Easa Module 1 - Mathematics
73
Click here to load reader
description
Basic Notes for Easa Module 1
Transcript of Easa Module 1 - Mathematics
Bo
ok
No
:IR
PA
RT
66
CA
TA
M7
Lu
fth
an
sa
Re
so
urc
e
Cw
mb
ran
S.
Wa
les
Fo
rT
rain
ing
Pu
rpo
se
sO
nly
Lu
fth
an
sa
19
95
Te
ch
nic
al
Tra
inin
gL
td
TrainingManual
Fundamentals
IRP
AR
T6
6C
AT
AM
1M
ath
em
ati
cs
LufthansaResource
TechnicalTraining
Fo
rtr
ain
ing
pu
rpo
se
sa
nd
inte
rna
lu
se
on
ly.
Co
py
rig
ht
by
Lu
fth
an
sa
Te
ch
nic
al
Tra
inin
gG
mb
H.
All
rig
hts
res
erv
ed
.N
op
art
so
fth
istr
ain
ing
ma
nu
al
ma
yb
es
old
or
rep
rod
uc
ed
ina
ny
form
wit
ho
ut
pe
rmis
sio
no
f:
LufthansaTechnicalTrainingGmbH
LufthansaBaseFrankfurt
D-6
05
46
Fra
nk
furt
/Ma
in
Te
l.+
49
69
/6
96
41
78
Fa
x+
49
69
/6
96
63
84
LufthansaBaseHamburg
We
gb
eim
Jä
ge
r1
93
D-2
23
35
Ha
mb
urg
Te
l.+
49
40
/5
07
02
41
3
Fa
x+
49
40
/5
07
04
74
6
LufthansaTechnicalTraining ForTrainingPurposesOnly
Pa
ge
:1
ATA
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
2
ARITHMETIC
General
Ari
thm
eti
cu
se
sre
al,
no
n--
ne
ga
tiv
en
um
be
rsa
nd
co
ns
ists
of
fou
ro
pe
rati
on
s,
ad
dit
ion
,s
ub
tra
cti
on
,m
ult
iplic
ati
on
an
dd
ivis
ion
.N
um
be
rsa
rere
pre
se
nte
db
ys
ym
bo
lsw
hic
ha
rec
alle
dd
igit
s.
Th
ere
are
ten
dig
its
wh
ich
are
1,
2,
3,
4,
5,
6,
7,
8,
9a
nd
0.
Th
es
ed
igit
sa
rec
om
bin
ed
tore
pre
se
nt
an
yv
alu
e.
ADDITION
Th
ep
roc
es
so
ffi
nd
ing
the
tota
lo
ftw
oo
rm
ore
nu
mb
ers
isc
alle
da
dd
itio
n.
Th
iso
pe
rati
on
isin
dic
ate
db
yth
ep
lus
(+)
sy
mb
ol.
Wh
en
nu
mb
ers
are
co
mb
ine
db
ya
dd
itio
n,
the
res
ult
ing
tota
lis
ca
lled
thesum
.W
he
na
dd
ing
wh
ole
nu
mb
ers
wh
os
eto
tal
ism
ore
tha
nn
ine
,it
isn
ec
es
sa
ryto
arr
an
ge
the
nu
mb
ers
inc
olu
mn
ss
oth
att
he
las
tdig
ito
fea
ch
nu
mb
er
isin
the
sa
me
co
lum
n.
Th
eu
nit
sc
olu
mn
co
nta
ins
the
va
lue
sz
ero
ton
ine
,th
ete
ns
co
lum
nc
on
tain
sm
ult
iple
so
fte
n,
up
ton
ine
ty,
an
dth
eh
un
dre
ds
co
lum
nc
on
sis
tso
fm
ult
iple
so
fo
ne
hu
nd
red
.
Example:
hu
nd
red
ste
ns
un
its
78
24
3+
46
27
83
To
ad
dth
es
um
of
the
ab
ov
e,
firs
ta
dd
the
un
its
co
lum
n,
8+
3+
2m
ak
es
13
.P
lac
eth
e3
inth
eu
nit
sc
olu
mn
of
the
an
sw
er
an
dc
arr
yth
e1
forw
ard
toth
ete
ns
co
lum
n.
Ad
din
gth
isw
eh
av
e1
+7
+4
+6
is1
8.
Pla
ce
the
8in
the
ten
sc
olu
mn
of
the
an
sw
er
an
dc
arr
yth
e1
forw
ard
toth
eh
un
dre
ds
co
lum
nw
hic
hw
en
ow
ad
d.
1+
2+
4is
7.
Pla
ce
the
7in
the
hu
nd
red
sc
olu
mn
of
the
an
sw
er.
We
se
eth
at
the
an
sw
er
(su
m)
toth
ea
dd
itio
nis
78
3.
Th
ep
roc
es
sis
ide
nti
ca
lif
an
yo
fth
en
um
be
rsin
clu
de
sa
de
cim
al
as
lon
ga
sth
ed
ec
ima
lp
oin
tsa
rea
rra
ng
ed
inth
es
am
ec
olu
mn
.
Example:
hu
nd
red
ste
ns
un
its
7.
82
4.
3+
46
.0
78
.3
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
3
SUBTRACTION
Th
ep
roc
es
so
ffi
nd
ing
thedifference
be
twe
en
two
nu
mb
ers
isk
no
wn
as
su
btr
ac
tio
na
nd
isin
dic
ate
db
yth
em
inu
s(-
-)s
ign
.S
ub
tra
cti
on
isa
cc
om
plis
he
db
yta
kin
gth
eq
ua
nti
tyo
fo
ne
nu
mb
er
aw
ay
fro
ma
no
the
rn
um
be
r.
To
fin
dth
ed
iffe
ren
ce
be
twe
en
two
nu
mb
ers
,a
rra
ng
eth
em
inth
es
am
em
an
ne
ru
se
dfo
ra
dd
itio
n.
Wit
ho
ne
nu
mb
er
at
the
top
an
dth
en
um
be
rto
be
su
btr
ac
ted
at
the
bo
tto
m,
alig
nth
ev
ert
ica
lco
lum
ns
so
the
las
td
igit
sa
rein
the
sa
me
co
lum
n.
Be
gin
nin
go
nth
eri
gh
t,s
ub
tra
ct
the
bo
tto
mn
um
be
rfr
om
the
top
.R
ep
ea
tth
isfo
re
ac
hc
olu
mn
.
Example:
hu
nd
red
ste
ns
un
its
44
3--
26
21
81
Pla
ce
26
2u
nd
er
44
3.
2fr
om
3le
av
es
1.
Wri
te1
inth
eu
nit
sc
olu
mn
of
the
an
sw
er.
6fr
om
4is
cle
arl
yim
po
ss
ible
,s
oth
e4
isin
cre
as
ed
inv
alu
eto
14
by
tak
ing
1fr
om
the
hu
nd
red
sc
olu
mn
lea
vin
g3
.1
4fr
om
6le
av
es
8.
Wri
te8
inth
ete
ns
co
lum
n.
Fin
ally
,3
fro
m2
inth
eh
un
dre
ds
co
lum
ns
lea
ve
s1
.
To
ch
ec
ka
su
btr
ac
tio
np
rob
lem
,y
ou
ca
na
dd
the
bo
tto
mn
um
be
rto
the
an
sw
er
an
dth
iss
ho
uld
eq
ua
lth
eto
pn
um
be
r.
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
4
MULTIPLICATION
Mu
ltip
lica
tio
nis
as
pe
cia
lfo
rmo
fre
pe
titi
ve
ad
dit
ion
.W
he
na
giv
en
nu
mb
er
isa
dd
ed
toit
se
lfa
sp
ec
ifie
dn
um
be
ro
fti
me
s,
the
pro
ce
ss
isc
alle
dm
ult
iplic
ati
on
.T
he
su
mo
f4
+4
+4
=1
2is
ex
pre
ss
ed
by
mu
ltip
lica
tio
na
s4
x3
=1
2.
Th
en
um
be
rs4
an
d3
are
ca
lledfactors
an
dth
ea
ns
we
r,1
2,
rep
res
en
tsth
eproduct.
Mu
ltip
lica
tio
nis
typ
ica
llyin
dic
ate
db
ya
nx
or
by
the
lac
ko
fa
ny
op
era
tio
ns
ign
.
On
eim
po
rta
nt
fac
tor
tore
me
mb
er
wh
en
mu
ltip
lyin
gis
tha
tth
eo
rde
rin
wh
ich
nu
mb
ers
are
mu
ltip
lied
do
es
no
tc
ha
ng
eth
ep
rod
uc
t.
Example:
3o
r4
x4
x3
12
12
Lik
ea
dd
itio
na
nd
su
btr
ac
tio
n,
wh
en
mu
ltip
lyin
gla
rge
nu
mb
ers
itis
imp
ort
an
tth
ey
are
alig
ne
dv
ert
ica
lly.
Example: 5
32
x2
41
06
40
21
28
12
,76
8
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
5
DIVISION
Ju
st
as
su
btr
ac
tio
nis
the
rev
ers
eo
fa
dd
itio
n,
div
isio
nis
the
rev
ers
eo
fm
ult
iplic
ati
on
.D
ivis
ion
isa
me
an
so
ffi
nd
ing
ou
th
ow
ma
ny
tim
es
an
um
be
ris
co
nta
ine
din
an
oth
er
nu
mb
er.
Th
en
um
be
rd
ivid
ed
isc
alle
ddividend
,th
en
um
be
ry
ou
are
div
idin
gb
yis
thedivisor,
an
dth
ere
su
ltis
thequotient.
Wit
hs
om
ed
ivis
ion
pro
ble
ms
,th
eq
uo
tie
nt
ma
yin
clu
de
are
ma
ind
er.
Are
ma
ind
er
rep
res
en
tsth
at
po
rtio
no
fth
ed
ivid
en
dth
at
ca
nn
ot
be
div
ide
db
yth
ed
ivis
or.
Div
isio
nis
ind
ica
ted
by
the
us
eo
fth
ed
ivis
ion
sig
n(÷
)w
ith
the
div
ide
nd
toth
ele
fta
nd
the
div
iso
rto
the
rig
ht
of
the
sig
n,
or
wit
hth
ed
ivid
en
din
sid
eth
es
ign
an
dth
ed
ivis
or
toth
ele
ft.
Div
isio
na
lso
isin
dic
ate
din
fra
cti
on
al
form
.
Fo
re
xa
mp
le,
inth
efr
ac
tio
n3 8
the
3is
the
div
ide
nd
an
dth
e8
isth
ed
ivis
or.
Wh
en
div
isio
nis
ca
rrie
do
ut,
the
qu
oti
en
tis
0.3
75
.
Th
ep
roc
es
so
fd
ivid
ing
larg
eq
ua
nti
tie
sis
pe
rfo
rme
db
yb
rea
kin
gth
ep
rob
lem
do
wn
into
as
eri
es
of
op
era
tio
ns
,e
ac
hre
su
ltin
gin
as
ing
led
igit
qu
oti
en
t.T
his
isb
es
till
us
tra
ted
by
ex
am
ple
.
Example:
div
ide
nd
div
iso
r
41
6÷
8=
52
or
52
8)
41
6
40 1
6
16
To
ch
ec
ka
div
isio
np
rob
lem
for
ac
cu
rac
y,m
ult
iply
the
qu
oti
en
tb
yth
eth
ed
ivis
or
an
da
dd
the
rem
ain
de
r(i
fa
ny
).If
the
op
era
tio
nis
ca
rrie
do
ut
pro
pe
rly,
the
res
ult
eq
ua
lsth
ed
ivid
en
d.
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
6
Calculate
thesum
ofthefollowingexamples:
Addition
i.0
.25
1+
10
.29
8
ii.1
8.0
98
+2
10
.09
9
iii.
0.0
25
+1
0.9
95
iv.
1.0
9+
1.2
+1
0.1
4
v.2
7.3
+0
.02
1+
68
.3
Subtraction
i.2
7.3
--4
.36
ii.2
1.7
6--
18
.51
iii.
32
.76
--2
0.0
86
iv.
10
.75
--1
9.9
99
--2
1.1
00
v.1
.09
--1
.2-
68
.3
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
7
AddingSignedNumbers
Wh
en
ad
din
gtw
oo
rm
ore
nu
mb
ers
wit
hth
es
am
es
ign
,ig
no
reth
es
ign
an
dfi
nd
the
su
mo
fth
ev
alu
es
an
dth
en
pla
ce
the
co
mm
on
sig
nin
fro
nt
of
the
an
sw
er.
Ino
the
rw
ord
s,a
dd
ing
two
or
mo
rep
os
itiv
en
um
be
rsa
lwa
ys
res
ult
sin
ap
os
itiv
es
um
,w
he
rea
sa
dd
ing
two
or
mo
ren
eg
ati
ve
nu
mb
ers
res
ult
sin
an
eg
ati
ve
su
m.
Wh
en
ad
din
ga
po
sit
ive
an
dn
eg
ati
ve
nu
mb
er,
fin
dth
ed
iffe
ren
ce
be
twe
en
the
two
nu
mb
ers
an
da
pp
ly(+
or
--)
of
the
larg
er
nu
mb
er.
Ino
the
rw
ord
s,
ad
din
gn
eg
ati
ve
nu
mb
er
isth
es
am
ea
ss
ub
tra
cti
ng
ap
os
itiv
en
um
be
r.T
he
res
ult
of
ad
din
go
rs
ub
tra
cti
ng
sig
ne
dn
um
be
rsis
ca
lled
alg
eb
raic
su
mo
fth
os
en
um
be
rs.
Ad
d2
5+
(--1
5)
25
25
+(-
-15
)o
r--
15
10
10
SubtractingSignedNumbers
Wh
en
su
btr
ac
tin
gn
um
be
rsw
ith
diff
ere
nt
sig
ns
,c
ha
ng
eth
eo
pe
rati
on
sig
nto
plu
sa
nd
ch
an
ge
the
sig
no
fth
es
ub
tra
he
nd
.O
nc
eth
isis
do
ne
,p
roc
ee
da
sy
ou
do
ina
dd
itio
n.
Fo
re
xa
mp
le+
3--
--4
isth
es
am
ea
s+
3+
+4
.T
he
reis
no
diff
ere
nc
eif
the
su
btr
ah
en
dis
larg
er
tha
nth
em
inu
en
d,
sin
ce
the
op
era
tio
nis
do
ne
as
tho
ug
hth
etw
oq
ua
nti
tie
sa
rea
dd
ed
.
Example:
Su
btr
ac
t4
8fr
om
--2
16
Step1:
Se
tu
pth
es
ub
tra
cti
on
pro
ble
m--
21
6--
48
Step2:
Ch
an
ge
the
op
era
tio
ns
ign
toa
plu
ss
ign
an
dc
ha
ng
eth
es
ign
of
the
su
btr
ah
en
d.
No
wa
dd
.
--2
16
+--
48
=--
26
4
MultiplyingSignedNumbers
Mu
ltip
lica
tio
no
fs
ign
ed
nu
mb
ers
isa
cc
om
plis
he
din
the
sa
me
ma
nn
er
as
mu
lti-
plic
ati
on
of
an
yo
the
rn
um
be
r.H
ow
ev
er,
aft
er
mu
ltip
lyin
g,
the
pro
du
ct
mu
st
be
giv
en
as
ign
.T
he
rea
reth
ree
rule
sto
follo
ww
he
nd
ete
rmin
ing
ap
rod
uc
tss
ign
.
1.
Th
ep
rod
uc
to
ftw
op
os
itiv
en
um
be
rsis
alw
ay
sp
os
itiv
e.
2.
Th
ep
rod
uc
to
ftw
on
eg
ati
ve
nu
mb
ers
isa
lwa
ys
po
sit
ive
.
3.
Th
ep
rod
uc
to
fa
po
sit
ive
an
da
ne
ga
tiv
en
um
be
ris
alw
ay
sn
eg
ati
ve
.
Example:
6x
2=
12
--6
x--
2=
12
(--6
)x
(--2
)=
12
(--6
)x
2=
--1
2
DividingSignedNumbers
Lik
em
ult
iply
ing
sig
ne
dn
um
be
rs,
div
isio
no
fs
ign
ed
nu
mb
ers
isa
cc
om
plis
he
din
the
sa
me
ma
nn
er
as
div
idin
ga
ny
oth
er
nu
mb
er.
Th
es
ign
of
the
qu
oti
en
tis
de
term
ine
du
sin
gth
eru
les
ide
nti
ca
lto
tho
se
us
ed
inm
ult
iplic
ati
on
.
Ex
am
ple
:
12÷
3=
41
2÷
(--3
)=
--4
(--1
2)÷
(--)
3=
4(-
-12
)÷
3=
--4
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
8
Calculate
thesum
ofthefollowingexamples:
i.--
8+
5
ii.--
7--
6--
3
iii.
8--
7--
15
iv.
--3
+5
+7
--4
--2
v.6
+4
--3
--5
--7
+2
vi.
8x
(--3
)
vii.
(--2
)x
(--5
)x
(--6
)
viii
.4
x(-
-3)
x(-
-2)
ix.
(--3
)x
(--4
)x
5
x.
--1
6÷
((--
2)
x(-
-4))
xi.
(15
x(-
-3)
x2
)÷
((--
5)
x(-
-6))
xii.
3+
5x
2
xiii
.(7
x5
)--
2+
(4x
6)
xiv
.(7
x5
)--
(12÷
4)
+3
xv.
11--
(9÷
3)
+7
xv
i.11
--(1
2÷
4)
+3
x(6
--2
)
xv
ii.(1
5÷
(4+
1))
--(
9x
3)
+7
(4+
3)
xv
iii.
10
--(1
2÷
6)
+3
(8--
3)
Question1.
16
ho
les
sp
ac
ed
48
mm
ap
art
are
tob
em
ark
ed
off
on
as
he
et
of
me
tal.
17
mm
isto
be
allo
we
db
etw
ee
nth
ec
en
tre
so
fth
eh
ole
sa
nd
the
ed
ge
of
the
me
tal.
Ca
lcu
late
the
tota
lle
ng
tho
fm
eta
lre
qu
ire
d.
Question2.
