1. Review for Term 1 Examination
Form 1 Mathematics
Chapter 0 – Chapter 4

2. Reminder Term 1 Examination Syllabus (p.1-185)
Chapter 0: Basic Mathematics
Chapter 1: Directed Numbers
Chapter 2: Using Algebra to Solve Problems (I)
Chapter 3: Percentages (I)
Chapter 4: Using Algebra to Solve Problems (II)
Date of Term 1 Examination
(Wed)
8:30 am – 9:30 am (1 hour)
Room 104

3. Ronald Hui Tak Sun Secondary School
Mathematics
Ronald Hui
Tak Sun Secondary School

4. Numbers Numbers (Chapter 0.1, page 2) Natural Number (自然數)
1, 2, 3, 4, 5, …
Whole Number (完整數)
0, 1, 2, 3, 4, 5, …
Even Number (偶數，雙數)
0, 2, 4, 6, 8, …
Odd Number (奇數，單數)
1, 3, 5, 7, 9, …
Ronald HUI

5. Calculation Arithmetic Operations “+” – Addition “–” – Subtraction
Add… to…, Plus, Sum
“–” – Subtraction
Subtract… from…, Minus, Difference
“x” – Multiplication
Multiply… by…, Times, Product
“” – Division
… is divided by…, Over, Quotient
Ronald HUI

6. Calculation order Please follow the order:
Brackets ( ), [ ], { } (小中大括號)
Multiplication / division first, then addition / subtraction (先乘除後加減)
From left to right (由左至右)
Ronald HUI

7. Numbers Prime numbers (質數) Multiples (倍數) Factors (因子)
2, 3, 5, 7, 11, 13, 17, 19, 23, …
Multiples (倍數)
Multiples of 2: 2, 4, 6, 8, …
Multiples of 3: 3, 6, 9, 12, …
Factors (因子)
2 is a factor of 2, 4, 6, 8, …
3 is a factor of 3, 6, 9, 12, …
Ronald HUI

8. Index Notation Factors of 18: 1, 2, 3, 6, 9, 18 18 = 2 x 9 = 2 x 3 x 3
Prime factors of 18: 2, 3
18 = 2 x 9
= 2 x 3 x 3
18 = 2 x  Index Notation
32 = 3 x 3 (指數記數法)
3 is base 底
2 is index 指數
Ronald HUI

9. H.C.F.
Case 1
18 = 2 x 3 x 3 = 2 x 32
24 = 2 x 2 x 2 x 3 = 23 x 3
HCF: = 2 x 3
Case 2
700 = 22 x 52 x 7
720 = 24 x 32 x 5
HCF is 20 = 22 x 5
Ronald HUI

10. L.C.M.
Case 1
4 = 2 x 2 = 22
6 = 2 x 3 = 2 x 3
LCM:12 = 2 x 2 x 3 = 22 x 3
Case 2
28 = 22 x 7
30 = 2 x 3 x 5
LCM: 420 = 22 x 3 x 5 x 7
Ronald HUI

11. H.C.F. and L.C.M.
If the numbers did not have common prime factors, their HCF is 1 and their LCM is the product of them
20 = 22 x 5
63 = 32 x 7
Their HCF is 1
Their LCM is 20 x 63 = 1260
Ronald HUI

12. Types of fractions Types of Fractions
Proper fraction (真分數)
Improper fraction (假分數)
Mixed number (帶分數)
The following fractions are equal
Ronald HUI

13. Addition, Subtraction Use improper fractions
Change to same denominators
Do operations on numerators
LCM of , 2, is 12
Ronald HUI

14. Multiplication, Division
Use improper fractions
Use reciprocal for division
Multiply numerators and denominators separately
Simplest Form!
Ronald HUI

15. Tools and units Select suitable units Length: km / m / cm / mm
Time: hr / min / sec
Weight: kg / g / lb
Temperature: C / F
Volume: L / ml
Ronald HUI

