Hybrid PBD 2021 Form 2 - Matematik

Matematik Tingkatan 2 Bab 2
Pentaksiran Sumatif
PT3



PRAKTIS PdPR Jawapan
Bab 2 PRAKTIS PdPR Bab 2

Bahagian A 7. Tentukan sama ada setiap yang berikut adalah
“Betul” atau “Salah”.
Determine whether each of the following is “True” or “False”.
1. x(–2y – 5) = [4 markah/ 4 marks]
A 2xy – 5 C 2xy – 5x
B –2xy – 5 D –2xy – 5x Jawapan / Answer :
2
2. (8 – 5p)(4q – 3r) = (c + d)(c – d) = c – d 2 Betul / True
A 20pq – 15pr – 32q – 24r
B –20pq + 15pr + 32r + 32p (c – d)(c + d) = c + d 2 Salah / False
2
C 32q + 15pr – 24r – 20pq
D 32q + 24r – 20pq – 15pr (c + d) = c + 2cd – d 2 Salah / False
2
2
3. Permudahkan.
Simplify. (c – d) = c – 2cd + d 2 Betul / True
2
2
(2m − 3) + 2(1 + 3m)
2
A 2m − 5m + 9 C 4m − 6m + 11 8. Senaraikan faktor sepunya bagi sebutan algebra
2
2
B 2m + 5m − 9 D 4m − 6m − 11 berikut.
2
2
List the common factors of the following algebraic terms.
4. Faktorkan. [4 markah / 4 marks]
Factorise.
3a − 6a − 45 Jawapan/ Answer :
2
A 3(a − 5)(a + 3) C 3(a + 5)(a − 3) 3ab, 9bc 2
B 3(a + 5)(a + 3) D 3(a − 5)(a − 3)
5. Rajah di bawah menunjukkan sebuah segi tiga
PQR.
The diagram below shows a triangle PQR. 1 3 3b b
P
Bahagian C
2x cm
9. (a) Faktorkan 2s − 4s + 2.
2
Factorise 2s − 4s + 2.
2
Q (2x + 7) cm R
[2 markah / 2 marks]
Cari luas, dalam cm , segi tiga itu. Jawapan / Answer :
2
Find the area, in cm , of the triangle. 2s − 4s + 2
2
2
A 2x + 7 C 4x + 14 =2(s − 2s + 1)
2
2
2
2
2
B 2x + 7x D 4x + 14x =2(s − 1)(s − 1)
Bahagian B
(b) Faktorkan setiap ungkapan algebra yang
6. Padankan setiap yang berikut. berikut.
Match each of the following. Factorise each of the following algebraic expressions.
[4 markah/ 4 marks] (i) 16 − 25q 2
Jawapan / Answer : [2 markah / 2 marks]
(g + h) 2 (g – h)(g + h) Jawapan / Answer :
16 − 25q 2
g – h 2 (g – 1)(g + 1)
2
= 4 − 5 q
2
2 2
=(4 − 5q)(4 + 5q)
g – 1 (g – h)(g – h)
2
g – 2gh + h 2 g + 2gh + h 2
2
2
24
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Bab 2.indd 24 9/3/21 5:13 PM

Matematik Tingkatan 2 Bab 2
(ii) 20pq − 5pr + 12pq − 3pr (b) Expand (5 + y)(4 − y).
[2 markah / 2 marks] Kembangkan (5 + y)(4 − y).
[2 markah / 2 marks]
Jawapan / Answer : Jawapan / Answer :
20pq − 5pr + 12pq − 3pr (5 + y)(4 − y)
= 5p(4q − r) + 3p(4q − r) = 20 − 5y + 4y − y 2
= (4q − r)(5p + 3p) = 20 − y − y 2
= 8p(4q − r)



(h − 6) ( j − 4)
(c) Ungkapkan − sebagai satu (c) Rajah di bawah menunjukkan beberapa
2h 6j
pecahan tunggal dalam bentuk termudah. jenis barang yang dijual di sebuah kedai.
(h − 6) ( j − 4) The diagrams below shows a few types of items sold in
Express 2h − 6j as a single fraction in its simplest a shop.
form.
[4 markah / 4 marks]
Jawapan / Answer :
3hj − 18j – hj + 4h
=
6hj
2hj − 18j + 4h Kemeja Beg tangan Kasut
= Shirt Handbag Shoes
6hj
2(hj − 9j + 2h) m
= Diberi harga kemeja ialah daripada harga
6hj 5
hj − 9j + 2h beg tangan. Manakala harga beg tangan
= 5
3hj ialah daripada harga kasut. Jika harga
m + 2
kasut ialah RM(m − 4), berapakah harga
2
kemeja tersebut? KBAT Mengaplikasi
10. (a) Permudahkan setiap yang berikut. Given that the price of shirt is m of the price of handbag.
Simplify each of the following. 5 5
2
(i)  r − 1  2 − r While the price of handbag is 2 m + 2 of the price of shoes.
4 7 [2 markah / 2 marks] If the price of shoes is RM(m − 4), how much is the price
of shirt?
Jawapan / Answer : [4 markah / 4 marks]
2