Inth
efi
rst
2h
ou
rso
fa
sh
ift
an
op
era
tor
ma
ke
s3
2s
old
ere
djo
ints
pe
rh
ou
r.In
the
ne
xt
3h
ou
rsth
eo
pe
rato
rm
ak
es
29
join
tsp
er
ho
ur.
Inth
efi
na
ltw
oh
ou
rs2
6jo
ints
are
ma
de
pe
rh
ou
r.H
ow
ma
ny
so
lde
red
join
tsa
rem
ad
ein
the
7h
ou
rs.
Question3.
Am
ac
hin
ist
ma
ke
s3
pa
rts
in1
5m
inu
tes
.H
ow
ma
ny
pa
rts
ca
nh
ep
rod
uc
ein
an
8h
ou
rs
hif
ta
llow
ing
20
min
ute
sfo
rs
tart
ing
an
d1
0m
inu
tes
for
fin
ish
ing
the
sh
ift.
Question4.
Th
ele
ng
tho
fa
me
tal
pla
teis
89
1m
m.
Riv
ets
are
pla
ce
d4
5m
ma
pa
rta
nd
the
dis
tan
ce
be
twe
en
the
ce
ntr
es
of
the
en
dri
ve
tsa
nd
the
ed
ge
of
the
pla
teis
18
mm
.H
ow
ma
ny
riv
ets
are
req
uir
ed
.
Question5.
32
pin
se
ac
h6
1m
mlo
ng
are
tob
etu
rne
din
ala
the
.If
2m
mis
allo
we
do
ne
ac
h
pin
for
pa
rtin
go
ff.
wh
at
tota
lle
ng
tho
fm
ate
ria
lis
req
uir
ed
tom
ak
eth
ep
ins
.
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
9
COMMONFRACTIONS
Introduction
Ac
om
mo
nfr
ac
tio
nre
pre
se
nts
ap
ort
ion
or
pa
rto
fa
qu
an
tity
.F
or
ex
am
ple
,if
a
nu
mb
er
isd
ivid
ed
into
thre
ee
qu
alp
art
s,
ea
ch
pa
rtis
on
e--
thir
d(1 3
)o
fth
en
um
be
r.
Afr
ac
tio
nc
on
sis
tso
ftw
on
um
be
rs,
on
ea
bo
ve
an
do
ne
be
low
alin
e,
orfraction
bar.
Th
efr
ac
tio
nb
ar
ind
ica
tes
div
isio
no
fth
eto
pn
um
be
r,o
rnumerator,
by
the
bo
tto
mn
um
be
r,o
rdenominator.
Fo
re
xa
mp
le,
the
fra
cti
on3 4
ind
ica
tes
tha
tth
ree
isd
ivid
ed
by
fou
rto
fin
dth
ed
ec
ima
le
qu
iva
len
to
f0
.75
.
Wh
en
afr
ac
tio
ns
nu
me
rato
ris
sm
alle
rth
an
the
de
no
min
ato
r,th
efr
ac
tio
nis
ca
lled
aproperfraction
.A
pro
pe
rfr
ac
tio
nis
alw
ay
sle
ss
tha
n1
.If
the
nu
me
rato
ris
larg
er
tha
nth
ed
en
om
ina
tor,
the
fra
cti
on
isc
alle
da
nim
properfraction
.In
this
sit
ua
tio
nth
efr
ac
tio
nis
gre
ate
rth
an
1.
Ifth
en
um
era
tor
an
dth
ed
en
om
ina
tor
are
ide
nti
ca
l,th
efr
ac
tio
nis
eq
ua
lto
1a
sa
ny
nu
mb
er
div
ide
db
yit
se
lfis
1.
Amixednumber
isth
ec
om
bin
ati
on
of
aw
ho
len
um
be
ra
nd
ap
rop
er
fra
cti
on
.
Mix
ed
nu
mb
ers
are
ex
pre
ss
ed
as
15 8
an
d2
99 16
an
da
rety
pic
ally
us
ed
inp
lac
eo
f
imp
rop
er
fra
cti
on
s.
Th
en
um
era
tor
an
dd
en
om
ina
tor
ofa
fra
cti
on
ca
nb
ec
ha
ng
ed
wit
ho
ut
ch
an
gin
gth
efr
ac
tio
ns
va
lue
.A
mix
ed
nu
mb
er
ca
nb
ec
on
ve
rte
din
toa
nim
pro
pe
rfr
ac
tio
na
nd
vic
ev
ers
a.
Example:
Co
nv
ert
82 3
=( 8
×3) +2
=26 3
Ex
pre
ss27 4
as
am
ixe
dn
um
be
r
27 4
=63 4
(sin
ce
27÷
4=
6re
ma
ind
er
3)
3
LowestTerm
s
Afr
ac
tio
nis
sa
idto
be
init
slowestterm
sw
he
nit
isim
po
ss
ible
tofi
nd
an
um
be
rw
hic
hw
illd
ivid
ee
xa
ctl
yin
tob
oth
its
nu
me
rato
ra
nd
de
no
min
ato
r.T
he
fra
cti
on
s5 7and1119
are
bo
thin
the
irlo
we
st
term
sb
ut
the
fra
cti
on6 10
isn
ot
init
slo
we
st
term
s
be
ca
us
eit
ca
nb
ere
du
ce
dto3 5
by
div
idin
gto
pa
nd
bo
tto
mn
um
be
rsb
y2
.
Ex
am
ple
:
Re
du
ce2135
toit
slo
we
st
term
s
2135
ise
qu
iva
len
tto21÷7and35÷7=3 5
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
10
ComparingtheSizeofFractions
Wh
en
the
va
lue
so
ftw
oo
rm
ore
fra
cti
on
sa
reto
be
co
mp
are
d,
the
ys
ho
uld
be
ex
pre
ss
ed
as
fra
cti
on
sw
ith
the
sa
me
de
no
min
ato
r.T
his
co
mm
on
de
no
min
ato
rs
ho
uld
be
the
low
es
tc
om
mo
nm
ult
iple
(LC
M)
of
the
de
no
min
ato
rso
fth
efr
ac
tio
ns
tob
ec
om
pa
red
.It
iss
om
eti
me
sc
alle
dth
elo
we
st
co
mm
on
de
no
min
ato
r(L
CD
).
Ex
am
ple
:
Arr
an
ge
the
fra
cti
on
s5 6,8 9and7 8
ino
rde
ro
fs
ize
be
gin
nin
gw
ith
the
sm
alle
st.
Th
eL
CM
of
the
de
no
min
ato
rs6
,8
,a
nd
9is
72
,i.
e.
the
low
es
tn
um
be
rw
hic
hth
ey
ca
na
llb
ed
ivid
ed
into
is7
2.
5 6is
eq
uiv
ale
nt
to( 5
×12)
( 6×12)=6072
8 9is
eq
uiv
ale
nt
to( 8
×8)
( 9×8)=6472
7 8is
eq
uiv
ale
nt
to( 7
×9)
( 8×9)=6372
Be
ca
us
ea
llth
efr
ac
tio
ns
ha
ve
be
en
ex
pre
ss
ed
wit
hth
es
am
ed
en
om
ina
tor
all
tha
tw
en
ee
dto
do
isto
co
mp
are
the
nu
me
rato
rs.
Th
ere
fore
the
ord
er
of
siz
eis
6072,6372and6472or5 6,7 8and8 9
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
11
AdditionofFractions
Tw
ofr
ac
tio
ns
wh
ich
ha
ve
the
sa
me
de
no
min
ato
rc
an
be
ad
de
dto
ge
the
rb
ya
dd
ing
the
irn
um
era
tors
.T
hu
s
3 11+5 11=( 3
+5)
11
=8 11
Wh
en
two
fra
cti
on
sh
av
ed
iffe
ren
td
en
om
ina
tors
the
yc
an
no
tb
ea
dd
ed
tog
eth
er
dir
ec
tly.
Ho
we
ve
r,if
we
ex
pre
ss
the
fra
cti
on
sw
ith
the
sa
me
de
no
min
ato
rth
ey
ca
nb
ea
dd
ed
.
Example:
Ad
d2 5and3 7
Th
elo
we
st
co
mm
on
de
no
min
ato
ro
f5
an
d7
is3
5
2 5+3 7=1435+1535
=( 14+15)
35
=2935
Wh
en
mix
ed
nu
mb
ers
are
tob
ea
dd
ed
tog
eth
er,
the
wh
ole
nu
mb
ers
an
dth
efr
ac
tio
ns
are
ad
de
ds
ep
ara
tely
.
Example:
Ad
d42 3and23 5
42 3+23 5=6+2 3+3 5
=6+1015+9 15
=6+1915
=6+14 15
=74 15
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
12
SubtractionofFractions
Ifth
efr
ac
tio
ns
tob
es
ub
tra
cte
dh
av
eth
es
am
ed
en
om
ina
tor
the
no
ne
nu
me
rato
rc
an
be
su
btr
ac
ted
fro
mth
eo
the
r.
Example:
9 16−5 16=4 16=1 4
Ifth
etw
ofr
ac
tio
ns
ha
ve
diff
ere
nt
de
no
min
ato
rsth
en
am
eth
od
sim
ilar
toth
at
for
ad
dit
ion
isu
se
d.
Example:
Su
btr
ac
t3 4
fro
m5 6
Th
elo
we
st
co
mm
on
de
no
min
ato
ris
12
5 6−3 4=1012−9 12
=( 10−9)
12
=1 12
Wh
en
mix
ed
nu
mb
ers
are
inv
olv
ed
firs
tsu
btr
ac
tth
ew
ho
len
um
be
rsa
nd
the
nd
ea
lw
ith
the
fra
cti
on
al
pa
rts
.
Example:
Su
btr
ac
t63 4−41 3
63 4−41 3=2+3 4−1 3
=2+( 9
−4)
12
=2+5 12
=25 12
Alt
ern
ati
ve
lyth
en
um
be
rsc
an
be
co
nv
ert
ed
into
imp
rop
er
fra
cti
on
sa
nd
the
nth
es
ub
tra
cti
on
ca
rrie
do
ut
as
be
fore
.
Example:
Su
btr
ac
t35 16−11 85316−9 8
5316−1816
3516
23 16
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
13
MultiplicationofFractions
Mu
ltip
lica
tio
no
ffr
ac
tio
ns
isp
erf
orm
ed
by
mu
ltip
lica
tio
nth
en
um
era
tors
of
ea
ch
fra
cti
on
tofo
rmth
ep
rod
uc
tn
um
era
tors
,a
nd
mu
ltip
lyin
gth
ein
div
idu
al
de
no
min
ato
rsto
form
the
pro
du
ct
de
no
min
ato
r.T
he
res
ult
ing
fra
cti
on
isth
en
red
uc
ed
toit
slo
we
st
term
s.
Example:
Mu
ltip
ly3 8by5 7
3 8×5 7=( 3
×5)
( 8×7)
=1556
Ifa
ny
fac
tors
are
co
mm
on
toa
nu
me
rato
ra
nd
ad
en
om
ina
tor
the
ys
ho
uld
be
ca
nc
elle
db
efo
rem
ult
iply
ing
.
Ex
am
ple
:
Fin
dth
ev
alu
eo
f2 3×5 7×2132
2 3×5 7×2132=( 1
×5×1)
( 1×1×16)
=5 16
Mix
ed
nu
mb
ers
mu
st
be
co
nv
ert
ed
into
improperfractions
be
fore
mu
ltip
lyin
g.
Example:
Mu
ltip
ly13 8×21 3
13 8×21 3=11 8×7 3
=( 11×7)
( 8×3)
=7724
=35 24
Inp
rob
lem
sw
ith
fra
cti
on
sth
ew
ord
“of”
isfr
eq
ue
ntl
yu
se
d.
Its
ho
uld
alw
ay
sb
eta
ke
na
sm
ea
nin
g“m
ult
iply
”.
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
14
DivisionofFractions
Div
isio
no
fc
om
mo
nfr
ac
tio
ns
isa
cc
om
plis
he
db
yin
ve
rtin
g,
or
turn
ing
ov
er,
the
div
iso
ra
nd
the
nm
ult
iply
ing
.Ho
we
ve
r,it
isim
po
rta
ntt
ha
tyo
uin
ve
rtth
ed
ivis
or
on
lya
nd
no
tth
ed
ivid
en
d.O
nc
eth
ed
ivis
or
isin
ve
rte
d,m
ult
iply
the
nu
me
rato
rsto
ob
tain
an
ew
nu
me
rato
r,m
ult
iply
the
de
no
min
ato
rsto
ob
tain
an
ew
de
no
min
ato
r,a
nd
red
uc
eth
eq
uo
tie
nt
toit
slo
we
st
term
s.
Example:
Div
ide3 5by7 8
3 5÷7 8
=3 5×8 7
=( 3
×8}
( 5×7)
=2435
Mix
ed
nu
mb
ers
mu
st
be
co
nv
ert
ed
into
improperfractions
be
fore
mu
ltip
lyin
g.
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
15
Solvethefollowingequations:
Co
nv
ert
the
follo
win
gm
ixe
dn
um
be
rsto
imp
rop
er
fra
cti
on
s:
i.26 7
ii.34 9
iii.213 5
iv.52125
v.21 7
Co
nv
ert
the
follo
win
gim
pro
pe
rfr
ac
tio
ns
tom
ixe
dn
um
be
rs:
i.11 3
ii.21 5
iii.53 7
iv.2104
v.99 8
Ad
dth
efo
llow
ing
fra
cti
on
s:
i.3 4+3 8
ii.1 8+2 3+5 12
iii.72 3+63 5
iv.33 8+52 7+43 4
v.2310+14 6
Su
btr
ac
tth
efo
llow
ing
fra
cti
on
s:
i.7 8−5 6
ii.33 8−11 4
iii.53 8−29 10
iv.21 5−32 5
v.13 4−22 5
Mu
ltip
lya
nd
sim
plif
yth
efo
llow
ing
fra
cti
on
s:
i.3 4×5 7
ii.2 9×12 3
iii.7 5×31 2
iv.33 4×13 5×11 8
v.3 4
of
16
Div
ide
an
ds
imp
lify
the
follo
win
gfr
ac
tio
ns
:
i.4 5÷11 3
ii.21 2÷33 4
iii.5÷51 5
iv.12 3÷�3 5
÷9 10�
v.28 9÷� 12 3+1 2�
Arr
an
ge
the
follo
win
gs
ets
of
fra
cti
on
sin
ord
er
of
siz
e:
i.1 2���5 6���2 3���7 12
ii.3 4���5 8���9 16���1732
iii.3 8���5 9���2 6���5 18
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
16
DECIM
ALS
Th
ete
rmd
ec
ima
lm
ea
ns
’re
late
dto
ten
’a
nd
the
ya
rea
me
tho
do
fre
pre
se
nti
ng
va
lue
sle
ss
tha
no
ne
as
an
alt
ern
ati
ve
tou
sin
gfr
ac
tio
ns
.D
ec
ima
ln
um
be
rsa
ree
as
ier
toa
dd
,s
ub
tra
ct
etc
tha
nc
om
mo
nfr
ac
tio
ns
.A
co
mm
on
fra
cti
on
ca
nb
ec
on
ve
rte
dto
ad
ec
ima
lfra
cti
on
by
div
idin
gth
en
um
era
tor
by
the
de
no
min
ato
r.F
or
ex
am
ple
,¾
isc
on
ve
rte
dto
ad
ec
ima
lb
yd
ivid
ing
the
3b
yth
e4
.
Th
ed
ec
ima
leq
uiv
ale
nt
of¾
is0
.75
.Im
pro
pe
rfr
ac
tio
ns
are
co
nv
ert
ed
tod
ec
ima
lsin
the
sa
me
ma
nn
er.
Ho
we
ve
r,w
ho
len
um
be
rsa
pp
ea
rto
the
left
of
the
de
cim
al
po
int.
Ina
de
cim
al,
ea
ch
dig
itre
pre
se
nts
am
ult
iple
of
ten
.T
he
firs
td
igit
rep
res
en
tste
nth
s,
the
se
co
nd
hu
nd
red
ths
,th
eth
ird
tho
us
an
dth
s.
Example:
0.5
isre
ad
as
fiv
ete
nth
s
0.0
5is
rea
da
sfi
ve
hu
nd
red
ths
0.0
05
isre
ad
as
fiv
eth
ou
sa
nd
ths
wh
en
wri
tin
gd
ec
ima
ls,
the
nu
mb
er
of
ze
ros
toth
eri
gh
to
fth
ed
ec
ima
ld
oe
sn
ot
aff
ec
tth
ev
alu
ea
slo
ng
as
no
oth
er
nu
mb
er
ex
ce
pt
ze
roa
pp
ea
rs.
Ino
the
rw
ord
s,
nu
me
ric
ally
,2
.5,
2.5
0a
nd
2.5
00
are
the
sa
me
.
Th
en
um
be
ro
fd
igit
sa
fte
rth
ed
ec
ima
lp
oin
ta
rec
alle
ddecim
alplaces
Examples:
27
.6o
ne
de
cim
al
po
int
27
.16
two
de
cim
al
po
ints
27
.02
6th
ree
de
cim
al
po
ints
an
ds
oo
n.
AddingDecim
als
Th
ea
dd
itio
no
fd
ec
ima
lsis
do
ne
inth
es
am
em
an
ne
ra
sth
ea
dd
itio
no
fw
ho
len
um
be
rs.
Ho
we
ve
r,c
are
mu
st
be
tak
en
toc
orr
ec
tly
alig
nth
ed
ec
ima
lp
oin
tsv
ert
ica
lly.
Ex
am
ple
:
Ad
dth
efo
llow
ing
25
.78
+5
.4+
0.2
37
rew
rite
wit
hth
ed
ec
ima
lp
oin
tsa
lign
ed
an
da
dd
.
25
.78
5.4
+0
.23
7
31
.41
7
On
ce
ev
ery
thin
gis
ad
de
d,
the
de
cim
alp
oin
tin
the
an
sw
er
isp
lac
ed
dir
ec
tly
be
low
the
oth
er
de
cim
al
po
ints
.
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
17
SubtractingDecim
als
Lik
ea
dd
ing
,s
ub
tra
cti
ng
de
cim
als
isd
on
ein
the
sa
me
ma
nn
er
as
wit
hw
ho
len
um
be
rs.
