Principal Invested Calculation with Compound Interest

This problem involves calculating the initial principal amount deposited in a bank account where interest is compounded yearly. We are given the amount accumulated after 3 years and after 6 years.

Let the principal amount be \( P \). Let the annual interest rate be \( r \). The formula for compound interest is:

\( A = P(1 + r)^n \)

where \( A \) is the amount after \( n \) years.

Using Given Information to Formulate Equations

We are given the following information:

• Amount at the end of the 3rd year, \( A_3 = 54000 \)
• Amount at the end of the 6th year, \( A_6 = 64800 \)

Using the compound interest formula, we can write these as:

• Equation 1: \( 54000 = P(1 + r)^3 \)
• Equation 2: \( 64800 = P(1 + r)^6 \)

Finding the Compound Factor over the Period

Notice that the period between the 3rd year and the 6th year is \( 6 - 3 = 3 \) years. The amount at the end of the 6th year is the amount at the end of the 3rd year compounded for another 3 years.

We can also see this by dividing Equation 2 by Equation 1:

\( \frac{A_6}{A_3} = \frac{P(1 + r)^6}{P(1 + r)^3} \)

\( \frac{64800}{54000} = (1 + r)^{6-3} \)

\( \frac{64800}{54000} = (1 + r)^3 \)

Calculating the Growth Factor

Now, let's calculate the value of \( (1 + r)^3 \):

\( (1 + r)^3 = \frac{64800}{54000} \)

We can simplify the fraction by dividing both the numerator and denominator by 100:

\( (1 + r)^3 = \frac{648}{540} \)

Both 648 and 540 are divisible by common factors. Let's divide by 108 (since \( 6 \times 108 = 648 \) and \( 5 \times 108 = 540 \)):

\( (1 + r)^3 = \frac{648 \div 108}{540 \div 108} = \frac{6}{5} \)

So, \( (1 + r)^3 = 1.2 \).

Determining the Principal Amount

We can substitute this value of \( (1 + r)^3 \) back into Equation 1:

\( 54000 = P(1 + r)^3 \)

\( 54000 = P \times 1.2 \)

To find the principal \( P \), we rearrange the equation:

\( P = \frac{54000}{1.2} \)

To divide by 1.2, we can write 1.2 as \( \frac{12}{10} \) or multiply the numerator and denominator by 10:

\( P = \frac{54000 \times 10}{1.2 \times 10} = \frac{540000}{12} \)

Now, perform the division:

\( P = 45000 \)

Therefore, the principal amount invested in the deposit was Rs. 45,000.