16. Form 1 Mathematics Chapter 1
Directed Numbers
Form 1 Mathematics
Chapter 1

17. Revision on Directed Numbers
What does the negative mean?
Money in bank:
+$1,100 – $950
Students in class (36 students in class)
Mon: – 2; Tue: – 1; Wed: 0; Thu: – 3; Fri:0
World time:
Sydney: +2; Rome: – 6; London: – 8; New York: – 13
Stairs in the building (up is “+”):
Go up 3 steps: +3; Go down 4 steps: – 4

18. Directed numbers on a number line
How do we write & say this? + – -6 -2 4 7
> > >
This called “Descending Order”
< < <
This called “Ascending Order”

19. Rules to Remember (p.50) ( + ) ( + ) = ( + ) ( – ) ( – ) = ( + )
( + ) ( + ) = ( + ) ( – ) ( – ) = ( + )
( + ) ( – ) = ( – ) ( – ) ( + ) = ( – )
正正得正 負負得正
正負得負 負正得負

20. Addition +3 + (+9) = 3 + 9 = 12 –7 + (+12) = –7 + 12 = 5=
( + ) ( + ) = ( + )
( – ) ( – ) = ( + )
( + ) ( – ) = ( – )
( – ) ( + ) = ( – )
= 12
–7 + (+12)
= –7 + 12
= 5

21. Addition +3 + (– 9) = 3 – 9 = – 6 – 6 + (– 7) = – 6 – 7 = – 13
= 3 – 9
( + ) ( + ) = ( + )
( – ) ( – ) = ( + )
( + ) ( – ) = ( – )
( – ) ( + ) = ( – )
= – 6
– 6 + (– 7)
= – 6 – 7
= – 13

22. Subtraction 7 – (+12) = 7 – 12 = – 5 – 3 – (+3) = – 3 – 3 = – 6
= 7 – 12
( + ) ( + ) = ( + )
( – ) ( – ) = ( + )
( + ) ( – ) = ( – )
( – ) ( + ) = ( – )
= – 5
– 3 – (+3)
= – 3 – 3
= – 6

23. Subtraction 7 – (– 7) = 7 + 7 = 14 – 3 – (–3) = – 3 + 3 = 0=
( + ) ( + ) = ( + )
( – ) ( – ) = ( + )
( + ) ( – ) = ( – )
( – ) ( + ) = ( – )
= 14
– 3 – (–3)
= – 3 + 3
= 0

24. Addition and Subtraction
(+ A) + (+ B) = A + B
(+ A) + (– B) = A – B
(– A) + (+ B) = – A + B
(– A) + (– B) = – A – B
( + ) ( + ) = ( + )
( – ) ( – ) = ( + )
( + ) ( – ) = ( – )
( – ) ( + ) = ( – )
(+ A) – (+ B) = A – B
(+ A) – (– B) = A + B
(– A) – (+ B) = – A – B
(– A) – (– B) = – A + B

25. Multiplication (+ 7)  (+ 7) = 7  7 = 49 (– 8)  3 = – 8  3 = – 24
= 7  7
( + ) ( + ) = ( + )
( – ) ( – ) = ( + )
( + ) ( – ) = ( – )
( – ) ( + ) = ( – )
= 49
(– 8)  3
= – 8  3
= – 24
Same as (– 8) + (– 8) + (– 8) = (– 24)

26. Multiplication 4  (– 2) = – (4  2) = – 8 (– 8)  (– 7) = + (8  7)
= – (4  2)
( + ) ( + ) = ( + )
( – ) ( – ) = ( + )
( + ) ( – ) = ( – )
( – ) ( + ) = ( – )
= – 8
(– 8)  (– 7)
= + (8  7)
= 56
Note: (– 8)  (– 7) = (– 8) (– 7)

