Slide 1

Quantitative Methods
Session 8– 31.07.12
Chapter 5 – Simple Interest & Compound Interest

Pranjoy Arup Das

• Interest is the payment to be made for the use of money.
• Suppose, Mr. A borrows a sum of money Rs. X from Mr. B, then Mr. A
will have to repay not only the sum of Rs. X but also some extra money
(say Rs. Y) to Mr. B, sort of like a charge or rent for using Mr. B’s money.
• Rs. X is termed as the principal (P), Rs. Y is the interest (I) for using
Mr.B’s money & Rs. (X + Y) is termed as the Amount(A) which Mr. A
will have to repay to Mr. B after a certain period of time (T) .
• The interest to be paid by the borrower Mr. A to the lender Mr. B is
fixed before or while lending the money. Interest is fixed as a
percentage of the principal and is called Rate of interest (R).

• Money is usually lent out for a fixed period of time (T) . Interest is
calculated on the basis of this time period at the agreed rate, which is
charged either yearly, half-yearly or quarterly. T is always calculated in
Years. So a loan period of 6 months will have to be taken as 0.5 years.

There are two kinds of interest Simple interest & Compound Interest
Simple interest is always calculated on the actual principal originally borrowed.
Throughout the entire time period, a fixed amount of interest is paid whenever
due which is either yearly, half yearly of quarterly
For Eg. Mr. A borrows Rs. 2000 from Mr. B for a period of 2 years at an interest
rate of 10% payable yearly. How much will A pay B after 2 years?
Compound interest is calculated on the increasing principal. The interest is
calculated on the original principal and then added to the original principal
which becomes the new principal for calculating the interest on next due date.
• Simple Interest:
Original Principal Rs. 2000
Interest at 1st year end = 10%of 2000
= Rs. 200
Interest at 2nd year end = 10%of 2000
= Rs. 200
So at the end of 2 years Mr. A has to
pay Rs. (2000 + 200+200)=
Rs. 2400/-

• Compound Interest:
Original principal Rs. 2000
Interest at 1st year end = 10% of 2000
= Rs. 200

New principal=Rs. 2000+200 = 2200
Interest at 2nd year end = 10% of 2200
= Rs. 220
So at the end of 2 years Mr. A has to
pay Rs. (2200 + 220) = Rs. 2420/-

Part 1 Simple Interest

• If a sum of money (principal) Rs. P is borrowed at the rate of R% simple
interest for T years, then
Simple interest amount for T years = 1st years int. + 2nd yrs int. +….+ Tth years int
= (R% of P) + (R% of P) + (R% of P) + …T times
= (R% of P) (1 + 1 + 1 +……………upto T)
= (R% of P) * T
=

I=
• Principal (P)+ Interest (I)= Total Amount(A) to be repaid after T years
=> P +
I
= A
=> P + (R*P*T)/100 = A

Eg. Mr. A borrows Rs. 5400/- from Mr. B under the conditions that :

a) The loan will be repaid after 3 years
b) The borrower will pay a simple interest of 8% per annum (per year)
1) How much interest will Mr. A have to pay at the end of 3 years?
2) What is the amount that Mr. A will have to pay at the end of 3years?
Solution:

Here the Principal = Rs. 5400 Time = 3years Rate of interest= 8%
• Interest at 1st year end = 8% of 5400
•
= Rs. 432
• Interest at 2nd year end = 8% of 5400
•
= Rs. 432
• Interest at 3nd year end = 8% of 5400
•
= Rs. 432
• Total interest at the end of the 3 year loan period
= Rs. (432+432+432) = Rs. 1296
• So at the end of 3 years Mr. A has to pay Rs. (5400 + 1296 )
•
= Rs. 6696

Alternative Solution: USING FORMULA OF SIMPLE INTEREST
Here Principal (P) = Rs.5400, Time (T) = 3 years , Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400* 3) / 100
I = Rs. 1296
The amount to be paid to Mr. B at the end of 3 years :
Amount = P + I
= 5400 + 1296
= Rs. 6696

Page 388 (RSA) Ex 1
SOLUTION 1: Please Note: ALWAYS CONVERT DAYS OR MONTHS TO YEARS
Here the Principal = Rs. 5400
Time = 10/12 years Rate of int. = 8%pa
• Interest for 1 year
= 8% of 5400
= Rs. 432
• That means Interest for 12 months = Rs. 432
•  Interest for 10 months will be: (10/12) * 432 = Rs. 360

• Total interest at the end of the 10 month loan period = Rs. 360
SOLUTION 2 :
Here the Principal (P) = 5400

Time (T) = 10/12 years Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400*
) / 100 = Rs.__________

Page 390 Exercise 21A, Pr no. 4
Solution:
Let the sum of money be Rs. x which is the principal (P)
It is given that interest on Rs. x @8% p.a. in 6 years = Rs. 8376
Here principal is Rs .x, Rate of interest (R ) is 8% p.a. and the time
(T) is 6 years
Interest on Rs. x @ 8% for 6 years (I) = (R*P*T)/100
= (8 * x * 6) /100 = Rs.12x/25
So that means , 12x/25 = 8376
=>
x = (8376 * 25) /12
=>
x = Rs. ___________

Page 388, Ex. 4 Solution:
Let the sum of money (P) be Rs. x
And let the rate of interest (R ) be y% p.a
In the first case Rs. x becomes Rs. 854 in 2 years @ y% p.a
Interest @ y% on Rs. x for 2 years (I) = (R*P*T)/100 = (y * x * 2) /100
= (2xy)/100
= Rs. xy/50
So, x + (xy/50) = 854
=> 50x + xy = 42700…………(i)
Similarly, in the 2nd case
Interest on Rs. x @ y% for 7/2 years = (y *x*7/2) / 100 = Rs.(7xy/200)

So, x + (7xy/200) = 969.50
=> 200x + 7xy = 193900…….(ii)
Eliminating xy from (i) & (ii) we can get the value of x and then y.

Page 389, Ex. 5 Solution:
Let the sum of money (P) be Rs. x
So after 8 years, this sum of money becomes Rs. 2x.
Since, Principal

x

+ Interest
= Amount
+ (Interest on Rs. x for 8 years) = 2x
=> (Interest on Rs. x for 8 years) = 2x – x = Rs. x

Let the rate of interest be R% p.a.
Here principal (P) = Rs. x , Interest (I)= Rs. x and Time (T)= 8 years
We know that I = (R * P* T) /100
 R = (I * 100) / (P*T)
= ( x* 100) / (x * 8)
= (100x) /(8x) = 100/8
= 25/2 = 12.5 %

Page 389, Ex. 6
Solution:
Let the loan @ 8% p.a. be Rs. x
 Loan @ 10% p.a. = Rs. (8000 – x)
Given that total annual interest = Rs. 714
That means, in 1 year:
Interest earned on Rs. x@ 8% + Interest earned @ 10% on Rs. (8000-x) = 714
=> {(8 * x * 1) / 100} + [{10 * (8000-x) * 1} / 100] = 714
=>
Solve for x.

(8x/100)

+

{(8000-x) / 10}

=

714

Session 9; Chapter 5; Part 2
Compound Interest

Page 401 Ex 1.
SOLUTION 1 : Without using formula.
• Here Principal (P)= Rs. 8000 , Rate ( R) = 10% p.a ., Time (T) = 3 years
• Interest at 1st year end @ 10% on 8000 for 1 year= (8000 * 10 * 1) /100
•
= Rs. 800
• New principal for 2nd year = Rs. 8000+800 = 8800
• Interest at 2nd year end @ 10% on 8800 for 1 year = (8800 * 10 * 1) /100
•
= Rs. 880
• New principal for 3rd year = Rs. 8800 + 880 = Rs. 9680
• Interest at 3rd year end @ 10% on 9680 for 1 year = (9680 * 10 * 1) /100
•
= Rs. 968
So at the end of 3 years Rs. 8000 will become Rs. (9680 + 968) = Rs. 10648/• So the compound interest (CI) = A – P = 10648 – 8000 = Rs. 2648
•
ALTERNATIVELY : CI = Rs. (800 + 880+ 968) = Rs. 2648

FORMULAS OF COMPOUND INTEREST:
• If a sum of money (principal) Rs. P is borrowed /lent at the rate of R%
compound interest for T years compounded yearly , then :
Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed / lent at the rate of R% compound interest for
compounded half yearly, then :

T years

Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed /lent at the rate of R% compound interest for
compounded quarterly, then :
Compound Interest (CI) on Rs. P @ R% for T years =

T years

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 8000 Rate ( R) = 10% p.a

Time (T) = 3 years

• Compound interest (CI) on Rs. 8000 at the end of 3 years @ 10% =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 4.
SOLUTION 1 : Without using formula.
Here Principal (P)= Rs. 25000, Time (T) = 1 year, Yearly rate of interest =12%
• The Interest is compounded half yearly
• So the half yearly rate of interest = 12 * ½ = 6% per half year
• Interest at 1st half year end = 6% of 25000
•
= Rs. 1500
• New principal for 2nd half year = Rs. 25000+1500 = Rs. 26500
• Interest at 2nd half year end = 6% of 26500
•
= Rs. 1590
• So at the end of 1 year Rs. 25000 will become Rs. (26500 +1590)
= Rs. 28090/• So the compound interest (CI) = 28090 – 25000 = Rs. 3090
• Or , CI = Rs. (1500 + 1590)= Rs. 3090

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 25000 Rate ( R) = 12% p.a Time (T) = 1 year
• Compound interest (CI) on Rs. 25000 at the end of 1 years @ 12% pa
compounded half yearly =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 5. SOLUTION : Without using formula.
Here Principal (P)= Rs. 8000, Time (T) = 9 months = 9/12 years = ¾ yrs,
• Note: 9 months = 3 quarters
• Yearly rate of interest = 20% pa
• The Interest is compounded quarterly . In a year there are 4 quarters
• So the per quarter rate of interest = 20/4 = 5% per quarter.
• Principal for 1st quarter = Rs. 8000
• Interest at 1st quarter end = 5% of 8000 = Rs. 400

• New principal for 2nd quarter = Rs. 8000+400 = Rs. 8400
• Interest at 2nd quarter end = 5% of 8400 = Rs. 420
• New principal for 3rd quarter = Rs. 8400+420 = Rs. 8820
• Interest at 3rd quarter end = 5% of 8820 = Rs. 441
• So at the end of 3 quarters or 9 months, Rs. 8000 will become
Rs. (8820 +441) = Rs. 9261/• So the compound interest (CI) = 9261 – 8000 = Rs. 1261
• Or, CI = Rs. (400 + 420+ 441)= Rs. 1261

Page 402 Ex 3. SOLUTION :
• Here Principal (P)= Rs. 5000, Time (T) = 3years,
• Rate of interest for 1st year (R1) = 5% pa
• Principal for 1st year = Rs. 5000
Interest at 1st year end

= 5% of 5000 = Rs. 250

• New principal for 2nd year = Rs. 5000 + 250 = Rs. 5250
• Rate of interest for 2nd year (R2) = 8% pa
Interest at 2nd year end = 8% of 5250 = Rs. 420

• New principal for 3rd year = Rs. 5250 + 420 = Rs. 5670
• Rate of interest for 3rd year (R3) = 10% pa
Interest at 3rd year end = 10% of 5670 = Rs. 567
• So the amount at the end of 3 years = Rs. (5670 +567) = Rs. 6237/• This means, at the end of 3 years, Rs. 5000 will become Rs. 6237/• Compound interest (CI) on Rs. 5000 after 3 years @ 5%,8% &10%=
6237 – 5000 = Rs. 1237
• Or, CI on Rs. 5000 after 3 years @ 5%,8% &10%= ( 250+ 420+ 567)= Rs. 1237

Page 402 Ex 6. SOLUTION :
• Here Principal (P) = a certain sum of money, CI = Rs. 3783,
Time (T) = 3years, Rate of interest (R) = 5% pa
• Let the principal (P) be Rs. x
• Assuming interest is being charged annually, We know that,
CI =
=> CI on Rs. x @ 5% for 3 years =

=> 3783 =
Find x then find SI on Rs. x @ 5% for 3 years = (x*5*3)/100


Slide 2

Quantitative Methods
Session 8– 31.07.12
Chapter 5 – Simple Interest & Compound Interest

Pranjoy Arup Das

• Interest is the payment to be made for the use of money.
• Suppose, Mr. A borrows a sum of money Rs. X from Mr. B, then Mr. A
will have to repay not only the sum of Rs. X but also some extra money
(say Rs. Y) to Mr. B, sort of like a charge or rent for using Mr. B’s money.
• Rs. X is termed as the principal (P), Rs. Y is the interest (I) for using
Mr.B’s money & Rs. (X + Y) is termed as the Amount(A) which Mr. A
will have to repay to Mr. B after a certain period of time (T) .
• The interest to be paid by the borrower Mr. A to the lender Mr. B is
fixed before or while lending the money. Interest is fixed as a
percentage of the principal and is called Rate of interest (R).

• Money is usually lent out for a fixed period of time (T) . Interest is
calculated on the basis of this time period at the agreed rate, which is
charged either yearly, half-yearly or quarterly. T is always calculated in
Years. So a loan period of 6 months will have to be taken as 0.5 years.

There are two kinds of interest Simple interest & Compound Interest
Simple interest is always calculated on the actual principal originally borrowed.
Throughout the entire time period, a fixed amount of interest is paid whenever
due which is either yearly, half yearly of quarterly
For Eg. Mr. A borrows Rs. 2000 from Mr. B for a period of 2 years at an interest
rate of 10% payable yearly. How much will A pay B after 2 years?
Compound interest is calculated on the increasing principal. The interest is
calculated on the original principal and then added to the original principal
which becomes the new principal for calculating the interest on next due date.
• Simple Interest:
Original Principal Rs. 2000
Interest at 1st year end = 10%of 2000
= Rs. 200
Interest at 2nd year end = 10%of 2000
= Rs. 200
So at the end of 2 years Mr. A has to
pay Rs. (2000 + 200+200)=
Rs. 2400/-

• Compound Interest:
Original principal Rs. 2000
Interest at 1st year end = 10% of 2000
= Rs. 200

New principal=Rs. 2000+200 = 2200
Interest at 2nd year end = 10% of 2200
= Rs. 220
So at the end of 2 years Mr. A has to
pay Rs. (2200 + 220) = Rs. 2420/-

Part 1 Simple Interest

• If a sum of money (principal) Rs. P is borrowed at the rate of R% simple
interest for T years, then
Simple interest amount for T years = 1st years int. + 2nd yrs int. +….+ Tth years int
= (R% of P) + (R% of P) + (R% of P) + …T times
= (R% of P) (1 + 1 + 1 +……………upto T)
= (R% of P) * T
=

I=
• Principal (P)+ Interest (I)= Total Amount(A) to be repaid after T years
=> P +
I
= A
=> P + (R*P*T)/100 = A

Eg. Mr. A borrows Rs. 5400/- from Mr. B under the conditions that :

a) The loan will be repaid after 3 years
b) The borrower will pay a simple interest of 8% per annum (per year)
1) How much interest will Mr. A have to pay at the end of 3 years?
2) What is the amount that Mr. A will have to pay at the end of 3years?
Solution:

Here the Principal = Rs. 5400 Time = 3years Rate of interest= 8%
• Interest at 1st year end = 8% of 5400
•
= Rs. 432
• Interest at 2nd year end = 8% of 5400
•
= Rs. 432
• Interest at 3nd year end = 8% of 5400
•
= Rs. 432
• Total interest at the end of the 3 year loan period
= Rs. (432+432+432) = Rs. 1296
• So at the end of 3 years Mr. A has to pay Rs. (5400 + 1296 )
•
= Rs. 6696

Alternative Solution: USING FORMULA OF SIMPLE INTEREST
Here Principal (P) = Rs.5400, Time (T) = 3 years , Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400* 3) / 100
I = Rs. 1296
The amount to be paid to Mr. B at the end of 3 years :
Amount = P + I
= 5400 + 1296
= Rs. 6696

Page 388 (RSA) Ex 1
SOLUTION 1: Please Note: ALWAYS CONVERT DAYS OR MONTHS TO YEARS
Here the Principal = Rs. 5400
Time = 10/12 years Rate of int. = 8%pa
• Interest for 1 year
= 8% of 5400
= Rs. 432
• That means Interest for 12 months = Rs. 432
•  Interest for 10 months will be: (10/12) * 432 = Rs. 360

• Total interest at the end of the 10 month loan period = Rs. 360
SOLUTION 2 :
Here the Principal (P) = 5400

Time (T) = 10/12 years Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400*
) / 100 = Rs.__________

Page 390 Exercise 21A, Pr no. 4
Solution:
Let the sum of money be Rs. x which is the principal (P)
It is given that interest on Rs. x @8% p.a. in 6 years = Rs. 8376
Here principal is Rs .x, Rate of interest (R ) is 8% p.a. and the time
(T) is 6 years
Interest on Rs. x @ 8% for 6 years (I) = (R*P*T)/100
= (8 * x * 6) /100 = Rs.12x/25
So that means , 12x/25 = 8376
=>
x = (8376 * 25) /12
=>
x = Rs. ___________

Page 388, Ex. 4 Solution:
Let the sum of money (P) be Rs. x
And let the rate of interest (R ) be y% p.a
In the first case Rs. x becomes Rs. 854 in 2 years @ y% p.a
Interest @ y% on Rs. x for 2 years (I) = (R*P*T)/100 = (y * x * 2) /100
= (2xy)/100
= Rs. xy/50
So, x + (xy/50) = 854
=> 50x + xy = 42700…………(i)
Similarly, in the 2nd case
Interest on Rs. x @ y% for 7/2 years = (y *x*7/2) / 100 = Rs.(7xy/200)

So, x + (7xy/200) = 969.50
=> 200x + 7xy = 193900…….(ii)
Eliminating xy from (i) & (ii) we can get the value of x and then y.

Page 389, Ex. 5 Solution:
Let the sum of money (P) be Rs. x
So after 8 years, this sum of money becomes Rs. 2x.
Since, Principal

x

+ Interest
= Amount
+ (Interest on Rs. x for 8 years) = 2x
=> (Interest on Rs. x for 8 years) = 2x – x = Rs. x

Let the rate of interest be R% p.a.
Here principal (P) = Rs. x , Interest (I)= Rs. x and Time (T)= 8 years
We know that I = (R * P* T) /100
 R = (I * 100) / (P*T)
= ( x* 100) / (x * 8)
= (100x) /(8x) = 100/8
= 25/2 = 12.5 %

Page 389, Ex. 6
Solution:
Let the loan @ 8% p.a. be Rs. x
 Loan @ 10% p.a. = Rs. (8000 – x)
Given that total annual interest = Rs. 714
That means, in 1 year:
Interest earned on Rs. x@ 8% + Interest earned @ 10% on Rs. (8000-x) = 714
=> {(8 * x * 1) / 100} + [{10 * (8000-x) * 1} / 100] = 714
=>
Solve for x.

(8x/100)

+

{(8000-x) / 10}

=

714

Session 9; Chapter 5; Part 2
Compound Interest

Page 401 Ex 1.
SOLUTION 1 : Without using formula.
• Here Principal (P)= Rs. 8000 , Rate ( R) = 10% p.a ., Time (T) = 3 years
• Interest at 1st year end @ 10% on 8000 for 1 year= (8000 * 10 * 1) /100
•
= Rs. 800
• New principal for 2nd year = Rs. 8000+800 = 8800
• Interest at 2nd year end @ 10% on 8800 for 1 year = (8800 * 10 * 1) /100
•
= Rs. 880
• New principal for 3rd year = Rs. 8800 + 880 = Rs. 9680
• Interest at 3rd year end @ 10% on 9680 for 1 year = (9680 * 10 * 1) /100
•
= Rs. 968
So at the end of 3 years Rs. 8000 will become Rs. (9680 + 968) = Rs. 10648/• So the compound interest (CI) = A – P = 10648 – 8000 = Rs. 2648
•
ALTERNATIVELY : CI = Rs. (800 + 880+ 968) = Rs. 2648

FORMULAS OF COMPOUND INTEREST:
• If a sum of money (principal) Rs. P is borrowed /lent at the rate of R%
compound interest for T years compounded yearly , then :
Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed / lent at the rate of R% compound interest for
compounded half yearly, then :

T years

Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed /lent at the rate of R% compound interest for
compounded quarterly, then :
Compound Interest (CI) on Rs. P @ R% for T years =

T years

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 8000 Rate ( R) = 10% p.a

Time (T) = 3 years

• Compound interest (CI) on Rs. 8000 at the end of 3 years @ 10% =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 4.
SOLUTION 1 : Without using formula.
Here Principal (P)= Rs. 25000, Time (T) = 1 year, Yearly rate of interest =12%
• The Interest is compounded half yearly
• So the half yearly rate of interest = 12 * ½ = 6% per half year
• Interest at 1st half year end = 6% of 25000
•
= Rs. 1500
• New principal for 2nd half year = Rs. 25000+1500 = Rs. 26500
• Interest at 2nd half year end = 6% of 26500
•
= Rs. 1590
• So at the end of 1 year Rs. 25000 will become Rs. (26500 +1590)
= Rs. 28090/• So the compound interest (CI) = 28090 – 25000 = Rs. 3090
• Or , CI = Rs. (1500 + 1590)= Rs. 3090

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 25000 Rate ( R) = 12% p.a Time (T) = 1 year
• Compound interest (CI) on Rs. 25000 at the end of 1 years @ 12% pa
compounded half yearly =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 5. SOLUTION : Without using formula.
Here Principal (P)= Rs. 8000, Time (T) = 9 months = 9/12 years = ¾ yrs,
• Note: 9 months = 3 quarters
• Yearly rate of interest = 20% pa
• The Interest is compounded quarterly . In a year there are 4 quarters
• So the per quarter rate of interest = 20/4 = 5% per quarter.
• Principal for 1st quarter = Rs. 8000
• Interest at 1st quarter end = 5% of 8000 = Rs. 400

• New principal for 2nd quarter = Rs. 8000+400 = Rs. 8400
• Interest at 2nd quarter end = 5% of 8400 = Rs. 420
• New principal for 3rd quarter = Rs. 8400+420 = Rs. 8820
• Interest at 3rd quarter end = 5% of 8820 = Rs. 441
• So at the end of 3 quarters or 9 months, Rs. 8000 will become
Rs. (8820 +441) = Rs. 9261/• So the compound interest (CI) = 9261 – 8000 = Rs. 1261
• Or, CI = Rs. (400 + 420+ 441)= Rs. 1261

Page 402 Ex 3. SOLUTION :
• Here Principal (P)= Rs. 5000, Time (T) = 3years,
• Rate of interest for 1st year (R1) = 5% pa
• Principal for 1st year = Rs. 5000
Interest at 1st year end

= 5% of 5000 = Rs. 250

• New principal for 2nd year = Rs. 5000 + 250 = Rs. 5250
• Rate of interest for 2nd year (R2) = 8% pa
Interest at 2nd year end = 8% of 5250 = Rs. 420

• New principal for 3rd year = Rs. 5250 + 420 = Rs. 5670
• Rate of interest for 3rd year (R3) = 10% pa
Interest at 3rd year end = 10% of 5670 = Rs. 567
• So the amount at the end of 3 years = Rs. (5670 +567) = Rs. 6237/• This means, at the end of 3 years, Rs. 5000 will become Rs. 6237/• Compound interest (CI) on Rs. 5000 after 3 years @ 5%,8% &10%=
6237 – 5000 = Rs. 1237
• Or, CI on Rs. 5000 after 3 years @ 5%,8% &10%= ( 250+ 420+ 567)= Rs. 1237

Page 402 Ex 6. SOLUTION :
• Here Principal (P) = a certain sum of money, CI = Rs. 3783,
Time (T) = 3years, Rate of interest (R) = 5% pa
• Let the principal (P) be Rs. x
• Assuming interest is being charged annually, We know that,
CI =
=> CI on Rs. x @ 5% for 3 years =

=> 3783 =
Find x then find SI on Rs. x @ 5% for 3 years = (x*5*3)/100


Slide 3

Quantitative Methods
Session 8– 31.07.12
Chapter 5 – Simple Interest & Compound Interest

Pranjoy Arup Das

• Interest is the payment to be made for the use of money.
• Suppose, Mr. A borrows a sum of money Rs. X from Mr. B, then Mr. A
will have to repay not only the sum of Rs. X but also some extra money
(say Rs. Y) to Mr. B, sort of like a charge or rent for using Mr. B’s money.
• Rs. X is termed as the principal (P), Rs. Y is the interest (I) for using
Mr.B’s money & Rs. (X + Y) is termed as the Amount(A) which Mr. A
will have to repay to Mr. B after a certain period of time (T) .
• The interest to be paid by the borrower Mr. A to the lender Mr. B is
fixed before or while lending the money. Interest is fixed as a
percentage of the principal and is called Rate of interest (R).

• Money is usually lent out for a fixed period of time (T) . Interest is
calculated on the basis of this time period at the agreed rate, which is
charged either yearly, half-yearly or quarterly. T is always calculated in
Years. So a loan period of 6 months will have to be taken as 0.5 years.

There are two kinds of interest Simple interest & Compound Interest
Simple interest is always calculated on the actual principal originally borrowed.
Throughout the entire time period, a fixed amount of interest is paid whenever
due which is either yearly, half yearly of quarterly
For Eg. Mr. A borrows Rs. 2000 from Mr. B for a period of 2 years at an interest
rate of 10% payable yearly. How much will A pay B after 2 years?
Compound interest is calculated on the increasing principal. The interest is
calculated on the original principal and then added to the original principal
which becomes the new principal for calculating the interest on next due date.
• Simple Interest:
Original Principal Rs. 2000
Interest at 1st year end = 10%of 2000
= Rs. 200
Interest at 2nd year end = 10%of 2000
= Rs. 200
So at the end of 2 years Mr. A has to
pay Rs. (2000 + 200+200)=
Rs. 2400/-

• Compound Interest:
Original principal Rs. 2000
Interest at 1st year end = 10% of 2000
= Rs. 200

New principal=Rs. 2000+200 = 2200
Interest at 2nd year end = 10% of 2200
= Rs. 220
So at the end of 2 years Mr. A has to
pay Rs. (2200 + 220) = Rs. 2420/-

Part 1 Simple Interest

• If a sum of money (principal) Rs. P is borrowed at the rate of R% simple
interest for T years, then
Simple interest amount for T years = 1st years int. + 2nd yrs int. +….+ Tth years int
= (R% of P) + (R% of P) + (R% of P) + …T times
= (R% of P) (1 + 1 + 1 +……………upto T)
= (R% of P) * T
=

I=
• Principal (P)+ Interest (I)= Total Amount(A) to be repaid after T years
=> P +
I
= A
=> P + (R*P*T)/100 = A

Eg. Mr. A borrows Rs. 5400/- from Mr. B under the conditions that :

a) The loan will be repaid after 3 years
b) The borrower will pay a simple interest of 8% per annum (per year)
1) How much interest will Mr. A have to pay at the end of 3 years?
2) What is the amount that Mr. A will have to pay at the end of 3years?
Solution:

Here the Principal = Rs. 5400 Time = 3years Rate of interest= 8%
• Interest at 1st year end = 8% of 5400
•
= Rs. 432
• Interest at 2nd year end = 8% of 5400
•
= Rs. 432
• Interest at 3nd year end = 8% of 5400
•
= Rs. 432
• Total interest at the end of the 3 year loan period
= Rs. (432+432+432) = Rs. 1296
• So at the end of 3 years Mr. A has to pay Rs. (5400 + 1296 )
•
= Rs. 6696

Alternative Solution: USING FORMULA OF SIMPLE INTEREST
Here Principal (P) = Rs.5400, Time (T) = 3 years , Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400* 3) / 100
I = Rs. 1296
The amount to be paid to Mr. B at the end of 3 years :
Amount = P + I
= 5400 + 1296
= Rs. 6696

Page 388 (RSA) Ex 1
SOLUTION 1: Please Note: ALWAYS CONVERT DAYS OR MONTHS TO YEARS
Here the Principal = Rs. 5400
Time = 10/12 years Rate of int. = 8%pa
• Interest for 1 year
= 8% of 5400
= Rs. 432
• That means Interest for 12 months = Rs. 432
•  Interest for 10 months will be: (10/12) * 432 = Rs. 360

• Total interest at the end of the 10 month loan period = Rs. 360
SOLUTION 2 :
Here the Principal (P) = 5400

Time (T) = 10/12 years Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400*
) / 100 = Rs.__________

Page 390 Exercise 21A, Pr no. 4
Solution:
Let the sum of money be Rs. x which is the principal (P)
It is given that interest on Rs. x @8% p.a. in 6 years = Rs. 8376
Here principal is Rs .x, Rate of interest (R ) is 8% p.a. and the time
(T) is 6 years
Interest on Rs. x @ 8% for 6 years (I) = (R*P*T)/100
= (8 * x * 6) /100 = Rs.12x/25
So that means , 12x/25 = 8376
=>
x = (8376 * 25) /12
=>
x = Rs. ___________

Page 388, Ex. 4 Solution:
Let the sum of money (P) be Rs. x
And let the rate of interest (R ) be y% p.a
In the first case Rs. x becomes Rs. 854 in 2 years @ y% p.a
Interest @ y% on Rs. x for 2 years (I) = (R*P*T)/100 = (y * x * 2) /100
= (2xy)/100
= Rs. xy/50
So, x + (xy/50) = 854
=> 50x + xy = 42700…………(i)
Similarly, in the 2nd case
Interest on Rs. x @ y% for 7/2 years = (y *x*7/2) / 100 = Rs.(7xy/200)

So, x + (7xy/200) = 969.50
=> 200x + 7xy = 193900…….(ii)
Eliminating xy from (i) & (ii) we can get the value of x and then y.

Page 389, Ex. 5 Solution:
Let the sum of money (P) be Rs. x
So after 8 years, this sum of money becomes Rs. 2x.
Since, Principal

x

+ Interest
= Amount
+ (Interest on Rs. x for 8 years) = 2x
=> (Interest on Rs. x for 8 years) = 2x – x = Rs. x

Let the rate of interest be R% p.a.
Here principal (P) = Rs. x , Interest (I)= Rs. x and Time (T)= 8 years
We know that I = (R * P* T) /100
 R = (I * 100) / (P*T)
= ( x* 100) / (x * 8)
= (100x) /(8x) = 100/8
= 25/2 = 12.5 %

Page 389, Ex. 6
Solution:
Let the loan @ 8% p.a. be Rs. x
 Loan @ 10% p.a. = Rs. (8000 – x)
Given that total annual interest = Rs. 714
That means, in 1 year:
Interest earned on Rs. x@ 8% + Interest earned @ 10% on Rs. (8000-x) = 714
=> {(8 * x * 1) / 100} + [{10 * (8000-x) * 1} / 100] = 714
=>
Solve for x.

