Business Mathematics
CHAPTER 1 : Fundamental Concept of Mathematics
Business Mathematics
CHAPTER 1 : Fundamental Concept of Mathematics
1.2
Sequence
➢1.2.1 Arithmetic Sequence
➢1.2.2 Geometric Sequence
❖ A Sequence is a list of things (usually numbers) that are in order.
1st term
3, 5, 7, 9, …
2nd term
3rd term
Three dots means goes on forever
(infinite)
4th term
(“term”, “element”, or “member” mean the same
thing)
3
When the sequence goes on forever it is called an infinite sequence,
otherwise it is a finite sequence
1, 2, 3, 4,... Is a very simple sequence (and it is an infinite sequence)
20, 25, 30, 35,... Is also an infinite sequence
1, 3, 5, 7 Is the sequence of the first 4 odd numbers (and is a finite sequnce)
4, 3, 2, 1 Is 4 to 1 backwards
1, 2, 4, 8, 16, 32,... Is an infinite sequence where every term doubles
a, b, c, d, e Is the sequence of the first 5 letters alphabetically
f, r, e, d Is the sequence of the letters in the name “fred”
0,1, 0, 1, 0, 1,... Is the sequence of alternating 0s and 1s (yes they are in order, It
is an alternating order in this case )
4
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...
This is the Fibonacci Sequence
The next number is found by adding the two numbers before it together:
•The 2 is found by adding the two numbers before it (1+1)
•The 21 is found by adding the two numbers before it (8+13)
Rule is Tn = Tn-1 + Tn-2
That rule is interesting because it depends on the values of the previous two terms.
Rules like that are called recursive formulas.
The Fibonacci Sequence is numbered from 0 onwards like this :
𝑻𝒏
n=
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
…
xn
0
1
1
2
3
5
8
13
21
34
55
89
144
233
377
…
5
A sequence usually has a Rule, which is a way to find the value of each term
Example : the sequence { 3, 5, 7, 9, …} starts at 3 and jumps 2 every time :
+2
+2
0
1
2
3
4
5
6
+2
7
8
+2
9 10
Saying "starts at 3 and jumps 2 every time" is fine, but it doesn't help us
calculate the:
10th term,
100th term, or
nth term, where n could be any term number we want.
So, we want a formula with "n" in it (where n is any term number)
6
So, What Can A Rule For {3, 5, 7, 9, ...} Be?
Test Rule : 2n
n
Term
Test Rule
1
3
2𝑛 = 2 × 1 = 2
2
5
2𝑛 = 2 × 2 = 4
3
7
2𝑛 = 2 × 3 = 6
too low by 1 every time
Test Rule : 2n + 1
n
Term
Test Rule
1
3
2𝑛 + 1 = 2 × 1 + 1 = 3
2
5
2𝑛 + 1 = 2 × 2 + 1 = 5
3
7
2𝑛 + 1 = 2 × 3 + 1 = 7
That Works!
So instead of saying "starts at 3 and jumps 2 every time"
we write this:
2n+1
7
8
In an Arithmetic Sequence the difference between one term and the
next is a constant.
Notation
To make it easier to use rules, we often use this special style:
𝑻𝒏
term number
term
General Term
a – first term
• 𝑻𝒏 term number
•
n is the term number
d – common difference
First term 𝑇1 = a
Second term 𝑇2 = a + d
Third term 𝑇3 = a + 2d
Fourth term 𝑇4 = a + 3d
𝑇𝑛 - nth term
9
Example :
1, 4, 7, 10, 13, 16, 19, 22, 25, …
This sequence has a difference of 3 between each number.
Its Rule is 𝑻𝒏= 3n - 2
In General we can write an arithmetic sequence like this:
{a, a+d, a+2d, a+3d, ... }
Tn = a + d(n-1)
(We use "n-1" because d is not used in the 1st term).
10
Example : Write the Rule, and calculate the 4th term for
3, 8, 13, 18, 23, 28, 33, 38, …
This sequence has a difference of 5 between each number.
The value of a and d are :
▪ a = 3 (the first term)
▪ d = 5 (“the common difference”)
The Rule can be calculated :
Tn = a + d(n − 1)
= 3 + 5(n − 1)
= 3 + 5n − 5
= 5n − 2
