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Review for Term 1 Examination
Form 1 Mathematics
Chapter 0 – Chapter 4
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
▫
▫
▫
12-12-12 (Wed)
8:30 am – 9:30 am (1 hour)
Room 104
Mathematics
Ronald Hui
Tak Sun Secondary School
Numbers

Numbers (Chapter 0.1, page 2)

Natural Number (自然數)


Whole Number (完整數)


0, 1, 2, 3, 4, 5, …
Even Number (偶數，雙數)


1, 2, 3, 4, 5, …
0, 2, 4, 6, 8, …
Odd Number (奇數，單數)

1, 3, 5, 7, 9, …
Ronald HUI
Calculation

Arithmetic Operations

“+” – Addition


“–” – Subtraction


Subtract… from…, Minus, Difference
“x” – Multiplication


Add… to…, Plus, Sum
Multiply… by…, Times, Product
“” – Division

… is divided by…, Over, Quotient
Ronald HUI
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
Numbers

Prime numbers (質數)


2, 3, 5, 7, 11, 13, 17, 19, 23, …
Multiples (倍數)
Multiples of 2:
 Multiples of 3:


2, 4, 6, 8, …
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
Index Notation

Factors of 18: 1, 2, 3, 6, 9, 18



Prime factors of 18: 2, 3
18 = 2 x 9
=2x3x3
18 = 2 x 32
 Index Notation
32 = 3 x 3
 3 is base 底
 2 is index 指數

Ronald HUI
(指數記數法)
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:
6=2x3


Case 2
700 = 22 x 52 x 7
 720 = 24 x 32 x 5
 HCF is 20 = 22 x 5

Ronald HUI
L.C.M.

Case 1
4 =2x2
= 22
 6 =2 x3
=2x3
 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
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
Types of fractions

Types of Fractions




4
7
Proper fraction (真分數)
Improper fraction (假分數)
Mixed number (帶分數)
7
4
3
1
4
The following fractions are equal
2 4
6 20



5 10 15 50
Ronald HUI
Addition, Subtraction



Use improper fractions
Change to same denominators
Do operations on numerators
2
1 7
2 5 7
2 
  
3
2 4
3 2 4
8 30 21
  
12 12 12
17

12
Ronald HUI
LCM of
3, 2, 4
is 12
Multiplication, Division



Use improper fractions
Use reciprocal for division
Multiply numerators and
denominators separately
2
1 7
2 5 4
2 
  
3
2 4
3 2 7
40 20


42 21
Ronald HUI
Simplest
Form!
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
Directed Numbers
Form 1 Mathematics
Chapter 1
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
Directed numbers on a number line
How do we write & say this?
–
+
-6
7
-2
> 0 > -2 > -6
-2 <
0 < 4 < 7
0
4
7
This called “Descending Order”
This called “Ascending Order”
Rules to Remember (p.50)
(+)(+)=(+)
(+)(–)=(–)
正正得正
正負得負
(–)(–)=(+)
(–)(+)=(–)
負負得正
負正得負
Addition
+3 + (+9)
= 3+9
= 12
–7 + (+12)
= –7 + 12
= 5
(+)(+)=(+)
(–)(–)=(+)
(+)(–)=(–)
(–)(+)=(–)
Addition
+3 + (– 9)
= 3–9
= –6
– 6 + (– 7)
= –6–7
= – 13
(+)(+)=(+)
(–)(–)=(+)
(+)(–)=(–)
(–)(+)=(–)
Subtraction
7 – (+12)
= 7 – 12
= –5
– 3 – (+3)
= –3–3
= –6
(+)(+)=(+)
(–)(–)=(+)
(+)(–)=(–)
(–)(+)=(–)
Subtraction
7 – (– 7)
= 7+7
= 14
– 3 – (–3)
= –3+3
= 0
(+)(+)=(+)
(–)(–)=(+)
(+)(–)=(–)
(–)(+)=(–)
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
Multiplication
(+ 7)  (+ 7)
= 77
= 49
(– 8)  3
= –83
= – 24
(+)(+)=(+)
(–)(–)=(+)
(+)(–)=(–)
(–)(+)=(–)
Same as (– 8) + (– 8) + (– 8) = (– 24)
Multiplication
4  (– 2)
= – (4  2)
= –8
(– 8)  (– 7)
= + (8  7)
= 56
Note: (– 8)  (– 7) = (– 8) (– 7)
(+)(+)=(+)
(–)(–)=(+)
(+)(–)=(–)
(–)(+)=(–)
Division
(+ 7)  (+ 7)
= 77
= 1
(– 9)  3
= –93
= –3
Note: (– 9) = (– 3) + (– 3) + (– 3)
(+)(+)=(+)
(–)(–)=(+)
(+)(–)=(–)
(–)(+)=(–)
Division
4  (– 2)
= – (4  2)
= –2
(– 12)  (– 6)
= + (12  6)
= 2
 12
12
Note :  12   6 
 2
6
6
(+)(+)=(+)
(–)(–)=(+)
(+)(–)=(–)
(–)(+)=(–)
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
(+)(+)=(+)
(–)(–)=(+)
(+)(–)=(–)
(–)(+)=(–)
Form 1 Mathematics
Chapter 2 and Chapter 4