(8x/100)

+

{(8000-x) / 10}

=

714

Session 9; Chapter 5; Part 2
Compound Interest

Page 401 Ex 1.
SOLUTION 1 : Without using formula.
• Here Principal (P)= Rs. 8000 , Rate ( R) = 10% p.a ., Time (T) = 3 years
• Interest at 1st year end @ 10% on 8000 for 1 year= (8000 * 10 * 1) /100
•
= Rs. 800
• New principal for 2nd year = Rs. 8000+800 = 8800
• Interest at 2nd year end @ 10% on 8800 for 1 year = (8800 * 10 * 1) /100
•
= Rs. 880
• New principal for 3rd year = Rs. 8800 + 880 = Rs. 9680
• Interest at 3rd year end @ 10% on 9680 for 1 year = (9680 * 10 * 1) /100
•
= Rs. 968
So at the end of 3 years Rs. 8000 will become Rs. (9680 + 968) = Rs. 10648/• So the compound interest (CI) = A – P = 10648 – 8000 = Rs. 2648
•
ALTERNATIVELY : CI = Rs. (800 + 880+ 968) = Rs. 2648

FORMULAS OF COMPOUND INTEREST:
• If a sum of money (principal) Rs. P is borrowed /lent at the rate of R%
compound interest for T years compounded yearly , then :
Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed / lent at the rate of R% compound interest for
compounded half yearly, then :

T years

Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed /lent at the rate of R% compound interest for
compounded quarterly, then :
Compound Interest (CI) on Rs. P @ R% for T years =

T years

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 8000 Rate ( R) = 10% p.a

Time (T) = 3 years

• Compound interest (CI) on Rs. 8000 at the end of 3 years @ 10% =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 4.
SOLUTION 1 : Without using formula.
Here Principal (P)= Rs. 25000, Time (T) = 1 year, Yearly rate of interest =12%
• The Interest is compounded half yearly
• So the half yearly rate of interest = 12 * ½ = 6% per half year
• Interest at 1st half year end = 6% of 25000
•
= Rs. 1500
• New principal for 2nd half year = Rs. 25000+1500 = Rs. 26500
• Interest at 2nd half year end = 6% of 26500
•
= Rs. 1590
• So at the end of 1 year Rs. 25000 will become Rs. (26500 +1590)
= Rs. 28090/• So the compound interest (CI) = 28090 – 25000 = Rs. 3090
• Or , CI = Rs. (1500 + 1590)= Rs. 3090

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 25000 Rate ( R) = 12% p.a Time (T) = 1 year
• Compound interest (CI) on Rs. 25000 at the end of 1 years @ 12% pa
compounded half yearly =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 5. SOLUTION : Without using formula.
Here Principal (P)= Rs. 8000, Time (T) = 9 months = 9/12 years = ¾ yrs,
• Note: 9 months = 3 quarters
• Yearly rate of interest = 20% pa
• The Interest is compounded quarterly . In a year there are 4 quarters
• So the per quarter rate of interest = 20/4 = 5% per quarter.
• Principal for 1st quarter = Rs. 8000
• Interest at 1st quarter end = 5% of 8000 = Rs. 400

• New principal for 2nd quarter = Rs. 8000+400 = Rs. 8400
• Interest at 2nd quarter end = 5% of 8400 = Rs. 420
• New principal for 3rd quarter = Rs. 8400+420 = Rs. 8820
• Interest at 3rd quarter end = 5% of 8820 = Rs. 441
• So at the end of 3 quarters or 9 months, Rs. 8000 will become
Rs. (8820 +441) = Rs. 9261/• So the compound interest (CI) = 9261 – 8000 = Rs. 1261
• Or, CI = Rs. (400 + 420+ 441)= Rs. 1261

Page 402 Ex 3. SOLUTION :
• Here Principal (P)= Rs. 5000, Time (T) = 3years,
• Rate of interest for 1st year (R1) = 5% pa
• Principal for 1st year = Rs. 5000
Interest at 1st year end

= 5% of 5000 = Rs. 250

• New principal for 2nd year = Rs. 5000 + 250 = Rs. 5250
• Rate of interest for 2nd year (R2) = 8% pa
Interest at 2nd year end = 8% of 5250 = Rs. 420

• New principal for 3rd year = Rs. 5250 + 420 = Rs. 5670
• Rate of interest for 3rd year (R3) = 10% pa
Interest at 3rd year end = 10% of 5670 = Rs. 567
• So the amount at the end of 3 years = Rs. (5670 +567) = Rs. 6237/• This means, at the end of 3 years, Rs. 5000 will become Rs. 6237/• Compound interest (CI) on Rs. 5000 after 3 years @ 5%,8% &10%=
6237 – 5000 = Rs. 1237
• Or, CI on Rs. 5000 after 3 years @ 5%,8% &10%= ( 250+ 420+ 567)= Rs. 1237

Page 402 Ex 6. SOLUTION :
• Here Principal (P) = a certain sum of money, CI = Rs. 3783,
Time (T) = 3years, Rate of interest (R) = 5% pa
• Let the principal (P) be Rs. x
• Assuming interest is being charged annually, We know that,
CI =
=> CI on Rs. x @ 5% for 3 years =

=> 3783 =
Find x then find SI on Rs. x @ 5% for 3 years = (x*5*3)/100


Slide 4

Quantitative Methods
Session 8– 31.07.12
Chapter 5 – Simple Interest & Compound Interest

Pranjoy Arup Das

• Interest is the payment to be made for the use of money.
• Suppose, Mr. A borrows a sum of money Rs. X from Mr. B, then Mr. A
will have to repay not only the sum of Rs. X but also some extra money
(say Rs. Y) to Mr. B, sort of like a charge or rent for using Mr. B’s money.
• Rs. X is termed as the principal (P), Rs. Y is the interest (I) for using
Mr.B’s money & Rs. (X + Y) is termed as the Amount(A) which Mr. A
will have to repay to Mr. B after a certain period of time (T) .
• The interest to be paid by the borrower Mr. A to the lender Mr. B is
fixed before or while lending the money. Interest is fixed as a
percentage of the principal and is called Rate of interest (R).

• Money is usually lent out for a fixed period of time (T) . Interest is
calculated on the basis of this time period at the agreed rate, which is
charged either yearly, half-yearly or quarterly. T is always calculated in
Years. So a loan period of 6 months will have to be taken as 0.5 years.

There are two kinds of interest Simple interest & Compound Interest
Simple interest is always calculated on the actual principal originally borrowed.
Throughout the entire time period, a fixed amount of interest is paid whenever
due which is either yearly, half yearly of quarterly
For Eg. Mr. A borrows Rs. 2000 from Mr. B for a period of 2 years at an interest
rate of 10% payable yearly. How much will A pay B after 2 years?
Compound interest is calculated on the increasing principal. The interest is
calculated on the original principal and then added to the original principal
which becomes the new principal for calculating the interest on next due date.
• Simple Interest:
Original Principal Rs. 2000
Interest at 1st year end = 10%of 2000
= Rs. 200
Interest at 2nd year end = 10%of 2000
= Rs. 200
So at the end of 2 years Mr. A has to
pay Rs. (2000 + 200+200)=
Rs. 2400/-

• Compound Interest:
Original principal Rs. 2000
Interest at 1st year end = 10% of 2000
= Rs. 200

New principal=Rs. 2000+200 = 2200
Interest at 2nd year end = 10% of 2200
= Rs. 220
So at the end of 2 years Mr. A has to
pay Rs. (2200 + 220) = Rs. 2420/-

Part 1 Simple Interest

• If a sum of money (principal) Rs. P is borrowed at the rate of R% simple
interest for T years, then
Simple interest amount for T years = 1st years int. + 2nd yrs int. +….+ Tth years int
= (R% of P) + (R% of P) + (R% of P) + …T times
= (R% of P) (1 + 1 + 1 +……………upto T)
= (R% of P) * T
=

I=
• Principal (P)+ Interest (I)= Total Amount(A) to be repaid after T years
=> P +
I
= A
=> P + (R*P*T)/100 = A

Eg. Mr. A borrows Rs. 5400/- from Mr. B under the conditions that :

a) The loan will be repaid after 3 years
b) The borrower will pay a simple interest of 8% per annum (per year)
1) How much interest will Mr. A have to pay at the end of 3 years?
2) What is the amount that Mr. A will have to pay at the end of 3years?
Solution:

Here the Principal = Rs. 5400 Time = 3years Rate of interest= 8%
• Interest at 1st year end = 8% of 5400
•
= Rs. 432
• Interest at 2nd year end = 8% of 5400
•
= Rs. 432
• Interest at 3nd year end = 8% of 5400
•
= Rs. 432
• Total interest at the end of the 3 year loan period
= Rs. (432+432+432) = Rs. 1296
• So at the end of 3 years Mr. A has to pay Rs. (5400 + 1296 )
•
= Rs. 6696

Alternative Solution: USING FORMULA OF SIMPLE INTEREST
Here Principal (P) = Rs.5400, Time (T) = 3 years , Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400* 3) / 100
I = Rs. 1296
The amount to be paid to Mr. B at the end of 3 years :
Amount = P + I
= 5400 + 1296
= Rs. 6696

Page 388 (RSA) Ex 1
SOLUTION 1: Please Note: ALWAYS CONVERT DAYS OR MONTHS TO YEARS
Here the Principal = Rs. 5400
Time = 10/12 years Rate of int. = 8%pa
• Interest for 1 year
= 8% of 5400
= Rs. 432
• That means Interest for 12 months = Rs. 432
•  Interest for 10 months will be: (10/12) * 432 = Rs. 360

• Total interest at the end of the 10 month loan period = Rs. 360
SOLUTION 2 :
Here the Principal (P) = 5400

Time (T) = 10/12 years Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400*
) / 100 = Rs.__________

Page 390 Exercise 21A, Pr no. 4
Solution:
Let the sum of money be Rs. x which is the principal (P)
It is given that interest on Rs. x @8% p.a. in 6 years = Rs. 8376
Here principal is Rs .x, Rate of interest (R ) is 8% p.a. and the time
(T) is 6 years
Interest on Rs. x @ 8% for 6 years (I) = (R*P*T)/100
= (8 * x * 6) /100 = Rs.12x/25
So that means , 12x/25 = 8376
=>
x = (8376 * 25) /12
=>
x = Rs. ___________

Page 388, Ex. 4 Solution:
Let the sum of money (P) be Rs. x
And let the rate of interest (R ) be y% p.a
In the first case Rs. x becomes Rs. 854 in 2 years @ y% p.a
Interest @ y% on Rs. x for 2 years (I) = (R*P*T)/100 = (y * x * 2) /100
= (2xy)/100
= Rs. xy/50
So, x + (xy/50) = 854
=> 50x + xy = 42700…………(i)
Similarly, in the 2nd case
Interest on Rs. x @ y% for 7/2 years = (y *x*7/2) / 100 = Rs.(7xy/200)

So, x + (7xy/200) = 969.50
=> 200x + 7xy = 193900…….(ii)
Eliminating xy from (i) & (ii) we can get the value of x and then y.

Page 389, Ex. 5 Solution:
Let the sum of money (P) be Rs. x
So after 8 years, this sum of money becomes Rs. 2x.
Since, Principal

x

+ Interest
= Amount
+ (Interest on Rs. x for 8 years) = 2x
=> (Interest on Rs. x for 8 years) = 2x – x = Rs. x

Let the rate of interest be R% p.a.
Here principal (P) = Rs. x , Interest (I)= Rs. x and Time (T)= 8 years
We know that I = (R * P* T) /100
 R = (I * 100) / (P*T)
= ( x* 100) / (x * 8)
= (100x) /(8x) = 100/8
= 25/2 = 12.5 %

Page 389, Ex. 6
Solution:
Let the loan @ 8% p.a. be Rs. x
 Loan @ 10% p.a. = Rs. (8000 – x)
Given that total annual interest = Rs. 714
That means, in 1 year:
Interest earned on Rs. x@ 8% + Interest earned @ 10% on Rs. (8000-x) = 714
=> {(8 * x * 1) / 100} + [{10 * (8000-x) * 1} / 100] = 714
=>
Solve for x.

(8x/100)

+

{(8000-x) / 10}

=

714

Session 9; Chapter 5; Part 2
Compound Interest

Page 401 Ex 1.
SOLUTION 1 : Without using formula.
• Here Principal (P)= Rs. 8000 , Rate ( R) = 10% p.a ., Time (T) = 3 years
• Interest at 1st year end @ 10% on 8000 for 1 year= (8000 * 10 * 1) /100
•
= Rs. 800
• New principal for 2nd year = Rs. 8000+800 = 8800
• Interest at 2nd year end @ 10% on 8800 for 1 year = (8800 * 10 * 1) /100
•
= Rs. 880
• New principal for 3rd year = Rs. 8800 + 880 = Rs. 9680
• Interest at 3rd year end @ 10% on 9680 for 1 year = (9680 * 10 * 1) /100
•
= Rs. 968
So at the end of 3 years Rs. 8000 will become Rs. (9680 + 968) = Rs. 10648/• So the compound interest (CI) = A – P = 10648 – 8000 = Rs. 2648
•
ALTERNATIVELY : CI = Rs. (800 + 880+ 968) = Rs. 2648

FORMULAS OF COMPOUND INTEREST:
• If a sum of money (principal) Rs. P is borrowed /lent at the rate of R%
compound interest for T years compounded yearly , then :
Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed / lent at the rate of R% compound interest for
compounded half yearly, then :

T years

Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed /lent at the rate of R% compound interest for
compounded quarterly, then :
Compound Interest (CI) on Rs. P @ R% for T years =

T years

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 8000 Rate ( R) = 10% p.a

Time (T) = 3 years

• Compound interest (CI) on Rs. 8000 at the end of 3 years @ 10% =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 4.
SOLUTION 1 : Without using formula.
Here Principal (P)= Rs. 25000, Time (T) = 1 year, Yearly rate of interest =12%
• The Interest is compounded half yearly
• So the half yearly rate of interest = 12 * ½ = 6% per half year
• Interest at 1st half year end = 6% of 25000
•
= Rs. 1500
• New principal for 2nd half year = Rs. 25000+1500 = Rs. 26500
• Interest at 2nd half year end = 6% of 26500
•
= Rs. 1590
• So at the end of 1 year Rs. 25000 will become Rs. (26500 +1590)
= Rs. 28090/• So the compound interest (CI) = 28090 – 25000 = Rs. 3090
• Or , CI = Rs. (1500 + 1590)= Rs. 3090

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 25000 Rate ( R) = 12% p.a Time (T) = 1 year
• Compound interest (CI) on Rs. 25000 at the end of 1 years @ 12% pa
compounded half yearly =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 5. SOLUTION : Without using formula.
Here Principal (P)= Rs. 8000, Time (T) = 9 months = 9/12 years = ¾ yrs,
• Note: 9 months = 3 quarters
• Yearly rate of interest = 20% pa
• The Interest is compounded quarterly . In a year there are 4 quarters
• So the per quarter rate of interest = 20/4 = 5% per quarter.
• Principal for 1st quarter = Rs. 8000
• Interest at 1st quarter end = 5% of 8000 = Rs. 400

• New principal for 2nd quarter = Rs. 8000+400 = Rs. 8400
• Interest at 2nd quarter end = 5% of 8400 = Rs. 420
• New principal for 3rd quarter = Rs. 8400+420 = Rs. 8820
• Interest at 3rd quarter end = 5% of 8820 = Rs. 441
• So at the end of 3 quarters or 9 months, Rs. 8000 will become
Rs. (8820 +441) = Rs. 9261/• So the compound interest (CI) = 9261 – 8000 = Rs. 1261
• Or, CI = Rs. (400 + 420+ 441)= Rs. 1261

Page 402 Ex 3. SOLUTION :
• Here Principal (P)= Rs. 5000, Time (T) = 3years,
• Rate of interest for 1st year (R1) = 5% pa
• Principal for 1st year = Rs. 5000
Interest at 1st year end

= 5% of 5000 = Rs. 250

• New principal for 2nd year = Rs. 5000 + 250 = Rs. 5250
• Rate of interest for 2nd year (R2) = 8% pa
Interest at 2nd year end = 8% of 5250 = Rs. 420

• New principal for 3rd year = Rs. 5250 + 420 = Rs. 5670
• Rate of interest for 3rd year (R3) = 10% pa
Interest at 3rd year end = 10% of 5670 = Rs. 567
• So the amount at the end of 3 years = Rs. (5670 +567) = Rs. 6237/• This means, at the end of 3 years, Rs. 5000 will become Rs. 6237/• Compound interest (CI) on Rs. 5000 after 3 years @ 5%,8% &10%=
6237 – 5000 = Rs. 1237
• Or, CI on Rs. 5000 after 3 years @ 5%,8% &10%= ( 250+ 420+ 567)= Rs. 1237

Page 402 Ex 6. SOLUTION :
• Here Principal (P) = a certain sum of money, CI = Rs. 3783,
Time (T) = 3years, Rate of interest (R) = 5% pa
• Let the principal (P) be Rs. x
• Assuming interest is being charged annually, We know that,
CI =
=> CI on Rs. x @ 5% for 3 years =

=> 3783 =
Find x then find SI on Rs. x @ 5% for 3 years = (x*5*3)/100


Slide 5

Quantitative Methods
Session 8– 31.07.12
Chapter 5 – Simple Interest & Compound Interest

Pranjoy Arup Das

• Interest is the payment to be made for the use of money.
• Suppose, Mr. A borrows a sum of money Rs. X from Mr. B, then Mr. A
will have to repay not only the sum of Rs. X but also some extra money
(say Rs. Y) to Mr. B, sort of like a charge or rent for using Mr. B’s money.
• Rs. X is termed as the principal (P), Rs. Y is the interest (I) for using
Mr.B’s money & Rs. (X + Y) is termed as the Amount(A) which Mr. A
will have to repay to Mr. B after a certain period of time (T) .
• The interest to be paid by the borrower Mr. A to the lender Mr. B is
fixed before or while lending the money. Interest is fixed as a
percentage of the principal and is called Rate of interest (R).

• Money is usually lent out for a fixed period of time (T) . Interest is
calculated on the basis of this time period at the agreed rate, which is
charged either yearly, half-yearly or quarterly. T is always calculated in
Years. So a loan period of 6 months will have to be taken as 0.5 years.

There are two kinds of interest Simple interest & Compound Interest
Simple interest is always calculated on the actual principal originally borrowed.
Throughout the entire time period, a fixed amount of interest is paid whenever
due which is either yearly, half yearly of quarterly
For Eg. Mr. A borrows Rs. 2000 from Mr. B for a period of 2 years at an interest
rate of 10% payable yearly. How much will A pay B after 2 years?
Compound interest is calculated on the increasing principal. The interest is
calculated on the original principal and then added to the original principal
which becomes the new principal for calculating the interest on next due date.
• Simple Interest:
Original Principal Rs. 2000
Interest at 1st year end = 10%of 2000
= Rs. 200
Interest at 2nd year end = 10%of 2000
= Rs. 200
So at the end of 2 years Mr. A has to
pay Rs. (2000 + 200+200)=
Rs. 2400/-

• Compound Interest:
Original principal Rs. 2000
Interest at 1st year end = 10% of 2000
= Rs. 200

New principal=Rs. 2000+200 = 2200
Interest at 2nd year end = 10% of 2200
= Rs. 220
So at the end of 2 years Mr. A has to
pay Rs. (2200 + 220) = Rs. 2420/-

Part 1 Simple Interest

• If a sum of money (principal) Rs. P is borrowed at the rate of R% simple
interest for T years, then
Simple interest amount for T years = 1st years int. + 2nd yrs int. +….+ Tth years int
= (R% of P) + (R% of P) + (R% of P) + …T times
= (R% of P) (1 + 1 + 1 +……………upto T)
= (R% of P) * T
=

I=
• Principal (P)+ Interest (I)= Total Amount(A) to be repaid after T years
=> P +
I
= A
=> P + (R*P*T)/100 = A

Eg. Mr. A borrows Rs. 5400/- from Mr. B under the conditions that :

a) The loan will be repaid after 3 years
b) The borrower will pay a simple interest of 8% per annum (per year)
1) How much interest will Mr. A have to pay at the end of 3 years?
2) What is the amount that Mr. A will have to pay at the end of 3years?
Solution:

Here the Principal = Rs. 5400 Time = 3years Rate of interest= 8%
• Interest at 1st year end = 8% of 5400
•
= Rs. 432
• Interest at 2nd year end = 8% of 5400
•
= Rs. 432
• Interest at 3nd year end = 8% of 5400
•
= Rs. 432
• Total interest at the end of the 3 year loan period
= Rs. (432+432+432) = Rs. 1296
• So at the end of 3 years Mr. A has to pay Rs. (5400 + 1296 )
•
= Rs. 6696

Alternative Solution: USING FORMULA OF SIMPLE INTEREST
Here Principal (P) = Rs.5400, Time (T) = 3 years , Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400* 3) / 100
I = Rs. 1296
The amount to be paid to Mr. B at the end of 3 years :
Amount = P + I
= 5400 + 1296
= Rs. 6696

Page 388 (RSA) Ex 1
SOLUTION 1: Please Note: ALWAYS CONVERT DAYS OR MONTHS TO YEARS
Here the Principal = Rs. 5400
Time = 10/12 years Rate of int. = 8%pa
• Interest for 1 year
= 8% of 5400
= Rs. 432
• That means Interest for 12 months = Rs. 432
•  Interest for 10 months will be: (10/12) * 432 = Rs. 360

• Total interest at the end of the 10 month loan period = Rs. 360
SOLUTION 2 :
Here the Principal (P) = 5400

Time (T) = 10/12 years Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400*
) / 100 = Rs.__________

Page 390 Exercise 21A, Pr no. 4
Solution:
Let the sum of money be Rs. x which is the principal (P)
It is given that interest on Rs. x @8% p.a. in 6 years = Rs. 8376
Here principal is Rs .x, Rate of interest (R ) is 8% p.a. and the time
(T) is 6 years
Interest on Rs. x @ 8% for 6 years (I) = (R*P*T)/100
= (8 * x * 6) /100 = Rs.12x/25
So that means , 12x/25 = 8376
=>
x = (8376 * 25) /12
=>
x = Rs. ___________

Page 388, Ex. 4 Solution:
Let the sum of money (P) be Rs. x
And let the rate of interest (R ) be y% p.a
In the first case Rs. x becomes Rs. 854 in 2 years @ y% p.a
Interest @ y% on Rs. x for 2 years (I) = (R*P*T)/100 = (y * x * 2) /100
= (2xy)/100
= Rs. xy/50
So, x + (xy/50) = 854
=> 50x + xy = 42700…………(i)
Similarly, in the 2nd case
Interest on Rs. x @ y% for 7/2 years = (y *x*7/2) / 100 = Rs.(7xy/200)

So, x + (7xy/200) = 969.50
=> 200x + 7xy = 193900…….(ii)
Eliminating xy from (i) & (ii) we can get the value of x and then y.

Page 389, Ex. 5 Solution:
Let the sum of money (P) be Rs. x
So after 8 years, this sum of money becomes Rs. 2x.
Since, Principal

x

+ Interest
= Amount
+ (Interest on Rs. x for 8 years) = 2x
=> (Interest on Rs. x for 8 years) = 2x – x = Rs. x

Let the rate of interest be R% p.a.
Here principal (P) = Rs. x , Interest (I)= Rs. x and Time (T)= 8 years
We know that I = (R * P* T) /100
 R = (I * 100) / (P*T)
= ( x* 100) / (x * 8)
= (100x) /(8x) = 100/8
= 25/2 = 12.5 %

Page 389, Ex. 6
Solution:
Let the loan @ 8% p.a. be Rs. x
 Loan @ 10% p.a. = Rs. (8000 – x)
Given that total annual interest = Rs. 714
That means, in 1 year:
Interest earned on Rs. x@ 8% + Interest earned @ 10% on Rs. (8000-x) = 714
=> {(8 * x * 1) / 100} + [{10 * (8000-x) * 1} / 100] = 714
=>
Solve for x.

(8x/100)

+

{(8000-x) / 10}

=

714

Session 9; Chapter 5; Part 2
Compound Interest

Page 401 Ex 1.
SOLUTION 1 : Without using formula.
• Here Principal (P)= Rs. 8000 , Rate ( R) = 10% p.a ., Time (T) = 3 years
• Interest at 1st year end @ 10% on 8000 for 1 year= (8000 * 10 * 1) /100
•
= Rs. 800
• New principal for 2nd year = Rs. 8000+800 = 8800
• Interest at 2nd year end @ 10% on 8800 for 1 year = (8800 * 10 * 1) /100
•
= Rs. 880
• New principal for 3rd year = Rs. 8800 + 880 = Rs. 9680
• Interest at 3rd year end @ 10% on 9680 for 1 year = (9680 * 10 * 1) /100
•
= Rs. 968
So at the end of 3 years Rs. 8000 will become Rs. (9680 + 968) = Rs. 10648/• So the compound interest (CI) = A – P = 10648 – 8000 = Rs. 2648
•
ALTERNATIVELY : CI = Rs. (800 + 880+ 968) = Rs. 2648

FORMULAS OF COMPOUND INTEREST:
• If a sum of money (principal) Rs. P is borrowed /lent at the rate of R%
compound interest for T years compounded yearly , then :
Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed / lent at the rate of R% compound interest for
compounded half yearly, then :

T years

Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed /lent at the rate of R% compound interest for
compounded quarterly, then :
Compound Interest (CI) on Rs. P @ R% for T years =

T years

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 8000 Rate ( R) = 10% p.a

Time (T) = 3 years

• Compound interest (CI) on Rs. 8000 at the end of 3 years @ 10% =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 4.
SOLUTION 1 : Without using formula.
Here Principal (P)= Rs. 25000, Time (T) = 1 year, Yearly rate of interest =12%
• The Interest is compounded half yearly
• So the half yearly rate of interest = 12 * ½ = 6% per half year
• Interest at 1st half year end = 6% of 25000
•
= Rs. 1500
• New principal for 2nd half year = Rs. 25000+1500 = Rs. 26500
• Interest at 2nd half year end = 6% of 26500
•
= Rs. 1590
• So at the end of 1 year Rs. 25000 will become Rs. (26500 +1590)
= Rs. 28090/• So the compound interest (CI) = 28090 – 25000 = Rs. 3090
• Or , CI = Rs. (1500 + 1590)= Rs. 3090

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 25000 Rate ( R) = 12% p.a Time (T) = 1 year
• Compound interest (CI) on Rs. 25000 at the end of 1 years @ 12% pa
compounded half yearly =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 5. SOLUTION : Without using formula.
Here Principal (P)= Rs. 8000, Time (T) = 9 months = 9/12 years = ¾ yrs,
• Note: 9 months = 3 quarters
• Yearly rate of interest = 20% pa
• The Interest is compounded quarterly . In a year there are 4 quarters
• So the per quarter rate of interest = 20/4 = 5% per quarter.
• Principal for 1st quarter = Rs. 8000
• Interest at 1st quarter end = 5% of 8000 = Rs. 400

• New principal for 2nd quarter = Rs. 8000+400 = Rs. 8400
• Interest at 2nd quarter end = 5% of 8400 = Rs. 420
• New principal for 3rd quarter = Rs. 8400+420 = Rs. 8820
• Interest at 3rd quarter end = 5% of 8820 = Rs. 441
• So at the end of 3 quarters or 9 months, Rs. 8000 will become
Rs. (8820 +441) = Rs. 9261/• So the compound interest (CI) = 9261 – 8000 = Rs. 1261
• Or, CI = Rs. (400 + 420+ 441)= Rs. 1261

Page 402 Ex 3. SOLUTION :
• Here Principal (P)= Rs. 5000, Time (T) = 3years,
• Rate of interest for 1st year (R1) = 5% pa
• Principal for 1st year = Rs. 5000
Interest at 1st year end

= 5% of 5000 = Rs. 250

• New principal for 2nd year = Rs. 5000 + 250 = Rs. 5250
• Rate of interest for 2nd year (R2) = 8% pa
Interest at 2nd year end = 8% of 5250 = Rs. 420

• New principal for 3rd year = Rs. 5250 + 420 = Rs. 5670
• Rate of interest for 3rd year (R3) = 10% pa
Interest at 3rd year end = 10% of 5670 = Rs. 567
• So the amount at the end of 3 years = Rs. (5670 +567) = Rs. 6237/• This means, at the end of 3 years, Rs. 5000 will become Rs. 6237/• Compound interest (CI) on Rs. 5000 after 3 years @ 5%,8% &10%=
6237 – 5000 = Rs. 1237
• Or, CI on Rs. 5000 after 3 years @ 5%,8% &10%= ( 250+ 420+ 567)= Rs. 1237

Page 402 Ex 6. SOLUTION :
• Here Principal (P) = a certain sum of money, CI = Rs. 3783,
Time (T) = 3years, Rate of interest (R) = 5% pa
• Let the principal (P) be Rs. x
• Assuming interest is being charged annually, We know that,
CI =
=> CI on Rs. x @ 5% for 3 years =

=> 3783 =
Find x then find SI on Rs. x @ 5% for 3 years = (x*5*3)/100


Slide 6

Quantitative Methods
Session 8– 31.07.12
Chapter 5 – Simple Interest & Compound Interest

Pranjoy Arup Das

• Interest is the payment to be made for the use of money.
• Suppose, Mr. A borrows a sum of money Rs. X from Mr. B, then Mr. A
will have to repay not only the sum of Rs. X but also some extra money
(say Rs. Y) to Mr. B, sort of like a charge or rent for using Mr. B’s money.
• Rs. X is termed as the principal (P), Rs. Y is the interest (I) for using
Mr.B’s money & Rs. (X + Y) is termed as the Amount(A) which Mr. A
will have to repay to Mr. B after a certain period of time (T) .
• The interest to be paid by the borrower Mr. A to the lender Mr. B is
fixed before or while lending the money. Interest is fixed as a
percentage of the principal and is called Rate of interest (R).

• Money is usually lent out for a fixed period of time (T) . Interest is
calculated on the basis of this time period at the agreed rate, which is
charged either yearly, half-yearly or quarterly. T is always calculated in
Years. So a loan period of 6 months will have to be taken as 0.5 years.