So, the 4th term is :
T4 = 5  4 − 2 = 18
11
Example 1
What is the fiftieth term of the arithmetic sequence 3, 7, 11, 15, …?
A 53
B 151
C 199
D 203
This sequence has a difference of 4 between each pair of numbers.
The values of a and d are :
a = 3 (the first term)
d = 4 (the “common difference”)
The Rule can be calculated :
Tn = a + d ( n − 1)
= 3 + 4 ( n − 1)
= 3 + 4n − 4
= 4n − 1
So, the 50th term is :
T50 = 4  50 − 1 = 200 − 1 = 199
12
Example 2
What is the twentieth term of the arithmetic sequence 21, 18, 15, 12, …?
A -39
B -36
C1
D 78
This sequence has a descending, so has a difference of -3 between each pair of numbers.
The values of a and d are :
a = 21 (the first term)
d = -3 (the “common difference”)
The Rule can be calculated :
Tn = a + d ( n − 1)
= 21 +
( -3 (n − 1) )
= 21-3n + 3
= 24 − 3n
So, the 20th term is :
T20 = 24 − 3  20 = 24 − 60 = −36
13
Example 3
When the first term of an arithmetic sequence is 4 and the tenth term is 67,
what is the common difference?
A 6
B 7
C 8
D 9
Let the first term be a and the common difference d
Use the formula for the nth term : Tn = a + d(n − 1)
The first term = 4  a = 4
equation (1)
The tenth term = 67  T10 = a + d (10 − 1) = 67
67 = a + 9d = 67 equation ( 2 )
Substitute a = 4 from (1) into (2)
 4 + 9d = 67
 9d = 63
 d = 63  9 = 7
The common difference is 7
14
Example 4
What is the thirty-second term of the arithmetic sequence -12, -7, -2, 3…?
A 143
B 148
C 53
D 167
This sequence has a difference of 5 between each pair of numbers.
The values of a and d are :
a = -12 (the first term)
d = 5 (the “common difference”)
The Rule can be calculated :
T5 = a + d(n − 1)
= -12 + 5(n − 1)
= -12 + 5n - 5
= 5n - 17
So, the 20th term is :
T32 = 5  32 − 17 = 160 − 17 = 143
15
Example 5
The fifth term of an arithmetic sequence is 11 and the tenth term is 41. What is the
common difference?
A -19
B -13
C -9
D 6
Let the first term be a and the common difference d
Use the formula for the nth term : Tn = a + d(n − 1)
The fifth term = 4  T5 = a + d(5 - 1) = 11  a + 4d = 11 (1)
The tenth term = 41  T10 = a + d(10 − 1) = 41  a + 9d = 41 (2 )
Substitute (1) from (2)
a + 9d = 41
a + 4d = 11
− − − − − − − Subtract
5d = 30
so d = 30  5 = 6
Substitute d = 6 into (1)  a + 4  6 = 11  a + 24 = 11  a = −13
The first term is -13
16
What to sum
Go to this value
4
To sum up the terms of this
arithmetic sequence:
Start at this value
n
= 1 + 2 + 3 + 4 = 10
n-1
a + (a + d ) + (a + 2d ) + (a + 3d ) + ...
n =1
 (a + kd ) = (2a + (n − 1)d )
It says “Sum up n where n goes from 1 to 4,
Answer = 10
k =0
 (called Sigma ) means "sum up"
17
Each term is the same !, and there “ ” of them so …
𝟐𝐒 = 𝐧 × (𝟐𝐚 + 𝐧 − 𝟏 𝐝)
Now, just divide by 2 and we get :
n
S=
× 2a + n − 1 d
2
First, we will call the whole sum “S” :
𝐒 = 𝐚 + 𝐚 + 𝐝 + ⋯ + 𝐚 + 𝐧 − 𝟐 𝐝 + (𝐚 + 𝐧 − 𝟏 𝐝)
n −1
n
(
)
 a + kd = (2a + (n − 1)d )
k =0
2
Next, rewrite S in reverse order :
𝐒 = 𝐚 + 𝐧 −𝟏 𝐝 + 𝐚 + 𝐧 − 𝟐 𝐝 + ⋯+ 𝐚 + 𝐝 +𝐚
Now add those two, term by term :
S =a
+ (𝐚 + 𝐝)
+ ⋯ + (𝐚 + 𝐧 − 𝟐 𝐝) +(𝐚 + 𝐧 − 𝟏 𝐝)
S = (𝐚 + 𝐧 − 𝟏 𝐝) + 𝐚 + 𝐧 − 𝟐 𝐝
+ ⋯ + (𝐚 + 𝐝)
+𝐚
__________________________________________________________________________
2S = (𝟐𝐚 + 𝐧 − 𝟏 𝐝) + 𝟐𝐚 + 𝐧 − 𝟐 𝐝
+ ⋯ + (𝟐𝐚 + 𝐧 − 𝟏 𝐝) +(𝟐𝐚 + 𝐧 − 𝟏 𝐝)
18
Example : Add up the first 10 terms of the arithmetic sequence :
{ 1, 4, 7, 10, 13, …}
The values of a and n are :
❑ a = 1 (the first term)
❑ d = 3 (the “common difference”) between terms
❑ n = 10 (how many terms to add up)
So
n −1
n
(
)
(2a + (n − 1)d )
a
+
kd
=

