Nota Ulangkaji SPM Matematik Tingkatan 4 dan Tingkatan 5

Tingkatan 4

Bab 1 Bentuk Piawai

1. Angka Bererti
2. Bentuk Piawai
3. Soalan Model SPM
1. Soalan Pendek (Kertas 1)

Bab 2 Ungkapan dan Persamaan Kuadratik

5. Soalan Model SPM
1. Soalan Panjang (Kertas 2)

Bab 3 Set

1. Set
2. Subset, Set Semesta, dan Set Pelengkap
3. Operasi ke atas Set
4. Soalan Model SPM
1. Soalan Pendek (Kertas 1)
2. Soalan Panjang (Kertas 2)

Bab 5 Garis Lurus

1. Kecerunan Garis Lurus
2. Kecerunan Garis Lurus dalam Sistem Koordinat Cartesan
3. Pintasan
4. Persamaan Garis Lurus
5. Garis Selari
6. Soalan Model SPM
1. Soalan Pendek (Kertas 1)
2. Soalan Panjang (Kertas 2)

Bab 6 Statistik III

1. Selang Kelas
2. Mod dan Min bagi Data Terkumpul
3. Histogram Selang Kelas Sama Saiz
4. Poligon Kekerapan
5. Kekerapan Longgokan
6. Sukatan Serakan
7. Soalan Model SPM
1. Soalan Panjang (Kertas 2)

Bab 7 Kebarangkalian I

1. Ruang Sampel
2. Peristiwa
3. Kebarangkalian suatu Peristiwa
4. Soalan Model SPM
1. Soalan pendek (Kertas 1)

Bab 8 Bulatan III

2. Sudut di antara Tangen dengan Perentas
3. Tangen Sepunya
4. Soalan Model SPM
1. Soalan pendek (Kertas 1)

Bab 9 Trigonometri II

1. Nilai Sin θ, Kos θ, dan Tan θ (0o ≤ θ ≤ 360o)
2. Graf Sinus, Kosinus, dan Tangen
3. Soalan Model SPM
1. Soalan Pendek (Kertas 1)

Bab 10 Sudut Dongakan dan Sudut Tunduk

1. Sudut Dongakan dan Sudut Tunduk
2. Menyelesaikan Masalah yang Melibatkan Sudut Dongakan dan Sudut Tunduk
3. Soalan Model SPM
1. Soalan Pendek (Kertas 1)

Bab 11 Garis dan Satah dalam Tiga Dimensi

1. Sudut di antara Garis dengan Satah
2. Sudut di antara Dua Satah

Tingkatan 5

Bab 1 Asas Nombor

1. Nombor dalam Asas Dua, Asas Lapan, dan Asas Lima
2. Soalan Model SPM
1. Soalan Pendek (Kertas 1)

Bab 2 Garf Fungsi II

1. Graf bagi beberapa Fungsi (Bahagian 1)
2. Graf bagi beberapa Fungsi (Bahagian 2)
3. Penyelesaian Persamaan dengan Kaedah Graf
4. Rantau Ketaksamaan dalam Dua Pemboleh Ubah
5. Soalan Model SPM
1. Soalan Pendek (Kertas 1)
2. Soalan Panjang (Kertas 2)

Bab 5 Ubahan

1. Ubahan Langsung
2. Ubahan Songsang
3. Ubahan Tercantum
4. Soalan Model SPM
1. Soalan Pendek (Kertas 1)

Bab 6 Kecerunan dan Luas di bawah Graf

1. Kuantiti yang diwakili oleh Kecerunan Graf
2. Kuantiti yang diwakili oleh Luas di Bawah Graf
3. Soalan Model SPM
1. Soalan Panjang (Kertas 2)

Bab 7 Kebarangkalian II

1. Kebarangkalian Suatu Peristiwa
2. Kebarangkalian Peristiwa Pelengkap
3. Kebarangkalian Peristiwa Bergabung
4. Soalan Model SPM
1. Soalan Pendek (Kertas 1)
2. Soalan Panjang (Kertas 2)

1. Bearing

Bab 9 Bumi sebagai Sfera

1. Longitud
2. Latitud
3. Kedudukan Tempat
5. Soalan Model SPM
1. Soalan Pendek (Kertas 1)
2. Soalan Panjang (Kertas 2)

Bab 10 Pelan dan Dongakan

1. Unjuran Ortogon
2. Pelan dan Dongakan
3. Soalan Model SPM
1. Soalan Panjang (Kertas 2)

1.2.2 Angles and Lines II, PT3 Practice

Question 6:
In Diagram below, PQRST is a straight line. Find the value of x.

Solution:

Question 7:
In Diagram below, find the value of y.

Solution:

Question 8:
In Diagram below, PSR and QST are straight lines.
Find the value of x.

Solution:
$\begin{array}{l}\angle UST+\angle STV={180}^{o}\\ \angle UST={180}^{o}-{116}^{o}={64}^{o}\\ \angle PST=\angle QSR\\ {x}^{o}+\angle UST={135}^{o}\\ {x}^{o}+\angle UST={135}^{o}\\ {x}^{o}={71}^{o}\\ x=71\end{array}$

Question 9:
In Diagram below, PWV is a straight line.