There are two kinds of interest Simple interest & Compound Interest
Simple interest is always calculated on the actual principal originally borrowed.
Throughout the entire time period, a fixed amount of interest is paid whenever
due which is either yearly, half yearly of quarterly
For Eg. Mr. A borrows Rs. 2000 from Mr. B for a period of 2 years at an interest
rate of 10% payable yearly. How much will A pay B after 2 years?
Compound interest is calculated on the increasing principal. The interest is
calculated on the original principal and then added to the original principal
which becomes the new principal for calculating the interest on next due date.
• Simple Interest:
Original Principal Rs. 2000
Interest at 1st year end = 10%of 2000
= Rs. 200
Interest at 2nd year end = 10%of 2000
= Rs. 200
So at the end of 2 years Mr. A has to
pay Rs. (2000 + 200+200)=
Rs. 2400/-

• Compound Interest:
Original principal Rs. 2000
Interest at 1st year end = 10% of 2000
= Rs. 200

New principal=Rs. 2000+200 = 2200
Interest at 2nd year end = 10% of 2200
= Rs. 220
So at the end of 2 years Mr. A has to
pay Rs. (2200 + 220) = Rs. 2420/-

Part 1 Simple Interest

• If a sum of money (principal) Rs. P is borrowed at the rate of R% simple
interest for T years, then
Simple interest amount for T years = 1st years int. + 2nd yrs int. +….+ Tth years int
= (R% of P) + (R% of P) + (R% of P) + …T times
= (R% of P) (1 + 1 + 1 +……………upto T)
= (R% of P) * T
=

I=
• Principal (P)+ Interest (I)= Total Amount(A) to be repaid after T years
=> P +
I
= A
=> P + (R*P*T)/100 = A

Eg. Mr. A borrows Rs. 5400/- from Mr. B under the conditions that :

a) The loan will be repaid after 3 years
b) The borrower will pay a simple interest of 8% per annum (per year)
1) How much interest will Mr. A have to pay at the end of 3 years?
2) What is the amount that Mr. A will have to pay at the end of 3years?
Solution:

Here the Principal = Rs. 5400 Time = 3years Rate of interest= 8%
• Interest at 1st year end = 8% of 5400
•
= Rs. 432
• Interest at 2nd year end = 8% of 5400
•
= Rs. 432
• Interest at 3nd year end = 8% of 5400
•
= Rs. 432
• Total interest at the end of the 3 year loan period
= Rs. (432+432+432) = Rs. 1296
• So at the end of 3 years Mr. A has to pay Rs. (5400 + 1296 )
•
= Rs. 6696

Alternative Solution: USING FORMULA OF SIMPLE INTEREST
Here Principal (P) = Rs.5400, Time (T) = 3 years , Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400* 3) / 100
I = Rs. 1296
The amount to be paid to Mr. B at the end of 3 years :
Amount = P + I
= 5400 + 1296
= Rs. 6696

Page 388 (RSA) Ex 1
SOLUTION 1: Please Note: ALWAYS CONVERT DAYS OR MONTHS TO YEARS
Here the Principal = Rs. 5400
Time = 10/12 years Rate of int. = 8%pa
• Interest for 1 year
= 8% of 5400
= Rs. 432
• That means Interest for 12 months = Rs. 432
•  Interest for 10 months will be: (10/12) * 432 = Rs. 360

• Total interest at the end of the 10 month loan period = Rs. 360
SOLUTION 2 :
Here the Principal (P) = 5400

Time (T) = 10/12 years Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400*
) / 100 = Rs.__________

Page 390 Exercise 21A, Pr no. 4
Solution:
Let the sum of money be Rs. x which is the principal (P)
It is given that interest on Rs. x @8% p.a. in 6 years = Rs. 8376
Here principal is Rs .x, Rate of interest (R ) is 8% p.a. and the time
(T) is 6 years
Interest on Rs. x @ 8% for 6 years (I) = (R*P*T)/100
= (8 * x * 6) /100 = Rs.12x/25
So that means , 12x/25 = 8376
=>
x = (8376 * 25) /12
=>
x = Rs. ___________

Page 388, Ex. 4 Solution:
Let the sum of money (P) be Rs. x
And let the rate of interest (R ) be y% p.a
In the first case Rs. x becomes Rs. 854 in 2 years @ y% p.a
Interest @ y% on Rs. x for 2 years (I) = (R*P*T)/100 = (y * x * 2) /100
= (2xy)/100
= Rs. xy/50
So, x + (xy/50) = 854
=> 50x + xy = 42700…………(i)
Similarly, in the 2nd case
Interest on Rs. x @ y% for 7/2 years = (y *x*7/2) / 100 = Rs.(7xy/200)

So, x + (7xy/200) = 969.50
=> 200x + 7xy = 193900…….(ii)
Eliminating xy from (i) & (ii) we can get the value of x and then y.

Page 389, Ex. 5 Solution:
Let the sum of money (P) be Rs. x
So after 8 years, this sum of money becomes Rs. 2x.
Since, Principal

x

+ Interest
= Amount
+ (Interest on Rs. x for 8 years) = 2x
=> (Interest on Rs. x for 8 years) = 2x – x = Rs. x

Let the rate of interest be R% p.a.
Here principal (P) = Rs. x , Interest (I)= Rs. x and Time (T)= 8 years
We know that I = (R * P* T) /100
 R = (I * 100) / (P*T)
= ( x* 100) / (x * 8)
= (100x) /(8x) = 100/8
= 25/2 = 12.5 %

Page 389, Ex. 6
Solution:
Let the loan @ 8% p.a. be Rs. x
 Loan @ 10% p.a. = Rs. (8000 – x)
Given that total annual interest = Rs. 714
That means, in 1 year:
Interest earned on Rs. x@ 8% + Interest earned @ 10% on Rs. (8000-x) = 714
=> {(8 * x * 1) / 100} + [{10 * (8000-x) * 1} / 100] = 714
=>
Solve for x.

(8x/100)

+

{(8000-x) / 10}

=

714

Session 9; Chapter 5; Part 2
Compound Interest

Page 401 Ex 1.
SOLUTION 1 : Without using formula.
• Here Principal (P)= Rs. 8000 , Rate ( R) = 10% p.a ., Time (T) = 3 years
• Interest at 1st year end @ 10% on 8000 for 1 year= (8000 * 10 * 1) /100
•
= Rs. 800
• New principal for 2nd year = Rs. 8000+800 = 8800
• Interest at 2nd year end @ 10% on 8800 for 1 year = (8800 * 10 * 1) /100
•
= Rs. 880
• New principal for 3rd year = Rs. 8800 + 880 = Rs. 9680
• Interest at 3rd year end @ 10% on 9680 for 1 year = (9680 * 10 * 1) /100
•
= Rs. 968
So at the end of 3 years Rs. 8000 will become Rs. (9680 + 968) = Rs. 10648/• So the compound interest (CI) = A – P = 10648 – 8000 = Rs. 2648
•
ALTERNATIVELY : CI = Rs. (800 + 880+ 968) = Rs. 2648

FORMULAS OF COMPOUND INTEREST:
• If a sum of money (principal) Rs. P is borrowed /lent at the rate of R%
compound interest for T years compounded yearly , then :
Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed / lent at the rate of R% compound interest for
compounded half yearly, then :

T years

Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed /lent at the rate of R% compound interest for
compounded quarterly, then :
Compound Interest (CI) on Rs. P @ R% for T years =

T years

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 8000 Rate ( R) = 10% p.a

Time (T) = 3 years

• Compound interest (CI) on Rs. 8000 at the end of 3 years @ 10% =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 4.
SOLUTION 1 : Without using formula.
Here Principal (P)= Rs. 25000, Time (T) = 1 year, Yearly rate of interest =12%
• The Interest is compounded half yearly
• So the half yearly rate of interest = 12 * ½ = 6% per half year
• Interest at 1st half year end = 6% of 25000
•
= Rs. 1500
• New principal for 2nd half year = Rs. 25000+1500 = Rs. 26500
• Interest at 2nd half year end = 6% of 26500
•
= Rs. 1590
• So at the end of 1 year Rs. 25000 will become Rs. (26500 +1590)
= Rs. 28090/• So the compound interest (CI) = 28090 – 25000 = Rs. 3090
• Or , CI = Rs. (1500 + 1590)= Rs. 3090

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 25000 Rate ( R) = 12% p.a Time (T) = 1 year
• Compound interest (CI) on Rs. 25000 at the end of 1 years @ 12% pa
compounded half yearly =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 5. SOLUTION : Without using formula.
Here Principal (P)= Rs. 8000, Time (T) = 9 months = 9/12 years = ¾ yrs,
• Note: 9 months = 3 quarters
• Yearly rate of interest = 20% pa
• The Interest is compounded quarterly . In a year there are 4 quarters
• So the per quarter rate of interest = 20/4 = 5% per quarter.
• Principal for 1st quarter = Rs. 8000
• Interest at 1st quarter end = 5% of 8000 = Rs. 400

• New principal for 2nd quarter = Rs. 8000+400 = Rs. 8400
• Interest at 2nd quarter end = 5% of 8400 = Rs. 420
• New principal for 3rd quarter = Rs. 8400+420 = Rs. 8820
• Interest at 3rd quarter end = 5% of 8820 = Rs. 441
• So at the end of 3 quarters or 9 months, Rs. 8000 will become
Rs. (8820 +441) = Rs. 9261/• So the compound interest (CI) = 9261 – 8000 = Rs. 1261
• Or, CI = Rs. (400 + 420+ 441)= Rs. 1261

Page 402 Ex 3. SOLUTION :
• Here Principal (P)= Rs. 5000, Time (T) = 3years,
• Rate of interest for 1st year (R1) = 5% pa
• Principal for 1st year = Rs. 5000
Interest at 1st year end

= 5% of 5000 = Rs. 250

• New principal for 2nd year = Rs. 5000 + 250 = Rs. 5250
• Rate of interest for 2nd year (R2) = 8% pa
Interest at 2nd year end = 8% of 5250 = Rs. 420

• New principal for 3rd year = Rs. 5250 + 420 = Rs. 5670
• Rate of interest for 3rd year (R3) = 10% pa
Interest at 3rd year end = 10% of 5670 = Rs. 567
• So the amount at the end of 3 years = Rs. (5670 +567) = Rs. 6237/• This means, at the end of 3 years, Rs. 5000 will become Rs. 6237/• Compound interest (CI) on Rs. 5000 after 3 years @ 5%,8% &10%=
6237 – 5000 = Rs. 1237
• Or, CI on Rs. 5000 after 3 years @ 5%,8% &10%= ( 250+ 420+ 567)= Rs. 1237

Page 402 Ex 6. SOLUTION :
• Here Principal (P) = a certain sum of money, CI = Rs. 3783,
Time (T) = 3years, Rate of interest (R) = 5% pa
• Let the principal (P) be Rs. x
• Assuming interest is being charged annually, We know that,
CI =
=> CI on Rs. x @ 5% for 3 years =

=> 3783 =
Find x then find SI on Rs. x @ 5% for 3 years = (x*5*3)/100


Slide 7

Quantitative Methods
Session 8– 31.07.12
Chapter 5 – Simple Interest & Compound Interest

Pranjoy Arup Das

• Interest is the payment to be made for the use of money.
• Suppose, Mr. A borrows a sum of money Rs. X from Mr. B, then Mr. A
will have to repay not only the sum of Rs. X but also some extra money
(say Rs. Y) to Mr. B, sort of like a charge or rent for using Mr. B’s money.
• Rs. X is termed as the principal (P), Rs. Y is the interest (I) for using
Mr.B’s money & Rs. (X + Y) is termed as the Amount(A) which Mr. A
will have to repay to Mr. B after a certain period of time (T) .
• The interest to be paid by the borrower Mr. A to the lender Mr. B is
fixed before or while lending the money. Interest is fixed as a
percentage of the principal and is called Rate of interest (R).

• Money is usually lent out for a fixed period of time (T) . Interest is
calculated on the basis of this time period at the agreed rate, which is
charged either yearly, half-yearly or quarterly. T is always calculated in
Years. So a loan period of 6 months will have to be taken as 0.5 years.

There are two kinds of interest Simple interest & Compound Interest
Simple interest is always calculated on the actual principal originally borrowed.
Throughout the entire time period, a fixed amount of interest is paid whenever
due which is either yearly, half yearly of quarterly
For Eg. Mr. A borrows Rs. 2000 from Mr. B for a period of 2 years at an interest
rate of 10% payable yearly. How much will A pay B after 2 years?
Compound interest is calculated on the increasing principal. The interest is
calculated on the original principal and then added to the original principal
which becomes the new principal for calculating the interest on next due date.
• Simple Interest:
Original Principal Rs. 2000
Interest at 1st year end = 10%of 2000
= Rs. 200
Interest at 2nd year end = 10%of 2000
= Rs. 200
So at the end of 2 years Mr. A has to
pay Rs. (2000 + 200+200)=
Rs. 2400/-

• Compound Interest:
Original principal Rs. 2000
Interest at 1st year end = 10% of 2000
= Rs. 200

New principal=Rs. 2000+200 = 2200
Interest at 2nd year end = 10% of 2200
= Rs. 220
So at the end of 2 years Mr. A has to
pay Rs. (2200 + 220) = Rs. 2420/-

Part 1 Simple Interest

• If a sum of money (principal) Rs. P is borrowed at the rate of R% simple
interest for T years, then
Simple interest amount for T years = 1st years int. + 2nd yrs int. +….+ Tth years int
= (R% of P) + (R% of P) + (R% of P) + …T times
= (R% of P) (1 + 1 + 1 +……………upto T)
= (R% of P) * T
=

I=
• Principal (P)+ Interest (I)= Total Amount(A) to be repaid after T years
=> P +
I
= A
=> P + (R*P*T)/100 = A

Eg. Mr. A borrows Rs. 5400/- from Mr. B under the conditions that :

a) The loan will be repaid after 3 years
b) The borrower will pay a simple interest of 8% per annum (per year)
1) How much interest will Mr. A have to pay at the end of 3 years?
2) What is the amount that Mr. A will have to pay at the end of 3years?
Solution:

Here the Principal = Rs. 5400 Time = 3years Rate of interest= 8%
• Interest at 1st year end = 8% of 5400
•
= Rs. 432
• Interest at 2nd year end = 8% of 5400
•
= Rs. 432
• Interest at 3nd year end = 8% of 5400
•
= Rs. 432
• Total interest at the end of the 3 year loan period
= Rs. (432+432+432) = Rs. 1296
• So at the end of 3 years Mr. A has to pay Rs. (5400 + 1296 )
•
= Rs. 6696

Alternative Solution: USING FORMULA OF SIMPLE INTEREST
Here Principal (P) = Rs.5400, Time (T) = 3 years , Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400* 3) / 100
I = Rs. 1296
The amount to be paid to Mr. B at the end of 3 years :
Amount = P + I
= 5400 + 1296
= Rs. 6696

Page 388 (RSA) Ex 1
SOLUTION 1: Please Note: ALWAYS CONVERT DAYS OR MONTHS TO YEARS
Here the Principal = Rs. 5400
Time = 10/12 years Rate of int. = 8%pa
• Interest for 1 year
= 8% of 5400
= Rs. 432
• That means Interest for 12 months = Rs. 432
•  Interest for 10 months will be: (10/12) * 432 = Rs. 360

• Total interest at the end of the 10 month loan period = Rs. 360
SOLUTION 2 :
Here the Principal (P) = 5400

Time (T) = 10/12 years Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400*
) / 100 = Rs.__________

Page 390 Exercise 21A, Pr no. 4
Solution:
Let the sum of money be Rs. x which is the principal (P)
It is given that interest on Rs. x @8% p.a. in 6 years = Rs. 8376
Here principal is Rs .x, Rate of interest (R ) is 8% p.a. and the time
(T) is 6 years
Interest on Rs. x @ 8% for 6 years (I) = (R*P*T)/100
= (8 * x * 6) /100 = Rs.12x/25
So that means , 12x/25 = 8376
=>
x = (8376 * 25) /12
=>
x = Rs. ___________

Page 388, Ex. 4 Solution:
Let the sum of money (P) be Rs. x
And let the rate of interest (R ) be y% p.a
In the first case Rs. x becomes Rs. 854 in 2 years @ y% p.a
Interest @ y% on Rs. x for 2 years (I) = (R*P*T)/100 = (y * x * 2) /100
= (2xy)/100
= Rs. xy/50
So, x + (xy/50) = 854
=> 50x + xy = 42700…………(i)
Similarly, in the 2nd case
Interest on Rs. x @ y% for 7/2 years = (y *x*7/2) / 100 = Rs.(7xy/200)

So, x + (7xy/200) = 969.50
=> 200x + 7xy = 193900…….(ii)
Eliminating xy from (i) & (ii) we can get the value of x and then y.

Page 389, Ex. 5 Solution:
Let the sum of money (P) be Rs. x
So after 8 years, this sum of money becomes Rs. 2x.
Since, Principal

x

+ Interest
= Amount
+ (Interest on Rs. x for 8 years) = 2x
=> (Interest on Rs. x for 8 years) = 2x – x = Rs. x

Let the rate of interest be R% p.a.
Here principal (P) = Rs. x , Interest (I)= Rs. x and Time (T)= 8 years
We know that I = (R * P* T) /100
 R = (I * 100) / (P*T)
= ( x* 100) / (x * 8)
= (100x) /(8x) = 100/8
= 25/2 = 12.5 %

Page 389, Ex. 6
Solution:
Let the loan @ 8% p.a. be Rs. x
 Loan @ 10% p.a. = Rs. (8000 – x)
Given that total annual interest = Rs. 714
That means, in 1 year:
Interest earned on Rs. x@ 8% + Interest earned @ 10% on Rs. (8000-x) = 714
=> {(8 * x * 1) / 100} + [{10 * (8000-x) * 1} / 100] = 714
=>
Solve for x.

(8x/100)

+

{(8000-x) / 10}

=

714

Session 9; Chapter 5; Part 2
Compound Interest

Page 401 Ex 1.
SOLUTION 1 : Without using formula.
• Here Principal (P)= Rs. 8000 , Rate ( R) = 10% p.a ., Time (T) = 3 years
• Interest at 1st year end @ 10% on 8000 for 1 year= (8000 * 10 * 1) /100
•
= Rs. 800
• New principal for 2nd year = Rs. 8000+800 = 8800
• Interest at 2nd year end @ 10% on 8800 for 1 year = (8800 * 10 * 1) /100
•
= Rs. 880
• New principal for 3rd year = Rs. 8800 + 880 = Rs. 9680
• Interest at 3rd year end @ 10% on 9680 for 1 year = (9680 * 10 * 1) /100
•
= Rs. 968
So at the end of 3 years Rs. 8000 will become Rs. (9680 + 968) = Rs. 10648/• So the compound interest (CI) = A – P = 10648 – 8000 = Rs. 2648
•
ALTERNATIVELY : CI = Rs. (800 + 880+ 968) = Rs. 2648

FORMULAS OF COMPOUND INTEREST:
• If a sum of money (principal) Rs. P is borrowed /lent at the rate of R%
compound interest for T years compounded yearly , then :
Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed / lent at the rate of R% compound interest for
compounded half yearly, then :

T years

Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed /lent at the rate of R% compound interest for
compounded quarterly, then :
Compound Interest (CI) on Rs. P @ R% for T years =

T years

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 8000 Rate ( R) = 10% p.a

Time (T) = 3 years

• Compound interest (CI) on Rs. 8000 at the end of 3 years @ 10% =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 4.
SOLUTION 1 : Without using formula.
Here Principal (P)= Rs. 25000, Time (T) = 1 year, Yearly rate of interest =12%
• The Interest is compounded half yearly
• So the half yearly rate of interest = 12 * ½ = 6% per half year
• Interest at 1st half year end = 6% of 25000
•
= Rs. 1500
• New principal for 2nd half year = Rs. 25000+1500 = Rs. 26500
• Interest at 2nd half year end = 6% of 26500
•
= Rs. 1590
• So at the end of 1 year Rs. 25000 will become Rs. (26500 +1590)
= Rs. 28090/• So the compound interest (CI) = 28090 – 25000 = Rs. 3090
• Or , CI = Rs. (1500 + 1590)= Rs. 3090

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 25000 Rate ( R) = 12% p.a Time (T) = 1 year
• Compound interest (CI) on Rs. 25000 at the end of 1 years @ 12% pa
compounded half yearly =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 5. SOLUTION : Without using formula.
Here Principal (P)= Rs. 8000, Time (T) = 9 months = 9/12 years = ¾ yrs,
• Note: 9 months = 3 quarters
• Yearly rate of interest = 20% pa
• The Interest is compounded quarterly . In a year there are 4 quarters
• So the per quarter rate of interest = 20/4 = 5% per quarter.
• Principal for 1st quarter = Rs. 8000
• Interest at 1st quarter end = 5% of 8000 = Rs. 400

• New principal for 2nd quarter = Rs. 8000+400 = Rs. 8400
• Interest at 2nd quarter end = 5% of 8400 = Rs. 420
• New principal for 3rd quarter = Rs. 8400+420 = Rs. 8820
• Interest at 3rd quarter end = 5% of 8820 = Rs. 441
• So at the end of 3 quarters or 9 months, Rs. 8000 will become
Rs. (8820 +441) = Rs. 9261/• So the compound interest (CI) = 9261 – 8000 = Rs. 1261
• Or, CI = Rs. (400 + 420+ 441)= Rs. 1261

Page 402 Ex 3. SOLUTION :
• Here Principal (P)= Rs. 5000, Time (T) = 3years,
• Rate of interest for 1st year (R1) = 5% pa
• Principal for 1st year = Rs. 5000
Interest at 1st year end

= 5% of 5000 = Rs. 250

• New principal for 2nd year = Rs. 5000 + 250 = Rs. 5250
• Rate of interest for 2nd year (R2) = 8% pa
Interest at 2nd year end = 8% of 5250 = Rs. 420

• New principal for 3rd year = Rs. 5250 + 420 = Rs. 5670
• Rate of interest for 3rd year (R3) = 10% pa
Interest at 3rd year end = 10% of 5670 = Rs. 567
• So the amount at the end of 3 years = Rs. (5670 +567) = Rs. 6237/• This means, at the end of 3 years, Rs. 5000 will become Rs. 6237/• Compound interest (CI) on Rs. 5000 after 3 years @ 5%,8% &10%=
6237 – 5000 = Rs. 1237
• Or, CI on Rs. 5000 after 3 years @ 5%,8% &10%= ( 250+ 420+ 567)= Rs. 1237

Page 402 Ex 6. SOLUTION :
• Here Principal (P) = a certain sum of money, CI = Rs. 3783,
Time (T) = 3years, Rate of interest (R) = 5% pa
• Let the principal (P) be Rs. x
• Assuming interest is being charged annually, We know that,
CI =
=> CI on Rs. x @ 5% for 3 years =

=> 3783 =
Find x then find SI on Rs. x @ 5% for 3 years = (x*5*3)/100


Slide 8

Quantitative Methods
Session 8– 31.07.12
Chapter 5 – Simple Interest & Compound Interest

Pranjoy Arup Das

• Interest is the payment to be made for the use of money.
• Suppose, Mr. A borrows a sum of money Rs. X from Mr. B, then Mr. A
will have to repay not only the sum of Rs. X but also some extra money
(say Rs. Y) to Mr. B, sort of like a charge or rent for using Mr. B’s money.
• Rs. X is termed as the principal (P), Rs. Y is the interest (I) for using
Mr.B’s money & Rs. (X + Y) is termed as the Amount(A) which Mr. A
will have to repay to Mr. B after a certain period of time (T) .
• The interest to be paid by the borrower Mr. A to the lender Mr. B is
fixed before or while lending the money. Interest is fixed as a
percentage of the principal and is called Rate of interest (R).

• Money is usually lent out for a fixed period of time (T) . Interest is
calculated on the basis of this time period at the agreed rate, which is
charged either yearly, half-yearly or quarterly. T is always calculated in
Years. So a loan period of 6 months will have to be taken as 0.5 years.

There are two kinds of interest Simple interest & Compound Interest
Simple interest is always calculated on the actual principal originally borrowed.
Throughout the entire time period, a fixed amount of interest is paid whenever
due which is either yearly, half yearly of quarterly
For Eg. Mr. A borrows Rs. 2000 from Mr. B for a period of 2 years at an interest
rate of 10% payable yearly. How much will A pay B after 2 years?
Compound interest is calculated on the increasing principal. The interest is
calculated on the original principal and then added to the original principal
which becomes the new principal for calculating the interest on next due date.
• Simple Interest:
Original Principal Rs. 2000
Interest at 1st year end = 10%of 2000
= Rs. 200
Interest at 2nd year end = 10%of 2000
= Rs. 200
So at the end of 2 years Mr. A has to
pay Rs. (2000 + 200+200)=
Rs. 2400/-

• Compound Interest:
Original principal Rs. 2000
Interest at 1st year end = 10% of 2000
= Rs. 200

New principal=Rs. 2000+200 = 2200
Interest at 2nd year end = 10% of 2200
= Rs. 220
So at the end of 2 years Mr. A has to
pay Rs. (2200 + 220) = Rs. 2420/-

Part 1 Simple Interest

• If a sum of money (principal) Rs. P is borrowed at the rate of R% simple
interest for T years, then
Simple interest amount for T years = 1st years int. + 2nd yrs int. +….+ Tth years int
= (R% of P) + (R% of P) + (R% of P) + …T times
= (R% of P) (1 + 1 + 1 +……………upto T)
= (R% of P) * T
=

I=
• Principal (P)+ Interest (I)= Total Amount(A) to be repaid after T years
=> P +
I
= A
=> P + (R*P*T)/100 = A

Eg. Mr. A borrows Rs. 5400/- from Mr. B under the conditions that :

a) The loan will be repaid after 3 years
b) The borrower will pay a simple interest of 8% per annum (per year)
1) How much interest will Mr. A have to pay at the end of 3 years?
2) What is the amount that Mr. A will have to pay at the end of 3years?
Solution:

Here the Principal = Rs. 5400 Time = 3years Rate of interest= 8%
• Interest at 1st year end = 8% of 5400
•
= Rs. 432
• Interest at 2nd year end = 8% of 5400
•
= Rs. 432
• Interest at 3nd year end = 8% of 5400
•
= Rs. 432
• Total interest at the end of the 3 year loan period
= Rs. (432+432+432) = Rs. 1296
• So at the end of 3 years Mr. A has to pay Rs. (5400 + 1296 )
•
= Rs. 6696

Alternative Solution: USING FORMULA OF SIMPLE INTEREST
Here Principal (P) = Rs.5400, Time (T) = 3 years , Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400* 3) / 100
I = Rs. 1296
The amount to be paid to Mr. B at the end of 3 years :
Amount = P + I
= 5400 + 1296
= Rs. 6696

Page 388 (RSA) Ex 1
SOLUTION 1: Please Note: ALWAYS CONVERT DAYS OR MONTHS TO YEARS
Here the Principal = Rs. 5400
Time = 10/12 years Rate of int. = 8%pa
• Interest for 1 year
= 8% of 5400
= Rs. 432
• That means Interest for 12 months = Rs. 432
•  Interest for 10 months will be: (10/12) * 432 = Rs. 360

• Total interest at the end of the 10 month loan period = Rs. 360
SOLUTION 2 :
Here the Principal (P) = 5400

Time (T) = 10/12 years Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400*
) / 100 = Rs.__________

Page 390 Exercise 21A, Pr no. 4
Solution:
Let the sum of money be Rs. x which is the principal (P)
It is given that interest on Rs. x @8% p.a. in 6 years = Rs. 8376
Here principal is Rs .x, Rate of interest (R ) is 8% p.a. and the time
(T) is 6 years
Interest on Rs. x @ 8% for 6 years (I) = (R*P*T)/100
= (8 * x * 6) /100 = Rs.12x/25
So that means , 12x/25 = 8376
=>
x = (8376 * 25) /12
=>
x = Rs. ___________

Page 388, Ex. 4 Solution:
Let the sum of money (P) be Rs. x
And let the rate of interest (R ) be y% p.a
In the first case Rs. x becomes Rs. 854 in 2 years @ y% p.a
Interest @ y% on Rs. x for 2 years (I) = (R*P*T)/100 = (y * x * 2) /100
= (2xy)/100
= Rs. xy/50
So, x + (xy/50) = 854
=> 50x + xy = 42700…………(i)
Similarly, in the 2nd case
Interest on Rs. x @ y% for 7/2 years = (y *x*7/2) / 100 = Rs.(7xy/200)

So, x + (7xy/200) = 969.50
=> 200x + 7xy = 193900…….(ii)
Eliminating xy from (i) & (ii) we can get the value of x and then y.

Page 389, Ex. 5 Solution:
Let the sum of money (P) be Rs. x
So after 8 years, this sum of money becomes Rs. 2x.
Since, Principal

x

+ Interest
= Amount
+ (Interest on Rs. x for 8 years) = 2x
=> (Interest on Rs. x for 8 years) = 2x – x = Rs. x

Let the rate of interest be R% p.a.
Here principal (P) = Rs. x , Interest (I)= Rs. x and Time (T)= 8 years
We know that I = (R * P* T) /100
 R = (I * 100) / (P*T)
= ( x* 100) / (x * 8)
= (100x) /(8x) = 100/8
= 25/2 = 12.5 %

Page 389, Ex. 6
Solution:
Let the loan @ 8% p.a. be Rs. x
 Loan @ 10% p.a. = Rs. (8000 – x)
Given that total annual interest = Rs. 714
That means, in 1 year:
Interest earned on Rs. x@ 8% + Interest earned @ 10% on Rs. (8000-x) = 714
=> {(8 * x * 1) / 100} + [{10 * (8000-x) * 1} / 100] = 714
=>
Solve for x.

(8x/100)

+

{(8000-x) / 10}

=

714

Session 9; Chapter 5; Part 2
Compound Interest

Page 401 Ex 1.
SOLUTION 1 : Without using formula.
• Here Principal (P)= Rs. 8000 , Rate ( R) = 10% p.a ., Time (T) = 3 years
• Interest at 1st year end @ 10% on 8000 for 1 year= (8000 * 10 * 1) /100
•
= Rs. 800
• New principal for 2nd year = Rs. 8000+800 = 8800
• Interest at 2nd year end @ 10% on 8800 for 1 year = (8800 * 10 * 1) /100
•
= Rs. 880
• New principal for 3rd year = Rs. 8800 + 880 = Rs. 9680
• Interest at 3rd year end @ 10% on 9680 for 1 year = (9680 * 10 * 1) /100
•
= Rs. 968
So at the end of 3 years Rs. 8000 will become Rs. (9680 + 968) = Rs. 10648/• So the compound interest (CI) = A – P = 10648 – 8000 = Rs. 2648
•
ALTERNATIVELY : CI = Rs. (800 + 880+ 968) = Rs. 2648

FORMULAS OF COMPOUND INTEREST:
• If a sum of money (principal) Rs. P is borrowed /lent at the rate of R%
compound interest for T years compounded yearly , then :
Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed / lent at the rate of R% compound interest for
compounded half yearly, then :

T years

Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed /lent at the rate of R% compound interest for
compounded quarterly, then :
Compound Interest (CI) on Rs. P @ R% for T years =

T years

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 8000 Rate ( R) = 10% p.a

Time (T) = 3 years

• Compound interest (CI) on Rs. 8000 at the end of 3 years @ 10% =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 4.
SOLUTION 1 : Without using formula.
Here Principal (P)= Rs. 25000, Time (T) = 1 year, Yearly rate of interest =12%
• The Interest is compounded half yearly
• So the half yearly rate of interest = 12 * ½ = 6% per half year
• Interest at 1st half year end = 6% of 25000
•
= Rs. 1500
• New principal for 2nd half year = Rs. 25000+1500 = Rs. 26500
• Interest at 2nd half year end = 6% of 26500
•
= Rs. 1590
• So at the end of 1 year Rs. 25000 will become Rs. (26500 +1590)
= Rs. 28090/• So the compound interest (CI) = 28090 – 25000 = Rs. 3090
• Or , CI = Rs. (1500 + 1590)= Rs. 3090

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 25000 Rate ( R) = 12% p.a Time (T) = 1 year
• Compound interest (CI) on Rs. 25000 at the end of 1 years @ 12% pa
compounded half yearly =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 5. SOLUTION : Without using formula.
Here Principal (P)= Rs. 8000, Time (T) = 9 months = 9/12 years = ¾ yrs,
• Note: 9 months = 3 quarters
• Yearly rate of interest = 20% pa
• The Interest is compounded quarterly . In a year there are 4 quarters
• So the per quarter rate of interest = 20/4 = 5% per quarter.
• Principal for 1st quarter = Rs. 8000
• Interest at 1st quarter end = 5% of 8000 = Rs. 400

• New principal for 2nd quarter = Rs. 8000+400 = Rs. 8400
• Interest at 2nd quarter end = 5% of 8400 = Rs. 420
• New principal for 3rd quarter = Rs. 8400+420 = Rs. 8820
• Interest at 3rd quarter end = 5% of 8820 = Rs. 441
• So at the end of 3 quarters or 9 months, Rs. 8000 will become
Rs. (8820 +441) = Rs. 9261/• So the compound interest (CI) = 9261 – 8000 = Rs. 1261
• Or, CI = Rs. (400 + 420+ 441)= Rs. 1261

Page 402 Ex 3. SOLUTION :
• Here Principal (P)= Rs. 5000, Time (T) = 3years,
• Rate of interest for 1st year (R1) = 5% pa
• Principal for 1st year = Rs. 5000
Interest at 1st year end

= 5% of 5000 = Rs. 250

• New principal for 2nd year = Rs. 5000 + 250 = Rs. 5250
• Rate of interest for 2nd year (R2) = 8% pa
Interest at 2nd year end = 8% of 5250 = Rs. 420

• New principal for 3rd year = Rs. 5250 + 420 = Rs. 5670
• Rate of interest for 3rd year (R3) = 10% pa
Interest at 3rd year end = 10% of 5670 = Rs. 567
• So the amount at the end of 3 years = Rs. (5670 +567) = Rs. 6237/• This means, at the end of 3 years, Rs. 5000 will become Rs. 6237/• Compound interest (CI) on Rs. 5000 after 3 years @ 5%,8% &10%=
6237 – 5000 = Rs. 1237
• Or, CI on Rs. 5000 after 3 years @ 5%,8% &10%= ( 250+ 420+ 567)= Rs. 1237

Page 402 Ex 6. SOLUTION :
• Here Principal (P) = a certain sum of money, CI = Rs. 3783,
Time (T) = 3years, Rate of interest (R) = 5% pa
• Let the principal (P) be Rs. x
• Assuming interest is being charged annually, We know that,
CI =
=> CI on Rs. x @ 5% for 3 years =

=> 3783 =
Find x then find SI on Rs. x @ 5% for 3 years = (x*5*3)/100


Slide 9

Quantitative Methods
Session 8– 31.07.12
Chapter 5 – Simple Interest & Compound Interest

Pranjoy Arup Das

• Interest is the payment to be made for the use of money.
• Suppose, Mr. A borrows a sum of money Rs. X from Mr. B, then Mr. A
will have to repay not only the sum of Rs. X but also some extra money
(say Rs. Y) to Mr. B, sort of like a charge or rent for using Mr. B’s money.
• Rs. X is termed as the principal (P), Rs. Y is the interest (I) for using
Mr.B’s money & Rs. (X + Y) is termed as the Amount(A) which Mr. A
will have to repay to Mr. B after a certain period of time (T) .
• The interest to be paid by the borrower Mr. A to the lender Mr. B is
fixed before or while lending the money. Interest is fixed as a
percentage of the principal and is called Rate of interest (R).