Ag
ain
,it
isim
po
rta
nt
tha
ty
ou
ke
ep
the
de
cim
al
po
ints
alig
ne
d.
Example:
Ify
ou
ha
ve
32
5.2
5k
ilos
of
ca
rgo
on
bo
ard
an
dre
mo
ve
30
.75
kilo
s,
ho
wm
uc
hc
arg
ore
ma
ins
?
32
5.2
5
--3
0.7
5
29
4.5
0
MultiplyingDecim
als
Wh
en
mu
ltip
lyin
gd
ec
ima
ls,
ign
ore
the
de
cim
al
po
ints
an
dm
ult
iply
the
res
ult
ing
wh
ole
nu
mb
ers
.O
nc
eth
ep
rod
uc
tis
ca
lcu
late
d,
co
un
tth
en
um
be
ro
fd
igit
sto
the
rig
ht
of
the
de
cim
al
po
int
inb
oth
the
mu
ltip
lier
an
dm
ult
iplic
an
d.
Th
isn
um
be
rre
pre
se
nts
the
nu
mb
er
of
pla
ce
sfr
om
the
left
the
de
cim
al
po
int
isp
lac
ed
inth
ep
rod
uc
t.
Ex
am
ple
:
26
.75
73
de
cim
al
x0
.32
2d
ec
ima
l
53
51
4
80
27
1
85
62
24
co
un
t5
de
cim
al
pla
ce
sto
the
left
of
the
dig
it4
8.5
62
24
DividingDecim
als
Wh
en
div
idin
gd
ec
ima
ls,
the
op
era
tio
nis
ca
rrie
do
ut
inth
es
am
em
an
ne
ra
sd
ivi-
sio
no
fw
ho
len
um
be
rs.
Ho
we
ve
r,to
en
su
rea
cc
ura
tep
lac
em
en
to
fd
ec
ima
lpo
int
inth
eq
uo
tie
nt,
two
rule
sa
pp
ly:
1.
Wh
en
the
div
iso
ris
aw
ho
len
um
be
r,th
ed
ec
ima
lp
oin
tin
the
qu
oti
en
ta
lign
sv
ert
ica
llyw
ith
the
de
cim
al
inth
ed
ivid
en
dw
he
nd
oin
glo
ng
div
isio
n.
2.
Wh
en
the
div
iso
ris
ad
ec
ima
lfr
ac
tio
n,
its
ho
uld
firs
tb
ec
on
ve
rte
dto
aw
ho
len
um
be
rb
ym
ov
ing
the
de
cim
al
po
int
toth
eri
gh
t.H
ow
ev
er,
wh
en
the
de
cim
al
inth
ed
ivis
or
ism
ov
ed
,th
ed
ec
ima
lin
the
div
ide
nd
mu
st
als
om
ov
ein
the
sa
me
dir
ec
tio
na
nd
the
sa
me
nu
mb
er
of
sp
ac
es
.
Ex
am
ple
:
Div
ide
37
.26
by
2.7
Mo
ve
the
de
cim
al
inth
ed
ivis
or
toth
eri
gh
tto
co
nv
ert
itto
aw
ho
len
um
be
r.
27
)3
7.2
6
Mo
ve
the
de
cim
al
inth
ed
ivid
en
dth
es
am
en
um
be
ro
fp
lac
es
toth
eri
gh
t.
27
)3
72
.6
13
.8
Div
ide
:2
7)
37
2.6
27
10
2
81
21
6
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
18
ConvertingDecim
als
toFractions
Alt
ho
ug
hd
ec
ima
lsa
rety
pic
ally
ea
sie
rto
wo
rkw
ith
,th
ere
are
tim
es
wh
en
the
us
eo
fa
fra
cti
on
ism
ore
pra
cti
ca
l.F
or
ex
am
ple
,w
he
nm
ea
su
rin
gs
om
eth
ing
,m
os
ts
ca
les
are
fra
cti
on
ali
nc
rem
en
ts.
Fo
rth
isre
as
on
itis
imp
ort
an
tth
aty
ou
kn
ow
ho
wto
co
nv
ert
ad
ec
ima
ln
um
be
rin
toa
fra
cti
on
.F
or
ex
am
ple
,0
.12
5is
rea
da
s1
25
tho
us
an
dth
s,
wh
ich
isw
ritt
en
as
12
5/1
00
0.
Th
isfr
ac
tio
nis
the
nre
du
ce
dto
its
low
es
tte
rms
.
Ex
am
ple
s:
0.800=800
1000=4 5
6.250=6+250
1000=61 4
0.037=37
1000
ConvertingFractionsto
Decim
als
To
co
nv
ert
afr
ac
tio
nin
toa
de
cim
alw
ed
ivid
eth
ed
en
om
ina
tor
into
the
nu
me
rato
r.
Co
nv
ert2732
tod
ec
ima
ls
2732=27÷32
=0
.84
37
5
Wh
en
we
ha
ve
mix
ed
nu
mb
ers
toc
on
ve
rtin
tod
ec
ima
lsw
en
ee
do
nly
de
al
wit
h
the
fra
cti
on
alp
art
.T
hu
sto
co
nv
ert29 16
into
de
cim
als
we
on
lyh
av
eto
de
alw
ith9 16
9 16=9÷16
=0
.56
25
Th
ed
ivis
ion
sh
ow
sth
at9 16=0.5625
an
dh
en
ce29 16
=2
.56
25
.
So
me
tim
es
afr
ac
tio
nw
illn
ot
div
ide
ou
te
xa
ctl
y.If
the
nu
mb
er
isre
cu
rrin
gth
ea
ns
we
rc
an
be
giv
en
to1
or
2d
ec
ima
lp
lac
es
or
tha
ts
pe
cif
ied
by
the
eq
ua
tio
n.
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
19
ConvertingFractionsto
Percentages
To
ch
an
ge
afr
ac
tio
nto
ap
erc
en
tag
ey
ou
mu
st
mu
ltip
lyb
y1
00
.
Example:
3 5a
sa
pe
rce
nta
ge
=3 5×100%
=( 3
×100)
5=
60
%
43 4
as
ap
erc
en
tag
e=19 4×100%1
=19004
=4
75
%
ConvertingPercentagesto
Fractions
To
ch
an
ge
ap
erc
en
tag
eto
afr
ac
tio
n,
div
ide
by
10
0%
.
Examples:
8%
as
afr
ac
tio
n=8%
100%
=8100=2 25
12½
%(1
2.5
)a
sa
fra
cti
on
=12.5%
100%
=25 2×1100=25200=1 8
ConvertingPercentagesto
Decim
als
To
co
nv
ert
ap
erc
en
tag
eto
ad
ec
ima
l,fi
rstl
y,c
on
ve
rtth
ep
erc
en
tag
eto
afr
ac
tio
n,
the
nth
efr
ac
tio
nto
ad
ec
ima
l.
Ex
am
ple
s:
65
%a
sa
fra
cti
on
=65100
,a
sa
de
cim
al
=0
.65
32½
%a
sa
fra
cti
on
=32.5100
,a
sa
de
cim
al
=0
.32
5
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
20
ConvertaDecim
alto
aPercentage
To
co
nv
ert
ad
ec
ima
lto
ap
erc
en
tag
e,f
irs
tly,
co
nv
ert
the
de
cim
alt
oa
fra
cti
on
,th
en
co
nv
ert
the
fra
cti
on
toa
pe
rce
nta
ge
.
Example:
0.0
21
as
afr
ac
tio
n=21
1000=2.1100
,a
sa
de
cim
al
=2
.1%
0.0
37
as
afr
ac
tio
n=37
1000=3.7100
,a
sa
de
cim
al
=3
.7%
0.4
3a
sa
fra
cti
on
=430
1000=43100
,a
sa
de
cim
al
=4
3%
ValuesofaPercentageofaQuantity
To
fin
dth
ev
alu
eo
fa
pe
rce
nta
ge
of
aq
ua
nti
ty,
firs
tly,
ex
pre
ss
the
pe
rce
nta
ge
as
afr
ac
tio
na
nd
mu
ltip
lyb
yth
eq
ua
nti
ty.
Examples:
4%
of
60
=4100×60
=240
100
=12 5
=22 5
3½
%o
f1
50
0=3.5100×1500
=5250
100
=52510
=1052
=521 2
ExpressingoneQuantity
asaPercentage
To
ex
pre
ss
on
eq
ua
nti
tya
sa
pe
rce
nta
ge
of
an
oth
er,
ma
ke
afr
ac
tio
no
fth
etw
oq
ua
nti
tie
sa
nd
mu
ltip
lyb
y1
00
.
Ex
am
ple
:
12
as
ap
erc
en
tag
eo
f5
0=1250×100
=2
4%
4a
sa
pe
rce
nta
ge
of
60
=4 60×100
=6
.67
%
3.2
as
ap
erc
en
tag
eo
f2
.4=3.22.4×100
=1
3.3
33
%o
r131 3%
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
21
Solvethefollowingequations:
Multiplication
i.5
.05
x1
3.8
ii.1
.27
x0
.87
1
iii.
--1
.01
x0
.89
iv.
27
.3x
--9
.31
v.1
.09
x1
04
x1
.2x
10
2
Division
i.2
33
.1÷
18
.5
ii.0
.12
54÷
0.0
57
iii.
0.6
87
5÷
22
iv.
24
.02
4÷
4.6
2
v.1
.09
x1
04÷
12
Convertthefollowingdecim
als
tofractionsin
theirlowestterm
s:
i.0
.2ii.
0.4
5iii
.0
.31
25
iv.
2.5
5v.
0.0
07
5v
i.2
.12
5
Fin
dth
ed
iffe
ren
ce
be
twe
en
i.1964
an
d0
.29
5ii.13 16
an
d1
.16
32
Convertthefollowingfractionsto
decim
als
(3decim
alplaces)
i.3 8
ii.1116
iii.2132
iv.15 8
v.27 16
Placethefollowingin
ascendingorderofsize;
i.1 5���0.167���3 20
ii.2 5���0.44���7 16
iii.1132���0.3594���0.3125
Expressthefollowingasapercentage%:
i.0
.43
ii.0
.02
5iii
.1
.25
iv.3 8
v.3 7
vi.1 12
vii.7 20
Expressthefollowingasfractions:
i.2
5%
ii.1
3%
iii.
4.5
%iv
.3
3%
Express:
i.3
0a
sa
pe
rce
nta
ge
of
50
ii.2
4a
sa
pe
rce
nta
ge
of
16
iii.
0.5
as
ap
erc
en
tag
eo
f1
2.5
iv.
3.2
as
ap
erc
en
tag
eo
f2
.4
v.0
.08
as
ap
erc
en
tag
eo
f0
.72
Calculate:
i.4
%o
f3
0
ii0
.8%
of
36
0
iii.
1.5
%o
f6
0
iv.
12
0%
of
75
v.8
0%
of
90
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
22
RATIO
&PROPORTION
Ratio
Ara
tio
isa
co
mp
ari
so
nb
etw
ee
ntw
os
imila
rq
ua
nti
tie
s.
Ifth
ele
ng
tho
fa
na
irc
raft
is7
5m
an
da
mo
de
lof
itis
1m
lon
gth
en
the
len
gth
of
the
mo
de
lis1 75
of
the
len
gth
of
the
air
cra
ft.
Inm
ak
ing
the
mo
de
lall
the
dim
en
sio
ns
of
the
air
cra
fta
rere
du
ce
din
the
rati
oo
f1
to7
5.
Th
era
tio
1to
75
isu
su
ally
wri
tte
n1
:7
5.
Ara
tio
ca
na
lso
be
wri
tte
na
sa
fra
cti
on
,a
sin
dic
ate
da
bo
ve
,a
nd
ara
tio
of
1:7
5
me
an
sth
es
am
ea
sth
efr
ac
tio
n1 75.
Be
fore
we
ca
ns
tate
ara
tio
the
un
its
mu
st
be
the
sa
me
.w
ec
an
sta
tea
rati
ob
etw
ee
n3
mm
an
d2
mp
rov
ide
dw
eb
rin
gb
oth
len
gth
sto
the
sa
me
un
its
.T
hu
sif
we
co
nv
ert
2m
to2
00
0m
mth
era
tio
be
twe
en
len
gth
sis
3:2
00
0.
Example:
Ex
pre
ss
the
follo
win
gra
tio
sa
sfr
ac
tio
ns
red
uc
ed
toth
eir
low
es
tte
rms
:
i.4
0m
mto
2.2
m
2.2
m=
22
00
40
:2
20
0=40
2200
=1 55
ii.8
00
gto
1.6
kg
1.6
kg
=1
60
0g
80
0:
16
00
=800
1600
=1 2
Ara
tio
pro
vid
es
am
ea
ns
of
co
mp
ari
ng
on
en
um
be
rto
an
oth
er.
Fo
re
xa
mp
le,
ifa
ne
ng
ine
turn
sa
t4
,00
0rp
ma
nd
the
pro
pe
ller
turn
sa
t2,4
00
rpm
,th
era
tio
of
the
two
sp
ee
ds
is4
,00
0to
2,4
00
,or
5to
3,w
he
nre
du
ce
dto
low
es
tte
rms
.Th
isre
lati
on
sh
ipc
an
als
ob
ee
xp
res
se
da
s5
/3o
r5
:3.
Th
eu
se
ofr
ati
os
isc
om
mo
nin
av
iati
on
.On
era
tio
yo
um
us
tbe
fam
ilia
rw
ith
isc
om
-p
res
sio
nra
tio
,w
hic
his
the
rati
oo
fc
ylin
de
rd
isp
lac
em
en
tw
he
nth
ep
isto
nis
at
bo
tto
mc
en
tre
toth
ec
ylin
de
rd
isp
lac
em
en
tw
he
nth
ep
isto
nis
at
top
ce
ntr
e.
Fo
re
xa
mp
le,
ifth
ev
olu
me
of
ac
ylin
de
rw
ith
the
pis
ton
at
bo
tto
mc
en
tre
is9
6c
ub
icin
ch
es
an
dth
ev
olu
me
wit
hth
ep
isto
na
tto
pc
en
tre
is1
2c
ub
icin
ch
es
,th
ec
om
pre
ss
ion
rati
ois
96
:12
or
8:1
wh
en
sim
plif
ied
.
An
oth
er
typ
ica
lra
tio
isth
at
of
diff
ere
nt
ge
ar
siz
es
,fo
re
xa
mp
le,
the
ge
ar
rati
oo
fa
dri
ve
ge
ar
wit
h1
5te
eth
toa
dri
ve
ng
ea
rw
ith
45
tee
this
15
:45
or
1:3
wh
en
red
uc
ed
.T
his
me
an
sth
at
for
ev
ery
on
eto
oth
on
the
dri
ve
ge
ar
the
rea
reth
ree
tee
tho
nth
ed
riv
en
ge
ar.
Ho
we
ve
r,w
he
nw
ork
ing
wit
hg
ea
rs,
the
rati
oo
fte
eth
iso
pp
os
ite
the
rati
oo
fre
vo
luti
on
s.
Ino
the
rw
ord
s,
sin
ce
the
dri
ve
ge
ar
ha
so
ne
thir
da
sm
an
yte
eth
as
the
dri
ve
ng
ea
r,th
ed
riv
eg
ea
rm
us
tc
om
ple
teth
ree
rev
olu
tio
ns
totu
rnth
ed
riv
en
ge
ar
on
ere
vo
luti
on
.Th
isre
su
lts
ina
rev
olu
tio
nra
tio
of3
:1,w
hic
his
op
po
sit
eth
era
tio
of
tee
th.
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
23
Proportion
Ap
rop
ort
ion
isa
sta
tem
en
to
fe
qu
alit
yb
etw
ee
ntw
oo
rm
ore
rati
os
an
dre
pre
se
nts
ac
on
ve
nie
nt
wa
yto
so
lve
pro
ble
ms
inv
olv
ing
rati
os
.F
or
ex
am
ple
,if
an
an
en
gin
eh
as
are
du
cti
on
ge
ar
rati
ob
etw
ee
nth
ec
ran
ks
ha
fta
nd
the
pro
pe
ller
of
3:2
an
dth
ee
ng
ine
istu
rnin
g2
,70
0rp
m,
wh
at
isth
es
pe
ed
of
the
pro
pe
ller?
Inth
isp
rob
lem
,le
t“X
”re
pre
se
nt
the
un
kn
ow
nv
alu
e,
wh
ich
inth
isc
as
eis
the
sp
ee
do
fth
e
pro
pe
ller.
Ne
xts
etu
pa
pro
po
rtio
na
lsta
tem
en
tus
ing
the
fra
cti
on
alf
orm3 2=2700
x.
To
so
lve
this
eq
ua
tio
n,
cro
ss
mu
ltip
lyto
arr
ive
at
the
eq
ua
tio
n3
x=
2x
2,7
00
,o
r5
,40
0.
tos
olv
efo
r(x
),d
ivid
e5
,40
0b
y3
.T
he
sp
ee
do
fth
ep
rop
elle
ris
1,8
00
rpm
.
3 2=( e
ngine---sp
eed)
( pro
peller---sp
eed)
3 2=2700
x
3x=5,400
x=1,800rp
m
Th
iss
am
ep
rop
ort
ion
ma
ya
lso
be
ex
pre
ss
ed
as
3:2
=2
,70
0:
X.
Th
efi
rst
an
dla
st
term
so
fth
ep
rop
ort
ion
are
ca
lledextremes
,a
nd
the
se
co
nd
an
dth
ird
term
sa
rec
alle
dth
emeans
.In
an
yp
rop
ort
ion
,th
ep
rod
uc
to
fth
ee
xtr
em
es
ise
qu
al
toth
ep
rod
uc
toft
he
me
an
s.I
nth
ise
xa
mp
le,m
ult
iply
the
ex
tre
me
sto
ge
t3x
,an
dm
ult
iply
the
me
an
sto
ge
t2
x2
,70
0o
r5
,40
0.
Th
isre
su
lts
inth
eid
en
tic
al
de
riv
ed
ea
rlie
r;3
x=
5,4
00
.
3:2
=e
ng
ine
sp
ee
d:
pro
pe
ller
sp
ee
d3
:2=
2,7
00
:x
3x
=2
:2
,70
0
3x
=5
,40
0
x=
1,8
00
rpm
.
