Applications of Compound Interest Formula: Definition, Uses

Applications of Compound Interest Formula: Compound interest is interest calculated on both the principal (original amount) and the interest already calculated. Every year, compound interest multiplies. We can find the applications of the compound interest formula in solving real-life problems mathematically. The various applications of compound interest formula include growth and decay, appreciation and depreciation, increase and decrease in commodity price, etc.

Compound interest keeps multiplying every year. The compound interest formula is represented as \(C.I = P{\left( {1 + \frac{{\frac{r}{n}}}{{100}}} \right)^{nt}} – P\). The compound Interest formula’s examples involve bank transactions, inflation in the profit and loss, etc. In this article, we have provided detailed information about the compound interest formulas, real-life applications of the compound interest formula, etc. Scroll down to learn more!

What is Compound Interest?

The extra or additional amount received from the borrower or paid to the lender is known as interest. We have two types of interests: simple interest and compound interest. In simple interest, the rate of interest on the principal amount remains constant throughout the given period. But, in compound interest, the interest keeps changing every year.

Compound interest is interest calculated on both principal amount and interest, compounded (multiplied) at regular intervals. The interest so far accumulated at regular intervals is added with the existing principal, and then the interest is taken for the new principal. The new principal is equalled to the sum of the initial principal and the interest calculated so far.
\(\rm{Compound}\,\rm{interest} = \rm{Interest}\,\rm{on}\,\rm{Principal} + \rm{Compounded}\,\rm{Interest}\,\rm{at}\,\rm{regular}\,\rm{intervals}\)

Compound Interest Formula

Compound interest is the interest calculated on both principal amount and interest, compounded (multiplied) at regular intervals. Compound interest keeps multiplying every year.
The formula for calculation of the compound interest is given as follows:
\(C.I = P{\left( {1 + \frac{{\frac{r}{n}}}{{100}}} \right)^{nt}} – P\)
Here,
1. \(P-\)Initial principal amount
2. \(C.I-\)Compound interest
3. \(r-\)Rate of interest per annum
4. \(t-\)Time period in years
5. \(n-\)Frequency of the number of times the interest is compounded annually.

Applications of Compound Interest Formula Example With Solution

The applications of the compound interest formula are its uses in solving real-life problems mathematically. The various applications of the compound interest formula are listed below:

1. Compound interest not annually (monthly)
2. Growth and decay in the population
3. Increase and decrease in commodity price
4. Increase and decay in the value of the item
5. Inflation in the profit and loss
6. Bank transactions

Compounding Interest Non-Annually

In our real life, we come across many situations where we find the interest compounded non-annually (monthly, quarterly, half-yearly, sometimes daily). These are discussed here:

Calculation of Interest Compounded Semi-annually or Half-Yearly

If the principal is compounded semi-annually or half–yearly, the total amount is given by
\(A = P{\left( {1 + \frac{{\frac{r}{2}}}{{100}}} \right)^{2t}}\)
And, interest is given as:
\(C.I = P{\left( {1 + \frac{{\frac{r}{2}}}{{100}}} \right)^{2t}} – P\)

Calculation of Interest Compounded Quarterly

If the principal is compounded quarterly, the total amount is given by
\(A = P{\left( {1 + \frac{{\frac{r}{4}}}{{100}}} \right)^{4t}}\)
And, interest is given as:
\(C.I = P{\left( {1 + \frac{{\frac{r}{4}}}{{100}}} \right)^{4t}} – P\)

Calculation of Interest Compounded Monthly

If the principal is compounded monthly, then the total amount is given by
\(A = P{\left( {1 + \frac{{\frac{r}{{12}}}}{{100}}} \right)^{12t}}\)
And, interest is given as:
\(C.I = P{\left( {1 + \frac{{\frac{r}{{12}}}}{{100}}} \right)^{12t}} – P\)

Calculation of Interest Compounded Daily

If the principal is compounded daily, then the total amount is given by
\(A = P{\left( {1 + \frac{{\frac{r}{{365}}}}{{100}}} \right)^{365t}}\)
And, interest is given as:
\(C.I = P{\left( {1 + \frac{{\frac{r}{{365}}}}{{100}}} \right)^{365t}} – P.\)

Calculation of Growth or Decay Using Compound Interest Formula

One of the main applications of the compound interest formula is the calculation of growth of the objects or population etc. We will analyse how the compound interest formula is used to find the growth using a simple example.