• Money is usually lent out for a fixed period of time (T) . Interest is
calculated on the basis of this time period at the agreed rate, which is
charged either yearly, half-yearly or quarterly. T is always calculated in
Years. So a loan period of 6 months will have to be taken as 0.5 years.

There are two kinds of interest Simple interest & Compound Interest
Simple interest is always calculated on the actual principal originally borrowed.
Throughout the entire time period, a fixed amount of interest is paid whenever
due which is either yearly, half yearly of quarterly
For Eg. Mr. A borrows Rs. 2000 from Mr. B for a period of 2 years at an interest
rate of 10% payable yearly. How much will A pay B after 2 years?
Compound interest is calculated on the increasing principal. The interest is
calculated on the original principal and then added to the original principal
which becomes the new principal for calculating the interest on next due date.
• Simple Interest:
Original Principal Rs. 2000
Interest at 1st year end = 10%of 2000
= Rs. 200
Interest at 2nd year end = 10%of 2000
= Rs. 200
So at the end of 2 years Mr. A has to
pay Rs. (2000 + 200+200)=
Rs. 2400/-

• Compound Interest:
Original principal Rs. 2000
Interest at 1st year end = 10% of 2000
= Rs. 200

New principal=Rs. 2000+200 = 2200
Interest at 2nd year end = 10% of 2200
= Rs. 220
So at the end of 2 years Mr. A has to
pay Rs. (2200 + 220) = Rs. 2420/-

Part 1 Simple Interest

• If a sum of money (principal) Rs. P is borrowed at the rate of R% simple
interest for T years, then
Simple interest amount for T years = 1st years int. + 2nd yrs int. +….+ Tth years int
= (R% of P) + (R% of P) + (R% of P) + …T times
= (R% of P) (1 + 1 + 1 +……………upto T)
= (R% of P) * T
=

I=
• Principal (P)+ Interest (I)= Total Amount(A) to be repaid after T years
=> P +
I
= A
=> P + (R*P*T)/100 = A

Eg. Mr. A borrows Rs. 5400/- from Mr. B under the conditions that :

a) The loan will be repaid after 3 years
b) The borrower will pay a simple interest of 8% per annum (per year)
1) How much interest will Mr. A have to pay at the end of 3 years?
2) What is the amount that Mr. A will have to pay at the end of 3years?
Solution:

Here the Principal = Rs. 5400 Time = 3years Rate of interest= 8%
• Interest at 1st year end = 8% of 5400
•
= Rs. 432
• Interest at 2nd year end = 8% of 5400
•
= Rs. 432
• Interest at 3nd year end = 8% of 5400
•
= Rs. 432
• Total interest at the end of the 3 year loan period
= Rs. (432+432+432) = Rs. 1296
• So at the end of 3 years Mr. A has to pay Rs. (5400 + 1296 )
•
= Rs. 6696

Alternative Solution: USING FORMULA OF SIMPLE INTEREST
Here Principal (P) = Rs.5400, Time (T) = 3 years , Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400* 3) / 100
I = Rs. 1296
The amount to be paid to Mr. B at the end of 3 years :
Amount = P + I
= 5400 + 1296
= Rs. 6696

Page 388 (RSA) Ex 1
SOLUTION 1: Please Note: ALWAYS CONVERT DAYS OR MONTHS TO YEARS
Here the Principal = Rs. 5400
Time = 10/12 years Rate of int. = 8%pa
• Interest for 1 year
= 8% of 5400
= Rs. 432
• That means Interest for 12 months = Rs. 432
•  Interest for 10 months will be: (10/12) * 432 = Rs. 360

• Total interest at the end of the 10 month loan period = Rs. 360
SOLUTION 2 :
Here the Principal (P) = 5400

Time (T) = 10/12 years Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400*
) / 100 = Rs.__________

Page 390 Exercise 21A, Pr no. 4
Solution:
Let the sum of money be Rs. x which is the principal (P)
It is given that interest on Rs. x @8% p.a. in 6 years = Rs. 8376
Here principal is Rs .x, Rate of interest (R ) is 8% p.a. and the time
(T) is 6 years
Interest on Rs. x @ 8% for 6 years (I) = (R*P*T)/100
= (8 * x * 6) /100 = Rs.12x/25
So that means , 12x/25 = 8376
=>
x = (8376 * 25) /12
=>
x = Rs. ___________

Page 388, Ex. 4 Solution:
Let the sum of money (P) be Rs. x
And let the rate of interest (R ) be y% p.a
In the first case Rs. x becomes Rs. 854 in 2 years @ y% p.a
Interest @ y% on Rs. x for 2 years (I) = (R*P*T)/100 = (y * x * 2) /100
= (2xy)/100
= Rs. xy/50
So, x + (xy/50) = 854
=> 50x + xy = 42700…………(i)
Similarly, in the 2nd case
Interest on Rs. x @ y% for 7/2 years = (y *x*7/2) / 100 = Rs.(7xy/200)

So, x + (7xy/200) = 969.50
=> 200x + 7xy = 193900…….(ii)
Eliminating xy from (i) & (ii) we can get the value of x and then y.

Page 389, Ex. 5 Solution:
Let the sum of money (P) be Rs. x
So after 8 years, this sum of money becomes Rs. 2x.
Since, Principal

x

+ Interest
= Amount
+ (Interest on Rs. x for 8 years) = 2x
=> (Interest on Rs. x for 8 years) = 2x – x = Rs. x

Let the rate of interest be R% p.a.
Here principal (P) = Rs. x , Interest (I)= Rs. x and Time (T)= 8 years
We know that I = (R * P* T) /100
 R = (I * 100) / (P*T)
= ( x* 100) / (x * 8)
= (100x) /(8x) = 100/8
= 25/2 = 12.5 %

Page 389, Ex. 6
Solution:
Let the loan @ 8% p.a. be Rs. x
 Loan @ 10% p.a. = Rs. (8000 – x)
Given that total annual interest = Rs. 714
That means, in 1 year:
Interest earned on Rs. x@ 8% + Interest earned @ 10% on Rs. (8000-x) = 714
=> {(8 * x * 1) / 100} + [{10 * (8000-x) * 1} / 100] = 714
=>
Solve for x.

(8x/100)

+

{(8000-x) / 10}

=

714

Session 9; Chapter 5; Part 2
Compound Interest

Page 401 Ex 1.
SOLUTION 1 : Without using formula.
• Here Principal (P)= Rs. 8000 , Rate ( R) = 10% p.a ., Time (T) = 3 years
• Interest at 1st year end @ 10% on 8000 for 1 year= (8000 * 10 * 1) /100
•
= Rs. 800
• New principal for 2nd year = Rs. 8000+800 = 8800
• Interest at 2nd year end @ 10% on 8800 for 1 year = (8800 * 10 * 1) /100
•
= Rs. 880
• New principal for 3rd year = Rs. 8800 + 880 = Rs. 9680
• Interest at 3rd year end @ 10% on 9680 for 1 year = (9680 * 10 * 1) /100
•
= Rs. 968
So at the end of 3 years Rs. 8000 will become Rs. (9680 + 968) = Rs. 10648/• So the compound interest (CI) = A – P = 10648 – 8000 = Rs. 2648
•
ALTERNATIVELY : CI = Rs. (800 + 880+ 968) = Rs. 2648

FORMULAS OF COMPOUND INTEREST:
• If a sum of money (principal) Rs. P is borrowed /lent at the rate of R%
compound interest for T years compounded yearly , then :
Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed / lent at the rate of R% compound interest for
compounded half yearly, then :

T years

Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed /lent at the rate of R% compound interest for
compounded quarterly, then :
Compound Interest (CI) on Rs. P @ R% for T years =

T years

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 8000 Rate ( R) = 10% p.a

Time (T) = 3 years

• Compound interest (CI) on Rs. 8000 at the end of 3 years @ 10% =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 4.
SOLUTION 1 : Without using formula.
Here Principal (P)= Rs. 25000, Time (T) = 1 year, Yearly rate of interest =12%
• The Interest is compounded half yearly
• So the half yearly rate of interest = 12 * ½ = 6% per half year
• Interest at 1st half year end = 6% of 25000
•
= Rs. 1500
• New principal for 2nd half year = Rs. 25000+1500 = Rs. 26500
• Interest at 2nd half year end = 6% of 26500
•
= Rs. 1590
• So at the end of 1 year Rs. 25000 will become Rs. (26500 +1590)
= Rs. 28090/• So the compound interest (CI) = 28090 – 25000 = Rs. 3090
• Or , CI = Rs. (1500 + 1590)= Rs. 3090

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 25000 Rate ( R) = 12% p.a Time (T) = 1 year
• Compound interest (CI) on Rs. 25000 at the end of 1 years @ 12% pa
compounded half yearly =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 5. SOLUTION : Without using formula.
Here Principal (P)= Rs. 8000, Time (T) = 9 months = 9/12 years = ¾ yrs,
• Note: 9 months = 3 quarters
• Yearly rate of interest = 20% pa
• The Interest is compounded quarterly . In a year there are 4 quarters
• So the per quarter rate of interest = 20/4 = 5% per quarter.
• Principal for 1st quarter = Rs. 8000
• Interest at 1st quarter end = 5% of 8000 = Rs. 400

• New principal for 2nd quarter = Rs. 8000+400 = Rs. 8400
• Interest at 2nd quarter end = 5% of 8400 = Rs. 420
• New principal for 3rd quarter = Rs. 8400+420 = Rs. 8820
• Interest at 3rd quarter end = 5% of 8820 = Rs. 441
• So at the end of 3 quarters or 9 months, Rs. 8000 will become
Rs. (8820 +441) = Rs. 9261/• So the compound interest (CI) = 9261 – 8000 = Rs. 1261
• Or, CI = Rs. (400 + 420+ 441)= Rs. 1261

Page 402 Ex 3. SOLUTION :
• Here Principal (P)= Rs. 5000, Time (T) = 3years,
• Rate of interest for 1st year (R1) = 5% pa
• Principal for 1st year = Rs. 5000
Interest at 1st year end

= 5% of 5000 = Rs. 250

• New principal for 2nd year = Rs. 5000 + 250 = Rs. 5250
• Rate of interest for 2nd year (R2) = 8% pa
Interest at 2nd year end = 8% of 5250 = Rs. 420

• New principal for 3rd year = Rs. 5250 + 420 = Rs. 5670
• Rate of interest for 3rd year (R3) = 10% pa
Interest at 3rd year end = 10% of 5670 = Rs. 567
• So the amount at the end of 3 years = Rs. (5670 +567) = Rs. 6237/• This means, at the end of 3 years, Rs. 5000 will become Rs. 6237/• Compound interest (CI) on Rs. 5000 after 3 years @ 5%,8% &10%=
6237 – 5000 = Rs. 1237
• Or, CI on Rs. 5000 after 3 years @ 5%,8% &10%= ( 250+ 420+ 567)= Rs. 1237

Page 402 Ex 6. SOLUTION :
• Here Principal (P) = a certain sum of money, CI = Rs. 3783,
Time (T) = 3years, Rate of interest (R) = 5% pa
• Let the principal (P) be Rs. x
• Assuming interest is being charged annually, We know that,
CI =
=> CI on Rs. x @ 5% for 3 years =

=> 3783 =
Find x then find SI on Rs. x @ 5% for 3 years = (x*5*3)/100


Slide 10

Quantitative Methods
Session 8– 31.07.12
Chapter 5 – Simple Interest & Compound Interest

Pranjoy Arup Das

• Interest is the payment to be made for the use of money.
• Suppose, Mr. A borrows a sum of money Rs. X from Mr. B, then Mr. A
will have to repay not only the sum of Rs. X but also some extra money
(say Rs. Y) to Mr. B, sort of like a charge or rent for using Mr. B’s money.
• Rs. X is termed as the principal (P), Rs. Y is the interest (I) for using
Mr.B’s money & Rs. (X + Y) is termed as the Amount(A) which Mr. A
will have to repay to Mr. B after a certain period of time (T) .
• The interest to be paid by the borrower Mr. A to the lender Mr. B is
fixed before or while lending the money. Interest is fixed as a
percentage of the principal and is called Rate of interest (R).

• Money is usually lent out for a fixed period of time (T) . Interest is
calculated on the basis of this time period at the agreed rate, which is
charged either yearly, half-yearly or quarterly. T is always calculated in
Years. So a loan period of 6 months will have to be taken as 0.5 years.

There are two kinds of interest Simple interest & Compound Interest
Simple interest is always calculated on the actual principal originally borrowed.
Throughout the entire time period, a fixed amount of interest is paid whenever
due which is either yearly, half yearly of quarterly
For Eg. Mr. A borrows Rs. 2000 from Mr. B for a period of 2 years at an interest
rate of 10% payable yearly. How much will A pay B after 2 years?
Compound interest is calculated on the increasing principal. The interest is
calculated on the original principal and then added to the original principal
which becomes the new principal for calculating the interest on next due date.
• Simple Interest:
Original Principal Rs. 2000
Interest at 1st year end = 10%of 2000
= Rs. 200
Interest at 2nd year end = 10%of 2000
= Rs. 200
So at the end of 2 years Mr. A has to
pay Rs. (2000 + 200+200)=
Rs. 2400/-

• Compound Interest:
Original principal Rs. 2000
Interest at 1st year end = 10% of 2000
= Rs. 200

New principal=Rs. 2000+200 = 2200
Interest at 2nd year end = 10% of 2200
= Rs. 220
So at the end of 2 years Mr. A has to
pay Rs. (2200 + 220) = Rs. 2420/-

Part 1 Simple Interest

• If a sum of money (principal) Rs. P is borrowed at the rate of R% simple
interest for T years, then
Simple interest amount for T years = 1st years int. + 2nd yrs int. +….+ Tth years int
= (R% of P) + (R% of P) + (R% of P) + …T times
= (R% of P) (1 + 1 + 1 +……………upto T)
= (R% of P) * T
=

I=
• Principal (P)+ Interest (I)= Total Amount(A) to be repaid after T years
=> P +
I
= A
=> P + (R*P*T)/100 = A

Eg. Mr. A borrows Rs. 5400/- from Mr. B under the conditions that :

a) The loan will be repaid after 3 years
b) The borrower will pay a simple interest of 8% per annum (per year)
1) How much interest will Mr. A have to pay at the end of 3 years?
2) What is the amount that Mr. A will have to pay at the end of 3years?
Solution:

Here the Principal = Rs. 5400 Time = 3years Rate of interest= 8%
• Interest at 1st year end = 8% of 5400
•
= Rs. 432
• Interest at 2nd year end = 8% of 5400
•
= Rs. 432
• Interest at 3nd year end = 8% of 5400
•
= Rs. 432
• Total interest at the end of the 3 year loan period
= Rs. (432+432+432) = Rs. 1296
• So at the end of 3 years Mr. A has to pay Rs. (5400 + 1296 )
•
= Rs. 6696

Alternative Solution: USING FORMULA OF SIMPLE INTEREST
Here Principal (P) = Rs.5400, Time (T) = 3 years , Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400* 3) / 100
I = Rs. 1296
The amount to be paid to Mr. B at the end of 3 years :
Amount = P + I
= 5400 + 1296
= Rs. 6696

Page 388 (RSA) Ex 1
SOLUTION 1: Please Note: ALWAYS CONVERT DAYS OR MONTHS TO YEARS
Here the Principal = Rs. 5400
Time = 10/12 years Rate of int. = 8%pa
• Interest for 1 year
= 8% of 5400
= Rs. 432
• That means Interest for 12 months = Rs. 432
•  Interest for 10 months will be: (10/12) * 432 = Rs. 360

• Total interest at the end of the 10 month loan period = Rs. 360
SOLUTION 2 :
Here the Principal (P) = 5400

Time (T) = 10/12 years Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400*
) / 100 = Rs.__________

Page 390 Exercise 21A, Pr no. 4
Solution:
Let the sum of money be Rs. x which is the principal (P)
It is given that interest on Rs. x @8% p.a. in 6 years = Rs. 8376
Here principal is Rs .x, Rate of interest (R ) is 8% p.a. and the time
(T) is 6 years
Interest on Rs. x @ 8% for 6 years (I) = (R*P*T)/100
= (8 * x * 6) /100 = Rs.12x/25
So that means , 12x/25 = 8376
=>
x = (8376 * 25) /12
=>
x = Rs. ___________

Page 388, Ex. 4 Solution:
Let the sum of money (P) be Rs. x
And let the rate of interest (R ) be y% p.a
In the first case Rs. x becomes Rs. 854 in 2 years @ y% p.a
Interest @ y% on Rs. x for 2 years (I) = (R*P*T)/100 = (y * x * 2) /100
= (2xy)/100
= Rs. xy/50
So, x + (xy/50) = 854
=> 50x + xy = 42700…………(i)
Similarly, in the 2nd case
Interest on Rs. x @ y% for 7/2 years = (y *x*7/2) / 100 = Rs.(7xy/200)

So, x + (7xy/200) = 969.50
=> 200x + 7xy = 193900…….(ii)
Eliminating xy from (i) & (ii) we can get the value of x and then y.

Page 389, Ex. 5 Solution:
Let the sum of money (P) be Rs. x
So after 8 years, this sum of money becomes Rs. 2x.
Since, Principal

x

+ Interest
= Amount
+ (Interest on Rs. x for 8 years) = 2x
=> (Interest on Rs. x for 8 years) = 2x – x = Rs. x

Let the rate of interest be R% p.a.
Here principal (P) = Rs. x , Interest (I)= Rs. x and Time (T)= 8 years
We know that I = (R * P* T) /100
 R = (I * 100) / (P*T)
= ( x* 100) / (x * 8)
= (100x) /(8x) = 100/8
= 25/2 = 12.5 %

Page 389, Ex. 6
Solution:
Let the loan @ 8% p.a. be Rs. x
 Loan @ 10% p.a. = Rs. (8000 – x)
Given that total annual interest = Rs. 714
That means, in 1 year:
Interest earned on Rs. x@ 8% + Interest earned @ 10% on Rs. (8000-x) = 714
=> {(8 * x * 1) / 100} + [{10 * (8000-x) * 1} / 100] = 714
=>
Solve for x.

(8x/100)

+

{(8000-x) / 10}

=

714

Session 9; Chapter 5; Part 2
Compound Interest

Page 401 Ex 1.
SOLUTION 1 : Without using formula.
• Here Principal (P)= Rs. 8000 , Rate ( R) = 10% p.a ., Time (T) = 3 years
• Interest at 1st year end @ 10% on 8000 for 1 year= (8000 * 10 * 1) /100
•
= Rs. 800
• New principal for 2nd year = Rs. 8000+800 = 8800
• Interest at 2nd year end @ 10% on 8800 for 1 year = (8800 * 10 * 1) /100
•
= Rs. 880
• New principal for 3rd year = Rs. 8800 + 880 = Rs. 9680
• Interest at 3rd year end @ 10% on 9680 for 1 year = (9680 * 10 * 1) /100
•
= Rs. 968
So at the end of 3 years Rs. 8000 will become Rs. (9680 + 968) = Rs. 10648/• So the compound interest (CI) = A – P = 10648 – 8000 = Rs. 2648
•
ALTERNATIVELY : CI = Rs. (800 + 880+ 968) = Rs. 2648

FORMULAS OF COMPOUND INTEREST:
• If a sum of money (principal) Rs. P is borrowed /lent at the rate of R%
compound interest for T years compounded yearly , then :
Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed / lent at the rate of R% compound interest for
compounded half yearly, then :

T years

Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed /lent at the rate of R% compound interest for
compounded quarterly, then :
Compound Interest (CI) on Rs. P @ R% for T years =

T years

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 8000 Rate ( R) = 10% p.a

Time (T) = 3 years

• Compound interest (CI) on Rs. 8000 at the end of 3 years @ 10% =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 4.
SOLUTION 1 : Without using formula.
Here Principal (P)= Rs. 25000, Time (T) = 1 year, Yearly rate of interest =12%
• The Interest is compounded half yearly
• So the half yearly rate of interest = 12 * ½ = 6% per half year
• Interest at 1st half year end = 6% of 25000
•
= Rs. 1500
• New principal for 2nd half year = Rs. 25000+1500 = Rs. 26500
• Interest at 2nd half year end = 6% of 26500
•
= Rs. 1590
• So at the end of 1 year Rs. 25000 will become Rs. (26500 +1590)
= Rs. 28090/• So the compound interest (CI) = 28090 – 25000 = Rs. 3090
• Or , CI = Rs. (1500 + 1590)= Rs. 3090

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 25000 Rate ( R) = 12% p.a Time (T) = 1 year
• Compound interest (CI) on Rs. 25000 at the end of 1 years @ 12% pa
compounded half yearly =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 5. SOLUTION : Without using formula.
Here Principal (P)= Rs. 8000, Time (T) = 9 months = 9/12 years = ¾ yrs,
• Note: 9 months = 3 quarters
• Yearly rate of interest = 20% pa
• The Interest is compounded quarterly . In a year there are 4 quarters
• So the per quarter rate of interest = 20/4 = 5% per quarter.
• Principal for 1st quarter = Rs. 8000
• Interest at 1st quarter end = 5% of 8000 = Rs. 400

• New principal for 2nd quarter = Rs. 8000+400 = Rs. 8400
• Interest at 2nd quarter end = 5% of 8400 = Rs. 420
• New principal for 3rd quarter = Rs. 8400+420 = Rs. 8820
• Interest at 3rd quarter end = 5% of 8820 = Rs. 441
• So at the end of 3 quarters or 9 months, Rs. 8000 will become
Rs. (8820 +441) = Rs. 9261/• So the compound interest (CI) = 9261 – 8000 = Rs. 1261
• Or, CI = Rs. (400 + 420+ 441)= Rs. 1261

Page 402 Ex 3. SOLUTION :
• Here Principal (P)= Rs. 5000, Time (T) = 3years,
• Rate of interest for 1st year (R1) = 5% pa
• Principal for 1st year = Rs. 5000
Interest at 1st year end

= 5% of 5000 = Rs. 250

• New principal for 2nd year = Rs. 5000 + 250 = Rs. 5250
• Rate of interest for 2nd year (R2) = 8% pa
Interest at 2nd year end = 8% of 5250 = Rs. 420

• New principal for 3rd year = Rs. 5250 + 420 = Rs. 5670
• Rate of interest for 3rd year (R3) = 10% pa
Interest at 3rd year end = 10% of 5670 = Rs. 567
• So the amount at the end of 3 years = Rs. (5670 +567) = Rs. 6237/• This means, at the end of 3 years, Rs. 5000 will become Rs. 6237/• Compound interest (CI) on Rs. 5000 after 3 years @ 5%,8% &10%=
6237 – 5000 = Rs. 1237
• Or, CI on Rs. 5000 after 3 years @ 5%,8% &10%= ( 250+ 420+ 567)= Rs. 1237

Page 402 Ex 6. SOLUTION :
• Here Principal (P) = a certain sum of money, CI = Rs. 3783,
Time (T) = 3years, Rate of interest (R) = 5% pa
• Let the principal (P) be Rs. x
• Assuming interest is being charged annually, We know that,
CI =
=> CI on Rs. x @ 5% for 3 years =

=> 3783 =
Find x then find SI on Rs. x @ 5% for 3 years = (x*5*3)/100


Slide 11

Quantitative Methods
Session 8– 31.07.12
Chapter 5 – Simple Interest & Compound Interest

Pranjoy Arup Das

• Interest is the payment to be made for the use of money.
• Suppose, Mr. A borrows a sum of money Rs. X from Mr. B, then Mr. A
will have to repay not only the sum of Rs. X but also some extra money
(say Rs. Y) to Mr. B, sort of like a charge or rent for using Mr. B’s money.
• Rs. X is termed as the principal (P), Rs. Y is the interest (I) for using
Mr.B’s money & Rs. (X + Y) is termed as the Amount(A) which Mr. A
will have to repay to Mr. B after a certain period of time (T) .
• The interest to be paid by the borrower Mr. A to the lender Mr. B is
fixed before or while lending the money. Interest is fixed as a
percentage of the principal and is called Rate of interest (R).

• Money is usually lent out for a fixed period of time (T) . Interest is
calculated on the basis of this time period at the agreed rate, which is
charged either yearly, half-yearly or quarterly. T is always calculated in
Years. So a loan period of 6 months will have to be taken as 0.5 years.

There are two kinds of interest Simple interest & Compound Interest
Simple interest is always calculated on the actual principal originally borrowed.
Throughout the entire time period, a fixed amount of interest is paid whenever
due which is either yearly, half yearly of quarterly
For Eg. Mr. A borrows Rs. 2000 from Mr. B for a period of 2 years at an interest
rate of 10% payable yearly. How much will A pay B after 2 years?
Compound interest is calculated on the increasing principal. The interest is
calculated on the original principal and then added to the original principal
which becomes the new principal for calculating the interest on next due date.
• Simple Interest:
Original Principal Rs. 2000
Interest at 1st year end = 10%of 2000
= Rs. 200
Interest at 2nd year end = 10%of 2000
= Rs. 200
So at the end of 2 years Mr. A has to
pay Rs. (2000 + 200+200)=
Rs. 2400/-

• Compound Interest:
Original principal Rs. 2000
Interest at 1st year end = 10% of 2000
= Rs. 200

New principal=Rs. 2000+200 = 2200
Interest at 2nd year end = 10% of 2200
= Rs. 220
So at the end of 2 years Mr. A has to
pay Rs. (2200 + 220) = Rs. 2420/-

Part 1 Simple Interest

• If a sum of money (principal) Rs. P is borrowed at the rate of R% simple
interest for T years, then
Simple interest amount for T years = 1st years int. + 2nd yrs int. +….+ Tth years int
= (R% of P) + (R% of P) + (R% of P) + …T times
= (R% of P) (1 + 1 + 1 +……………upto T)
= (R% of P) * T
=

I=
• Principal (P)+ Interest (I)= Total Amount(A) to be repaid after T years
=> P +
I
= A
=> P + (R*P*T)/100 = A

Eg. Mr. A borrows Rs. 5400/- from Mr. B under the conditions that :

a) The loan will be repaid after 3 years
b) The borrower will pay a simple interest of 8% per annum (per year)
1) How much interest will Mr. A have to pay at the end of 3 years?
2) What is the amount that Mr. A will have to pay at the end of 3years?
Solution:

Here the Principal = Rs. 5400 Time = 3years Rate of interest= 8%
• Interest at 1st year end = 8% of 5400
•
= Rs. 432
• Interest at 2nd year end = 8% of 5400
•
= Rs. 432
• Interest at 3nd year end = 8% of 5400
•
= Rs. 432
• Total interest at the end of the 3 year loan period
= Rs. (432+432+432) = Rs. 1296
• So at the end of 3 years Mr. A has to pay Rs. (5400 + 1296 )
•
= Rs. 6696

Alternative Solution: USING FORMULA OF SIMPLE INTEREST
Here Principal (P) = Rs.5400, Time (T) = 3 years , Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400* 3) / 100
I = Rs. 1296
The amount to be paid to Mr. B at the end of 3 years :
Amount = P + I
= 5400 + 1296
= Rs. 6696

Page 388 (RSA) Ex 1
SOLUTION 1: Please Note: ALWAYS CONVERT DAYS OR MONTHS TO YEARS
Here the Principal = Rs. 5400
Time = 10/12 years Rate of int. = 8%pa
• Interest for 1 year
= 8% of 5400
= Rs. 432
• That means Interest for 12 months = Rs. 432
•  Interest for 10 months will be: (10/12) * 432 = Rs. 360

• Total interest at the end of the 10 month loan period = Rs. 360
SOLUTION 2 :
Here the Principal (P) = 5400

Time (T) = 10/12 years Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400*
) / 100 = Rs.__________

Page 390 Exercise 21A, Pr no. 4
Solution:
Let the sum of money be Rs. x which is the principal (P)
It is given that interest on Rs. x @8% p.a. in 6 years = Rs. 8376
Here principal is Rs .x, Rate of interest (R ) is 8% p.a. and the time
(T) is 6 years
Interest on Rs. x @ 8% for 6 years (I) = (R*P*T)/100
= (8 * x * 6) /100 = Rs.12x/25
So that means , 12x/25 = 8376
=>
x = (8376 * 25) /12
=>
x = Rs. ___________

Page 388, Ex. 4 Solution:
Let the sum of money (P) be Rs. x
And let the rate of interest (R ) be y% p.a
In the first case Rs. x becomes Rs. 854 in 2 years @ y% p.a
Interest @ y% on Rs. x for 2 years (I) = (R*P*T)/100 = (y * x * 2) /100
= (2xy)/100
= Rs. xy/50
So, x + (xy/50) = 854
=> 50x + xy = 42700…………(i)
Similarly, in the 2nd case
Interest on Rs. x @ y% for 7/2 years = (y *x*7/2) / 100 = Rs.(7xy/200)

So, x + (7xy/200) = 969.50
=> 200x + 7xy = 193900…….(ii)
Eliminating xy from (i) & (ii) we can get the value of x and then y.