= r − 1  r − 1  − r Tip KBATKBAT
KBAT
7
4
4
1 1 2 Tulis hubungan antara harga bagi setiap barang.
2
= r − r + − r Write the relationship between the prices of each item.
2 16 7
11 1
= r − r + Jawapan / Answer :
2
14 16
Harga beg tangan (RM)
The price of handbag (RM)
5
= × (m − 4)
2
9m – 36 3(m + 2) m + 2
2
(ii) ÷
m – 4 m – 2 5
2
[2 markah / 2 marks] = m + 2 × (m − 2)(m + 2)
Jawapan / Answer : = 5(m − 2)
9m – 36 3(m + 2) Harga kemeja (RM)
2
÷
2
m – 4 m – 2 The price of shirt (RM)
9(m – 2)(m + 2) m – 2 m
= × = 5 × 5(m − 2)
(m – 2)(m + 2) 3(m + 2)
9(m – 2) = m(m − 2)
= = m − 2m
2
3(m + 2)
3(m – 2)
=
(m + 2)
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Bab 2.indd 25 9/3/21 5:13 PM

Rajah di sebelah menunjukkan sebidang tanah berbentuk segi empat tepat A 48 – 2x F (2x) m B
yang dijadikan sebuah ladang lembu. Panjang tanah tersebut ialah 56 m dan Kolam
lebarnya ialah 48 m. Modul HEBAT M17 (2x) m Pond
E
The diagram shows a rectangular land made as a cow farm. The length of the land is 56 m and the
width is 48 m. Ladang lembu
Cow farm
(a) Hitung luas, dalam m , ladang lembu itu.
2
Calculate the area, in m , of the cow farm. 56 – 2x
2
Jalan berturap
Luas ladang lembu/ Area of the cow farm Paved road
= Luas ABCD – Luas AFE – Luas CDE
Area of ABCD – Area of AFE – Area of CDE D C
 
= 56 × 48 –  1 × 2x × (48 – 2x) – 1 × 48 × (56 – 2x) 
2
2
= 2 688 – x(48 – 2x) – 24(56 – 2x)
= 2 688 – 48x + 2x – 1 344 + 48x (a) Luas ladang lembu
2
= (2x + 1 344) m 2 = luas segi empat tepat – luas
2
kolam – luas jalan berturap
Area of cow farm
(b) Pemilik tanah itu akan membina pagar sepanjang EC. Nyatakan panjang, = area of rectangle – area of pond –
dalam m, pagar yang diperlukan. area of paved road
The owner of the land will build a fence along EC. State the length, in m, of the fence needed. (b) Gunakan teorem Pythagoras
untuk mencari panjang EC.
EC = ED + DC 2 Use the Pythagoras theorem to find
2
2
the length of EC.
+ 48
(56 – 2
x
)
2
EC =  2 (c) Bilangan pagar yang
=  + 2 304 digunakan boleh ditentukan
3 136 – 224
+ 4
x
x
2
dengan jumlah perimeter
– 224
2
4
=  + 5 440 dibahagi jumlah panjang pagar
x
x
=  dan jaraknya.
+ 1 360
x
– 56
2
x
4(
The number of fences needed can
x
x
2
– 56
= (2 + 1 360) m be determined using the total
perimeter divided by the total fence
and the distance.
6
(c) Jika pemilik tanah itu ingin memagari ABCD dengan keadaan setiap panjang pagar itu ialah m dan berjarak
1 m antara satu sama lain, hitung bilangan pagar yang diperlukan. 5y
2y
If the owner wants to fence ABCD such that the length of the fence is 6 m and 1 m away from each other, calculate the number of the
fences needed. 5y 2y
Perimeter ABCD/ Perimeter of ABCD = 2(56) + 2(48)
= 208 m
Bilangan pagar yang diperlukan/ Number of fences needed
= 208 = 208
6
1
17
 5y + 2y   10y 
= 208 = 208 × 10y
1 × 5
6 × 2
 5y × 2 + 2y × 5  17
= 122 6 y
17
Kuiz 2
Praktis TIMSS/PISA
26
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Bab 2.indd 26 9/3/21 5:13 PM

BAB


3 Rumus Algebra



Algebraic Formulae




3.1 Rumus Algebra Nota Buku Teks
PBD
PBD
PBD
Algebraic Formulae Visual ms. 45 – 50
FOKUS TOPIK

Rumus algebra/ Algebraic formula • Ungkapan algebra – gabungan dua atau lebih sebutan algebra
Ungkapan algebra menggunakan operasi.
Algebraic expression Algebraic expression – combination of two or more algebraic terms using operations.
Perkara rumus a = 5b + c • Rumus algebra – ungkapan algebra yang ditulis dalam bentuk
Subject of formula persamaan dengan menghubungkan beberapa pemboleh ubah.
Pemboleh ubah/ Variables Algebraic formula – an algebraic expression which is written in the form of an equation
by relating few variables.
Tip Penting

Perkara rumus/ Subject of formula:
• Nilai pekali mesti bersamaan 1./ The coefficient must equal to 1.
• Diungkapkan dalam sebutan pemboleh ubah lain./ Expressed in terms of other variables.