A money lender has \(\rm{Rs}. 34,000\). He grows his money at \(5\%\) annual rate of interest. Now, he wants to know his total wealth at the end of three years from now. We can find his total wealth after three years (growth) using the compound interest formula as follows:

Let us assume the total wealth given today is the principal \(P = \rm{RS}. 34,000.\)
And, the growth per year is the rate of interest \(\%r = 5\%.\)
We need to find the total wealth at the end of the next three years, so consider \(n=3\) years.
The amount calculated using the compound interest formula gives his total wealth after three years.
So, the amount obtained by the compound interest formula is given by
\(A = P (1 + \frac{r}{100})^n\)
\( \Rightarrow A = 34,000{\left( {1 + \frac{5}{{100}}} \right)^3} = 34,000{\left( {\frac{{105}}{{100}}} \right)^3}\)
\(⇒ A = \rm{Rs}. 39,359.25\)
Hence, his total wealth after three years is given by \(\rm{Rs}. 39,359.25.\)

Similarly, we can find the depreciation by replacing the plus sign with the minus sign in the above formula.

Applications of Compound Interest Formula – Growth or Increase

We know that growth per year is called the rate of growth. We can find the growth or total growth (where the rate of growth for n years is constant) using the compound interest formula as given below:
Final number \(= \rm{initial}\,\rm{number} \times (1 + \frac {r}{100})^3\)
Let the rate of growth is different for different years. Let \(r_1\%\) is the rate of growth for the first year and \(r_2\%\) is the rate of growth for the second year. Then, the total growth after two years is given by using the compound interest formula as follows:

Final value \(= \rm{initial}\,\rm{value} \times (1 + \frac {r_1}{100})(1 + \frac {r_2}{100})\)

Applications of Compound Interest Formula – Decay or Decrease or Depreciation

We know that the decay or decrease of the value per year is called the rate of decay. We can find the total value (where the rate of decay for \(n\) years is constant) by using the compound interest formula as shown below:

Value after \(n\) years \(= \rm{present}\,\rm{value}\times (1 – \frac {r}{100})^n\)
Let the rate of decay is different for different years. Let \(r_1\%\) is the rate of decay for the first year and \(r_2\%\) is the rate of decay for the second year. Then, the total value after two years is given by using the compound interest formula as follows:
Value after two years \(= \rm{Present}\,\rm{value} \times (1 – \frac {r_1}{100})(1 – \frac {r_2}{100}).\)

Solved Examples – Applications of Compound Interest Formula

Q.1. The present population of town \(A\) is \(10,000\). If the population of town \(A\) was increased by \(10\%\) every year. Find the population of town \(A\) after three years.
Ans: Given the initial population of town \(A = 10000\).
Rate of growth \(r = 10\%\)
Let \(n = 3\) years
The population of town \(A\) is found by using the compound interest formula as follows:
\(\rm{Final}\,\rm{population} = \rm{intial}\,\rm{population} × (1 + \frac{r}{100})^n\)
\(⟹ \rm{Final}\,\rm{population}\,\rm{after}\,\rm{three}\,\rm{years} = 10000 × (1 + \frac{10}{100})^3 = 13310\)
Hence, the total population of town \(A\) after three years is \(13310\).

Q.2. The count of a certain breed of bacteria was found to increase at the rate of \(2\%\) per hour. Find the bacteria at the end of \(2\) hours if the count was initially \(600000\).
Ans: Given, the initial population \((P)=600000\), rate \((R) = 2\%\).
Since the population of bacteria increases at the rate of \(2\%\) per hour, we use the formula
\(A = P(1 + \frac{R}{100})^n\)
Thus, the population at the end of \(2\) hours \(= 600000 (1 + \frac{2}{100})^2\)
\(= 600000 (1 + 0.02)^2\)
\(= 624240\)
Hence, the count of bacteria after two hours is \(624240\).