Page 389, Ex. 5 Solution:
Let the sum of money (P) be Rs. x
So after 8 years, this sum of money becomes Rs. 2x.
Since, Principal

x

+ Interest
= Amount
+ (Interest on Rs. x for 8 years) = 2x
=> (Interest on Rs. x for 8 years) = 2x – x = Rs. x

Let the rate of interest be R% p.a.
Here principal (P) = Rs. x , Interest (I)= Rs. x and Time (T)= 8 years
We know that I = (R * P* T) /100
 R = (I * 100) / (P*T)
= ( x* 100) / (x * 8)
= (100x) /(8x) = 100/8
= 25/2 = 12.5 %

Page 389, Ex. 6
Solution:
Let the loan @ 8% p.a. be Rs. x
 Loan @ 10% p.a. = Rs. (8000 – x)
Given that total annual interest = Rs. 714
That means, in 1 year:
Interest earned on Rs. x@ 8% + Interest earned @ 10% on Rs. (8000-x) = 714
=> {(8 * x * 1) / 100} + [{10 * (8000-x) * 1} / 100] = 714
=>
Solve for x.

(8x/100)

+

{(8000-x) / 10}

=

714

Session 9; Chapter 5; Part 2
Compound Interest

Page 401 Ex 1.
SOLUTION 1 : Without using formula.
• Here Principal (P)= Rs. 8000 , Rate ( R) = 10% p.a ., Time (T) = 3 years
• Interest at 1st year end @ 10% on 8000 for 1 year= (8000 * 10 * 1) /100
•
= Rs. 800
• New principal for 2nd year = Rs. 8000+800 = 8800
• Interest at 2nd year end @ 10% on 8800 for 1 year = (8800 * 10 * 1) /100
•
= Rs. 880
• New principal for 3rd year = Rs. 8800 + 880 = Rs. 9680
• Interest at 3rd year end @ 10% on 9680 for 1 year = (9680 * 10 * 1) /100
•
= Rs. 968
So at the end of 3 years Rs. 8000 will become Rs. (9680 + 968) = Rs. 10648/• So the compound interest (CI) = A – P = 10648 – 8000 = Rs. 2648
•
ALTERNATIVELY : CI = Rs. (800 + 880+ 968) = Rs. 2648

FORMULAS OF COMPOUND INTEREST:
• If a sum of money (principal) Rs. P is borrowed /lent at the rate of R%
compound interest for T years compounded yearly , then :
Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed / lent at the rate of R% compound interest for
compounded half yearly, then :

T years

Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed /lent at the rate of R% compound interest for
compounded quarterly, then :
Compound Interest (CI) on Rs. P @ R% for T years =

T years

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 8000 Rate ( R) = 10% p.a

Time (T) = 3 years

• Compound interest (CI) on Rs. 8000 at the end of 3 years @ 10% =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 4.
SOLUTION 1 : Without using formula.
Here Principal (P)= Rs. 25000, Time (T) = 1 year, Yearly rate of interest =12%
• The Interest is compounded half yearly
• So the half yearly rate of interest = 12 * ½ = 6% per half year
• Interest at 1st half year end = 6% of 25000
•
= Rs. 1500
• New principal for 2nd half year = Rs. 25000+1500 = Rs. 26500
• Interest at 2nd half year end = 6% of 26500
•
= Rs. 1590
• So at the end of 1 year Rs. 25000 will become Rs. (26500 +1590)
= Rs. 28090/• So the compound interest (CI) = 28090 – 25000 = Rs. 3090
• Or , CI = Rs. (1500 + 1590)= Rs. 3090

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 25000 Rate ( R) = 12% p.a Time (T) = 1 year
• Compound interest (CI) on Rs. 25000 at the end of 1 years @ 12% pa
compounded half yearly =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 5. SOLUTION : Without using formula.
Here Principal (P)= Rs. 8000, Time (T) = 9 months = 9/12 years = ¾ yrs,
• Note: 9 months = 3 quarters
• Yearly rate of interest = 20% pa
• The Interest is compounded quarterly . In a year there are 4 quarters
• So the per quarter rate of interest = 20/4 = 5% per quarter.
• Principal for 1st quarter = Rs. 8000
• Interest at 1st quarter end = 5% of 8000 = Rs. 400

• New principal for 2nd quarter = Rs. 8000+400 = Rs. 8400
• Interest at 2nd quarter end = 5% of 8400 = Rs. 420
• New principal for 3rd quarter = Rs. 8400+420 = Rs. 8820
• Interest at 3rd quarter end = 5% of 8820 = Rs. 441
• So at the end of 3 quarters or 9 months, Rs. 8000 will become
Rs. (8820 +441) = Rs. 9261/• So the compound interest (CI) = 9261 – 8000 = Rs. 1261
• Or, CI = Rs. (400 + 420+ 441)= Rs. 1261

Page 402 Ex 3. SOLUTION :
• Here Principal (P)= Rs. 5000, Time (T) = 3years,
• Rate of interest for 1st year (R1) = 5% pa
• Principal for 1st year = Rs. 5000
Interest at 1st year end

= 5% of 5000 = Rs. 250

• New principal for 2nd year = Rs. 5000 + 250 = Rs. 5250
• Rate of interest for 2nd year (R2) = 8% pa
Interest at 2nd year end = 8% of 5250 = Rs. 420

• New principal for 3rd year = Rs. 5250 + 420 = Rs. 5670
• Rate of interest for 3rd year (R3) = 10% pa
Interest at 3rd year end = 10% of 5670 = Rs. 567
• So the amount at the end of 3 years = Rs. (5670 +567) = Rs. 6237/• This means, at the end of 3 years, Rs. 5000 will become Rs. 6237/• Compound interest (CI) on Rs. 5000 after 3 years @ 5%,8% &10%=
6237 – 5000 = Rs. 1237
• Or, CI on Rs. 5000 after 3 years @ 5%,8% &10%= ( 250+ 420+ 567)= Rs. 1237

Page 402 Ex 6. SOLUTION :
• Here Principal (P) = a certain sum of money, CI = Rs. 3783,
Time (T) = 3years, Rate of interest (R) = 5% pa
• Let the principal (P) be Rs. x
• Assuming interest is being charged annually, We know that,
CI =
=> CI on Rs. x @ 5% for 3 years =

=> 3783 =
Find x then find SI on Rs. x @ 5% for 3 years = (x*5*3)/100


Slide 12

Quantitative Methods
Session 8– 31.07.12
Chapter 5 – Simple Interest & Compound Interest

Pranjoy Arup Das

• Interest is the payment to be made for the use of money.
• Suppose, Mr. A borrows a sum of money Rs. X from Mr. B, then Mr. A
will have to repay not only the sum of Rs. X but also some extra money
(say Rs. Y) to Mr. B, sort of like a charge or rent for using Mr. B’s money.
• Rs. X is termed as the principal (P), Rs. Y is the interest (I) for using
Mr.B’s money & Rs. (X + Y) is termed as the Amount(A) which Mr. A
will have to repay to Mr. B after a certain period of time (T) .
• The interest to be paid by the borrower Mr. A to the lender Mr. B is
fixed before or while lending the money. Interest is fixed as a
percentage of the principal and is called Rate of interest (R).

• Money is usually lent out for a fixed period of time (T) . Interest is
calculated on the basis of this time period at the agreed rate, which is
charged either yearly, half-yearly or quarterly. T is always calculated in
Years. So a loan period of 6 months will have to be taken as 0.5 years.

There are two kinds of interest Simple interest & Compound Interest
Simple interest is always calculated on the actual principal originally borrowed.
Throughout the entire time period, a fixed amount of interest is paid whenever
due which is either yearly, half yearly of quarterly
For Eg. Mr. A borrows Rs. 2000 from Mr. B for a period of 2 years at an interest
rate of 10% payable yearly. How much will A pay B after 2 years?
Compound interest is calculated on the increasing principal. The interest is
calculated on the original principal and then added to the original principal
which becomes the new principal for calculating the interest on next due date.
• Simple Interest:
Original Principal Rs. 2000
Interest at 1st year end = 10%of 2000
= Rs. 200
Interest at 2nd year end = 10%of 2000
= Rs. 200
So at the end of 2 years Mr. A has to
pay Rs. (2000 + 200+200)=
Rs. 2400/-

• Compound Interest:
Original principal Rs. 2000
Interest at 1st year end = 10% of 2000
= Rs. 200

New principal=Rs. 2000+200 = 2200
Interest at 2nd year end = 10% of 2200
= Rs. 220
So at the end of 2 years Mr. A has to
pay Rs. (2200 + 220) = Rs. 2420/-

Part 1 Simple Interest

• If a sum of money (principal) Rs. P is borrowed at the rate of R% simple
interest for T years, then
Simple interest amount for T years = 1st years int. + 2nd yrs int. +….+ Tth years int
= (R% of P) + (R% of P) + (R% of P) + …T times
= (R% of P) (1 + 1 + 1 +……………upto T)
= (R% of P) * T
=

I=
• Principal (P)+ Interest (I)= Total Amount(A) to be repaid after T years
=> P +
I
= A
=> P + (R*P*T)/100 = A

Eg. Mr. A borrows Rs. 5400/- from Mr. B under the conditions that :

a) The loan will be repaid after 3 years
b) The borrower will pay a simple interest of 8% per annum (per year)
1) How much interest will Mr. A have to pay at the end of 3 years?
2) What is the amount that Mr. A will have to pay at the end of 3years?
Solution:

Here the Principal = Rs. 5400 Time = 3years Rate of interest= 8%
• Interest at 1st year end = 8% of 5400
•
= Rs. 432
• Interest at 2nd year end = 8% of 5400
•
= Rs. 432
• Interest at 3nd year end = 8% of 5400
•
= Rs. 432
• Total interest at the end of the 3 year loan period
= Rs. (432+432+432) = Rs. 1296
• So at the end of 3 years Mr. A has to pay Rs. (5400 + 1296 )
•
= Rs. 6696

Alternative Solution: USING FORMULA OF SIMPLE INTEREST
Here Principal (P) = Rs.5400, Time (T) = 3 years , Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400* 3) / 100
I = Rs. 1296
The amount to be paid to Mr. B at the end of 3 years :
Amount = P + I
= 5400 + 1296
= Rs. 6696

Page 388 (RSA) Ex 1
SOLUTION 1: Please Note: ALWAYS CONVERT DAYS OR MONTHS TO YEARS
Here the Principal = Rs. 5400
Time = 10/12 years Rate of int. = 8%pa
• Interest for 1 year
= 8% of 5400
= Rs. 432
• That means Interest for 12 months = Rs. 432
•  Interest for 10 months will be: (10/12) * 432 = Rs. 360

• Total interest at the end of the 10 month loan period = Rs. 360
SOLUTION 2 :
Here the Principal (P) = 5400

Time (T) = 10/12 years Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400*
) / 100 = Rs.__________

Page 390 Exercise 21A, Pr no. 4
Solution:
Let the sum of money be Rs. x which is the principal (P)
It is given that interest on Rs. x @8% p.a. in 6 years = Rs. 8376
Here principal is Rs .x, Rate of interest (R ) is 8% p.a. and the time
(T) is 6 years
Interest on Rs. x @ 8% for 6 years (I) = (R*P*T)/100
= (8 * x * 6) /100 = Rs.12x/25
So that means , 12x/25 = 8376
=>
x = (8376 * 25) /12
=>
x = Rs. ___________

Page 388, Ex. 4 Solution:
Let the sum of money (P) be Rs. x
And let the rate of interest (R ) be y% p.a
In the first case Rs. x becomes Rs. 854 in 2 years @ y% p.a
Interest @ y% on Rs. x for 2 years (I) = (R*P*T)/100 = (y * x * 2) /100
= (2xy)/100
= Rs. xy/50
So, x + (xy/50) = 854
=> 50x + xy = 42700…………(i)
Similarly, in the 2nd case
Interest on Rs. x @ y% for 7/2 years = (y *x*7/2) / 100 = Rs.(7xy/200)

So, x + (7xy/200) = 969.50
=> 200x + 7xy = 193900…….(ii)
Eliminating xy from (i) & (ii) we can get the value of x and then y.

Page 389, Ex. 5 Solution:
Let the sum of money (P) be Rs. x
So after 8 years, this sum of money becomes Rs. 2x.
Since, Principal

x

+ Interest
= Amount
+ (Interest on Rs. x for 8 years) = 2x
=> (Interest on Rs. x for 8 years) = 2x – x = Rs. x

Let the rate of interest be R% p.a.
Here principal (P) = Rs. x , Interest (I)= Rs. x and Time (T)= 8 years
We know that I = (R * P* T) /100
 R = (I * 100) / (P*T)
= ( x* 100) / (x * 8)
= (100x) /(8x) = 100/8
= 25/2 = 12.5 %

Page 389, Ex. 6
Solution:
Let the loan @ 8% p.a. be Rs. x
 Loan @ 10% p.a. = Rs. (8000 – x)
Given that total annual interest = Rs. 714
That means, in 1 year:
Interest earned on Rs. x@ 8% + Interest earned @ 10% on Rs. (8000-x) = 714
=> {(8 * x * 1) / 100} + [{10 * (8000-x) * 1} / 100] = 714
=>
Solve for x.

(8x/100)

+

{(8000-x) / 10}

=

714

Session 9; Chapter 5; Part 2
Compound Interest

Page 401 Ex 1.
SOLUTION 1 : Without using formula.
• Here Principal (P)= Rs. 8000 , Rate ( R) = 10% p.a ., Time (T) = 3 years
• Interest at 1st year end @ 10% on 8000 for 1 year= (8000 * 10 * 1) /100
•
= Rs. 800
• New principal for 2nd year = Rs. 8000+800 = 8800
• Interest at 2nd year end @ 10% on 8800 for 1 year = (8800 * 10 * 1) /100
•
= Rs. 880
• New principal for 3rd year = Rs. 8800 + 880 = Rs. 9680
• Interest at 3rd year end @ 10% on 9680 for 1 year = (9680 * 10 * 1) /100
•
= Rs. 968
So at the end of 3 years Rs. 8000 will become Rs. (9680 + 968) = Rs. 10648/• So the compound interest (CI) = A – P = 10648 – 8000 = Rs. 2648
•
ALTERNATIVELY : CI = Rs. (800 + 880+ 968) = Rs. 2648

FORMULAS OF COMPOUND INTEREST:
• If a sum of money (principal) Rs. P is borrowed /lent at the rate of R%
compound interest for T years compounded yearly , then :
Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed / lent at the rate of R% compound interest for
compounded half yearly, then :

T years

Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed /lent at the rate of R% compound interest for
compounded quarterly, then :
Compound Interest (CI) on Rs. P @ R% for T years =

T years

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 8000 Rate ( R) = 10% p.a

Time (T) = 3 years

• Compound interest (CI) on Rs. 8000 at the end of 3 years @ 10% =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 4.
SOLUTION 1 : Without using formula.
Here Principal (P)= Rs. 25000, Time (T) = 1 year, Yearly rate of interest =12%
• The Interest is compounded half yearly
• So the half yearly rate of interest = 12 * ½ = 6% per half year
• Interest at 1st half year end = 6% of 25000
•
= Rs. 1500
• New principal for 2nd half year = Rs. 25000+1500 = Rs. 26500
• Interest at 2nd half year end = 6% of 26500
•
= Rs. 1590
• So at the end of 1 year Rs. 25000 will become Rs. (26500 +1590)
= Rs. 28090/• So the compound interest (CI) = 28090 – 25000 = Rs. 3090
• Or , CI = Rs. (1500 + 1590)= Rs. 3090

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 25000 Rate ( R) = 12% p.a Time (T) = 1 year
• Compound interest (CI) on Rs. 25000 at the end of 1 years @ 12% pa
compounded half yearly =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 5. SOLUTION : Without using formula.
Here Principal (P)= Rs. 8000, Time (T) = 9 months = 9/12 years = ¾ yrs,
• Note: 9 months = 3 quarters
• Yearly rate of interest = 20% pa
• The Interest is compounded quarterly . In a year there are 4 quarters
• So the per quarter rate of interest = 20/4 = 5% per quarter.
• Principal for 1st quarter = Rs. 8000
• Interest at 1st quarter end = 5% of 8000 = Rs. 400

• New principal for 2nd quarter = Rs. 8000+400 = Rs. 8400
• Interest at 2nd quarter end = 5% of 8400 = Rs. 420
• New principal for 3rd quarter = Rs. 8400+420 = Rs. 8820
• Interest at 3rd quarter end = 5% of 8820 = Rs. 441
• So at the end of 3 quarters or 9 months, Rs. 8000 will become
Rs. (8820 +441) = Rs. 9261/• So the compound interest (CI) = 9261 – 8000 = Rs. 1261
• Or, CI = Rs. (400 + 420+ 441)= Rs. 1261

Page 402 Ex 3. SOLUTION :
• Here Principal (P)= Rs. 5000, Time (T) = 3years,
• Rate of interest for 1st year (R1) = 5% pa
• Principal for 1st year = Rs. 5000
Interest at 1st year end

= 5% of 5000 = Rs. 250

• New principal for 2nd year = Rs. 5000 + 250 = Rs. 5250
• Rate of interest for 2nd year (R2) = 8% pa
Interest at 2nd year end = 8% of 5250 = Rs. 420

• New principal for 3rd year = Rs. 5250 + 420 = Rs. 5670
• Rate of interest for 3rd year (R3) = 10% pa
Interest at 3rd year end = 10% of 5670 = Rs. 567
• So the amount at the end of 3 years = Rs. (5670 +567) = Rs. 6237/• This means, at the end of 3 years, Rs. 5000 will become Rs. 6237/• Compound interest (CI) on Rs. 5000 after 3 years @ 5%,8% &10%=
6237 – 5000 = Rs. 1237
• Or, CI on Rs. 5000 after 3 years @ 5%,8% &10%= ( 250+ 420+ 567)= Rs. 1237

Page 402 Ex 6. SOLUTION :
• Here Principal (P) = a certain sum of money, CI = Rs. 3783,
Time (T) = 3years, Rate of interest (R) = 5% pa
• Let the principal (P) be Rs. x
• Assuming interest is being charged annually, We know that,
CI =
=> CI on Rs. x @ 5% for 3 years =

=> 3783 =
Find x then find SI on Rs. x @ 5% for 3 years = (x*5*3)/100


Slide 13

Quantitative Methods
Session 8– 31.07.12
Chapter 5 – Simple Interest & Compound Interest

Pranjoy Arup Das

• Interest is the payment to be made for the use of money.
• Suppose, Mr. A borrows a sum of money Rs. X from Mr. B, then Mr. A
will have to repay not only the sum of Rs. X but also some extra money
(say Rs. Y) to Mr. B, sort of like a charge or rent for using Mr. B’s money.
• Rs. X is termed as the principal (P), Rs. Y is the interest (I) for using
Mr.B’s money & Rs. (X + Y) is termed as the Amount(A) which Mr. A
will have to repay to Mr. B after a certain period of time (T) .
• The interest to be paid by the borrower Mr. A to the lender Mr. B is
fixed before or while lending the money. Interest is fixed as a
percentage of the principal and is called Rate of interest (R).

• Money is usually lent out for a fixed period of time (T) . Interest is
calculated on the basis of this time period at the agreed rate, which is
charged either yearly, half-yearly or quarterly. T is always calculated in
Years. So a loan period of 6 months will have to be taken as 0.5 years.

There are two kinds of interest Simple interest & Compound Interest
Simple interest is always calculated on the actual principal originally borrowed.
Throughout the entire time period, a fixed amount of interest is paid whenever
due which is either yearly, half yearly of quarterly
For Eg. Mr. A borrows Rs. 2000 from Mr. B for a period of 2 years at an interest
rate of 10% payable yearly. How much will A pay B after 2 years?
Compound interest is calculated on the increasing principal. The interest is
calculated on the original principal and then added to the original principal
which becomes the new principal for calculating the interest on next due date.
• Simple Interest:
Original Principal Rs. 2000
Interest at 1st year end = 10%of 2000
= Rs. 200
Interest at 2nd year end = 10%of 2000
= Rs. 200
So at the end of 2 years Mr. A has to
pay Rs. (2000 + 200+200)=
Rs. 2400/-

• Compound Interest:
Original principal Rs. 2000
Interest at 1st year end = 10% of 2000
= Rs. 200

New principal=Rs. 2000+200 = 2200
Interest at 2nd year end = 10% of 2200
= Rs. 220
So at the end of 2 years Mr. A has to
pay Rs. (2200 + 220) = Rs. 2420/-

Part 1 Simple Interest

• If a sum of money (principal) Rs. P is borrowed at the rate of R% simple
interest for T years, then
Simple interest amount for T years = 1st years int. + 2nd yrs int. +….+ Tth years int
= (R% of P) + (R% of P) + (R% of P) + …T times
= (R% of P) (1 + 1 + 1 +……………upto T)
= (R% of P) * T
=

I=
• Principal (P)+ Interest (I)= Total Amount(A) to be repaid after T years
=> P +
I
= A
=> P + (R*P*T)/100 = A

Eg. Mr. A borrows Rs. 5400/- from Mr. B under the conditions that :

a) The loan will be repaid after 3 years
b) The borrower will pay a simple interest of 8% per annum (per year)
1) How much interest will Mr. A have to pay at the end of 3 years?
2) What is the amount that Mr. A will have to pay at the end of 3years?
Solution:

Here the Principal = Rs. 5400 Time = 3years Rate of interest= 8%
• Interest at 1st year end = 8% of 5400
•
= Rs. 432
• Interest at 2nd year end = 8% of 5400
•
= Rs. 432
• Interest at 3nd year end = 8% of 5400
•
= Rs. 432
• Total interest at the end of the 3 year loan period
= Rs. (432+432+432) = Rs. 1296
• So at the end of 3 years Mr. A has to pay Rs. (5400 + 1296 )
•
= Rs. 6696

Alternative Solution: USING FORMULA OF SIMPLE INTEREST
Here Principal (P) = Rs.5400, Time (T) = 3 years , Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400* 3) / 100
I = Rs. 1296
The amount to be paid to Mr. B at the end of 3 years :
Amount = P + I
= 5400 + 1296
= Rs. 6696

Page 388 (RSA) Ex 1
SOLUTION 1: Please Note: ALWAYS CONVERT DAYS OR MONTHS TO YEARS
Here the Principal = Rs. 5400
Time = 10/12 years Rate of int. = 8%pa
• Interest for 1 year
= 8% of 5400
= Rs. 432
• That means Interest for 12 months = Rs. 432
•  Interest for 10 months will be: (10/12) * 432 = Rs. 360

• Total interest at the end of the 10 month loan period = Rs. 360
SOLUTION 2 :
Here the Principal (P) = 5400

Time (T) = 10/12 years Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400*
) / 100 = Rs.__________

Page 390 Exercise 21A, Pr no. 4
Solution:
Let the sum of money be Rs. x which is the principal (P)
It is given that interest on Rs. x @8% p.a. in 6 years = Rs. 8376
Here principal is Rs .x, Rate of interest (R ) is 8% p.a. and the time
(T) is 6 years
Interest on Rs. x @ 8% for 6 years (I) = (R*P*T)/100
= (8 * x * 6) /100 = Rs.12x/25
So that means , 12x/25 = 8376
=>
x = (8376 * 25) /12
=>
x = Rs. ___________

Page 388, Ex. 4 Solution:
Let the sum of money (P) be Rs. x
And let the rate of interest (R ) be y% p.a
In the first case Rs. x becomes Rs. 854 in 2 years @ y% p.a
Interest @ y% on Rs. x for 2 years (I) = (R*P*T)/100 = (y * x * 2) /100
= (2xy)/100
= Rs. xy/50
So, x + (xy/50) = 854
=> 50x + xy = 42700…………(i)
Similarly, in the 2nd case
Interest on Rs. x @ y% for 7/2 years = (y *x*7/2) / 100 = Rs.(7xy/200)

So, x + (7xy/200) = 969.50
=> 200x + 7xy = 193900…….(ii)
Eliminating xy from (i) & (ii) we can get the value of x and then y.

Page 389, Ex. 5 Solution:
Let the sum of money (P) be Rs. x
So after 8 years, this sum of money becomes Rs. 2x.
Since, Principal

x

+ Interest
= Amount
+ (Interest on Rs. x for 8 years) = 2x
=> (Interest on Rs. x for 8 years) = 2x – x = Rs. x

Let the rate of interest be R% p.a.
Here principal (P) = Rs. x , Interest (I)= Rs. x and Time (T)= 8 years
We know that I = (R * P* T) /100
 R = (I * 100) / (P*T)
= ( x* 100) / (x * 8)
= (100x) /(8x) = 100/8
= 25/2 = 12.5 %

Page 389, Ex. 6
Solution:
Let the loan @ 8% p.a. be Rs. x
 Loan @ 10% p.a. = Rs. (8000 – x)
Given that total annual interest = Rs. 714
That means, in 1 year:
Interest earned on Rs. x@ 8% + Interest earned @ 10% on Rs. (8000-x) = 714
=> {(8 * x * 1) / 100} + [{10 * (8000-x) * 1} / 100] = 714
=>
Solve for x.

(8x/100)

+

{(8000-x) / 10}

=

714

Session 9; Chapter 5; Part 2
Compound Interest

Page 401 Ex 1.
SOLUTION 1 : Without using formula.
• Here Principal (P)= Rs. 8000 , Rate ( R) = 10% p.a ., Time (T) = 3 years
• Interest at 1st year end @ 10% on 8000 for 1 year= (8000 * 10 * 1) /100
•
= Rs. 800
• New principal for 2nd year = Rs. 8000+800 = 8800
• Interest at 2nd year end @ 10% on 8800 for 1 year = (8800 * 10 * 1) /100
•
= Rs. 880
• New principal for 3rd year = Rs. 8800 + 880 = Rs. 9680
• Interest at 3rd year end @ 10% on 9680 for 1 year = (9680 * 10 * 1) /100
•
= Rs. 968
So at the end of 3 years Rs. 8000 will become Rs. (9680 + 968) = Rs. 10648/• So the compound interest (CI) = A – P = 10648 – 8000 = Rs. 2648
•
ALTERNATIVELY : CI = Rs. (800 + 880+ 968) = Rs. 2648

FORMULAS OF COMPOUND INTEREST:
• If a sum of money (principal) Rs. P is borrowed /lent at the rate of R%
compound interest for T years compounded yearly , then :
Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed / lent at the rate of R% compound interest for
compounded half yearly, then :

T years

Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed /lent at the rate of R% compound interest for
compounded quarterly, then :
Compound Interest (CI) on Rs. P @ R% for T years =

T years

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 8000 Rate ( R) = 10% p.a

Time (T) = 3 years

• Compound interest (CI) on Rs. 8000 at the end of 3 years @ 10% =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 4.
SOLUTION 1 : Without using formula.
Here Principal (P)= Rs. 25000, Time (T) = 1 year, Yearly rate of interest =12%
• The Interest is compounded half yearly
• So the half yearly rate of interest = 12 * ½ = 6% per half year
• Interest at 1st half year end = 6% of 25000
•
= Rs. 1500
• New principal for 2nd half year = Rs. 25000+1500 = Rs. 26500
• Interest at 2nd half year end = 6% of 26500
•
= Rs. 1590
• So at the end of 1 year Rs. 25000 will become Rs. (26500 +1590)
= Rs. 28090/• So the compound interest (CI) = 28090 – 25000 = Rs. 3090
• Or , CI = Rs. (1500 + 1590)= Rs. 3090

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 25000 Rate ( R) = 12% p.a Time (T) = 1 year
• Compound interest (CI) on Rs. 25000 at the end of 1 years @ 12% pa
compounded half yearly =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 5. SOLUTION : Without using formula.
Here Principal (P)= Rs. 8000, Time (T) = 9 months = 9/12 years = ¾ yrs,
• Note: 9 months = 3 quarters
• Yearly rate of interest = 20% pa
• The Interest is compounded quarterly . In a year there are 4 quarters
• So the per quarter rate of interest = 20/4 = 5% per quarter.
• Principal for 1st quarter = Rs. 8000
• Interest at 1st quarter end = 5% of 8000 = Rs. 400

• New principal for 2nd quarter = Rs. 8000+400 = Rs. 8400
• Interest at 2nd quarter end = 5% of 8400 = Rs. 420
• New principal for 3rd quarter = Rs. 8400+420 = Rs. 8820
• Interest at 3rd quarter end = 5% of 8820 = Rs. 441
• So at the end of 3 quarters or 9 months, Rs. 8000 will become
Rs. (8820 +441) = Rs. 9261/• So the compound interest (CI) = 9261 – 8000 = Rs. 1261
• Or, CI = Rs. (400 + 420+ 441)= Rs. 1261

Page 402 Ex 3. SOLUTION :
• Here Principal (P)= Rs. 5000, Time (T) = 3years,
• Rate of interest for 1st year (R1) = 5% pa
• Principal for 1st year = Rs. 5000
Interest at 1st year end

= 5% of 5000 = Rs. 250

• New principal for 2nd year = Rs. 5000 + 250 = Rs. 5250
• Rate of interest for 2nd year (R2) = 8% pa
Interest at 2nd year end = 8% of 5250 = Rs. 420

• New principal for 3rd year = Rs. 5250 + 420 = Rs. 5670
• Rate of interest for 3rd year (R3) = 10% pa
Interest at 3rd year end = 10% of 5670 = Rs. 567
• So the amount at the end of 3 years = Rs. (5670 +567) = Rs. 6237/• This means, at the end of 3 years, Rs. 5000 will become Rs. 6237/• Compound interest (CI) on Rs. 5000 after 3 years @ 5%,8% &10%=
6237 – 5000 = Rs. 1237
• Or, CI on Rs. 5000 after 3 years @ 5%,8% &10%= ( 250+ 420+ 567)= Rs. 1237

Page 402 Ex 6. SOLUTION :
• Here Principal (P) = a certain sum of money, CI = Rs. 3783,
Time (T) = 3years, Rate of interest (R) = 5% pa
• Let the principal (P) be Rs. x
• Assuming interest is being charged annually, We know that,
CI =
=> CI on Rs. x @ 5% for 3 years =

=> 3783 =
Find x then find SI on Rs. x @ 5% for 3 years = (x*5*3)/100


Slide 14

Quantitative Methods
Session 8– 31.07.12
Chapter 5 – Simple Interest & Compound Interest

Pranjoy Arup Das

• Interest is the payment to be made for the use of money.
• Suppose, Mr. A borrows a sum of money Rs. X from Mr. B, then Mr. A
will have to repay not only the sum of Rs. X but also some extra money
(say Rs. Y) to Mr. B, sort of like a charge or rent for using Mr. B’s money.
• Rs. X is termed as the principal (P), Rs. Y is the interest (I) for using
Mr.B’s money & Rs. (X + Y) is termed as the Amount(A) which Mr. A
will have to repay to Mr. B after a certain period of time (T) .
• The interest to be paid by the borrower Mr. A to the lender Mr. B is
fixed before or while lending the money. Interest is fixed as a
percentage of the principal and is called Rate of interest (R).

• Money is usually lent out for a fixed period of time (T) . Interest is
calculated on the basis of this time period at the agreed rate, which is
charged either yearly, half-yearly or quarterly. T is always calculated in
Years. So a loan period of 6 months will have to be taken as 0.5 years.

