Simple & compound interest

Kes’s Pratibha college of commerce and computer
studies,chinchwad,pune-44
Department: Mathematics
Faculty Name: Vidya Pankaj Bhoge

Subject Name:
Business Mathematics
Topic Name:
Simple and compound interest

• Course content:
Definition: Simple and compound Interest
Formula: Simple and Compound Interest
Problems on simple and Compound Interest.

Objectives of the Topic:
-Define the concept of interest and show how it relates to the time
value of money.
-Distinguish between simple and compound interest and
demonstrate how to calculate each.
-Show how the total interest charge on loan contracts for either
simple or compound interest is determined.
-Using either formulas or compound interest tables, compute the
future and present values of a single payment.
-To understand difference between simple and Compound interest.

Outcome:
Student’s are able to understand:
-The concepts Related to simple and Compound Interest.
- To solve any example of Simple and Compound
Interest.
-Difference between simple and Compound Interest.
-How to use term such as principal amount, rate of
interest, term in year in real life.

Beneficiaries:
-If you want to get the most return on money you save or invest, you
want compound interest. The two types of interest are simple and
compound. Simple interest is paid only on the money you save or
invest (the principle), while compound interest is paid on your
principle plus on the interest you have already earned.
-Compound interest is beneficial if you are the one receiving the
interest, if you are the one paying compound interest on a loan or
credit card, then it's costing you a lot of money, as interest is charged
on interest.
-simple interest and compound interest are basic financial concepts,
becoming thoroughly familiar with them may help you make more
informed decisions when taking out a loan or investing.

Simple Interest:
Terminology:
Interest:
When a person borrows some amount of money from
other person or institution, the borrower has to pay
some charge to the lender for the use of the money.
This charge is called interest.

The Interest depends on two things
i) The period for which the money is borrowed.
ii) the rate of interest.

Principal:
The sum borrowed is called principal.

Term:
Time for which it is borrowed is called term.
Principal amount:
The total sum returned by the borrower i.e.
principal together with interest is called amount.
Simple interest:
Simple interest is calculated on the principal, or
original, amount of a loan.

Formula:
Simple Interest:
I =
𝑃.𝑟.𝑛
100
where,
P: denotes the principal,
n: denotes term in years,
r : denotes rate of interest %per
annum(p.c.p.a),
I: denotes simple interest,

Total amount is obtained by adding interest to the
principal,
thus,
A = P + I
= P+
𝑃.𝑟.𝑛
100
=P(1+
𝑟.𝑛
100
)

Example:
Find the simple Interest on Rs.2530 for 9 years at
12%p.a?
Solution:
Given that:
P=Rs.2530
r= 12%p.a
n= 9 years
I=?
We know that, the formula of Simple Interest
I =
𝑃.𝑟.𝑛
100

I =
𝑃.𝑟.𝑛
100
=
2530𝑋12𝑋9
100
=
273240
100
= 2,732.4Rs

Example:
A sum of Rs 4,800 amounted to Rs 6,240 in a certain period. If the rate of
simple interest is 12%p.a.Find period?
 Solution:
 Given that:
A= Rs 6,240 P=Rs 4,800 ,r=12% p.a ,n=?
We use formula,
A=P(1+
𝑟.𝑛
100
)
6,240 =4,800 (1+
12𝑋𝑛
100
)
If we divided both sides by 4,800,we get
6,240
4,800
=(1+
12𝑋𝑛
100
)

13
10
= 1+
12𝑋𝑛
100
13
10
− 1=
12𝑋𝑛
100
3
10
=
12𝑋𝑛
100
3𝑋100
10𝑋12
= n
n=
5
2
thus, the period is 2.5 years.

Example:
A sum of money doubles itself in 10 years. Find the rate of simple interest?
Solution:
Given that:
A=2P; n=6,
To find r.
Now ,we use formula
A=P(1+
𝑟.𝑛
100
)
2P=P(1+
𝑟.𝑛
100
)

2 = 1+
𝑟.10
100
2-1=
𝑟.10
100
1=
𝑟.10
100
r =
100
10
r =10
Thus, the rate of interest is 10%p.a

Example:
what sum will amount to Rs 3,296 in 4 months at 9%p.a.
simple interest?
Solution:
Given that
A=Rs 3,296 ,n=4 month=n=
4
12
=
1
3
; r=9
To find P.
We use formula
A=P(1+
𝑟.𝑛
100
)

