1WI
MRTEMflTIK (I4,+9)
FIIBII4RT INTBTIMEN PEPTHIKSHHN SPh,I MULflI Tru{UN
MHTH PETRJHBHN
l
!
f
RUMUS MATEMATIK
MATHEMATICAL FORMULAE
Rumus-rumus berikut boleh membantu anda menjawab soalan. Simbol-simbol yang diberi adalah yang biasa digunakan.
Thefollowingformulae mqt be helpful in answering the questions. The synbols given are the ones commonly used.
NOMBORDA}{ OPERASI
NUMBERSAND OPERATIONS
L a* x ail -
am+n
2 Q*+An-Am-t
3 (a*)n:
onn
n
.!4 a" : (a*)
s
Faedah mudah
6
Faedah kompaun
7
Jumlah bayarun balik
/
Simple interest,
/
I:
Prt
Compound interest,
/
MV:
Tbtal repayment, A
\t
*
t)"'
P+Prt
PERKAITAN DANALGEBRA
RE LATIONSIIIP AND AL GE B RA
Jarak / Distance
:''l @, -i$i(y, - r$
Titik tengah I Midpoint,
3
(x,
y): (t?,
Yt+ Yz\
2)
jarak
Laju purata: )umlah
Jumlah masa
Average speed
:
Total distance
Total time
lz- lt
xz- xt
m:m:
pintasan-y
pintffiat-x
v-interceot
__!_____L
x-intercePt
o':;;*(1:)
1l
SUKATAN DAl\[ GEOMETRI
MEAS URE MENT AND GE OME TRY
I
Teorem Pythagoras I \,thagoras Theorem,
2
Hasil tambah sudut pedalaman poligon I
: (n- 2) x 180'
3
Lilitan bulatan
: r.d:2nj
:
Circumference of circle = nd
4
U
-
:
2nj
Arc length :
2nr
Luas
sekor
1tj,
Luas lelayang
e
g
360"
:
e
360o
:
e
360'
:
+"hasil
L
Luas trapezium
darab panjang dua pepenjuru
x product of two diagonals
:
Area oftrapeziu*
9
_
360'
Area of sector
Area of kite =
"2
8
2nr
nrz
Panjanglengkok
TEr2
7
Sum of interior angles of a polygon
Luas bulatan: nj2
Area of circle
.
e : d + b2
Luas pernrukaan
+"hasil
:
tambah dua sisi selari x tinggi
L x sum of two parallel
sides
x height
silinder: 2nj2 + 2njt
:2nr2 * Znrh
Surface area ofcylinder
10
Luas pennukaan
kor:
Surface area ofcone
11
:
nj2 + njs
fir2
*
nrs
Luas permukaan sfera
:4fij'
Surface area of sphere
4nr2
:
12
Isi padu prisma: luas keratan rentas x tinggi
I/olume of prism : arca of cross section x height
13
Isi padu siltnder: njzt
Volume of cylinder
:
nr2h
ru
14
kon:!-xi?t
Isipadu
Tolume of cone
15
: Lnr'h
sfera:lnf
Isipadu
Volurne of sphere
16 Isi padu
:
*"r'
piramid:
+
Volume of pyramid:
r
+
luas tapak x tinggi
,. base area x height
t7 Faktorskala.k-PA'
,PA
Scalefactor,
l8
k:
Luas imej :
Area of image
PA'
PA
fri x luas objek
: H x area of object
STATISTIK DAN KEBARANGKALIAN
S TATIS
TI CS AND P ROBAB ILITY
tMinlMean.T:
.N *
\.r
Mrnl
Mean,T:t4
T
varians / variance, oz
:L(x.l,)2 :U- -*
NN
C :Z'frf-- D'
4
Varians
5
Sisihan piawai I standard deviation,
I
variance,
:Y --*
LTU
6:
6 Sisihan piawai / Standard deviation, o:
7
8
PU\:
W:
W
re--r
tr;
n(A)
r,(S)
P(A',):1-P(A)
lv
7 6\ RoJ=
,{s
BAHAGIANA
=_L
R
'^/s
I 2i -7x-15=0
2=k
t1[$
(2x+3)(x-s)=0
3_
x=__,x=5
k=-6
---!3r/s
R=
2 (a)
+
53618
647s
6
(b)
62308
- --:-=
'{s
3{s
(b) Ungkapkan dalam asas sepuluh dahulu:21016
Express in base ten
-3
=2
S=8
first:2l0ro
Kemudian ungkapkan kepada asas enam: 550u
(a) -3+7=4
Then, express in base six: 5505
(a) a(e
(b) x=2
-N=2a
y
K=3
-2=8
Y=10
61 lltz+
aX3) + 24 +
=24+ 24+
+02
7
+ 4)(6)
7
48
-2x=22
-2(2) =22
"=12
=96m
(a)
Kos perubatan selepas deduktibel
All(BUC)
Medical cost after deiluctible
=RNI22 000
= RM21 400
-
RM600
Kos yang ditanggung syarikat insurans
