Finding the Principal Sum with Compound Interest

This problem involves calculating the initial sum of money (principal) given the compound interest earned over a specific period at a certain rate.

Understanding Compound Interest

Compound interest is calculated on the initial principal and also on the accumulated interest of previous periods. The formula for the amount (A) after compounding is:

\( A = P \left(1 + \frac{r}{100}\right)^n \)

Where:

• \( P \) is the principal amount
• \( r \) is the annual rate of interest
• \( n \) is the number of years

The compound interest (CI) is the difference between the amount and the principal:

\( CI = A - P \)

Substituting the formula for A, we get:

\( CI = P \left(1 + \frac{r}{100}\right)^n - P \)

\( CI = P \left[ \left(1 + \frac{r}{100}\right)^n - 1 \right] \)

Applying the Formula to the Given Problem

We are given:

• Compound Interest (CI) = Rs. 3,762
• Rate of Interest (r) = 9% p.a.
• Time (n) = 2 years

We need to find the Principal (P).

Using the compound interest formula:

\( 3762 = P \left[ \left(1 + \frac{9}{100}\right)^2 - 1 \right] \)

Let's simplify the expression inside the brackets:

\( 1 + \frac{9}{100} = 1 + 0.09 = 1.09 \)

Now, square this value:

\( (1.09)^2 = 1.09 \times 1.09 = 1.1881 \)

Subtract 1 from the result:

\( 1.1881 - 1 = 0.1881 \)

Substitute this back into the CI equation:

\( 3762 = P \times 0.1881 \)

To find P, we rearrange the equation:

\( P = \frac{3762}{0.1881} \)

Calculation of the Principal

Let's perform the division:

\( P = \frac{3762}{0.1881} \)

To make the division easier, we can remove the decimal from the denominator by multiplying both the numerator and denominator by 10000 (since there are 4 decimal places):

\( P = \frac{3762 \times 10000}{0.1881 \times 10000} = \frac{37620000}{1881} \)

Now, divide 37620000 by 1881:

\( 37620000 \div 1881 = 20000 \)

So, the principal amount (sum) is Rs. 20,000.

Verification

Let's verify if a principal of Rs. 20,000 at 9% p.a. for 2 years yields a CI of Rs. 3,762.

Amount after 2 years \( A = 20000 \left(1 + \frac{9}{100}\right)^2 = 20000 (1.09)^2 = 20000 \times 1.1881 = 23762 \)

Compound Interest \( CI = A - P = 23762 - 20000 = 3762 \)

This matches the given compound interest, confirming our calculation.

Final Answer

The sum of money is Rs. 20,000.

Revision Table: Compound Interest Concepts

Additional Information: Compound Interest Calculations

Compound interest is a powerful concept in finance. It is used in various applications like bank deposits, loans, and investments.

• The frequency of compounding (annually, semi-annually, quarterly, monthly) affects the final amount. If interest is compounded more frequently than annually, the formula changes slightly: \( A = P \left(1 + \frac{r}{nk}\right)^{nk} \), where \( nk \) is the number of times interest is compounded per year, and \( k \) is the number of years.
• For this problem, compounding is annual (\( nk=1 \times n \)), so the simplified formula \( A = P \left(1 + \frac{r}{100}\right)^n \) is used.
• Understanding the difference between simple and compound interest is crucial. Compound interest grows faster than simple interest over time because the interest itself earns interest.