3,296=P(1+
9𝑋
1
3
100
)
3,296=P(1+
3
100
)
3,296=P(
103
100
)
P=
100𝑋3,296
103
P=3200rs

EX:Find the simple interest on Rs 1,250 for 2
1
2
years at
12 % p.a?
Solution:
Given that:
P= RS1,250 r=12% p.a,
n=2
1
2
years=(2+
1
2
)𝑦𝑒𝑎𝑟𝑠 =
2𝑋2+1
2
=
5
2
years
We use Formula,
I=
𝑃𝑋𝑟𝑋𝑛
100
=
1250𝑋12𝑋
5
2
100
=RS.375

EX: In what period a sum of Rs 10,000 will earn a
simple interest of Rs 3200 at 8 %p.a?
Solution:
P= Rs10,000, S.I=Rs3200 ,r=8%p.a n=?
we use formula,
S.I=
100
3200 =
10,000𝑋8𝑋𝑛
100
n=
3200𝑋100
10000𝑋8
=4
4 years is required period.

Compound Interest:
When interest is added to the principal at the end of each period,
and the total so obtained is treated as the principal for the next
period, then the interest so obtained is called “Compound
Interest”.
For Example:
Suppose Rasiklal starts an industry by taking a loan of Rs
50,000 from a bank at 16% p.a.Bank charges interest
quarterly. But,Rasiklal is not able to pay interest at the end
of each quareter.Then the bank will add the interest at the
end of each quarter to the principal.

Let us find How much Rasiklal has to pay towards interest at the end
of 1st year.
Interest I=
𝑃.𝑟.𝑛
100
Interest on rs 50,000 for 1st quarter I=
𝑃.𝑟.𝑛
100
=
𝟓𝟎,𝟎𝟎𝟎𝑿𝟒𝑿𝟏
100
=2,000Rs
Principal for 2nd quarter P+I=50,000+2,000=52,000Rs
Interest on rs 52,000 for 2st quarter I=
𝑃.𝑟.𝑛
100
=
𝟓𝟐,𝟎𝟎𝟎𝑿𝟒𝑿𝟏
100
=2,080Rs
Principal for 3rd quarter P+I=52,000+2,080=54,080Rs
Interest on rs 54,080 for 3rd quarter I=
𝑃.𝑟.𝑛
100
=
𝟓𝟒,𝟎𝟖𝟎𝑿𝟒𝑿𝟏
100
=2,163.20Rs
Principal for 4th quarter P+I=54,080+2,163.20=56,243.20Rs
Interest on rs 56,243.20 for
4thquarter
I=
𝑃.𝑟.𝑛
100
=
𝟓𝟔,𝟐𝟒𝟑.𝟐𝟎𝑿𝟒𝑿𝟏
100
=2249.73Rs
Total Interest that Rasiklal has to pay =2,000+2,080+2,163.20+2,249.73
=8,492.93

Compound Interest:
Formula:
If P is the principal, r the rate of interest p.a. and n the period in
years then amount A at the end of n years is given by:
A=P(1+
𝑟
100
)n
Then the compound interest is given by
C.I=A-P
=P(1+
𝑟
100
)n-P
=P (1+
𝑟
100
)n-1
Thus,
C.I= P (1+
𝑟
100
)n-1

EX:Find Compound Interest on RS 5000 for 5th year
when the rate of interest is 9%p.a?
Given that :
P=Rs5000. ,r= 9% , n= 5years
In=
𝑃𝑋𝑟
100
(1+
𝑟
100
)n-1
635.2111

1)Compounded Yearly:
A=P(1+
𝑟
100
)n
2) If compounded Half yearly
A=P(1+
𝑟/2
100
)2Xn
3) If compounded Quarterly
A=P(1+
𝑟/4
100
)nX4
4) If compounded Monthly
A=P(1+
𝑟/12
100
)nX12

Compound interest on a given sum in the nth year:
In=
𝑃𝑋𝑟
100
(1+
𝑟
100
)n-1
EX:Find Compound Interest on RS 4000 for 5th year when
the rate of interest is 8%p.a?
Solution:
In=
𝑃𝑋𝑟
100
(1+
𝑟
100
)n-1
In=
4000𝑋8
100
(1+
8
100
)5-1
In=320 (1+
8
100
)4
In=320 (1+0.08)4
In=320X1.36048896=Rs 435.37.

