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2.2.1 Number Patterns and Sequences, PT3 Practice

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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.
    19     13     K      1      M     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

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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

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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

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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

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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)

Symbol of Element

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A symbol of element is the chemical symbol written in short form to represent a particular element. Some elements are represented by the first letter of its name. Examples:
Element Symbol
Fluorine F
Hydrogen H
Iodine I
Nitrogen N
Oxygen O
Phosphorus P
Sulphur S
Carbon C
Vanadium V
If there are two or more elements that have mane start with the same alphabet letter, a second letter is added to differentiate between these elements. The second letter used is always lowercase. Examples:
Elements Symbol
Bromine Br
Calcium Ca
Chlorine Cl
Chromium Cr
Magnesium Mg
Manganese Mn
Neon Ne
Nickel Ni
Silicon Si
Helium He
Argon Ar
Aluminium Al
Zinc Zn
Platinum Pt
Some elements are represented by their Latin names. Example:
Elements Latin Name Symbol
Copper Cuprum
Cu
Iron Ferrum
Fe
Lead Plumbum
Pb
Mercury Hydrargyrum
Hg
Potassium Kalium
K
Silver Argentum
Ag
Sodium Natrium
Na
Tin Stannum
Sn
(Notes: You MUST Memorise the symbol for all these 31 elements)

Brownian Motion

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Brownian Motion

  1. Brownian motion is the physical phenomenon that tiny particles immersed in a fluid move about randomly.
  2. A fluid can be a liquid or a gas.
  3. Brownian movement, an example of diffusion, supports the kinetic theory of matter.
  4. Examples of Brownian movement are
    1. movement of smoke particles in air
    2. movement of pollen grains in water

Diffusion in Gas

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Diffusion in Gas

(Diffusion in Gas)
Observation
The brown colour bromine vapour spreads evenly throughout the gas jar in a few minutes
  • Bromine vapour is made of tiny and discrete molecules that move randomly to fill up space.
  • Bromine vapour moves randomly and diffuses in all directions in air from areas of higher concentration to areas of lower concentration.
Conclusion : The rate of diffusion is highest in gas and lowest in solid.

Diffusion in Liquid

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Diffusion in Liquid

(Diffusion in Liquid)
Observation
The purple colour of potassium manganate(VII) fills up the entire test tube after a few hours
  • Diffusion has taken place in the liquid.
  • The rate of diffusion of the particles in water is faster than the diffusion rate of particles in solid.
  • The occurrence of diffusion proves that potassium permanganate(VII) consist of tiny and discrete particles.

Diffusion in Solid

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Diffusion in Solid

Diffusion in Solid
Observation
The blue colour of copper(II) sulphate fills up the entire test tube after a few days
  • Copper(II) sulphate crystals are made of copper(II) ions and sulphate ions which are tiny and discrete.
  • The particles in the copper(II) sulphate crystal will separate to become ions and diffuse randomly upwards until the whole agar turns blue.