k =0
2
becomes
10 −1
10
(
)
(2 1 + (10 − 1)  3)
1
+
k

3
=

k =0
2
Example 6
What is the sum of the first sixteen term of the arithmetic sequence 1,
5, 9, 13, …?
A 62
B 496
C 528
D 992
1, 5, 9, 13, …
The values of a, d and n are :
❑ a = 1 (the first term)
❑ d = 4 (the “common difference”) between terms
❑ n = 16 (how many terms to add up)
Use the formula Sn =
n
( 2a + (n − 1) d)
2
16
( 2  1 + 15  4 )
2
= 8 ( 2 + 60 )
Therefore S16 =
= 8  62
= 496
20
Example 7
What is the sum of the first thirty terms of the arithmetic sequence
50, 45, 40, 35, …?
A -675
B 275
C 675
D 3,675
50, 45, 40, 35, …
The values of a, d and n are :
❑ a = 50 (the first term)
❑ d = -5 (the “common difference”) between terms
❑ n = 30 (how many terms to add up)
Use the formula Sn =
n
( 2a + (n − 1) d)
2
30
( 2  50 + 29  -5 )
2
= 15 (100 − 145 )
Therefore S30 =
= 15  −45
= -675
21
Example 8
What is the sum of the eleventh to twentieth terms (inclusive) of the arithmetic
sequence 7, 12, 17, 22, …?
A 154
B 295
C 795
D 1,090
7, 12, 17, 22, …
The values of a, d and n are :
❑ a = 7 (the first term)
❑ d = 5 (the “common difference”) between terms
To find the sum of the eleventh to twentieth terms we subtract the sum of the first
ten terms from the sum of the first 20 terms
So we use the formula Sn =
n
( 2a + (n − 1) d ) twice to find S10 and S20
2
10
(2  7 + (10 − 1) 5) = 5(14 + 45) = 5  59 = 295
2
20
(2  7 + (20 − 1) 5) = 10(14 + 95) = 10 109 = 1,090
S20 =
2
S10 =
Therefore the sum of the eleventh to twentieth terms = 1,090 – 295 = 795
22
Example 9
The eleventh term of arithmetic sequence is 30 and the
sum of the first eleven terms is 55. What is the common
difference?
A 5
B6
C 8
D 10
The eleventh term , T11 = a + d (11 − 1)
(Equation1)
a + 10d = 30
11
( 2a + (11 − 1) d)
2
11
55= ( 2a + 10d ) (Equation 2)
2
55=11( a + d )
The sum of the first eleven terms, 55=
5=a + 5d
Subtract (Equation 2) from (Equation 1)
Let the first term be a and the common difference d
Use the formula for the n’th term : Tn = a + d(n − 1)
and the formula for the sum to n terms Sn =
n
(2a + (n − 1)d )
2
a + 10d = 30
a + 5d = 5
Subtract
5d = 25
So d = 25 ÷ 5 = 5
23
Business Mathematics
CHAPTER 1 : Fundamental Concept of Mathematics
25
Geometric Sequences
In a Geometric Sequence each term is found by multiplying the
previous term by a constant.
Example :
2, 4, 8, 16, 32, 64, 128, 256, …
This sequence has a factor of 2 between each number.
Its Rule
𝑻𝒏 = 𝟐 𝒏
In General we can write a geometric sequence like this:
{a, ar, ar2, ar3, ... }
where:
•a is the first term, and
•r is the factor between the terms (called the "common
ratio")
Note : r should not be 0
• when r = 0, we get the sequence {a, 0, 0, …} which is not geometric
26
Example :
10, 30, 90, 270, 810, 2430, …
This sequence has a factor of 3 between each number.
The values of a and r are :
• a = 10 (the first term)
• r = 3 (the “common ratio”)
The Rule for any term is :
Tn = 10  3(n −1)
So, the 4th term is :
T4 = 10  3(4-1) = 10  33 = 10  27 = 270
and the 10th term is :
T10 = 10  3(10-1) = 10  39 = 10 19683 = 196830
27
Tn = ar(n-1)
(We use "n-1" because ar0 is the 1st term)
Example :
A geometric sequence can also have smaller
and smaller values :
Example :
{1, 2, 4, 8, …}
4, 2, 1, 0.5, 0.25, …
The sequence starts at 1 and doubles
each time, so
The sequence has a factor of 0.5 (a
half between each number.)
• a = 1 (the first term)
• r = 2 (the “common ratio” between
terms is a doubling)
Its Rule is
T4 = 4  (0.5)
n −1
And we get :
a, ar, ar 2,ar 3,...


