# Math's: Percent: Interest: Simple and Compound, Rate of Growth and Depreciation

Get unlimited access to the best preparation resource for CBSE : fully solved questions with step-by-step explanation- practice your way to success.

Download PDF of This Page (Size: 203K) ↧

## Interest

Interest is an extra payment from a borrower or to a depositor over and above the original amount borrowed or deposited.

The money originally borrowed or deposited is called the Principal (P), and the extra money paid is called the Interest (I). The total money paid back, that is, the sum of Principal and the Interest is called the Amount.

**Interest are of two types-**

Simple Interest- Simple Interest is only calculated on the principle sum of loan or deposit. The amount of interest remains the same every year.

Compound Interest- Compound Interest is firstly calculated on the principle and then afterwards on the accumulated sum (principal+ interest).

1. Prerna deposited at a bank at an interest rate of per year. How much interest will she earn at the end of 3 years?

Solution-

Prerna will earn at the end of 3 years.

2. Zeba deposited at a bank at an interest rate of 10% compounded annually. Find the amount at the end of 3 years?

Solution-

Zeba’s Amount at the end of 3 years will be .

## Rate of Growth and Depreciation

Growth mean increase in the value or number of something like population, plants, viruses etc. whereas depreciation means decrease in the value of an article such as machinery, crops, motorcycles etc.

If is the value of an article in the beginning and is its value after ’ conversion periods and the rate of growth/depreciation for the period be denoted by , then we can write,

in case of growth, and

in case of depreciation

If the rate of growth/depreciation varies for each conversion period, then,

for growth, and

for depreciation.

3. The population of a city is 9765625. What will be its population after 3 years, if the rate of growth of population is 4% per year?

Solution- Here and

∴

Hence, the population of that city after years will be

4. The cost of machinery is today. What will be the value of machinery after years if it depreciates by p.a.?

Solution- Here and

∴

Hence, the value of the machinery after years will be .