There are two kinds of interest Simple interest & Compound Interest
Simple interest is always calculated on the actual principal originally borrowed.
Throughout the entire time period, a fixed amount of interest is paid whenever
due which is either yearly, half yearly of quarterly
For Eg. Mr. A borrows Rs. 2000 from Mr. B for a period of 2 years at an interest
rate of 10% payable yearly. How much will A pay B after 2 years?
Compound interest is calculated on the increasing principal. The interest is
calculated on the original principal and then added to the original principal
which becomes the new principal for calculating the interest on next due date.
• Simple Interest:
Original Principal Rs. 2000
Interest at 1st year end = 10%of 2000
= Rs. 200
Interest at 2nd year end = 10%of 2000
= Rs. 200
So at the end of 2 years Mr. A has to
pay Rs. (2000 + 200+200)=
Rs. 2400/-

• Compound Interest:
Original principal Rs. 2000
Interest at 1st year end = 10% of 2000
= Rs. 200

New principal=Rs. 2000+200 = 2200
Interest at 2nd year end = 10% of 2200
= Rs. 220
So at the end of 2 years Mr. A has to
pay Rs. (2200 + 220) = Rs. 2420/-

Part 1 Simple Interest

• If a sum of money (principal) Rs. P is borrowed at the rate of R% simple
interest for T years, then
Simple interest amount for T years = 1st years int. + 2nd yrs int. +….+ Tth years int
= (R% of P) + (R% of P) + (R% of P) + …T times
= (R% of P) (1 + 1 + 1 +……………upto T)
= (R% of P) * T
=

I=
• Principal (P)+ Interest (I)= Total Amount(A) to be repaid after T years
=> P +
I
= A
=> P + (R*P*T)/100 = A

Eg. Mr. A borrows Rs. 5400/- from Mr. B under the conditions that :

a) The loan will be repaid after 3 years
b) The borrower will pay a simple interest of 8% per annum (per year)
1) How much interest will Mr. A have to pay at the end of 3 years?
2) What is the amount that Mr. A will have to pay at the end of 3years?
Solution:

Here the Principal = Rs. 5400 Time = 3years Rate of interest= 8%
• Interest at 1st year end = 8% of 5400
•
= Rs. 432
• Interest at 2nd year end = 8% of 5400
•
= Rs. 432
• Interest at 3nd year end = 8% of 5400
•
= Rs. 432
• Total interest at the end of the 3 year loan period
= Rs. (432+432+432) = Rs. 1296
• So at the end of 3 years Mr. A has to pay Rs. (5400 + 1296 )
•
= Rs. 6696

Alternative Solution: USING FORMULA OF SIMPLE INTEREST
Here Principal (P) = Rs.5400, Time (T) = 3 years , Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400* 3) / 100
I = Rs. 1296
The amount to be paid to Mr. B at the end of 3 years :
Amount = P + I
= 5400 + 1296
= Rs. 6696

Page 388 (RSA) Ex 1
SOLUTION 1: Please Note: ALWAYS CONVERT DAYS OR MONTHS TO YEARS
Here the Principal = Rs. 5400
Time = 10/12 years Rate of int. = 8%pa
• Interest for 1 year
= 8% of 5400
= Rs. 432
• That means Interest for 12 months = Rs. 432
•  Interest for 10 months will be: (10/12) * 432 = Rs. 360

• Total interest at the end of the 10 month loan period = Rs. 360
SOLUTION 2 :
Here the Principal (P) = 5400

Time (T) = 10/12 years Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400*
) / 100 = Rs.__________

Page 390 Exercise 21A, Pr no. 4
Solution:
Let the sum of money be Rs. x which is the principal (P)
It is given that interest on Rs. x @8% p.a. in 6 years = Rs. 8376
Here principal is Rs .x, Rate of interest (R ) is 8% p.a. and the time
(T) is 6 years
Interest on Rs. x @ 8% for 6 years (I) = (R*P*T)/100
= (8 * x * 6) /100 = Rs.12x/25
So that means , 12x/25 = 8376
=>
x = (8376 * 25) /12
=>
x = Rs. ___________

Page 388, Ex. 4 Solution:
Let the sum of money (P) be Rs. x
And let the rate of interest (R ) be y% p.a
In the first case Rs. x becomes Rs. 854 in 2 years @ y% p.a
Interest @ y% on Rs. x for 2 years (I) = (R*P*T)/100 = (y * x * 2) /100
= (2xy)/100
= Rs. xy/50
So, x + (xy/50) = 854
=> 50x + xy = 42700…………(i)
Similarly, in the 2nd case
Interest on Rs. x @ y% for 7/2 years = (y *x*7/2) / 100 = Rs.(7xy/200)

So, x + (7xy/200) = 969.50
=> 200x + 7xy = 193900…….(ii)
Eliminating xy from (i) & (ii) we can get the value of x and then y.

Page 389, Ex. 5 Solution:
Let the sum of money (P) be Rs. x
So after 8 years, this sum of money becomes Rs. 2x.
Since, Principal

x

+ Interest
= Amount
+ (Interest on Rs. x for 8 years) = 2x
=> (Interest on Rs. x for 8 years) = 2x – x = Rs. x

Let the rate of interest be R% p.a.
Here principal (P) = Rs. x , Interest (I)= Rs. x and Time (T)= 8 years
We know that I = (R * P* T) /100
 R = (I * 100) / (P*T)
= ( x* 100) / (x * 8)
= (100x) /(8x) = 100/8
= 25/2 = 12.5 %

Page 389, Ex. 6
Solution:
Let the loan @ 8% p.a. be Rs. x
 Loan @ 10% p.a. = Rs. (8000 – x)
Given that total annual interest = Rs. 714
That means, in 1 year:
Interest earned on Rs. x@ 8% + Interest earned @ 10% on Rs. (8000-x) = 714
=> {(8 * x * 1) / 100} + [{10 * (8000-x) * 1} / 100] = 714
=>
Solve for x.

(8x/100)

+

{(8000-x) / 10}

=

714

Session 9; Chapter 5; Part 2
Compound Interest

Page 401 Ex 1.
SOLUTION 1 : Without using formula.
• Here Principal (P)= Rs. 8000 , Rate ( R) = 10% p.a ., Time (T) = 3 years
• Interest at 1st year end @ 10% on 8000 for 1 year= (8000 * 10 * 1) /100
•
= Rs. 800
• New principal for 2nd year = Rs. 8000+800 = 8800
• Interest at 2nd year end @ 10% on 8800 for 1 year = (8800 * 10 * 1) /100
•
= Rs. 880
• New principal for 3rd year = Rs. 8800 + 880 = Rs. 9680
• Interest at 3rd year end @ 10% on 9680 for 1 year = (9680 * 10 * 1) /100
•
= Rs. 968
So at the end of 3 years Rs. 8000 will become Rs. (9680 + 968) = Rs. 10648/• So the compound interest (CI) = A – P = 10648 – 8000 = Rs. 2648
•
ALTERNATIVELY : CI = Rs. (800 + 880+ 968) = Rs. 2648

FORMULAS OF COMPOUND INTEREST:
• If a sum of money (principal) Rs. P is borrowed /lent at the rate of R%
compound interest for T years compounded yearly , then :
Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed / lent at the rate of R% compound interest for
compounded half yearly, then :

T years

Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed /lent at the rate of R% compound interest for
compounded quarterly, then :
Compound Interest (CI) on Rs. P @ R% for T years =

T years

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 8000 Rate ( R) = 10% p.a

Time (T) = 3 years

• Compound interest (CI) on Rs. 8000 at the end of 3 years @ 10% =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 4.
SOLUTION 1 : Without using formula.
Here Principal (P)= Rs. 25000, Time (T) = 1 year, Yearly rate of interest =12%
• The Interest is compounded half yearly
• So the half yearly rate of interest = 12 * ½ = 6% per half year
• Interest at 1st half year end = 6% of 25000
•
= Rs. 1500
• New principal for 2nd half year = Rs. 25000+1500 = Rs. 26500
• Interest at 2nd half year end = 6% of 26500
•
= Rs. 1590
• So at the end of 1 year Rs. 25000 will become Rs. (26500 +1590)
= Rs. 28090/• So the compound interest (CI) = 28090 – 25000 = Rs. 3090
• Or , CI = Rs. (1500 + 1590)= Rs. 3090

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 25000 Rate ( R) = 12% p.a Time (T) = 1 year
• Compound interest (CI) on Rs. 25000 at the end of 1 years @ 12% pa
compounded half yearly =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 5. SOLUTION : Without using formula.
Here Principal (P)= Rs. 8000, Time (T) = 9 months = 9/12 years = ¾ yrs,
• Note: 9 months = 3 quarters
• Yearly rate of interest = 20% pa
• The Interest is compounded quarterly . In a year there are 4 quarters
• So the per quarter rate of interest = 20/4 = 5% per quarter.
• Principal for 1st quarter = Rs. 8000
• Interest at 1st quarter end = 5% of 8000 = Rs. 400

• New principal for 2nd quarter = Rs. 8000+400 = Rs. 8400
• Interest at 2nd quarter end = 5% of 8400 = Rs. 420
• New principal for 3rd quarter = Rs. 8400+420 = Rs. 8820
• Interest at 3rd quarter end = 5% of 8820 = Rs. 441
• So at the end of 3 quarters or 9 months, Rs. 8000 will become
Rs. (8820 +441) = Rs. 9261/• So the compound interest (CI) = 9261 – 8000 = Rs. 1261
• Or, CI = Rs. (400 + 420+ 441)= Rs. 1261

Page 402 Ex 3. SOLUTION :
• Here Principal (P)= Rs. 5000, Time (T) = 3years,
• Rate of interest for 1st year (R1) = 5% pa
• Principal for 1st year = Rs. 5000
Interest at 1st year end

= 5% of 5000 = Rs. 250

• New principal for 2nd year = Rs. 5000 + 250 = Rs. 5250
• Rate of interest for 2nd year (R2) = 8% pa
Interest at 2nd year end = 8% of 5250 = Rs. 420

• New principal for 3rd year = Rs. 5250 + 420 = Rs. 5670
• Rate of interest for 3rd year (R3) = 10% pa
Interest at 3rd year end = 10% of 5670 = Rs. 567
• So the amount at the end of 3 years = Rs. (5670 +567) = Rs. 6237/• This means, at the end of 3 years, Rs. 5000 will become Rs. 6237/• Compound interest (CI) on Rs. 5000 after 3 years @ 5%,8% &10%=
6237 – 5000 = Rs. 1237
• Or, CI on Rs. 5000 after 3 years @ 5%,8% &10%= ( 250+ 420+ 567)= Rs. 1237

Page 402 Ex 6. SOLUTION :
• Here Principal (P) = a certain sum of money, CI = Rs. 3783,
Time (T) = 3years, Rate of interest (R) = 5% pa
• Let the principal (P) be Rs. x
• Assuming interest is being charged annually, We know that,
CI =
=> CI on Rs. x @ 5% for 3 years =

=> 3783 =
Find x then find SI on Rs. x @ 5% for 3 years = (x*5*3)/100


Slide 15

Quantitative Methods
Session 8– 31.07.12
Chapter 5 – Simple Interest & Compound Interest

Pranjoy Arup Das

• Interest is the payment to be made for the use of money.
• Suppose, Mr. A borrows a sum of money Rs. X from Mr. B, then Mr. A
will have to repay not only the sum of Rs. X but also some extra money
(say Rs. Y) to Mr. B, sort of like a charge or rent for using Mr. B’s money.
• Rs. X is termed as the principal (P), Rs. Y is the interest (I) for using
Mr.B’s money & Rs. (X + Y) is termed as the Amount(A) which Mr. A
will have to repay to Mr. B after a certain period of time (T) .
• The interest to be paid by the borrower Mr. A to the lender Mr. B is
fixed before or while lending the money. Interest is fixed as a
percentage of the principal and is called Rate of interest (R).

• Money is usually lent out for a fixed period of time (T) . Interest is
calculated on the basis of this time period at the agreed rate, which is
charged either yearly, half-yearly or quarterly. T is always calculated in
Years. So a loan period of 6 months will have to be taken as 0.5 years.

There are two kinds of interest Simple interest & Compound Interest
Simple interest is always calculated on the actual principal originally borrowed.
Throughout the entire time period, a fixed amount of interest is paid whenever
due which is either yearly, half yearly of quarterly
For Eg. Mr. A borrows Rs. 2000 from Mr. B for a period of 2 years at an interest
rate of 10% payable yearly. How much will A pay B after 2 years?
Compound interest is calculated on the increasing principal. The interest is
calculated on the original principal and then added to the original principal
which becomes the new principal for calculating the interest on next due date.
• Simple Interest:
Original Principal Rs. 2000
Interest at 1st year end = 10%of 2000
= Rs. 200
Interest at 2nd year end = 10%of 2000
= Rs. 200
So at the end of 2 years Mr. A has to
pay Rs. (2000 + 200+200)=
Rs. 2400/-

• Compound Interest:
Original principal Rs. 2000
Interest at 1st year end = 10% of 2000
= Rs. 200

New principal=Rs. 2000+200 = 2200
Interest at 2nd year end = 10% of 2200
= Rs. 220
So at the end of 2 years Mr. A has to
pay Rs. (2200 + 220) = Rs. 2420/-

Part 1 Simple Interest

• If a sum of money (principal) Rs. P is borrowed at the rate of R% simple
interest for T years, then
Simple interest amount for T years = 1st years int. + 2nd yrs int. +….+ Tth years int
= (R% of P) + (R% of P) + (R% of P) + …T times
= (R% of P) (1 + 1 + 1 +……………upto T)
= (R% of P) * T
=

I=
• Principal (P)+ Interest (I)= Total Amount(A) to be repaid after T years
=> P +
I
= A
=> P + (R*P*T)/100 = A

Eg. Mr. A borrows Rs. 5400/- from Mr. B under the conditions that :

a) The loan will be repaid after 3 years
b) The borrower will pay a simple interest of 8% per annum (per year)
1) How much interest will Mr. A have to pay at the end of 3 years?
2) What is the amount that Mr. A will have to pay at the end of 3years?
Solution:

Here the Principal = Rs. 5400 Time = 3years Rate of interest= 8%
• Interest at 1st year end = 8% of 5400
•
= Rs. 432
• Interest at 2nd year end = 8% of 5400
•
= Rs. 432
• Interest at 3nd year end = 8% of 5400
•
= Rs. 432
• Total interest at the end of the 3 year loan period
= Rs. (432+432+432) = Rs. 1296
• So at the end of 3 years Mr. A has to pay Rs. (5400 + 1296 )
•
= Rs. 6696

Alternative Solution: USING FORMULA OF SIMPLE INTEREST
Here Principal (P) = Rs.5400, Time (T) = 3 years , Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400* 3) / 100
I = Rs. 1296
The amount to be paid to Mr. B at the end of 3 years :
Amount = P + I
= 5400 + 1296
= Rs. 6696

Page 388 (RSA) Ex 1
SOLUTION 1: Please Note: ALWAYS CONVERT DAYS OR MONTHS TO YEARS
Here the Principal = Rs. 5400
Time = 10/12 years Rate of int. = 8%pa
• Interest for 1 year
= 8% of 5400
= Rs. 432
• That means Interest for 12 months = Rs. 432
•  Interest for 10 months will be: (10/12) * 432 = Rs. 360

• Total interest at the end of the 10 month loan period = Rs. 360
SOLUTION 2 :
Here the Principal (P) = 5400

Time (T) = 10/12 years Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400*
) / 100 = Rs.__________

Page 390 Exercise 21A, Pr no. 4
Solution:
Let the sum of money be Rs. x which is the principal (P)
It is given that interest on Rs. x @8% p.a. in 6 years = Rs. 8376
Here principal is Rs .x, Rate of interest (R ) is 8% p.a. and the time
(T) is 6 years
Interest on Rs. x @ 8% for 6 years (I) = (R*P*T)/100
= (8 * x * 6) /100 = Rs.12x/25
So that means , 12x/25 = 8376
=>
x = (8376 * 25) /12
=>
x = Rs. ___________

Page 388, Ex. 4 Solution:
Let the sum of money (P) be Rs. x
And let the rate of interest (R ) be y% p.a
In the first case Rs. x becomes Rs. 854 in 2 years @ y% p.a
Interest @ y% on Rs. x for 2 years (I) = (R*P*T)/100 = (y * x * 2) /100
= (2xy)/100
= Rs. xy/50
So, x + (xy/50) = 854
=> 50x + xy = 42700…………(i)
Similarly, in the 2nd case
Interest on Rs. x @ y% for 7/2 years = (y *x*7/2) / 100 = Rs.(7xy/200)

So, x + (7xy/200) = 969.50
=> 200x + 7xy = 193900…….(ii)
Eliminating xy from (i) & (ii) we can get the value of x and then y.

Page 389, Ex. 5 Solution:
Let the sum of money (P) be Rs. x
So after 8 years, this sum of money becomes Rs. 2x.
Since, Principal

x

+ Interest
= Amount
+ (Interest on Rs. x for 8 years) = 2x
=> (Interest on Rs. x for 8 years) = 2x – x = Rs. x

Let the rate of interest be R% p.a.
Here principal (P) = Rs. x , Interest (I)= Rs. x and Time (T)= 8 years
We know that I = (R * P* T) /100
 R = (I * 100) / (P*T)
= ( x* 100) / (x * 8)
= (100x) /(8x) = 100/8
= 25/2 = 12.5 %

Page 389, Ex. 6
Solution:
Let the loan @ 8% p.a. be Rs. x
 Loan @ 10% p.a. = Rs. (8000 – x)
Given that total annual interest = Rs. 714
That means, in 1 year:
Interest earned on Rs. x@ 8% + Interest earned @ 10% on Rs. (8000-x) = 714
=> {(8 * x * 1) / 100} + [{10 * (8000-x) * 1} / 100] = 714
=>
Solve for x.

(8x/100)

+

{(8000-x) / 10}

=

714

Session 9; Chapter 5; Part 2
Compound Interest

Page 401 Ex 1.
SOLUTION 1 : Without using formula.
• Here Principal (P)= Rs. 8000 , Rate ( R) = 10% p.a ., Time (T) = 3 years
• Interest at 1st year end @ 10% on 8000 for 1 year= (8000 * 10 * 1) /100
•
= Rs. 800
• New principal for 2nd year = Rs. 8000+800 = 8800
• Interest at 2nd year end @ 10% on 8800 for 1 year = (8800 * 10 * 1) /100
•
= Rs. 880
• New principal for 3rd year = Rs. 8800 + 880 = Rs. 9680
• Interest at 3rd year end @ 10% on 9680 for 1 year = (9680 * 10 * 1) /100
•
= Rs. 968
So at the end of 3 years Rs. 8000 will become Rs. (9680 + 968) = Rs. 10648/• So the compound interest (CI) = A – P = 10648 – 8000 = Rs. 2648
•
ALTERNATIVELY : CI = Rs. (800 + 880+ 968) = Rs. 2648

FORMULAS OF COMPOUND INTEREST:
• If a sum of money (principal) Rs. P is borrowed /lent at the rate of R%
compound interest for T years compounded yearly , then :
Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed / lent at the rate of R% compound interest for
compounded half yearly, then :

T years

Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed /lent at the rate of R% compound interest for
compounded quarterly, then :
Compound Interest (CI) on Rs. P @ R% for T years =

T years

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 8000 Rate ( R) = 10% p.a

Time (T) = 3 years

• Compound interest (CI) on Rs. 8000 at the end of 3 years @ 10% =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 4.
SOLUTION 1 : Without using formula.
Here Principal (P)= Rs. 25000, Time (T) = 1 year, Yearly rate of interest =12%
• The Interest is compounded half yearly
• So the half yearly rate of interest = 12 * ½ = 6% per half year
• Interest at 1st half year end = 6% of 25000
•
= Rs. 1500
• New principal for 2nd half year = Rs. 25000+1500 = Rs. 26500
• Interest at 2nd half year end = 6% of 26500
•
= Rs. 1590
• So at the end of 1 year Rs. 25000 will become Rs. (26500 +1590)
= Rs. 28090/• So the compound interest (CI) = 28090 – 25000 = Rs. 3090
• Or , CI = Rs. (1500 + 1590)= Rs. 3090

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 25000 Rate ( R) = 12% p.a Time (T) = 1 year
• Compound interest (CI) on Rs. 25000 at the end of 1 years @ 12% pa
compounded half yearly =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 5. SOLUTION : Without using formula.
Here Principal (P)= Rs. 8000, Time (T) = 9 months = 9/12 years = ¾ yrs,
• Note: 9 months = 3 quarters
• Yearly rate of interest = 20% pa
• The Interest is compounded quarterly . In a year there are 4 quarters
• So the per quarter rate of interest = 20/4 = 5% per quarter.
• Principal for 1st quarter = Rs. 8000
• Interest at 1st quarter end = 5% of 8000 = Rs. 400

• New principal for 2nd quarter = Rs. 8000+400 = Rs. 8400
• Interest at 2nd quarter end = 5% of 8400 = Rs. 420
• New principal for 3rd quarter = Rs. 8400+420 = Rs. 8820
• Interest at 3rd quarter end = 5% of 8820 = Rs. 441
• So at the end of 3 quarters or 9 months, Rs. 8000 will become
Rs. (8820 +441) = Rs. 9261/• So the compound interest (CI) = 9261 – 8000 = Rs. 1261
• Or, CI = Rs. (400 + 420+ 441)= Rs. 1261

Page 402 Ex 3. SOLUTION :
• Here Principal (P)= Rs. 5000, Time (T) = 3years,
• Rate of interest for 1st year (R1) = 5% pa
• Principal for 1st year = Rs. 5000
Interest at 1st year end

= 5% of 5000 = Rs. 250

• New principal for 2nd year = Rs. 5000 + 250 = Rs. 5250
• Rate of interest for 2nd year (R2) = 8% pa
Interest at 2nd year end = 8% of 5250 = Rs. 420

• New principal for 3rd year = Rs. 5250 + 420 = Rs. 5670
• Rate of interest for 3rd year (R3) = 10% pa
Interest at 3rd year end = 10% of 5670 = Rs. 567
• So the amount at the end of 3 years = Rs. (5670 +567) = Rs. 6237/• This means, at the end of 3 years, Rs. 5000 will become Rs. 6237/• Compound interest (CI) on Rs. 5000 after 3 years @ 5%,8% &10%=
6237 – 5000 = Rs. 1237
• Or, CI on Rs. 5000 after 3 years @ 5%,8% &10%= ( 250+ 420+ 567)= Rs. 1237

Page 402 Ex 6. SOLUTION :
• Here Principal (P) = a certain sum of money, CI = Rs. 3783,
Time (T) = 3years, Rate of interest (R) = 5% pa
• Let the principal (P) be Rs. x
• Assuming interest is being charged annually, We know that,
CI =
=> CI on Rs. x @ 5% for 3 years =

=> 3783 =
Find x then find SI on Rs. x @ 5% for 3 years = (x*5*3)/100


Slide 16

Quantitative Methods
Session 8– 31.07.12
Chapter 5 – Simple Interest & Compound Interest

Pranjoy Arup Das

• Interest is the payment to be made for the use of money.
• Suppose, Mr. A borrows a sum of money Rs. X from Mr. B, then Mr. A
will have to repay not only the sum of Rs. X but also some extra money
(say Rs. Y) to Mr. B, sort of like a charge or rent for using Mr. B’s money.
• Rs. X is termed as the principal (P), Rs. Y is the interest (I) for using
Mr.B’s money & Rs. (X + Y) is termed as the Amount(A) which Mr. A
will have to repay to Mr. B after a certain period of time (T) .
• The interest to be paid by the borrower Mr. A to the lender Mr. B is
fixed before or while lending the money. Interest is fixed as a
percentage of the principal and is called Rate of interest (R).

• Money is usually lent out for a fixed period of time (T) . Interest is
calculated on the basis of this time period at the agreed rate, which is
charged either yearly, half-yearly or quarterly. T is always calculated in
Years. So a loan period of 6 months will have to be taken as 0.5 years.

There are two kinds of interest Simple interest & Compound Interest
Simple interest is always calculated on the actual principal originally borrowed.
Throughout the entire time period, a fixed amount of interest is paid whenever
due which is either yearly, half yearly of quarterly
For Eg. Mr. A borrows Rs. 2000 from Mr. B for a period of 2 years at an interest
rate of 10% payable yearly. How much will A pay B after 2 years?
Compound interest is calculated on the increasing principal. The interest is
calculated on the original principal and then added to the original principal
which becomes the new principal for calculating the interest on next due date.
• Simple Interest:
Original Principal Rs. 2000
Interest at 1st year end = 10%of 2000
= Rs. 200
Interest at 2nd year end = 10%of 2000
= Rs. 200
So at the end of 2 years Mr. A has to
pay Rs. (2000 + 200+200)=
Rs. 2400/-

• Compound Interest:
Original principal Rs. 2000
Interest at 1st year end = 10% of 2000
= Rs. 200

New principal=Rs. 2000+200 = 2200
Interest at 2nd year end = 10% of 2200
= Rs. 220
So at the end of 2 years Mr. A has to
pay Rs. (2200 + 220) = Rs. 2420/-

Part 1 Simple Interest

• If a sum of money (principal) Rs. P is borrowed at the rate of R% simple
interest for T years, then
Simple interest amount for T years = 1st years int. + 2nd yrs int. +….+ Tth years int
= (R% of P) + (R% of P) + (R% of P) + …T times
= (R% of P) (1 + 1 + 1 +……………upto T)
= (R% of P) * T
=

I=
• Principal (P)+ Interest (I)= Total Amount(A) to be repaid after T years
=> P +
I
= A
=> P + (R*P*T)/100 = A

Eg. Mr. A borrows Rs. 5400/- from Mr. B under the conditions that :

a) The loan will be repaid after 3 years
b) The borrower will pay a simple interest of 8% per annum (per year)
1) How much interest will Mr. A have to pay at the end of 3 years?
2) What is the amount that Mr. A will have to pay at the end of 3years?
Solution:

Here the Principal = Rs. 5400 Time = 3years Rate of interest= 8%
• Interest at 1st year end = 8% of 5400
•
= Rs. 432
• Interest at 2nd year end = 8% of 5400
•
= Rs. 432
• Interest at 3nd year end = 8% of 5400
•
= Rs. 432
• Total interest at the end of the 3 year loan period
= Rs. (432+432+432) = Rs. 1296
• So at the end of 3 years Mr. A has to pay Rs. (5400 + 1296 )
•
= Rs. 6696

Alternative Solution: USING FORMULA OF SIMPLE INTEREST
Here Principal (P) = Rs.5400, Time (T) = 3 years , Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400* 3) / 100
I = Rs. 1296
The amount to be paid to Mr. B at the end of 3 years :
Amount = P + I
= 5400 + 1296
= Rs. 6696

Page 388 (RSA) Ex 1
SOLUTION 1: Please Note: ALWAYS CONVERT DAYS OR MONTHS TO YEARS
Here the Principal = Rs. 5400
Time = 10/12 years Rate of int. = 8%pa
• Interest for 1 year
= 8% of 5400
= Rs. 432
• That means Interest for 12 months = Rs. 432
•  Interest for 10 months will be: (10/12) * 432 = Rs. 360

• Total interest at the end of the 10 month loan period = Rs. 360
SOLUTION 2 :
Here the Principal (P) = 5400

Time (T) = 10/12 years Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400*
) / 100 = Rs.__________

Page 390 Exercise 21A, Pr no. 4
Solution:
Let the sum of money be Rs. x which is the principal (P)
It is given that interest on Rs. x @8% p.a. in 6 years = Rs. 8376
Here principal is Rs .x, Rate of interest (R ) is 8% p.a. and the time
(T) is 6 years
Interest on Rs. x @ 8% for 6 years (I) = (R*P*T)/100
= (8 * x * 6) /100 = Rs.12x/25
So that means , 12x/25 = 8376
=>
x = (8376 * 25) /12
=>
x = Rs. ___________

Page 388, Ex. 4 Solution:
Let the sum of money (P) be Rs. x
And let the rate of interest (R ) be y% p.a
In the first case Rs. x becomes Rs. 854 in 2 years @ y% p.a
Interest @ y% on Rs. x for 2 years (I) = (R*P*T)/100 = (y * x * 2) /100
= (2xy)/100
= Rs. xy/50
So, x + (xy/50) = 854
=> 50x + xy = 42700…………(i)
Similarly, in the 2nd case
Interest on Rs. x @ y% for 7/2 years = (y *x*7/2) / 100 = Rs.(7xy/200)

So, x + (7xy/200) = 969.50
=> 200x + 7xy = 193900…….(ii)
Eliminating xy from (i) & (ii) we can get the value of x and then y.

Page 389, Ex. 5 Solution:
Let the sum of money (P) be Rs. x
So after 8 years, this sum of money becomes Rs. 2x.
Since, Principal

x

+ Interest
= Amount
+ (Interest on Rs. x for 8 years) = 2x
=> (Interest on Rs. x for 8 years) = 2x – x = Rs. x

Let the rate of interest be R% p.a.
Here principal (P) = Rs. x , Interest (I)= Rs. x and Time (T)= 8 years
We know that I = (R * P* T) /100
 R = (I * 100) / (P*T)
= ( x* 100) / (x * 8)
= (100x) /(8x) = 100/8
= 25/2 = 12.5 %

Page 389, Ex. 6
Solution:
Let the loan @ 8% p.a. be Rs. x
 Loan @ 10% p.a. = Rs. (8000 – x)
Given that total annual interest = Rs. 714
That means, in 1 year:
Interest earned on Rs. x@ 8% + Interest earned @ 10% on Rs. (8000-x) = 714
=> {(8 * x * 1) / 100} + [{10 * (8000-x) * 1} / 100] = 714
=>
Solve for x.

