ICSE Class 7 Maths Chapter 11 Simple Interest

Chapter 11: Simple Interest

Terms Used

1. Simple Interest: It is the money, which the lender gets from the borrower, in consideration of the sum (money borrowed) used by the borrower. The simple interest (S.I.) and the interest (I) mean the same.

2. Principal (P): It is the sum (money) which the lender gives to the borrower.

3. Rate or Rate of Interest (R): It is the interest for a fixed period on every \(₹ 100\).

Example: (i) Rate of interest is 18% per year means, on \(₹ 100\) the interest in one year is \(₹ 18\).

(ii) Rate is 1.5% per month means, the interest of one month on \(₹ 100\) is \(₹ 1.5\), that is, \(₹ 1.50\).

4. Time Period (T): It is the time for which the sum (principal) is borrowed or lent.

Questions on interest involve four quantities: the Principal (P), the Rate of Interest (R), the Time Period (T) and the Interest (I) of that period. And, all these are related to each other as:

\[Interest = \frac{Principal \times Rate \times Time}{100} \text{, that is, } I = \frac{P \times R \times T}{100}\]

The formula \(I = \frac{P \times R \times T}{100}\) can also be expressed as:

(i) \(P = \frac{I \times 100}{R \times T}\) (ii) \(R = \frac{I \times 100}{P \times T}\) (iii) \(T = \frac{I \times 100}{P \times R}\)

5. Amount (A): It is the sum of the Principal and the Interest on it.

\[Amount = Principal + Interest \text{, that is, } A = P + I\]

Example 1

Find the amount of a loan of \(₹ 3,000\) at 4% per year and for 5 years.

Solution:

Given: \(P = ₹ 3,000\), \(R = 4\%\) and \(T = 5\) years.

\[Interest, I = \frac{P \times R \times T}{100} = ₹ \frac{3,000 \times 4 \times 5}{100} = ₹ 600\]

And, \(Amount = P + I = ₹ 3,000 + ₹ 600 = ₹ 3,600\)

Teacher's Note

When you borrow money from a bank for a car or home, the bank charges simple or compound interest on the principal amount you borrowed, which is why understanding these calculations helps you make informed financial decisions.

Example 2

Find the interest on \(₹ 800\) at 6 percent per month for 9 months.

Solution:

Note: Time is always taken according to the percentage rate, that is, if the rate is given per annum, that is, per year, the time must be taken in years and if the rate is per month, the time must be taken in months.

Since, in this example, Rate = 6% per month. Therefore, Time should be taken in months.

Now, \(P = ₹ 800\), \(R = 6\%\) per month and \(T = 9\) months.

\[\Rightarrow I = \frac{P \times R \times T}{100} = ₹ \frac{800 \times 6 \times 9}{100} = ₹ 432\]

Example 3

Find the S.I. on \(₹ 900\) lent on August 10 and received back on October 22 of the same year, the rate being \(8\frac{3}{4}\%\) per annum (p.a.).

Solution:

Given: \(P = ₹ 900\), \(R = 8\frac{3}{4}\% \text{ p.a. } = \frac{35}{4}\% \text{ p.a.}\)

\[T = \frac{21}{31-10} \text{ (August)} + 30 \text{ (September)} + 22 \text{ (October)} = 73 \text{ days} = \frac{73}{365} \text{ years} = \frac{1}{5} \text{ years}\]

[As rate is per annum]

\[\therefore S.I. = \frac{P \times R \times T}{100} = ₹ \frac{900 \times 35 \times 1}{100 \times 4 \times 5} = ₹ 15.75\]

For finding time, the starting date is not included. In example 2, given above, the starting date is 10th August. Therefore, from total number of days in August, 10 days are subtracted. But the last date, that is, October 22 is included.

Teacher's Note

Banks and lending institutions use this method to calculate the exact interest owed when loans span across multiple months, which is why it's important to understand how to count days correctly in financial transactions.

Example 4

(i) What sum will earn an interest of \(₹ 480\) in 3 years, at 16% per year?

(ii) In what time will \(₹ 2,100\) fetch an interest of \(₹ 525\) at 5% p.a.?

(iii) At what rate percent per year will a sum of money double itself in 10 years?

Solution:

(i) Given: \(I = ₹ 480\), \(T = 3\) years and \(R = 16\%\)

\[\therefore Sum (P) = \frac{I \times 100}{R \times T} = ₹ \frac{480 \times 100}{16 \times 3} = ₹ 1,000\]

(ii) Given: \(P = ₹ 2,100\), \(I = ₹ 525\) and \(R = 5\%\)

\[\therefore Time (T) = \frac{I \times 100}{P \times R} = \frac{525 \times 100}{2,100 \times 5} \text{ years } = 5 \text{ years}\]

(iii) Given: \(P = ₹ 100\), \(A = ₹ 200\)

Then, \(I = ₹ 200 - ₹ 100 = ₹ 100\)

\[\therefore Rate\% (R) = \frac{I \times 100}{P \times T} = \frac{100 \times 100}{100 \times 10} = 10\%\]

Example 5

What sum of money will amount to \(₹ 992\) at 4% in 6 years?