Q.3. A town had \(10,000\) residents in \(2010\). Its population decays at a rate of \(10\%\) per annum. What will be its total population in \(2015\)?
Ans: The population of the town decreases by \(10\%\) every year. Thus, it has a new population every year.
The population for the next year is calculated on the current year population. For the decrease, we have the formula
\(A = P (1 – \frac {r}{100})^n\)
Therefore, the population at the end of \(5\) years
\(= 10000(1 – \frac{10}{100})^5\)
\(= 10000(1 – 0.1)^5\)
\( = 5904.9\)
\( = 5904\) (Approx).

Q.4. A sum of \(\rm{Rs}.5000\) is borrowed at the rate is \(8\%\). The amount was compounded monthly. Find the compound interest for two years?
Ans: Given, principal \((P) = \rm{Rs}.5000\)
Rate of interest \((r) = 8\%\)
Time period \((T) = 2\) years
Compound interest after two years is
\(= \rm{Principal}(1 + \frac{r}{12 × 100})^{12 × \rm{time}} – \rm{Principal}\)
\( = 5000(1 + \frac{8}{1200})^{12 × 2} – 5000\)
\( = 5000 ( 1 + \frac{2}{300})^{24} – 5000\)
\( = 5000 × 1.1712 – 5000\)
\( = 5856 – 5000 = \rm{Rs}.856\)
The monthly compound interest for two years is \(\rm{Rs}.856\)

Q.5. Calculate the compound interest to be paid on a loan of \(\rm{Rs}.2000\) for \(\frac{3}{2}\) years at \(10\%\) per annum compounded half-yearly?
Ans: Principal, \(P = \rm{Rs}.2000\),
Time, \(T = \frac{3}{2}\rm{years} = 3 × \frac{1}{2}\rm{years} = 3\,\rm{half}\,\rm{years}\).
Rate, \(r = \frac{10\%}{2} = 5\%,\,2\)
Amount, \(A\) can be given as:
\(A = P (1 + \frac{r}{100})^n\)
\(⟹ A = 2000 × (1 + \frac{5}{100})^3 = 2000 ×(\frac{21}{20})^3 = \rm{Rs}.2315.25\)
So, \(CI = A – P = \rm{Rs}.2315.25 – \rm{Rs}.2000 = \rm{Rs}.315.\)

Summary:

This article discussed what compound interest is and the formula used to calculate compound interest. We have studied the applications of the compound interest formula with the help of solved examples.
We have studied the growth and decay, increasing and decreasing the value or population using the compound interest formula.

Learn Important Compound Interest Formula

Frequently Asked Questions

We have provided some frequently asked questions on Applications of Compound Interest Formula here:

Q.1. What is the use of the compound interest formula, and who benefits from this?
Ans: The compound interest is used to get more money as interest. The investors benefit from the compound interest since the interest received is on the principal plus on the interest they already earned.

Q.2. How do you calculate the compound interest if the interest is compounded every month?
Ans: If the principal is compounded monthly (\(12\) months in a year), the interest is given by
\({\text{Compound}\;\text{interest}} = {\text{Principal}}{\left( {1 + \frac{{\frac{{\text{r}}}{{12}}}}{{100}}} \right)^{12 \times {\text{time}}}} – {\text{Principal}}.\)

Q.3. What is the compound interest formula used for the growth or increase?
Ans: \({\text{Final}\;\text{growth}} = {\text{initial}\;\text{growth}} \times {\left( {1 + \frac{r}{{100}}} \right)^n}\)

Q.4. What is the main disadvantage of the compound interest?
Ans: The main disadvantage of compound interest is that it is more expensive, and we need to pay more as an interest to the lender if we have taken some amount from the lender.

Q.5. What are the applications of the compound interest formula?
Ans: Some of the applications of the compound interest formula are
1. Growth and decay
2. Appreciation and depreciation
3. Increase or decrease the value