1. Tulis satu rumus algebra berdasarkan situasi berikut. SP 3.1.1 TP2
Write an algebraic formula based on the following situations.

Hasil tambah bagi 6p dan 3q ialah S. Tulis satu rumus bagi S. S = 6p + 3q
The sum of 6p and 3q is S. Write a formula for S.

(a) Sofea membeli 40 kg beras dan menggunakan y kg beras itu setiap hari.
Selepas 2 minggu, baki beras yang tinggal ialah x kg. Tulis satu rumus bagi x. x = 40 – 14y
Sofea bought 40 kg of rice and used y kg of it every day. After 2 weeks, x kg of rice is left. Write a
formula for x.
(b) Kuasa tiga bagi suatu nombor P ialah Q. Tulis satu rumus bagi Q. Q = P 3
The cube of a number P is Q. Write a formula for Q.
(c) Sebuku roti berharga RM2.40 dan sebotol jem berharga RM7.50. Jian Wei
membayar RMJ untuk p buku roti dan q botol jem. Tulis satu rumus bagi J. J = 2.4p + 7.5q
A loaf of bread costs RM2.40 and a bottle of jam costs RM7.50. Jian Wei paid RMJ for p loaf of
bread and q bottle of jam. Write a formula for J.


2. Lengkapkan peta titi di bawah. SP 3.1.1 TP1 i-Think Peta Titi
i-Think
i-Think
Complete the bridge map below.
Perkara rumus P = mcθ m = y – 3
2
Subject A = 2πr + 2πrh t x + 3
seperti seperti
Faktor penghubung as as
Relating factor A P m
3. Tanda (✓) bagi pemboleh ubah dalam kurungan yang ditulis sebagai perkara rumus. SP 3.1.1 TP1
Mark (✓) for the variable in brackets that is written as subject of formula.
(a) S = 3a + 2b – 4c [S] (b) p = pq + 2q [p] 1 1 1
(c) + = [c]
a b c
✓




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Bab 3.indd 27 9/3/21 5:13 PM

Matematik Tingkatan 2 Bab 3
4. Ungkapkan pemboleh ubah dalam kurungan sebagai perkara rumus. SP 3.1.2 TP3
Express the variable in the brackets as the subject of the formula.


1
1
(i) p = 4x + 7 [x] (ii) 1 + = [g]
a 3 g
Kumpulkan perkara rumus yang
p = 4x + 7 dikehendaki pada sebelah kiri. 1 + = 1
1
4x + 7 = p Group the desired subject of formula on the left. a 3 g
1
4x + 7 – 7 = p – 7 Samakan pekali. 1 = + 1
4x ÷ 4 = (p – 7) ÷ 4 Equalise the denominators. g a 3
p – 7 1 3 + a
x = =
4 Lakukan pendaraban silang. g 3a
Perform the cross-multiplication.
g(3 + a) = 3a
Lakukan konsep kesamaan. 1 1
Perform the equality concept. g(3 + a) × = 3a ×
3 + a 3 + a
3a
g =
3 + a
2
(a) h = 8p + 7 [p] (b) x = z + 2y [y] (c) 3x = y – z [z]
2
2
8p + 7 = h x = z + 2y
2
2
8p + 7 – 7 = h – 7 z + 2y = x 2 y – z = 3x
2
2
8p = h – 7 z – z + 2y = x – z 2 y – – z = 3x – y
2
2
y
2
8p ÷ 8 = (h – 7) ÷ 8 2y ÷ 2 = (x – z ) ÷ 2 2 2 2
2
2
h – 7 x – z 2 1 1 y 2 1
2
p = y = –z × –1 = 3x – × –1
8 2 2
y
z = – 3x
2






4 – 3w 2(y – 2) (f) 9k = 3(5g – h) [h]
(d) J = [w] (e) = y [y]
w 3p
4 – 3w 2(y – 2) 3(5g – h) = 9k
J = = y
1
w 3p 3(5g – h) × = 9k × 1
4 – 3w 3 3
J × w = × w 2(y – 2) × 3p = y × 3p
w 3p 5g – h = 3k
Jw = 4 – 3w 2y – 4 – 3py + 4 = 3py – 3py + 4 5g – h – 5g = 3k – 5g
Jw + 3w = 4 – 3w + 3w 2y – 3py = 4 –h = 3k – 5g
Jw + 3w = 4 1 1
w(J + 3) = 4 y(2 – 3p) = 4 –h × –1 = (3k – 5g) × –1
1 1 y(2 – 3p) × 1 = 4 × 1
w(J + 3) × = 4 × 2 – 3p 2 – 3p h = –3k + 5g
J + 3 J + 3
4 y = 4
w = 2 – 3p
J + 3










28
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Bab 3.indd 28 9/3/21 5:13 PM

Matematik Tingkatan 2 Bab 3
5. Ungkapkan pemboleh ubah dalam kurungan sebagai perkara rumus. SP 3.1.2 TP3
Express the variable in the brackets as the subject of the formula.