= 1, 1 2, 1 22 , 1 23 ,...
= 1, 2, 4, 8,...
28
Example 10
What is the eleventh term of the geometric
sequence 3, 6, 12, 24, …?
A 66
B 768
C 1536
D 3072
This sequence has factor of 2 between
each pair of numbers.
The values of a and r are :
• a = 3 (the first term)
• r = 2 (the “common ratio”)
The Rule for any term is :
Tn = ar n -1
Therefore,
T11 = 3  210
= 3 1024
= 3072
29
Example 11
1
What is the ninth term of the geometric sequence
81, 27, 9, 3, …?
A
C
1
81
1
243
B
1
27
D 1
This sequence has a factor of 3 between each
pair of numbers.
The value of a and r are :
▪ a = 81 (the first term)
1
▪ r = (the “common ratio”)
3
9
The Rule for any term is :
Tn = ar (n −1)
Therefore 𝑇𝑛
8
 1
T9 = 81  
3
1
= 81
6561
81
=
6561
1
=
81
30
Example 12
The first term of a geometric sequence is 5 and their
sixth term is 160. What is the common ratio?
A 0.5
B2
C 3
D4
Let the first term be a and the common ratio r
We know the first term a = 5
Use the formula for the n’th term :
Tn = ar n −1
T6 = ar 6−1
The sixth term
= 160
ar 5 = 160
When a = 5
Therefore,
5r 5 = 160
r 5 = 160  5
= 32
r = 5 32
=2
31
Example 13
The fourth term
The fourth term of a geometric sequence is 27 and the
seventh term is 1. What is the first term?
1
A
3
B 81
C 243
D 729
T4 = ar 4−1 = 27
ar 3 = 27 equation (1)
The seventh term
T7 = ar 7−1 = 1
ar 6 = 1 equation ( 2 )
Divide (2) by (1)
1
27
1
r=
3
1
Substitute r = into equation (1),
3
r3 =
Let the first term be a and the common ratio r
3
Use the formula for the n’th term :
Tn = ar n −1
 1
a×   = 27
3
1
a× = 27
27
a = 27×27 = 729
The first term is 729
32
First, call the whole sum “S” : S = a + ar + ar 2 + ... + ar (n − 2 ) + ar (n −1)
Equation 1
Next, multiply S by r : S  r = ar + ar 2 + ar 3 + ... + ar (n−1) + ar n
Equation 2
S = a + ar + ar 2 + ... + ar (n −1)
- S  r = − ar − ar 2 − ... − ar (n −1) − ar n
______________________________
S - S r = a
− ar n
S - S  r = a - ar n
Factor out S and a : 𝑆 1 − 𝑟 = 𝑎(1 − 𝑟 𝑛 )
Divide by 1 − 𝑟 ∶ 𝑆 =
𝑎(1−𝑟 𝑛 )
(1−𝑟)
1− rn
ar = a 
 1− r
( )
n −1
k
k =0



Well … when r is less than 1, then 𝑟 𝑛 goes to zero and we get :
 1 
(
)
ar
=
a



1− r

k
k =0


33
Summing a Geometric Series
When we need to sum a Geometric Sequence, there is a
handy formula.
To sum:
a + ar + ar2 + ... + ar(n-1)
Each term is ark, where k starts at 0 and goes up to n-1
1− rn
ar = a 
 1− r
( )
n −1
k
k =0



a is the first term
r is the "common ratio" between
terms
n is the number of terms
34
R must be between ( but not including) -1 and 1
and r should not be 0 because we get the sequence {a, 0, 0, …} which isn’t geometric
Example : Add up ALL
the terms of the
Geometric Sequence
that halves each time :
1 1 1 1 
 , , , ,...
 2 4 8 16 
We have :
1
▪ a = 2 ( the first term )
1
▪ r = 2 ( halves each time)