27. Division (+ 7)  (+ 7) = 7  7 = 1 (– 9)  3 = – 9  3 = – 3
= 7  7
( + ) ( + ) = ( + )
( – ) ( – ) = ( + )
( + ) ( – ) = ( – )
( – ) ( + ) = ( – )
= 1
(– 9)  3
= – 9  3
= – 3
Note: (– 9) = (– 3) + (– 3) + (– 3)

28. Division 4  (– 2) = – (4  2) = – 2 (– 12)  (– 6) = + (12  6) = 2
= – (4  2)
( + ) ( + ) = ( + )
( – ) ( – ) = ( + )
( + ) ( – ) = ( – )
( – ) ( + ) = ( – )
= – 2
(– 12)  (– 6)
= + (12  6)
= 2

29. Multiplication and Division
(+ A)  (+ B) = A  B
(+ A)  (– B) = – (A  B)
(– A)  (+ B) = – (A  B)
(– A)  (– B) = A  B
( + ) ( + ) = ( + )
( – ) ( – ) = ( + )
( + ) ( – ) = ( – )
( – ) ( + ) = ( – )
(+ A)  (+ B) = A  B
(+ A)  (– B) = – (A  B)
(– A)  (+ B) = – (A  B)
(– A)  (– B) = A  B

30. Using Algebra to Solve Problems
Form 1 Mathematics
Chapter 2 and Chapter 4

31. Algebraic Expression (代數式)
Pay attention to the followings:
Consider A as a variable (變數)
1  A = A
A + A + A = 3A
A  A  A = A3
3A  4A = 12A2
Ronald HUI

32. Like Terms and Unlike Terms
When we work on the followings, we should put together the like terms and then simplify!
For examples:
1. A +2B +3A +4B = A +3A +2B +4B
= 4A + 6B
2. A -2B –3A +4B = A -3A -2B +4B
= (A-3A) + (-2B+4B)
= (-2A) + (2B)
= -2A +2B
Ronald HUI

33. Equations If a = b Then a + c = b + c Addition of equality (等量相加)
X – 7 = 2
X = 2 +7
X = 9
X = 2 +7
If a = b
Then a + c = b + c
Ronald HUI

34. Equations If a = b Then a - c = b - c Subtraction of equality (等量相減)
X + 7 = 12
X = 12 -7
X = 5
X = 12 -7
If a = b
Then a - c = b - c
Ronald HUI

35. Equations If a = b Then ac = bc Multiplication of equality (等量相乘)
X  5 = 4
X 5  5 = 4  5
X = 20
X = 4  5
If a = b
Then ac = bc
Ronald HUI

36. Equations If a = b Then a  c = b  c (but c  0)
Division of equality (等量相除)
5X = 20
5X  5 = 20  5
X = 4
X = 20  5
If a = b
Then a  c = b  c
(but c  0)
Ronald HUI

37. Equations a(b + c) = ab + ac (a + b)c = ac + bc Distributive Law (分配律)
5 (X+2) = 20
5 (X) + 5 (2) = 20
5X+10 = 20
5X = 10
X = 2
a(b + c) = ab + ac
(a + b)c = ac + bc
Ronald HUI

38. Equations -a(b + c) = -ab - ac -a(b - c) = -ab + ac
Distributive Law (分配律)
-5 (X-2) = 20
(-5) (X) – (-5) (2) = 20
-5X+10 = 20
-5X = 10
X = -2
-a(b + c) = -ab - ac
-a(b - c) = -ab + ac
Ronald HUI

39. Equations If a  c = b  d Then ad = bc (but c0 and d0)
Cross Method (交义相乘)
If a  c = b  d
Then ad = bc
(but c0 and d0)
Ronald HUI

40. Forming Equations
Use a letter x to represent unknown number 設 x 為變數 (即想求的答案)
Follow the question and form an equation 根據問題，製造算式
Solve for x 算出 x 的值
Write answer in words (with units!) 寫出答案 (包括單位)
Ronald HUI

41. Forming Equations Page 99 Question 3:
If the sum of two consecutive natural numbers is 37, find the smaller number.
Step 1: Let the smaller number be x.
Step 2: Then, the larger number will be x + 1
So, x + (x + 1) = 37
Step 3: x + x + 1 = 37
2x = 36
x = 18
Step 4: Therefore, the smaller number is 18.
Checking: = 37!
Ronald HUI

42. Forming Equations Page 99 Question 5:
The number of candies Susan has is represented by the algebraic expression x, where x stands for the number of boxes of candies she buys, If Susan has 36 candies after the purchases, find the number of boxes of candies she buys.
Step 1: Given x is the number of boxes.
Step 2: Then, x = 36
Step 3: x = 30
x = 3
Step 4: Therefore, the number of boxes is 3.
or Susan buys 3 boxes of candies.
Ronald HUI

43. Forming Equations Page 99 Question 8:
There are three $100 notes and some $10 coins inside a bag. If there are altogether $460, how many $10 coins are there?
Step 1: Let there are x $10 coins.
Step 2: Then, 3 ($100) + x ($10) = $460
Step 3: x = 460
10x = 160
x = 16
Step 4:  There are 16 $10 coins.
Ronald HUI

44. Inequality (不等式) “>”: Greater than “<”: Less than
“”: Greater than or equal to (or Not less than)
A combination of “>” and “=”
“”: Less than or equal to (or Not greater than)
A combination of “<“ and “=”
Ronald HUI

45. Inequality (不等式)
P.166 Question 9: Add 5 to 4 times of a number p and the sum is greater than 33.
Add 5  “+5”
4 times of a number p  “4p”
The sum  “4p+5”
Then, the inequality is  “4p+5 > 33”
Ronald HUI

46. Inequality (不等式)
P.166 Question 11: When the sum of a number s and 3 is multiplied by 2, the product is smaller than -10.
The sum of s and 3  “s+3”
Multiplied by 2  “2”
The product  “(s+3) 2” or 2(s+3)
The inequality is  “2(s+3) < -10”
Write 3 numbers  by guessing!
Ronald HUI

47. Inequality (不等式)
P.166 Question 13a: Mike has 2 packs of $1.9 stamps, 1 pack of $2.4 stamps and y packs of $1.4 stamps. Each pack contains 10 pieces of stamps. Write an inequality…
2 packs of $1.9  2  $1.9  10 = $38
1 pack of $2.4  1  $2.4  10 = $24
y packs of $1.4  y  $1.4  10 = $14y
Total value  $38+$24+$14y
The inequality is  $38+$24+$14y <$100
or y+62 < 100
Ronald HUI

48. Inequality (不等式)
P.166 Question 13b: Find the value of y if the total value of the stamps that Mike has is $76.
Total value  $38+$24+$14y
The equation is  $38+$24+$14y = $76
or y+62 = 76
14y = 14
y = 1
Therefore, y = 1.
or “Mike has one pack of $1.4 stamps.”
Ronald HUI

49. Formula (方程式) A formula is an equation with e.g. A = (U + L)  H  2
At least 1 variables on the right of equal sign.
Only 1 variable (Subject) on the left of equal sign.
e.g. A = (U + L)  H  2
A is the subject of the formula
U, L and H are the variables of the formula
Do you know what this formula means?
Ronald HUI

50. Formula (方程式)
P.169 Question 18: v = u + gt (u = -8, g = 10, t = 3, v = ?)
v = u + gt
= (-8) + (10) (3)  Method of substitution =
= 22
Ronald HUI

51. Formula (方程式)
P.169 Question 20: In the formula S=88+0.5t , if S=120, find the value of t.
S = t
(120) = t
= 0.5t
32 = 0.5t
32  2 = t
64 = t
t = 64
Ronald HUI