(i) x y = 36 [x] (ii)  = h [m]
2
m
4
Tip Penting
x y = 36 4m = h Kuasa dua
2
1 1 2 Square
m
x y × = 36 × 1 2 = h 2
4
2
y y x x 2
36 4m = h 2
2
x = 1 1
y 4m × = h × Punca kuasa dua
2
Songsangan bagi kuasa 4 4 Square root
 dua bagi nilai itu. m = h Kuasa tiga
x
 = 36 dua ialah punca kuasa 2
2
y
The inverse of square is the
  square root of the value. 4 Songsangan bagi Cube
36
punca kuasa dua ialah
x = x x 3
y kuasa dua bagi nilai itu.
The inverse of a square root
6 is the square of the value. Punca kuasa tiga
x = Cube root
 y
(a) p = rq + 2sq 2 [q] k 1
2
(b) x = 5 y  [y] (c) V = s²h [s]
3
rq + 2sq = p
2
2
q (r + 2s) = p 1  y 2 2 2 1 s²h = V
2
k
3
1 1 5 = x
q (r + 2s) × = p × 1 3 3
2
r + 2s r + 2s 25k s²h × = V ×
p = x 2 3 h h
q = y
2
r + 2s 3V
²
s
25k 1 1  =
2
p × = x × h
2
 = y 25k 25k
q
r + 2s 3V
1 x 2 s =
p y = 25k h
q =
r + 2s 25k
y =
x 2
2
2
4 1 (f) p = h – 9 [h]
(d) V = pr 3 [r] (e) e =  [f ]
3 f
h – 9 = p 2
2
4 pr = V 1 h – 9 + 9 = p + 9
3
2
2
3  = e
f
2
2
4 pr × 3 = V × 3 2 h = p + 9
3

1
h
p
+ 9
2
3 4p 4p 1 2 = e 2  = 
2
3V f 2
+ 9
p
r = 1 h = 
3
4p f = e 2
 f = e 1 2
3 3 3V

r
 = 3
4p
3V

r = 3 4p
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Bab 3.indd 29 9/3/21 5:13 PM

Matematik Tingkatan 2 Bab 3
6. Cari nilai bagi setiap yang berikut. SP 3.1.3 TP3 E-pop Quiz
Find the value of each of the following.

(a) Diberi k = y – 3h , cari nilai k apabila y = 4 dan

Diberi a = b + c , cari nilai b apabila a = 13 dan h = –7.
2
2
2
c = 5. Given k = y – 3h , find the value of k when y = 4 and h = –7.

Given a = b + c , find the value of b when a = 13 and c = 5.
2
2
2

a = b + c 2 Gantikan nilai a dan c k = y – 3h
2
2


13 = b + 5 dalam persamaan. = 4 – 3(–7)
2
2
2
Substitute the values of a
and c into the equation.
169 = b + 25 = 4 + 21
2


b = 169 – 25
2

= 25

144
b =  Ungkapkan b sebagai perkara = 5
rumus dan selesaikan.
= 12 Express b as the subject of formula
and solve it.
1
1
2
(b) Diberi x = 3(2 – y ) , cari nilai x apabila y = –2 dan (c) Diberi + 1 = m , cari nilai k apabila h = 2 dan
z
k
h
z = –8. m = 3.
3(2 – y ) 1 1 1
2
Given x = , find the value of x when y = –2 and Given + = , find the value of k when h = 2 and m = 3.
z h k m
z = –8.
1 + 1 = 1
2
x = 3(2 – y ) h k m
z 1 + 1 = 1
2
= 3[2 – (–2) ] 2 k 3
–8 1 = 1 – 1
k
3
2
= 3(–2)
–8 1 = – 1
k
6
= 6
8 k = – 6
3
=
4

(d) Diberi 4 + p = t, cari nilai p apabila q = –2 dan (e) Diberi x = y² z, cari nilai z apabila x = 64 dan
5q
y = 2.
t = –1. Given x = y² z , find the value of z when x = 64 and y = 2.

4 + p
Given = t, find the value of p when q = –2 and t = –1.
5q x = y² z

4 + p = t 64 = 2² z

5q

4 z = 64
4 + p = –1
5(–2)  64
z =
4

4 + p = –1(–10) = 16
= 10
p = 10 – 4 z = 16 2
= 6 = 256
30
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Bab 3.indd 30 9/3/21 5:13 PM

Matematik Tingkatan 2 Bab 3
7. Selesaikan setiap yang berikut. SP 3.1.4
E-pop Quiz
Solve each of the following.