Solution:

Let the sum (principal) be \(₹ 100\).

\[\therefore Interest = \frac{P \times R \times T}{100} = ₹ \frac{100 \times 4 \times 6}{100} = ₹ 24\]

And, amount (A) = P + I = \(₹ 100 + ₹ 24 = ₹ 124\)

\[\therefore When A = ₹ 124; P = ₹ 100\]

[Applying Unitary method]

\[\Rightarrow When A = ₹ 992; P = ₹ \frac{100}{124} \times 992 = ₹ 800\]

Alternative Method:

\[A = P + I \Rightarrow A = P + \frac{P \times R \times T}{100} \text{, that is, } ₹ 992 = P + \frac{P \times 4 \times 6}{100}\]

\[\Rightarrow ₹ 992 = \frac{100 P + 24 P}{100} \text{, that is, } ₹ 992 \times 100 = 124 P\]

\[\Rightarrow P = ₹ \frac{992 \times 100}{124} = ₹ 800\]

Teacher's Note

When you save money in a bank account, the bank calculates the final amount (principal plus interest) you'll have after a certain period, helping you plan for future goals like buying something or paying for education.

Example 6

How long will it take \(₹ 1,500\) to become \(₹ 2,040\) at 8% per annum simple interest?

Solution:

Given: \(P = ₹ 1,500\); \(A = ₹ 2,040\) and \(R = 8\%\)

\[\therefore I = A - P = ₹ 2,040 - ₹ 1,500 = ₹ 540\]

And, so, \[Time (T) = \frac{I \times 100}{P \times R} = \frac{540 \times 100}{1,500 \times 8} = \frac{9}{2} \text{ years } = 4\frac{1}{2} \text{ years}\]

\[4\frac{1}{2} \text{ years } = 4 \text{ years } + \frac{1}{2} \text{ year}\]

\[= 4 \text{ years } + \frac{1}{2} \times 12 \text{ months}\]

\[= 4 \text{ years } + 6 \text{ months}\]

or 4 years and 6 months

Example 7

A invests \(₹ 8,000\) and B invests \(₹ 11,000\) at the same rate of interest per annum. If at the end of 3 years, B gets \(₹ 720\) more interest than A; find the rate of interest.

Solution:

Let the rate of interest = R% per annum.

For A: \(P = ₹ 8,000\) and \(T = 3\) years

\[\therefore I = \frac{P \times R \times T}{100} = \frac{₹ 8,000 \times R \times 3}{100} = ₹ 240 R\]

For B: \(P = ₹ 11,000\) and \(T = 3\) Years

\[\therefore I = \frac{₹ 11,000 \times R \times 3}{100} = ₹ 330 R\]

Since, B gets \(₹ 720\) more interest than A

\[\therefore 330 R - 240 R = 720\]

\[\Rightarrow 90 R = 720 \text{ and } R = \frac{720}{90} = 8\]

\[\Rightarrow Rate \text{ of } interest = 8\%\]

Alternative method:

Since, A invests \(₹ 8,000\) and B invests \(₹ 11,000\)

\[\therefore B \text{ invests } (₹ 11,000 - ₹ 8,000) = ₹ 3,000 \text{ more than } A\]

\[\Rightarrow Interest \text{ of } 3 \text{ years on } ₹ 3,000 = ₹ 720\]

\[\therefore \frac{3000 \times R \times 3}{100} = 720\]

\[\Rightarrow 90 R = 720 \text{ and } R = \frac{720}{90} = 8\]

\[\therefore Rate \text{ of } interest = 8\%\]

Teacher's Note

Different people invest different amounts of money at the same interest rate, so comparing how much interest each person earns helps us understand who earned more and at what rate the money grows.

Exercise 11

1. Find the S.I. and the amount on:

(i) \(₹ 150\) for 4 years at 5% per year.

(ii) \(₹ 350\) for \(3\frac{1}{2}\) years at 8% p.a.

(iii) \(₹ 620\) for 4 months at 8 p per rupee per month.

(iv) \(₹ 3,380\) for 30 months at \(4\frac{1}{2}\%\) p.a.

(v) \(₹ 600\) from July 12 to Dec. 5 at 10% p.a.

(vi) \(₹ 850\) from 10th March to 3rd August at \(2\frac{1}{2}\%\) p.a.

(vii) \(₹ 225\) for 3 years 9 months at 16% p.a.

2. On what sum of money does the S.I. for 10 years at 5% become \(₹ 1,600\)?

3. Find the time in which \(₹ 2,000\) will amount to \(₹ 2,330\) at 11% p.a.?

4. In what time will a sum of money double itself at 8% p.a.

5. In how many years will \(₹ 870\) amount to \(₹ 1,044\), the rate of interest being \(2\frac{1}{2}\%\) p.a.?

6. Find the rate percent, if the S.I. on \(₹ 275\) in 2 years is \(₹ 22\).

7. Find the sum which will amount to \(₹ 700\) in 5 years at 8% p.a.

8. What is the rate of interest, if \(₹ 3,750\) amounts to \(₹ 4,650\) in 4 years?

9. In 4 years, \(₹ 6,000\) amounts to \(₹ 8,000\). In what time will \(₹ 525\) amount to \(₹ 700\) at the same rate?

10. The interest on a sum of money at the end of \(2\frac{1}{2}\) years is \(\frac{4}{5}\) of the sum. What is the rate percent?

11. What sum of money lent out at 5% for 3 years will produce the same interest as \(₹ 900\) lent out at 4% for 5 years?

12. A sum of \(₹ 1,780\) becomes \(₹ 2,136\) in 4 years. Find: (i) the rate of interest. (ii) the sum that will become \(₹ 810\) in 7 years at the same rate of interest?

13. A sum amounts to \(₹ 2,652\) in 6 years at 5% p.a. simple interest. Find: (i) the sum (ii) the time in which the same sum will double itself at the same rate of interest.

14. P and Q invest \(₹ 36,000\) and \(₹ 25,000\) respectively at the same rate of interest per year. If at the end of 4 years, P gets \(₹ 3,080\) more interest than Q; find the rate of interest.

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