(a) Which line is perpendicular to line PWV?
(b) State the value of  ∠ RWU.

Solution:
(a) SW

(b) ∠ RWU = 13o + 29o + 20o = 62o

Question 10:
In Diagram below, UVW is a straight line.

(a) Which line is parallel to line TU?
(b) State the value of  ∠ QVS.

Solution:
(a) QV

(b) ∠ QVS = 8o + 18o = 26o

12.2.2 Solid Geometry (II), PT3 Focus Practice

                         
12.2.2 Solid Geometry (II), PT3 Focus Practice
 
  Question 5:
  Sphere below has a surface area of 221.76 cm2.
 
 
 Calculate its radius.
                          $\left(\pi =\frac{22}{7}\right)$                        
 
  Solution:
  Surface area of sphere = 4πr2
                          $\begin{array}{l}4\pi {r}^{2}=221.76\\ 4×\frac{22}{7}×{r}^{2}=221.76\\ {r}^{2}=\frac{221.76×7}{4×22}\\ {r}^{2}=17.64\\ r=\sqrt{17.64}\\ r=4.2\text{cm}\end{array}$                      
 
 
 
                         
Question 6:
  Diagram below shows a right pyramid with a square base.
 
 Given the height of the pyramid is 4 cm.
  Calculate the total surface area, in cm2, of the right pyramid.
 
  Solution:
 h2 = 32 + 42
  = 9 + 16
  = 25
  h = √25   = 5 cm2
 
  Total surface area of the right pyramid
  = Base area + 4 (Area of triangle)
  = (6 × 6) + 4 × 4 (½ × 6 × 5)
  = 36 + 60
  = 96 cm2
 
 
 
         
Question 7:
 
Diagram below shows a prism.
 
 Draw to full scale, the net of the prism on the grid in the answer space. The grid has equal squares with sides of 1 unit.
 
 Solution:
 
 
 
 

8.2.2 Coordinates, PT3 Focus Practice

Question 6:
The point M (x, 4), is the midpoint of the line joining straight line Q (-2, -3) and R (14, y).
The value of x and y are

Solution:

$\begin{array}{l}x=\frac{-2+14}{2}\\ x=\frac{12}{2}\\ x=6\\ \\ 4=\frac{-3+y}{2}\\ 8=-3+y\\ y=11\end{array}$

Question 7:
In diagram below, PQR is a right-angled triangle. The sides QR and PQ are parallel to the y-axis and the x-axis respectively. The length of QR = 6 units.

Given that M is the midpoint of PR, then the coordinates of M are

Solution:
x-coordinate of R = 3
y-coordinate of R = 1 + 6 = 7
R = (3, 7)

Question 8:
Given points (–2, 8) and (10, 8), find the length of PQ.

Solution:

Question 9:
In diagram below, ABC is an isosceles triangle.

Find
(a) the value of k,
(b) the length of BC.

Solution:

Question 10:
Diagram below shows a rhombus PQRS drawn on a Cartesian plane. PS is parallel to x-axis.

Given the perimeter of PQRS is 40 units, find the coordinates of point R.

Solution:

4.2.2 Linear Equations I, PT3 Practice

Question 6:
Solve each of the following equations:

Solution:
(a)
$\begin{array}{l}5a-16=a\\ 5a-a=16\\ 4a=16\\ a=4\end{array}$

(b)
$\begin{array}{l}6+\frac{2}{3}\left(-9p+12\right)=p\\ 6-6p+8=p\\ -6p-p=-8-6\\ -7p=-14\\ p=2\end{array}$

Question 7::
Solve each of the following equations:

Solution:
(a)
$\begin{array}{l}a-2=\frac{a}{5}\\ 5a-2=a\\ 4a=2\\ a=2\end{array}$

(b)
$\begin{array}{l}\frac{b-3}{4}=\frac{2+b}{5}\\ 5b-15=8+4b\\ 5b-4b=8+15\\ b=23\end{array}$

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Question 8:
Solve each of the following equations:

Solution:
(a)
$\begin{array}{l}x=-24-x\\ 2x=-24\\ x=-12\end{array}$

(b)

Question 9:
Solve each of the following equations:

Solution:
(a)
$\begin{array}{l}5p=8p-9\\ 5p-8p=-9\\ -3p=-9\\ p=3\end{array}$

(b)
$\begin{array}{l}3q=\frac{20-13q}{4}\\ 12q=20-13q\\ 12q+13q=20\\ 25q=20\\ q=\frac{4}{5}\end{array}$

2.2.2 Number Patterns and Sequences, PT3 Practice

               
Question 6:
 How many prime numbers are there between 10 and 40?
 

 Solution:
 
Prime numbers between 10 and 40
 
= 11, 13, 17, 19, 23, 29, 31, 37
There are 8 prime numbers between 10 and 40.
 
 
 
               
Question 7:
 
Diagram below shows a sequence of prime numbers.
               Find the value of x + y.
 