(a) Pak Rahman membeli x kg ikan dan y kg udang (b) Hazimah berumur k tahun dan adik lelakinya
masing-masing berharga RM2.50 per kg dan berumur 9 tahun lebih muda daripadanya.
RM3 per kg. Dia dapat menjual semula semua Jumlah umur mereka ialah T. TP4
ikan pada harga RM4 per kg dan udang pada Hazimah's age is k years old and her brother's age is 9 years
harga RM5 per kg. Bentukkan satu rumus untuk younger than her. Their total age is T.
jumlah keuntungan, p daripada semua jualannya. (i) Tulis satu rumus algebra yang mewakili
Pak Rahman bought x kg of fishes and y kg of prawns at jumlah umur mereka.
RM2.50 per kg and RM3 per kg respectively. He managed to Write and algebraic formula to represent their total age.
sell all the fishes at RM4 per kg and prawns at RM5 per kg.
Form a formula for the total profit, p, from the sales. TP4 T = k + (k – 9)
= 2k – 9
Keuntungan = Hasil jualan – Jumlah kos
Profit = Total sales – Total cost (ii) Hitung umur Hazimah jika jumlah umur
p = (4x + 5y) – (2.5x + 3y) mereka ialah 37.
= 4x + 5y – 2.5x – 3y Calculate Hazimah's age if their total age is 37.
= 1.5x + 2y T = 2k – 9
37 = 2k – 9
2k = 37 + 9
= 46
k = 23
p = (4 – 2.5)x + (5 – 3)y
= 1.5x + 2y




(c) Rajah di sebelah menunjukkan sebidang P Q R
tanah PRTU yang berbentuk segi empat
tepat. Kawasan segi tiga bersudut tegak QRS
digunakan untuk menanam pokok-pokok
bunga. Kawasan selebihnya diliputi dengan x m S
batu kerikil. Diberi nisbah panjang PQ kepada
panjang QR ialah 1 : 2, panjang RS dan ST
adalah sama dan luas kawasan yang diliputi
oleh batu kerikil ialah L m . U y m T
2
The diagram shows a rectangular plot of land PRTU. The right-angled triangular region QRS is used for planting flowers. The
remaining region is covered with pebbles. Given the ratio of the length of PQ to the length of QR is 1 : 2, the lengths of RS and ST are
the same and the area of the region covered with pebbles is L m . TP4
2
(i) Ungkapkan L dalam sebutan x dan y.
Express L in terms of x and y.
(ii) Cari nilai y apabila L = 20 dan x = 3.
Find the value of y when L = 20 and x = 3.

2
5
(i) QR = L = Luas PRTU – Luas QRS (ii) L = xy
PR 3 Area of PRTU – Area of QRS 6
2
5
QR = PR 1 2y x 20 = (3)y
1 21 2
3 = xy – 2 3 2 6
2
5
= y xy 20 = y
3 = xy –
6 2
5
= xy y = 2 × 20
6 5
= 8





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Bab 3.indd 31 9/3/21 5:13 PM

Matematik Tingkatan 2 Bab 3
(d) Jadual di bawah menunjukkan harga sewa kereta mengikut jam. TP5 KBAT Menganalisis
The table below shows the prices of car rental according to hour.
Jam/ Hour Harga per jam (RM) / Price per hour (RM)
5 jam pertama / First 5 hours 22
Jam seterusnya / The next hours 18

(i) Jiha ingin menyewa kereta antara 7 jam sehingga 10 jam. Diberi jumlah bayaran yang dikenakan
kepada Jiha ialah RMK dan dia telah menyewa kereta selama n jam. Ungkapkan K dalam sebutan n.
Jiha wants to rent a car between 7 hours to 10 hours. Given that the total payment charged to Jiha is RMK and she has rented
the car for n hours. Express K in terms of n.
Jumlah bayaran = Bayaran 5 jam pertama + Bayaran baki jam seterusnya
Total payment = Payment for the first 5 hours + Payment for the remaining hours
K = 22 × 5 + 18(n – 5)
= 110 + 18(n – 5)


(ii) Jika Jiha telah menyewa selama 9 jam, hitung jumlah bayaran sewa kereta.
If Jiha has rented for 9 hours, calculate the total payment of the car rental.
K = 110 + 18(n – 5)
= 110 + 18(9 – 5)
= 110 + 18(4)
= 182
Maka, jumlah bayaran sewa kereta ialah RM182.
Thus, the total payment of the car rental is RM182.

(e) Jumlah markah Lim, Aina dan Siva dalam suatu ujian Matematik ialah J. Markah mereka merupakan
tiga nombor ganjil yang berturutan. TP6 KBAT Mengaplikasi
The total marks of Lim, Aina and Siva in a Mathematics test is J. Their marks are three consecutive odd numbers.

