**Test links for this topic at the bottom of the page; scroll to the bottom of the page to take the tests.**

Read Compound Interest, part-1 here before you begin with part-2.

In the first article of our series, we covered the basics of Compound interest, including what does it stand-for. We gave a few tooltips and formulas that are important for the concept and help you grasp the topic better. In the second article of the series, we help you master the topic by providing a series of tooltips providing various formulae and tools that you can use to solve Compound Interest questions.

**Compound Interest Tooltip 6: Formula for compound interest when compounding in a year but time is in fraction**

The formula to calculate compound interest when the time given is in fractions is as follows:

A =P[(1+R/100)^{real part}{1+(Fraction part x R/100)}]

Where

A: Amount at the end of the time period

P: Principal amount

R: Rate of interest

Real Part and Fraction part: For example, the time given is two and half years, then real part would be two and half years.

**Example 1: **

**Manpreet gave Rs. 1000 to Richa for 1 year 6 months. Then find the amount paid by Richa to Manpreet after this duration if rate of interest is 5% per annum compounded annually?**

**Solution :**

We have a time of 1 years and 6 months = (1 year + 1/2 years)

So to find the amount we will use

A =P[(1+R/100)^{real part}{1+(Fraction part x R/100)}]

A = 1000[(1+5/100)^{1}{1+{(6/12 x 5)/100]

A= P[(21/20) x (41/40)]

A = Rs. 1076.25

**Compound Interest Tooltip 7: Concept of Equal Installments**

How does the concept of equal installments work for Compound interests?

Well, in this case, the problem basically tells us that a certain sum of money is borrowed on compound interest for a certain period and it is returned with the help of equal installments. Lets us derive a formula for these installments.

Let us derive a formula where the amount is returned in two equal installments for a time period of two years.

Assume P to be the principal and r the rate of interest.

Step 1: P(1+r/100)=P1 (Amount for one year)

Step 2: New Principal

Now let X be the first installment. After giving the first installment, the principal value will change and the new principal will be = P1 – X

P2 = P1 –X (1+R/100)

Step 3: Amount and Interest for the second year

Now the interest charged will be charged on this amount.

Amount at the end of second year: [P(1+r/100)-X][1+r/100]

Step 4: Since the installments are equal, this new amount has to equal X.

Hence,

[P(1+r/100)-X][1+r/100]=X

On solving, we have

P [(1+R/100)^{2}-X (1+R/100)] = X

P [(1+R/100)^{2}] = X+X (1+R/100)

Divide both sides by (1+r/100)^{2}

So we left with

**P= X/(1+r/100) ^{2} + X/(1+r/100)**

**Generalizing the formula for EQUAL INSTALLMENTS **

**P= X/(1+r/100) ^{n} + …………………….X/(1+r/100)**

Where x is the installment and n is number of installment

**Example 2: **

**Richa borrowed a sum of Rs. 4800 from Ankita as a loan . She promised Ankita that she will pay it back in two equal installments .If the rate of Interest be 5% per annum compounded annually, find the amount of each installment**

**Solution:**

Given that principal value = 4800

Rate =5%

Two equal installments annually = 2 years

Appling the formula = X/(1+r/100)^{n}…………………….X/(1+r/100)

So we have here two equal installments so

P= X/(1+r/100)^{2} + X/(1+r/100)

4800=X/(1+5/100)^{2} + X/(1+5/100)

On simplifying

We have x= Rs. 2581.46

So the amount of each installment is 2581.46

**Compound Interest Tooltip 8: Application of Compound Interest for concepts of population growth. **

**Case 1: When population growing in a constant rate**

If the rate growth of population increased with a constant, rate then the population after T years will be = P (1+R/100)^{t}

In fact, this is nothing else but an application of the fundamentals of compound interest.

It is actually similar to finding the compound amount after time T years

Net population after T years = = P (1+R/100)^{t}

Net population increase = P [(1+R/100)^{t}– 1]

**Example 3:**

**The population of Chandigarh is increasing at a rate of 4% per annum. The population of Chandigarh is 450000, find the population in 3 years hence.**

Solution:

P = 450000

Rate of increasing = 4%

Time =3 years

Therefore, the total population will be

=> T = P (1+R/100)^{3}

=> T = 450000(1+4/100)^{3}

=> T = 506188

**Case 2: When the population growing with different rates and for different intervals of time **

If the rate growth of population increased with different, rate and for different intervals of time then the population after T years will be =

P (1+R1/100)^{t1 }x (1+R2/100)^{t2}…………………………….. (1+RN/100)^{tn}

Let us take an example for this concept.

**Example 4:**

**The population of Chandigarh increased at a rate of 1% for first year, the rate for second year is 2%, and for third year, it is 3%. Then what will be the population after 3 years if present population of Chandigarh is 45000?**

**Solution:**

Since the rate growth of population increased with different rates for the three difference years, the population after T years will be =

P (1+R1/100)^{t1 }x (1+R2/100)^{t }x (1+R/100)^{tn}

45000(1+1/100)^{1} x (1+2/100)^{1} x (1+3/100)^{1}=47749.77

**Case 3: When the population is decreasing with rate R**

Population after a time period of T years=P (1-R/100)^{t}

Where T is the total population

““““““`R is rate at which the population is decreasing

**Example 5:**

The population of Chandigarh is increases at a rate of 1% for first year, it decreases at the rate of 2% for the second year and for third year it again increases at the rate of 3%. Then what will be the population after 3 years if present population of Chandigarh is 45000.

Solution:

Since the rate growth of population is increasing first and then decreasing for the second year and again it increases for third year, then the population after T years will be =

Present Population: P (1+1/100)^{t1 }x (1-2/100)^{t }x (1+3/100)^{t3}

Present Population: 45000(1+1/100)^{1} x (1-2/100)^{1} x (1+3/100)^{1}=45877.23

**Compound Interest Tooltip 8: ****Negative Compound Interest**

As we can see from the last case above, it is not necessary that there is always an increase in any quantity or amount. There can also be a reduction in the amount of something. This reduction is actually called the rate of depreciation, especially in the financial world. In this case, we do nothing else but take the interest rate to be negative. The formula for this is as follows:

Let P be the original amount.

Let P1 be the new amount at the end of t years.

P1 = P (1-R/100)^{t}

Here R is the rate of interest (negative rate).

Always remember, the rate of depreciation is nothing else but negative rate of interest.

**Example 6:**

**Manpreet bought a new car. The value of the car is Rs. 45000. If rate of depreciation is 10% per annum then what will be the value of the car after 2 years **

**Solution:**

P = 45000

Rate of depreciation = 10%

T = 2 years

Therefore, the value will be after 2 years

= P (1 – R/100)^{t}

= 45000(1-10/100)^{2}

= 36450