 Solution:
 
Value of x + y
 
= 19 + 23
 = 42

 
               
Question 8:
 
The following data shows a sequence of prime numbers in ascending order.
               Find the value of x, y and of z.
 
 Solution:
 
x = 11, y = 31, z = 37
 
 
 
       
Question 9:
 
The following data shows a sequence of prime numbers in ascending order.
           Find the value of p, q and of r.
 
 Solution:
 
p = 59, q = 71, r = 79
 
 

 
 
       
Question 10:
 
Diagram below shows several number cards.
            List three prime numbers from the diagram.
 
 Solution:
 
Prime number card from the diagram = 5, 29 and 97.
 
 

 
 

2.2.1 Number Patterns and Sequences, PT3 Practice

Question 1:
Diagram below is part of a number line.
What is the value of P and of Q?

Solution:
Each gradation on the number line represents 4 units. Thus P = –12 and Q = 12.

Question 2:
Diagram below shows a sequence of numbers. K and M represent two numbers.
What are the values of K and M?

Solution:

K
= 13 – 6 = 7
M = 1 – 6 = –5

Question 3:
Diagram below shows five integers.

Find the sum of the largest integer and the smallest integer.

Solution:
The largest integer = 1
The smallest integer = –10
Sum of the largest integer and the smallest integer
= 1 + (–10)
= –9

1.2.3 Whole Numbers, PT3 Practice

      
Question 11:
 Britney rents a stall to sell cakes. The rental of the stall is RM500 per month. The cost for baking a cake is RM5 per box and she sells it at RM10 per box. What is the minimum box of cakes must Britney sell each month to make a profit?
 
 Solution:
 
Rental of the stall = RM500 monthly
 Cost of baking a cake per box = RM5
 Selling price per box = RM10
 Profit per box
 = RM10 – RM5
 = RM5
 
 The number of boxes of cakes to accommodate the cost
 = RM500 ÷ RM5
 = 100
 
 Hence, the minimum box of cakes must Britney sell each month to make a profit = 101
 
 
 
 

1.2.2 Whole Numbers, PT3 Practice

Question 6:
Solve 4 (8 + 14)(37 – 18) + 46

Solution:
4 (8 + 14)(37 – 18) + 46
= 4(22)(19) + 46
= 4 × 22 × 19 + 46
= 1672 + 46
= 1718

Question 7:
Calculate the value of
113 + 6(47 – 4 × 9)

Solution:
113 + 6(47 – 4 × 9)
= 113 + 6[47 – (4 × 9)]
= 113 + 6(47 – 36)
= 113 + 6(11)
= 113 + 66
= 179

Question 8:
Solve 128 + (9 × 4) – 318 ÷ 6

Solution:
128 + (9 × 4) – 318 ÷ 6
= 128 + 36 – 53
= 164 – 53
= 111

Question 9:
Calculate the value of
62 × 15 – 40 + 68 ÷ 17.

Solution:
(62 × 15) – 40 + (68 ÷ 17)
= 930 – 40 + 4
= 894

Question 10:
402 – ( ) = 78 – 6 × 8 ÷ 16.
Find the missing number in the box.

Solution:
402 – ( ) = 78 – 6 × 8 ÷ 16
402 – ( ) = 78 – 48 ÷ 16
402 – ( ) = 78 – 3
402 – ( ) = 75
( ) = 402 – 75
( ) = 327

1.2.1 Whole Numbers, PT3 Practice

     
Question 1:
 Solve 8 × 2 ÷ 4 + 12.
 

 Solution:
 
8 × 2 ÷ 4 + 12
 = 16 ÷ 4 + 12
 = 4 + 12
 = 16
 
 
 
      
Question 2:
 Solve 7(6 + 5) – 3(13 – 9).
 

 Solution:
 
7(6 + 5) – 3(13 – 9)
 = 7(11) – 3(4)
 = 77 – 12
 = 65
 
 
 
     
Question 3:
 Solve 64 + (136 – 87) × (28 ÷ 4).
 

 Solution:
 
64 + (136 – 87) × (28 ÷ 4)
 = 64 + 49 × 7
 = 64 + 343
 = 407
 
 
 
    
Question 4:
 Solve 84 ÷ 4 – 9 + 6 × 7
 

 Solution:
 
84 ÷ 4 – 9 + 6 × 7
 = 21 – 9 + 42
 = 12 + 42
 = 54
 
 
 
    
Question 5:
 Solve 92 + 13 × 5 – 42
 

 Solution:
 
92 + 13 × 5 – 42
 = 92 + (13 × 5) – 42
 = 92 + 65 – 42
 = 157 – 42
 = 115
 
 
 
 

Element and Compound

Matter can be divided into elements and  compounds.

Elements

1. An element is a substance that consists of only one type of atom.
2. Element can be either atoms or molecules.
Example:
(Both the iron and oxygen are element because they consist of only one type of atoms)
Compounds
1. A compound is a substance composed of molecules made up of atoms of two or more elements.
2. A compound is made up of either molecules or ions.
Example:
(Both the sodium chloride and carbon dioxide are compound because they consist of more than one type of atoms)