Example:
Find the compound interest on Rs 10,000 for 4 years at 5
% p.a.
Solution:
Given that:
P=Rs 10,000 , r=5, n=4, to find C.I.
The amount of the end of 4 years is given by,
A=P(1+
𝑟
100
)n
A =10,000(1+
5
100
)4
A =10,000(
105
100
)4

A=10,000(1.05)4
= Rs 12,155
C.I=A-P
=12,155-10,000
=Rs 2,155
Example:
Find the compound interest on Rs 12,000 for 4 years at 6 %
p.a.
Solution:
P=12,000RS,n=4year,,r=%p.a
Firstly Calculate A,
A=P(1+
𝑟
100
)n

Example:
Find the compound Interest on Rs 8,000 at the rate of 8% p.a for 2
years compounded half yearly.
 Solution:
 Given that:
P= Rs 8,000 , r=8% p.a ,n=2
To find C.I
We use formula,
A=P(1+
𝑟/2
100
)2Xn
A=8,000(1+
4
100
)2X2
A=8,000(1+0.04)4
A=8000X1.16985856
A=9358.86
C.I=A-P =9358.86-8000 =1358.86Rs

EX: Find difference between Compound Interest
and simple interest on RS 5,00 for 2 years at 10%
p.a compounded yearly?
Solution:
Given that:
P= 500 ,n= 2 ,r=10%p.a
S.I=
100
=
500𝑋10𝑋2
100
=100

A=P(1+
𝑟
100
)n
A=500(1+
10
100
)2
A=500(1+
1
10
)2
A=500(1+0.1 )2
A=500(1.1 )2
A=Rs 605
C.I=A-P
C.I=605-500=105
Therefore,
C.I-S.I=105-100=5

EX: If the rate of interest is 10.5%p.a what is the effective
rate if compounding is done (i)Quarterly ii)Half Yearly?
Solution:
r=10.5%p.a, P=100Rs ,n=1 year.
i)When compounded interest is quarterly.
If compounded Quarterly
A=P(1+
𝑟/4
100
)nX4
A=100(1+
10.5/4
100
)4
A=100(1+
2.625
100
)4
A=100(1+0.02625 )4
A=100(1.02625 )4
A=110.92

Therefore,Effective rate of interest is 10.92%.
ii)When compounded interest is Half yearly,
If compounded Half yearly
A=P(1+
𝑟/2
100
)2Xn
r=10.5%p.a, P=100Rs ,n=1 year.
A=P(1+
𝑟/2
100
)2X1
A=P(1+
10.5/2
100
)2
A=P(1+
5.25
100
)2
A=110.78
C.I=A-P=10.78
Therefore,Effective rate of interest is 10.78%.

Simple Annuities:
Definition:
An Annuity is a series of payments made at equal intervals
.The payments may be equal or different .
When payment are equal ,the annunity is called “Simple
Annunity”.
The time intervel between two successive payments is called
period of the annuinity.
The period may be a month,a quarter,six months or even an
years.
Perpetuity:
In some annuities,the payments are made forever.
For EX: academic prizes etc, such an annuity is called
perpetuity.

Immediate Annuity and Annuity Due:
If payments of an annuity are made at the
beginning of each period then the annuity is called
annuity Due and When payments are made at the
end of each period ,then the annuity is called
Immediate annuity.
For EX:
Recurring Deposit,House rent etc.

If A denotes the amount of an immediate annuty then
A=
𝑥
𝑖
{(1+i)n -1}
where,
A: Amount of Immediate Annuity
x:Periodic Installment
n:Number of Installments
i:Rate of interest per rupees per period.
EX: Find the amount of an immediate annuity of Rs 15000/-
per year payable 12 years at 10 %p.a?
Solution:
x=Periodic installment=15000,n=12years,
i=10%p.a=10/100=0.1.

2-1-21
x=Periodic installment=15000,n=12years,
i=10%p.a=10/100=0.1.
A=
𝑥
𝑖
{(1+i)n -1}
A=
15000
0.1
{(1+0.1)12 -1}
A=1,50000{3.1383 -1}
A=RS.320745.

EX:Find the amount of an annuity of Rs400 payable quarterly for 3
years at 16 %p.a.
solution:here,Installment x=400
periodis 1 quarter.
16%p.a means 4%per quarter i.e 4 piase per rupee.
thus ,i=4/100=0.04
n:number of installments=3X4=12
A=
𝑥
𝑖
X{(1+i)n -1}
A=
400
0.04
X{(1+0.04)12 -1}
A=10,000X{1.60103 -1}
A=RS.60103.

Formulae for Amount & Present value of simple
Immediate Annuity:
P=
𝑥
𝑖
{1-(1+i)-n}
where,
P: Present value of Immediate Annuity
x:Periodic Installment
n:Number of Installments

EX: A man deposits his provident fund of R 2,00,000 in a charity trust at 5 %p.a
and settles to withdraw at Rs 15000/- per month for his expenses.If he begins to
spend from first year and goes on spending at this rate,show that he will not be
able to withdraw same amount in 23rd year?
Solution:
Suppose that the person can withdraw rs 15000 for n years,after which his
balance will be less than R15000.here,present value P=2,00,000.
To find n?
Annual installment x=15000
Rate per rupees p.a i=5/100=0.05
We have,
P=
𝑥
𝑖
{1-(1+i)-n}
2,00,000=
15000
0.05
{1-(1+0.05)-n}
2,00,000=300000{1-(1.05)-n}
0.667=1-(1.05)-n}
(1.05)-n=1-0.667

(1.05)-n =0.333
By taking log on both side,we get
log( (1.05)-n )=log(0.333)
-nXlog(1.05)=-0.4814
-nX0.021189=-0.4814
n=0.4814/0.021189=22.71
n= approximately 23 years.

Equated Monthly Instalments(EMI):
The repayment is generally made in monthly instalment
over period of 1 year,two year etc .This Monthly instalment
of repayment is called “EMI”.
There are two ways by which banks or housing finance
companies charge interest.
i)Interest on reducing Balance :
P=
𝑥
𝑖
{1-(1+i)-n}
where,
P: Present value of Immediate Annuity
x:Monthly Installment
n:Total Number of Month s in the terms.
i:Rate of interest per rupees per month.(i/100X12)

ii)Flat interest rate:
i)A=P(1+
𝑟𝑛
100
)
ii) EMI=
𝐴
𝐾
where,K =Number of Month.
EX: Calculate EMI for 1lacks at 5% flat rate over 20Year?
Solution:
Given that
P=1,00,000,r=5,n=20,K=12X20=240Month
A=P(1+
𝑟𝑛
100
)=1,00,000(1+
5𝑋20
100
)=2,00,000
EMI=
𝐴
𝐾
=
2,00,000
240
=Rs 833.33

By using Method Reducing Balance:
Let Amount of loan P=100,000 and flat rate of interest
be 5%.
Let 20 years then EMI turns out to be 833.33 as seen
above .
Let us Calculate EMI on reducing balanec, at the same
rate and for same term
i=5/100X12=5/1200,n=20Years=240month
P=
𝑥
𝑖
{1-(1+i)-n}
1,00,000=
𝑥
5/1200
{1-(1+5/1200)-240}
1,00,000=
𝑥
0.0041667
{1-(1+0.00416667)-240}

1,00,000=
𝑥
0.0041667
{1-(1.00416667)-240}
416.6667=x(1-0.3686)
416.6667=x0.6314
x=416.6667/0.6314
x=659.909249
x=660.
Thus,Borrowing money on reducing balance is highly beneficial.
EX: Find EMI on loan of Rs 3,00,000 to be paid in 4 years at
12%p.a on the outstanding amount at the begining of each Month?
Solution:
P=Rs3,00,000 n=4years= 4X12=48month i=12/100X12=0.01

P=
𝑥
𝑖
{1-(1+i)-n} hint:(1.01)^-48=0.6203
3,00,000=
𝑥
0.01
{1-(1+0.01)-48}
x=7900.9745.
EX: A two wheeler manufacturing company sells a motor
cycle costing Rs 44,000 on instalment basis by charging
EMI on Rs 4500 for 1 year.Find Flat rate of interest?
Solution:
ii)Flat interest rate:
i)A=P(1+
𝑟𝑛
100
)
ii) EMI=
𝐴
𝐾
where,K =Number of Month.

A=4500X12=54000,P=44,000,r?,n=1year
A=P(1+
𝑟𝑛
100
)
54000=44000(1+
𝑟
100
)
Divide 44000 on both sides, we get
54
44
=1+
𝑟
100
54
44
−1=
𝑟
100
=
54−44
44
=
𝑟
100
=
10
44
=
𝑟
100
=r=10X100/44=22.7

Simple & compound interest

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