PVUDPNF JT XSJUUFO PO UIF CSBODI

Tip Exercise 24.1

3FNFNCFS
GPS 1 "OJUB IBT GPVS DBSET ćFZ BSF ZFMMPX
SFE
HSFFO BOE CMVF 4IF ESBXT B DBSE BU SBOEPN BOE
JOEFQFOEFOU FWFOUT UIFO UPTTFT B DPJO %SBX B USFF EJBHSBN UP TIPX BMM QPTTJCMF PVUDPNFT
1 " BOE UIFO #
1 "

¨ 1 #
BOE GPS NVUVBMMZ 2 ćF TQJOOFS TIPXO IBT OVNCFST PO BO JOOFS DJSDMF BOE MFUUFST PO BO PVUFS SJOH 8IFO TQVO

FYDMVTJWF FWFOUT JU HJWFT B SFTVMU DPOTJTUJOH PG B OVNCFS BOE B MFUUFS %SBX B USFF EJBHSBN UP TIPX BMM QPTTJCMF
1 " PS #
1 "
1 #
PVUDPNFT XIFO ZPV TQJO JU

H A
G1 2B
F4 3C
D
E

3 " UJO DPOUBJOT FJHIU HSFFO DPVOUFST BOE GPVS ZFMMPX DPVOUFST 0OF DPVOUFS JT ESBXO BOE OPU
SFQMBDFE
BOE UIFO BOPUIFS JT ESBXO GSPN UIF CBH

(a) %SBX B USFF EJBHSBN UP TIPX UIF QPTTJCMF PVUDPNFT GPS ESBXJOH UXP DPVOUFST
(b) -BCFM UIF CSBODIFT UP TIPX UIF QSPCBCJMJUZ PG FBDI FWFOU

24.2 Calculating probability from tree diagrams
t 5P EFUFSNJOF UIF QSPCBCJMJUZ PG B DPNCJOBUJPO PG PVUDPNFT
NVMUJQMZ BMPOH FBDI PG UIF DPOTFDVUJWF CSBODIFT *G

TFWFSBM DPNCJOBUJPOT TBUJTGZ UIF TBNF PVUDPNF DPOEJUJPOT
UIFO BEE UIF QSPCBCJMJUJFT PG UIF EJČFSFOU QBUIT

t ćF TVN PG BMM UIF QSPCBCJMJUJFT PO B TFU PG CSBODIFT NVTU FRVBM POF

Exercise 24.2

1 8IFO B DPMPVSFE CBMM JT ESBXO GSPN B CBH
B DPJO JT UPTTFE PODF PS UXJDF
EFQFOEJOH PO UIF
DPMPVS PG UIF CBMM ESBXO ćFSF BSF UISFF CMVF CBMMT
UXP ZFMMPX CBMMT BOE B CMBDL CBMM JO UIF
CBH ćF USFF EJBHSBN TIPXT UIF QPTTJCMF PVUDPNFT

Unit 6: Data handling 169

24 Probability using tree diagrams

Blue H
Yellow
Black TH
HT
TH
HT

T

(a) $PQZ BOE MBCFM UIF EJBHSBN UP TIPX UIF QSPCBCJMJUZ PG FBDI FWFOU "TTVNF UIF ESBX PG
UIF CBMMT JT SBOEPN BOE UIF DPJO JT GBJS

(b) $BMDVMBUF UIF QSPCBCJMJUZ PG B CMVF CBMM BOE B IFBE
(c) $BMDVMBUF UIF QSPCBCJMJUZ PG B ZFMMPX CBMM BOE UXP IFBET
(d) $BMDVMBUF UIF QSPCBCJMJUZ UIBU ZPV XJMM OPU HFU IFBET BU BMM

2 ćF USFF EJBHSBN CFMPX TIPXT UIF QPTTJCMF PVUDPNFT XIFO UXP DPJOT BSF UPTTFE

First Second
toss
toss

1 H
2

1 H
2
1T

2

1 H
2

1 T
2

1T

2

(a) $PQZ BOE UIF USFF EJBHSBN UP TIPX UIF QPTTJCMF PVUDPNFT XIFO B UIJSE DPJO JT UPTTFE
(b) $BMDVMBUF UIF QSPCBCJMJUZ PG UPTTJOH UISFF IFBET
(c) $BMDVMBUF UIF QSPCBCJMJUZ PG HFUUJOH BU MFBTU UXP UBJMT
(d) $BMDVMBUF UIF QSPCBCJMJUZ PG HFUUJOH GFXFS IFBET UIBO UBJMT
(e) $BMDVMBUF UIF QSPCBCJMJUZ PG HFUUJOH BO FRVBM OVNCFS PG IFBET BOE UBJMT

Mixed exercise 1 %JOFP JT QMBZJOH B HBNF XIFSF TIF SPMMT B OPSNBM TJY TJEFE EJF BOE UPTTFT B DPJO *G UIF TDPSF
PO UIF EJF JT FWFO
TIF UPTTFT UIF DPJO PODF *G UIF TDPSF PO UIF EJF JT PEE
TIF UPTTFT UIF DPJO
Tip UXJDF

ćJT JT BO FYBNQMF PG (a) %SBX B USFF EJBHSBN UP TIPX BMM QPTTJCMF PVUDPNFT
DPOEJUJPOBM QSPCBCJMJUZ (b) "TTVNJOH UIF EJF BOE UIF DPJO BSF GBJS BOE BMM PVUDPNFT BSF FRVBMMZ MJLFMZ
MBCFM UIF
ćF ĕSTU DPJO JT SFNPWFE
BOE OPU SFQMBDFE
MFBWJOH CSBODIFT XJUI UIF DPSSFDU QSPCBCJMJUJFT
GFXFS QPTTJCMF PVUDPNFT (c) 8IBU JT UIF QSPCBCJMJUZ PG PCUBJOJOH UXP UBJMT
GPS UIF TFDPOE DPJO JO (d) 8IBU JT UIF QSPCBCJMJUZ PG PCUBJOJOH B ĕWF
B IFBE BOE B UBJM JO BOZ PSEFS

FBDI DBTF
2 " CBH DPOUBJOT UXP D BOE ĕWF D DPJOT :PV BSF BTLFE UP ESBX B DPJO GSPN UIF CBH BU
SBOEPN XJUIPVU SFQMBDJOH BOZ VOUJM ZPV HFU B D DPJO

(a) %SBX B QSPCBCJMJUZ USFF UP TIPX UIF QPTTJCMF PVUDPNFT
(b) -BCFM UIF CSBODIFT UP TIPX UIF QSPCBCJMJUZ PG FBDI FWFOU
(c) $BMDVMBUF UIF QSPCBCJMJUZ PG HFUUJOH B D PO ZPVS ĕSTU ESBX
(d) 8IBU JT UIF QSPCBCJMJUZ PG ESBXJOH UIF UXP D DPJOT CFGPSF ZPV ESBX B D DPJO
(e) * G ZPV ESBX UXP D DPJOT PO ZPVS ĕSTU UXP ESBXT
XIBU JT UIF QSPCBCJMJUZ PG HFUUJOH B D

DPJO PO ZPVS UIJSE ESBX 8IZ

170 Unit 6: Data handling

Answers Exercise 1.3 (s) 4.03 (t) 6.87
(u) 6.61 (v) 3.90
1 (a) 2, 3, 5, 7 (w) −19.10 (x) 20.19
(b) 53, 59
Chapter 1 (c) 97, 101, 103 Exercise 1.7

Exercise 1.1 2 (a) 2 × 2 × 3 × 3 (i) (ii) (iii)
(b) 5 × 13
1 Number Natural Integer Prime Fraction (c) 2 × 2 × 2 × 2 × 2 × 2 1 (a) 5.65 5.7 6
(d) 2 × 2 × 3 × 7
−0.2 ✓ (e) 2 × 2 × 2 × 2 × 5 (b) 9.88 9.9 10
(f) 2 × 2 × 2 × 5 × 5 × 5
(g) 2 × 5 × 127 (c) 12.87 12.9 13
(h) 13 × 151
−57 ✓ (d) 0.01 0.0 0

3.142 ✓ (e) 10.10 10.1 10

0 (f) 45.44 45.4 45
0.3˙ ✓
(g) 14.00 14.0 14

3 (a) LCM = 378, HCF = 1 (h) 26.00 26.0 26
(b) LCM = 255, HCF = 5
1 ✓✓ (c) LCM = 864, HCF = 3 2 (a) 53 200 (b) 713 000
(d) LCM = 848, HCF = 1 (c) 17.4 (d) 0.00728
51 ✓ ✓ (e) LCM = 24 264, HCF = 2
(f) LCM = 2574, HCF = 6
10 270 ✓ ✓ (g) LCM = 35 200, HCF = 2 3 (a) 36 (b) 5.2
(h) LCM = 17 325, HCF = 5 (c) 12 000 (d) 0.0088
− 1 ✓ (e) 430 000 (f) 120
4 (g) 0.0046 (h) 10

2✓ Mixed exercise
7

9 ✓ ✓✓ Exercise 1.4

11 ✓ ✓ ✓ 1 square: 121, 144, 169, 196, 225, 256, 289 1 natural: 24, 17
✓✓ cube: 125, 216
rational: − 3 , 0.65, 3 1 , 0 66
3 512 2 (a) 7 (b) 5 (c) 14
(e) 3 (f) 25 42
2 (a) 121, 144, 169, 196, . . . (d) 10
integer: 24, –12, 0, 17
3 prime: 17
4
(b) 1 , 1 , 2 , 2 , etc. (g) (h) 5 (i) 2 2 (a) two are prime: 2 and 3
4 6 7 9 (b) 2 × 2 × 3 × 3
(j) 5 (k) 1 3 (l) 12 (c) 2 and 36 (d) 36
(c) 83, 89, 97, 101, . . . 4
(d) 2, 3, 5, 7
(m) –5 (n) 5 (o) 6 3 (a) 2 × 2 × 7 × 7
(b) 3 × 3 × 5 × 41
6 (c) 2 × 2 × 3 × 3 × 5 × 7 × 7

Exercise 1.2 A 3 (a) 23 cm (b) 529 cm2

1 (a) 18 (b) 36 (c) 90 Exercise 1.5
(d) 24 (e) 36 (f) 24
(g) 72 (h) 96 1 −3 °C 4 14th and 28th March

2 (a) 6 (b) 18 (c) 9 2 (a) −2°C (b) −9°C (c) −12°C 5 (a) true (b) true
(d) 3 (e) 10 (f) 1 (c) false (d) false
(g) 12 (h) 50
3 (a) 4 (b) 7 (c) −1 6 (a) 5 (b) 5
(d) −2 (e) −3 (c) 64 (d) 145

Exercise 1.2 B Exercise 1.6 7 (a) 16.07 (b) 9.79 (c) 13.51
(d) 11.01 (e) 0.12 (f) −7.74
1 18 m 1 (a) 26 (b) 66
(c) 23.2 (d) 15.66
2 120 shoppers (e) 3.39 (f) 2.44 8 (a) 1240 (b) 0.765
(g) 3.83 (h) 2.15 (c) 0.0238 (d) 31.5
(i) 1.76 (j) 2.79
3 20 students (k) 7.82 (l) 0.21 9 (a) It is not strictly possible – to have
(m) 8.04 (n) 1.09 a tile of 790 cm2, the builders
4 after 420 seconds; 21 laps (Francesca), (o) 8.78 (p) 304.82 must be using rounded values.
5 laps (Ayuba) and 4 laps (Claire) (q) 94.78 (r) 0.63

