Compound Interest

Ankur Kulhari

Interest (I): It is the extra money paid by the borrower to the owner (lender) as a form of compensation for the use of the money borrowed
Compound Interest (CI)

CI = P(1+R/100)T – P
A = P(1+R/100)T


P= Principle or sum {It is the money borrowed or lent out for a time period T}
R=Rate at which interest is charged on P per annum {Thus, 6 p.c. means that Rs.6 is the interest on Rs.100 in one year.}
A = Amount {The Addition of Interest and Principle i.e. A = (I + P)}

Compound Interest is based on the principal amount and the accumulated interest.
Compound Interest questions can be designed in many ways. Lets discuss all one by one:

Case 1: When compound Interest is calculated on yearly basis.
Compound Interest is calculated at the end of one year duration and
now for next year CI+P = next year principle.
Let principle = P, time = T years and rate = R% per annum and
Let A be the total amount at the end of T years, then

A = P (1+R/100)T
CI = P(1+R/100)T – P {Important to note}


Example 1. An amount of Rs.8000 is invested in a FD scheme for 2 years at compound interest rate 5% per annum. How much interest will be incurred on maturity of the FD?


I = 8000(1+5100)2 – 8000
= 8000(105100)2 – 8000
= 8000 [105100)2 -1] = 8000 [1.103-1] = 8000*0.103
= 824


Case 2: When compound interest is calculated bi-yearly (6 months).

Compound Interest is calculated at the end of 6 months duration from investment and now for 6 months CI+P = next year principle.
If the annual rate is R% per annum and is to be calculated for T years, then in this case,
rate = (R/2) half-yearly and,
time = (2T) 2 half-years).

CI = P [1+(R/2)/100]T/2 – P


Example 2. An amount of Rs.15,000 is invested at 10% per annum for one year. If the interest is compounded half-yearly, then what amount will be received at the end of the year?


A = 15000 [1+(10/2)/100]2
= 15000 [105/100]2
= 15000 X 1.103 (Click to see short trick for finding square quickly)
= 16545


Case 3: When compound interest is reckoned quarterly.

In this case, rate = (R/4)% quarterly and,
time = (4T) quarter years.

A= P[1+(R/4)/100]4T


Example 3. Find the compound interest on Rs. 15,000 for 9 months at 16% per annum compounded quarterly.


CI = 15000[1+(16/4)/100]4X(9/12) -15000
= 15000[104/100]3 – 15000
= 15000 [1.120-1] (click here to learn quick cubes)
= 15000 X 0.120
= 1800


Case 4: When rates are different for different years, say for 1st, 2nd, 3rd year R1, R2, R3 respectively. Then


A = P (1+R1/100) (1+R2/100) (1+R3/100)


Example 4. An amount of 200000 is deposited in bank under a scheme which provides 10% rate 1st years, 20% rate 2nd year and 5% rate 3rd year. After 3rd year what amount will be returned?


A = 200000 (1+10/100)(1+20/100)(1+5/100)
A= 200000(110/100)(120/100)(105/100)
A= 277200


Some Tricks to Solve easily


Trick 1: If a sum of money becomes “n” times in “T years” at simple interest, then the rate of interest per annum can be given by:




Trick 2: If an amount P1 is lent out at simple interest of R1% per annum and another amount P2 at simple interest rate of R2% per annum, then the rate of interest for the whole sum can be given by:

R=(P1.R1 + P2.R2)/(P1+P2)


If P1 and P2 are lent at rate R so that
SI1+SI2= SI then,
SI1+SI2= (P1+P2)R.T/100
P1.R1.T/100 + P2.R2.T/100 = (P1+P2)R.T/100
R=(P1.R1 + P2.R2)/(P1+P2)


Trick 3: In what time will the simple interest becomes “n times” of the principal P at “R %” per annum:

T =n x 100/R




Trick 4: If a certain sum of money is lent out in n parts in such a manner that equal sum of money is obtained at simple interest on each part where interest rates are R1, R2, … , Rn respectively and time periods are T1, T2, … , Tn respectively, then the ratio in which the sum will be divided in n parts can be given by
1/R1 . T1 : 1/R2 . T2 : ……….. : 1/Rn . Tn


P1=SI.100/R1.T1, P2=SI.100/R2.T2, P3=SI.100/R3.T3………….., Pn=SI.100/Rn.Tn
P1 : P2 : P3 : ……………………. : Pn
SI.100/R1 . T1 : SI.100/R2 . T2 : SI.100/R3 . T3 : ………….. : SI.100/Rn . Tn


Difference between CI and SI:
Shortcut Trick:

For two years:


CI-SI = P(R/100)2


For three years:


CI-SI = 3P(R/100)2 + P (R/100)3



For two years:

CI-SI =[P(1+R/100)2 – P ]- 2PR/100
= [P(1+2R/100+R2/(100 . 100)) – P] -2PR/100
= [P + 2PR/100 + PR2/(100 . 100)-P] – 2PR/100
= [2PR/100 + PR2/(100 . 100)] – 2PR/100
= P(R/100)2

For three years:

Similarly we can calculate the difference for three years:
CI-SI = [P(1+R/100)3 – P ]- 3PR/100
= [P((1+R/100) (1+2R/100+R2/(100 . 100)) – P] -3PR/100
= [P+2PR/100+PR2/(100 . 100)+PR/100+2PR2/1002+PR3/1003-P] – 3PR/100
= [3PR/100+3PR2/1002+PR3/1003] – 3PR/100
= 3P(R/100)2 + P (R/100)3

Comments 3

    1. Post

      Generally in competitive exams upto 3 years is asked. But we can devise the same for n also in the similar fashion.

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