(i) Tulis satu rumus bagi jumlah markah tiga orang murid itu dalam sebutan n.
Write a formula for the total marks of the three students in terms of n.
Katakan nombor ganjil yang pertama = n
Let the first odd number = n
Tiga markah itu:
The three marks:
n, n + 2, n + 4
J = n + (n + 2) + (n + 4)
J = 3n + 6

(ii) Tentukan julat markah tiga orang murid itu.
Determine the range of the marks of the three students.
Julat / Range Julat = Markah tertinggi – Markah terendah
= (n + 4) – n Range = Highest mark – Lowest mark
= 4


(iii) Cari jumlah markah tiga orang murid itu apabila markah tertinggi antara mereka ialah 89.
Find the total marks of the three students when the highest mark among them is 89.

Apabila / When J = 3(85) + 6
n + 4 = 89 = 261
n = 85


Kriteria Kejayaan: ............................................................................................ .
Saya berjaya
• membentuk rumus berdasarkan suatu situasi.
• menukar perkara rumus bagi suatu persamaan algebra.
• menentukan nilai suatu pemboleh ubah apabila nilai pemboleh ubah lain diberi.
• menyelesaikan masalah yang melibatkan rumus.

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Bab 3.indd 32 9/3/21 5:13 PM

Matematik Tingkatan 2 Bab 3

8. Lakukan projek di bawah. TP6
Carry out the project below.
Project-Based Learning
Pr ojek
Projek

Objektif aktiviti: Meningkatkan pengetahuan dan pemahaman murid mengenai penggunaan
Activity objective: konsep rumus algebra dalam kehidupan seharian.
Improve knowledge and understanding of the usage of the concept of algebraic formulae in daily
life.

Pernyataan masalah: Bagaimanakah ukuran dan isi padu bagi suatu gabungan pepejal geometri
Problem statement: boleh ditentukan?
How can measurements and volume of a composite solid geometry be determined?

Pencarian fakta: Menentukan isi padu pepejal geometri dengan menggunakan kaedah
Fact finding: saintifik dan rumus algebra.
Determine the volume of a solid geometry by using scientific method and algebraic formula.

Konsep yang Kaedah sesaran air dan isi padu pepejal geometri.
diaplikasikan: Water displacement method and volume of solid geometry.
Concepts applied:

Pelan tindakan / Action plan :
(a) Kerja secara berpasukan iaitu seramai dua atau tiga orang murid sepasukan. Setiap pasukan perlu
mencari atau membawa sebarang gabungan pepejal geometri ke kelas untuk mencari isi padu dan
ukuran pepejal tersebut.
Work as a team which consists of two or three students in a team. Each team needs to search or bring any composite solid to the
class to find the volume and measurement of the solid.
(b) Setiap pasukan perlu mencari isi padu pepejal geometri yang dipilih dengan menggunakan kaedah
sesaran air.
Each team has to find the volume of the chosen solid geometry by using water displacement method.
(c) Catat isi padu pepejal geometri dan ukur sisi/jejari bagi pepejal tersebut.
Record the volume of the solid geometry and measure its sides/radius.

Penyelesaian: Imbas kod QR (Isi Padu) untuk merujuk rumus-rumus isi padu pepejal
Solution: geometri yang berkaitan. Ganti nilai-nilai di dalam rumus tersebut dan
bandingkan dengan jawapan anda di (c).
Scan the QR code (Volume) to refer the formulae of volume of the related solid geometry. Substitute
the values in the formula and compare to your answer in (c).

Pembentangan: Bentang hasil dapatan dengan menggunakan Microsoft Power Point.
Presentation: Present the findings using Microsoft Power Point.


Kaedah Sesaran Air Isi Padu
Water Displacement Volume
Method
VIDEO INFO


















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Bab 3.indd 33 9/3/21 5:13 PM

Pentaksiran Sumatif
PT3




PRAKTIS PdPR Jawapan
Bab 3 PRAKTIS PdPR Bab 3

Bahagian A
Bahagian B

1. Nyatakan bilangan sebutan bagi
5xy + 6x − 3y − 7. 5. Ungkapkan m sebagai perkara rumus.
State the number of terms of 5xy + 6x − 3y − 7. Express m as the subject of the formula.
A 1 (i) 10 = 3m − 2n
B 2 [2 markah / 2 marks]
C 3 Jawapan / Answer :
D 4
10 + 2n
m =
3
1
2
2. Diberi V = pr h, maka r ialah n 2
3 (ii) =
1 3m 9
2
Given that V = pr h, then r is [2 markah / 2 marks]
3
A 3V − ph Jawapan / Answer :
3V
B  m = 3n
ph
2
C 1 3V 2 2 6. (a) Padankan ungkapan algebra berikut apabila
ph
V
D 1 3ph 2 2 n digunakan sebagai perkara rumus.
Match the following algebraic expressions when n is used
as the subject of the formula.
[2 markah / 2 marks]
Jawapan / Answer :
3. Diberi pq = 25t , ungkapkan t dalam sebutan p
2
dan q. 2 – m n = –2m
Given pq = 25t , express t in terms of p and q. n = p + 4 4 – p
2
A t = pq − 25
B t = pq + 25
pq 2 – m
C t = n =
5 4n – pn = –2m (p + 4) 2
pq
D t =  
5 (b) Diberi 7r − 3 = −2pq. Tandakan (✓) pada
pernyataan yang betul dan (✗) pada
pernyataan yang salah jika perkara rumus
2
4. Diberi ab c = 2a + 5c. Cari nilai c apabila a = −5 adalah p dan r.
2
dan b = 7. Given that 7r − 3 = −2pq. Mark (✓) against the correct
2
Given ab c = 2a + 5c. Find the value of c when a = −5 and b = 7. statement and (✗) against the incorrect statement if the
2
A 5 subjects of the formula are p and r.
1
B – [2 markah / 2 marks]
5
1 Jawapan / Answer :
C
5 (i)
D –5 p = 3 – 7r ✓
2q
(ii)
r = 2pq + 3 ✗
7