52. Formula (方程式)
P.169 Question 22a: It is given that V=Ah3. If V = 40 and h = 6, find the value of A.
V = Ah  3
(40) = A (6)  3
40  3 = 6A
120 = 6A
120  6 = A
20 = A
A = 20
Ronald HUI

53. Formula (方程式)
P.169 Question 22b: It is given that V=Ah3. Find the value of h such that A = 5 and the value of V is half of that given in (a)
In (a), V = 40. So, in (b), V = 20.
V = Ah  3
(20) = (5) h  3
20  3 = 5h
60 = 5h
60  5 = h
h = 12
Ronald HUI

54. Sequence (數列) Consider a sequence: 2, 4, 6, 8, 10, …
Can you guess what are the next numbers?
What is the 10th term?
What is the 100th term?
What is the nth term?
1, 2, 3, 4, 5, …, 10, …, 100, …, n (第幾個)
2, 4, 6, 8, 10, …, 20, …, 200, …, 2n (答案)
Ronald HUI

55. Sequence (數列) A Sequence is a group of numbers with number pattern.
Each number in the sequence is called a Term.
The first one in the sequence is called the First Term.
In the sequence, 2, 4, 6, 8, 10, …
2 is the first term, 4 is the 2nd term, 6 is …
The general term (通項) is 2n or an=2n
Ronald HUI

56. Sequence (數列) Given a general term: an=2n-1 What is the first 5 terms?
The first 5 terms are:
a1 = 2(1)-1 = 1
a2 = 2(2)-1 = 3
a3 = 2(3)-1 = 5
a4 = 2(4)-1 = 7
a5 = 2(5)-1 = 9
What is the 17th term?
The 17th term is a17 = 2(17)-1 = 33
Ronald HUI

57. Sequences (數列)
P.176 Questions 8-10: Find the first 3 terms of the sequence:
8. an=n+4
a1=1+4=5; a2=2+4=6; a3=3+4=7
9. an=n/3
a1=1/3; a2=2/3; a3=3/3=1
10. an=20-3n
a1=20-3(1)=17; a2=20-3(2)=14; a3=20-3(3)=11
Ronald HUI

58. Sequences (數列)
P.176 Question 14: The general term of a sequence is 9(n+6). Find the first term, the 6th term and the 10th term of the sequence.
General term: an = 9(n+6)
First term = a1 = 9(1+6) = 9 (7) = 63
6th term = a6 = 9(6+6) = 9 (12) = 108
10th term = a10 = 9(10+6) = 9 (16) = 144
Ronald HUI

59. Sequences (數列)
P.176 Question 15: The general term an of a sequence is n(n+3). Find the first 5 terms and the 20th term of the sequence.
General term: an = (n)(n+3)
1st term = a1 = (1)(1+3) = 1(4) = 4
2nd term = a2 = (2)(2+3) = 2(5) = 10
3rd term = a3 = (3)(3+3) = 3(6) = 18
4th term = a4 = (4)(4+3) = 4(7) = 28
5th term = a5 = (5)(5+3) = 5(8) = 40
20th term = a20 = (20)(20+3) = 20(23)=460
Ronald HUI

60. Sequences (數列)
P.176 Question 17b: Consider the sequence 6, 12, 18, 24, 30, … Use an algebraic expression to represent the general term an of the sequence
n = 1, 2, 3, 4, 5, ……, n
an = 6, 12, 18, 24, 30, ……, ??
an = 6n
Ronald HUI

61. Functions (函數)
A function is an input-process-output relationship between numbers.
For each input, there is one (and only one) output.
Sequence an = 2n is a function.
Formula A = r2 is a function.
Ronald HUI