DirectProportion
If5
litre
so
fo
ilh
as
am
as
so
f4
kg
,th
en
10
litre
so
fth
es
am
eo
ilw
illh
av
ea
ma
ss
of
8k
g.
Th
at
is,
ifw
ed
ou
ble
the
qu
an
tity
of
oil
its
ma
ss
isa
lso
do
ub
led
.N
ow
2½
litre
so
fo
ilw
illh
av
ea
ma
ss
of
2k
g.
Th
at
isif
we
ha
lve
the
qu
an
tity
of
oil
we
ha
lve
its
ma
ss
.T
his
isa
ne
xa
mp
leo
fd
ire
ct
pro
po
rtio
n.
As
the
qu
an
tity
of
oil
inc
rea
se
sth
em
as
sin
cre
as
es
inth
es
am
ep
rop
ort
ion
.A
sth
eq
ua
nti
tyo
fo
ild
ec
rea
se
sth
em
as
sd
ec
rea
se
sin
the
sa
me
pro
po
rtio
n.
Example:
Th
ee
lec
tric
alr
es
ista
nc
eo
fa
wir
e1
50
mm
lon
gis
2O
hm
s.
Fin
dth
ere
sis
tan
ce
of
as
imila
rw
ire
wh
ich
is1
mlo
ng
.
Th
ele
ng
ths
oft
he
two
wir
es
are
inc
rea
se
din
the
rati
oo
f10
00
:15
0.
Th
ere
sis
tan
ce
will
als
oin
cre
as
ein
the
rati
o1
00
0:1
50
.
Th
us
res
ista
nc
eo
fw
ire
1m
lon
g=2×1000
150
=1
3.3
Oh
ms
InverseProportion
Am
oto
rc
ar
will
tra
ve
l30
km
in1
ho
ur
ifit
ss
pe
ed
is3
0k
mp
er
ho
ur.
Ifit
ss
pe
ed
isin
cre
as
ed
to6
0k
mp
er
ho
ur
the
tim
eta
ke
nto
tra
ve
l30
km
will
be½
ho
ur.
Th
at
isw
he
nth
es
pe
ed
isd
ou
ble
dth
eti
me
tak
en
ish
alv
ed
.T
his
isa
ne
xa
mp
leo
fin
ve
rse
pro
po
rtio
n.
Wh
en
we
mu
ltip
lyth
es
pe
ed
by
2w
ed
ivid
ed
the
tim
eta
ke
nb
y2
.
Example:
Tw
op
ulle
ys
of
15
0m
ma
nd
50
mm
dia
me
ter
are
co
nn
ec
ted
by
ab
elt
.If
the
larg
er
pu
lley
rev
olv
es
at
80
rev
/min
fin
dth
es
pe
ed
of
the
sm
alle
rp
ulle
y.
Be
ca
us
eth
eb
elt
mo
ve
sth
ec
irc
um
fere
nc
eo
fe
ac
hp
ulle
yth
es
am
ed
ista
nc
ew
ec
an
ima
gin
eth
at
the
sm
alle
rp
ulle
ym
us
tb
ere
vo
lvin
gfa
ste
rth
an
the
larg
er
pu
lley
be
ca
us
eit
’sc
irc
um
fere
nc
eis
sh
ort
er.
So
we
ca
ns
ee
tha
tth
ero
tati
on
al
sp
ee
da
nd
dia
me
ter
are
inin
ve
rse
pro
po
rtio
nto
ea
ch
oth
er.
ie.T
he
pu
lley
dia
me
ters
ha
ve
the
rela
tio
ns
hip
1:3
wh
ilst
the
sp
ee
ds
ha
ve
the
rela
tio
ns
hip
3:1
.
Th
ere
fore
Sp
ee
do
fs
ma
ller
pu
lley
=80×3 1=
24
0re
v/m
in.
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
24
ProportionalParts
Th
ed
iag
ram
be
low
sh
ow
sth
elin
eA
Bw
ho
se
len
gth
rep
res
en
ts1
0m
div
ide
din
totw
op
art
sin
the
rati
o2
:3.
Fro
mth
ed
iag
ram
the
line
ha
sb
ee
nd
ivid
ed
into
ato
tal
of5
pa
rts
.Th
ele
ng
thA
Cc
on
tain
s2
pa
rts
an
dth
ele
ng
thB
Cc
on
tain
s3
pa
rts
.Ea
ch
pa
rtis
2m
lon
g,
he
nc
eA
Cis
4m
lon
ga
nd
BC
is6
mlo
ng
.
We
co
uld
tac
kle
the
pro
ble
ma
sfo
llow
s;
To
tal
nu
mb
er
of
pa
rts
=2
+3
=5
Le
ng
tho
fe
ac
hp
art
=10 5
=2
m
Le
ng
tho
fA
C=
2x
2=
4m
Le
ng
tho
fB
C=
3x
2=
6m
Example:
Ac
ert
ain
bra
ss
ism
ad
eb
ya
lloy
ing
co
pp
er
an
dz
inc
inth
era
tio
of
7:3
.H
ow
mu
ch
co
pp
er
mu
st
be
mix
ed
wit
h3
0g
of
zin
c.
3p
art
sh
av
ea
ma
ss
of
30
g
1p
art
ha
sa
ma
ss
of
10
g
7p
art
sh
av
ea
ma
ss
of
70
g
Th
ere
fore
,M
as
so
fc
op
pe
rn
ee
de
d=
70
g.
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
25
Solvethefollowingequations:
Ex
pre
ss
the
follo
win
gra
tio
sa
sfr
ac
tio
ns
red
uc
ed
toth
eir
low
es
tte
rms
;
i.1
5g
to2
kg
ii.2
1ft
to9
inc
he
siii
.2
0c
mto
10
0m
miv
.4
00
mto
3k
m
Fin
dth
em
iss
ing
va
lue
;
i.3
:4=
6:x
ii.2
0:1
=x
:3.2
iii.
24
0:4
00
=x
:1iv
.1
:2.6
=x
:13
v.1
8:x
=2
:1
Fiv
em
en
bu
ilda
wa
llta
ke
20
da
ys
toc
om
ple
teit
.H
ow
lon
gw
ou
ldit
tak
e4
me
nto
co
mp
lete
it.
4p
eo
ple
ca
nc
lea
na
no
ffic
ein
6h
ou
rs.
Ho
wm
an
yp
eo
ple
wo
uld
be
ne
ed
ed
toc
lea
nth
eo
ffic
ein
4h
ou
rs.
8p
eo
ple
tak
e5
ho
urs
toc
ha
ng
ea
ne
ng
ine
.H
ow
lon
gw
ou
ldit
tak
e4
pe
op
leto
do
this
wo
rk.
An
en
gin
ee
rin
gc
om
pa
ny
em
plo
y1
2m
en
tofa
bri
ca
tea
nu
mb
er
of
co
nta
ine
rs.
Th
ey
tak
e9
da
ys
toc
om
ple
teth
ew
ork
.If
the
co
mp
an
yh
ad
em
plo
ye
d8
me
n,h
ow
lon
gw
ou
ldit
ha
ve
tak
en
.
Atr
ain
tra
ve
ls2
00
km
in4
ho
urs
.If
ittr
av
els
at
the
sa
me
rate
,h
ow
lon
gw
illit
tak
eto
co
mp
lete
ajo
urn
ey
of
35
0k
m.
Ab
ar
of
me
tal1
0.5
mlo
ng
isto
be
cu
tin
toth
ree
pa
rts
inth
era
tio
of1 2:13 4:3
.F
ind
the
len
gth
of
ea
ch
pa
rt.
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
26
POWERSANDROOTS
Powers
Wh
en
an
um
be
ris
mu
ltip
lied
by
its
elf
,it
iss
aid
tob
era
ise
dto
ag
ive
np
ow
er.
Fo
re
xa
mp
le,
6x
6is
ex
pre
ss
ed
as
62,
6x
6x
6is
ex
pre
ss
ed
as
63
etc
.In
this
ex
am
ple
the
nu
mb
er
6is
refe
rre
dto
as
the
ba
se
nu
mb
er
an
dth
es
ma
lln
um
be
rs2
an
3a
rere
ferr
ed
toa
sth
ee
xp
on
en
ts.
Ifth
ee
xp
on
en
tis
ap
os
itiv
en
um
be
rth
en
the
ba
se
ism
ult
iplie
db
yit
se
lfa
ss
ho
wn
ab
ov
e.
Ex
am
ple
:
32
isre
ad
3s
qu
are
do
r3
toth
ep
ow
er
of
2.
23
isre
ad
2c
ub
ed
or
2to
the
po
we
ro
fth
ree
.
Ifth
ee
xp
on
en
tis
an
eg
ati
ve
nu
mb
er
the
nth
ere
cip
roc
al
of
the
nu
mb
er
ism
ult
i-p
lied
by
its
elf
.
Example:
2--
3is
rea
d2
toth
ep
ow
er
of
min
us
3.
Th
ism
ea
ns
tha
tth
ere
cip
roc
al
of
two
ism
ult
iplie
da
sb
elo
w.
1 2×1 2×1 2=1 8
Ifth
ee
xp
on
en
th
as
no
sig
nth
en
itis
as
su
me
dto
be
po
sit
ive
.
Roots
Th
ero
ot
of
an
um
be
ris
tha
tv
alu
ew
hic
h,
wh
en
mu
ltip
lied
by
its
elf
ac
ert
ain
nu
mb
er
of
tim
es
,p
rod
uc
es
tha
tn
um
be
r.F
or
ex
am
ple
,4
isa
roo
to
f1
6b
ec
au
se
wh
en
mu
ltip
lied
by
its
elf
,th
ep
rod
uc
tis
16
.H
ow
ev
er,
4is
als
oa
roo
to
f6
4b
ec
au
se
4x
4x
4=
64
.
Th
es
ym
bo
lu
se
dto
ind
ica
tea
roo
tis
the
rad
ica
ls
ign
(x
)p
lac
ed
ov
er
the
nu
mb
er.
Ifo
nly
the
rad
ica
ls
ign
ap
pe
ars
ov
er
an
um
be
r,it
ind
ica
tes
yo
ua
reto
ex
tra
ct
thesquare
root
or
se
co
nd
roo
to
fth
en
um
be
ru
nd
er
the
sig
n.
Ifth
era
dic
al
sig
na
pp
ea
rsw
ith
an
ind
ex
nu
mb
er
ne
xt
toit
this
ind
ica
tes
the
roo
tw
hic
his
to
be
tak
en
.F
or
ex
am
ple64
3 in
dic
ate
sth
at
the
cu
be
roo
to
rth
ird
roo
tis
tob
eta
ke
n.
So
64
3 =
4b
ec
au
se
4x
4x
4=
64
an
d16
2 =
4b
ec
au
se
4x
4=
16
No
teth
at
the
2is
ge
ne
rally
no
tu
se
ds
oif
the
rad
ica
ls
ign
ap
pe
ars
wit
hn
on
um
be
rth
en
itis
as
su
me
dto
be
2(s
qu
are
roo
t).
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
27
Indices
Base,Index&Power
Th
eq
ua
nti
ty2×2×2×2
ma
yb
ew
ritt
en
as24.
No
w24
isc
alle
dth
efo
urt
hp
ow
er
of
the
ba
se
2.
Th
en
um
be
r4
,w
hic
hg
ive
sth
en
um
be
ro
f2
sto
be
mu
ltip
lied
tog
eth
er
isc
alle
dth
ein
de
x(p
lura
l:
Ind
ice
s).
Sim
ilarl
ya×
a×
a=
a3
He
rea3
isth
eth
ird
po
we
ro
fth
eb
as
ea,
an
dth
ein
de
xis
3.
Th
us
inth
ise
xp
res
sio
n
xnxn
isc
alle
dth
en
thp
ow
er
of
xx
isc
alle
dth
eb
as
e,
an
dn
isc
alle
dth
ein
de
x.R
em
em
be
rth
at,
ina
lge
bra
,le
tte
rss
uc
ha
sa
inth
ea
bo
ve
ex
pre
ss
ion
me
rely
rep
res
en
tn
um
be
rs.
He
nc
eth
ela
ws
of
ari
thm
eti
ca
pp
lys
tric
tly
toa
lge
bra
icte
rms
as
we
lla
s
nu
mb
ers
.Th
ee
xp
res
sio
n1 2
isc
alle
dth
ere
cip
roc
al
of
2,
Sim
ilarl
yth
ee
xp
res
sio
n1 p
isc
alle
dth
ere
cip
roc
al
of
plik
ew
ise
the
ex
pre
ss
ion
1 xnis
ca
lled
the
rec
ipro
ca
lo
fxn
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
28
LawsofIndices
1.Multiplication
Ifp
ow
ers
are
mu
ltip
lied
tog
eth
er
the
nw
ec
an
se
eth
efo
llow
ing
.
23×24
ise
qu
al
to(2
x2
x2
)x
(2x
2x
2x
2)
so23×24=27
Multiplicationofpowers
whichhavethesamebasecanbesim
plifiedby
addingthepowers
together.
2.Division
Ifp
ow
ers
are
tob
ed
ivid
ed
,it
ca
nb
es
ee
nth
at25
23
ise
qu
al
to( 2
x2x2
x2x2)
( 2x2
x2)
Th
isfr
ac
tio
nc
an
be
red
uc
ed
to2
x2
wh
ich
ise
qu
al
to22
so25
23
=25−3
=22
Divisionofpowers
whichhavethesamebasecanbecarriedoutbysub-
tractingoneindexfrom
theother.
3.Powers
ofPowers
Ifa
po
we
ris
tob
eit
se
lfra
ise
dto
ap
ow
er
the
nw
eh
av
eth
efo
llow
ing
.
� 23�4
ise
qu
al
to23
x23
x23
x23
or
(2
x2
x2
)x
(2
x2
x2
)x
(2
x2
x2
)x
(2
x2
x2
)
or
23x4
or212
Raisingapowerto
apowercanbeachievedbymultiplyingtheindices
together.
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
29
4.Numbers
raisedto
thepowerofzero
orone.
Th
em
os
td
iffic
ult
ind
ice
sto
vis
ua
lise
are
x0a
nd
x1.
Ifw
ec
on
sid
er( x
×x×
x×
x)
( x×
x×
x×
x)w
eh
av
ea
va
lue
of
1a
sa
ny
nu
mb
er
div
ide
db
y
its
elf
is1
.
Th
isfr
ac
tio
nc
an
als
ob
ee
xp
res
se
da
sx4 x4
or
x0.
So
:
Anynumberraisedto
thepowerofzero
equals
one.
Ifw
ec
on
sid
er( x
×x×
x×
x)
( x×
x×
x)w
eh
av
ea
va
lue
wh
ich
isx1
or
x.S
o
Anynumberraisedto
thepowerofoneis
equalto
itself.
Negativeindices
Ifw
ec
on
sid
er
( x×
x×
x)
( x×
x×
x×
x×
x)w
eh
av
ea
va
lue
of1 x2
wh
ich
ca
nb
ee
xp
res
se
da
sx3
−5
or
x−2
Ifabaseis
raisedto
anegativepowerthevalueis
equalto
thereciprocal
ofthebaseraisedto
apositivepower.
FractionalIndices
Th
ec
ub
ero
ot
of
5(w
ritt
en
as53
)is
the
nu
mb
er
wh
ich
,w
he
nm
ult
iplie
db
yit
se
lfth
ree
tim
es
,g
ive
s5
.
53
×53
×53
=5
bu
tw
ea
lso
kn
ow
tha
t513×513×513=513
+13
+13=5
Co
mp
ari
ng
the
se
ex
pre
ss
ion
s
53
=513
Sim
ilarl
yth
efo
urt
hro
ot
of
ba
se
d(
wri
tte
na
sd4
)is
the
nu
mb
er
wh
ich
,w
he
nm
ult
iplie
db
yit
se
lffo
ur
tim
es
,g
ive
sd.
d4
×d4
×64
×d4
=d
Bu
tw
ea
lso
kn
ow
tha
td14×
d14×
d14×
d14=
d14
+14
+14
+14=
d
Co
mp
ari
ng
the
se
ex
pre
ss
ion
sd4
=d14
Th
ela
wis
:
Afractionalindexrepresents
aroot,thedenominatoroftheindex
denotestherootto
betaken.
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
30
Solvethefollowingequations:
Fin
dth
ev
alu
es
of
the
follo
win
g
i.82
ii.24
iii.33
iv.25
v.16
v
i.144
v
ii.169
viii
.83
ix.27
3 x
.216
3
Sim
plif
yth
efo
llow
ing
,g
ivin
ge
ac
ha
ns
we
ra
sa
po
we
r
i.25×26
ii.a×
a2×
a5
iii.n8÷
n5
iv.105×103÷104
v.z4×
z2×
z−3
vi.32×3−3÷33
vii.
� 93�4
viii
.� t×
t3�2
ix.�1 73�4
Fin
dth
ev
alu
eo
fth
efo
llow
ing
i.8
11
/4ii.
82
/3iii
.1
63
/4iv
.9
2.5
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
31
TRANSPOSITIONOFFORMULAE
Th
efo
rmu
lay=
ax+
bh
as
ya
sit
ssubject.
By
rea
rra
ng
ing
this
form
ula
we
co
uld
ma
ke
xth
es
ub
jec
t.W
ea
reth
en
sa
idto
ha
ve
tra
ns
po
se
dth
efo
rmu
lato
ma
ke
xth
es
ub
jec
t.
Th
eru
les
for
tra
ns
form
ing
afo
rmu
laa
re:
1.
Re
mo
ve
sq
ua
rero
ots
or
oth
er
roo
ts.
2.
Ge
tri
do
ffr
ac
tio
ns
.
3.
Cle
ar
bra
ck
ets
.
4.
Co
llec
tto
ge
the
rth
ete
rms
co
nta
inin
gth
ere
qu
ire
ds
ub
jec
t.
5.
Fa
cto
ris
eif
ne
ce
ss
ary
.
6.
Iso
late
the
req
uir
ed
su
bje
ct.
Th
es
es
tep
ss
ho
uld
be
pe
rfo
rme
din
the
ord
er
giv
en
.