5 (a) 1024 cm2 (b) 200 tiles (b) 28.1 cm
(c) 110 tiles

Answers 171

Chapter 2 (c) −8x 3 + 4x 2 + 2x (d) x + 3y 2 (a) 17 (b) 65
2 2 (c) 15 (d) –163
(b) 4
Exercise 2.1 (e) –2x2 + 8x (f) 3x2 – 6x 3 (a) 1
(e) 2
1 (a) 3(x + 2) (b) 6(x − 1) (g) –5x2 – 6x 4 (h) 3 (c) –4
(f) 2
(c) 2(11 + x) (d) 18x Exercise 2.5 A (d) 3 2
(e) 3x2 + 4 (f) x2 + 8
(g) 1 − x (h) x + 1 1 (a) x6 (b) 3x 4y 4
y2
5 3 (g) 2

(i) 4 + 3x (j) 12 – 5x (c) 2x2 (d) xy 10 Mixed exercise
3y
2 (a) p + 5 (b) p − 4 (c) 4p 1 (a) x + 12 (b) x − 4
(c) 5x
3 (a) $ x (e) 5x9 (f) x 7y 3 (d) x
3 2y3 (e) 4x 3
(g) 12 − x
(b) $ x , $ 2x , and $ 2x (g) 50 x 3 (h) 49 (f ) x
9 9 3 27 y 25x3 y 4

Exercise 2.2 (i) x 7y 8x10 y 3 (h) x 3 − x
3
(j) −14
9
1 (a) 54 cm2 (b) 1.875 m2 x16 3125x 4 y2 2 (a) −6 (b) 24 (c)
(c) 110.25 cm2 (d) 8 cm2 y16 16
(k) (l) 3 (a) −2 (b) 2 (c) 5
(d) 7 3
2 –104 x8 x5
2 (a) y2 (b) y4 (e) −2

3 17 8 1 4 (a) 9a + b
x5 y7 x9
4 17.75 (c) (d) (b) x 2 + 3x − 2

y16 y 22 (c) −4a 4b + 6a 2b3
x 22 2x4
5 (a) 6 (b) 91 (e) (f ) (d) −7x + 4

Exercise 2.3 1 8 (e) 4x
y
3 (a) x 2 (b) x15
1 (a) 3x 2 − 2x + 3 (f ) 5x − 5y
2
(b) 4x 2y − 2xy (c) x1 1
6 5 (a) 11x − 3
(d) x 9

(c) 5ab − 4ac 1 (b) 6x 2 + 15x − 8

(d) 4x 2 + 5x − y − 5 (e) 8x3 (f) 2x3 y 3 (c) −2x 2 + 5x + 12

(e) −30mn (f) 6x 2y 1 4 x3 y 1 or x3 (d) −x 3 + 3x 2 − x + 5
2 y
(g) 6xy 3 (h) −4x 3y (g) x y (h)

(i) 4b (j) 1 6 (a) 5x5 (b) 15
4y 6
1
(k) 3b 9m (i) x3 (j) x −2 y −4 or x2 y4 (c) 1 (d) 16x4y8
4 x4
(l)

(m) 20 y (n) 3x2 (k) y −2 or 1 (e) 64 x 9 (f) x 9y 8
3x y y2 y15

(o) 2y2 (p) y2 2 (b) x2 (g) 27x4 (h) xy 6
x2 2 2
4 (a) x 3 4y3

(q) 15a2 (r) −14 y 7 (d) x2 1 x1
4 5 3 2
(c) y 3

3 1 29 1 7 (a) 5xy (b)
4 4 16
x 27 x 2 (e) x y 2 (f ) x y or− − 1 29 y
6y 10 x9
(s) (t) x 4 y 16 (c) x −9 y or

1 5 5
3 3
Exercise 2.4 Exercise 2.5 B (d) 2x y− or 2y3

1 (a) –1296 (b) –1 (c) 8 1
(d) 2 (e) 1
3 x3
4
1 (a) 2x 2 − 4x (b) xy − 3x (f) 1 Chapter 3
(c) −2x − 2 (d) −3x + 2
(e) −2x 2 + 6x (f) 3x + 1 625 Exercise 3.1 A
(g) x 3 − 2x 2 − x (h) x 2 + x + 2
(g) 32 (h) 4 (i) 3 1 (a) (i) 150° (ii) 180° (iii) 135°
(b) x 2 + xy (j) 3 (b) 45°
x 2
2 2
2 (a) x2 +

172 Answers

(c) (i) 810° (ii) 72° (f) x = 58° (base ∠s isosceles and 5 (a) ∠MNP = 42°
∠s in triangle); y = 26° (b) ∠MNO = 104°
(d) quarter to one or 12:45 (ext ∠s equals sum int opps) (c) ∠PON = 56°

2 No. If the acute angle is ≤ 45° it will (g) x = 33° (base ∠s isosceles then ext Exercise 3.4
produce an acute or right angle. ∠s equals sum int opps)

3 Yes. The smallest obtuse angle is 91° (h) x = 45° (co-int ∠s then ∠s in 1 (a) (i) 1080° (ii) 135°
and the largest is 179°. Half of those triangle) (b) (i) 1440° (ii) 144°
will range from 45.5° to 89.5°, all of (c) (i) 2340° (ii) 156°
which are acute. (i) x = 45° (base ∠s isosceles);
y = 75° (base ∠s isosceles)
4 (a) 45° 2 900 = 128.57°
(b) (90 − x)° 2 (a) x = 36°; so A = 36° and B = 72° 7
(c) x° (b) x = 40°; so A = 80°; B = 40° and
∠ACD = 120° 3 20 sides
(c) x = 60°
5 (a) 135° (b) 90° (d) x = 72° 4 (a) 165.6° (b) 360 = 25 sides
(c) (180 − x)° (d) x° 14.4
(e) (90 + x)° (f) (90 − x)° 3 ∠B = 34°; ∠C = 68°
5 (a) x = 156°
Exercise 3.3 (b) x = 85°
(c) x = 113°; y = 104°

Exercise 3.1 B 1 (a) square, rhombus Exercise 3.5
(b) rectangle, square
1 ∠QON = 48°, so a = 48° (vertically (c) square, rectangle 1
opposite) (d) square, rectangle, rhombus,
parallelogram D M (d) tangent
2 (a) ∠EOD = 41° (∠s on line), so (e) square, rectangle
x = 41° (vertically opposite) (f) square, rectangle, parallelogram, sector
rhombus O 50° (a)
(b) x = 20° (∠s round point) (g) square, rhombus, kite
(h) rhombus, square, kite diameter
(i) rhombus, square, kite (b) chord
P
Exercise 3.1 C ( ) major E N
arc ()
1 (a) x = 85° (co-int ∠s);
y = 72° (alt ∠s) 2 45° 45° Exercise 3.6
45° 45°
(b) x = 99° (co-int ∠s); y = 123°
(∠ABF = 123°, co-int ∠s then 1, 2 student’s own diagrams
vertically opposite)
45° 45° 3 2.1 cm
(c) x = 72° (∠BFE = 72°, then alt ∠s); 45° 45°
y = 43° (∠s in triangle BCJ) 4 (a) scalene (b) yes
27° 63°
(d) x = 45° (∠s round a point); 27° 63° 5 6.6 cm
y = 90° (co-int ∠s)
27° Mixed exercise
2 (a) x = 112° 27°
(b) x = 45° ( STQ corr ∠s then 63° 1 (a) x = 113°
vertically opposite) 63° (b) x = 41°
(c) x = 90° (∠ECD and ∠ACD co-int (c) x = 66°
∠s then ∠s round a point) 3 (a) x = 69° (d) x = 74°; y = 106°; z = 37°
(d) x = 18° (∠DFE co-int with ∠CDF (b) x = 64° (e) x = 46°; y = 104°
then ∠BFE co-int with ∠ABF) (c) x = 52° (f) x = 110°; y = 124°
(e) x = 85° (d) x = 115° (g) x = 40°; y = 70°; z = 70°
(e) x = 30°; 2x = 60°; 3x = 90° (h) x = 35°; y = 55°
Exercise 3.2 (f) a = 44°; b = 68°; c = 44°; d = e = 68°
2 (a) x = 60 + 60 + 120 = 240°
1 (a) 103° ( s in triangle) 4 (a) ∠Q + ∠R = 210° (b) x = 90 + 90 + 135 = 315°
(b) 51° (ext ∠ equals sum int opps) (b) ∠R = 140° (c) x = 80°
(c) 68° (ext ∠ equals sum int opps) (c) ∠Q = 70°
(d) 53° (base ∠s isosceles)
(e) 60° (equilateral triangle)

Answers 173

3 (a) (i) radius (ii) chord 3 (a) Score Frequency 4 (a) student’s own chart
0−29 1 (b) student’s own chart
(iii) diameter 30−39 1
40−49 7
(b) AO, DO, OC, OB 50−59 19 5 (a) cars (b) 17% (c) 20
60−69 12 (d) handcarts and bicycles
(c) 24.8 cm 70−79 6
80−100 4
(d) student’s own diagram

4 student’s own diagram 6 (a) student’s own chart

5 student’s own diagram (b) 6 (c) 50 (d) C

Chapter 4 7 (a) 26° C
(b) May–Nov
Exercise 4.1 (b) 10 (c) 2 (d) 26 (c) northern hemisphere
(d) No
1 (a) gender, eye colour, hair colour (e) There are very few marks at the (e) 35 mm
(b) height, shoe size, mass, number of (f) April
brothers/sisters low and high end of the scale. (g) none
(c) shoe size, number of brothers/
sisters 4 (a)
(d) height, mass
(e) possible answers include: gender, Eye colour Brown Blue Green Mixed exercise
eye colour, hair colour − collected
by observation; height, Male 401 1 (a) survey or questionnaire
mass − collected by measuring; (b) discrete; you cannot have half a
shoe size, number of siblings − Female 212 child
collected by survey, questionnaire (c) quantitative; it can be counted
(b) (d)
Exercise 4.2
Hair colour Brown Black Blonde No. of
2 1 children in 0 1 2 3 4 5 6
Male 2 4 0 family

Female 1 Frequency 7 10 11 12 5 2 1

1 Phone Tally Frequency No. of brothers/ 0 1 2 3 4 (e) student’s own chart
calls sisters (f) student’s own chart
1/ 1 2 1
2 // 2 Male 011 1 0
3 // 2
4 //// 5 Female 211 2 student’s own pictogram
5 //// //// 9
6 //// // 7 (c) student’s own sentences 3 (a) compound bar chart
7 //// / 6 (b) It shows how many people, out of
8 /// 3 Exercise 4.3 every 100, have a mobile phone
9 /// 3 and how many have a land line
10 // 2 1 (a) pictogram phone.
(b) number of students in each year (c) No. The figures are percentages.
group in a school (d) Canada, USA and Denmark
(c) 30 students (e) Germany, UK, Sweden and Italy
(d) half a stick figure (f) Denmark
(e) 225 (g) own opinion with reason
(f) Year 11; 285
2 (a) (g) rounded; unlikely the year groups 4 (a) Student’s own chart
will all be multiples of 15 (b) Student’s own chart
(c) Student’s own chart
No. of 2 student’s own pictogram
mosquitoes 0 1 2 3 4 5 6
5 (a) Value of car over time
3 (a) number of boys and girls in 15
Value ($ 000)
Frequency 7 6 9 7 8 7 6 Class 10A 10

(b) It is impossible to say; frequency (b) 18 (c) 30 5
is very similar for all numbers of
mosquitoes. (d) the favourite sports of students in 0

Class 10A, separated by gender

(e) athletics 2008
2009
(f) athletics 2010
2011
2012
2013

(g) 9 Time (years)