62. Form 1 Mathematics Chapter 3
Percentages (I)
Form 1 Mathematics
Chapter 3

63. What is Percentage? (p.110)
“%” is the symbol of percentage. Per cent means “per one hundred”.
e.g.

64. Simple Percentage Problems
P.116 Q2(c): 125% of $365.2
P.117 Q10(c): 132 g is 88% of t g
Remark: same units!

65. Simple Percentage Problems
P.118 Q boys and 700 girls took an examination. 15% of boys and 5% of girls failed. Find the total number of students passed.
Number of boys failed = 1100  15% = 165
Number of girls failed = 700  5% = 35
Number of students failed = = 200
Number of students = = 1800
Number of students passed = = 1600

66. Simple Percentage Problems
P.118 Q boys and 700 girls took an examination. 15% of boys and 5% of girls failed. Find the total number of students passed.
Number of boys passed = 1100  (1 – 15%) = 1100  (85%) = 935
Number of girls passed = 700  (1 – 5%) = 700  (95%) = 665
Number of students passed = = 1600

67. Percentage Increase (p.119)
New value > Original value
Last year, Andy was 150cm tall. His height is increased by 20% this year. His height now is

68. Percentage Decrease (p.122)
New value < Original value
The price of a car was $160,000 last month. If it is reduced by 15% this month, the new price

69. Percentage Change (p.126)
We can summarize the two formulae into one and call percentage change.
New value > Original value  + (increase)
New value < Original value  – (decrease)
The temperature was dropped from 30C to 27C in the evening. The percentage change is

70. Percentage Change
P.130 Q19. James had $5000 in his savings account last year. This year, he has $10000.
Change %
James then takes out 40% of his savings. The amount now is = $10000  (1 – 40%) = $6000
New change %

71. Profit (盈利, p.131)
Selling Price (售價) > Cost Price (成本價)

72. Loss (虧蝕, p.134)
Selling Price (售價) < Cost Price (成本價)

73. Profit and Loss
A shop sold a car at the profit % of 20%. If the profit was $32000, what was the cost of the car? What was the selling price?
Profit or Loss?
Let n be the Cost Price. Then,
So, the Selling Price is

74. Profit and Loss
P.138 Q22. A merchant bought a chair for $500 and a desk for $750. He sold the chair at a loss of 30% and the desk at a profit of 24%.
Selling price of chair
Selling price of desk
Total selling prices = $350 + $930 = $1280
Total cost prices = $500 + $750 = $1250
So, the profit % is

75. Discount (折扣, p.138) Western style vs. Chinese style 10% discount 9折

76. Discount
Marked Price > Selling Price

77. Discount
A pair of scorer shoes is sold at 25% discount at a marked price of $650. Can Vincent buy the shoes if he has $500?
Marked price = $650
Selling price of the shoes
Since Vincent got $500, he can buy the shoes.

78. Discount
A golden ring is sold at 30% discount with a selling price of $441.What is the marked price?
Selling price = $441.
Let n be Marked price. Then,
So, the Marked price is $630.

79. Discount
P.142 Q10. In shop A, the marked price of a MD is $1920 and the selling price is $1248. In shop B, the prices are $1760 and $1320 respectively.
A’s discount %
B’s discount %
So, B’s discount is smaller.

80. Simple Interest (單利息, p.144)
Amount(A) = Principal(P) + Interest(I)
Interest(I) = Principal(P)  Interest rate(r%)  Time(T)
Formula:

81. Simple Interest
If $40000 is deposited in a bank for 5 years at 3%p.a., find the simple interest and the amount.
P=$40000 r=3 T=5 (years)
Simple interest (I) is
So, the amount (A) is

82. Simple Interest
How long will $20000 amount to $30000 at 5%p.a. simple interest?
P=$20000 r=5 A=30000
I = $30000-$20000=$10000
Let T be the time. Then,
The time required is 10 years.

83. Simple Interest
P.151 Q14. Judy borrows $ for 10 years. Bank A’s interest is 5% p.a. and B is 4.5% p.a.
A’s interest
B’s interest
The difference is $50000.
She pays $50000 less in Bank B from in Bank A.

84. Ronald HUI
Good luck!!!