Compound Interest
The difference between the amount and the money borrowed is called the compound interest for given period of time
1) Let principal =P; time =n years; and rate = r% per annum and let A be the total amount at the end of n years, then
A = P*[1+ (r/100)]^{n};
CI = {P*[1+ (r/100)]^{n} -1}
2) When compound interest reckoned half yearly, then r% become r/2% and time n become 2n;
A= P*[1+ (r/2*100)]^{2n}
3) For quarterly
A= P*[1+ (r/4*100)]^{4n}
4) The difference between compound interest and simple interest over two years is given by
Pr^{2}/100^{2}or P(r/100)2
5) The difference between compound interest and simple interest over three years is given by
P(r/100)^{2}*{(r/100)+3}
6) When Rates are different for different years, say R1%, R2%, R3% for 1st, 2nd and 3rd year respectively, Then total amount is given by
P ((1 + R^{1})/100) ((1 + R^{2})/100) ((1 + R^{2})/100)
7) Present worth of Rs. x due n years hence is given by
x/(1+R/100)
1). Interest is compounded half-yearly, therefore,
Example:
Find the compound interest on Rs. 20,000 in 2 years at 4 % per annum, the interest being compounded half-yearly.
Solution:
Principal = Rs. 20000, Rate = 2 % per half-year, Time = 2 years = 4 half- years
Amount=Rs.21648.64
Compound Interest = Total amount – Principal
= 21648.64 – 20000
= Rs. 1648.64
2). If interest is compounded annually,
Example:
Find compound interest on Rs. 8500 at 4 % per annum for 2 years, compounded annually.
Solution:
We are given:
Principal = Rs. 8500, Rate = 4 % per annum, Time = 2 years
= Rs. 9193.6
Compound Interest = Total amount – Principal
= 9193.6 – 8500
= 693.6
Compound Interest = Rs. 693.6
3). When Rates are different for different years, say R1%, R2%, R3% for 1st, 2nd and 3rd year respectively. Then, Amount (= Principal + Compound interest) = P(1 + R1/100)(1 + R2/100)(1 + R3/100).
Example:
Find the compound interest on a principal amount of Rs.5000 after 2 years, if the rate of interest for the 1st year is 2% and for the 2nd year is 4%.
Solution:
Here R1 = 2% R2 = 4% and p = Rs.5000, we have to find CI (compound interest).
CI = 5000(1 + 2/100)(1 + 4/100) – 5000
= 5000 x (102/100)(104/100) – 5000
= 5000 x (51/50) x (52/50) – 5000
= 5000 x (51 x 52/2500) – 5000
= 5000 x (2652 / 2500) – 5000
= 5304 – 5000 = 304
Hence the required compound interest is Rs.304.
4). When compound interest is reckoned half-yearly.
If the annual rate is r% per annum and is to be calculated for n years, then in this case, rate = (n/2%) half-yearly and time = (2n) half-yearly.
Form the above we get
Example:
Sam investment Rs.15,000 @ 10% per annum for one year. If the interest is compounded half-yearly, then the amount received by Sam at the end of the year will be.
Solution:
P = Rs. 15000; R = 10% p.a = 5% half-year, T = 1 year = 2 half year
Amount = Rs
= Rs.16537.50
If the simple interest for certain sum for 2yrs at the annual rate of interest R% is SI. Then,
Compound interest (CI) = SI (1+r/200) (no. of years =2)
5). If the simple interest for a certain sum for 2 yr at 5%pa is 200, then what will be the compound interest for same sum for same period and the same rate of interest?
Sol:
Si =200 r=5%
Ci =200(1+5/200) =200*(205/200) =205
If a certain sum at compound interest becomes x times n_{1}^yr and y times n_{2}^yr then,
X^{1/N1} = Y^{1/N2}
6). If an amount at compound interest becomes twice in 5yr, then in how many years, it will be 16 times at the same rate of interest?
2^{1/5 }= 16 ^{1/x2}
=2^{4*1/x2}
1/5 = 4/x_{2}
X_{2} = 5*4 =20yrs
If a certain sum at compound interest amounts to A_{1} in n yrs and A_{2} in (n+1) yrs,
then
Rate of compound interest =(A_{2} – A_{1})/A_{1} *100%
Sum = A_{1} (A_{1} /A_{2})^{n}
7). A sum of money invested at compound interest amounts to 800 in 2yr and 840 in 3yrs .Find the rate of interest and the sum.
A_{1} =800 ; A_{2 }=840,
Rate of interest = (840-800)/800 *100% =40/8 =5%
Sum = 800 *(800/840)^{2} =320000/441 = Rs.725.62
If the populations of a city P and it increases with the rate of R% per annum, then
- Populations after n yr = p(1+R/100)^{n}
- Populations n yr ago = p / (1+R/100)^{n}
8). The population of a city A is 5000. It increases by 10% in 1^{st} year. It decreases by 20% in the 2^{nd} yr because of some reason. In the 3^{rd} yr, the population increases by 30%. What will be the [population of area A at the end of 3yrs?
=5000(1+10/100)(1-20/100)(1+30/100)
= 500*(11/10)*(4/5)*(13/10) = 5720
Difference between ci and si 2yr =pr^{2} /100^{2 }
9). The difference between c.i and s.i for 2yr at the rate of 5% per annum is 5 .then the sum
5 = p (5/100)^{2} = Rs.2000
Rate of interest (no .of years =2)
(for only ci)
2% = 4.04%
3% = 6.09%
4% = 8. 16%
5% = 10.25%
6% = 12.36%
7% = 14.49%
8% = 16.64%
9% = 18.81%
10%= 20.00
+ 1.00 =21%
10). What is the Compound interest for Rs. 1500 at 5% rate of interest for 2 years?
1500*(10.25/100) =153.75
Difference between the compound interest and the simple interest
Example:
If the difference between the compound interest and the simple interest on a certain sum of money at 5% per annum for 3 years is Rs. 1220. What is the sum?
Solution: