Solving Compound Interest Problems to Find the Principal Sum

This problem involves calculating the initial principal sum based on the amounts accumulated after different time periods under compound interest compounded annually. We are given the amount after 3 years and the amount after 6 years and need to find the original sum (principal).

Understanding Compound Interest

Compound interest is calculated on the initial principal and also on the accumulated interest of previous periods. The formula for compound interest when compounded annually is:

\(A = P(1 + \frac{R}{100})^t\)

Where:

• \(A\) is the amount after \(t\) years
• \(P\) is the principal sum
• \(R\) is the annual rate of interest
• \(t\) is the time in years

Setting Up the Equations

We are given two pieces of information:

1. Amount after 3 years is Rs. 18,600.
2. Amount after 6 years is Rs. 27,900.

Using the compound interest formula, we can write two equations:

Equation 1 (for t=3 years):

\(18600 = P(1 + \frac{R}{100})^3\)

Equation 2 (for t=6 years):

\(27900 = P(1 + \frac{R}{100})^6\)

Finding the Growth Factor

To eliminate the principal \(P\) and find the growth factor \((1 + \frac{R}{100})\), we can divide Equation 2 by Equation 1:

\(\frac{\text{Amount after 6 years}}{\text{Amount after 3 years}} = \frac{P(1 + \frac{R}{100})^6}{P(1 + \frac{R}{100})^3}\)

\(\frac{27900}{18600} = (1 + \frac{R}{100})^{6-3}\)

\(\frac{279}{186} = (1 + \frac{R}{100})^3\)

Let's simplify the fraction \(\frac{279}{186}\):

• Both numbers are divisible by 3: \(279 \div 3 = 93\), \(186 \div 3 = 62\). The fraction becomes \(\frac{93}{62}\).
• Both 93 and 62 are divisible by 31: \(93 \div 31 = 3\), \(62 \div 31 = 2\). The fraction simplifies to \(\frac{3}{2}\).

So, we have:

\((1 + \frac{R}{100})^3 = \frac{3}{2}\)

Calculating the Principal Sum

Now we can substitute the value of \((1 + \frac{R}{100})^3\) back into Equation 1:

\(18600 = P(1 + \frac{R}{100})^3\)

\(18600 = P \times \frac{3}{2}\)

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

\(P = 18600 \times \frac{2}{3}\)

\(P = (18600 \div 3) \times 2\)

\(P = 6200 \times 2\)

\(P = 12400\)

Final Answer

The principal sum is Rs. 12,400.

The principal amount grows to Rs. 18,600 in 3 years and then from Rs. 18,600 to Rs. 27,900 in the next 3 years. The ratio of amounts after equal intervals of time under compound interest is constant. The ratio from year 3 to year 6 (a 3-year interval) is \(\frac{27900}{18600} = \frac{3}{2}\). This means the amount multiplies by \(\frac{3}{2}\) every 3 years. To find the principal (amount at year 0), we divide the amount at year 3 by the same ratio:

Principal = Amount at Year 3 \(\div\) Ratio

Principal = \(18600 \div \frac{3}{2}\)

Principal = \(18600 \times \frac{2}{3}\)

Principal = \(12400\)

Revision Table: Compound Interest Concepts

Additional Information on Compound Interest Calculations

When solving compound interest problems, especially those involving amounts at different time intervals, understanding the multiplicative nature of compound growth is key. If the amounts are given at times \(t_1\) and \(t_2\) where \(t_2 > t_1\), the growth factor over the period \((t_2 - t_1)\) years is \(\frac{A_{t_2}}{A_{t_1}}\). If \((t_2 - t_1)\) is a multiple of some base period (like 3 years in this case), say \(k \times \Delta t\), then the growth factor over \(\Delta t\) is the \(k\)-th root of \(\frac{A_{t_2}}{A_{t_1}}\).

In this problem, the intervals are 3 years (from 0 to 3) and 3 years (from 3 to 6). The amount grew by a factor of \(\frac{27900}{18600} = \frac{3}{2}\) in the second 3-year period. Therefore, it must have grown by the same factor \(\frac{3}{2}\) in the first 3-year period (from principal to the amount at year 3). This confirms our method of dividing the amount at year 3 by the growth factor over 3 years to find the principal.