ICSE Class 8 Maths Chapter 09 Simple and Compound Interest PDF Download

Chapter 9: Simple And Compound Interest

In the previous classes, you have already learnt about simple interest and other related terms. You have also solved many problems on simple interest. In this chapter, we shall refresh those concepts and solve slightly tougher problems and shall also learn about compound interest.

Simple Interest

Principal. The money borrowed (lent or invested) is called principal.

Interest. The additional money paid by the borrower to the moneylender in lieu of the money used is called interest.

Amount. The total money paid by the borrower to the moneylender is called amount. Thus, amount = principal + interest.

Rate. It is the interest paid on ₹ 100 for specified period.

For example:

(i) Rate of \(6\frac{1}{4}\)% per annum means that the interest paid on ₹ 100 for one year is ₹ \(6\frac{1}{4}\).

(ii) Rate of 1.25% per month means that the interest paid on ₹ 100 for one month is ₹ 1.25.

(iii) Rate of 2.5% per quarterly means that the interest paid on ₹ 100 for 3 months is ₹ 2.5.

However, if the time period for the interest rate is not given, then we shall take the time period as one year.

Time. It is the time for which the money is borrowed (or invested).

Simple interest. It is the interest calculated on the original money (principal) at given rate of interest for any given time.

Simple interest is given by the formula:

\[\text{Simple interest} = \frac{\text{Principal} \times \text{Rate} \times \text{Time}}{100}\]

In solving problems on simple interest, remember the following:

If P denotes the principal, R the rate of interest, T the time for which the money is borrowed (or invested), I (or S.I.) the simple interest and A the amount, then

\[I = \frac{P \times R \times T}{100}\]

\[P = \frac{I \times 100}{R \times T}, \quad R = \frac{I \times 100}{P \times T}, \quad T = \frac{I \times 100}{P \times R}\]

\[A = P + I = P + \frac{P \times R \times T}{100} = \left(1 + \frac{R \times T}{100}\right)P\]

Teacher's Note

Simple interest is used in everyday banking when you open a savings account - the bank calculates interest on your fixed deposit based on the principal amount you deposit.

For counting the time between two given dates, only one of the two dates is counted (either first or last). Usually, we exclude the date of start and include the date of return.

For converting the time in days into years, always divide by 365, whether it is a leap year or not.

The time must be taken in accordance with the interest rate percent. Thus, if the interest rate is per month then time must be taken in months.

Example 1.

Find the simple interest on ₹ 7850 at 7.5% per annum for 3 years 4 months. Also find the amount.

Solution. Here, P (principal) = ₹ 7850, R (rate of interest) = 7.5% p.a.,

T (time) = 3 years 4 months = \(3\frac{4}{12}\) years = \(3\frac{1}{3}\) years = \(\frac{10}{3}\) years.

\[\therefore I \text{ (simple interest)} = \frac{P \times R \times T}{100} = ₹ \frac{7850 \times 7.5 \times \frac{10}{3}}{100}\]

\[= ₹ (785 \times 2.5) = ₹ 1962.50\]

Amount = P + I = ₹ 7850 + ₹ 1962.50 = ₹ 9812.50.

Example 2.

Find the simple interest on ₹ 15840 at \(6\frac{2}{3}\)% per annum from 18th October 2011 to 12th March 2012. Also find the amount.

Solution. Here, P = ₹ 15840, R = \(6\frac{2}{3}\)% p.a. = \(\frac{20}{3}\)% p.a.,

T = October November December January February March

13 + 30 + 31 + 31 + 29 + 12

(31 - 18) (leap year)

= 146 days = \(\frac{146}{365}\) years = \(\frac{2}{5}\) years

\[\therefore I \text{ (simple interest)} = \frac{P \times R \times T}{100} = ₹ \frac{15840 \times \frac{20}{3} \times \frac{2}{5}}{100}\]

\[= ₹ \frac{1584 \times 4}{15} = ₹ 422.40\]

Amount = P + I = ₹ 15840 + ₹ 422.40 = ₹ 16262.40.

Example 3.

What sum of money will fetch ₹ 661.50 as simple interest in one year 9 months at \(6\frac{2}{3}\)% per annum?

Solution. Let the sum of money (principal) be ₹ P.

Interest = ₹ 661.50 = ₹ \(\frac{1323}{2}\), rate = \(6\frac{2}{3}\)% p.a. = \(\frac{20}{3}\)% p.a.,

time = 1 year 9 months = \(1\frac{9}{12}\) years = \(1\frac{3}{4}\) years = \(\frac{7}{4}\) years.

Using \(P = \frac{I \times 100}{R \times T}\), we get

\[P = \frac{\frac{1323}{2} \times 100}{\frac{20}{3} \times \frac{7}{4}} = \frac{1323}{2} \times 100 \times \frac{3}{20} \times \frac{4}{7} = \frac{1323 \times 5 \times 6}{7} = 5670\]

Hence, the required sum of money = ₹ 5670.

Teacher's Note

When borrowing money for education or starting a small business, understanding how to calculate the principal from the interest helps you determine how much you should borrow.