# Applications of Compound Interest Formula

There are **some situations where we could use the formula for calculation of amount in CI**.

Here are a few:

(i) **Increase (or decrease) in population**.

(ii) The **growth of a bacteria if the rate of growth is known**.

(iii) The **value of an item, if its price increases or decreases in the intermediate years**.

**Example 9: The population of a city was 20,000 in the year 1997. It increased at the rate of 5% p.a. Find the population at the end of the year 2000**.

**Solution:**

- Increase at 5% =
5 100 × 20000 = - Population in 1999 = 20000 +
= (Treat as the Principal for the 2nd year ) - Population in 2000 =
+ = (Treat as the Principal for the 3rd year) - At the end of 2000 the population =
+ =

**So, at the end of the year 2000, the population is approximately = **

Aruna asked what is to be done if there is a decrease. The teacher then considered the following example.

**Example 10: A TV was bought at a price of Rs. 21,000. After one year the value of the TV was depreciated by 5% (Depreciation means reduction of value due to use and age of the item). Find the value of the TV after one year.**

**Solution:**

- TV was bought at a price of Rs. 21,000.
- Reduction = 5% of ₹
per year - Hence, reduction = ₹
- Value at the end of 1 year = ₹
– ₹ = ₹ - We have found the answer.

### Try these

1.**A machinery worth Rs. 10,500 depreciated by 5%. Find its value after one year**.

To find the value of a machinery after depreciation, we can use the formula for depreciation:

The formula is: V =

Where: P is the **initial value of the machinery** (Rs. 10,500),

r is the **rate of depreciation** (5%),

V is the **value after depreciation**

t is the **number of years**

Substitute the values into the formula:

V =

= 10500 × (1 -

**So, the value of the machinery after one year will be Rs. 9,975**.

2.**Find the population of a city after 2 years, which is at present 12 lakh, if the rate of increase is 4 %**.

To find the population of a city after 2 years given a 4% annual rate of increase, we can use the formula for compound interest, as population growth in this context follows the same principle.

The formula is: P =

Where:

r is the rate of increase (

t is the time in years (

P is the population after t years.

P = 1200000 ×

= 1200000 ×

**So, the population after 2 years will be approximately 12,97,920**.