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


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

Addition of equality (等量相加)
X–7=2
X -7 +7 = 2 +7
If
a
= b
X=9
Then a + c = b + c
X–7=2
X = 2 +7
X=9
Ronald HUI

Subtraction of equality (等量相減)
X + 7 = 12
X +7 - 7 = 12 -7
If
a
= b
X=5
Then a - c = b - c
X + 7 = 12
X = 12 -7
X=5
Ronald HUI

Multiplication of equality (等量相乘)
X5=4
X 5  5 = 4  5
If
a = b
X = 20
Then ac = bc
X5=4
X=45
X = 20
Ronald HUI

Division of equality (等量相除)
5X = 20
5X  5 = 20  5
If
a
= b
X=4
Then a  c = b  c
(but c  0)
5X = 20
X = 20  5
X=4
Ronald HUI

Distributive Law (分配律)
5 (X+2) = 20
5 (X) + 5 (2) = 20
a(b + c) = ab + ac
5X+10 = 20
(a + b)c = ac + bc
5X = 10
X=2
Ronald HUI
Distributive Law (分配律)
-5 (X-2) = 20
(-5) (X) – (-5) (2) = 20
-a(b + c) = -ab - ac
-5X+10 = 20 -a(b - c) = -ab + ac
-5X = 10
X = -2

Ronald HUI

Cross Method (交义相乘)
3x 7

4 5
5 3 x   7  4
15 x  28
28
x
15
If
a  c = b  d
Then
ad = bc
(but c0 and d0)
Ronald HUI




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

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:
18 + 19 = 37!
Ronald HUI

Page 99 Question 5:
◦ The number of candies Susan has is represented by
the algebraic expression 6 + 10x, 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, 6 + 10x = 36
◦ Step 3:
10x = 30
x=3
◦ Step 4: Therefore, the number of boxes is 3.
or Susan buys 3 boxes of candies.
Ronald HUI

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:
300 + 10x = 460
10x = 160
x = 16
◦ Step 4:  There are 16 $10 coins.
Ronald HUI

“>”: 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





P.166 Question 9: Add 5 to 4 times of a
number p and the sum is greater than 33.
Add 5
4 times of a number p
The sum
Then, the inequality is




“+5”
“4p”
“4p+5”
“4p+5 > 33”
Ronald HUI

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






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
1 pack of $2.4
y packs of $1.4
Total value
The inequality is





2  $1.9  10 = $38
1  $2.4  10 = $24
y  $1.4  10 = $14y
$38+$24+$14y
$38+$24+$14y <$100
or
14y+62 < 100
Ronald HUI





P.166 Question 13b: Find the value of y if the
total value of the stamps that Mike has is $76.
Total value
The equation is
 $38+$24+$14y
 $38+$24+$14y = $76
or
14y+62 = 76
14y = 14
y=1
Therefore, y = 1.
or “Mike has one pack of $1.4 stamps.”
Ronald HUI

A formula is an equation with
◦ 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





P.169 Question 18: v = u + gt
(u = -8, g = 10, t = 3, v = ?)
v = u + gt
= (-8) + (10) (3)  Method of substitution
= -8 + 30
= 22
Ronald HUI

P.169 Question 20: In the formula
S=88+0.5t , if S=120, find the value of t.
S = 88 + 0.5t
(120) = 88 + 0.5t
120 - 88 = 0.5t
32 = 0.5t
32  2 = t
64 = t
t = 64
Ronald HUI

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


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







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






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



Given a general term: an=2n-1
What is the first 5 terms?
The first 5 terms are:
◦
◦
◦
◦
◦


a1
a2
a3
a4
a5
=
=
=
=
=
2(1)-1
2(2)-1
2(3)-1
2(4)-1
2(5)-1
=
=
=
=
=
1
3
5
7
9
What is the 17th term?
The 17th term is a17 = 2(17)-1 = 33
Ronald HUI




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





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:
First term =
6th term =
10th term =
an = 9(n+6)
a1 = 9(1+6) = 9 (7) = 63
a6 = 9(6+6) = 9 (12) = 108
a10 = 9(10+6) = 9 (16) = 144
Ronald HUI








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

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




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
Percentages (I)
Form 1 Mathematics
Chapter 3
What is Percentage? (p.110)
•
•
“%” is the symbol of percentage. Per cent
means “per one hundred”.
e.g.
81
25 1
90
81% 
100
25% 
100

4
90% 
100
 0 .9
1 1
100
 100% 
%  20%
5 5
5
3 3
300
75
1
 100% 
%  %  37 %  37.5%
8 8
8
2
2
35%  0.35
45.67%  0.4567 3.14%  0.0314
0.74  74%
0.125  12.5%
Simple Percentage Problems
• P.116 Q2(c): 125% of $365.2
365.2 125
3652 5
913
$365.2 125%  $