 1  1 k  1  1 
    = 



 2 1 
k =0  2  2  
1− 
 2

1
1
2
=
=1
1
2
1
1
1
Yes … adding (2)+(4)+(8)+… equals exactly 1.
35
Example : Sum the first 4 terms of
10, 30, 90, 270, 810, 2430, …
This sequence has a factor of 3 between each number.
The values of a, r and n are :
▪ a = 10 ( the first term )
▪ r = 3 ( the “common ratio” )
▪ n = 4 (we want to sum the first 4 terms)
So
1− rn
ar = a 
 1− r
( )
n −1
k
k =0



becomes
 1 − 34 
 = 400
10  3 = 10

k =0
 1− 3 
4 −1
(
k
)
You can check it yourself :
10 + 30 + 90 + 270 = 400
And, yes, it is easier to just add them in this example, as there are only 4
terms. But imagine adding 50 terms … then the formula is much easier
36
Example :
… doubling the grains of rice on each square …
… how many grains of rice in total?
So we have :
• a = 1 (the first term)
• r = 2 (doubles each time)
• n = 64 (64 squares on a chess board)
1− rn
ar = a 
 1− r
( )
n −1
k
k =0



 1 − 264 

ar = 1

k =0
 1− 2 
1 − 264
=
-1
= 264 − 1
= 18, 446, 744, 073, 709, 551, 651
64 −1
( )
k
On the page Binary Digits we give an example of grains
of rice on a chess board. The question is asked :
When we place rice on a chess board :
▪ 1 grain on the first square,
▪ 2 grains on the second square,
▪ 4 grains on the third and so on,
37
Example 14
What is the sum of the first eight term of the
geometric sequence 5, 15, 45, …?
A 320
B 5465
C 16400
D 32800
 r n −1 

Use the formula Sn = a 
 r −1 
 38 − 1 

Therefore, Sn = 5
 3 −1 
 6561 − 1 
= 5

 3 −1 
= 5  6560  2
= 16400
5, 15, 45, …
This sequence has a factor of 3 between each pair of number.
The values of a, r and n are :
▪ a = 5 ( the first term )
▪ r = 3 ( the “common ratio” )
▪ n = 8 (we want to sum the first 8 terms)
38
Example 15
What is the ninth term of the geometric sequence :
1 − 1 1 − 1 
 , , , ,...
 3 9 27 81 
A
1
2
B
4
9
C
1
3
D
1
4
1 − 1 1 − 1 
 , , , ,...
 3 9 27 81 
We known :
1
▪ a = 3 (the first term)
▪ r=
−1
3
(the “common ratio”)
Use the formula for the sum to infinity of a
geometric sequence :
 1 
S = a

 1- r 
Therefore,
1
3
S9 =
 −1 
1−  
 3 
1
1 4
= 3 = 
4 3 3
3
1
=
4
39
Example 16
What is the sum of the first nine terms of the
geometric sequence 20, 10, 5, …?
Give your answer as a decimal correct to 1
decimal place.
A 39.9
B 40.0
C 59.9
D 79.8
20, 10, 5, …
 1 - rn 
Use the formula Sn = a 

