In this topic we will learn to solve problems which are concerned with finding the interest on the amount that is originally lent, borrowed or invested as well as on the accumulated interest of previous periods.
The first section in this topic is an introduction to compound interest with an explanation of the terminologies involved, the formula used to compute compound interest. You will learn about annual and non-annual compounding. And then we have shared some important tips on compound interest. The exercises contain solved question and answers on compound interest. When solving exercises, we recommend you to try solving the problems on your own, that way you will understand better and you can define your own approach to solving problems on compound interest.
Introduction to Simple Interest
Compound Interest is the extra amount that is calculated on the initial principal and also on the accumulated interest of previous periods of a deposit or loan. This addition of interest to the principal is called compounding. Compound Interest is denoted by CI.
Principal: The amount borrowed (or invested) is called the Principal generally denoted by P.
Amount: Denoted by A is the final value of principal with interest. (Normally it is the future value)
Rate of interest: Is the annual simple interest rate. Denoted by r.
Interest: Denoted by I. Is the extra amount by which the principal increases.
Time period: Is the time interval for which the money is borrowed (or invested). Denoted by n.
Compound Interest problem involves two rates depending on which the amount A is calculated
- Annual rate :
The amount A at the end of ‘t’ time periods at a rate of interest of r% p.a is given by
A = P(1 + $\frac{r}{100})^t$.
- The rate per compounding period, i = $\frac rm$
In this case, the amount A is given by
A = P(1 + $\frac{i}{100)^n}$
where i = $\frac rm$ (i is the interest rate per period)
n = mt (n is the total number of compounding periods)
m – number of compounding periods per year
t – time in years
The compounded interest CI = A – P.
Effective rate of interest
If an amount is compounded yearly once at a certain rate of interest, that interest is called as nominal interest. But if the interest is compounded more than once (Ex. half yearly or quarterly compounding) the rate of interest exceeds the per annum nominal interest rate. This exceeded rate of interest is called as Effective rate of interest.
Important Points:
- Unlike Simple Interest, the Compound Interest earned every year is not constant.
- The principal is same at the beginning of every year.
- Number of years for an amount to double itself at CI r% pa is approx. given by $\frac{72}{r}$% years.
The error in the answer increases if r<5% or r>20%
- When Rates are different for different years, say R1%, R2%, R3% for 1st, 2nd and 3rd year respectively.
Amount = P(1+$\frac{R1}{100})(1+\frac{R2}{100})(1+\frac{R3}{100})$
Problems
Compound Interest - Exercise - Blog
14 questions
- What will be the rate of interest per annum if an amount of Rs. 40000 amounts to Rs. 46305 in 3 years?
- 3%
- 5%
- 7%
- 9%
Show the answer
Answer: 5%
A = 46305, P = 40000, t = 3
A = P (1 + $\frac{r}{100})^t$
46305 = 40000(1+$\frac{r}{100})^3$
(1+$\frac{r}{100})^3= \frac{46305}{40000}$
(1 +$\frac{r}{100})^3 =\frac{9261}{8000}$
1 +$\frac{r}{100} = \frac{21}{20}$
$\frac{r}{100} =\frac{1}{20}$
r = 5
Hence the rate of interest is 5% pa
- What will be the CI for Rs. 5000 after 2 years at 10% rate of interest compounded semi-annually?
- 1077.5
- 6077.5
- 5560.5
- 560.5
Show the answer
Answer: 1077.5
A = P(1 +$\frac{i}{100})^n$ , i =$\frac{r}{m}$, n = mt
Given, P = 5000, t = 2, r = 10, m = 2
∴ i =$\frac{10}{2}$ = 5 and n = 2x2 = 4
A = 5000 (1 +$\frac{5}{100})^4$
A = 5000 x 1.2155 = 6077.5
CI = A - P = 6077.5 - 5000 = Rs. 1077.5
- How much will Rs. 20000 amount to after one and half year at 4% pa compounded annually?
- 1200
- 1400
- 21200
- 21400
Show the answer
Answer: 21200
P = 20000, t =$\frac32$, r = 4% p.a, A = ?
A = 20000(1 + $\frac{4}{100})^\frac32$
A = 20000(1.1248$)^\frac12$
A = 20000 x 1.06 = 21200
- What is the principal if it amounts to Rs. 82680 after 2 years such that the rate of interest for first year is 4% and second year is 6% compounded annually?
- Rs. 75000
- Rs. 80000
- Rs. 10000
- Rs. 5000
Show the answer
Answer: Rs. 75000
A = 82680, P = ?, t = 2, r1 = 4% pa and r2 = 6% pa
A = P (1 +$\frac{r1}{100}$)(1+$\frac{r2}{100}$)
82680 = P (1.04)(1.06)
82680 = P x 1.1024
P = 75000
Hence the principal is Rs. 75000
- What is the effective annual rate of interest of 12% pa compounded semi-annually for 1 year?
- 11.24%
- 13.39%
- 10.24%
- 12.36%
Show the answer
Answer: 12.36%
Let us consider an amount of Rs. 1000
Then, P = 1000, r = 12%, m = 2 and t = 1, n = m x t = 2, i = r/m = 6%
A = P (1 +$\frac{i}{100})^n$
= 1000(1 + $\frac{6}{100})^2$
= 1123.6
Effective rate of interest =$\frac{A - P}{P}$ x 100 = 12.36%$
- Find the least number of years for which a sum of money will be more than doubled at 12% rate of interest pa.
- 10%
- 5
- 3
- 6
Show the answer
Answer: 6
Number of years for an amount to double itself at CI r% pa is approx. given by $\frac{72}{r}$ years.
Here, its given that r = 12% pa
Hence the amount doubles itself in approximately $\frac{72}{12}$ = 6 years
- Population of a country increases at the rate of 7% in every 6 months. Approximately after how much time will the population double itself.
- 5 years
- 10 years
- 20 years
- 25 years
Show the answer
Answer: 5 years
Number of time periods for an amount to double itself at CI r% per time period is approx. given by $\frac{72}{r}$ time periods .
Here r = 5% in every 6 months
Time taken to double the population = $\frac{72}{7} $= 10 time periods.
Since each time period is for 6 months, the population will double itself in at least $\frac{10}{2}$ = 5 years.
- What is the difference in interest paid by two customers on a loan of Rs. 10k such that one customer takes the loan at 10% p.a for 2 years compounded annually and the other customer takes the loan at the same rate of interest compounded semi-annually.
- 33.05
- 24.24
- 55.06
- 47.74
Show the answer
Answer: 55.06
First customer
A = 10000(1 +$\frac{10}{100})^2$
= 12100
Interest paid by first customer = Rs. 12100 - Rs. 10000 = Rs. 2100
Second Customer
i =$\frac{10}{2}$ = 5, n = 2 * 2 = 4
A = 10000(1 + $\frac{5}{100})^4$
= 12155.06
Interest paid by second customer = Rs. 12155.06 - Rs. 10000 = Rs. 2155.06
Difference in interest amount paid by two customers is 2155.06 - 2100 = 55.06
- How much will Rs. 10k amount to after 15 months at 4% pa compounded quarterly?
- Rs. 10510.1
- Rs. 10100.1
- Rs. 10410.5
- Rs. 15010.5
Show the answer
Answer: Rs. 10510.1
P = 10000, r = 4% pa , m = 4, t =$\frac{5}{4}$ , i = 1%, n = 5
A = P (1 +$\frac{i}{100})^n$
= 10000 (1 +$\frac{1}{100})^5$
= Rs. 10510.1
- An amount of Rs. 1k generates an interest of Rs. 100 after 1 year. This interest is added to the account after the first year. Find the total interest earned at the end of second year.
- Rs. 200
- Rs. 400
- Rs. 210
- Rs. 420
Show the answer
Answer: Rs. 210
Since, the interest earned at the end of second year also includes interest on the interest earned in first year, the total interest earned at the end of 2 years is CI.
First find the rate of interest:
r % of 1000 = 100
=>$\frac{r}{100}$ * 1000 = 100
=> r = 10
Hence the rate of interest is 10% pa
Amount after 2 years at 10%pa compounded annually is
A = 1000 (1 +$\frac{10}{100})^2$
= 1210
Interest earned after 2 years = Rs. 1210 - Rs. 1000 = Rs. 210
- The interest obtained on a certain sum after 2 years at SI was Rs. 720 and at CI was Rs. 774. Find the rate of interest.
- 12% p.a
- 15% p.a
- 6% p.a
- 10% p.a
Show the answer
Answer: 15% p.a
SI is same for every year. Hence SI per year =$\frac{720}{2}$ = 360.
CI at the end of first year is same as SI at the end of first year. But the difference in CI after 2 years and SI after 2 years is due to the interest on interest earned at the first year.
Hence 774 - 720 = 54.
54 is the interest earned on interest at the end of first year which is 360.
i.e r% of 360 = 54
r = 15
Hence the rate of interest is 15% p.a
- Difference between the CI and SI on a sum of Rs. 500 compounded annually at the end of two years is Rs. 5. Find the rate of interest pa.
- 10% p.a
- 5% p.a
- 15% p.a
- 9% p.a
Show the answer
Answer: 10% p.a
Difference between CI and SI = Interest on the Interest at the end of first year
5 =$\frac{r}{100}$ * (500 *$\frac{r}{100}$) [Interest at the end of first year =$\frac{r}{100}$ x 500]
r$^2$ = 100
r = 10
Hence the rate of interest pa = 10%
- The interest obtained on a certain sum at 4% p.a SI in 2 years is Rs. 864. Find the CI on the same sum at the same rate of interest for 2 years.
- Rs. 11681.2
- Rs. 1681.2
- Rs. 10881.2
- Rs. 881.2
Show the answer
Answer: Rs. 881.2
First find the sum using SI
864 =$\frac{\text{P x 4 x 2}}{100}$
P = Rs. 10800
Next,find the amount by applying CI on the principal
A = P (1 +$\frac{r}{100})^n$
= 10800 (1 +$\frac{4}{100})^2$
= 11681.2
CI = 11681.2 - 10800 = Rs. 881.2
- If a sum of Rs. 1000 amounts to Rs. 1728 in 3 years at certain rate of CI compounded annually. Find the SI on the same sum at the same rate of interest for the same time period.
- Rs. 900
- Rs. 300
- Rs. 1000
- Rs. 600
Show the answer
Answer: Rs. 600
Rate of interest is obtained using the formula for CI
1728 = 1000(1 +$\frac{r}{100})^3$
(1 +$\frac{r}{100})^3$ = 1.728
1 +$\frac{r}{100}$ = 1.2
r = 20%
SI for Rs. 1000 at 20% p.a in 3 years is
SI =$\frac{\text{1000 x 20 x 3}}{100}$
= Rs. 600