(b) 49.6%
(c) $3900

174 Answers

Chapter 5 3 (a) 83% (b) 60% (c) 7% (g) 0.000028 (h) 94 000 000
(d) 37.5% (e) 125% (f) 250% (i) 0.00245
Exercise 5.1

4 (a) 60 kg (b) $24 Exercise 5.4 B

1 (a) x = 65 (b) x = 168 (c) x = 55 (c) 150 litres (d) 55 ml 1 (a) 6.56 × 10−17
(d) x = 117 (e) x = 48 (f) x = 104 (b) 1.28 × 10−14
(e) $64 (f) £19.50 (c) 1.44 × 1013
Exercise 5.2 (d) 1.58 × 10−20
(g) 18 km (h) 0.2 grams (e) 5.04 × 1018
(f) 1.98 × 1012
(i) $2.08 (j) 475 cubic metres (g) 1.52 × 1017
(h) 2.29 × 108
1 (a) 25 (b) 17 (c) 59 5 (a) +20% (b) −10% (i) 4.50 × 10−3
(c) +53.3% (d) +3.3%
8 11 5 (e) −28.3% (f) +33.3%
(g) +2566.7%
(d) 15 (e) 59 (f) 25

4 49

2 (a) 108 (b) 63 (c) 14 6 (a) $54.72 (b) $945
5 13 (c) $32.28 (d) $40 236
(e) $98.55 (f) $99.68 2 (a) 12 × 1030
(d) 28 (e) 3 (f ) 6 (b) 4.5 × 1011
5 19 (c) 3.375 × 1036
(d) 1.32 × 10−11
(g) 120 (h) 3 (i) 72 7 (a) $58.48 (b) $520 (e) 2 × 1026
14 (c) $83.16 (d) $19 882 (f) 2.67 × 105
(e) $76.93 (f) $45.24 (g) 1.2 × 102
(j) 3 (k) 233 (l) 7 (h) 2 × 10−3
50 4 (i) 2.09 × 10−8
Exercise 5.3 B
13 19 19
3 (a) 24 (b) 60 (c) 21 1 28 595 tickets

(d) 35 (e) 183 (f ) 161 2 1800 shares 3 (a) the Sun (b) 6.051 × 106
6 56 20

(g) 18 (h) 41 (i) 29 3 $129 375 4 (a) 500 seconds = 5 × 102 seconds
(b) 19 166.67 seconds = 1.92 × 104
65 40 21 seconds

(j) −5 (k) −10 (l) −26 4 21.95%
6 3 9
5 $15 696
Exercise 5.5
4 (a) 24 (b) 96 (c) 7 6 $6228
7 96
10 10 9 1 (a) 4 × 5 = 20
(d) 27 (e) 9 (f ) 14 7 2.5 grams (b) 70 × 5 = 350
(c) 1000 × 7 = 7000
(g) 2 (h) 3 (i) 4 8 7 = 28% increase, so $7 more is better (d) 42 ÷ 6 = 7
5 25

5 (a) 38 (b) 4 (c) 39 Exercise 5.3 C 2 (a) 20 (b) 3
9 5 7 (c) 12 (d) 243
19 5 215 1 $50
(d) 4 (e) 12 (f ) 72

(g) 0 (h) 11 (i) 187 2 (a) 1200 (b) 960 Mixed exercise
170
9 3 $150
1 (a) 40 (b) 6
6 (a) $525 (b) $375 4 (a) $12 (b) 27 750 (c) $114 885 (c) 22 (d) 72

7 (a) 300 Exercise 5.4 A 4 2 2
(b) 6 hours 56 min
1 (a) 4.5 × 104 2 (a) 5 (b) 3 (c) 3
(c) 8 × 10
Exercise 5.3 A (e) 4.19 × 106 (b) 8 × 105 1 (b) 63 5
(g) 6.5 × 10−3 (d) 2.345 × 106
(i) 4.5 × 10−4 (f) 3.2 × 1010 3 (a) 6 31 (c) 3
(k) 6.75 × 10−3 (h) 9 × 10−3
1 (a) 16.7% (b) 62.5% (c) 29.8% (j) 8 × 10−7 3 (e) 48 71
(d) 30% (e) 4% (f) 47% 2 (a) 2500 (l) 4.5 × 10−10
(g) 112% (h) 207% (c) 426 500 (d) 44 334 (f) 6
(e) 0.00000915 (b) 39 000
(d) 0.00001045 361 (h) 45 (i) 68
(f) 0.000000001 15
(g) 16

2 (a) 1 (b) 1 (c) 49 (j) 14
8 2 50 9

(d) 3 (e) 11 4 (a) $760 (b) $40 000
5 50

Answers 175

5 $10 000 (e) x = 17 (f ) x = 23 = 3 5 (f) a = d + bc (g) a = cd − b
6 6 2
6 (a) 720 (b) 11 779
(g) x = −1 (h) x = 9 = 4 1 (h) a = de − c (i) a = e d
7 67.9% 2 2 b bc

8 (a) 5.9 × 109 km (i) x = − 1 (j) x = 9 (j) a = ef − d (k) a = c( f b de)
(b) 5.753 × 109 km 3 bc

9 (a) 9.4637 × 1012 km (k) x = 16 = 1133 (l) x = 10 d (e c) d
(b) 1.6 × 10−5 light years 13 c
(c) 3.975 × 1013 km −11 (l) a = b (m) a b
(m) x = 42 (n) x = 2

Exercise 6.3 (n) a d b
c

Chapter 6 1 (a) 3 (b) 8 (c) 5 Exercise 6.4 B
(d) a (e) 3y (f) 5ab
Exercise 6.1 (g) 4xy (h) pq (i) 7ab 1 (a) b P l (b) b = 35.5 cm
(j) xy2z (k) ab3 (l) 3xy 2

1 (a) −2x −2y (b) −5a + 5b 2 (a) 12(x + 4) (b) 2(1 + 4y) 2 (a) r = C (b) 9 cm (c) 46 cm
(c) 6x − 3y (d) 8x − 4xy (c) 4(a − 4) (d) x(3 − y)
(e) −2x2 − 6xy (f) −9x + 9 (e) a(b + 5) (f) 3(x − 5y) 2π
(g) 12 − 6a (h) 3 − 4x − y (g) 8xz(3y − 1) (h) 3b(3a − 4c)
(i) 3 (j) −3x − 7 (i) 2y(3x − 2z) (j) 2x(7 − 13y) 3 use 2A ; b = 3.8 cm
(k) 2x2 − 2xy (l) −3x2 + 6xy h

4 (a) (i) 70 kg (ii) 12 kg

(b) 11 656 kg

2 (a) 14x − 2y − 9x 3 (a) x(x + 8) (b) a(12 − a) (c) T 70P = B
(b) −5xy + 10x (c) x(9x + 4) (d) 2x(11 − 8x)
(c) 6x − 6y − 2xy (e) 2b(3ab + 4) (f) 18xy(1 − 2x) 12
(d) −2x − 6y − xy (g) 3x(2 − 3x) (h) 2xy 2(7x − 3)
(e) 12xy − 14 − y + 3x (i) 3abc2(3c –ab) (j) x(4x − 7y) (d) 960 kg
(f) 4x2 − 2x2y − 3y (k) b2(3a − 4c) (l) 7ab(2a − 3b)
(g) −2x2 + 2x + 5 5 (a) t= m (b) 6 seconds
(h) 6x2 + 4y − 8xy 5
(i) –5x – 3
4 (a) (3 + y)(x + 4) Mixed exercise
(b) (y − 3)(x + 5)
Exercise 6.2 (c) (a + 2b)(3 − 2a) 1 (a) x = −3 (b) x = −6
(d) (2a − b)(4a − 3) (c) x = 9 (d) x = −6
(e) (2 − y)(x + 1) (e) x = 2 (f) x = −13
(f) (x − 3)(x + 4) (g) x = 1.5 (h) x = 5
(g) (2 + y)(9 − x)
1 (a) x = 3 (b) x = 4 (h) (2b − c)(4a + 1)
(i) (x − 6)(3x − 5)
(c) x = 9 = 4 1 (d) x = 4 (j) (x − y)(x − 2) 2 (a) x = m+r (b) x = mq − p
2 2 (k) (2x + 3)(3x + y) np n
(l) (x − y)(4 − 3x)
(e) x = 36 = 18 = 3 3 (f ) x=5 3 (a) 3x + 2 (b) −12x 2 + 8x
10 5 5 (c) −8x + 4y − 6 (d) −16y + 2y2
(e) 11x + 5 (f) 5x 2 + 7x − 4
(g) x = 2 (h) x = −5 (g) −2x 2 + 11x (h) 10x 2 + 24x
(i) x = 4
(j) x = − 3 = −1 1
2 2

(k) x = 11 = 5 1 (l) x = 3 Exercise 6.4 A 4 (a) 4(x − 2) (b) 3(4x − y)
2 2
(c) −2(x + 2) (d) 3x(y − 8)
1 m = D
2 (a) x = 10 (b) x = −2 k (e) 7xy(2xy + 1) (f) (x − y)(2 + x)

(c) x = − 8 = −2 2 (d) x = 4 = 1 1 2 c = y − mx (g) (4 + 3x)(x − 3)
3 3 3 3
(h) 4x(x + y)(x − 2)

(e) x = 8 (f ) x= 1 3 b = P + c 5 (a) 4(x − 7) = 4x − 28
4 a (b) 2x(x + 9) = 2x 2 + 18x
(c) 4x(4x + 3y) = 16x 2 + 12xy
(g) x = −4 (h) x = −9 (d) 19x(x + 2y) = 19x 2 + 38xy

(i) x = −10 (j) x = −13 4 b= axc

(k) x = −34 (l) x = 20 = 1173 5 (a) a = c − b (b) a = 2c + 3b 6 (a) x = 15°
13 (b) x = 26°
3 (a) x = 18 cd (d) a = d b c (c) x = 30°
(c) x = 24 (b) x = 27 (c) a= b

(d) x = −44 (e) a = bc − d (or a = −d + bc)

176 Answers

Chapter 7 Exercise 7.2 Exercise 7.3 A

Exercise 7.1 A 1 (a) cube 1 (a) 2.56 mm2 (b) 523.2 m2
(b) cuboid (c) 13.5 cm2 (d) 401.92 mm2
(c) square-based pyramid
1 (a) 120 mm (b) 45 cm (d) octahedron 2 (a) 384 cm2 (b) 8 cm
(c) 128 mm (d) 98 mm
(e) 36.2 cm (f) 233 mm 2 (a) cuboid 3 (a) 340 cm2 (b) 153 000 cm2
(b) triangular prism (c) 4 tins
2 (a) 15.7 m (b) 44.0 cm (c) cylinder
(c) 54.0 mm (d) 21.6 m 4 (a) 90 000 mm3 (b) 60 cm3
(e) 18.8 m (f) 151 mm 3 The following are examples; there are (c) 20 410 mm3 (d) 1120 cm3
(g) 24.4 cm other possible nets. (e) 960 cm3 (f) 5.76 m3
(a) (g) 1800 cm3 (h) 1.95 m3
3 90 m
(b)
4 164 × 45.50 = $7462 5 332.5 cm3
(c)
5 9 cm each 6 (a) 224 m3 (b) 44 people
(d)
6 about 88 cm 7 211.95 m3

7 (a) 197.82 cm (b) 219.8 cm 8 various answers − for example:

Exercise 7.1 B Volume (mm3) 64 000 64 000 64 000 64 000
Length (mm) 80 50 100 50
1 (a) 332.5 cm2 (b) 1.53 m2 Breadth (mm) 40 64 80 80
(c) 399 cm2 (d) 150 cm2 Height (mm) 20 20 8 16
(e) 59.5 cm2 (f) 71.5 cm2
(g) 2296 mm2 (h) 243 cm2 Exercise 7.3 B
(i) 7261.92 cm2
(b) 153.86 mm2 1 (a) 5.28 cm3 (b) 33 493.33 m3
2 (a) 7850 mm2 (d) 17.44 cm2 (c) 25.2 cm3 (d) 169.56 cm3
(c) 7693 mm2 (e) 65 111.04 mm3
(e) 167.47 cm2 (b) 82 cm2
(d) 581.5 cm2 2 (a) (i) 1.08 × 1012 km3
3 (a) 288 cm2 (f) 39 cm2 (ii) 5.10 × 108 km2
(c) 373.5 cm2 (h) 4000 cm2
(e) 366 cm2 (b) 1.48 × 108 km2
(g) 272.93 cm2 (b) 90 cm2
(i) 5639.63 cm2 (d) 61.2 cm2 Mixed exercise
(f) 2562.5 cm2
4 (a) 30 cm2 1 (a) 346.19 cm2 (b) 65.94 cm
(c) 33.6 cm2
(e) 720 cm2 2 4.55 cm

5 11.1 m2 3 (a) 2000 mm2 (b) 33 000 mm2
(c) 40 cm2 (d) 80 cm2
6 70 mm = 7 cm (e) 106 cm2 (f) 35 cm2
(g) 175.84 cm2

Exercise 7.1 C 4 15 m

1 (a) 43.96 mm (b) 47.1 cm 5 (a) cuboid is smaller
(c) 8.37 mm (b) 732.67 cm2 (b) 14 240 mm3
(c) student’s own diagram
2 6668.31 km (d) cylinder 7536 mm2, cuboid
9000 mm2
3 (a) 75.36 cm2
(c) 92.34 mm2 6 42

4 61.33 cm2 7 volume cone = 23.55 cm3
volume pyramid = 30 cm3
difference = 6.45 cm3

Answers 177

8 volume 3 balls = 1144.53 cm3 Exercise 8.4 3 (a) 5 (b) 28 (c) 40
volume tube = 1860.39 cm3
space = 715.86 cm3 1 17 153 153

Chapter 8 H H T (d) 40
T HH HT
HT TT 153

Exercise 8.1 (a) 3 (b) 1 (e) The four situations represent all
the possible outcomes, so they
4 4 must add up to one.

1 (a) red = 3 , white = 9 , green = 17 2 (a) Mixed exercise
10 25 50

(b) 30% (c) 1 (d) 1 Yellow 1 (a) 10 000 (b) heads 0.4083;
3 12
1 1, 1 1, 2 3 tails 0.5917
2 (a) 0.61, 0.22, 0.11, 0.05, 0.01 Green 2 2, 1 2, 2 1, 3 1
(b) (i) highly likely 3 3, 1 3, 2 2, 3 (c)
(ii) unlikely 3, 3 (d) c2ould be − probability of the tails
(iii) highly unlikely
outcome is higher than the heads

outcome for a great many tosses

Exercise 8.2 (b) 9 (c) 1 (d) 1 2 (a) 1 (b) 2 (c) 1
3 (a) 2 5 10
3 3
9 9
1 (a) 1, 2, 3, 4, 5, 6, 7, 8, 9 or 10 (d) 0 (e) 10 (f ) 10

(b) (i) 1 = 0.1 (ii) 1 (iii) 3 Snack (g) 1
10 10 2

(iv) 3 (v) 2 (vi) 1 cola, cola, cola, 3 (a) 1 (b) 7, 1
10 5 2 biscuit cake muffin
36 6
(vii) 3 (viii) 9 (ix) 0 fruit fruit fruit
10 10 juice, juice, juice, (c) 1 (d) 1
biscuit cake muffin
2 (a) 2 Drink 2 6
(b) water, water, water,
5 3 biscuit cake muffin 4 (a)
5
no sugar; probability = $1 $1 $1 50c 50c $5 20c 20c

3 (a) 1 (b) 1 $5 6 6 6 5.5 5.5 10 5.2 5.2

42 $5 6 6 6 5.5 5.5 10 5.2 5.2

(c) 1 (b) 1 2 $5 6 6 6 5.5 5.5 10 5.2 5.2
2
9 (c)
1 1 $5 6 6 6 5.5 5.5 10 5.2 5.2
3 2 3

4 (a) or 0.33 (b) or 0.5 $2 3 3 3 2.5 2.5 7 2.2 2.2

(c) 1 or 0.16 Exercise 8.5 50c 1.5 1.5 1.5 1 1 5.5 0.7 0.7

6 50c 1.5 1.5 1.5 1 1 5.5 0.7 0.7

5 (a) 7 (b) 1 (c) 2 1 (a) (b) 3 (c) 1 (d) 35
20 2 5
C AU A 14 4 56
(d) 3 (e) 1 L CA CU CA
10 5 C LA LU LA
13 T CA CU CA
6 40 T TA TU TA 1
TA TU TA
Exercise 8.3 (b) 4 58
(c) 1 (d) 14 61
15
5 15 3
2 (a) 1
1 0.73 (b) 2 1 Chapter 9
15
2 5 15 (c) Exercise 9.1
8 (d) 1
(e) 7 45
3 (a) 0.16 (b) 0.84 (c) 0.6 30 1 (a) 17, 19, 21 (add 2)
(d) strawberry 63, lime 66, lemon 54, 15 2 (b) 121, 132, 143 (add 11)
blackberry 69, apple 48 (g) 3 (c) 8, 4, 2 (divide by 2)
(f ) (d) 40, 48, 56 (add 8)
10 (e) −10, −12, −14 (subtract 2)
5 (f) 2, 4, 8 (multiply by 2)

4 (a) 0.6 (b) 0.97 (c) 11
(d) 114

178 Answers

(g) 11, 16, 22 (add one more each 3 (a) {} 4 (a) 23 (b) 286
time than added to previous (b) {1, 3, 5, 6, 7, 9, 11, 12, 13, 15, 18} 99 999
term) (c) {1, 3, 5, 7, 9, 11, 13, 15}
(d) {1, 2, 4, 5, 6, 7, 8, 10, 11, 12, 13, 5 (a)
(h) 21, 26, 31 (add 5) 14, 16, 17, 18, 19, 20}
(e) {2, 4, 6, 8, 10, 12}
2 (a) 7, 9, 11, 13 (f) {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, BC
13, 15, 18} 25 16 18
(b) 37, 32, 27, 22
1 1 1
(c) 1, 2 , 4 , 8

(d) 5, 11, 23, 47 4 (a) {–2, –1, 0, 1, 2}
(b) {1, 2, 3, 4, 5}
(e) 100, 47, 20.5, 7.25 21

3 (a) 5, 7, 9 T35 = 73 5 (a) {x: x is even, x ≤ 10} (b) 21 (c) 16 1
(b) 1, 4, 9 T35 = 1225 (b) {x: x is square numbers, x ≤ 25}
(c) 5, 11, 17 T35 = 209 17 41 (iii)
(d) 0, 7, 26 T35 = 42 874
(e) 0, 2, 6 T35 = 1190 (d) (i) (ii) 5
(f) 1, −1, −3 T35 = −67
Exercise 9.3 B 40 80

1 abdfgjk (iv) 59 (v) 21

80 80

4 (a) 8n − 6 (b) 1594 (c) 30th P C Chapter 10
(d) T18 = 138 and T19 = 146, so 139 is p
not a term h te Exercise 10.1
i m
s r
c
y 1 (a) x −1 0 1 2 3
5 (a) 2n + 5 T50 = 105 y4 5678
(b) 3 − 8n T50 = −397 lnoquvwxz
(c) 6n − 4 T50 = 296
(d) n2 T50 = 2500 2 (a) 9 (b) x −1 0 1 2 3
(e) 1.2n + 1.1 T50 = 61.1 (b) 20 y 1 −1 −3 −5 −7
(c) {c, h, i, s, y}
6 (a) (d) {c, e, h, i, m, p, r, s, t, y} (c) x −1 0 1 2 3
(e) {a, b, d, f, g, j, k, l, n, o, q, u, v, w, y9 7531
n1 2 3 4 5 6 x, z}
Tn 6 11 6 21 26 31 (f) {c, h, i, s, y} (d) x −1 0 1 2 3
y −1 −2 −3 −4 −5
(b) Tn = 5n + 1 3 (a)
(c) 496 (d) 55th (e) x 4 4 4 4 4
y −1 0 1 2 3
Exercise 9.2

1 (a) 3 16, 12, 0.090090009. . . MS
78 – x x 36 – x
(b) 45, 3 90, π, 8 (in fact, any five values of y are correct)
(f) x −1 0 1 2 3
2 (a) 4 (b) 74 (c) 79 y −2 −2 −2 −2 −2
9 99 90
7
(d) 103 (e) 943 (f ) 928 (b) 21 (g) x −1 0 1 2 3
900 999 4995 y 1.5 −0.5 −2.5 −4.5 −6.5
(c) 0.57

Exercise 9.3 A (h) x −1 0 1 2 3
y −1.2 −0.8 −0.4 0 0.4
1 (a) false (b) true (c) false Mixed exercise
(d) true (e) false (f) true
(g) false (h) true 1 (a) 5n − 4 T120 = 596 (i) x −1 0 1 2 3
(b) 26 − 6n T120 = −694 y −1 −0.5 0 0.5 1
2 (a) The set of even number from two (c) 3n − 1 T120 = 359
to twelve. (j) x −1 0 1 2 3
2 (a) –4, –2, 0, 2, 4, 8 y 0.5 −0.5 −1.5 −2.5 −3.5
(b) 6 (b) 174
(c) {2} (c) T46 2 student’s graphs of values above
(d) {2, 4, 6, 8}
(e) {2} 3 0.213231234. . ., 2, 4π
(f) {10, 12}
3 y=x−2

Answers 179

4 (a) no (b) yes (c) yes (h) x2 7xy + 2 y2 (m) (x 3)(x − 4)
(n) (x + 4)(x − 3)
(d) no (e) no (f) no (i) 2 4 2 3 (o) (x + 9)(x − 6)

(g) yes (horizontal lines) (j) x2 + x − 132
(k) 1 − 1 x2
(h) yes (vertical lines) 2 (a) 5(x 2)(x + 1)
4 (b) 3(x 4)(x − 2)
5 (a) m = 1 (b) m = −1 (c) 3 ( 3)(x −1)
(l) −3x2 + 11x − 6 (d) 5(x 2)(x −1)
(c) m = −1 (d) m = 6 (e) x (x + 10)(x + 2)
7 (m) −12x2 + 14x − 4 (f) x2 y (x + 2)(x 1)
(g) x (x + 7)(x − 2)
(e) m = 2 (f) m = 0 2 (a) x2 + 8x 16 (h) 3(x 3)(x − 2)
(b) x2 6x 9 (i) −2(x + 4)(x − 6)
(g) undefined (c) x2 + 10x + 25 (j) 2(x 7)(x − 8)
(d) y2 4 y 4
(h) m = 1 (e) x2 + 2xy + y2
16 (f) 4x2 4xy + y2
(g) 9 2 12x 4
6 (a) m = 3, c = −4 (h) 4x2 12 y 9 y2
(i) 4x2 20x 25
(b) m = −1, c = −1 (j) 16x2 − 48x + 36
(k) 9 6x + x2
(c) m = − 1 , c = 5 (l) 16 − 16x + 4x2
2 (m) 36 − 36 y + 9 y2
(d) m = 1, c = 0
3 (a) x2 − 25
(e) m = 1 , c = 1 (b) 4x2 25 3 (a) (x + 3)(x − 3)
2 4 (c) 49 y2 − 9
(d) x4 y4 (b) (4 x)(4 − x)
(f ) m = 4 , c = −2 (e) 9 2 16
5 (f) x6 y4 (c) (x + 5)(x − 5)
(g) 16x4 y4 − 4z 4
(g) m is undefined, c = 7 (h) 4x8 4 y2 (d) (7 )(7 − x)
(i) 16x2 y4 − 25y2 (e) (3x 2y)(3x 2y)
(h) m = −3, c = 0 (j) 64x6 y4 − 49z 4 (f) (9 2x)(9 2x)
(g) (x + 3y)(x 3y)
(i) m = − 1 , c = + 14 Exercise 10.2 B (h) (11y + 12x)(11y 12x)
3 3 (i) (4x 7 y)(4x 7 y)
1 (a) (x + 2)(x + 2) (j) 2(x 3)(x − 3)
(j) m = −1, c = −4 (b) (x + 4)(x + 3)
(c) (x + 3)(x + 3) (k) 2(10 )(10 − x)
(k) m = 1, c = −4 (d) (x + 1)(x + 4)
(e) (x + 3)(x + 5) (l) (x2 + y)(x2 y)
(l) m = −2, c = 5 (f) (x 1)(x − 8) (m) (5 8 )(5 − 8 )
(g) (x 5)(x − 3)
(m) m = –2, c = −20 (h) (x 1)(x − 3) (n) (xy + 10)(xy −10)
(i) (x 26)(x −1)
7 (a) y x (b) y 1 x (j) (x 8)(x + 1)
(d) y 2 2x (k) (x + 5)(x − 2)
(c) y = 2 5 − 1 (l) (x + 4)(x − 8)

(e) y 1 x − 1 (f) y 2x +1
2
1
(g) x = 2 (h) y x + 2
(j) y +3 4
(i) y = −2x = x

(k) y = 3x − 2 (l) y = x − 3

8 (a) x = 2, y = –6 (o) ⎛5 + 8w ⎞ ⎛ 5 − 8w ⎞
(b) x = 6, y = 3 ⎝⎜ y2 z ⎠⎟ ⎝⎜ y2 z ⎟⎠
(c) x = –4, y = 6
(d) x = 10, y = 10 ((p) 5x5 1)(5x5 1)
(e) x = −25, y = –5
( )( )(q) 1 9 2 3 1 9 2 3

9 (a) 1 (b) 1 (c) –1 Exercise 10.2 C
(d) 2 (e) 0 1
(f ) 2 1 (a) x = 0 or x = 3
(b) x = –2 or x = 2
10 (a) (0, 0) (b) (–1.5, 0.5) (c) x = 0 or x = 2
(d) x = 0 or x = − 2
(c) (–2, 3) (d) (5, 10)
3
(e) (0.5, –3) (f) (1, 1.5)
(e) x = –1 or x = 1
Exercise 10.2 A (b) x2 x − 6 (f) x = − 7 or x = 7
(d) x2 + 2x 35
1 (a) x2 + 5x 6 (f) 2x2 x 1 22
(c) x2 + 12x + 35
(e) x2 4x 3 (g) x = − 1 or x = 1
(g) y2 9 y 14
22

(h) x = –4 or x = –2
(i) x = –4 or x = –1

180 Answers

(j) x = 5 or x = –1 (b) Caroline’s distance at 7 km/h (k) 10x − 45
(k) x = 5 or x = –4 y (l) −2x3 + 16x2 − 8x
(l) x = –10 or x = 2 (m) 2x3 8x2 + 16x
(m) x = 5 or x = 3 45
(n) x = 20 or x = –3 9 (a) a(a + 2)(a − 2)
(o) x = 7 or x = 8 40
(p) x = 10 (b) (x2 + 1)(x 1)(x 1)
(q) x = 2 35
(c) (x 2)(x + 1)
30
Distance (d) (x 1)(x −1)
25
Mixed exercise (e) (2x 3y + 2z)(2x − 3y 2z)
20
1 (a) y 1 x (f) (x + 12)(x + 4)
2 15
(g) ⎛ x2 + x⎞ ⎛ x2 − x ⎞
x −1 0 2 3 10 ⎝ 2⎠ ⎝ 2 ⎠

y −0.5 0 1 1.5 5 (h) (x + 1)(x − 6)
x
(b) y 1 x + 3 (i) 4(x 3)(x − 4)
2 0 2 46
Time (j) 2(x 3)(x − 4)
x −1 0 2 3
y 3.5 3 2 1.5 (c) 7 (d) 7 (k) 5(1+ 2 8 )(1 − 2 8 )
(e) (i) 3 hours (ii) 1 h 24 min
(c) y = 2 (l) 3(x 3)(x + 2)
(iii) 48 min
x −1 0 2 3 (f) (i) 21 km (ii) 17.5 km 10 (a) x = –5 or x = –1
y 2 222 (b) x = –2 or x = 2
(d) y 2x − 4 = 0 (iii) 5.25 km (c) x = 2 or x = 1
x −1 0 2 3 (g) 7 hours (d) x = –1
y 2 4 8 10 (e) x = 5 or x = –1
7 (f) x = 2
All four plotted on the same graph.
(a) (b) Chapter 11
(i) 1 (0.5, 6.5)
2 (a) m = −2, c = −1 (ii) 2 Exercise 11.1 A
(iii) –1 (0, 5)
(b) m = 1, c = −6 (1, 3)

(c) m = 1, c = 8 1 1 (a) 5 cm (b) 17 cm
2 (c) 12 mm (d) 10 cm
(d) m = 0, c = − (e) 1.09 cm (f) 0.45 cm
(g) 8.49 cm (h) 6.11 cm
(e) m = − 2 , c = 2
3

(f) m = −1, c = 0 (iv) − 4 (–0.5, 3) 2 (a) 55.7 mm (b) 14.4 cm
3 (c) 5.29 cm (d) 10.9 mm
3 (a) y x − 3 (b) y 2 x + 1 (e) 9.85 cm (f) 9.33 cm
(d) y 3 2 (v) undefined (–1.5, 0.25)
(c) y x − 2 4
(e) y 2x − 3 5 x − 3
(g) y = 2
(f) y x + 2 8 (a) x2 − 16x + 64 3 (a) no (b) yes (c) no
(b) 2x2 2 (d) yes
(h) x = −4

4 A 0, B 1, C 2, D 1, E 4 (c) 9 2 12xy 4 y2 4 (a) 32 = 5.66

5 (a) y 2x − 6 (b) y = 7 (d) 1 − 12 y 36 y2 (b) 18 = 4.24
(d) x = −10
(c) y = 4 x + 4 (f) y = −3 (e) 9 2 4 (c) 32 = 5.66
3
(f) 4x2 20x 25 (d) 180 = 13.4
(e) 3
(e) y x (g) 9 4 y2 6x2 y + 1 (f) 45 = 6.71

6 (a) t 0 2 4 6 (h) x2 + xy + 1 y2 Exercise 11.1 B
D 0 14 28 42 4

(i) x2 − 1 1 20 mm
4 2 44 cm

(j) 1 − 4
x2

Answers 181

3 height = 86.6 mm, area = 4330 mm2 3 ∠CDB = ∠DBE = ∠BEA (alt ∠ equal) Chapter 12
4 13 m and 15 m ∠BAE = ∠CBD (corr ∠ equal)
5 0.7 m AE = BD (given) Exercise 12.1
Δ BAE is congruent to Δ CBD (ASA)
AB = BC 1

Exercise 11.2 ∴ AB = 1 AC (a) (b) (c) (d) (e) (f)

1 (a) 2.24 cm (b) 6 mm 2 mean 6.14 27.44 13.08 5 4.89 5.22

(c) 7.5 mm (d) 6.4 cm median 6 27 13 5 5 5

(e) y = 6.67 cm, z = 4.8 cm Mixed exercise mode 6 27 and 12 no 4 6
38 mode

(f) x = 5.59 cm, y = 13.6 cm 1 (a) sketch (b) 130 m

(g) x = 9 cm, y = 24 cm 2 102 = 62 + 82 ∴ 6 ABC is right-angled 2 (a) (iii) and (vi)
(converse Pythagoras) (b) sensible answer from student,
(h) x = 50 cm, y = 20 cm e.g. different sets can still add up
to the same total as another set. If
2 ∠ABC = ∠ADE (corr ∠ are equal) 3 (a) 18 = 4.24 divided by the same number they
∠ACB = ∠AED (corr ∠ are equal) (b) 20 = 4.47 will have the same mean.
∠A = ∠A (common) (c) 8 2.83
∴ 6 ABC is similar to 6 ADE (d) 5
(e) 3.5
3 25.5 m 3 255

Exercise 11.3 4 15

1 (a) x = 18 cm 4 P = 2250 mm 5 Need to know how many cows there
(b) x = 27 cm, y = 16 cm are to work out mean litres of milk
5 (a) x = 3.5 cm produced per cow.
(b) x = 63°, y = 87°
2 9:4 (c) x = 12 cm 6 (a) 2.78 (b) 1

3 (a) 254.48 cm2 6 (a) 4 : 1 7 (a) $20.40 (b) $6 (c) $10
(b) 529 mm2 (b) 1 : 9 (d) 2 (only the category B workers)
(e) The mean is between $20 and $40
4 (a) x = 2 cm (b) x = 15 cm 7 18 cm2 so the statement is true.

5 28 000 cm3 8 23 750 mm2 Exercise 12.2

6 (a) 5 : 1 (b) 25 : 1 9 (a) 3 cm
(c) 125 : 1
(b) height = 12 cm, area of base = 1 (a) mean = 4.3, median = 5,
Exercise 11.4 256 cm2 mode = 2 and 5.
The data is bimodal and the lower
1 (a) 6ABC is congruent to 6ZXY (SAS) 10 (a) 6ABC is congruent to 6HGI (RHS) mode (2) is not representative of
(b) although the two triangles look (b) 6 ABC is congruent to 6 DEF the data.
to be the same size, there are no (ASA)
marks to confirm this to be the (c) 6ACB is congruent to 6EDF (SAS) (b) mean = 3.15, median = 2,
case so we cannot be certain they (d) 6CAB is congruent to 6GIH (SAS) mode = 2.
are congruent The mean is not representative
(c) 6XYZ is congruent to 6ONM (ASA) 11 5.63 m of the data because it is too high.
(d) 6DEF is congruent to 6RQP (ASA) This is because there are some
(e) 6ACB is congruent to 6ACD (RHS) 12 (a) 140 mm values in the data set that are
(f) 6 PMN is congruent to 6 NOP (SSS much higher than the others.
or SAS) 68 mm 560 mm (This gives a big range, and when
(g) 6 PRQ is congruent to 6 ZYX (SAS) 420 mm the range is big, the mean is
(h) 6ABC is congruent to 6EDC (ASA) generally not representative.)

2 PO = PO (common) (c) mean = 17.67, median = 17,
MP = NO (given) no mode.
MO = NP (diagonals of isosceles There is no mode, so this cannot be
trapezium are equal) representative of the data. The mean
Δ MPO is congruent to Δ NOP (SSS) and median are similar, so they are
both representative of the data.

140 mm
(b) 156 mm

182 Answers

2 (a) mean = 12.8, median = 15, (b) 36.74 ≈ 37 6 (a) 19
mode = 17, range = 19 (c) 30 ≤ m < 40 (b) 5
(d) 30 ≤ m < 40 (c) Q1 = 18, Q3 = 23, IQR = 5
(b) mode too high, mean not reliable (d) fairly consistent, so data not
as range is large 2 spread out

3 (a) Runner B has the faster mean Words per Mid Frequency f × midpoint 7 (a) 15 scores fell into the bottom 60%
time; he or she also achieved the minute (w) point of marks
faster time, so would technically
be beating Runner A. 31 ≤ w < 36 33.5 40 1340 (b) Less than half students scored
above the 60th percentile, so you
(b) A is more consistent with a range 36 ≤ w < 41 38.5 70 2695 can assume the marks overall
of only 2 seconds (B has a range were pretty low.
of 3.8 seconds). 41 ≤ w < 46 43.5 80 3480

Exercise 12.3 46 ≤ w < 51 48.5 90 4365 Chapter 13
51 ≤ w < 55 53.5 60 3210
1
Score Frequency Score × 55 ≤ w < 60 58.5 20 1170 Exercise 13.1
frequency Total 360 16 260
1 student’s own diagrams
(fx)
0 6 0 (a) 45.17 2 (a) 2600 m (b) 230 mm
1 6 6 (b) 46 ≤ w < 51 (c) 820 cm (d) 2450.809 km
2 10 20 (c) 41 ≤ w < 46 (e) 20 mm (f) 0.157 m
3 11 33 (d) 29
4 5 20 3 (a) 9080 g (b) 49 340 g
5 1 5 Exercise 12.5 (c) 500 g (d) 0.068 kg
6 1 6 (e) 0.0152 kg (f) 2.3 tonne
Total 40 90 1 (a) Q1 = 47, Q2 = 55.5, Q3 = 63,
(b) 3 IQR = 16 4 (a) 19 km 100 m
(a) 2.25 (c) 2 (b) 9015 cm 15 cm
(d) 6 (b) Q1 = 8, Q2 = 15, Q3 = 17, IQR = 9 (c) 435 mm 2 mm
(c) Q1 = 0.7, Q2 = 1.05, Q3 = 1.4, (d) 492 cm 63 cm
2 Data set A (e) 635 m 35 m
IQR = 0.7 (f) 580 500 cm 500 cm
(d) Q1 = 1, Q2 = 2.5, Q3 = 4, IQR = 3
BC 5 (a) 1200 mm2 (b) 900 mm2
Mixed exercise (c) 16 420 mm2 (d) 370 000 m2
mean 3.5 46.14 4.12 (e) 0.009441 km2 (f) 423 000 mm2
1 (a) mean 6.4, median 6,
median 3 40 4.5 mode 6, range 6

mode 3 and 5 40 6.5 (b) mean 2.6, median 2, 6 (a) 69 000 mm3
mode 2, range 5 (b) 19 000 mm3
Exercise 12.4 (c) 30 040 mm3
(c) mean 13.8, median 12.8, (d) 4 815 000 cm3
1 (a) no mode, range 11.9 (e) 0.103 cm3
(f) 0.0000469 m3
Marks 2 (a) 19 (b) 9 and 10 (c) 5.66
(m) Mid Frequency Frequency × 7 220 m
3 C − although B’s mean is bigger it
point (f ) midpoint has a larger range. C’s smaller range
suggests that its mean is probably
0 ≤ m < 10 5 2 10 more representative. 8 110 cm

10 ≤ m < 20 15 5 75 4 (a) 4.82 cm3 (b) 5 cm3 9 42 cm
(c) 5 cm3
20 ≤ m < 30 25 13 325 10 88 (round down as you cannot have
5 (a) 36.47 years part of a box)
30 ≤ m < 40 35 16 560 (b) 40 ≤ a < 50
(c) 30 ≤ a < 40
40 ≤ m < 50 45 14 630 (d) don’t know the actual ages

50 ≤ m < 60 55 13 715

Total 63 2315

Answers 183

Exercise 13.2 3 (a) 27.52 m2 Mixed exercise
(b) upper bound: 28.0575 m2
1 lower bound: 26.9875 m2 1 (a) 2700 m (b) 690 mm
(c) 6000 kg (d) 0.0235 kg
Name Time Time Lunch (a) (b) 4 (a) 195.5 cm ≤ h < 196.5 cm (e) 263 000 mg (f) 29 250 ml
in 93.5 kg ≤ m < 94.5 kg (g) 0.24 l (h) 1000 mm2
out Hours Daily (i) 0.006428 km2 (j) 7 900 000 cm3
(k) 29 000 000 m3 (l) 0.168 cm3
worked earnings

Half (b) maximum speed =

Dawoot 1 past 9 past 3 hour 7 1 hours $55.88 greatest distance = 405
4 five 4 2 shortest time 33.5
2 23 min 45 s
= 12.09 m/s
Nadira 8:17 5:30 1 hour 8h $64.94 3 2 h 19 min 55 s
a.m. p.m. 2 43 min $64.82 5 (a) upper bound of area: 15.5563 cm2
$60.59 lower bound of area: 14.9963 cm2
John 08:23 17:50 45 min 8 h $71.64 4 1.615 m ≤ h < 1.625 m
42 min (b) upper bound of hypotenuse:
8.0910 cm 5 (a) No, that is lower than the lower
Robyn 7:22 4:30 1 hour 8 h lower bound of hypotenuse: bound of 45
a.m. p.m. 8 min 7.9514 cm
(b) Yes, that is within the bounds
Mari 08:08 18:30 45 min 9 h
37 min
6 (a) 3.605 cm to 3.615 cm
2 6 h 25 min Exercise 13.4 2.565 cm to 2.575 cm

3 20 min 1 (a) 1 unit = 100 000 rupiah (b) lower bound of area: 9.246825 cm2
(b) (i) 250 000 upper bound of area:
4 (a) 5 h 47 min (b) 10 h 26 min (ii) 500 000 9.308625 cm2
(c) 12 h 12 min (d) 14 h 30 min (iii) 2 500 000
(c) (i) Aus $80 (c) lower: 9.25 cm2, upper: 9.31 cm2
(ii) Aus $640
5 (a) 09:00 (b) 1 hour (c) 09:30 7 (a) conversion graph showing litres
(d) 30 minutes
(e) It would arrive late at Peron Place against gallons (conversion factor)
at 10:54 and at Marquez Lane
at 11:19. (b) (i) 45l (ii) 112.5l

2 (a) temperature in degrees F against (c) (i) 3.33 gallons
temperature in degrees C
(ii) 26.67 gallons
(b) (i) 32 °F (ii) 50 °F
(iii) 210 °F (d) (i) 48.3 km/gal and 67.62 km/gal

Exercise 13.3 (c) oven could be marked in (ii) 10.62 km/l and 14.87 km/l
Fahrenheit, but of course she
1 The upper bound is ‘inexact’ so 42.5 could also have experienced a 8 €892.06
in table means < 42.5 power failure or other practical
problem 9 (a) US$1 = IR49.81 (b) 99 620
Upper bound Lower bound (c) US$249.95
(d) Fahrenheit scale as 50 °C is hot,
(a) 42.5 41.5 not cold 10 £4239.13

(b) 13325.5 13324.5 Chapter 14

(c) 450 350 3 (a) 9 kg (b) 45 kg

(d) 12.245 12.235 (c) (i) 20 kg (ii) 35 kg Exercise 14.1 A

(e) 11.495 11.485 (iii) 145 lb 1 (a) y
5
(f) 2.55 2.45 Exercise 13.5 4

(g) 395 385 2x + 3 = y3
x–y=0
(h) 1.1325 1.1315 1 (a) (i) US$1 = ¥76.16 2
(ii) £1 = NZ$1.99
2 (a) 71.5 ≤ h < 72.5 (iii) €1 = IR69.10 1 x
(b) Yes, it is less than 72.5 (although (iv) Can$1 = €0.71 –5 –4 –3 –2 –1 0 1 2 3 45
it would be impossible to measure (v) ¥1 = £0.01
to that accuracy). (vi) R1 = US$0.12 –1

(b) (i) 2490.50 (ii) 41 460 –2
(iii) 7540.15
(–3,–3) –3
(c) (i) 9139.20 (ii) 52 820 –4
(iii) 145 632
–5

x = −3, y = −3

184 Answers

(b) 5 y 2 (a) (A) y x + 4 (c)
(B) y = 3 x + 6
4 –4 –3.5 –3 –2 –1 0 1
2
3 4x+2 = y (d)
2 (C) y 4x − 2
1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3
1 (D) y x
x (E) y x − 4 (e)

–5 –4 –3 –2 –1–10 1 2 3 3 4 5 (F) y = −2 –7 –6 –5 –4 –3 –2 –1 2 0 1
(–1, –2) –2 – 2y =
(b) (i) (−2, 6) (ii) (−2, −2)
x (f )
(iii) (4, 0)
–5 –4 –3 –2 –1 0 1 2 3 4 5
–3 3 (a) x = 4, y = 2
(b) x = −1, y = 2
–4 (c) x = 0, y = 4 (g)
–5 (d) x = 3, y = 1
–4 –3.5 –3 –2 –1
x = −1, y = −2 0 1

(c) y 4 (a) x = 6, y = −1 (h)
5
0.25 0.75 1.25 1.75
4 y = –2x + 2 (b) x = 1, y = 2
0 0.5 1 1.5 2
3
2 (c) x = 18, y = −8 (i)
1x 1−131
(d) x = 1, y = –2 –1 0 1 2 3 4 5 6 7
–5–4 –3–2–10 1 2 3 4 5 6 7 8 9 10 (e) x = 3, y =

–2 (f) x = 3, y = 7

–3 (3,–4) 2 (a) 3, 4
–4 (b) –2, –1, 0, 1, 2, 3
5 (a) x = 2, y = 1 (c) 2
–5 2y + 3x –1 = 0 (b) x = 2, y = 2 (d) –1, 0, 1, 2, 3
(c) x = 2, y = −1 (e) 0, 1, 2, 3, 4
–6 (d) x = 3, y = 2 (f) 4, 5, 6, 7, 8, 9
–7 (e) x = 3, y = 2.5
(f) x = 4, y = 2
–8 (g) x = 5, y = 3
–9 (h) x = 0.5, y = −0.5
(i) x = −9, y = −2
–10

x = 3, y = −4

(d) 5 y 3 (a) x ≤ –2 (b) x > 11
(c) x ≥ 4 (d) x ≤ 52
y–x=4 4 (e) x ≥ 9 (f) x > 73
(g) x > 9 1
3 x 6 a chocolate bar costs $1.20 and a box (h) e ≥ –3
2 of gums $0.75 4
1

–10–9–8–7–6–5–4–3–2–1 0 1 2 3 4 5

(–5.5, 1.5) –2 (i) r < 111 (j) x ≥ 45

–3 7 number of students is 7 2
–4

–5 8 6 quarters and 6 dimes (k) g < 9 1 (l) y ≤ 10
–6
4 11
–7 y = –x –7 Exercise 14.1 B
–8 (m) n > 31 (n) n > –2
1 (a) x = –3, y = 4 (o) x ≤ −6 1
–9 (b) x = –1, y = 1
–10 28
3
x = −5.5, y = −1.5
(c) x = 3, y = –4
(e) 5 y (d) x = 15, y = 30 Exercise 14.3
(e) x = 4, y = 2
y = –2x + 34 (f) x = 15, y = 12
3x + 3y = 3
3 77 1 (a)

2 2 (a) c + d = 15 y
(b) 3 desks and 12 chairs 5
1
x 4

–5 –4 –3 –2 –1–10 12 3 4 5 3
(2, –1)
–2 2

–3 Exercise 14.2 1
–5 –4 –3 –2 ––110
–4 1 (a) x
12345
–5 –5 –4 –3 –2 –1 0 1 2 3 4 5

x = 2, y = −1 (b) –2 y < 4x + 1
–3
–5 –4 –3 –2 –1 0 1 2 3 4 5
–4

–5

Answers 185

(b) (c) 3 (a) y ≥ 2x − 1
(b) y < –4x + 3
–5 –4 –3 –2 y y (c) y > 1 x
5 5
4 y ≥ –2x + 5 2
3 4
2 (d) y ≥ –x – 3
1 3
4
––110 1 2 3 4 2
–2
x 1 x
5 –5 –4 –3 –2 ––110 12345

–2

–3 –3 3x – y ≥ 3
–4 –4
–5
–5

(c) (d)
y
y
5 5
4 y > –x + 1
3 2x < 3y – 6 4
2 3
1 5 (a)
2
–5 –4 –3 –2 ––110 1 2 3 4 x
–2 5 1 x
–5 –4 –3 –2 ––110 12345

–3 –2
–4
–5 –3

–4

2 (a) –5

y (e)
5
y
4 5

3 4

2 3 (b) (–2, 2), (–2, 1), (–1, 2), (–1, –1),
(0, 1)
1 x 2
–5 –4 –3 –2 ––110 12345
–2 < y ≤ 4 1 x Exercise 14.4
–5 –4 –3 –2 ––110 12345
–2 1

2y ≤ –4 –3 –2 y
–4 5
–3

–5 –4 4

(b) –5 3
y
(f ) 2
5
4 1y 1 x
3 x –5 –4 –3 –2 ––110 12345
2
1 –5 –4 –3 –2 ––110 1 2 3 4 5 6 7 8 9 10

x – y < 3 –2 –2
4 2
–3
–3
x –4
2345
–5 –4 –3 –2 ––110 1 –5 –4
–2 x–y>7
–3 –6 –5
–4
–5 –7

–8

–9

–10

186 Answers

2 (a) Exercise 14.5 (g) x = –7.20 or –16.80
8y (h) x = –3.70 or –7.30
1 (a) (x + 3)2 − 5 (i) x = 4.19 or –22.19
7 (b) (x − 2)2 + 3 (j) x = 4.32 or –2.32
(c) (x + 7)2 − 5 (k) x = 8.58 or –0.58
6 (d) (x − 6)2 − 6 (l) x = 11.05 or –9.05
(e) (x + 5)2 − 8
5 (f) (x + 11)2 + 20 3 (a) x = 5.46 or –1.46
(g) (x + 12)2 − 23 (b) x = 1.67 or –3.5
4 (h) (x − 8)2 − 7 (c) x = 1.31 or 0.19
(i) (x + 9)2 + 12 (d) x = 5 or –2.5
3 (j) (x −1)2 + 9 (e) x = 2.25 or 1
(k) (x − 4)2 − 21 (f) x = 1.77 or –1.27
2 (l) (x + 10)2 −17 (g) x = 0.80 or –0.14
(h) x = 10.65 or 5.35
1 2 (a) x = 6, x = –1 (i) x = 9.91 or –0.91
x (b) x = 3, x = –2 (j) x = 0.5 or –3
(c) x = 3, x = 1 (k) x = –0.67 or –1.5
–1–10 1 2 3 4 5 6 7 8 (d) x = 7, x = –1 (l) x = 1 or –0.75
(e) x = 15.81, x = 0.19
(b) greatest posible value is at the (f) x = –6.85, x = –0.15
point (2, 5) so 2y + x = 12 (g) x = 0.11, x = –9.11
(h) x = –3.70, x = –7.30
3 let x = number of chocolate fudge (i) x = 11.05, x = –9.05
cakes and y = number of vanilla fudge
cakes 3 (a) x = 2, x = –0.5
(b) x = 3, x = 1
60 y (c) x = 2.53, x = –0.53 4 numbers are 57 and 58 or –58 and –57
(d) x = 3, x = –0.5
50 (e) x = 7.47, x = –1.47 5 dimensions are 12 cm × 18 cm
x ≤ 30 (f) x = –2.27, x = 1.77
x + y ≤ 40 Exercise 14.7
40profit 3x + 2y Exercise 14.6
1 (a) (2x − 3)(x + 1)
30 1 (a) x = 10 or 4 (b) (3x +1)2
20 y ≤ 20 (b) x = 6 or –20 (c) (2x − 3)2
(c) x = 1 or –6 (d) (2x + 1)(3x − 5)
10 (d) x = 3 or 5 (e) (4x − 3)(x + 1)
x (e) x = 4 or –1 (f) (7x − 1)(2x − 7)
(f) x = 2 (g) (x + 5)(3x − 4)
–1 0 10 20 30 40 50 60 (g) x = 2 or –6 (h) (2x − 1)(3x + 7)
–1 (h) x = –5 (i) (3x + 5)(x − 5)
(i) x = 2 or –4 (j) (3x − 11)(x + 6)
to maximise profit x = 30 and y = 10 (j) x = 6 or –2 (k) (5x + 3)(3x − 5)
maximum profit = $110 (k) x = 4 or 5 (l) (8x + 1)(x + 3)
(l) x = 5 or –8
4 let x = number of litres of orange and
y = number of litres of lemon 2 (a) x = –0.76 or – 5.24
(b) x = 1.56 or –2.56
2000 y orange = x lemon = y (c) x = –4.76 or –9.24
(d) x = 7.12 or –1.12
1900 (e) x = –2.17 or –7.83
(f) x = –0.15 or –6.85
1800 2 (a) (2x + 3)2

1700

1600 x ≤ 1800 (b) 1 (2x 1)(2x 1) or
1500 1800 2
1500 x ≥ 1000
1400

1300 ⎛ 1 ⎞ ( )
⎝⎜ 2 ⎠
1200 x + x+

1100

1000 profit x + 1.9y (c) 2(5x + 2)2
900
800 (d) (2x + y)(3x − 5y)
700
(e) (4x2 − 3)(x2 + 1)
600
500 y ≤ 600 (f) 2(2x − 1)(3x + 1)

400

300 (g) 2(1 − 2x)(1 + x)

200 y ≤ 1x (h) (x + 3)(x + 2)
100 3
x

0 200 400 600 800 1000 1200 (i) (3x + 8)(x − 4)

maximum income ($2940) when (j) 2(3x − 11)(x + 6)
x = 1800 and y = 600
(k) (4x − 3)(x + 1)

(l) (8x + y)(x + 3y)

Answers 187

Exercise 14.8 Mixed exercise 10 (a) x y
x+y
1 (a) x 1 x = 2, y = −5
7 (b) 3 (b) 2 – x
(d) 2x
(c) 1 2 x = −2, y = 5 5 2p
5 10 p2
7 3 $5000 at 5% and $10 000 at 8% (c)

(e) 8z (f) 2 4 (a) x > –5 (b) x ≤ –1 (d) 3xy
y 2
3 5 y
(g) 2x (h) 1 5 (e) xy3z
5ab 4
3y 4

(i) 2b 3 (f) 45a2 + 16a − 50
3a 30a
2
2 (a) x + 3 (b) a b
x ab 1 x (g) (x + 2)(x −1)
–5 –4 –3 –2 ––110 12345 x(x + 4)
(c) 2x + y (d) 3 1
x x−5 x = –2 –2 x=3 (h) 1
–3 2x 1
(e) a + 2 (f) a b
a ab –4 (i) −6 − 1)

(g) x + 5 (h) x2 − 3 –5 (2x + 1)(x + 5)(
x+2
6 Chapter 15
(i) 7
the integer solutions are: Exercise 15.1
3 least, x = –1, y = 2, giving x + y = 1.5

3 (a) x2 y2 2 1 (a) length = 10 cm, width = 7 cm
6 (b) greatest, x = 2, y = 4, giving x + y = 5 (b) length = 14.4 cm, width = 7 cm

6 2
7 (a) x(x − 2y)
(c) 35 (d) 5xy 2 (a) length = 9.14 cm, width = 5.5 cm
a2 2 (b) (a2 + b)(a2 − b) (b) Yes. The length of the mini
(c) (x − 5)(x + 11) pitch = width of standard pitch
(e) a (f) (b )(b −1) (d) (2y – 1)(y + 7) and 2 × width of mini pitch <
c (e) –2(x + 1)(2x − 3) length of standard pitch. It is
(h) 4 − x (f) (x − 6)(x + 3) possible to have two mini pitches
(g) 1 3 8 (a) a = 1.30 or –2.30 on a standard pitch so, with
42 (b) x = 1.33 or –1 three standard pitches, up to six
(c) x = 3 or –5 matches could take place at the
(i) −2 − 3x (d) x = 0.67 or 1 same time.
2x +1 (e) x = 1 or –0.625
(f) x = 1
4 (a) 3x 2y (b) 13p + 2 9 (a) x = –2 or 0.4 3 (a) 1 or 1
xy 8p 200 150
q ± q2 − 4 pr
(c) 29 (d) 7 (b) x = (b) (i) and (ii) diagram drawn using
10 p p+1 student’s scale including × for net
2p posts
(e) 57 (f ) 18 − 5m
6 4 (a) student’s scale drawing − diagonal
(x + 2)(x 1) distance = 11.3 m

(g) (x 9 17 (b) using Pythagoras’ theorem

3)( 2)(x −1)

(h) 17 − x 4) Exercise 15.2

x(x − 3)( 1 (a) 090°
(b) 225°
(i) (2x 5 3x 1) (c) 315°

3)(x +1)(x 2 (a) 250° (b) 310°

(c) 135°

188 Answers

3 student’s drawing (d) sin D = 63, cos D = 16, tan D = 63 5 (a) θ = 66.4° (b) θ = 43.2°
(a) 223° (b) 065° (c) θ = 60.7° (d) θ = 31.1°
(c) 11 km (d) 13 km 65 65 16
6 (a) a = 2.9 m (b) z = 3.7 cm
Exercise 15.3 (e) sin E = 84 , cos E = 13, tan E = 84 (c) m = 9.2 m (d) r = 5.0 cm

85 85 13

1 10 (a) 45° (b) 64° (c) 57° Exercise 15.7
(d) 60° (e) 30° (f) 27°
hypotenuse
opp(A) (a) (b) (c) (d) 11 4.86 m
adj(A) czf q
ay g p 12 (a) x = 30° and y = 4.69 cm 1 (a) 11.2 cm2 (b) 25.4 cm2
bxe r (b) x = 3 m and y = 53.1° (c) 17.0 cm2 (d) 70.4 cm2
(c) x = 48.2° (e) 3.8 cm2 (f) 24.6 cm2
(d) x = 22.9° and y = 8.90 cm
2 (a) opp(30°) = x cm 2 44.9 cm2
adj(60°) = x cm
opp(60°) = adj(30°) = y cm Exercise 15.4 3 (a) x = 115° and area = 31.7 cm2
(b) x = 108° and area = 43.0 cm2
(b) adj(40°) = q cm 1 1.68 m (c) x = 122.2° and area = 16.3 cm2
opp(50°) = q cm 2 (a) 30°
opp(40°) = adj(50°) = p cm (b) 33°

Exercise 15.5 Exercise 15.8

3 (a) 0.65 (b) 1.43 (c) 5.14 1 (a) –cos 68° (b) sin 24° 1 (a) 90° (b) 5 cm
(d) 0.41 (e) 0 (c) cos 105° (d) sin 55° (c) 36.9° (d) 4 cm
(e) cos 135° (f) sin 35° (e) 9.85 cm (f) 66.0°
4 (a) tan A = 3 (b) tan x = 2 , tan y = 3 (g) –cos 60° (h) sin 82° (g) 10.3 cm (h) 16.9°
3 2 (i) –cos 125° (j) –cos 50°
4 1
d
(c) tan 55° = , tan B = d 2 (a) 21.2 m (b) 10.6 m
(c) 21.2 m (d) 18.4 m
(d) tan y = 5 , ∠X = (90 − y)°, tan X = 12 2 (a) θ = 15° or 165°
12 5 (b) θ = 124°
(c) θ = 52° or 128°
(e) AC = 2 cm, tan B = 4 , tan C = 3 (d) θ = 70° 3 (a) 9.80 cm (b) 24.1°
3 4 (e) θ = 20° or 160°
(f) θ = 48°
5 (a) x = 1.40 cm (g) θ =39° Mixed exercise
(b) y = 19.29 m (h) θ =110°
(c) c = 3.32 cm (i) θ =7° or 173° 1 (a) x = 37.6°
(d) a = 13 m (j) θ =84° or 96° (b) x = 44.0°
(e) x = 35.70 cm (c) x = 71.4°

6 (a) 26.6° (b) 40.9° (c) 51.3° 2 35.3°
(d) 85.2° (e) 14.0° (f) 40.9°
(g) 79.7° (h) 44.1° 3 AB = 300 m
Exercise 15.6
4 AC ≈ 41.6 cm

7 (a) 16° (b) 46° (c) 49° 1 (a) x = 40.4° (b) x = 18.5 5 ≈ 16 m
(d) 23° (e) 38° (c) x = 61.2° (d) x = 18.2
(e) x = 7.1 (f) x = 28.1°
8 (a) hyp = y, adj(θ) = z, cosθ = z (g) x = 22.5° (h) x = 8.3 lookout 50°
(i) x = 14.0° (L)
y
25 m
(b) hyp = c, adj(θ) = b, cosθ = b
2 (a) x = 6.3 cm (b) x = 6.8 cm base of lookout 5 m swimmer shark
c (c) x = 4.4 cm (d) x = 13.2 cm (B) (W) (S)
(e) x = 8.7 cm (f) x = 10.8 cm
(c) hyp = c, adj(θ) = a, cosθ = a 6 RS = 591 m (c) A
3 (a) θ = 82.4° (b) θ = 23.1°
c (c) θ = 33.3° (d) θ = 28.9°
(e) θ = 38.4° (f) θ = 82.5°
(d) hyp = p, adj(θ) = r, cosθ = r
4 cos B = a2 + c2 − b2
p 2ac

(e) hyp = x, adj(θ) = z, cosθ = z

x

9 (a) sin A = 7, cos A = 12 tan A = 7 Chapter 16
13 12
13,

(b) sin B = 5 , cos B = 19.6, tan B = 10 Exercise 16.1

11 22 19.6

(c) sinC = 3, cosC = 4, tanC = 3 1 (a) E (b) C
(d) D (e) B
554

Answers 189

2 (a) student’s own line (line should go 3 (a) $930.75 (b) $1083.75 (b) $92.74
close to (160, 4.2) and (175, 5.55)); (c) $765 (d) $1179.38 (c) student’s own graph; a comment
answers (b) and (c) depend on
student’s best fit line 4 $1203.40 such as, the amount of compound
interest increases faster than the
(b) ≈ 4.7 m 5 $542.75 simple interest
(c) between 175 cm and 185 cm
(d) fairly strongly positive 6 (a) $625 (b) $25 (c) $506.50 5 $862.50
(e) taller athletes can jump further
Exercise 17.2 6 $3360

3 (a) distance (m) 1 (a) $7.50 (b) $160 (c) $210 7 (a) $1335, $2225
(b) (d) $448 (e) $343.75 (b) $1950, $3250
(c) $18 000, $30 000
600 (d) 2 5 years
8 (a) $4818 (b) 120%
500 3 2.8%
9 $425
Distance (m) 400 4 $2800 more 10 $211.20
11 $43.36 (each)
300 12 $204

200 5 $2281 more

100 6 (a) $7.50 (b) $187.73
(c) $225.75 (d) $574.55
0 (e) $346.08 Chapter 18
0 6 8 10 12 14
Age (years) 7 $562.75 Exercise 18.1

(c) weak positive 8 $27 085.85 1
(e) 12 years old
(f) Not very reliable because Exercise 17.3 x −3 −2 −1 01 2 3
2 + 2 −7 −2 1 2 1 −2 −7
correlation is very weak 1 (a) $100 (b) $200 (c) $340 (a)
(g) 600 m (d) $900

Mixed exercise (b) y x2 − 3 6 1 −2 −3 −2 1 6

1 (a) the number of accidents for 2 $300 (c) y x2 − 2 −11 −6 −3 −2 −3 −6 −11
different speeds
3 $500
(b) average speed
(c) answers to (c) depend on 4 $64.41 (d) y x2 − 3 −12 −7 −4 −3 −4 −7 −12

student’s best fit line 5 (a) $179.10 (e) y = x2 + 1 9.5 4.5 1.5 0.5 1.5 4.5 9.5
(i) ≈ 35 accidents (b) $40.04 2
(ii) < 40 km/h (c) $963.90
(d) strong positive
(e) there are more accidents when Mixed exercise 12 y
vehicles are travelling at a higher
average speed 1 (a) 12 h (b) 40 h (c) 25 1 h 10 (e)
2 (a) $1190 (b) $1386 2 8

2 (a) there a strong negative correlation (c) $1232
at first, but this becomes weaker as
the cars get older 3 (a) $62 808 (b) $4149.02 6 (b)

(b) 0−2 years 4 (a) 4
(c) it stabilises around the $6000 level
(d) 2−3 years Years Simple Compound 2 x
(e) $5000−$8000 interest interest –4 –2 0 24
1 300
2 300 609 –2
3 600 927.27
Chapter 17 4 900 1125.09 –4
5 1200 1592.74
Exercise 17.1 6 1500 1940.52 –6
7 1800 2298.74 (a)
1 $19.26 8 2100 2667.70
2 $25 560 2400 –8

–10 (c)
–12 (d)

190 Answers

2 (a) y = 3 + x2 (b) y = x2 + 2 (c) 6 m (d)
(c) y = x2 (d) y = −x2 + 3 (d) just short of 4 seconds
(e) y = −x2 − 4 (e) 3 seconds y
y = − 8 8
3 (a) x 7
6
Exercise 18.2
5

4

x −2 −1 0 1 2 3 4 5 1 (a) 3
y x2 − 3x + 2 12 6 2 0 0 2 6 12
y 2
10
9 1
8
y 7 –8 –7 –6 –5 –4 –3 –2 –10 1 2 3 4 5 6 7 8x
14 6
12 y = x2 − 3x + 2 5 xy = 5 –2
10 4 –3
8 x 3 –4 y = − 8
6 234 56 2 –5 x
4 1 –6
2 x –7
–8 –6 –4 –2 0 8 10 –8

246 (e)

–2
–3
–4 y
8
–5
–6 y = − 4 7
xy = 5 –7 x 6

–8 5

–3 –2 –1–20 1 –9 4
(b)
3

(b) 2

y 1

x −3 −2 −1 0 1 2 3 16 –8 –7 –6 –5 –4 –3 –2 –10 1 2 3 4 5 6 7 8x
y x2 − 2x − 1 14 7 2 −1 −2 −1 2
14 –2
–3
12 –4
–5
10 –6
–7
8 –8 y = − 4
x
y 6 16
15 y = x2 − 2x + 1 x
10 4 y =
2

–16–14–12–10–8–6–4––220 2 4 6 8 10 12 14 16 x

5 –4 2 (a)

–4 –3 –2 –1 0 x –6 Length 1 2 3 4 6 8 12 24
–5 1 2 34
–8 Width 24 12 8 6 4 3 2 1
(c)
–10 (b)

–12 24 y
22
y = 16 –14 20
x –16 18
x −2 −1 0 1 2 3 4 5 6 16
y x2 + 4x + 1 −11 −4 1 4 5 4 1 −4 −11 (c) 14
12
y 10
8
6y 10 6
9 4
4 8 2Width (m)
7 0x
2 6
–4 –2 0 5 0 2 4 6 8 10 12 14 16 18 20 22 24
x 4 Length (m)
–2 2 4 68 3
2 xy = 9 (c) the curve represents all the
possible measurements for the
–4 1x rectangle with an area of 24 m2

–10 –8 –6 –4 –2 0 2 4 6 8 10 (d) ≈ 3.4 m

–6 –2

–8 –3
–4
–5
–10 –6
–12 y = x2 + 4x + 1 –7
–8
4 (a) 8 m xy = 9 –9
(b) 2 seconds

Answers 191

Exercise 18.3 Exercise 18.4 2 (a)

1 (a) −2 or 3 (b) −1 or 2 1 (a) 8y 8y
(c) −3 or 4 7
y = 2x + 2 y = 2x2 + 3x − 2 7
6
y=x+2
5
6
2 (a) 4
5
3
3y 4
2
3
4 1x
2 2 y = x
1 –8–7–6–5–4–3–2–10 1 2 3 4 5 6 7 8
1 x
–4 –3 –2 –1–10
x –8–7–6–5–4–3–2–10 1 2 3 4 5 6 7 8 –2
123
–2 –3
–3
–4 –4
–5
–6 –5
–7
–2 –8 –6

–3 (1, 4) and (−2, −2) –7

–8

–4 (−2, 0) and (1, 3)

(b) (i) −2 or 1 (b) 12 y (b)
(ii) ≈ −1.6 or 0.6
(iii) ≈ −2.6 or 1.6 8y

11 7

y 10 6

3 (a) 14 y = x2 − x −6 9 y = x2 + 2x 5
13 4
8
3
12 7 2 y = −x + 4

11 6 1x

10 5 y = x2 + 2x − 3 –8–7–6–5–4–3–2–10 1 2 3 4 5 6 7 8
4
9
3 –2

8 2 –3

7 1x –4

6 –8–7–6–5–4–3–2–10 1 2 3 4 5 6 7 8 –5

5 –2 –6

4 –3 –7

3 –4 y = −x + 1 –8
–5
2 (−4, 8) and (1, 3)

1x –6 (c)

–5 –4 –3 –2 –10 1 2 3 4 5 –7 8y

–8

–2 7

–3 (−4, 5) and (1, 0)

–4 (c) 6 y = −2x2 + 2x + 4
5
–5 y
6
–6 4

–7 y = −x2 + 4 5 3 3
4 x
(b) (i) 0 or 1 y= 2
(ii) −2 or 3
(iii) −3 or 4 3 1x

2 –3 –2 –1 0 1 2 3 4

y 1x

14 y = x2 − x − 6 –6 –5 –4 –3 –2 –10 1 2 3 4 5 6 –2
13
–3
12 –2 –4 y = −2x + 4
–3
11 –4
–5
10 –6 (0, 4) and (2, 0)

9 (d)

8

7 (iii) y = 6 4y
6 3
y =1 x
5 ≈ (−2.3, −1.3), (1, 3) and 2 2
4 1
(1.3, 2.3) y = −0.5x2 + 1x + 1.5
3 x

2 –8–7–6–5–4–3–2–10 1 2 3 4 5 6 7 8
1 (ii) y = 0 x
–2
–4 –2 –10 1 2 3 4 5
–3

–2 –4

–3 –5

–4 –6

–5 (i) y = –6 –7

–6 –8

–7 (−3, −6) and (1, 2)

–8

192 Answers