Examples:
i.T
ran
sp
os
eth
efo
rmu
laF
=m
ato
ma
ke
ath
es
ub
jec
t.
Step1
.D
ivid
eb
oth
sid
es
by
m.
the
n,
F m=
ma
m
or
F m=
ao
ra=
F m
ii.T
ran
sp
os
ex=
y bto
ma
ke
bth
es
ub
jec
t
Step1.
Mu
ltip
lyb
oth
sid
es
by
b.
the
n,
x×
b=
y b×
b
bx=
yo
ry=
bx
Transposethefollowing:
C=
πd
for
d
S=
πdn
for
d
I=
PRT
for
R
v2=2gh
for
h
x=
a yfo
ry
P=
RT V
for
T
S=
ts Tfo
rt
M I=
E Rfo
rR
GY l=
T Jfo
rJ
v=
u+
at
for
t
n=
p+
crfo
rr
y=
ax+
bfo
rx
y=
x 5+17
for
x
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
32
AREAS
Th
ea
rea
of
ap
lan
efi
gu
reis
me
as
ure
db
ys
ee
ing
ho
wm
an
ys
qu
are
un
its
itc
on
tain
s.
1s
qu
are
me
tre
isth
ea
rea
co
nta
ine
din
as
qu
are
me
tre
isth
ea
rea
co
nta
ine
din
as
qu
are
ha
vin
ga
sid
eo
f1
me
tre
;1
sq
ua
rec
en
tim
etr
eis
the
are
ac
on
tain
ed
ina
sq
ua
reh
av
ing
as
ide
of
1c
en
tim
etr
e,
etc
.T
he
sta
nd
ard
ab
bre
via
tio
ns
are 1
sq
ua
rem
etr
e1
m2
1s
qu
are
ce
nti
me
tre
1c
m2
1s
qu
are
mill
ime
tre
1m
m2
1s
qu
are
inc
h1
in2
1s
qu
are
foo
t1
ft2
1s
qu
are
ya
rd1
yd
2
Th
efo
llow
ing
pro
vid
es
the
form
ula
efo
ra
rea
sa
nd
pe
rim
ete
rso
fs
imp
leg
eo
-m
etr
ica
ls
ha
pe
s.
Re
cta
ng
le
Are
a=
l×b
Pe
rim
ete
r=2l+2b
Tri
an
gle
Are
a=1 2×
b×
h
Cir
cle
Are
a=πr2
Cir
cu
mfe
ren
ce
=2πr=
πd
π=3.142
or22 7
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
33
VOLUMES
Th
ev
olu
me
of
as
olid
fig
ure
isfo
un
db
ys
ee
ing
ho
wm
an
yc
ub
icu
nit
sit
co
nta
ins
.1
cu
bic
me
tre
isth
ev
olu
me
co
nta
ine
din
sid
ea
cu
be
ha
vin
ga
ne
dg
e1
me
tre
lon
g;
1c
ub
icc
en
tim
etr
eis
the
vo
lum
ec
on
tain
ed
ins
ide
ac
ub
eh
av
ing
an
ed
ge
1c
en
tim
etr
elo
ng
,etc
.Th
es
tan
da
rda
bb
rev
iati
on
sfo
ru
nit
so
fvo
lum
ea
rea
sfo
llow
s:
1s
qu
are
me
tre
1m
3
1s
qu
are
ce
nti
me
tre
1c
m3
1s
qu
are
mill
ime
tre
1m
m3
1s
qu
are
inc
h1
in3
1s
qu
are
foo
t1
ft3
1s
qu
are
ya
rd1
yd
3
Th
efo
llow
ing
fig
ure
sg
ive
the
form
ula
efo
rth
ev
olu
me
sa
nd
su
rfa
ce
are
as
of
so
lidfi
gu
res
.
An
ys
olid
ha
vin
ga
un
ifo
rmc
ros
s--
se
cti
on
;
Vo
lum
e=
Cro
ss
--s
ec
tio
na
la
rea
xL
en
gth
of
so
lid
Su
rfa
ce
Are
a=
La
tera
lS
urf
ac
e+
En
ds
i.e
.(p
eri
me
ter
of
cro
ss
--s
ec
tio
nx
Le
ng
tho
fS
olid
)+
(To
tal
are
ao
fe
nd
s)
Cy
lind
er
Vo
lum
e=πr2
h
Su
rfa
ce
Are
a=2πr(h+
r)
Sp
he
re
Vo
lum
e=4 3πr3
Su
rfa
ce
Are
a=4πr2
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
34
Solvethefollowingproblems:
i.F
ind
the
are
aa
nd
pe
rim
ete
ro
fa
rec
tan
gle
wh
os
ele
ng
this
12
inc
he
sa
nd
wid
this
7in
ch
es
.
ii.A
ca
rpe
th
as
an
are
ao
f3
6m
2.
Ifit
iss
qu
are
wh
at
len
gth
of
sid
eh
as
the
ca
rpe
t?
iii.
Atr
ian
gle
ha
sa
ba
se
of
7c
ma
nd
an
alt
itu
de
of
3c
m.
Ca
lcu
late
its
are
a.
iv.
Th
ea
rea
of
atr
ian
gle
is4
0ft
2.
Its
ba
se
is8
ftlo
ng
.C
alc
ula
teit
sv
ert
ica
lh
eig
ht.
v.C
alc
ula
teth
ev
olu
me
of
am
eta
lp
ipe
wh
os
ein
sid
ed
iam
ete
ris
6c
ma
nd
wh
os
eo
uts
ide
dia
me
ter
is8
cm
,if
it2
0c
mlo
ng
.
vi.
Are
cta
ng
ula
rta
nk
is2
.7c
mlo
ng
,1
.8c
mw
ide
an
d3
.2c
mh
igh
.H
ow
ma
ny
litre
so
fw
ate
rw
illit
ho
ldw
he
nfu
ll?
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
35
CONVERSIONS
Length
1in
.=
2.5
4c
m
1m
=3
9.3
7in
.o
r3
.28
1ft
.
1ft
.=
0.3
04
8m
Volume
1im
pg
al
=4
,54
6lit
res
1U
Sg
al
=3
.78
5lit
res
Force&Weight
1N
=0
.22
48
lb1
lb
Power
1H
P=
55
0ft
.lb
/se
c
1H
P=
74
6W
1W
=1
J/s
ec
1W
=0
.73
8ft
.lb
/se
c
1B
tu/h
r=
0.2
93
W
Temperature
10F
=((
9/5
)x0C
)+
32
10C
=0F
-3
2x
(5/9
)0C
=K
+2
73
.15
Pressure
1a
tm=
76
0m
mH
g
1a
tm=
29
.92
inH
g
1a
tm=
14
.7lb
/in
2
1P
a=
0.0
00
14
5lb
/in
2
1b
ar
=1
4.5
lb/i
n2
1b
ar
=1
00
,00
0P
a
FundamentalConstant
g=
9.8
N/k
g
OtherUsefulData
1lit
rew
ate
r=
1k
g
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
36
Convertthefollowingweights
andmeasures:
i.C
on
ve
rt6
mto
fee
t.
ii.C
on
ve
rt2
5U
Sg
allo
ns
tolit
res
.
iii.
Co
nv
ert
25
4in
ch
es
toc
m.
iv.
Co
nv
ert
4.5
litre
toU
Sg
allo
ns
v.C
on
ve
rt3
50
imp
eri
al
ga
llon
sto
litre
s.
vi.
Co
nv
ert
the
follo
win
gto˚F
--2
0˚C
--5˚C
37˚C
88˚C
vii.
Co
nv
ert
the
follo
win
gto˚C
--4
0˚F
16˚F
10
0˚F
21
5˚F
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
37
TEST
Wo
rko
ut
the
va
lue
of
the
follo
win
g:
1.
7+
4x
3=
2.
5x
4--
3x
6+
5=
3.
10
--1
2÷
6+
3(8
--3
)=
4.
53
=
5.2 5+3 7
=
6.5 6−3 4
=
7.3 8×5 7
=
8.3 5÷7 8
=
9.
div
ide
74
.52
by
8.1
=
10
.m
ult
iply
20
.3x
17
.4=
11.
Co
nv
ert
0.8
00
toa
fra
cti
on
=
12
.c
on
ve
rt3 5
toa
pe
rce
nta
ge
=
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.2
ALGEBRA
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6
ALGEBRA
Introduction
Th
em
eth
od
so
fa
lge
bra
are
an
ex
ten
sio
no
fth
os
eu
se
din
ari
thm
eti
c.
Ina
lge
bra
we
us
ele
tte
rsa
nd
sy
mb
ols
as
we
lla
sn
um
be
rsto
rep
res
en
tva
lue
s.W
he
nw
ew
rite
tha
ta
su
mo
fm
on
ey
is£
50
we
are
ma
kin
gaparticularstatement
bu
tif
we
wri
tea
su
mo
fmo
ne
yis
£P
we
are
ma
kin
gageneralstatement.
Th
isg
en
era
lsta
tem
en
tw
illc
ov
er
an
yn
um
be
rw
ec
are
tos
ub
sti
tute
forP
.
USEOFSYMBOLS
Ate
ch
nic
ian
oft
en
ha
sto
ind
ica
teth
at
ce
rta
inq
ua
nti
tie
so
rm
ea
su
rem
en
tsh
av
eto
be
ad
de
d,
su
btr
ac
ted
,m
ult
iplie
do
rd
ivid
ed
.F
req
ue
ntl
yth
ish
as
tob
ed
on
ew
ith
ou
tu
sin
ga
ctu
al
nu
mb
ers
.
Th
es
tate
me
nt:
Are
ao
fa
rec
tan
gle
=le
ng
thx
bre
ad
th
isa
pe
rfe
ctl
yg
en
era
lsta
tem
en
tw
hic
ha
pp
lies
toa
llre
cta
ng
les
.If
we
us
es
ym
bo
lsw
eo
bta
ina
mu
ch
sh
ort
er
sta
tem
en
t.
ifA
=th
ea
rea
of
the
rec
tan
gle
l=
the
len
gth
of
the
rec
tan
gle
an
db
=th
eb
rea
dth
of
the
rec
tan
gle
the
nth
es
tate
me
nt
be
co
me
s:
A=
lx
b
Kn
ow
ing
wh
at
the
sy
mb
ols
A,
lan
db
sta
nd
for,
this
sta
tem
en
tc
on
ve
ys
as
mu
ch
info
rma
tio
na
sth
efi
rst
sta
tem
en
t.T
ofi
nd
the
are
ao
fa
pa
rtic
ula
rre
cta
ng
lew
ere
pla
ce
the
sy
mb
ols
lan
db
by
the
ac
tua
ldim
en
sio
ns
of
the
rec
tan
gle
,fi
rstm
ak
ing
su
reth
at
lan
db
ha
ve
the
sa
me
un
its
.T
ofi
nd
the
are
ao
fa
rec
tan
gle
wh
os
ele
ng
this
50
mm
an
dw
ho
se
bre
ad
this
30
mm
we
pu
tl
=5
0m
ma
nd
b=
30
mm
.
A=
lx
b=
50
x3
0=
15
00
mm
2
Ma
ny
ve
rba
ls
tate
me
nts
ca
nb
etr
an
sla
ted
into
sy
mb
ols
as
the
follo
win
gs
tate
me
nts
sh
ow
: Th
ed
iffe
ren
ce
of
two
nu
mb
ers
=x
--y
Tw
on
um
be
rsm
ult
iplie
dto
ge
the
r=
ax
b
On
en
um
be
rd
ivid
ed
by
an
oth
er
=p÷
q
SUBSTITUTION
Th
ep
roc
es
so
ffi
nd
ing
the
nu
me
ric
al
va
lue
of
an
alg
eb
raic
ex
pre
ss
ion
for
giv
en
va
lue
so
fth
es
ym
bo
lsth
at
ap
pe
ar
init
isc
alle
ds
ub
sti
tuti
on
.
Ex
am
ple
:
Ifx
=3
,y
=4
an
dz
=5
fin
dth
ev
alu
eo
f:
( 3y+2z)
( x+
z)=( 3
×4)+( 2
×5)
3+
5
( 12+10)
8=22 8
=2
.75
or
2¾
.
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.2
ALGEBRA
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6
Solvethefollowingsubstitutionequations:
Ifa
=2
,b
=3
an
dc
=5
.F
ind
the
va
lue
so
fth
efo
llow
ing
.
i.a
+7
ii.9
ciii
.3
bc
iv.
4c
+6
bv.
a+
2b
+5
c
vi.
8c
--4
bv
ii.abc6
viii
.5
a+
9b
+8
c
a+
b+
c
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.2
ALGEBRA
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6
ADDITION&SUBTRACTIONOFALGEBRAIC
TERMS
Liketerm
sa
ren
um
eri
ca
lm
ult
iple
so
fth
es
am
ea
lge
bra
icq
ua
nti
ty.
7x
,5
xa
nd
--3
x
are
thre
elik
ete
rms
.
An
ex
pre
ss
ion
co
ns
isti
ng
of
like
term
sc
an
be
red
uc
ed
toa
sin
gle
term
by
ad
din
go
rs
ub
tra
cti
ng
the
nu
me
ric
al
co
eff
icie
nts
.
7x
--5
x+
3x
=(7
--5
+3
)x
=5
x
3b
2+
7b
2=
(3+
7)
b2
=1
0b
2
--3
y--
5y
=(-
-3--
5)
y=
--8
y
q--
3q
=(1
--3
)q
=--
2q
On
lylik
ete
rms
ca
nb
ea
dd
ed
or
su
btr
ac
ted
.T
hu
s7
a+
3b
--2
cis
an
ex
pre
ss
ion
co
nta
inin
gth
reeunliketerm
sa
nd
itc
an
no
tbe
sim
plif
ied
an
yfu
rth
er.
Sim
ilarl
yw
ith
8a
2b
+7
ab
3--
6a
2b
2w
hic
ha
rea
llu
nlik
ete
rms
.It
isp
os
sib
leto
ha
ve
se
ve
rals
ets
of
like
term
sin
an
ex
pre
ss
ion
an
de
ac
hs
et
ca
nth
en
be
sim
plif
ied
.
8x
+3
y--
4z
--5
x+
7z
--2
y+
2z
=(8
--5
)x+
(3--
2)y
+(-
-4+
7+
2)z
=3
x+
y+
5z
MULTIPLICATION&DIVISIONSIGNS
Wh
en
us
ing
sy
mb
ols
mu
ltip
lica
tio
ns
ign
sa
ren
ea
rly
alw
ay
so
mit
ted
an
dl
xb
be
co
me
slb
.O
fco
urs
eth
es
am
es
ch
em
ec
an
no
ta
pp
lyto
nu
mb
ers
an
dw
ec
an
no
tw
rite
9x
6a
s9
6.
Th
em
ult
iplic
ati
on
sig
nc
an
,h
ow
ev
er,
be
om
itte
dw
he
na
sy
mb
ol
an
da
nu
mb
er
are
tob
em
ult
iplie
dto
ge
the
r.T
hu
s5
xm
isw
ritt
en
5m
.T
he
sy
ste
mm
ay
be
ex
ten
de
dto
thre
eo
rm
ore
qu
an
titi
es
an
dh
en
cePxLxAxN
isw
ritt
en
PLAN.
Th
es
ym
bo
lsn
ee
dn
ot
be
wri
tte
nin
an
ys
pe
cia
lo
rde
rb
ec
au
se
the
ord
er
inw
hic
hn
um
be
rsa
rem
ult
iplie
dto
ge
the
ris
un
imp
ort
an
t.T
hu
sPLAN
isth
es
am
ea
sLANP
orNAPL.
Itis
us
ua
l,h
ow
ev
er,
tow
rite
nu
mb
ers
be
fore
sy
mb
ols
,th
at
is,
itis
be
tte
rto
wri
te8
xy
tha
nx
y8
or
x8
y.In
alg
eb
raic
ex
pre
ss
ion
sth
en
um
be
rin
fro
nt
of
the
sy
mb
ols
isc
alle
dth
ecoefficient.
Th
us
inth
ee
xp
res
sio
n8
xth
ec
oe
ffic
ien
to
fx
is8
.
Th
ed
ivis
ion
sig
n÷
iss
eld
om
us
ed
ina
lge
bra
an
dit
ism
ore
co
nv
en
ien
tto
wri
te
P÷
qin
the
fra
cti
on
al
form
p q
Ex
am
ple
:
( lp)
( 2πR)
=lp÷2πR
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.2
ALGEBRA
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6
MULTIPLICATION&DIVISIONOFALGEBRAIC
QUANTITIES
Th
eru
les
are
ex
ac
tly
the
sa
me
as
tho
se
us
ed
wit
hn
um
be
rs.
(+x)(+
y)=
+(x
y)=
+xy
=xy
5x×3y=5×3×
x×
y=15xy
(x)(−
y)=
−(x
y)=
−xy
(2x)(−3y)
=−(2
x)(3
y)=
−6xy
(−4x)(2
y)=
−(4
x)(2
y)=
−8xy
(−3x)(−2y)
=+(3
x)(2
y)=6xy
( +x)
( +y)
=+
x y=
x y( −3x)
2y
=−3x2y
( −5x)
( −6y)
=+5x6y=5x6y
4x
( −3y)
=−4x3y
Wh
en
mu
ltip
lyin
ge
xp
res
sio
ns
co
nta
inin
gth
es
am
es
ym
bo
ls,
ind
ice
sa
reu
se
d:
m×
m=
m2
3m
×5m
=3×
m×5×
m=15m2
(−m)×
m2=( −
m)×
m×
m=
−m3
5m2n×3m
n3=5×
m×
m×
n×3×
m×
n×
n×
n=15m3n4
3m
n×
� −2n2� =3×
m×
n×(−2)×
n×
n=
−6m
n3
Wh
en
div
idin
ga
lge
bra
ice
xp
res
sio
ns
,c
an
ce
llati
on
be
twe
en
nu
me
rato
ra
nd
de
no
min
ato
ris
oft
en
po
ss
ible
,c
an
ce
llin
gis
eq
uiv
ale
nt
tod
ivid
ing
bo
thn
um
era
tor
an
dd
en
om
ina
tor
by
the
sa
me
qu
an
tity
:
pq p=( p
×q)
p=
q
( 3p2q)
( 6pq2)=( 3
×p×
p×
q)
( 6×
p×
q×
q)=3p
6q=
p 2q
( 18x2y2z)
( 6xy
z)=( 18×
x×
x×
y×
y×
z)
( 6×
x×
y×
z)=3xy
Re
me
mb
er
the
wo
rdB
OD
MA
Sw
hic
hg
ive
sth
ein
itia
lle
tte
rso
fth
ec
orr
ec
ts
eq
ue
nc
ei.
e.
Bra
ck
ets
,O
f,D
ivis
ion
,M
ult
iply
,A
dd
,S
ub
tra
ct.
Th
us
2x2+� 12x4−3x4� ÷3x2−
x2=2x2+9x4÷3x2−
x2
=2x2+3x2−
x2
=5x2−
x2
=4x2
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.2
ALGEBRA
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6
Sim
plify
thefollowing:
1)
i.7
x+
11x
ii.7
x--
5x
iii.
3x
--6
xiv
.--
2x
--4
x
v.--
8x
+3
xv
i.--
2x
+7
xv
ii.5
m+
13
m--
6m
viii
.6
b2
--4
b2
+3
b2
ix.
6a
b--
3a
b--
2a
b
x.
14
xy
+5
xy
--7
xy
+2
xy
xi.
--5
x+
7x
--3
x--
2x
xii.
3x
--2
y+
4z
--2
x--
3y
+5
z+
6x
+2
y--
3z
xiii
.3
a2b
+2
ab
3+
4a
2b
2--
5a
b3
+11
b4
+6
a2b
xiv
.p
q+
2.1
qr
--2
.2rq
+8
qp
2)
i.2
zx
5y
ii.3
ax
3b
iii.
3x
4m
iv.¼
qx
16
pv.
zx
(y)
vi.
(--3
a)
x(-
-2b
)v
ii.8
mx
(--3
n)
viii
.(-
-4a
)x
3b
ix.
8p
x(-
-q)
x(-
-3r)
x.
3a
x(-
-4b
)x
(--c
)x
5d
xi.
ax
ax
ii.3
mx
(--3
m)
xiii
.8
mn
x(-
-3m
2n
3)
xiv
.7
ab
x(-
-3a
2)
xv.
m2n
x(-
-mn
)x
5m
2n
2x
vi.
5a
2x
(--3
b)
x5
ab
3)
i.1
2x÷
6ii.
4a÷
(--7
b)
iii.
(--5
a)÷
8b
iv.
4a÷
2b
v.4
ab÷
2a
vi.
12
x2y
z2÷
4x
z2
vii.
(--1
2a
2b
)÷
6a
viii
.8
a2b
c2÷
4a
c2
ix7
a2b
2÷
3a
b
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.2
ALGEBRA
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6
BRACKETS
Bra
ck
ets
are
us
ed
toin
dic
ate
the
ord
er
inw
hic
hc
ert
ain
op
era
tio
ns
mu
st
tak
ep
lac
e.
Wh
en
rem
ov
ing
bra
ck
ets
ea
ch
term
wit
hin
the
bra
ck
et
ism
ult
iplie
db
yth
eq
ua
nti
tyo
uts
ide
the
bra
ck
et:
3(x+
y)=3x+3y
5(2
x+3y)
=5×2x+5×3y=10x+15y
4(a−2b)=4×
a−4×2b=4a−8b
m(a+
b)=
ma+
mb
3x(2p+3q)=3x×2p+3x×3q=6px+9qx
4a(2
a+
b)=4a×2a+4a×
b=8a2+4ab
Wh
en
ab
rac
ke
th
as
am
inu
ss
ign
infr
on
to
fit
,th
es
ign
so
fa
llth
ete
rms
ins
ide
the
bra
ck
et
are
ch
an
ge
dw
he
nth
eb
rac
ke
tis
rem
ov
ed
.T
he
rea
so
nfo
rth
isru
lem
ay
be
se
en
fro
mth
efo
llow
ing
ex
am
ple
s:
−3(2
x−5y)
=(−3)×2x+(−3)×
−5y=
−6x+15y
−(m
+n)=
−m−
n−(p−
q)=
−p+
q
−2(p+3q)=
−2p−6q
Wh
en
sim
plif
yin
ge
xp
res
sio
ns
co
nta
inin
gb
rac
ke
tsfi
rst
rem
ov
eth
eb
rac
ke
tsa
nd
the
na
dd
the
like
term
sto
ge
the
r.
(3x+7y)−(4
x+3y)
=3x+7y−4x−3y=
−x+4y
3(2
x+3y)−(x+5y)
=6x+9y−
x−5y=5x+4y
x(a+
b)−
x(a+3b)=
ax+
bx−
ax−3bx=
−2bx
2(5
a+3b)+3(a−2b)=10a+6b+3a−6b=13a
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.2
ALGEBRA
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6
Removethebrackets
inthefollowing:
i.3
(x+
4)
ii.2
(a+
b)
iii.
3(3
x--
2y
)iv
.½
(x--
1)
v.5
(2p
--3
q)
vi.
7(a
--3
m)
vii.
--(a
+b
)v
iii.
--(a
--2
b)
ix.
--(3
p--
3q
)x
.--
4(x
+3
)
xi.
--2
(2x
--5
)x
ii.--
5(4
--3
x)
xiii
.2
k(k
--5
)x
iv.
--3
y(3
x+
4)
xv.
4x
y(a
b--
ac
+d
)x
vi.
3x
2(x
2--
2x
y+
y2)
xv
ii.--
7p
(2p
2--
p+
1)
Removethebrackets
andsim
plify:
i.3
(x
+1
)+
2(x
+4
)ii.
5(2
a+
4)
--3
(4a
+2
)iii
.3
(x+
4)
--(2
x+
5)
iv.
4(1
--2
x)
--3
(3x
--4
)v.
5(2
x--
y)
--3
(x+
2y
)v
i.½
(y--
1)
+¾
(2y
--3
)
vii.
--(4
a+
5b
--3
c)
--2
(2a
--3
b--
4c
)
viii
.2
x(x
--5
)--
x(x
--2
)--
3x
(x--
5)
ix.
3(a
--b
)--
2(2
a--
3b
)+
4(a
--3
b)
Findtheproducts
ofthefollowing:
i.(x
+4
)(x
+5
)ii.
(2x
+5
)(x
+3
)iii
.(5
x+
1)
(2x
+3
)
iv.
(7x
+2
)(3
x+
2)
v.(x
--4
)(x
--2
)v
i.(2
x--
1)
(x--
4)
vii.
(2x
--4
)(3
x--
2)
viii
.(x
--2
)(x
+7
)ix
.(2
x+
5)
(x--
2)
x.
(3x
+4
y)
(2x
--3
y)
xi.
(2x
+3
)2
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.2
ALGEBRA
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6
ALGEBRAIC
FRACTIONS
Sin
ce
alg
eb
raic
ex
pre
ss
ion
sc
on
tain
sy
mb
ols
(or
lett
ers
)w
hic
hre
pre
se
ntn
um
be
rsa
llth
eru
leo
fo
pe
rati
on
sw
ith
nu
mb
ers
als
oa
pp
lyto
alg
eb
raic
term
s,
inc
lud
ing
fra
cti
on
s.
Th
us
1 1 a=1÷1 a=1×
a 1=( 1
×a)
1=
a
an
d
a b÷
c d=
a b×
d c=
ad bc
an
d ( x+
y)
1
( x−
y)=( x
+y)
÷1
( x−
y)=( x
+y)
×( x
−y)
1=( x
+y)( x
−y)
Yo
us
ho
uld
no
tein
the
las
te
xa
mp
leh
ow
we
pu
tb
rac
ke
tsro
un
dx+
ya
nd
x−
yto
rem
ind
us
tha
tth
ey
mu
st
be
tre
ate
da
ss
ing
lee
xp
res
sio
ns
,o
the
rwis
ew
em
ay
ha
ve
be
en
tem
pte
dto
ha
nd
leth
ete
rms
xa
nd
yo
nth
eir
ow
n.
Adding&SubtractingAlgebraic
Fractions
Co
ns
ide
rth
ee
xp
res
sio
na b+
c dw
hic
his
the
ad
dit
ion
of
two
fra
cti
on
al
term
s.
Ifw
ew
ish
toe
xp
res
sth
es
um
oft
he
se
fra
cti
on
sa
so
ne
sin
gle
fra
cti
on
the
nw
eu
se
the
sa
me
tec
hn
iqu
ea
sfo
rn
um
be
rfr
ac
tio
ns
.
Fir
st
fin
dth
elo
we
st
co
mm
on
de
no
min
ato
r.T
his
isth
eL
CM
ofb
an
dd
wh
ich
isbd
.e
ac
hfr
ac
tio
nis
the
ne
xp
res
se
dw
ith
bd
as
the
de
no
min
ato
r.
Ex
am
ple
:
a b=( a
×d)
( b×
d)=
ad
bd
an
dc d=( c
×b)
( d×
b)=
cb bd
an
da
dd
ing
the
se
ne
wfr
ac
tio
ns
we
ha
ve
:
a b+
c d=
ad
bd+
cb bd=( a
d+
cb)
bd
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.2
ALGEBRA
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6
Multiplication&DivisionofAlgebraic
Fractions
As
wit
ho
rdin
ary
ari
thm
eti
cfr
ac
tio
ns
,n
um
era
tors
ca
nb
em
ult
iplie
dto
ge
the
r,a
sc
an
de
no
min
ato
rs,
ino
rde
rto
form
as
ing
lefr
ac
tio
n.
Ex
am
ple
;
a b×
c d=( a
×c)
( b×
d)
or
3x2y×
p 4q×
r2 s=( 3
x×
p×
r2)
( 2y×4q×
s)
Fa
cto
rsw
hic
ha
rec
om
mo
nto
bo
thn
um
era
tor
an
dd
en
om
ina
tor
ma
yb
ec
an
ce
lled
.it
isim
po
rta
nt
tore
alis
eth
at
this
ca
nc
elli
ng
me
an
sd
ivid
ing
the
nu
me
rato
ra
nd
de
no
min
ato
rb
yth
es
am
eq
ua
nti
ty.
Ex
am
ple
:
8ab
3m
n×9m
n2
4ab2
=( 8
×a×
b×9×
n×
n×
m)
( 3×
m×
n×4×
a×
b×
b)
=6n b
( 5x2y)
8ab3÷10xy
( 4a2b)
=( 5
x2y)
8ab3×( 4
a2b)
10xy
=( 5
×x×
x×
y×4a×
a×
b)
( 8×
a×
b×
b×
b×10×
x×
y)
=ax4b2
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.2
ALGEBRA
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6
Sim
plify
thefollowing:
i.x 3+
x 4+
x 5ii.5a12−7a18
iii.2 q−3 2q
iv.3 y−5 3y+4 5y
v.3 5p−2 3q
vi.3x−4y
5z
vii.1−2x 5+
x 8v
iii.1 x+1 y
ix.3m
−( 2
m+
n)
7x
.( a
−b)
ab
x.6a b2×
b 3a2
xii.9x2
6y2×
y3 x3x
iii.6pq
4rs
×8s2 3p
xiv
.6ab c×
ad 2b×8cd2
4bc
xv.2z2
3ac2×6a2
5zy2×10c3
3y3
xv
i.ab2
bc2÷
a2
bc3
xv
ii.6ab5c
d÷4a2
7bd
xv
iii.3pq
5rs
÷p2
15s2
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.2
ALGEBRA
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6
LINEAREQUATIONS
An
ari
thm
eti
ca
lqu
an
tity
ha
sa
de
fin
ite
va
lue
,s
uc
ha
s9
3,
3.7
3o
r3 4.
An
alg
eb
raic
qu
an
tity
,h
ow
ev
er,
giv
en
by
alg
eb
raic
ex
pre
ss
ion
ss
uc
ha
sx−( −3)
or
x2,
rep
res
en
tsm
an
ya
mo
un
tsd
ep
en
din
go
nth
ev
alu
eg
ive
nto
x.
Equations
As
tate
me
nt
of
the
typ
ex−3=5
isc
alle
da
ne
qu
ati
on
.
Th
ism
ea
ns
tha
tth
eq
ua
nti
tyo
nth
ele
ft--
ha
nd
sid
eo
fth
ee
qu
ati
on
ise
qu
alt
oth
eq
ua
nti
tyo
nth
eri
gh
t--h
an
ds
ide
.W
ec
an
se
eth
at,
un
like
an
ide
nti
ty,
the
reis
on
lyo
ne
va
lue
ofx
tha
tw
ills
ati
sfy
the
eq
ua
tio
n,
or
ma
ke
the
left
--h
an
ds
ide
eq
ua
lto
the
rig
ht--
ha
nd
sid
e.
Th
ep
roc
es
so
ffi
nd
ing
x=8
isc
alle
ds
olv
ing
the
eq
ua
tio
n,
an
dth
ev
alu
e8
isk
no
wn
as
thesolutionorroot
of
the
eq
ua
tio
n.
SolvingLinearEquations
Lin
ea
re
qu
ati
on
sc
on
tain
on
lyth
efi
rst
po
we
ro
fth
eu
nk
no
wn
qu
an
tity
.
7t−5=4t+7
an
d5x 3=( 2
x+5)
2
are
bo
the
xa
mp
les
of
line
ar
eq
ua
tio
ns
.
Inth
ep
roc
es
so
fs
olv
ing
an
eq
ua
tio
nth
ea
pp
ea
ran
ce
so
fth
ee
qu
ati
on
ma
yb
ec
on
sid
era
ble
alt
ere
db
utt
he
va
lue
so
nb
oth
sid
es
mu
str
em
ain
the
sa
me
.We
mu
st
ma
inta
inth
ise
qu
alit
y,a
nd
he
nc
ew
ha
tev
er
we
do
too
ne
sid
eo
fth
ee
qu
ati
on
we
mu
st
do
ex
ac
tly
the
sa
me
toth
eo
the
rs
ide
.
Aft
er
an
eq
ua
tio
nis
so
lve
d,
the
so
luti
on
sh
ou
ldb
ec
he
ck
ed
by
su
bs
titu
tin
gth
ere
-s
ult
ine
ac
hs
ide
of
the
eq
ua
tio
ns
ep
ara
tely
.If
ea
ch
sid
eo
fth
ee
qu
ati
on
the
nh
as
the
sa
me
va
lue
the
so
luti
on
isc
orr
ec
t.In
the
de
tail
wh
ich
follo
ws
,LH
Sm
ea
ns
left
--h
an
ds
ide
an
dR
HS
me
an
sri
gh
t--h
an
ds
ide
.
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.2
ALGEBRA
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6
EquationRequiringMultiplication&Division
Ex
am
ple
:
So
lve
the
eq
ua
tio
nx 6=3
Mu
ltip
lye
ac
hs
ide
by
6,
we
ge
t
x 6×6=3×6
x=18
Ch
ec
k:
wh
en
x=18
,L
HS
=18 6
,R
HS
=3
EquationsRequiringAddition&Subtraction
Ex
am
ple
:
So
lve
the
eq
ua
tio
nx−4=8
Ifw
ea
dd
4to
ea
ch
sid
e,
we
ge
t
x−4+4=8+4
x=12
Th
eo
pe
rati
on
of
ad
din
g4
toe
ac
hs
ide
isth
es
am
ea
str
an
sfe
rrin
g--
4to
the
RH
Sb
ut
ins
od
oin
gth
es
ign
isc
ha
ng
ed
fro
ma
min
us
toa
plu
s.
x−4=8
x=8+4
x=12
Ch
ec
k:
wh
en
x=12
,L
HS
=1
2--
4=
8,
RH
S=
8
Solvethefollowingequations:
i.x
+3
=8
ii.x
--4
=6
iii.
2x
=8
iv.
2x
--7
=9
v.5
x+
3=
18
vi.
3x
--7
=x
--5
vii.
9--
2x
=3
x+
7v
iii.
4x
--3
=6
x--
9
ix.
5x
--8
=3
x+
2x
.2
(x+
1)
=9
xi.
5(x
--3
)=
12
xii.
3(2
x--
1)
+4
(2x
+5
)=
40
xiii
.7
(2--
3x
)=
3(5
x--
1)
xiv
.x 2+
x 3=10
xv.3x+3 8=2+2x 3
xv
i.2x 5=
x 8+1 2
xv
ii.( x
+3)
2=( x
−3)
3
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.3
GEOMETRY
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6
GEOMETRY
COORDINATES&GRAPHS
Coordinates
Co
ord
ina
tes
are
nu
mb
ers
wh
ich
are
us
ed
tore
pre
se
nt
ap
art
icu
lar
po
int
on
ag
rap
h.
Co
ord
ina
tea
xe
sc
on
sis
to
fa
ho
riz
on
tal
line
(xa
xis
)a
nd
av
ert
ica
llin
e(y
ax
is).
Th
ep
oin
to
fin
ters
ec
tio
no
fth
es
etw
olin
es
isc
alle
dth
eo
rig
in(d
en
ote
db
yth
ele
tte
r“O
“).
Alo
ng
the
xa
nd
ya
xe
sw
ec
an
ma
rko
ffu
nit
so
fm
ea
su
rem
en
t(n
ot
ne
ce
ss
ari
lyth
es
am
eo
nb
oth
ax
es
).T
he
ori
gin
tak
es
the
va
lue
ze
roo
nb
oth
ax
es
.Th
ex
ax
ista
ke
sp
os
itiv
ev
alu
es
toth
eri
gh
to
fth
eo
rig
ina
nd
ne
ga
tiv
ev
alu
es
toth
ele
fto
fth
eo
rig
in.
Th
ey
ax
ista
ke
sp
os
itiv
ev
alu
es
ab
ov
eth
eo
rig
ina
nd
ne
ga
tiv
ev
alu
es
be
low
the
ori
gin
.
An
yp
oin
to
nth
isd
iag
ram
ca
nb
ed
efi
ne
db
yit
sc
oo
rdin
ate
s(c
on
sis
tin
go
ftw
on
um
be
rs).