The cost borne by the insurance company
(b)
cu(AnB)'
75 ,RM2l
100
400
= RM16 0s0
Kos yang ditanggung Puan Amanda
The cost borne by Puan Arnanda
25 *RM21
= 100
400 + RM600
= RMs 950
l0 |umlah pendapatan
tahunan
Total annualincome
(a)
r
= RM117 600 - RMl4 640
= RMl02 960
= 1@, b), (a, b), (a, b), (a, c), (a, d), (b, c), (b, c), (c, d),
(d, d)I
n(E) =e
Pendapatan bercukai
Taxed income
(b) >d(E) =2(e)
= RM102 960 - RM21 000
= RM79 460
=18
BAHAGIANB
r1
(a)
Palsu
False
(b)
a-d<3-d
(c) 4n3 + n2
n = 1,2,3, ...
(d)
y+x-7=O
@ Global Mediastreet Sdn. Bhd. (762284-U)
(6-2)x180"
6
l2
-
RM2 500
(e)
likax *3, maka 3x + 4 +
Ifx+3,then3x+4*73.
L2
ls
Iikax = 3, maka 3x + 4= 13.
Ifx=3,then3x+4=73.
(a)
(i) Lokus X ialah bulatan dengan pusat O dan jejari 5 cm.
Locus
X
with center O and radius of 5 cm.
is a circle
(ii) Garis lurus
13.
BD lstraight line BD
(b)
(a) 23
(b) Qt--9'Qz=22
22-9
=13
\.
(c) Meanltwin,T
n
15.72
2#=sa23
Yariancelvariant=
)l
"...
BAHAGTANC
n' - (i)'
16 (a) (i)
x=
-5423 - 8.722
54.16
2+
1+3+5+5+11+10
= s.29
(d) 7.36
13
13
(ii) MinlMean=
18
=
x-2=tt
(b)
(a)
(b) Grafdilukis betul dengan skala seragam untuk0
<f<
10
dan0 < h<120.
Graph conecty drawn with uniform
scale
for 0 < t <
10 and
0</,<120.
ls n
Ri )u
't*
rs
(c)
(c)
(i)
(i,
la
2</<3.5;6.5<t<7.5
112.5<h<114.5
(a)
(i) (-4, 11)
(ii) (-s,7)
(b)
(r) Pembesaran dengan faktor skala 3 dan pus at A(2,
Enlargement with scale factor of 3 and centre at A(2, 3) .
(ii) btas ABCD|Area of ABCD
= 32 x Luas WXYZlAreaof WXYZ
=32 x 40
AI rd
ft*
l0-
t2
13-
15
5
2
10
20
2
5
t0
50
3
8
24
192
4
11
44
484
t2
t44
t2
1
f=
15
lfr = 100 >ff =seo
MinlMean=100 =o.oz
15
3)
.
Sisihan piawarl Standard deviation=
e;
= 3.8s
Yafiars I Vari an c e = 3.8 52
= 14.8
= 360
(ii)
Luas rantau berTorekl Area of shaded region
=360-40
= 320
@ Global Mediastreet Sdn. Bhd. (762284-U)
)tu
(i)
1-3
4-6
7 -9
246
Si
l3
Sisihan piawai Batrisyia adalah lebih daripada sisihan
piawai Anna. Data Batrisyia lebih terserak daripada
data Anna. Serakan bagi jumlah masa panggilan yang
dibuat oleh Batrisyia adalah lebih terserak daripada
serakan bagi junrlah masa panggilan yang dibuat oleh
Anna. (Lain-lain jawapan yang munasabah)
Isi p4du silinder I Volum e of cylinder
22
=_x
7
Batrbyials standard deliation is more than Anna\. Batrisyiai
ilata is more disperse than Annu's, The dispersion of total time
of voice calls by Bafiiryia is wtiler than Atna
(Any other reasonable answer)
Isi padu kon/Tolume of cone
122^)
=-x-x5'x/
37
76=y+I0
=66
(l + 10) +.1= too
Isi padu hemisferalVolume of henisphere
l=45
t422
=-x-x-x
237
x=55
NfalAlpha = 55
Lebihan isi padu lExcess volume = 198 - 66
=75.43
Luas permukaan kotakl Area of the box
=30x30
= 900
-
56.57
2xZxr=8.8
7
Luas bulatanlArea of circle
r=l.4cm
22 l5--)
=-x
/
Panjang kuboi d,l Length of cuboid
=1.4x2x4
ll.2
= 707.14
=
Luas segi tiga sarnalArea ofequilaural triangle
Lebar kuboid/Width of cuboid
=!xa0x25.98
=389.7
2
=1.4x2
kotak AlfalThe area ofremainingbox (Alpha)
=2.8
= 192.86
Panjang lengkokl Length
Luas lebihan kotakBeta/The area of remainingbox (Beta)
of arc
=!xe.s
4
= 510.3
(c) Alfa.