(8x/100)

+

{(8000-x) / 10}

=

714

Session 9; Chapter 5; Part 2
Compound Interest

Page 401 Ex 1.
SOLUTION 1 : Without using formula.
• Here Principal (P)= Rs. 8000 , Rate ( R) = 10% p.a ., Time (T) = 3 years
• Interest at 1st year end @ 10% on 8000 for 1 year= (8000 * 10 * 1) /100
•
= Rs. 800
• New principal for 2nd year = Rs. 8000+800 = 8800
• Interest at 2nd year end @ 10% on 8800 for 1 year = (8800 * 10 * 1) /100
•
= Rs. 880
• New principal for 3rd year = Rs. 8800 + 880 = Rs. 9680
• Interest at 3rd year end @ 10% on 9680 for 1 year = (9680 * 10 * 1) /100
•
= Rs. 968
So at the end of 3 years Rs. 8000 will become Rs. (9680 + 968) = Rs. 10648/• So the compound interest (CI) = A – P = 10648 – 8000 = Rs. 2648
•
ALTERNATIVELY : CI = Rs. (800 + 880+ 968) = Rs. 2648

FORMULAS OF COMPOUND INTEREST:
• If a sum of money (principal) Rs. P is borrowed /lent at the rate of R%
compound interest for T years compounded yearly , then :
Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed / lent at the rate of R% compound interest for
compounded half yearly, then :

T years

Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed /lent at the rate of R% compound interest for
compounded quarterly, then :
Compound Interest (CI) on Rs. P @ R% for T years =

T years

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 8000 Rate ( R) = 10% p.a

Time (T) = 3 years

• Compound interest (CI) on Rs. 8000 at the end of 3 years @ 10% =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 4.
SOLUTION 1 : Without using formula.
Here Principal (P)= Rs. 25000, Time (T) = 1 year, Yearly rate of interest =12%
• The Interest is compounded half yearly
• So the half yearly rate of interest = 12 * ½ = 6% per half year
• Interest at 1st half year end = 6% of 25000
•
= Rs. 1500
• New principal for 2nd half year = Rs. 25000+1500 = Rs. 26500
• Interest at 2nd half year end = 6% of 26500
•
= Rs. 1590
• So at the end of 1 year Rs. 25000 will become Rs. (26500 +1590)
= Rs. 28090/• So the compound interest (CI) = 28090 – 25000 = Rs. 3090
• Or , CI = Rs. (1500 + 1590)= Rs. 3090

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 25000 Rate ( R) = 12% p.a Time (T) = 1 year
• Compound interest (CI) on Rs. 25000 at the end of 1 years @ 12% pa
compounded half yearly =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 5. SOLUTION : Without using formula.
Here Principal (P)= Rs. 8000, Time (T) = 9 months = 9/12 years = ¾ yrs,
• Note: 9 months = 3 quarters
• Yearly rate of interest = 20% pa
• The Interest is compounded quarterly . In a year there are 4 quarters
• So the per quarter rate of interest = 20/4 = 5% per quarter.
• Principal for 1st quarter = Rs. 8000
• Interest at 1st quarter end = 5% of 8000 = Rs. 400

• New principal for 2nd quarter = Rs. 8000+400 = Rs. 8400
• Interest at 2nd quarter end = 5% of 8400 = Rs. 420
• New principal for 3rd quarter = Rs. 8400+420 = Rs. 8820
• Interest at 3rd quarter end = 5% of 8820 = Rs. 441
• So at the end of 3 quarters or 9 months, Rs. 8000 will become
Rs. (8820 +441) = Rs. 9261/• So the compound interest (CI) = 9261 – 8000 = Rs. 1261
• Or, CI = Rs. (400 + 420+ 441)= Rs. 1261

Page 402 Ex 3. SOLUTION :
• Here Principal (P)= Rs. 5000, Time (T) = 3years,
• Rate of interest for 1st year (R1) = 5% pa
• Principal for 1st year = Rs. 5000
Interest at 1st year end

= 5% of 5000 = Rs. 250

• New principal for 2nd year = Rs. 5000 + 250 = Rs. 5250
• Rate of interest for 2nd year (R2) = 8% pa
Interest at 2nd year end = 8% of 5250 = Rs. 420

• New principal for 3rd year = Rs. 5250 + 420 = Rs. 5670
• Rate of interest for 3rd year (R3) = 10% pa
Interest at 3rd year end = 10% of 5670 = Rs. 567
• So the amount at the end of 3 years = Rs. (5670 +567) = Rs. 6237/• This means, at the end of 3 years, Rs. 5000 will become Rs. 6237/• Compound interest (CI) on Rs. 5000 after 3 years @ 5%,8% &10%=
6237 – 5000 = Rs. 1237
• Or, CI on Rs. 5000 after 3 years @ 5%,8% &10%= ( 250+ 420+ 567)= Rs. 1237

Page 402 Ex 6. SOLUTION :
• Here Principal (P) = a certain sum of money, CI = Rs. 3783,
Time (T) = 3years, Rate of interest (R) = 5% pa
• Let the principal (P) be Rs. x
• Assuming interest is being charged annually, We know that,
CI =
=> CI on Rs. x @ 5% for 3 years =

=> 3783 =
Find x then find SI on Rs. x @ 5% for 3 years = (x*5*3)/100


Slide 17

Quantitative Methods
Session 8– 31.07.12
Chapter 5 – Simple Interest & Compound Interest

Pranjoy Arup Das

• Interest is the payment to be made for the use of money.
• Suppose, Mr. A borrows a sum of money Rs. X from Mr. B, then Mr. A
will have to repay not only the sum of Rs. X but also some extra money
(say Rs. Y) to Mr. B, sort of like a charge or rent for using Mr. B’s money.
• Rs. X is termed as the principal (P), Rs. Y is the interest (I) for using
Mr.B’s money & Rs. (X + Y) is termed as the Amount(A) which Mr. A
will have to repay to Mr. B after a certain period of time (T) .
• The interest to be paid by the borrower Mr. A to the lender Mr. B is
fixed before or while lending the money. Interest is fixed as a
percentage of the principal and is called Rate of interest (R).

• Money is usually lent out for a fixed period of time (T) . Interest is
calculated on the basis of this time period at the agreed rate, which is
charged either yearly, half-yearly or quarterly. T is always calculated in
Years. So a loan period of 6 months will have to be taken as 0.5 years.

There are two kinds of interest Simple interest & Compound Interest
Simple interest is always calculated on the actual principal originally borrowed.
Throughout the entire time period, a fixed amount of interest is paid whenever
due which is either yearly, half yearly of quarterly
For Eg. Mr. A borrows Rs. 2000 from Mr. B for a period of 2 years at an interest
rate of 10% payable yearly. How much will A pay B after 2 years?
Compound interest is calculated on the increasing principal. The interest is
calculated on the original principal and then added to the original principal
which becomes the new principal for calculating the interest on next due date.
• Simple Interest:
Original Principal Rs. 2000
Interest at 1st year end = 10%of 2000
= Rs. 200
Interest at 2nd year end = 10%of 2000
= Rs. 200
So at the end of 2 years Mr. A has to
pay Rs. (2000 + 200+200)=
Rs. 2400/-

• Compound Interest:
Original principal Rs. 2000
Interest at 1st year end = 10% of 2000
= Rs. 200

New principal=Rs. 2000+200 = 2200
Interest at 2nd year end = 10% of 2200
= Rs. 220
So at the end of 2 years Mr. A has to
pay Rs. (2200 + 220) = Rs. 2420/-

Part 1 Simple Interest

• If a sum of money (principal) Rs. P is borrowed at the rate of R% simple
interest for T years, then
Simple interest amount for T years = 1st years int. + 2nd yrs int. +….+ Tth years int
= (R% of P) + (R% of P) + (R% of P) + …T times
= (R% of P) (1 + 1 + 1 +……………upto T)
= (R% of P) * T
=

I=
• Principal (P)+ Interest (I)= Total Amount(A) to be repaid after T years
=> P +
I
= A
=> P + (R*P*T)/100 = A

Eg. Mr. A borrows Rs. 5400/- from Mr. B under the conditions that :

a) The loan will be repaid after 3 years
b) The borrower will pay a simple interest of 8% per annum (per year)
1) How much interest will Mr. A have to pay at the end of 3 years?
2) What is the amount that Mr. A will have to pay at the end of 3years?
Solution:

Here the Principal = Rs. 5400 Time = 3years Rate of interest= 8%
• Interest at 1st year end = 8% of 5400
•
= Rs. 432
• Interest at 2nd year end = 8% of 5400
•
= Rs. 432
• Interest at 3nd year end = 8% of 5400
•
= Rs. 432
• Total interest at the end of the 3 year loan period
= Rs. (432+432+432) = Rs. 1296
• So at the end of 3 years Mr. A has to pay Rs. (5400 + 1296 )
•
= Rs. 6696

Alternative Solution: USING FORMULA OF SIMPLE INTEREST
Here Principal (P) = Rs.5400, Time (T) = 3 years , Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400* 3) / 100
I = Rs. 1296
The amount to be paid to Mr. B at the end of 3 years :
Amount = P + I
= 5400 + 1296
= Rs. 6696

Page 388 (RSA) Ex 1
SOLUTION 1: Please Note: ALWAYS CONVERT DAYS OR MONTHS TO YEARS
Here the Principal = Rs. 5400
Time = 10/12 years Rate of int. = 8%pa
• Interest for 1 year
= 8% of 5400
= Rs. 432
• That means Interest for 12 months = Rs. 432
•  Interest for 10 months will be: (10/12) * 432 = Rs. 360

• Total interest at the end of the 10 month loan period = Rs. 360
SOLUTION 2 :
Here the Principal (P) = 5400

Time (T) = 10/12 years Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400*
) / 100 = Rs.__________

Page 390 Exercise 21A, Pr no. 4
Solution:
Let the sum of money be Rs. x which is the principal (P)
It is given that interest on Rs. x @8% p.a. in 6 years = Rs. 8376
Here principal is Rs .x, Rate of interest (R ) is 8% p.a. and the time
(T) is 6 years
Interest on Rs. x @ 8% for 6 years (I) = (R*P*T)/100
= (8 * x * 6) /100 = Rs.12x/25
So that means , 12x/25 = 8376
=>
x = (8376 * 25) /12
=>
x = Rs. ___________

Page 388, Ex. 4 Solution:
Let the sum of money (P) be Rs. x
And let the rate of interest (R ) be y% p.a
In the first case Rs. x becomes Rs. 854 in 2 years @ y% p.a
Interest @ y% on Rs. x for 2 years (I) = (R*P*T)/100 = (y * x * 2) /100
= (2xy)/100
= Rs. xy/50
So, x + (xy/50) = 854
=> 50x + xy = 42700…………(i)
Similarly, in the 2nd case
Interest on Rs. x @ y% for 7/2 years = (y *x*7/2) / 100 = Rs.(7xy/200)

So, x + (7xy/200) = 969.50
=> 200x + 7xy = 193900…….(ii)
Eliminating xy from (i) & (ii) we can get the value of x and then y.

Page 389, Ex. 5 Solution:
Let the sum of money (P) be Rs. x
So after 8 years, this sum of money becomes Rs. 2x.
Since, Principal

x

+ Interest
= Amount
+ (Interest on Rs. x for 8 years) = 2x
=> (Interest on Rs. x for 8 years) = 2x – x = Rs. x

Let the rate of interest be R% p.a.
Here principal (P) = Rs. x , Interest (I)= Rs. x and Time (T)= 8 years
We know that I = (R * P* T) /100
 R = (I * 100) / (P*T)
= ( x* 100) / (x * 8)
= (100x) /(8x) = 100/8
= 25/2 = 12.5 %

Page 389, Ex. 6
Solution:
Let the loan @ 8% p.a. be Rs. x
 Loan @ 10% p.a. = Rs. (8000 – x)
Given that total annual interest = Rs. 714
That means, in 1 year:
Interest earned on Rs. x@ 8% + Interest earned @ 10% on Rs. (8000-x) = 714
=> {(8 * x * 1) / 100} + [{10 * (8000-x) * 1} / 100] = 714
=>
Solve for x.

(8x/100)

+

{(8000-x) / 10}

=

714

Session 9; Chapter 5; Part 2
Compound Interest

Page 401 Ex 1.
SOLUTION 1 : Without using formula.
• Here Principal (P)= Rs. 8000 , Rate ( R) = 10% p.a ., Time (T) = 3 years
• Interest at 1st year end @ 10% on 8000 for 1 year= (8000 * 10 * 1) /100
•
= Rs. 800
• New principal for 2nd year = Rs. 8000+800 = 8800
• Interest at 2nd year end @ 10% on 8800 for 1 year = (8800 * 10 * 1) /100
•
= Rs. 880
• New principal for 3rd year = Rs. 8800 + 880 = Rs. 9680
• Interest at 3rd year end @ 10% on 9680 for 1 year = (9680 * 10 * 1) /100
•
= Rs. 968
So at the end of 3 years Rs. 8000 will become Rs. (9680 + 968) = Rs. 10648/• So the compound interest (CI) = A – P = 10648 – 8000 = Rs. 2648
•
ALTERNATIVELY : CI = Rs. (800 + 880+ 968) = Rs. 2648

FORMULAS OF COMPOUND INTEREST:
• If a sum of money (principal) Rs. P is borrowed /lent at the rate of R%
compound interest for T years compounded yearly , then :
Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed / lent at the rate of R% compound interest for
compounded half yearly, then :

T years

Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed /lent at the rate of R% compound interest for
compounded quarterly, then :
Compound Interest (CI) on Rs. P @ R% for T years =

T years

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 8000 Rate ( R) = 10% p.a

Time (T) = 3 years

• Compound interest (CI) on Rs. 8000 at the end of 3 years @ 10% =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 4.
SOLUTION 1 : Without using formula.
Here Principal (P)= Rs. 25000, Time (T) = 1 year, Yearly rate of interest =12%
• The Interest is compounded half yearly
• So the half yearly rate of interest = 12 * ½ = 6% per half year
• Interest at 1st half year end = 6% of 25000
•
= Rs. 1500
• New principal for 2nd half year = Rs. 25000+1500 = Rs. 26500
• Interest at 2nd half year end = 6% of 26500
•
= Rs. 1590
• So at the end of 1 year Rs. 25000 will become Rs. (26500 +1590)
= Rs. 28090/• So the compound interest (CI) = 28090 – 25000 = Rs. 3090
• Or , CI = Rs. (1500 + 1590)= Rs. 3090

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 25000 Rate ( R) = 12% p.a Time (T) = 1 year
• Compound interest (CI) on Rs. 25000 at the end of 1 years @ 12% pa
compounded half yearly =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 5. SOLUTION : Without using formula.
Here Principal (P)= Rs. 8000, Time (T) = 9 months = 9/12 years = ¾ yrs,
• Note: 9 months = 3 quarters
• Yearly rate of interest = 20% pa
• The Interest is compounded quarterly . In a year there are 4 quarters
• So the per quarter rate of interest = 20/4 = 5% per quarter.
• Principal for 1st quarter = Rs. 8000
• Interest at 1st quarter end = 5% of 8000 = Rs. 400

• New principal for 2nd quarter = Rs. 8000+400 = Rs. 8400
• Interest at 2nd quarter end = 5% of 8400 = Rs. 420
• New principal for 3rd quarter = Rs. 8400+420 = Rs. 8820
• Interest at 3rd quarter end = 5% of 8820 = Rs. 441
• So at the end of 3 quarters or 9 months, Rs. 8000 will become
Rs. (8820 +441) = Rs. 9261/• So the compound interest (CI) = 9261 – 8000 = Rs. 1261
• Or, CI = Rs. (400 + 420+ 441)= Rs. 1261

Page 402 Ex 3. SOLUTION :
• Here Principal (P)= Rs. 5000, Time (T) = 3years,
• Rate of interest for 1st year (R1) = 5% pa
• Principal for 1st year = Rs. 5000
Interest at 1st year end

= 5% of 5000 = Rs. 250

• New principal for 2nd year = Rs. 5000 + 250 = Rs. 5250
• Rate of interest for 2nd year (R2) = 8% pa
Interest at 2nd year end = 8% of 5250 = Rs. 420

• New principal for 3rd year = Rs. 5250 + 420 = Rs. 5670
• Rate of interest for 3rd year (R3) = 10% pa
Interest at 3rd year end = 10% of 5670 = Rs. 567
• So the amount at the end of 3 years = Rs. (5670 +567) = Rs. 6237/• This means, at the end of 3 years, Rs. 5000 will become Rs. 6237/• Compound interest (CI) on Rs. 5000 after 3 years @ 5%,8% &10%=
6237 – 5000 = Rs. 1237
• Or, CI on Rs. 5000 after 3 years @ 5%,8% &10%= ( 250+ 420+ 567)= Rs. 1237

Page 402 Ex 6. SOLUTION :
• Here Principal (P) = a certain sum of money, CI = Rs. 3783,
Time (T) = 3years, Rate of interest (R) = 5% pa
• Let the principal (P) be Rs. x
• Assuming interest is being charged annually, We know that,
CI =
=> CI on Rs. x @ 5% for 3 years =

=> 3783 =
Find x then find SI on Rs. x @ 5% for 3 years = (x*5*3)/100


Slide 18

Quantitative Methods
Session 8– 31.07.12
Chapter 5 – Simple Interest & Compound Interest

Pranjoy Arup Das

• Interest is the payment to be made for the use of money.
• Suppose, Mr. A borrows a sum of money Rs. X from Mr. B, then Mr. A
will have to repay not only the sum of Rs. X but also some extra money
(say Rs. Y) to Mr. B, sort of like a charge or rent for using Mr. B’s money.
• Rs. X is termed as the principal (P), Rs. Y is the interest (I) for using
Mr.B’s money & Rs. (X + Y) is termed as the Amount(A) which Mr. A
will have to repay to Mr. B after a certain period of time (T) .
• The interest to be paid by the borrower Mr. A to the lender Mr. B is
fixed before or while lending the money. Interest is fixed as a
percentage of the principal and is called Rate of interest (R).

• Money is usually lent out for a fixed period of time (T) . Interest is
calculated on the basis of this time period at the agreed rate, which is
charged either yearly, half-yearly or quarterly. T is always calculated in
Years. So a loan period of 6 months will have to be taken as 0.5 years.

There are two kinds of interest Simple interest & Compound Interest
Simple interest is always calculated on the actual principal originally borrowed.
Throughout the entire time period, a fixed amount of interest is paid whenever
due which is either yearly, half yearly of quarterly
For Eg. Mr. A borrows Rs. 2000 from Mr. B for a period of 2 years at an interest
rate of 10% payable yearly. How much will A pay B after 2 years?
Compound interest is calculated on the increasing principal. The interest is
calculated on the original principal and then added to the original principal
which becomes the new principal for calculating the interest on next due date.
• Simple Interest:
Original Principal Rs. 2000
Interest at 1st year end = 10%of 2000
= Rs. 200
Interest at 2nd year end = 10%of 2000
= Rs. 200
So at the end of 2 years Mr. A has to
pay Rs. (2000 + 200+200)=
Rs. 2400/-

• Compound Interest:
Original principal Rs. 2000
Interest at 1st year end = 10% of 2000
= Rs. 200

New principal=Rs. 2000+200 = 2200
Interest at 2nd year end = 10% of 2200
= Rs. 220
So at the end of 2 years Mr. A has to
pay Rs. (2200 + 220) = Rs. 2420/-

Part 1 Simple Interest

• If a sum of money (principal) Rs. P is borrowed at the rate of R% simple
interest for T years, then
Simple interest amount for T years = 1st years int. + 2nd yrs int. +….+ Tth years int
= (R% of P) + (R% of P) + (R% of P) + …T times
= (R% of P) (1 + 1 + 1 +……………upto T)
= (R% of P) * T
=

I=
• Principal (P)+ Interest (I)= Total Amount(A) to be repaid after T years
=> P +
I
= A
=> P + (R*P*T)/100 = A

Eg. Mr. A borrows Rs. 5400/- from Mr. B under the conditions that :

a) The loan will be repaid after 3 years
b) The borrower will pay a simple interest of 8% per annum (per year)
1) How much interest will Mr. A have to pay at the end of 3 years?
2) What is the amount that Mr. A will have to pay at the end of 3years?
Solution:

Here the Principal = Rs. 5400 Time = 3years Rate of interest= 8%
• Interest at 1st year end = 8% of 5400
•
= Rs. 432
• Interest at 2nd year end = 8% of 5400
•
= Rs. 432
• Interest at 3nd year end = 8% of 5400
•
= Rs. 432
• Total interest at the end of the 3 year loan period
= Rs. (432+432+432) = Rs. 1296
• So at the end of 3 years Mr. A has to pay Rs. (5400 + 1296 )
•
= Rs. 6696

Alternative Solution: USING FORMULA OF SIMPLE INTEREST
Here Principal (P) = Rs.5400, Time (T) = 3 years , Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400* 3) / 100
I = Rs. 1296
The amount to be paid to Mr. B at the end of 3 years :
Amount = P + I
= 5400 + 1296
= Rs. 6696

Page 388 (RSA) Ex 1
SOLUTION 1: Please Note: ALWAYS CONVERT DAYS OR MONTHS TO YEARS
Here the Principal = Rs. 5400
Time = 10/12 years Rate of int. = 8%pa
• Interest for 1 year
= 8% of 5400
= Rs. 432
• That means Interest for 12 months = Rs. 432
•  Interest for 10 months will be: (10/12) * 432 = Rs. 360

• Total interest at the end of the 10 month loan period = Rs. 360
SOLUTION 2 :
Here the Principal (P) = 5400

Time (T) = 10/12 years Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400*
) / 100 = Rs.__________

Page 390 Exercise 21A, Pr no. 4
Solution:
Let the sum of money be Rs. x which is the principal (P)
It is given that interest on Rs. x @8% p.a. in 6 years = Rs. 8376
Here principal is Rs .x, Rate of interest (R ) is 8% p.a. and the time
(T) is 6 years
Interest on Rs. x @ 8% for 6 years (I) = (R*P*T)/100
= (8 * x * 6) /100 = Rs.12x/25
So that means , 12x/25 = 8376
=>
x = (8376 * 25) /12
=>
x = Rs. ___________

Page 388, Ex. 4 Solution:
Let the sum of money (P) be Rs. x
And let the rate of interest (R ) be y% p.a
In the first case Rs. x becomes Rs. 854 in 2 years @ y% p.a
Interest @ y% on Rs. x for 2 years (I) = (R*P*T)/100 = (y * x * 2) /100
= (2xy)/100
= Rs. xy/50
So, x + (xy/50) = 854
=> 50x + xy = 42700…………(i)
Similarly, in the 2nd case
Interest on Rs. x @ y% for 7/2 years = (y *x*7/2) / 100 = Rs.(7xy/200)

So, x + (7xy/200) = 969.50
=> 200x + 7xy = 193900…….(ii)
Eliminating xy from (i) & (ii) we can get the value of x and then y.

Page 389, Ex. 5 Solution:
Let the sum of money (P) be Rs. x
So after 8 years, this sum of money becomes Rs. 2x.
Since, Principal

x

+ Interest
= Amount
+ (Interest on Rs. x for 8 years) = 2x
=> (Interest on Rs. x for 8 years) = 2x – x = Rs. x

Let the rate of interest be R% p.a.
Here principal (P) = Rs. x , Interest (I)= Rs. x and Time (T)= 8 years
We know that I = (R * P* T) /100
 R = (I * 100) / (P*T)
= ( x* 100) / (x * 8)
= (100x) /(8x) = 100/8
= 25/2 = 12.5 %

Page 389, Ex. 6
Solution:
Let the loan @ 8% p.a. be Rs. x
 Loan @ 10% p.a. = Rs. (8000 – x)
Given that total annual interest = Rs. 714
That means, in 1 year:
Interest earned on Rs. x@ 8% + Interest earned @ 10% on Rs. (8000-x) = 714
=> {(8 * x * 1) / 100} + [{10 * (8000-x) * 1} / 100] = 714
=>
Solve for x.

(8x/100)

+

{(8000-x) / 10}

=

714

Session 9; Chapter 5; Part 2
Compound Interest

Page 401 Ex 1.
SOLUTION 1 : Without using formula.
• Here Principal (P)= Rs. 8000 , Rate ( R) = 10% p.a ., Time (T) = 3 years
• Interest at 1st year end @ 10% on 8000 for 1 year= (8000 * 10 * 1) /100
•
= Rs. 800
• New principal for 2nd year = Rs. 8000+800 = 8800
• Interest at 2nd year end @ 10% on 8800 for 1 year = (8800 * 10 * 1) /100
•
= Rs. 880
• New principal for 3rd year = Rs. 8800 + 880 = Rs. 9680
• Interest at 3rd year end @ 10% on 9680 for 1 year = (9680 * 10 * 1) /100
•
= Rs. 968
So at the end of 3 years Rs. 8000 will become Rs. (9680 + 968) = Rs. 10648/• So the compound interest (CI) = A – P = 10648 – 8000 = Rs. 2648
•
ALTERNATIVELY : CI = Rs. (800 + 880+ 968) = Rs. 2648

FORMULAS OF COMPOUND INTEREST:
• If a sum of money (principal) Rs. P is borrowed /lent at the rate of R%
compound interest for T years compounded yearly , then :
Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed / lent at the rate of R% compound interest for
compounded half yearly, then :

T years

Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed /lent at the rate of R% compound interest for
compounded quarterly, then :
Compound Interest (CI) on Rs. P @ R% for T years =

T years

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 8000 Rate ( R) = 10% p.a

Time (T) = 3 years

• Compound interest (CI) on Rs. 8000 at the end of 3 years @ 10% =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 4.
SOLUTION 1 : Without using formula.
Here Principal (P)= Rs. 25000, Time (T) = 1 year, Yearly rate of interest =12%
• The Interest is compounded half yearly
• So the half yearly rate of interest = 12 * ½ = 6% per half year
• Interest at 1st half year end = 6% of 25000
•
= Rs. 1500
• New principal for 2nd half year = Rs. 25000+1500 = Rs. 26500
• Interest at 2nd half year end = 6% of 26500
•
= Rs. 1590
• So at the end of 1 year Rs. 25000 will become Rs. (26500 +1590)
= Rs. 28090/• So the compound interest (CI) = 28090 – 25000 = Rs. 3090
• Or , CI = Rs. (1500 + 1590)= Rs. 3090

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 25000 Rate ( R) = 12% p.a Time (T) = 1 year
• Compound interest (CI) on Rs. 25000 at the end of 1 years @ 12% pa
compounded half yearly =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 5. SOLUTION : Without using formula.
Here Principal (P)= Rs. 8000, Time (T) = 9 months = 9/12 years = ¾ yrs,
• Note: 9 months = 3 quarters
• Yearly rate of interest = 20% pa
• The Interest is compounded quarterly . In a year there are 4 quarters
• So the per quarter rate of interest = 20/4 = 5% per quarter.
• Principal for 1st quarter = Rs. 8000
• Interest at 1st quarter end = 5% of 8000 = Rs. 400

• New principal for 2nd quarter = Rs. 8000+400 = Rs. 8400
• Interest at 2nd quarter end = 5% of 8400 = Rs. 420
• New principal for 3rd quarter = Rs. 8400+420 = Rs. 8820
• Interest at 3rd quarter end = 5% of 8820 = Rs. 441
• So at the end of 3 quarters or 9 months, Rs. 8000 will become
Rs. (8820 +441) = Rs. 9261/• So the compound interest (CI) = 9261 – 8000 = Rs. 1261
• Or, CI = Rs. (400 + 420+ 441)= Rs. 1261

Page 402 Ex 3. SOLUTION :
• Here Principal (P)= Rs. 5000, Time (T) = 3years,
• Rate of interest for 1st year (R1) = 5% pa
• Principal for 1st year = Rs. 5000
Interest at 1st year end

= 5% of 5000 = Rs. 250

• New principal for 2nd year = Rs. 5000 + 250 = Rs. 5250
• Rate of interest for 2nd year (R2) = 8% pa
Interest at 2nd year end = 8% of 5250 = Rs. 420

• New principal for 3rd year = Rs. 5250 + 420 = Rs. 5670
• Rate of interest for 3rd year (R3) = 10% pa
Interest at 3rd year end = 10% of 5670 = Rs. 567
• So the amount at the end of 3 years = Rs. (5670 +567) = Rs. 6237/• This means, at the end of 3 years, Rs. 5000 will become Rs. 6237/• Compound interest (CI) on Rs. 5000 after 3 years @ 5%,8% &10%=
6237 – 5000 = Rs. 1237
• Or, CI on Rs. 5000 after 3 years @ 5%,8% &10%= ( 250+ 420+ 567)= Rs. 1237

Page 402 Ex 6. SOLUTION :
• Here Principal (P) = a certain sum of money, CI = Rs. 3783,
Time (T) = 3years, Rate of interest (R) = 5% pa
• Let the principal (P) be Rs. x
• Assuming interest is being charged annually, We know that,
CI =
=> CI on Rs. x @ 5% for 3 years =

=> 3783 =
Find x then find SI on Rs. x @ 5% for 3 years = (x*5*3)/100


Slide 19

Quantitative Methods
Session 8– 31.07.12
Chapter 5 – Simple Interest & Compound Interest

Pranjoy Arup Das

• Interest is the payment to be made for the use of money.
• Suppose, Mr. A borrows a sum of money Rs. X from Mr. B, then Mr. A
will have to repay not only the sum of Rs. X but also some extra money
(say Rs. Y) to Mr. B, sort of like a charge or rent for using Mr. B’s money.
• Rs. X is termed as the principal (P), Rs. Y is the interest (I) for using
Mr.B’s money & Rs. (X + Y) is termed as the Amount(A) which Mr. A
will have to repay to Mr. B after a certain period of time (T) .
• The interest to be paid by the borrower Mr. A to the lender Mr. B is
fixed before or while lending the money. Interest is fixed as a
percentage of the principal and is called Rate of interest (R).

• Money is usually lent out for a fixed period of time (T) . Interest is
calculated on the basis of this time period at the agreed rate, which is
charged either yearly, half-yearly or quarterly. T is always calculated in
Years. So a loan period of 6 months will have to be taken as 0.5 years.

There are two kinds of interest Simple interest & Compound Interest
Simple interest is always calculated on the actual principal originally borrowed.
Throughout the entire time period, a fixed amount of interest is paid whenever
due which is either yearly, half yearly of quarterly
For Eg. Mr. A borrows Rs. 2000 from Mr. B for a period of 2 years at an interest
rate of 10% payable yearly. How much will A pay B after 2 years?
Compound interest is calculated on the increasing principal. The interest is
calculated on the original principal and then added to the original principal
which becomes the new principal for calculating the interest on next due date.
• Simple Interest:
Original Principal Rs. 2000
Interest at 1st year end = 10%of 2000
= Rs. 200
Interest at 2nd year end = 10%of 2000
= Rs. 200
So at the end of 2 years Mr. A has to
pay Rs. (2000 + 200+200)=
Rs. 2400/-

• Compound Interest:
Original principal Rs. 2000
Interest at 1st year end = 10% of 2000
= Rs. 200

New principal=Rs. 2000+200 = 2200
Interest at 2nd year end = 10% of 2200
= Rs. 220
So at the end of 2 years Mr. A has to
pay Rs. (2200 + 220) = Rs. 2420/-

Part 1 Simple Interest

• If a sum of money (principal) Rs. P is borrowed at the rate of R% simple
interest for T years, then
Simple interest amount for T years = 1st years int. + 2nd yrs int. +….+ Tth years int
= (R% of P) + (R% of P) + (R% of P) + …T times
= (R% of P) (1 + 1 + 1 +……………upto T)
= (R% of P) * T
=

I=
• Principal (P)+ Interest (I)= Total Amount(A) to be repaid after T years
=> P +
I
= A
=> P + (R*P*T)/100 = A

Eg. Mr. A borrows Rs. 5400/- from Mr. B under the conditions that :

a) The loan will be repaid after 3 years
b) The borrower will pay a simple interest of 8% per annum (per year)
1) How much interest will Mr. A have to pay at the end of 3 years?
2) What is the amount that Mr. A will have to pay at the end of 3years?
Solution:

Here the Principal = Rs. 5400 Time = 3years Rate of interest= 8%
• Interest at 1st year end = 8% of 5400
•
= Rs. 432
• Interest at 2nd year end = 8% of 5400
•
= Rs. 432
• Interest at 3nd year end = 8% of 5400
•
= Rs. 432
• Total interest at the end of the 3 year loan period
= Rs. (432+432+432) = Rs. 1296
• So at the end of 3 years Mr. A has to pay Rs. (5400 + 1296 )
•
= Rs. 6696

Alternative Solution: USING FORMULA OF SIMPLE INTEREST
Here Principal (P) = Rs.5400, Time (T) = 3 years , Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400* 3) / 100
I = Rs. 1296
The amount to be paid to Mr. B at the end of 3 years :
Amount = P + I
= 5400 + 1296
= Rs. 6696

Page 388 (RSA) Ex 1
SOLUTION 1: Please Note: ALWAYS CONVERT DAYS OR MONTHS TO YEARS
Here the Principal = Rs. 5400
Time = 10/12 years Rate of int. = 8%pa
• Interest for 1 year
= 8% of 5400
= Rs. 432
• That means Interest for 12 months = Rs. 432
•  Interest for 10 months will be: (10/12) * 432 = Rs. 360

• Total interest at the end of the 10 month loan period = Rs. 360
SOLUTION 2 :
Here the Principal (P) = 5400

Time (T) = 10/12 years Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400*
) / 100 = Rs.__________

Page 390 Exercise 21A, Pr no. 4
Solution:
Let the sum of money be Rs. x which is the principal (P)
It is given that interest on Rs. x @8% p.a. in 6 years = Rs. 8376
Here principal is Rs .x, Rate of interest (R ) is 8% p.a. and the time
(T) is 6 years
Interest on Rs. x @ 8% for 6 years (I) = (R*P*T)/100
= (8 * x * 6) /100 = Rs.12x/25
So that means , 12x/25 = 8376
=>
x = (8376 * 25) /12
=>
x = Rs. ___________

Page 388, Ex. 4 Solution:
Let the sum of money (P) be Rs. x
And let the rate of interest (R ) be y% p.a
In the first case Rs. x becomes Rs. 854 in 2 years @ y% p.a
Interest @ y% on Rs. x for 2 years (I) = (R*P*T)/100 = (y * x * 2) /100
= (2xy)/100
= Rs. xy/50
So, x + (xy/50) = 854
=> 50x + xy = 42700…………(i)
Similarly, in the 2nd case
Interest on Rs. x @ y% for 7/2 years = (y *x*7/2) / 100 = Rs.(7xy/200)

So, x + (7xy/200) = 969.50
=> 200x + 7xy = 193900…….(ii)
Eliminating xy from (i) & (ii) we can get the value of x and then y.