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Bab 3.indd 34 9/3/21 5:14 PM

Matematik Tingkatan 2 Bab 3
7. (a) Lengkapkan jadual di bawah. (ii) Jika nilai y ialah RM2, hitung nilai x.
Complete the following table. If the value of y is RM2, calculate the value of x.
[2 markah/ 2 marks] [2 markah / 2 marks]
Jawapan/ Answer : Jawapan / Answer :
16x + 6(2) = 36
Ungkapan Perkara rumus 16x + 12 = 36
algebra Subject of the formula 16x = 36 – 12
Algebraic expression
16x = 24
(i) k + 5 = H H x = RM1.50
2
(ii) L = 4pj 2 L

(b) Jamie mempunyai sekeping foto seperti
+
h
i
(b) Diberi   2 = g + 2. Tandakan (✓) ditunjukkan dalam rajah di bawah. Panjang
5 foto ini adalah 4.5 cm lebih daripada
bagi pernyataan yang betul dan (✗) bagi lebarnya. KBAT Mengaplikasi
pernyataan yang salah. Jamie has a photo as shown in the diagram below. The
  2 length of the photo is 4.5 cm longer than the width.
h + i
Given 5 = g + 2. Mark (✓) against the correct
statement and (✗) against the incorrect statement.
[2 markah/ 2 marks]
Jawapan/ Answer :
(i) y
h = 600, jika i = 5 dan g = 3
h = 600, if i = 5 and g = 3 ✓

(ii) g = 5, jika h = 20 dan i = 5
g = 5, if h = 20 and i = 5 ✗ (i) Bentukkan satu rumus algebra bagi
perimeter foto tersebut.
Form an algebraic formula for the perimeter of the
Bahagian c photo. [2 markah/ 2 marks]
Jawapan/ Answer :
8. (a) Jadual di bawah menunjukkan harga dua Perimeter foto
jenis buah-buahan. Perimeter of the photo
The table below shows the prices of two types of fruits. p = (2 × y) + 2(y + 4.5)
p = 4y + 9
Buah A Buah B
Fruit A Fruit B
RM8x / kg RM2y / kg
(ii) Berapakah luas foto jika perimeter, p
Edwin membeli 2 kg buah A dan 3 kg buah B ialah 69 cm?
dengan harga RM36. What is the area of the photo if the perimeter is
Edwin bought 2 kg of fruit A and 3 kg of fruit B for RM36. 69 cm?
[4 markah/ 4 marks]
(i) Berdasarkan situasi di atas, tulis suatu
ungkapan algebra yang mewakili Tip KBATKBAT
KBAT
jumlah harga buah-buahan yang dibeli Cari nilai y kemudian hitung luas gambar.
oleh Edwin. Find the value of y then calculate the area of the photo.
Based on the situation above, write an algebraic
expression that represents the total price of fruits Jawapan/ Answer :
bought by Edwin. p = 4y + 9
[2 markah / 2 marks]
Jawapan / Answer : 69 = 4y + 9
2(8x) + 3(2y) = 36 4y = 69 – 9
60
16x + 6y = 36 y = 4
y = 15 cm

Luas foto/ Area of photo = (15)(15 + 4.5)
= 292.5 cm
2




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Bab 3.indd 35 9/3/21 5:14 PM