t'
= 56.57
Beta/Beta = 45
Luas lebihan
x,/
= 198
17 (a) x+y =100
(b)
^j
3-
[Penerangan yang boleh diterima]
Alp ha. [Reas onabl e explanation
-1''
]
Perimeter r.antau tidak berlor eklPerimeter of non-shaded region
= 2(11.2) + 2(2.8) + t6(2.2)
= 63.2 crn
(a) {(N, il), (N, S), (N; a), (a, ll),
(S, a), (a,N), (a,u),(a,S)l
BAHAGIANA
L y<6
"122=!6
5y>3x+5
y) -x+9
(b)
2 -2i-7x+4=0
(a)
2*+7x-4=0
=#"RMs000
x= | ,x= -4
2'
= RM500
Harga secawan teh/Price of a cup of tea = RMx
Harga segelas jus manggaiPrice of a glass of mango juice = RNIy
fumlah terkumpuUA ccumulated. amount
= RM500 x 12
= RM6 000
x+y- / ...(1'
4y-x=
18 ...@
(b) Aliran tunai bulanan En. Vicky
O*@
5Y
Mr. ViclE's monthly cashflow
= RMs 000 + RM750 - RM500
=25
/=5
(a) Laluan png muugkin dari
(1) BenailTrue
(ii)
- RMl200 - RMI
000
= RM3 0s0
x=2
a (a)
Simpanan bulanan untuk dana kecemasan
d
Mo nthly s avlngs fo r em ergen cy fun
(2x-t)(x+4)=o
3
(u, S), (u, a), (S, .tQ, (s, a),
Possible paths
Palsulra/se
A-G-F-E=Z+4+6=L2
A-G-C-E=2+4+7=13
A-G-C-F-E=2+4+5+6=17
A-G-F-C-E=2+4+5+7=18
(b) Implikasi l:likam3 = -64,ma.kam= -4
Implikasi2: Iikam= -4, maka m3 = -64
L; If m3 = -64, then m = -4
Implication 2: If m = -4, then m3 = -64
Implicatinn
Maka, A-G-F-E ialah laluan terpendek dengan jarak
12meten
(c) (9-2)x180"=1260n
@ Global Mediastreet Sdn.Bhd..
bucu A ke bucu E:
from vertex A to lrertex E
(7622U-U)
Thus, A-G-F-E is the shortat path with the distance 12 metres,
l4
r0
(a)
(a)
#=o+.s
60
t=140
(a) t(e) - 2m(-3)
=0
^=-
a
2
(b) (i) 30r + sjy = 2 79s
50x + 4OY
=2220
so\/x\ _ /2 leo\
(ii)
' ' /30
\s0 aol\r/ \2220)
|
lx\_
(iii)
\y/ - go(ao) x=18
/=33
140 -s0\/21e0\
so(so)
\-so
to )\zzzo )
2(18) + 3(33) = RMl3s
t4 (a) (:,6)
O) (i) (a)
V ialah pantulan pada garis lurus AB.
Y
is a reflection at straight kne AB.
(b) U ialah pembesaran pada F, dengan faktor skah
lJ
(ii)
90 =
is a enlargement ar
(I
\J/
)'*
E with
scal"
|orto, of
l.
fuas objeklAreaofobject
Ltas objeWArea
of objecr
= 8lO
Luas rantau berloreWArea
of shadedregion
=810-90
=720
3
15 tal
-2
(b) r=3
(c):=z(S)+c
--
1
l=2x-7
r=-7)
(d)
o-\?'/
/2.