Page 389, Ex. 5 Solution:
Let the sum of money (P) be Rs. x
So after 8 years, this sum of money becomes Rs. 2x.
Since, Principal

x

+ Interest
= Amount
+ (Interest on Rs. x for 8 years) = 2x
=> (Interest on Rs. x for 8 years) = 2x – x = Rs. x

Let the rate of interest be R% p.a.
Here principal (P) = Rs. x , Interest (I)= Rs. x and Time (T)= 8 years
We know that I = (R * P* T) /100
 R = (I * 100) / (P*T)
= ( x* 100) / (x * 8)
= (100x) /(8x) = 100/8
= 25/2 = 12.5 %

Page 389, Ex. 6
Solution:
Let the loan @ 8% p.a. be Rs. x
 Loan @ 10% p.a. = Rs. (8000 – x)
Given that total annual interest = Rs. 714
That means, in 1 year:
Interest earned on Rs. x@ 8% + Interest earned @ 10% on Rs. (8000-x) = 714
=> {(8 * x * 1) / 100} + [{10 * (8000-x) * 1} / 100] = 714
=>
Solve for x.

(8x/100)

+

{(8000-x) / 10}

=

714

Session 9; Chapter 5; Part 2
Compound Interest

Page 401 Ex 1.
SOLUTION 1 : Without using formula.
• Here Principal (P)= Rs. 8000 , Rate ( R) = 10% p.a ., Time (T) = 3 years
• Interest at 1st year end @ 10% on 8000 for 1 year= (8000 * 10 * 1) /100
•
= Rs. 800
• New principal for 2nd year = Rs. 8000+800 = 8800
• Interest at 2nd year end @ 10% on 8800 for 1 year = (8800 * 10 * 1) /100
•
= Rs. 880
• New principal for 3rd year = Rs. 8800 + 880 = Rs. 9680
• Interest at 3rd year end @ 10% on 9680 for 1 year = (9680 * 10 * 1) /100
•
= Rs. 968
So at the end of 3 years Rs. 8000 will become Rs. (9680 + 968) = Rs. 10648/• So the compound interest (CI) = A – P = 10648 – 8000 = Rs. 2648
•
ALTERNATIVELY : CI = Rs. (800 + 880+ 968) = Rs. 2648

FORMULAS OF COMPOUND INTEREST:
• If a sum of money (principal) Rs. P is borrowed /lent at the rate of R%
compound interest for T years compounded yearly , then :
Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed / lent at the rate of R% compound interest for
compounded half yearly, then :

T years

Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed /lent at the rate of R% compound interest for
compounded quarterly, then :
Compound Interest (CI) on Rs. P @ R% for T years =

T years

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 8000 Rate ( R) = 10% p.a

Time (T) = 3 years

• Compound interest (CI) on Rs. 8000 at the end of 3 years @ 10% =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 4.
SOLUTION 1 : Without using formula.
Here Principal (P)= Rs. 25000, Time (T) = 1 year, Yearly rate of interest =12%
• The Interest is compounded half yearly
• So the half yearly rate of interest = 12 * ½ = 6% per half year
• Interest at 1st half year end = 6% of 25000
•
= Rs. 1500
• New principal for 2nd half year = Rs. 25000+1500 = Rs. 26500
• Interest at 2nd half year end = 6% of 26500
•
= Rs. 1590
• So at the end of 1 year Rs. 25000 will become Rs. (26500 +1590)
= Rs. 28090/• So the compound interest (CI) = 28090 – 25000 = Rs. 3090
• Or , CI = Rs. (1500 + 1590)= Rs. 3090

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 25000 Rate ( R) = 12% p.a Time (T) = 1 year
• Compound interest (CI) on Rs. 25000 at the end of 1 years @ 12% pa
compounded half yearly =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 5. SOLUTION : Without using formula.
Here Principal (P)= Rs. 8000, Time (T) = 9 months = 9/12 years = ¾ yrs,
• Note: 9 months = 3 quarters
• Yearly rate of interest = 20% pa
• The Interest is compounded quarterly . In a year there are 4 quarters
• So the per quarter rate of interest = 20/4 = 5% per quarter.
• Principal for 1st quarter = Rs. 8000
• Interest at 1st quarter end = 5% of 8000 = Rs. 400

• New principal for 2nd quarter = Rs. 8000+400 = Rs. 8400
• Interest at 2nd quarter end = 5% of 8400 = Rs. 420
• New principal for 3rd quarter = Rs. 8400+420 = Rs. 8820
• Interest at 3rd quarter end = 5% of 8820 = Rs. 441
• So at the end of 3 quarters or 9 months, Rs. 8000 will become
Rs. (8820 +441) = Rs. 9261/• So the compound interest (CI) = 9261 – 8000 = Rs. 1261
• Or, CI = Rs. (400 + 420+ 441)= Rs. 1261

Page 402 Ex 3. SOLUTION :
• Here Principal (P)= Rs. 5000, Time (T) = 3years,
• Rate of interest for 1st year (R1) = 5% pa
• Principal for 1st year = Rs. 5000
Interest at 1st year end

= 5% of 5000 = Rs. 250

• New principal for 2nd year = Rs. 5000 + 250 = Rs. 5250
• Rate of interest for 2nd year (R2) = 8% pa
Interest at 2nd year end = 8% of 5250 = Rs. 420

• New principal for 3rd year = Rs. 5250 + 420 = Rs. 5670
• Rate of interest for 3rd year (R3) = 10% pa
Interest at 3rd year end = 10% of 5670 = Rs. 567
• So the amount at the end of 3 years = Rs. (5670 +567) = Rs. 6237/• This means, at the end of 3 years, Rs. 5000 will become Rs. 6237/• Compound interest (CI) on Rs. 5000 after 3 years @ 5%,8% &10%=
6237 – 5000 = Rs. 1237
• Or, CI on Rs. 5000 after 3 years @ 5%,8% &10%= ( 250+ 420+ 567)= Rs. 1237

Page 402 Ex 6. SOLUTION :
• Here Principal (P) = a certain sum of money, CI = Rs. 3783,
Time (T) = 3years, Rate of interest (R) = 5% pa
• Let the principal (P) be Rs. x
• Assuming interest is being charged annually, We know that,
CI =
=> CI on Rs. x @ 5% for 3 years =

=> 3783 =
Find x then find SI on Rs. x @ 5% for 3 years = (x*5*3)/100


Slide 20

Quantitative Methods
Session 8– 31.07.12
Chapter 5 – Simple Interest & Compound Interest

Pranjoy Arup Das

• Interest is the payment to be made for the use of money.
• Suppose, Mr. A borrows a sum of money Rs. X from Mr. B, then Mr. A
will have to repay not only the sum of Rs. X but also some extra money
(say Rs. Y) to Mr. B, sort of like a charge or rent for using Mr. B’s money.
• Rs. X is termed as the principal (P), Rs. Y is the interest (I) for using
Mr.B’s money & Rs. (X + Y) is termed as the Amount(A) which Mr. A
will have to repay to Mr. B after a certain period of time (T) .
• The interest to be paid by the borrower Mr. A to the lender Mr. B is
fixed before or while lending the money. Interest is fixed as a
percentage of the principal and is called Rate of interest (R).

• Money is usually lent out for a fixed period of time (T) . Interest is
calculated on the basis of this time period at the agreed rate, which is
charged either yearly, half-yearly or quarterly. T is always calculated in
Years. So a loan period of 6 months will have to be taken as 0.5 years.

There are two kinds of interest Simple interest & Compound Interest
Simple interest is always calculated on the actual principal originally borrowed.
Throughout the entire time period, a fixed amount of interest is paid whenever
due which is either yearly, half yearly of quarterly
For Eg. Mr. A borrows Rs. 2000 from Mr. B for a period of 2 years at an interest
rate of 10% payable yearly. How much will A pay B after 2 years?
Compound interest is calculated on the increasing principal. The interest is
calculated on the original principal and then added to the original principal
which becomes the new principal for calculating the interest on next due date.
• Simple Interest:
Original Principal Rs. 2000
Interest at 1st year end = 10%of 2000
= Rs. 200
Interest at 2nd year end = 10%of 2000
= Rs. 200
So at the end of 2 years Mr. A has to
pay Rs. (2000 + 200+200)=
Rs. 2400/-

• Compound Interest:
Original principal Rs. 2000
Interest at 1st year end = 10% of 2000
= Rs. 200

New principal=Rs. 2000+200 = 2200
Interest at 2nd year end = 10% of 2200
= Rs. 220
So at the end of 2 years Mr. A has to
pay Rs. (2200 + 220) = Rs. 2420/-

Part 1 Simple Interest

• If a sum of money (principal) Rs. P is borrowed at the rate of R% simple
interest for T years, then
Simple interest amount for T years = 1st years int. + 2nd yrs int. +….+ Tth years int
= (R% of P) + (R% of P) + (R% of P) + …T times
= (R% of P) (1 + 1 + 1 +……………upto T)
= (R% of P) * T
=

I=
• Principal (P)+ Interest (I)= Total Amount(A) to be repaid after T years
=> P +
I
= A
=> P + (R*P*T)/100 = A

Eg. Mr. A borrows Rs. 5400/- from Mr. B under the conditions that :

a) The loan will be repaid after 3 years
b) The borrower will pay a simple interest of 8% per annum (per year)
1) How much interest will Mr. A have to pay at the end of 3 years?
2) What is the amount that Mr. A will have to pay at the end of 3years?
Solution:

Here the Principal = Rs. 5400 Time = 3years Rate of interest= 8%
• Interest at 1st year end = 8% of 5400
•
= Rs. 432
• Interest at 2nd year end = 8% of 5400
•
= Rs. 432
• Interest at 3nd year end = 8% of 5400
•
= Rs. 432
• Total interest at the end of the 3 year loan period
= Rs. (432+432+432) = Rs. 1296
• So at the end of 3 years Mr. A has to pay Rs. (5400 + 1296 )
•
= Rs. 6696

Alternative Solution: USING FORMULA OF SIMPLE INTEREST
Here Principal (P) = Rs.5400, Time (T) = 3 years , Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400* 3) / 100
I = Rs. 1296
The amount to be paid to Mr. B at the end of 3 years :
Amount = P + I
= 5400 + 1296
= Rs. 6696

Page 388 (RSA) Ex 1
SOLUTION 1: Please Note: ALWAYS CONVERT DAYS OR MONTHS TO YEARS
Here the Principal = Rs. 5400
Time = 10/12 years Rate of int. = 8%pa
• Interest for 1 year
= 8% of 5400
= Rs. 432
• That means Interest for 12 months = Rs. 432
•  Interest for 10 months will be: (10/12) * 432 = Rs. 360

• Total interest at the end of the 10 month loan period = Rs. 360
SOLUTION 2 :
Here the Principal (P) = 5400

Time (T) = 10/12 years Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400*
) / 100 = Rs.__________

Page 390 Exercise 21A, Pr no. 4
Solution:
Let the sum of money be Rs. x which is the principal (P)
It is given that interest on Rs. x @8% p.a. in 6 years = Rs. 8376
Here principal is Rs .x, Rate of interest (R ) is 8% p.a. and the time
(T) is 6 years
Interest on Rs. x @ 8% for 6 years (I) = (R*P*T)/100
= (8 * x * 6) /100 = Rs.12x/25
So that means , 12x/25 = 8376
=>
x = (8376 * 25) /12
=>
x = Rs. ___________

Page 388, Ex. 4 Solution:
Let the sum of money (P) be Rs. x
And let the rate of interest (R ) be y% p.a
In the first case Rs. x becomes Rs. 854 in 2 years @ y% p.a
Interest @ y% on Rs. x for 2 years (I) = (R*P*T)/100 = (y * x * 2) /100
= (2xy)/100
= Rs. xy/50
So, x + (xy/50) = 854
=> 50x + xy = 42700…………(i)
Similarly, in the 2nd case
Interest on Rs. x @ y% for 7/2 years = (y *x*7/2) / 100 = Rs.(7xy/200)

So, x + (7xy/200) = 969.50
=> 200x + 7xy = 193900…….(ii)
Eliminating xy from (i) & (ii) we can get the value of x and then y.

Page 389, Ex. 5 Solution:
Let the sum of money (P) be Rs. x
So after 8 years, this sum of money becomes Rs. 2x.
Since, Principal

x

+ Interest
= Amount
+ (Interest on Rs. x for 8 years) = 2x
=> (Interest on Rs. x for 8 years) = 2x – x = Rs. x

Let the rate of interest be R% p.a.
Here principal (P) = Rs. x , Interest (I)= Rs. x and Time (T)= 8 years
We know that I = (R * P* T) /100
 R = (I * 100) / (P*T)
= ( x* 100) / (x * 8)
= (100x) /(8x) = 100/8
= 25/2 = 12.5 %

Page 389, Ex. 6
Solution:
Let the loan @ 8% p.a. be Rs. x
 Loan @ 10% p.a. = Rs. (8000 – x)
Given that total annual interest = Rs. 714
That means, in 1 year:
Interest earned on Rs. x@ 8% + Interest earned @ 10% on Rs. (8000-x) = 714
=> {(8 * x * 1) / 100} + [{10 * (8000-x) * 1} / 100] = 714
=>
Solve for x.

(8x/100)

+

{(8000-x) / 10}

=

714

Session 9; Chapter 5; Part 2
Compound Interest

Page 401 Ex 1.
SOLUTION 1 : Without using formula.
• Here Principal (P)= Rs. 8000 , Rate ( R) = 10% p.a ., Time (T) = 3 years
• Interest at 1st year end @ 10% on 8000 for 1 year= (8000 * 10 * 1) /100
•
= Rs. 800
• New principal for 2nd year = Rs. 8000+800 = 8800
• Interest at 2nd year end @ 10% on 8800 for 1 year = (8800 * 10 * 1) /100
•
= Rs. 880
• New principal for 3rd year = Rs. 8800 + 880 = Rs. 9680
• Interest at 3rd year end @ 10% on 9680 for 1 year = (9680 * 10 * 1) /100
•
= Rs. 968
So at the end of 3 years Rs. 8000 will become Rs. (9680 + 968) = Rs. 10648/• So the compound interest (CI) = A – P = 10648 – 8000 = Rs. 2648
•
ALTERNATIVELY : CI = Rs. (800 + 880+ 968) = Rs. 2648

FORMULAS OF COMPOUND INTEREST:
• If a sum of money (principal) Rs. P is borrowed /lent at the rate of R%
compound interest for T years compounded yearly , then :
Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed / lent at the rate of R% compound interest for
compounded half yearly, then :

T years

Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed /lent at the rate of R% compound interest for
compounded quarterly, then :
Compound Interest (CI) on Rs. P @ R% for T years =

T years

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 8000 Rate ( R) = 10% p.a

Time (T) = 3 years

• Compound interest (CI) on Rs. 8000 at the end of 3 years @ 10% =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 4.
SOLUTION 1 : Without using formula.
Here Principal (P)= Rs. 25000, Time (T) = 1 year, Yearly rate of interest =12%
• The Interest is compounded half yearly
• So the half yearly rate of interest = 12 * ½ = 6% per half year
• Interest at 1st half year end = 6% of 25000
•
= Rs. 1500
• New principal for 2nd half year = Rs. 25000+1500 = Rs. 26500
• Interest at 2nd half year end = 6% of 26500
•
= Rs. 1590
• So at the end of 1 year Rs. 25000 will become Rs. (26500 +1590)
= Rs. 28090/• So the compound interest (CI) = 28090 – 25000 = Rs. 3090
• Or , CI = Rs. (1500 + 1590)= Rs. 3090

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 25000 Rate ( R) = 12% p.a Time (T) = 1 year
• Compound interest (CI) on Rs. 25000 at the end of 1 years @ 12% pa
compounded half yearly =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 5. SOLUTION : Without using formula.
Here Principal (P)= Rs. 8000, Time (T) = 9 months = 9/12 years = ¾ yrs,
• Note: 9 months = 3 quarters
• Yearly rate of interest = 20% pa
• The Interest is compounded quarterly . In a year there are 4 quarters
• So the per quarter rate of interest = 20/4 = 5% per quarter.
• Principal for 1st quarter = Rs. 8000
• Interest at 1st quarter end = 5% of 8000 = Rs. 400

• New principal for 2nd quarter = Rs. 8000+400 = Rs. 8400
• Interest at 2nd quarter end = 5% of 8400 = Rs. 420
• New principal for 3rd quarter = Rs. 8400+420 = Rs. 8820
• Interest at 3rd quarter end = 5% of 8820 = Rs. 441
• So at the end of 3 quarters or 9 months, Rs. 8000 will become
Rs. (8820 +441) = Rs. 9261/• So the compound interest (CI) = 9261 – 8000 = Rs. 1261
• Or, CI = Rs. (400 + 420+ 441)= Rs. 1261

Page 402 Ex 3. SOLUTION :
• Here Principal (P)= Rs. 5000, Time (T) = 3years,
• Rate of interest for 1st year (R1) = 5% pa
• Principal for 1st year = Rs. 5000
Interest at 1st year end

= 5% of 5000 = Rs. 250

• New principal for 2nd year = Rs. 5000 + 250 = Rs. 5250
• Rate of interest for 2nd year (R2) = 8% pa
Interest at 2nd year end = 8% of 5250 = Rs. 420

• New principal for 3rd year = Rs. 5250 + 420 = Rs. 5670
• Rate of interest for 3rd year (R3) = 10% pa
Interest at 3rd year end = 10% of 5670 = Rs. 567
• So the amount at the end of 3 years = Rs. (5670 +567) = Rs. 6237/• This means, at the end of 3 years, Rs. 5000 will become Rs. 6237/• Compound interest (CI) on Rs. 5000 after 3 years @ 5%,8% &10%=
6237 – 5000 = Rs. 1237
• Or, CI on Rs. 5000 after 3 years @ 5%,8% &10%= ( 250+ 420+ 567)= Rs. 1237

Page 402 Ex 6. SOLUTION :
• Here Principal (P) = a certain sum of money, CI = Rs. 3783,
Time (T) = 3years, Rate of interest (R) = 5% pa
• Let the principal (P) be Rs. x
• Assuming interest is being charged annually, We know that,
CI =
=> CI on Rs. x @ 5% for 3 years =

=> 3783 =
Find x then find SI on Rs. x @ 5% for 3 years = (x*5*3)/100


Slide 21

Quantitative Methods
Session 8– 31.07.12
Chapter 5 – Simple Interest & Compound Interest

Pranjoy Arup Das

• Interest is the payment to be made for the use of money.
• Suppose, Mr. A borrows a sum of money Rs. X from Mr. B, then Mr. A
will have to repay not only the sum of Rs. X but also some extra money
(say Rs. Y) to Mr. B, sort of like a charge or rent for using Mr. B’s money.
• Rs. X is termed as the principal (P), Rs. Y is the interest (I) for using
Mr.B’s money & Rs. (X + Y) is termed as the Amount(A) which Mr. A
will have to repay to Mr. B after a certain period of time (T) .
• The interest to be paid by the borrower Mr. A to the lender Mr. B is
fixed before or while lending the money. Interest is fixed as a
percentage of the principal and is called Rate of interest (R).

• Money is usually lent out for a fixed period of time (T) . Interest is
calculated on the basis of this time period at the agreed rate, which is
charged either yearly, half-yearly or quarterly. T is always calculated in
Years. So a loan period of 6 months will have to be taken as 0.5 years.

There are two kinds of interest Simple interest & Compound Interest
Simple interest is always calculated on the actual principal originally borrowed.
Throughout the entire time period, a fixed amount of interest is paid whenever
due which is either yearly, half yearly of quarterly
For Eg. Mr. A borrows Rs. 2000 from Mr. B for a period of 2 years at an interest
rate of 10% payable yearly. How much will A pay B after 2 years?
Compound interest is calculated on the increasing principal. The interest is
calculated on the original principal and then added to the original principal
which becomes the new principal for calculating the interest on next due date.
• Simple Interest:
Original Principal Rs. 2000
Interest at 1st year end = 10%of 2000
= Rs. 200
Interest at 2nd year end = 10%of 2000
= Rs. 200
So at the end of 2 years Mr. A has to
pay Rs. (2000 + 200+200)=
Rs. 2400/-

• Compound Interest:
Original principal Rs. 2000
Interest at 1st year end = 10% of 2000
= Rs. 200

New principal=Rs. 2000+200 = 2200
Interest at 2nd year end = 10% of 2200
= Rs. 220
So at the end of 2 years Mr. A has to
pay Rs. (2200 + 220) = Rs. 2420/-

Part 1 Simple Interest

• If a sum of money (principal) Rs. P is borrowed at the rate of R% simple
interest for T years, then
Simple interest amount for T years = 1st years int. + 2nd yrs int. +….+ Tth years int
= (R% of P) + (R% of P) + (R% of P) + …T times
= (R% of P) (1 + 1 + 1 +……………upto T)
= (R% of P) * T
=

I=
• Principal (P)+ Interest (I)= Total Amount(A) to be repaid after T years
=> P +
I
= A
=> P + (R*P*T)/100 = A

Eg. Mr. A borrows Rs. 5400/- from Mr. B under the conditions that :

a) The loan will be repaid after 3 years
b) The borrower will pay a simple interest of 8% per annum (per year)
1) How much interest will Mr. A have to pay at the end of 3 years?
2) What is the amount that Mr. A will have to pay at the end of 3years?
Solution:

Here the Principal = Rs. 5400 Time = 3years Rate of interest= 8%
• Interest at 1st year end = 8% of 5400
•
= Rs. 432
• Interest at 2nd year end = 8% of 5400
•
= Rs. 432
• Interest at 3nd year end = 8% of 5400
•
= Rs. 432
• Total interest at the end of the 3 year loan period
= Rs. (432+432+432) = Rs. 1296
• So at the end of 3 years Mr. A has to pay Rs. (5400 + 1296 )
•
= Rs. 6696

Alternative Solution: USING FORMULA OF SIMPLE INTEREST
Here Principal (P) = Rs.5400, Time (T) = 3 years , Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400* 3) / 100
I = Rs. 1296
The amount to be paid to Mr. B at the end of 3 years :
Amount = P + I
= 5400 + 1296
= Rs. 6696

Page 388 (RSA) Ex 1
SOLUTION 1: Please Note: ALWAYS CONVERT DAYS OR MONTHS TO YEARS
Here the Principal = Rs. 5400
Time = 10/12 years Rate of int. = 8%pa
• Interest for 1 year
= 8% of 5400
= Rs. 432
• That means Interest for 12 months = Rs. 432
•  Interest for 10 months will be: (10/12) * 432 = Rs. 360

• Total interest at the end of the 10 month loan period = Rs. 360
SOLUTION 2 :
Here the Principal (P) = 5400

Time (T) = 10/12 years Rate of int.(R) =8%pa

So the interest to be paid at the end of the loan period of 3 years :
Simple Interest (I) = (R*P*T) / 100
= (8 * 5400*
) / 100 = Rs.__________

Page 390 Exercise 21A, Pr no. 4
Solution:
Let the sum of money be Rs. x which is the principal (P)
It is given that interest on Rs. x @8% p.a. in 6 years = Rs. 8376
Here principal is Rs .x, Rate of interest (R ) is 8% p.a. and the time
(T) is 6 years
Interest on Rs. x @ 8% for 6 years (I) = (R*P*T)/100
= (8 * x * 6) /100 = Rs.12x/25
So that means , 12x/25 = 8376
=>
x = (8376 * 25) /12
=>
x = Rs. ___________

Page 388, Ex. 4 Solution:
Let the sum of money (P) be Rs. x
And let the rate of interest (R ) be y% p.a
In the first case Rs. x becomes Rs. 854 in 2 years @ y% p.a
Interest @ y% on Rs. x for 2 years (I) = (R*P*T)/100 = (y * x * 2) /100
= (2xy)/100
= Rs. xy/50
So, x + (xy/50) = 854
=> 50x + xy = 42700…………(i)
Similarly, in the 2nd case
Interest on Rs. x @ y% for 7/2 years = (y *x*7/2) / 100 = Rs.(7xy/200)

So, x + (7xy/200) = 969.50
=> 200x + 7xy = 193900…….(ii)
Eliminating xy from (i) & (ii) we can get the value of x and then y.

Page 389, Ex. 5 Solution:
Let the sum of money (P) be Rs. x
So after 8 years, this sum of money becomes Rs. 2x.
Since, Principal

x

+ Interest
= Amount
+ (Interest on Rs. x for 8 years) = 2x
=> (Interest on Rs. x for 8 years) = 2x – x = Rs. x

Let the rate of interest be R% p.a.
Here principal (P) = Rs. x , Interest (I)= Rs. x and Time (T)= 8 years
We know that I = (R * P* T) /100
 R = (I * 100) / (P*T)
= ( x* 100) / (x * 8)
= (100x) /(8x) = 100/8
= 25/2 = 12.5 %

Page 389, Ex. 6
Solution:
Let the loan @ 8% p.a. be Rs. x
 Loan @ 10% p.a. = Rs. (8000 – x)
Given that total annual interest = Rs. 714
That means, in 1 year:
Interest earned on Rs. x@ 8% + Interest earned @ 10% on Rs. (8000-x) = 714
=> {(8 * x * 1) / 100} + [{10 * (8000-x) * 1} / 100] = 714
=>
Solve for x.

(8x/100)

+

{(8000-x) / 10}

=

714

Session 9; Chapter 5; Part 2
Compound Interest

Page 401 Ex 1.
SOLUTION 1 : Without using formula.
• Here Principal (P)= Rs. 8000 , Rate ( R) = 10% p.a ., Time (T) = 3 years
• Interest at 1st year end @ 10% on 8000 for 1 year= (8000 * 10 * 1) /100
•
= Rs. 800
• New principal for 2nd year = Rs. 8000+800 = 8800
• Interest at 2nd year end @ 10% on 8800 for 1 year = (8800 * 10 * 1) /100
•
= Rs. 880
• New principal for 3rd year = Rs. 8800 + 880 = Rs. 9680
• Interest at 3rd year end @ 10% on 9680 for 1 year = (9680 * 10 * 1) /100
•
= Rs. 968
So at the end of 3 years Rs. 8000 will become Rs. (9680 + 968) = Rs. 10648/• So the compound interest (CI) = A – P = 10648 – 8000 = Rs. 2648
•
ALTERNATIVELY : CI = Rs. (800 + 880+ 968) = Rs. 2648

FORMULAS OF COMPOUND INTEREST:
• If a sum of money (principal) Rs. P is borrowed /lent at the rate of R%
compound interest for T years compounded yearly , then :
Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed / lent at the rate of R% compound interest for
compounded half yearly, then :

T years

Compound Interest (CI) on Rs. P @ R% for T years =

• If Rs. P is borrowed /lent at the rate of R% compound interest for
compounded quarterly, then :
Compound Interest (CI) on Rs. P @ R% for T years =

T years

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 8000 Rate ( R) = 10% p.a

Time (T) = 3 years

• Compound interest (CI) on Rs. 8000 at the end of 3 years @ 10% =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 4.
SOLUTION 1 : Without using formula.
Here Principal (P)= Rs. 25000, Time (T) = 1 year, Yearly rate of interest =12%
• The Interest is compounded half yearly
• So the half yearly rate of interest = 12 * ½ = 6% per half year
• Interest at 1st half year end = 6% of 25000
•
= Rs. 1500
• New principal for 2nd half year = Rs. 25000+1500 = Rs. 26500
• Interest at 2nd half year end = 6% of 26500
•
= Rs. 1590
• So at the end of 1 year Rs. 25000 will become Rs. (26500 +1590)
= Rs. 28090/• So the compound interest (CI) = 28090 – 25000 = Rs. 3090
• Or , CI = Rs. (1500 + 1590)= Rs. 3090

SOLUTION 2 : Using formula.
• Here Principal (P)= Rs. 25000 Rate ( R) = 12% p.a Time (T) = 1 year
• Compound interest (CI) on Rs. 25000 at the end of 1 years @ 12% pa
compounded half yearly =
CI =
=
=
=
=
=
= Rs. _____________

Page 402 Ex 5. SOLUTION : Without using formula.
Here Principal (P)= Rs. 8000, Time (T) = 9 months = 9/12 years = ¾ yrs,
• Note: 9 months = 3 quarters
• Yearly rate of interest = 20% pa
• The Interest is compounded quarterly . In a year there are 4 quarters
• So the per quarter rate of interest = 20/4 = 5% per quarter.
• Principal for 1st quarter = Rs. 8000
• Interest at 1st quarter end = 5% of 8000 = Rs. 400

• New principal for 2nd quarter = Rs. 8000+400 = Rs. 8400
• Interest at 2nd quarter end = 5% of 8400 = Rs. 420
• New principal for 3rd quarter = Rs. 8400+420 = Rs. 8820
• Interest at 3rd quarter end = 5% of 8820 = Rs. 441
• So at the end of 3 quarters or 9 months, Rs. 8000 will become
Rs. (8820 +441) = Rs. 9261/• So the compound interest (CI) = 9261 – 8000 = Rs. 1261
• Or, CI = Rs. (400 + 420+ 441)= Rs. 1261

Page 402 Ex 3. SOLUTION :
• Here Principal (P)= Rs. 5000, Time (T) = 3years,
• Rate of interest for 1st year (R1) = 5% pa
• Principal for 1st year = Rs. 5000
Interest at 1st year end

= 5% of 5000 = Rs. 250

• New principal for 2nd year = Rs. 5000 + 250 = Rs. 5250
• Rate of interest for 2nd year (R2) = 8% pa
Interest at 2nd year end = 8% of 5250 = Rs. 420

• New principal for 3rd year = Rs. 5250 + 420 = Rs. 5670
• Rate of interest for 3rd year (R3) = 10% pa
Interest at 3rd year end = 10% of 5670 = Rs. 567
• So the amount at the end of 3 years = Rs. (5670 +567) = Rs. 6237/• This means, at the end of 3 years, Rs. 5000 will become Rs. 6237/• Compound interest (CI) on Rs. 5000 after 3 years @ 5%,8% &10%=
6237 – 5000 = Rs. 1237
• Or, CI on Rs. 5000 after 3 years @ 5%,8% &10%= ( 250+ 420+ 567)= Rs. 1237

Page 402 Ex 6. SOLUTION :
• Here Principal (P) = a certain sum of money, CI = Rs. 3783,
Time (T) = 3years, Rate of interest (R) = 5% pa
• Let the principal (P) be Rs. x
• Assuming interest is being charged annually, We know that,
CI =
=> CI on Rs. x @ 5% for 3 years =

=> 3783 =
Find x then find SI on Rs. x @ 5% for 3 years = (x*5*3)/100