Matematik Tingkatan 2 Bab 3
9. (a) (i) Permudahkan: (c) Selesaikan setiap yang berikut.
Simplify: Solve each of the following.
8s + 5 – 10s (i) Tiga orang murid Tingkatan 5 Delta
[1 markah/ 1 mark] menggunakan 1 l cat untuk mengecat
Jawapan/ Answer : anggaran 7 m mural di kantin sekolah.
2
–2s + 5 Jadual di bawah menunjukkan
hubungan di antara isi padu cat yang
digunakan dengan luas kawasan mural
yang dicat.
Three students from Form 5 Delta used 1 l of paint
(ii) Nyatakan bilangan sebutan bagi to paint approximately 7 m of mural at the school
2
12ab + 7 – y. canteen. The table shows the relationship between
State the number of terms of 12ab + 7 – y. the volume of paint used and the area of mural
[1 markah/ 1 mark] painted.
Jawapan/ Answer : Luas kawasan
3 Jumlah cat yang mural yang
digunakan, P(l) dicat, L (m )
2
Total paint used, P(l) Area of mural
2
painted, L (m )
1 7
7 – m 2 14
(b) Diberi = p.
2n 3 21
7 – m
Given 2n = p. 4 28
(i) Ungkapkan m dalam sebutan n dan p. 5 35
Express m in terms of n and p. Bentukkan satu rumus untuk
[2 markah/ 2 marks] mewakilkan hubungan itu.
Jawapan / Answer : Form a formula to represent the relationship.
7 – m = p [1 markah/ 1 mark]
2n Jawapan/ Answer :
7 – m = p 2 L = 7P
2n
7 – m = 2np 2 (ii) Perimeter sebuah dekagon sekata,
–m = 2np – 7 P, adalah sama dengan perimeter
2
m = 7 – 2np 2 sebuah pentagon sekata bersisi a cm.
Bentukkan rumus algebra bagi situasi
itu.
The perimeter of a regular decagon, P, is equal to
the perimeter of a regular pentagon with side a cm.
Form an algebraic formula for the situation.
[1 markah/ 1 mark]
Jawapan/ Answer :
(ii) Cari nilai m apabila n = –3 dan p = 2. P = 5a
Find the value of m if n = –3 and p = 2.
[2 markah/ 2 marks] (iii) Aina mendapat x markah dalam ujian
Jawapan / Answer : Matematik manakala Lisa mendapat
m = 7 – 2np 2 y markah lebih tinggi daripada Aina.
= 7 – 2(–3)(2) 2 Bentukkan rumus algebra bagi jumlah
= 7 – (–24) markah mereka, L.
= 31 Aina gets x marks for Mathematics test while Lisa
gets y marks higher than Aina. Form an algebraic
formula for their total marks, L.
[2 markah/ 2 marks]
Jawapan/ Answer :
L = x + (x + y)
= 2x + y




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Bab 3.indd 36 9/3/21 5:14 PM

Matematik Tingkatan 2 Bab 3










Sebanyak 300 keping tiket konsert akan dijual di Stadium Nasional Bukit Jalil. Rajah di
bawah menunjukkan harga tiket untuk orang dewasa dan kanak-kanak.
About 300 concert tickets will be sold at Stadium Nasional Bukit Jalil. The diagram below shows the prices of
adult ticket and child ticket. Gunakan satu pemboleh ubah
untuk mewakili bilangan tiket
kanak-kanak atau bilangan
KONSERT KONSERT tiket dewasa.
KONSERT
KONSERT
CONCER T CONCER T Use one variable to represent the
CONCERT
CONCERT
number of tickets for kids or the
Dewasa/
Kanak-kanak/
Dewasa/ TIKET/TICKET Kanak-kanak/ number of tickets for adult.
Child
Adult
Adult Child
TIKET/TICKET
RM12
RM10
TIKET/TICKET
TIKET/TICKET
RM12 RM10
(a) Dengan hanya mewakilkan satu pemboleh ubah, tulis satu rumus algebra untuk mewakili jumlah tiket, T yang
dapat dijual.
By only representing one variable, write an algebraic formula to represent the total tickets, T can be sold.
Katakan d = jumlah penonton dewasa atau Katakan k = jumlah penonton kanak-kanak
Let d = total adult spectators or Let k = total child spectators
T = 12d + 10(300 – d) T = 10k + 12(300 – k)

(b) (i) Hitung jumlah harga tiket konsert, P, dalam RM, yang dapat dijual jika bilangan tiket yang dapat dijual
untuk kanak-kanak ialah 35 keping.
Calculate the total price, P, of tickets sold for the concert, in RM, if the number of tickets sold for child is 35 tickets.
P = 10k + 12(300 – k)
= 10(35) + 12(300 – 35)
= 350 + 3 180
= 3 530



(ii) Oleh kerana sambutan hangat daripada orang ramai, semua tiket habis dijual dan pihak penganjur
menambahkan 100 tiket untuk dewasa dan 50 tiket untuk kanak-kanak. Hitung jumlah harga tiket, dalam
RM, jika semua tiket telah dijual.
Due to a very great response from public, all tickets are sold out and the organiser decided to add another additional 100 tickets for
adults and 50 tickets for child. Calculate the total price of tickets, in RM, if all tickets are sold.
Tiket asal kanak-kanak/ Original child tickets Tambahan tiket untuk dewasa
= 35 Additional tickets for adult
= 265 + 100
Maka, tiket asal dewasa/ Thus, original adult tickets = 365
= 300 –35
= 265 Jumlah tiket yang dijual
Total tickets sold
Tambahan tiket untuk kanak-kanak = 85(10) + 365(12)
Additional tickets for child = 850 + 4 380
= 35 + 50 = RM5 230
= 85

Kuiz 3


Praktis TIMSS/PISA




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Bab 3.indd 37 9/3/21 5:14 PM