In general, when a borrower borrows money from a lender, they promise to repay it after a time period. At the end of the specified period, the borrower must repay the borrowed funds and some additional funds for using the lender’s funds. This extra money is referred to as interest. There are two types of interest such as simple interest and compound interest. This article will discuss the Relation between Compound and Simple Interest.

Simple interest is a fixed proportion of the principal amount borrowed or lent that is paid or received over a period of time. Whereas, Compound interest accrues and is added to the accumulated interest of previous periods, so borrowers must pay interest on interest as well as principal.

This is an important concept in Maths, as students will not only make use of it in their academics but also in their daily lives when they become working individuals. Continue reading to know more.

Simple Interest and Compound Interest

Interest is the extra money paid by institutions such as banks, post offices on money deposited with them. To calculate simple interest and compound interest, we use the following terms:

Principal: The money borrowed is called the principal amount of the sum.

Amount: The total money paid back by the borrower to the lender is called the amount.

Thus, \({\rm{Amount}} = {\rm{Principal}} + {\rm{Interest}}\)

Rate: The interest of \(₹100\) for \(1\) year is known as the rate per anum.

Time: Time is the period for which the sum is given to the borrower.

Simple Interest (SI)

If interest is calculated uniformly on the original principal throughout the loan period, it is called simple interest.

If, \(P=\) Principal, \(R=\) Rate of interest per annum and \(T=\) time,

Then the simple interest formula is given by:

\(S . I .=\frac{P \times T \times R}{100}\)

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Compound Interest (CI)

The interest accrued during the first unit of time is added to the original principal, and the amount obtained is taken as the principal for the second unit of time. The amount of this principle at the end of the second unit of the time becomes the principal of the third unit of time and so on. 

The required compound interest is the difference between the final amount and the principal.

\(\text {Compound Interest} = \text {Final Amount} – \text {Original Principal}\)

\(C.I. =A-P \ldots(i)\)

\(A=P\left(1+\frac{R}{100}\right)^{n} \ldots .(ii)\)

Now, substituting \(A=P\left(1+\frac{R}{100}\right)^{n}\) in equation (i), we have,

\(C . I .=P\left(1+\frac{R}{100}\right)^{n}-P\)

Here, \(P=\) Principal, \(R=\) Rate of interest per annum \(n=\) Number of years \(A=\) Final amount.

Relation Between Simple Interest and Compound Interest

There are two ways to calculate interest on a loan. Simple interest is easier to calculate than compound interest due to some fundamental relation between the two. The following example will tell us the relation between simple and compound interest.

Example: Let us say Rimi has borrowed (principal) \(₹100\) at \(5 \%\) per annum for \(3\) years at simple interest. How much will she pay if she pays the final amount of \(1^{\text {st }}\) year/ \(2^{\text {nd }}\) year \(/ 3^{\text {rd }}\) year?
Solution: For \(1^{\text {st }}\) year the principal amount is \(P= ₹100\)
\(\text {Simple Interest}=₹ \frac{100 \times 5 \times 1}{100}\)
\(=₹5\)
Amount \(=₹100+₹5\)
\(=₹105\)
For \(2^{\text {nd }}\) year the principal amount is \(P=₹100\)
\(\text {Simple Interest}=₹\frac{100 \times 5 \times 1}{100}\)
\(=₹5\)
Amount \(=₹105+₹5\)
\(=₹110\)
For \(3^{\text {rd }}\) year the principal amount is \(P=₹100\)
\(\text {Simple Interest}=₹\frac{100 \times 5 \times 1}{100}\)
\(=₹5\)
Amount \(=₹110+₹5\)
\(=₹115\)
So, at the end of \(3\) years, Rimi will pay an amount of \(₹115\).

Example: Suppose Rimi has borrowed (principal) \(₹100\) at \(5 \%\) per annum for \(3\) years at compound interest. How much will she pay if she pays the final amount of \(1^{\text {st }}\) year \(/ 2^{\text {nd }}\) year/ \(3^{\text {rd }}\) year?
Solution: For \(1^{\text {st }}\) year the principal amount is \(P=₹100\)
\(\text {Interest} =₹\frac{100 \times 5 \times 1}{100}\)
\(=₹5\)
Amount \(=₹100+₹5\)
\(=₹105\)
For \(2^{\text {nd }}\) year the principal amount is \(P=₹105\)
\(\text {Interest} =₹\frac{105 \times 5 \times 1}{100}\)
\(=₹5.25\)
Amount \(=₹105+₹5.25\)
\(=₹110.25\)
For \(3^{\text {rd }}\) year the principal amount is \(P=₹110.25\)
\(\text {Interest} =₹\frac{110.25 \times 5 \times 1}{100}\)
\(=₹5.5125\)
Amount \(=₹110.25+₹5.5125\)
\(=₹115.7625\)
From this method, we can also calculate compound interest after \(3\) years. Compound Interest \(=\) Interest of \(1^{\text {st }}\) year \(+\) Interest of \(2^{\text {nd }}\) year \(+\) Interest of \(3^{\text {rd }}\) year \(=₹5+₹5.25+₹5.5125=₹15.7625\)
Compound interest in \(3\) years
\(=\) Amount at the end of \(3\) years \(-\) original sum (Principal for \(1^{\text {st }}\) year)
\(=₹115.7625-₹100=₹15.7625\).