 1- r 
  1 9 
1−   
2 

Therefore, S9 = 20   
1
 1


2


  1 
1− 

512


= 20

1 
1

2 

511 1

512 2
511 1
= 20 

512 2
2555
=
64
= 39.9 to 1 decimal place
= 20 
1
This sequence has a factor of 2 between each pair of numbers.
The values of a, r and n are :
▪ a = 20 ( the first term )
𝟏
▪ r = 𝟐 ( the “common ratio” )
▪ n = 9 (we want to sum the first 9 terms)
40
Business Mathematics
CHAPTER 1 : Fundamental Concept of Mathematics
2.1 Simple Interest
2.1.1 Introduction to Interest
2.1.2 Simple Interest Formula – Years or Months
42
Interest: how much is paid for the use of money (as a percent, or an amount)
How Much does it Cost to Borrow Money?
Different places charge different
amounts at different times!
Money is Not Free to Borrow
People can always find a use for
money, so it costs to borrow
money.
But they usually charge this way
As a percent (per year) of the
amount borrowed
It is called INTEREST
43
Example: Amir Borrow RM1,000 from the Bank
Amir wants to borrow RM1,000. The local bank says "10% Interest".
So to borrow the RM1,000 for 1 year will cost:
RM1,000 × 10% = RM100
In this case the "Interest" is RM100, and the "Interest Rate" is 10%
TODAY
NEXT YEAR
RM1000
RM1000
Amir
Bank
Amir
Bank
RM100
interest
Amir Borrows RM1,000, but has to pay back RM1,100
Note : I am showing a full year loan, but banks often want you to pay back
the loan in small monthly amounts, and they also charge extra fees too!
44
principal
borrower
TODAY
NEXT YEAR
RM1000
RM1000
Amir
Lender
Bank
The Loan
Amir
Bank
RM100
interest
The Repayment
Amir is the Borrower, the Bank is the Lender
The Principal of the Loan is RM1,000
The Interest is RM100
The important part of the word “Interest” is Inter-meaning between (we see interin words like interior and interval), because the interest happens between the
start and end of the loan.
45
Example:
Amir borrows RM1,000 for 5
Years, at 10% simple
interest:
Example:
Amir borrows RM1,000 for 7
Years, at 6% simple
interest:
• Interest = RM1,000 × 10%
x 5 Years = RM500
• Plus the Principal of
RM1,000 means Amir needs
to pay RM1,500 after 5
Years
• Interest = RM1,000 × 6% x
7 Years = RM420
• Plus the Principal of
RM1,000 means Amir needs
to pay RM1,420 after 7
Years
46
There is a formula for simple interest
I = Prt
where
❖
❖
❖
❖
I = interest
P = amount borrowed (called “Principal”)
r = interest rate
t = time
Example :
Ila borrowed RM3000 for 4 years at 5% interest rate, how
much interest is that?
I = Prt
I=
RM3,000 × 5% × 4
years
I = RM3000 × 0.05 × 4
I = RM600
47
Example 17
Jerry borrowed RM4000 for 5 years at 6% simple
interest rate. How much interest is that?
A RM800
B RM1000
C RM1200
D RM1500
where
❖
❖
❖
❖
I = interest
P = Principal = RM400
r = interest rate = 6% = 0.06
t = time = 5 years
I = RM4000 × 0.06 × 5 = 1200
Use
I = Prt
where
❖
❖
❖
❖
I = interest
P = amount borrowed (called “Principal”)
r = interest rate
t = time
48
Example 18
Jerry borrowed RM7000 at a simple interest rate of
3% per year. After a certain number of years he
had paid RM840 in interest altogether. How many
years was that?
A 2 years
C
1
32
Use
years
B 3 years
D 4 years
❖
❖
❖
❖
I = interest = RM840
P = Principal = RM7000
r = interest rate = 3% = 0.03
t = time
So, 840 = 7000 × 0.03 × t
Simplify : 840 = 210 × 𝑡
Rearrange : 𝑡 = 840 ÷ 210
Solve : t = 4
I = Prt
where
❖
❖
❖
❖
where
I = interest
P = amount borrowed (called “Principal”)
r = interest rate
t = time
The time was 4 years
49
Example 19
Sam borrowed RM4500 for 2 years and had to pay
RM630 at a simple interest at the end of that time.
What rate of interest did he pay?
A 6%
B 7%
C 8%
D 9%
Use
where
❖
❖
❖
❖
I = interest = RM630
P = Principal = RM4500
r = interest rate
t = time = 2 years
So, 630 = 4500 × r × 2
Simplify : 630 = 9000 × 𝑟
Rearrange : r= 630 ÷ 9000
Solve : r = 0.07 = 7%
I = Prt
where
❖
❖
❖
❖
I = interest
P = amount borrowed (called “Principal”)
r = interest rate
t = time
50
Example 20
1
Julie borrowed RM3500 for 3 years at 7 % annual
2
simple interest rate. How much interest is that?
A RM375
B RM787.50
C RM810
D RM812.50
where
❖
❖
❖
❖
I = interest
P = Principal = RM350
1
r = interest rate = 7 2% = 0.075
t = time = 3 years
Therefore, I = RM3500 × 0.075 ×
3 = 787.50
Use
I = Prt
where
❖
❖
❖
❖
I = interest
P = amount borrowed (called “Principal”)
r = interest rate
t = time
51
Example 21
Dan borrowed RM2000 for 6 month at 12% annual
simple interest rate. How much interest is that?
A RM120
B RM144
C RM1200
D RM1440
where
❖
❖
❖
❖
I = interest
P = Principal = RM2000
r = interest rate = 12% = 0.12
t = time = 6 month = 0.5 year
I = RM200 × 0.12 × 0.5 = RM120
Use
I = Prt
I=RM2000
where
❖
❖
❖
❖
I = interest
P = amount borrowed (called “Principal”)
r = interest rate
t = time
52
53