(e)
"2
r\
a
BAHAGIANC
16
(a) fi\ 25 xx=100
100
Simpanan Ridhwan/Ridhwan\ savings = RM400
(c) (i) t1<l<
(ii) 1.6<r<
12 (a)
(b)
7o xx-1oo
12
100
1.8
Harga hadiah/
(b)
- 90 = 60 km
(c) Laiutspeed=rd*
(i)
22
lenama AlBrand e = L *
'27
*
3.52
x
x
3.52
x7
7
= 134.75
60
= _51.43 km
-51.43 km
= RM142.86
(iii) RM42.86
30minitlminutes
1s0
Presentls prtce
1
jr
= 89.83
h-1
@ Global Mediastreet Sdn. Bhd. (762284-U)
''2-
lenamaBlBranaA=i*12
ls
].r
(ii)
Jenama AlBrandA
=J!- * n+.ZS
= 101.06
lerama B I BranO, =
#x
BAHAGIANA
I (a) 10,
89.83
= 85.34
|enama A. Kerana kandungannya melebihi jenama B.
(b)
4226
(c)
724,
2 MV=4r*i)"'
Brand A. Because it has more volume than brand B.
p = nuiz ooo, r=
(c)
fenama A. Kerana kandungannya melebihi jenama B.
(atau lain-lain penjelasan yang munasabah)
3 (a)
|umlah bayarutT bahk/ Totat repayrnent
= 50 000 + (50 000 x 0.05 x 3)
= RM5T s00
o
+a+
(ii)
-: [l
(b) (i)
Ansuran bulananl Monthlv instalment
(ii)
57 500
3x12
=RML597.22
(b) (i) Jumlah petdapatanlTotalincome
Implikasi: fika k ialah gandaan bagi 8, maka k ialah
gandaan bagi 4.
Implication: If k is the multiple of 8, then k is the multiple of 4.
Songsangan: Jika ft bukan gandaan bagi 8, maka ft
bukan gandaan bagi 4.
Inverse: If k k not the multiple of 8, then k is not the multiple
of4.
(Anc)uB
(b)
= RM5 500
Simpanan tetap bulanan/ Monthly fxed savings
(a) Titik minimuml Minilnum point: (2,
10
= r00 x5s00
of symmetry,
x=2
(b)
fumlah perbelanjaarLl Total expenses
= RMI 597 + RM500 + RM100 + RM100 + RM200
=RMZ497
Baki pendapat
an I Incom e b alance
RMss0
-
f(x)=x2+4x+c
-s=(2)r+4(2)+c
c=-21
Min/Mean
RM300
V=
k - 2) + k + k + (sk4
Lebihan pendapatanl Surplus of income
=RM46s0 -PM2497
o=
= RM2 153
k2
or
a=
21, _
1
+k2+l(+(5kk=2
Puan Azah akan mengalami lebihan pendapatan
2t
"10
dalam pelan kewangannya, sebanyak RM2 153 (rujuk
pengiraan di atas)
Yes.
9
1
Puan Azah will experience surplus of iflcome in her financial
plan, because her surplus income is RM2 1 53 (refer to the above
calculations)
45
1\*/2
b\(-L,
'\10 el \10"r\e)
Ya. Puan Azah akan mencapai pelan kewangannya untuk
membeli televisyen pintar dalam masa tiga bulan. Kerana
lebihan pendapatannya adalah sebanyak RM2 153 setiap
bulan dan simpanan bulanannya adalah sebanyak RM550
setiap bulan.
2
45
aP
(-)
lain-lain s eb ab y ang munas ab ah)
Puan Azah will achieve her plan to buy a smart teleyision in a
period of three months.
Because her surplus income isRMz153 per month andher monthly
savings isRM550 per month.
(at au
- 3q= to
4p + 104 = -16
-t3q
Yes.
= 26
q= -2
P=I
Sudut sepadan rtjukanl Corresponding
= kos/cos-r 0.2588
= 75"
(or any other relevant reasons)
@ Global Medinstreet Sdn. Bhd. (762284-U)
2)
-zk=o
Ya.
(c)
-l)
Persamaan paksi simetri/Eqa ation of axis
= RM550
-
-o
(a) (PUR)nQ
=5000+500
= RM5 s00
= RM4 6s0
)
= RMl3 513.95
bayaranbalk/Total repayment
= s0 000 + (50 000 x 0.05 x 5)
= RM62 s00
_
3
o'04 \rzx:)
(a) (i) fumlah
(ii)
= o.on, n = 2. t =
MV = RMI2 oooit +
"''-'-"'-"""\''
2
Brand A. Because it has more volume thnn brand B.