Thus, we can find the amount for the \(1^{\text {st }}\) year, \(2^{\text {nd }}\) year, and \(3^{\text {rd }}\) year for compound interest using the simple interest formula, which seems lengthy.

To save time and to avoid lengthy calculations, we use a unique formula for compound interest. We know that, for compound interest, \(A=P\left(1+\frac{R}{100}\right)^{n}\) When \(P=\) Principal, \(R=\) Rate of interest per annum and \(n=\) Number of years and \(A=\) Final amount.

Here, \(p=100, R=5 \%\)
When \(n=1\)
\(A=100\left(1+\frac{5}{100}\right)^{1}=₹105\)
Thus, the amount of the \(1^{\text {st }}\) year is \(₹105\).
When \(n=2\),
\(A=100\left(1+\frac{5}{100}\right)^{2}=₹110.25\)
Thus, the amount of the \(2^{\text {nd }}\) year is \(₹110.25\).
When \(n=3\),
\(A=100\left(1+\frac{5}{100}\right)^{3}=₹115.7625\)
Thus, the amount of the \(3^{\text {rd }}\) year is \(₹115.7625\).
So, the compound interest is \(₹115.7625-₹100=₹15.7625( C.I.=A-P)\)

We can conclude that in the \(1^{\text {st }}\) year, the interest and the principal amount are equal for simple interest and the compound interest method. In the \(2^{\text {nd }}\) year, we can see the principal amount and interest remain unchanged for the simple interest method, but the interest and principal amount changed for the compound interest method. In the same way, in the \(3^{\text {rd }}\) year, the interest and the principal amount remain unchanged at simple interest, and they changed at compound interest.

In other words, we can say that for the simple interest method, the principal amount remains the same, and for the compound interest method, the amount at the end of the year is the principal amount of the following year. For the simple interest method, the interest charged every year is the same, and for compound interest, the interest changes every year.

Solved examples on Relation between Compound and Simple Interest

Q.1. Calculate the difference between the compound interest and the simple interest on \(₹4000\) at \(8 \%\) per annum and in \(2\) years.
Ans: Principal \(P=₹4000\); Rate \(R=8 \%\) and Time \(T=2\) years
Therefore, simple interest \(=₹\frac{4000 \times 8 \times 2}{100}=₹640\)
For \(C.I.\): Principal for first-year \(=₹4000\)
Interest on it \(=₹\frac{4000 \times 8 \times 1}{100}=₹320\)
Amount \(=₹4000+₹320=₹4320\)
Therefore, principal for the second year \(=₹4320\)
Interest on it \(=₹\frac{4320 \times 8 \times 1}{100}=₹345.60\)
Therefore, \(C.I.\) for \(2\) years \(=₹320+₹345.60\)
\(=₹665.60\)
Required difference between \(C.I.\) and \(S.I. = C.I.- S.I.\)
\(=₹665.60 – ₹640\)
\(=₹25.60\)

Q.2. A sum of \(₹500\) is lent for at the rate of \(20 \%\) compounded annually. Find the interest after one year.
Ans:
\(P=\) Principal \(=₹500, R=\) Rate of Interest \(=20 \%\) per annum, \(T=\) Time \(=1\) year.
For \(1^{s t}\) year: Compound Interest \(=\) Simple Interest
So, we can apply here the formula of simple interest to find the interest after \(1\) year.
Let \(I\) be the interest. Then, S.I. \(=\frac{p \times R \times T}{100}\)
\(\Rightarrow \text {Interest} =\frac{500 \times 20 \times 1}{100}=5 \times 20=₹100\)
Hence, the interest is \(₹100\).

Q.3. A sum of \(₹10000\) is borrowed at the rate is \(8 \%\). The amount was compounded annually. Find the compound interest for two years?
Ans:
Given, principal \((P)=₹10000\)
Rate of interest \((r)=8 \%\)
Time period \((T)=2\) years
Compound interest after two years is :
\(=\) Principal \(\left(1+\frac{r}{100}\right)^{T}-\) Principal
\(=\left[10000\left(1+\frac{8}{100}\right)^{2}\right]-10000\)
\(=\left[10000\left(1+\frac{8}{100}\right)^{2}\right]-10000\)
\(=\left[10000\left(\frac{108}{100}\right)^{2}\right]-10000\)
\(=11664-10000=₹1664\)
The compound interest for two years is \(₹1664\).