Th
efi
rst,
the
xc
oo
rdin
ate
,d
efi
ne
sth
eh
ori
zo
nta
ldis
tan
ce
of
the
po
int
fro
mth
ey
ax
is,
the
se
co
nd
,th
ey
co
ord
ina
te,
de
fin
es
the
ve
rtic
ald
ista
nc
eo
fth
ep
oin
tfr
om
the
xa
xis
.
Ing
en
era
l,a
po
int
isd
efi
ne
db
yit
sc
oo
rdin
ate
sw
hic
ha
rew
ritt
en
inth
efo
rm(x
,y),
the
xc
oo
rdin
ate
alw
ay
sw
ritt
en
firs
t.T
he
co
ord
ina
tes
are
alw
ay
sw
ritt
en
inb
rac
ke
tsw
ith
ac
om
ma
be
twe
en
the
mto
av
oid
co
nfu
sio
n.
Example:
Th
ep
oin
t(3
,2
)m
ay
be
plo
tte
do
nth
ec
oo
rdin
ate
ax
es
as
follo
ws
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.3
GEOMETRY
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6
Graphs
An
eq
ua
tio
nin
vo
lvin
gtw
ov
ari
ab
les
ca
nb
ere
pre
se
nte
d,
on
co
ord
ina
tea
xe
s,
by
me
an
so
fa
gra
ph
.T
he
line
ar
an
dq
ua
dra
tic
eq
ua
tio
ns
co
ns
ide
red
inth
ela
st
ch
ap
ter
ca
nb
ere
pre
se
nte
da
slin
es
on
ag
rap
h.
Fo
ra
giv
en
ran
ge
of
va
lue
so
fx
,th
ec
orr
es
po
nd
ing
yv
alu
es
ca
nb
ec
alc
ula
ted
fro
mth
ee
qu
ati
on
be
ing
co
ns
ide
red
.T
he
po
ints
ob
tain
ed
ca
nth
en
be
plo
tte
da
nd
join
ed
tog
eth
er
tofo
rmth
eg
rap
h.
Be
fore
plo
ttin
gth
ep
oin
tso
na
gra
ph
,th
ea
xe
sm
us
tb
ed
raw
nin
aw
ay
tha
tta
ke
sin
toa
cc
ou
nt
the
ran
ge
of
the
x--
va
lue
sa
nd
the
ran
ge
of
the
y--
va
lue
s.
Ifg
rap
hp
ap
er
isu
se
d(w
hic
his
de
sir
ab
le)
yo
us
ho
uld
us
ea
sc
ale
tha
tin
vo
lve
sa
se
ns
ible
nu
mb
er
of
un
its
pe
rs
qu
are
i.e
.y
ou
sh
ou
ldu
se
ste
ps
of,
for
ex
am
ple
,1
,2
,5
or
10
etc
.u
nit
sp
er
sq
ua
red
ep
en
din
go
nth
eq
ue
sti
on
.Y
ou
sh
ou
lda
vo
idu
sin
gs
tep
sa
lon
gth
ea
xe
so
f,fo
re
xa
mp
le7
or
9u
nit
sp
er
sq
ua
rea
sth
isc
an
co
mp
lica
teth
eg
rap
hu
nn
ec
es
sa
rily
.
Example:
Dra
wth
eg
rap
ho
fy
=2
x+
1b
etw
ee
nx
=0
an
dx
=5
By
tak
ing
the
xv
alu
es
0,1
,2,.
....
...5
,we
ca
nc
alc
ula
teth
ec
orr
es
po
nd
ing
yv
alu
es
,a
ss
ho
wn
be
low
,b
yfi
rst
ev
alu
ati
ng
the
co
mp
on
en
tp
art
so
fth
ee
qu
ati
on
.
x:
01
23
45
2x
02
46
81
0
+1
11
11
11
y:
13
57
911
We
the
np
lot
the
po
ints
ob
tain
ed
,e
ac
hp
oin
tb
ein
gd
efi
ne
db
yit
sx
co
ord
ina
tea
nd
its
co
rre
sp
on
din
gy
co
ord
ina
te.
Th
ep
oin
tsa
reth
en
join
ed
tog
eth
er
toth
eg
rap
h.
Th
ev
alu
eo
fy
the
refo
red
ep
en
ds
on
the
va
lue
allo
ca
ted
tox
.W
eth
ere
fore
ca
lly
thedependentvariable
.S
inc
ew
ec
an
giv
ex
an
yv
alu
e,
we
ca
llx
the
independentvariable
.It
isu
su
alt
om
ark
the
va
lue
so
fth
ein
de
pe
nd
en
tv
ari
ab
lea
lon
gth
eh
ori
zo
nta
lax
is(x
).T
he
de
pe
nd
en
tv
ari
ab
lev
alu
es
are
ma
rke
do
ffa
lon
gth
ev
ert
ica
la
xis
(y).
Eq
ua
tio
ns
of
the
typ
ey
=2
x+
1,
wh
ere
the
hig
he
st
po
we
rso
fth
ev
ari
ab
les
,x
an
dy,
are
the
firs
ta
rec
alle
de
qu
ati
on
so
fth
efirstdegree
.A
lle
qu
ati
on
so
fth
isty
pe
giv
eg
rap
hs
wh
ich
are
str
aig
ht
line
sa
nd
he
nc
eth
ey
are
oft
en
ca
lled
linearequations
.In
ord
er
tod
raw
gra
ph
so
flin
ea
re
qu
ati
on
sw
en
ee
do
nly
tak
etw
op
oin
ts,
ho
we
ve
rth
ree
po
ints
are
ad
vis
ab
le.
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.3
GEOMETRY
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6
Drawgraphsofthefollowingfunctionstakingvaluesofxbetw
een--3and
4.
i.y
=2
x+
5
ii.y
=3
x--
5
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.3
GEOMETRY
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6
TheStraightLineGraph
As
tra
igh
tlin
eis
de
fin
ed
as
the
sh
ort
es
td
ista
nc
eb
etw
ee
ntw
op
oin
ts.
Th
ee
qu
ati
on
of
as
tra
igh
tlin
eis
giv
en
by
:
y=
mx
+c
Wh
ere
mre
pre
se
nts
the
gra
die
nt
of
the
line
an
dc
isth
ep
oin
tw
he
reth
elin
ec
ros
se
sth
ey
ax
is(t
he
yin
terc
ep
t).
Th
ep
oin
tw
he
reth
elin
ec
ros
se
sth
ex
ax
isis
ca
lled
the
xin
terc
ep
t.
Gra
die
ntc
an
be
de
fin
ed
as
the
inc
rea
se
alo
ng
the
ya
xis
co
mp
are
dto
the
inc
rea
se
alo
ng
the
xa
xis
.In
the
dia
gra
mb
elo
wit
ca
nb
es
ee
nth
at
as
the
va
lue
of
xin
cre
as
es
by
1,
the
va
lue
of
yin
cre
as
es
by
2s
oth
eg
rad
ien
tis
2.
Inth
ed
iag
ram
on
the
rig
ht
itc
an
be
se
en
tha
ta
sth
ev
alu
eo
fx
inc
rea
se
sb
y1
the
va
lue
of
yd
ec
rea
se
sb
y3
.T
his
de
cre
as
eis
rep
res
en
ted
ma
the
ma
tic
ally
as
an
inc
rea
se
of
-3s
oth
eg
rad
ien
tis
-3.
m=
Gra
die
nt
of
the
line
c=
Inte
rce
pt
on
the
ya
xis
No
te:
inth
ise
xa
mp
lem
=2
an
dc
=0
,w
he
ne
ve
rc
=0
the
line
will
pa
ss
thro
ug
hth
eo
rig
in.
Example:
Inth
ise
xa
mp
lem
=--
3a
nd
c=
6
As
c=
6,
we
kn
ow
tha
tth
islin
ec
uts
the
ya
xis
at
y=
6(t
his
ca
nb
ev
eri
fie
db
ys
ub
sti
tuti
ng
x=
0in
toth
ee
qu
ati
on
of
the
line
,a
sx
=0
alo
ng
the
ya
xis
)
Sim
ilarl
y,a
sy
=0
alo
ng
the
xa
xis
,w
ec
an
su
bs
titu
tey
=0
into
the
eq
ua
tio
no
fth
elin
eto
fin
dw
he
reth
ein
ters
ec
tsw
ith
the
ax
is(t
he
inte
rce
pt)
.
we
ha
ve
,w
he
n
y=
0
6--
3x
=0
3x
=6
x=
2
He
nc
eth
elin
ec
uts
the
xa
xis
at
x=
2.
We
ca
nn
ow
sa
yth
at
the
yin
terc
ep
t=6
an
dth
ex
inte
rce
pt
=2
.
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.3
GEOMETRY
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6
Example:
Inth
ise
xa
mp
le,
m=
4a
nd
c=
--2
.
We
kn
ow
,im
me
dia
tely
tha
tth
ein
terc
ep
tis
--2
(th
ev
alu
eo
fc
).T
ofi
nd
the
xin
terc
ep
t,w
es
ub
sti
tute
y=
0in
toth
ee
qu
ati
on
of
the
line
.
0=
--2
+4
4x
=2
x=
0.5
He
nc
eth
ex
inte
rce
pt
isx
=0
.5.
As
tra
igh
tlin
ep
ara
llelt
oth
ex
ax
ista
ke
sth
efo
rmy
=c
on
sta
nt.
Sim
ilarl
y,a
str
aig
ht
line
pa
ralle
lto
the
ya
xis
tak
es
the
form
x=
co
ns
tan
t.
Th
es
ec
as
ea
reill
us
tra
ted
be
low
:
LufthansaResource
TechnicalTraining
MATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
55
ANSWERSTOQUESTIONS
From
Page5
Calculate
thesum
ofthefollowingexamples:
Addition
i.0
.25
1+
10
.29
8=
10
.54
9
ii.1
8.0
98
+2
10
.09
9=
22
8.1
97
iii.
0.0
25
+1
0.9
95
=11
.02
0
iv.
1.0
9+
1.2
+1
0.1
4=
12
.43
v.2
7.3
+0
.02
1+
68
.3=
95
.62
1
Subtraction
i.2
7.3
--4
.36
=2
2.9
3
ii.2
1.7
6--
18
.51
=3
.25
iii.
32
.76
--2
0.0
86
=1
2.6
74
iv.
10
.75
--1
9.9
99
--2
1.1
00
=-3
0.3
49
v.1
.09
--1
.2-
68
.3=
-68
.41
MATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
56
From
Page7
Calculate
thesum
ofthefollowingexamples:
i.--
8+
5=
-3
ii.--
7--
6--
3=
-16
iii.
8--
7--
15
=-1
4
iv.
--3
+5
+7
--4
--2
=3
v.6
+4
--3
--5
--7
+2
=-3
vi.
8x
(--3
)=-2
4
vii.
(--2
)x
(--5
)x
(--6
)=-6
0
viii
.4
x(-
-3)
x(-
-2)=
24
ix.
(--3
)x
(--4
)x
5=
60
x.
--1
6÷
((--
2)
x(-
-4))
=-2
xi.
(15
x(-
-3)
x2
)÷
((--
5)
x(-
-6))
=-3
xii.
3+
5x
2=
13
xiii
.(7
x5
)--
2+
(4x
6)=
57
xiv
.(7
x5
)--
(12÷
4)
+3
=4
xv.
11--
(9÷
3)
+7
=1
5
xv
i.11
--(1
2÷
4)
+3
x(6
--2
)=
20
xv
ii.(1
5÷
(4+
1))
--(9
x3
)+
7(4
+3
)=2
5
xv
iii.
10
--(1
2÷
6)
+3
(8--
3)=
23
Question1.
16
ho
les
sp
ac
ed
48
mm
ap
art
are
tob
em
ark
ed
off
on
as
he
et
of
me
tal.
17
mm
isto
be
allo
we
db
etw
ee
nth
ec
en
tre
so
fth
eh
ole
sa
nd
the
ed
ge
of
the
me
tal.
Ca
lcu
late
the
tota
lle
ng
tho
fm
eta
lre
qu
ire
d.
=7
54
mm
Question2.
Inth
efi
rst
2h
ou
rso
fa
sh
ift
an
op
era
tor
ma
ke
s3
2s
old
ere
djo
ints
pe
rh
ou
r.In
the
ne
xt
3h
ou
rsth
eo
pe
rato
rm
ak
es
29
join
tsp
er
ho
ur.
Inth
efi
na
ltw
oh
ou
rs2
6jo
ints
are
ma
de
pe
rh
ou
r.H
ow
ma
ny
so
lde
red
join
tsa
rem
ad
ein
the
7h
ou
rs.=
21
3
Question3.
Am
ac
hin
ist
ma
ke
s3
pa
rts
in1
5m
inu
tes
.H
ow
ma
ny
pa
rts
ca
nh
ep
rod
uc
ein
an
8h
ou
rs
hif
ta
llow
ing
20
min
ute
sfo
rs
tart
ing
an
d1
0m
inu
tes
for
fin
ish
ing
the
sh
ift.
=9
0
Question4.
Th
ele
ng
tho
fa
me
tal
pla
teis
89
1m
m.
Riv
ets
are
pla
ce
d4
5m
ma
pa
rta
nd
the
dis
tan
ce
be
twe
en
the
ce
ntr
es
of
the
en
dri
ve
tsa
nd
the
ed
ge
of
the
pla
teis
18
mm
.H
ow
ma
ny
riv
ets
are
req
uir
ed
.=1
9
Question5.
32
pin
se
ac
h6
1m
mlo
ng
are
tob
etu
rne
din
ala
the
.If
2m
mis
allo
we
do
ne
ac
hp
info
rp
art
ing
off
.w
ha
tto
tal
len
gth
of
ma
teri
al
isre
qu
ire
dto
ma
ke
the
pin
s.
=2
01
6m
m
MATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
57
From
Page14
Solvethefollowingequations:
Co
nv
ert
the
follo
win
gm
ixe
dn
um
be
rsto
imp
rop
er
fra
cti
on
s:
i.26 7
=20 7
ii.34 9
=31 9
iii.213 5
=1085
iv.52125
=14625
v.21 7=15 7
Co
nv
ert
the
follo
win
gim
pro
pe
rfr
ac
tio
ns
tom
ixe
dn
um
be
rs:
i.11 3
=32 3
ii.21 5
=41 5
iii.53 7
=74 7
iv.2104
=511 2
v.99 8
=123 8
Ad
dth
efo
llow
ing
fra
cti
on
s:
i.3 4+3 8
=11 8
ii.1 8+2 3+5 12
=15 24
iii.72 3+63 5
=144 15
iv.33 8+52 7+43 4
=132356
v.2310+14 6
=41930
Su
btr
ac
tth
efo
llow
ing
fra
cti
on
s:
i.7 8−5 6
=11 24
ii.33 8−11 4
=21 8
iii.53 8−29 10
=21940
iv.21 5−32 5
=4 5
v.13 4−22 5
=−13
20
Mu
ltip
lya
nd
sim
plif
yth
efo
llow
ing
fra
cti
on
s:
i.3 4×5 7
=1528
ii.2 9×12 3
=1027
iii.7 5×31 2
=49 10
iv.33 4×13 5×11 8
=63 4
v.3 4
of
16
=1
2
Div
ide
an
ds
imp
lify
the
follo
win
gfr
ac
tio
ns
:
i.4 5÷11 3
=3 5
ii.21 2÷33 4
=2 3
iii.5÷51 5
=2526
iv.12 3÷�3 5
÷9 10� =21 2
v.28 9÷� 12 3+1 2� =11 3
Arr
an
ge
the
follo
win
gs
ets
of
fra
cti
on
sin
ord
er
of
siz
e:
i.1 2���5 6���2 3���7 12
=1 2���7 12���2 3���5 6
ii.3 4���5 8���9 16���1732
=1732���9 16���3 4���5 8
iii.3 8���5 9���2 6���5 18=5 18���2 6���3 8���5 9
MATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
58
From
page20
Solvethefollowingequations:
Multiplication
i.5
.05
x1
3.8
=6
9.6
9
ii.1
.27
x0
.87
1=
1.1
06
17
iii.
--1
.01
x0
.89
=-0
.89
89
iv.
27
.3x
--9
.31
=-2
54
.16
3
v.1
.09
x1
04
x1
.2x
10
2=
13
08
00
0
Division
i.2
33
.1÷
18
.5=
12
.6
ii.0
.12
54÷
0.0
57
=2
.2
iii.
0.6
87
5÷
22
=0
.03
12
5
iv.
24
.02
4÷
4.6
2=
5.2
v.1
.09
x1
04÷
12
=9
08
.33
4
Convertthefollowingdecim
als
tofractionsin
theirlowestterm
s:
i.0
.2=1 5
ii.0
.45
=9 20
iii.
0.3
12
5=5 16
iv.
2.5
5=21120
v.0
.00
75
=3400
vi.
2.1
25
-=21 8
Fin
dth
ed
iffe
ren
ce
be
twe
en
i.1964
an
d0
.29
5=
0.0
01
87
5
ii.13 16
an
d1
.16
32
=0
.97
57
Convertthefollowingfractionsto
decim
als
(3decim
alplaces)
i.3 8
=0
.37
5ii.1116
=0
.68
8iii
.2132
=0
.65
6iv
.15 8
=1
.62
5v.27 16=
2.4
38
Placethefollowingin
ascendingorderofsize;
i.1 5���0.167���3 20
=1 5���0.167���3 20
ii.2 5���0.44���7 16
=2 5���7 16���0.44
iii.1132���0.3594���0.3125
=0.3125���1132���0.3594
Expressthefollowingasapercentage%:
i.0
.43
=4
3%
ii.0
.02
5=
2.5
%iii
.1
.25
=1
25
%iv
.3 8
=3
7.5
%
v.3 7
=4
2.8
6%
vi.1 12
=8
.34
%v
ii.7 20
=3
5%
Expressthefollowingasfractions:
i.2
5%
=1
/4ii.
13
%=
13
/10
0iii
.4
.5%
=9
/20
0iv
.3
3%
=3
3/1
00
Express:
i.3
0a
sa
pe
rce
nta
ge
of
50
=6
0%
ii.2
4a
sa
pe
rce
nta
ge
of
16
=1
50
%
iii.
0.5
as
ap
erc
en
tag
eo
f1
2.5
=4
%
iv.
3.2
as
ap
erc
en
tag
eo
f2
.4=
13
3%
v.0
.08
as
ap
erc
en
tag
eo
f0
.72
=11
.12
%
Calculate:
i.4
%o
f3
0=
11
/5o
r1
.2
ii0
.8%
of
36
0=
22
2/2
5o
r2
.88
iii.
1.5
%o
f6
0=
9/1
0o
r0
.9
iv.
12
0%
of
75
=9
33
/4o
r9
3.7
5
v.8
0%
of
90
=7
2
iv.
12
0%
of
75
v.8
0%
of
90
MATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
59
From
Page24
Solvethefollowingequations:
Ex
pre
ss
the
follo
win
gra
tio
sa
sfr
ac
tio
ns
red
uc
ed
toth
eir
low
es
tte
rms
;
i.1
5g
to2
kg
=3
:40
0ii.
21
ftto
9in
ch
es
=2
8:1
iii.2
0c
mto
10
0m
m=
2:1
iv.4
00
mto
3k
m=
2:1
5
Fin
dth
em
iss
ing
va
lue
;
i.3
:4=
6:8
ii.2
0:1
=6
4:3
.2iii
.2
40
:40
0=
0.6
:1iv
.1
:2.6
=5
:13
v.1
8:9
=2
:1
Fiv
em
en
bu
ilda
wa
llta
ke
20
da
ys
toc
om
ple
teit
.H
ow
lon
gw
ou
ldit
tak
e4
me
nto
co
mp
lete
it.
=2
5d
ay
s
4p
eo
ple
ca
nc
lea
na
no
ffic
ein
6h
ou
rs.
Ho
wm
an
yp
eo
ple
wo
uld
be
ne
ed
ed
toc
lea
nth
eo
ffic
ein
4h
ou
rs.
6p
eo
ple
8p
eo
ple
tak
e5
ho
urs
toc
ha
ng
ea
ne
ng
ine
.H
ow
lon
gw
ou
ldit
tak
e4
pe
op
leto
do
this
wo
rk.
10
Ho
urs
An
en
gin
ee
rin
gc
om
pa
ny
em
plo
y1
2m
en
tofa
bri
ca
tea
nu
mb
er
of
co
nta
ine
rs.
Th
ey
tak
e9
da
ys
toc
om
ple
teth
ew
ork
.If
the
co
mp
an
yh
ad
em
plo
ye
d8
me
n,h
ow
lon
gw
ou
ldit
ha
ve
tak
en
.1
3.5
Da
ys
Atr
ain
tra
ve
ls2
00
km
in4
ho
urs
.If
ittr
av
els
at
the
sa
me
rate
,h
ow
lon
gw
illit
tak
eto
co
mp
lete
ajo
urn
ey
of
35
0k
m.
7h
ou
rs
Ab
ar
of
me
tal1
0.5
mlo
ng
isto
be
cu
tin
toth
ree
pa
rts
inth
era
tio
of1 2:13 4:3
.F
ind
the
len
gth
of
ea
ch
pa
rt.
1m
,3
.5m
,6
m
MATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
60
From
Page29
Solvethefollowingequations:
Fin
dth
ev
alu
es
of
the
follo
win
g
i.82
=6
4ii.24
=1
6iii
.33
=2
7iv
.25
=3
2
v.16
=
4v
i.144
=
12
vii.169
=
13
viii
.83
=2
ix.27
3 =
3x
.216
3 =
6
Sim
plif
yth
efo
llow
ing
,g
ivin
ge
ac
ha
ns
we
ra
sa
po
we
r
i.25×26
=2
11
ii.a×
a2×
a5
=a8
iii.n5÷
n3
=n3
iv.105×103÷104
=104
v.z4×
z2×
z−3
=z3
vi.32×3−3÷33
=3−4
vii.
� 93�4
=97
viii
.� t×
t3�2 =
t6
ix.�1 73�4
=7
Fin
dth
ev
alu
eo
fth
efo
llow
ing
i.8
11
/4=
3ii.
82
/3=
4iii
.1
63
/4=
8iv
.9
2.5
=2
43
MATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
61
From
Page30
Transposethefollowing:
C=
πd
so
d=
c π
S=
πdn
so
d=
s( π
n)
I=
PRT
so
R=
I PT
v2=2gh
so
h=
v2 2g
x=
a ys
oy=
a x
P=
RT V
so
T=
PV R
S=
ts Ts
ot=
ST s
M I=
E Rs
oR=
IE M
GY l=
T Js
oJ=
Tl
GY
v=
u+
at
so
t=
v−
ua
n=
p+
crs
or=
n−
pc
y=
ax+
bs
ox=
y−
ba
y=
x 5+17
so
x=5y−17
MATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
62
From
Page33
Solvethefollowingproblems:
i.F
ind
the
are
aa
nd
pe
rim
ete
ro
fa
rec
tan
gle
wh
os
ele
ng
this
12
inc
he
sa
nd
wid
this
7in
ch
es
.a
rea
=8
4in
2p
eri
me
ter
=3
8in
ch
es
ii.A
ca
rpe
th
as
an
are
ao
f3
6m
2.
Ifit
iss
qu
are
wh
at
len
gth
of
sid
eh
as
the
ca
rpe
t?6
m
iii.
Atr
ian
gle
ha
sa
ba
se
of
7c
ma
nd
an
alt
itu
de
of
3c
m.
Ca
lcu
late
its
are
a.
=1
0.5
cm
iv.
Th
ea
rea
of
atr
ian
gle
is4
0ft
2.
Its
ba
se
is8
ftlo
ng
.C
alc
ula
teit
sv
ert
ica
lh
eig
ht.
=1
0ft
v.C
alc
ula
teth
ev
olu
me
of
am
eta
lp
ipe
wh
os
ein
sid
ed
iam
ete
ris
6c
ma
nd
wh
os
eo
uts
ide
dia
me
ter
is8
cm
,if
it2
0c
mlo
ng
.4
39
.88
cm
2
vi.
Are
cta
ng
ula
rta
nk
is2
.7c
mlo
ng
,1
.8c
mw
ide
an
d3
.2c
mh
igh
.H
ow
ma
ny
litre
so
fw
ate
rw
illit
ho
ldw
he
nfu
ll?0
.01
55
52
litre
s
MATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
63
From
Page35
Convertthefollowingweights
andmeasures:
i.C
on
ve
rt6
mto
fee
t.=
19
.68
6
ii.C
on
ve
rt2
5U
Sg
allo
ns
tolit
res
.=
94
.62
5
iii.
Co
nv
ert
25
4in
ch
es
toc
m.
=6
45
.16
iv.
Co
nv
ert
4.5
litre
toU
Sg
allo
ns
=1
.19
US
Ga
llon
s
v.C
on
ve
rt3
50
imp
eri
al
ga
llon
sto
litre
s.
=1
59
1.1
Lit
res
vi.
Co
nv
ert
the
follo
win
gto˚F
--2
0˚C
=-4
--5˚C
=2
3
37˚C
=9
8.6
88˚C
=1
90
.4
vii.
Co
nv
ert
the
follo
win
gto˚C
--4
0˚F
=-4
0
16˚F
=-8
.89
10
0˚F
=3
7.7
8
21
5˚F
=1
01
.67
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
64
TEST
Wo
rko
ut
the
va
lue
of
the
follo
win
g:
1.
7+
4x
3=
19
2.
5x
4--
3x
6+
5=
7
3.
10
--1
2÷
6+
3(8
--3
)=
23
4.
53
=1
25
5.2 5+3 7
=2935
6.5 6−3 4
=1 9
7.3 8×5 7
=1556
8.3 5÷7 8
=2435
9.
div
ide
74
.52
by
8.1
=9
.2
10
.m
ult
iply
20
.3x
17
.4=
35
3.2
2
11.
Co
nv
ert
0.8
00
toa
fra
cti
on
=4 5
12
.c
on
ve
rt3 5
toa
pe
rce
nta
ge
=6
0%
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
65
Fro
mP
ag
e3
8
Solvethefollowingsubstitutionequations:
Ifa
=2
,b
=3
an
dc
=5
.F
ind
the
va
lue
so
fth
efo
llow
ing
.
i.a
+7
=9
ii.9
c=
45
iii.
3b
c=
45
iv.
4c
+6
b=
38
v.a
+2
b+
5c
=3
3
vi.
8c
--4
b=
28
vii.
abc6
=5
viii
.5
a+
9b
+8
c=
7.7
a+
b+
c
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
66
From
Page41
Sim
plify
thefollowing:
1)
i.7
x+
11x
=1
8x
ii.7
x--
5x
=2
xiii
.3
x--
6x
=-3
xiv
.--
2x
--4
x=
-6x
v.--
8x
+3
x=
-5x
vi.
--2
x+
7x
=5
xv
ii.5
m+
13
m--
6m
=1
2m
viii
.6
b2
--4
b2
+3
b2
=1
3b
2ix
.6
ab
--3
ab
--2
ab
=a
b
x.
14
xy
+5
xy
--7
xy
+2
xy
=1
4x
yx
i.--
5x
+7
x--
3x
--2
x=
-3x
xii.
3x
--2
y+
4z
--2
x--
3y
+5
z+
6x
+2
y--
3z
=7
x-
7y
+6
z
xiii
.3
a2b
+2
ab
3+
4a
2b
2--
5a
b3
+11
b4
+6
a2b
=9
a2b
-3a
b3+
4a
2b
2+
11b
4
xiv
.p
q+
2.1
qr
--2
.2rq
+8
qp
=9
pq
+4
.3q
r
2)
i.2
zx
5y
=1
0y
zii.
3a
x3
b=
9a
biii
.3
x4
m=
12
miv
.¼
qx
16
p=
4p
q
v.z
x(y
)=
xy
zv
i.(-
-3a
)x
(--2
b)
=6
ab
vii.
8m
x(-
-3n
)=
-24
mn
viii
.(-
-4a
)x
3b
=-1
2a
bix
.8
px
(--q
)x
(--3
r)=
24
pq
r
x.
3a
x(-
-4b
)x
(--c
)x
5d
=1
2a
bc
xi.
ax
a=
a2
xii.
3m
x(-
-3m
)=
-9m
2
xiii
.8
mn
x(-
-3m
2n
3)
=--
24
m3n
4x
iv.
7a
bx
(--3
a2)
=-2
1a
3b
xv.
m2n
x(-
-mn
)x
5m
2n
2=
-5m
5n
4x
vi.
5a
2x
(--3
b)
x5
ab
=-7
5a
3b
2
3)
i.1
2x÷
6=
2x
ii.4
a÷
(--7
b)
=−4 7a b
iii.
(--5
a)÷
8b
=−58b
iv.
4a÷
2b
=2a b
v.4
ab÷
2a
vi.
12
x2y
z2÷
4x
z2
=3
xy
vii.
(--1
2a
2b
)÷
6a
=-2
ab
viii
.8
a2b
c2÷
4a
c2
=2
ab
ix7
a2b
2÷
3a
b=
21
a3b
3
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
67
From
Page43
Removethebrackets
inthefollowing:
i.3
(x+
4)
=3
x+
12
ii.2
(a+
b)
=2
a+
2b
iii.
3(3
x--
2y
)=
9x
-6y
iv.½
(x--
1)
=½
x-½
v.5
(2p
--3
q)
=1
0p
-1
5q
vi.
7(a
--3
m)
=7
a-2
1m
vii.
--(a
+b
)=
-a-b
viii
.--
(a--
2b
)-a
+2
bix
.--
(3p
--3
q)
=-3
p+
3q
x.
--4
(x+
3)
=-4
x-1
2x
i.--
2(2
x--
5)
=-4
x+
10
xii.
--5
(4--
3x
)=
-20
+1
5x
xiii
.2
k(k
--5
)=
2k
2--
10
kx
iv.
--3
y(3
x+
4)
=-9
xy
-12
y
xv.
4x
y(a
b--
ac
+d
)=
4a
bx
y-4
ac
xy
+4
dx
y
xv
i.3
x2(x
2--
2x
y+
y2)
=3
x4
--6
x3y
+3
x2y
2
xv
ii.--
7p
(2p
2--
p+
1)
=-1
4p
3-7
p2
--7
p
Removethebrackets
andsim
plify:
i.3
(x
+1
)+
2(x
+4
)=
5x
+11
ii.5
(2a
+4
)--
3(4
a+
2)
=-2
a-1
5
iii.
3(x
+4
)--
(2x
+5
)=
x+
7iv
.4
(1--
2x
)--
3(3
x--
4)
=1
6--
17
x
v.5
(2x
--y
)--
3(x
+2
y)
=7
x-3
yv
i.½
(y--
1)
+¾
(2y
--3
)=
-11
/4+
2y
vii.
--(4
a+
5b
--3
c)
--2
(2a
--3
b--
4c
)=
-8a
+11
b+
5c
viii
.2
x(x
--5
)--
x(x
--2
)--
3x
(x--
5)
=-2
x2+
7x
ix.
3(a
--b
)--
2(2
a--
3b
)+
4(a
--3
b)
=3
a-9
b
Findtheproducts
ofthefollowing:
i.(x
+4
)(x
+5
)=
x2+
9x
+2
0ii.
(2x
+5
)(x
+3
)=
2x
2+
8x
+1
5
iii.
(5x
+1
)(2
x+
3)
=1
0x
2+
17
x+
3
iv.
(7x
+2
)(3
x+
2)
=2
1x
2+
20
x+
4v.
(x--
4)
(x--
2)
=x
2--
6x
+8
vi.
(2x
--1
)(x
--4
)=
2x
2--
9x
+4
vii.
(2x
--4
)(3
x--
2)
=6
x2
--1
6x
+8
viii
.(x
--2
)(x
+7
)=
x2+
5x
-14
ix.
(2x
+5
)(x
--2
)2
x2+
x-1
0
x.
(3x
+4
y)
(2x
--3
y)
=6
x2
--6
xy
-12
y2
xi.
(2x
+3
)2=
4x
2+
12
x+
9
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
68
From
Page46
Sim
plify
thefollowing:
i.x 3+
x 4+
x 5=47x60
ii.5a12−7a18
=a 36
iii.2 q−3 2q
=1 2q
iv.3 y−5 3y+4 5y
=3215y
v.3 5p−2 3q
=9q−10p
15pq
vi.3x−4y
5z
=15zx
−4y
5z
vii.1−2x 5+
x 8=1−21x40
viii
.1 x+1 y
=y+
xxy
ix.3m
−( 2
m+
n)
7=19m
+n
7x
.( a
−b)
ab
xi.6a b2×
b 3a2
=3 ab
xii.9x2
6y2×
y3 x3=3y
2x
xiii
.6pq
4rs
×8s2 3p
=4qs r
xiv
.6ab c×
ad 2b×8cd2
4bc
=6a2
bd3 c
xv.2z2
3ac2×6a2
5zy2×10c3
3y3
=8acx3y5
xv
i.ab2
bc2÷
a2
bc3
=cb2
ax
vii.6ab5c
d÷4a2
7bd
=21b2
10ac
xv
iii.3pq
5rs
÷p2
15s2
=9qs
pr
LufthansaResource
TechnicalTraining
LufthansaTechnicalTraining ForTrainingPurposesOnlyMATHEMATICS
M1.1
ARITHMETIC
IRPART66
M1
AlJ
CW
MA
ug
us
t2
00
6P
ag
e:
69
From
Page48
Solvethefollowingequations:
i.x
+3
=8
so
x=
5ii.
x--
4=
6s
ox
=1
0iii
.2
x=
8s
ox
=4
iv.
2x
--7
=9
so
x=
8v.
5x
+3
=1
8s
ox
=3
vi.
3x
--7
=x
--5
so
x=
1
vii.
9--
2x
=3
x+
7s
ox
=2
/5v
iii.
4x
--3
=6
x--
9s
ox
=3
ix.
5x
--8
=3
x+
2s
ox
=5
x.
2(x
+1
)=
9s
ox
=3
1/2
xi.
5(x
--3
)=
12
so
x=
52
/5x
ii.3
(2x
--1
)+
4(2
x+
5)
=4
0s
ox
=2
3/1
4
xiii
.7
(2--
3x
)=
3(5
x--
1)
so
x=
17
/36
xiv
.x 2+
x 3=10
so
x=
12
xv.3x+3 8=2+2x 3
=3956
xv
i.2x 5=
x 8+1 2
so
x=19 11
xv
ii.( x
+3)
2=( x
−3)
3s
ox
=-6
LufthansaResource
TechnicalTraining
TABLEOFFIGURES
LufthansaResource
TechnicalTraining
Pa
ge
i
ATA
1ARITHMETIC
2.................................
AD
DIT
ION
2.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
SU
BT
RA
CT
ION
3.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
MU
LTIP
LIC
AT
ION
4.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
DIV
ISIO
N5
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
CO
MM
ON
FR
AC
TIO
NS
9.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
DE
CIM
AL
S1
6.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
RA
TIO
&P
RO
PO
RT
ION
22
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
PO
WE
RS
AN
DR
OO
TS
26
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
TR
AN
SP
OS
ITIO
NO
FF
OR
MU
LA
E3
1.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
AR
EA
S3
2.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
VO
LU
ME
S3
3.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
CO
NV
ER
SIO
NS
35
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
TE
ST
37
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
ALGEBRA
38
...................................
US
EO
FS
YM
BO
LS
38
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
SU
BS
TIT
UT
ION
38
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
AD
DIT
ION
&S
UB
TR
AC
TIO
NO
FA
LG
EB
RA
ICT
ER
MS
40
..
..
..
..
..
.
MU
LTIP
LIC
AT
ION
&D
IVIS
ION
SIG
NS
40
..
..
..
..
..
..
..
..
..
..
..
..
..
MU
LTIP
LIC
AT
ION
&D
IVIS
ION
OF
AL
GE
BR
AIC
QU
AN
TIT
IES
41
..
..
.
BR
AC
KE
TS
43
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
AL
GE
BR
AIC
FR
AC
TIO
NS
45
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
LIN
EA
RE
QU
AT
ION
S4
8.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
GEOMETRY
50
..................................
CO
OR
DIN
AT
ES
&G
RA
PH
S5
0.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
AN
SW
ER
ST
OQ
UE
ST
ION
S5
5.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
TE
ST
64
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
