ICSE Class 6 Maths Chapter 11 Interest

Chapter 11: Interest (Simple Interest)

Basic Concept

Sometimes, when in need, we borrow money. This money could be borrowed from a bank or a money lender. While returning this money after using it for a certain time (period), we pay some additional money with the sum (money) borrowed.

This additional money that is paid for having used the money borrowed is called Interest.

The money we borrow is called Principal or Sum and the total money we return is called Amount, such that:

Amount = Principal + Interest i.e. A = P + I

Rate Percent (R)

It is the interest on ₹ 100 for a specific period of time (in general, for one year).

Examples

1. Rate of interest is 6% per annum means ₹ 6 is the interest on ₹ 100 for one year.

2. Rate of interest is 1% per month means ₹ 1 is the interest on ₹ 100 for one month.

Also, rate of interest = 1% per month = 1% × 12 months = 12% per annum (per year)

3. In the same way, if the rate of interest semi-annually is 4%, ₹ 4 is the interest on ₹ 100 for half-a-year, i.e. for six months

Also, rate of interest = 4% semi-annually = 2 × 4% annually = 8% per annum

Time (T)

It is the period for which the money is borrowed (taken) or lent (given).

Calculating Interest

The value of interest depends on three factors:

(i) Principal (P) (ii) Rate of Interest (R) (iii) Time (T)

And it is calculated by using the formula:

Interest = $\frac{\text{Principal × Rate × Time}}{100}$ i.e. I = $\frac{\text{P × R × T}}{100}$

Interest and simple interest mean the same.

Example 1

Find the interest on ₹ 800 for 3 years at 9% per annum.

Solution

Here, Principal (P) = ₹ 800, Rate (R) = 9% and Time (T) = 3 years.

Interest = $\frac{\text{P × R × T}}{100}$ = $\frac{\text{₹ 800 × 9 × 3}}{100}$ = ₹ 216 (Ans.)

Example 2

Find the simple interest on ₹ 2,500 at 1.5% per month for $1\frac{1}{2}$ years.

Solution

Given P = ₹ 2,500 and R = 1.5% per month,

and time (T) = $1\frac{1}{2}$ years = $\frac{3}{2}$ × 12 months = 18 months,

Interest = $\frac{\text{P × R × T}}{100}$ = $\frac{\text{₹ 2,500 × 1.5 × 18}}{100}$ = ₹ 675 (Ans.)

When the rate of interest is taken per month, the time must also be in months.

Alternative Method

Given P = ₹ 2,500 and R = 1.5% per month,

= 1.5 × 12% per year = 18% per year,

and time T = $1\frac{1}{2}$ years = $\frac{3}{2}$ years

Interest I = $\frac{\text{P × R × T}}{100}$ = $\frac{\text{₹ 2,500 × 18 × 3}}{100\times 2}$ = ₹ 675 (Ans.)

When the rate of interest is taken per year, the time must also be in years.

Teacher's Note

Simple interest is used in everyday banking for savings accounts and personal loans. Understanding how interest accumulates helps you make better financial decisions about borrowing and saving money.

Exercise 11 (A)

1. Find the interest (simple interest) on:

(i) ₹ 200 for 3 years at 6% per annum (p.a.).

(ii) ₹ 800 for 9 months at 1.5 percent per month.

(iii) ₹ 2,000 for 10 months at 12% per year.

(iv) ₹ 460 for 8 months at 5 paise per rupee per month. (5 paise per rupee = 5%)

(v) ₹ 2,450 for 3 years at 12 paise per rupee per year.

2. Rohit borrowed ₹ 4,000 from his friend and agreed to pay him interest at the rate of 15% per year. Find:

(i) the interest to be paid by Rohit in 2 years

(ii) the amount that Rohit must pay at the end of the 2nd year in order to clear his debt.

3. Sheela deposited ₹ 3,600 in a bank for 3 years. If the bank pays interest on this deposit at the rate of 10 percent per annum, find how much money will Sheela get from the bank at the end of 3 years.

4. John lends ₹ 15,000 for 3 years at 8% per annum, and Rahul lends ₹ 25,000 for the same time at 5% per annum. Find:

(i) the interest earned by John in 3 years.

(ii) the interest earned by Rahul in 3 years

(iii) the amount each gets in 3 years.

(iv) the difference of their interests.

(v) the difference of amounts they finally get.

5. A man borrows ₹ 750 at 10% per annum, ₹ 1,200 at 8% per annum, and ₹ 2,000 at 6% per annum. Find the total interest paid by him in 4 years.

Also, find (i) the total sum borrowed and (ii) the total amount the man has to pay at the end of 4 years to clear his debt.

Teacher's Note

These problems reflect real-life borrowing scenarios where understanding interest calculations helps you determine the true cost of loans and make informed financial choices.

Inverse Problems On Simple Interest

The formula Interest = $\frac{\text{Principal × Rate × Time}}{100}$

can be re-written as

(i) Principal = $\frac{\text{100 × Interest}}{\text{Rate × Time}}$, i.e. P = $\frac{\text{100 × I}}{\text{R × T}}$

(ii) Rate% = $\frac{\text{100 × Interest}}{\text{Principal × Time}}$%, i.e. R% = $\frac{\text{100 × I}}{\text{P × T}}$%

(iii) Time = $\frac{\text{100 × Interest}}{\text{Principal × Rate}}$, i.e. T = $\frac{\text{100 × I}}{\text{P × T}}$

When Principal Is Required

Example 3

The interest on a certain loan for 5 years at 6% was ₹ 120. What was the loan?

Solution

Given Rate = 6%, Time = 5 years and Interest = ₹ 120

Principal (Loan) = $\frac{\text{100 × Interest}}{\text{Rate × Time}}$ = $\frac{\text{100 × ₹ 120}}{6\times 5}$ = ₹ 400 (Ans.)

Example 4

Find the principal that will amount to ₹ 1,300 in $2\frac{1}{2}$ years at 12% per annum.

Solution

Let the principal be ₹ 100.

Interest I = $\frac{\text{P × R × T}}{100}$ = $\frac{\text{₹ 100 × 12 × 5}}{100\times 2}$ = ₹ 30

Amount = ₹ 100 + ₹ 30 = ₹ 130

Amount = Principal + Interest

Applying unitary method, we get:

When amount = ₹ 130, principal = ₹ 100

When amount = ₹ 1, principal = ₹ $\frac{100}{130}$

When amount = ₹ 1,300, principal = ₹ $\frac{100}{130}$ × 1,300 = ₹ 1,000 (Ans.)

Alternative (algebraic) Method

Let the principal be ₹ x

Interest I = $\frac{\text{P × R × T}}{100}$ = $\frac{\text{₹ x × 12 × 5}}{100\times 2}$ = ₹ $\frac{\text{3x}}{10}$

Since Principal + Interest = Amount

x + $\frac{\text{3x}}{10}$ = 1300, i.e. $\frac{\text{10x + 3x}}{10}$ = 1300

$\frac{\text{13x}}{10}$ = 1300 and x = 1300 × $\frac{10}{13}$, i.e. x = 1000

Principal = ₹ 1,000 (Ans.)

Teacher's Note

Inverse problems require working backwards from the final amount to find the original principal borrowed, which is essential for understanding loan agreements and retirement planning.

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