8.2.2 Coordinates, PT3 Focus Practice

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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:

x= −2+14 2 x= 12 2 x=6 4= −3+y 2 8=−3+y y=11


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)

P( 1,1 ),R( 3,7 ) Coordinates of M =( 1+3 2 , 1+7 2 ) =( 2,4 )


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

Solution:
Length of PQ = [ 10−( −2 ) ] 2 + ( 8−8 ) 2 = ( 14 ) 2 +0 =14 units


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

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

Solution:
( a ) For an isosceles triangle,  y−coordinate of C is the midpoint of straight line AB. 2+k 2 =−3 2+k=−6   k=−8 ( b ) B=( −2,−8 ) BC= [ 10−( −2 ) ] 2 + [ −3−( −8 ) ] 2  = 12 2 + 5 2  =13 units


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:
All sides of rhombus have the same length, therefore length of each side= 40 4 =10 units PQ=10 ( 9− x 1 ) 2 + ( 7−( −1 ) ) 2 = 10 2 81−18 x 1 + x 1 2 +64=100 x 1 2 −18 x 1 +45=0 ( x 1 −3 )( x 1 −15 )=0 x 1 =3,15 x 1 =3 Q=( 3,−1 ),R=( x 2 ,−1 ) QR=10 ( x 2 −3 ) 2 + [ −1−( −1 ) ] 2 = 10 2 x 2 2 −6 x 2 +9+0=100 x 2 2 −6 x 2 −91=0 ( x 2 +7 )( x 2 −13 )=0 x 2 =−7,13 x 2 =13 ∴R=( 13,−1 )



4.2.2 Linear Equations I, PT3 Practice

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Question 6:
Solve each of the following equations:
(a)  5 a − 16 = a (b)  6 + 2 3 ( − 9 p + 12 ) = p


Solution:
(a)
5 a − 16 = a 5 a − a = 16 4 a = 16 a = 4

(b)
6 + 2 3 ( − 9 p + 12 ) = p 6 − 6 p + 8 = p − 6 p − p = − 8 − 6 − 7 p = − 14 p = 2


Question 7::
Solve each of the following equations:
(a)  a − 2 = a 5 (b)  b − 3 4 = 2 + b 5

Solution:
(a)
a − 2 = a 5 5 a − 2 = a 4 a = 2 a = 2

(b)
b − 3 4 = 2 + b 5 5 b − 15 = 8 + 4 b 5 b − 4 b = 8 + 15 b = 23

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Question 8:
Solve each of the following equations:
(a)  x = − 24 − x (b)  y + 5 4 ( 4 − 2 y ) = − 4

Solution:
(a)
x = − 24 − x 2 x = − 24 x = − 12

(b)
y + 5 4 ( 4 − 2 y ) = − 4 4 y + 20 − 10 y = − 16    ( × 4 ) − 6 y = − 16 − 20 − 6 y = − 36 y = 6


Question 9:
Solve each of the following equations:
(a)  5 p = 8 p − 9 (b)  3 q = 20 − 13 q 4

Solution:
(a)
5 p = 8 p − 9 5 p − 8 p = − 9 − 3 p = − 9 p = 3

(b)
3 q = 20 − 13 q 4 12 q = 20 − 13 q 12 q + 13 q = 20 25 q = 20 q = 4 5

2.2.2 Number Patterns and Sequences, PT3 Practice

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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.
13, 17, x, y, 29 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.
 5, 7, x, 13, 17, 19, 23, 29, y, z  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.
  p, 61, 67, q, 73, r, 83  Find the value of p, q and of r.

Solution:
p = 59, q = 71, r = 79



Question 10:
Diagram below shows several number cards.
    49     39     29         87     97     77          5      15     25     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

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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