Q.4. In how many years will \(₹750\) amount to \(₹900\) at \(4 \%\) per annum?
Ans: Here, \(P=₹750, A=₹900, R=4 \%\) per annum.
Let \(₹750\) amount to \(₹900\) at \(4 \%\) per annum in \(T\) years.
Now, \(\text {Interest} = \text {Amount} – \text {Principal}\)
\(=₹900-₹750=₹150\)
\(I=\frac{P \times R \times T}{100}\)
\(\Rightarrow 150=\frac{750 \times 4 \times T}{100}\)
\(\Rightarrow T=\frac{150 \times 100}{750 \times 4}=5\) years
Thus, \(₹750\) amounts to \(₹900\) at \(4 \%\) per annum in \(5\) years.

Q.5. Find the compound interest increased on \(₹16000\) in \(3\) years, when the interest rates for successive years are \(10 \%, 12 \%\) and \(15 \%\), respectively.
Ans: For the first year: Principal \(P=₹16000\); Rate \(R=10 \%\) and Time \(T=1\) year
Therefore, Interest \(=₹\frac{16000 \times 10 \times 1}{100}=₹1600\)
And amount \(=₹16000+₹1600=₹17600\)
For the second year: Principal \(P=₹17600\); Rate \(R=12 \%\) and Time \(T=1\) year
Therefore, Interest \(=₹\frac{17600 \times 12 \times 1}{100}=₹2112\)
And amount \(=₹17600+₹2112=₹19712\)
For the third year: Principal \(P=₹19712\); Rate \(R=15 \%\) and Time \(T=1\) year
Therefore, Interest \(=₹\frac{19712 \times 15 \times 1}{100}=₹2956.80\)
And amount \(=₹19712+₹2956.80=₹22668.80\)
Therefore, \(C.I.\) accrued \(=\) Final amount \(-\) Initial principal
\(=₹22668.80-₹16000=₹6668.80\).

Summary

This article has discussed simple interest, compound interest, and the relation between simple interest and compound interest. We learned that by using simple interest formula, we could understand the concept of compound interest. We solved some examples related to simple and compound interest and related to the relation between them.

FAQs

We have provided some frequently asked questions on Relation between Compound and Simple Interest here:

Q.1. What is simple interest and compound interest?
Ans: If interest is calculated uniformly on the original principal throughout the loan period, it is called simple interest.
If the interest accrued during the first unit of time is added to the original principal, it is compound interest. The amount obtained is taken as the principal for the second unit of time. The amount of this principle at the end of the second unit of the time becomes the principal of the third unit of time and so on.

Q.2. What is the simplest way to understand compound interest?
Ans: The easiest way to understand compound interest is by calculating the simple interest and amount for the first year and considering that as the principal for the second year, and proceeding further.

Q.3. What is the formula for simple and compound interest?
Ans: If \(P=\) Principal, \(R=\) Rate of interest per annum and \(T=\) time, then the simple interest formula is given by
\(S . I .=\frac{P \times T \times R}{100}\)
The compound interest formula is given by:
\(C . I=P\left(1+\frac{R}{100}\right)^{n}-P\)
Here, \(P=\) Principal, \(R=\) Rate of interest per annum and \(n=\) Number of years

Q.4. What is the formula for the difference between SI and CI for 2 years?
Ans: The formula for difference between \(S.I.\) and \(C.I.\) for \(2\) years is \(P\left(\frac{r}{100}\right)^{2}\).

Q.5: What is the difference between simple interest and compound interest, for example?
Ans: Let us see the difference between simple interest and compound interest using an example.
We have taken the sum borrowed (principal) \(=₹1000\) at \(10 \%\) per annum for \(2\) years.

At simple interest	At compound interest
For \(1^{\text {st }}\) year \(P=₹1000\) \(I=₹\frac{1000 \times 10 \times 1}{100}\) \(=₹100(S.I.)\) Amount \(=₹1000+₹100\) \(=₹1100\)	\(P=₹1000\) \(I=₹\frac{1000 \times 10 \times 1}{100}\) \(=₹100(C.I.)\) Amount \(=₹1000+₹100\) \(=₹1100\)
For \(1^{\text {st }}\) year : \(C.I.=S.I.\)
For \((2^{nd}\) year \(P=₹1000\) \(I=₹\frac{1000 \times 10 \times 1}{100}\) \(=₹100(S.I.)\) Amount \(=₹1000+₹100\) \(=₹1200\)	\(P=₹1100\) \(I=₹\frac{1100 \times 10 \times 1}{100}\) \(=₹110(C.I.)\) Amount \(=₹1000+₹110\) \(=₹1210\)
For \(2^{\text {nd }}\) year: \(C . I .\) is more than the \(S . I\).

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