(or any other reasonable explanation)
17
fr
l6
reference angle
[
[
- 75") atanl or (180'
= (180"
(iii) / 3 \2
+ 75")
lt/=
= 105" atailor25l"
Luas PQRS/Area of
ro
lqns
-
Luas PQRS/-Area o/PQRS = 22.5
ro
(a)e=#
$1{
14 (a) (0
27
s=T
(b)
2702\
sr=nT
/=*
s101s2025
1
(ii)
BAHAGIANB
tt (a) 4t + 2(-6) = 0
|
(b)
(0
/2 -3\
ot(2)_13xr1-s
-l-2 3 \
\s
3O saatlseconds
(ii) I20-44
7)
10-0
=8ms-2
-71
(iii)
/=5
1
Jarak/osra
nce
(c) 4x+3Y=Zg
2x + 5Y
(i
=25
rs (a)
3H)=ffi)
(;)=En
+nr(1
;)G")
@)
#6=-+
l=6
b)*=#=+
1
3=t(6)+c
3
2
y= 33
4-*-T
(c)
-: = f(8)
+c
c=-9
,
=
1* -9,
Pintasan-rix -intercEt Y = 0
?
o =:_nx _ 9
x=
13 (a)
PQ
L2
='{G1
="'h[i
PQ=12
(b) (i)
PantulanPadagatisx= 6
Reflection at the line x -- 6
(ii)
Pembesaran dengan faktor skah
Enlargement with
@ Globat Mediastreet Sdn. Bhd.
scale
faaor
I
(7622U-U)
]
(t)
y<**,
(n)
y> -x+2
<*t,r*_u
x=5,y=3
n
;(2s)(r00)
=1250m
x=3
(b)
I
Jarak/orcra /,ce =
pada pusat S(7, 2)
at centre S(7,2)
t7
=
+ a0)(120)
;(30
=4200m
Masa/rme (s)
BAHAGTANC
(b) Tidak, kerana julat nilai bagi Kedai Kek A adalah lebih
tinggi daripada Kedai Kek B. Hasil jualan kek di Kedai Kek
B adalah lebih baik berbanding Kedai Kek A kerana nilai
sisihan piawai bagi Kedai Kek B adalah lebih kecil daripada
Kedai KekA.
16 (a) (i) 1 :3
'25-
111)
x
=
x=
r
RMz 400
75
RM2 400
75
t5
No, because the range value of Bakery A is higher than Bakery B.
The sales reyenue of cakes in Bakery B is better than that ofBakery A
because the standard deviation value of Bakery B is smaller than that
= RM800
(b) (i) 28x + -x
24y = 3 200
of Bakery
A.
25x+27Y=3080
/28 24\/x\ /3200\
\zs zz)\y)=
\e oso]
/x\
\y)=
x=80,y=40
eB)eD
|
127 -24\t3200\
- er)er\-zs zs /\r oso/
(30)(30)(30)
=27 000
cm3
Alia perlu guna acuan kek berbnetuk silinder
yang
mempunyai isi padu 28 631.48 cm3 berbanding acuan kek
berbentukkubus dengan isipadu2T 000 cm3.
Alia needs to use a cylindrical cake mold that has a volume of
28 637.48 cm3 compared to cube-shaped one that has volume of
Tidak bersetuju, sebab harga yang ditawar oleh
A lebih murah pada harga RM240
pembekal
27
berbanding dengan pembekal B pada harga RM360.
Disagree because the price ofered by Supplier A is much cheaper
at RM240 comltared to Supplier B! RM360.
(c) Bilangan jubin lNumber of
tiles-
(20
x lgoX9-l
= 640
10Q) cm'z
(5u x 5u) cm'
Cukup kerana bilik cikgu hanya memerlukan 640 keping
jubin sahaja.
Enough because the teacher\ room only needs 640 pieces of tiles.
t7
(a)
\iatlRange=657
-
452
\iatlRange=655-453
= 205
MirtlMean
x_
10841
20
=202
MitlMean
n-
10 738
20
= 542.05
= s36.90
Sisihan piawai
Sisihan piawai
Standard deviation
Standard deviation
5 956 t29
20
@ Global Medinstreet Sdn. Bhd. (762284-U)
J8
OO0 cm3.