$
 $
 $456.5
1
100
10 4
2
• P.117 Q10(c): 132 g is 88% of t g
• Remark: same units!
t  88%  132
t  132  88%
100
t  132 
88
100
t  3
2
t  150
Simple Percentage Problems
• P.118 Q14. 1100 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 = 165+35 = 200
Number of students = 1100+700 = 1800
Number of students passed = 1800-200 = 1600
Simple Percentage Problems
• P.118 Q14. 1100 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 = 935-665 = 1600
Percentage Increase (p.119)
•
New value > Original value
Increase  New value  Original value
Increase
Increase % 
100%
Original value
New value  Original value  1  increase %
•
Last year, Andy was 150cm tall. His height is
increased by 20% this year. His height now is
 150 cm  1  20%  150 cm  1.2  180 cm
Percentage Decrease (p.122)
•
New value < Original value
Decrease  Original value  New value
Decrease
Percentage decrease 
100%
Original value
New value  Original value  1  Percentage decrease 
•
The price of a car was $160,000 last month. If
it is reduced by 15% this month, the new price
 $160,000  1 15%  $160,000  0.85  $136,000
Percentage Change (p.126)
•
We can summarize the two formulae into one
and call percentage change.
Change % 
New value  Original value
100%
Original value
•
•
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
27  30
3

100% 
100%  10%
30
30
Percentage Change
• P.130 Q19. James had $5000 in his savings
account last year. This year, he has $10000.
• Change % 
10000  5000
5000
100% 
100%  100%
5000
5000
• James then takes out 40% of his savings. The
amount now is = $10000  (1 – 40%) = $6000
• New change % 
6000  5000
1000
100% 
100%  20%
5000
5000
Profit (盈利, p.131)
•
Selling Price (售價) > Cost Price (成本價)
Profit  Selling Price  Cost Price
Profit
Profit % 
100%
Cost Price
Profit  Cost Price  Profit %
Selling Price  Cost Price  1  Profit %
Loss (虧蝕, p.134)
•
Selling Price (售價) < Cost Price (成本價)
Loss  Cost Price  Selling Price
Loss
Loss % 
100%
Cost Price
Loss  Cost Price  Loss %
Selling Price  Cost Price  1  Loss %
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  Cost  profit %
• Profit or Loss?
• Let n be the Cost Price. Then, n  20%  $32,000
• So, the Selling Price is
 $160,000  $32,000  $192,000
1
n   $32,000
5
n  $32,000  5
n  $160,000
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  $500  1  30%  $500  70%  $350
Selling price of desk  $750  1  24%  $750 124%  $930
Total selling prices = $350 + $930 = $1280
Total cost prices = $500 + $750 = $1250
So, the profit % is
1280  1250
30

100% 
100%  0.024 100%  2.4%
1250
1250
Discount (折扣, p.138)
• Western style vs. Chinese style
• 10% discount
• 20% discount
• 70% discount
9折
8折
3折
• 12% discount
• 25% discount
• 5% discount
88折
75折
95折
Discount
• Marked Price > Selling Price
Discount  Marked Price  Selling Price
Discount
Discount % 
100%
Marked Price
Discount  Marked Price  Discount %
Selling Price  Marked Price  1  Discount %
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?
MP  1  discount %
• Marked price = $650
• Selling price of the shoes  $650  1  25% 
 $650  75%
 $487.5
• Since Vincent got $500, he can buy the shoes.
Discount
• A golden ring is sold at 30% discount with a
selling price of $441.What is the marked price?
MP  1  discount %
• Selling price = $441.
• Let n be Marked price. Then, n  1  30%   $441
7
n   $441
10
• So, the Marked price is $630.
10
n  $441
7
n  $630
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.
1920  1248
672
100% 
100%  35%
• A’s discount % 
1920
1920
• B’s discount % 
1760  1320
440
100% 
100%  25%
1760
1760
• So, B’s discount is smaller.
Simple Interest (單利息, p.144)
• Amount(A) = Principal(P) + Interest(I)
• Interest(I) = Principal(P)
 Interest rate(r%)  Time(T)
• Formula:
I  P  r% T
PrT
I
100
A PI
PrT
A P
100
 rT 
A  P  1 

 100 
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  $40000  3%  5
I  P  r%T
3
 $40000 
5
100
 $6000
• So, the amount (A) is  $40000  $6000  $46000
Simple Interest
• How long will $20000 amount to $30000 at
5%p.a. simple interest?
I  P  r%T
• P=$20000
r=5 A=30000
• I = $30000-$20000=$10000
• Let T be the time. Then, $20000  5%  T  $10000
• The time required is 10 years.
$10000 5
T

$20000 100
1 100
T 
2 5
T  10
Simple Interest
• P.151 Q14. Judy borrows $1000000 for 10 years.
Bank A’s interest is 5% p.a. and B is 4.5% p.a.
• A’s interest  $1000000  5%10  $500000
• B’s interest  $1000000  4.5%10  $450000
• The difference is $50000.
• She pays $50000 less in Bank B from in Bank A.
Ronald HUI
Good luck!!!