Understanding the Compound Interest Problem

The question asks for the difference in the compound interest earned during the second year and the third year for a specific principal amount and interest rate. We are given:

• Principal (P) = Rs. 15,000
• Rate of Interest (R) = 16% per annum
• Compounding Frequency: Annually

We need to calculate the interest earned specifically in the second year and the interest earned specifically in the third year, and then find the difference between these two amounts.

Calculating Compound Interest Year by Year

Compound interest for a specific year is calculated on the amount accumulated at the end of the previous year. The formula for the amount (A) after n years with principal P and rate R is:

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

The interest earned in the \(n\)-th year is the difference between the amount at the end of \(n\) years and the amount at the end of \((n-1)\) years.

\(CI_{n\text{-th year}} = A_n - A_{n-1}\)

Step-by-Step Calculation

Step 1: Calculate Amount and CI at the end of Year 1

Amount at the end of Year 1 (\(A_1\)):

\(A_1 = P \left(1 + \frac{R}{100}\right)^1\)

\(A_1 = 15000 \left(1 + \frac{16}{100}\right)^1\)

\(A_1 = 15000 (1 + 0.16)\)

\(A_1 = 15000 \times 1.16\)

\(A_1 = 17400\)

Compound Interest for Year 1 (\(CI_1\)):

\(CI_1 = A_1 - P\)

\(CI_1 = 17400 - 15000\)

\(CI_1 = 2400\)

Step 2: Calculate Amount and CI at the end of Year 2

Amount at the end of Year 2 (\(A_2\)):

\(A_2 = P \left(1 + \frac{R}{100}\right)^2\)

\(A_2 = 15000 \left(1 + \frac{16}{100}\right)^2\)

\(A_2 = 15000 (1.16)^2\)

\(A_2 = 15000 \times 1.3456\)

\(A_2 = 20184\)

Compound Interest for Year 2 (interest earned during the second year) (\(CI_{\text{2nd year}}\)):

\(CI_{\text{2nd year}} = A_2 - A_1\)

\(CI_{\text{2nd year}} = 20184 - 17400\)

\(CI_{\text{2nd year}} = 2784\)

Alternatively, the interest for the second year is 16% of the amount at the end of the first year:

\(CI_{\text{2nd year}} = A_1 \times \frac{R}{100}\)

\(CI_{\text{2nd year}} = 17400 \times \frac{16}{100}\)

\(CI_{\text{2nd year}} = 17400 \times 0.16\)

\(CI_{\text{2nd year}} = 2784\)

Step 3: Calculate Amount and CI at the end of Year 3

Amount at the end of Year 3 (\(A_3\)):

\(A_3 = P \left(1 + \frac{R}{100}\right)^3\)

\(A_3 = 15000 \left(1 + \frac{16}{100}\right)^3\)

\(A_3 = 15000 (1.16)^3\)

\(A_3 = 15000 \times 1.560896\)

\(A_3 = 23413.44\)

Compound Interest for Year 3 (interest earned during the third year) (\(CI_{\text{3rd year}}\)):

\(CI_{\text{3rd year}} = A_3 - A_2\)

\(CI_{\text{3rd year}} = 23413.44 - 20184\)

\(CI_{\text{3rd year}} = 3229.44\)

Alternatively, the interest for the third year is 16% of the amount at the end of the second year:

\(CI_{\text{3rd year}} = A_2 \times \frac{R}{100}\)

\(CI_{\text{3rd year}} = 20184 \times \frac{16}{100}\)

\(CI_{\text{3rd year}} = 20184 \times 0.16\)

\(CI_{\text{3rd year}} = 3229.44\)

Step 4: Calculate the Difference

Difference between the compound interest for the second year and the third year is:

\(\text{Difference} = CI_{\text{3rd year}} - CI_{\text{2nd year}}\)

\(\text{Difference} = 3229.44 - 2784\)

\(\text{Difference} = 445.44\)

Summary of Calculations

Difference in CI = CI for 3rd Year - CI for 2nd Year

Difference in CI = Rs. 3229.44 - Rs. 2784 = Rs. 445.44

The difference between the compound interest for the second year and the third year is Rs. 445.44.

Revision Table: Key Compound Interest Concepts

Additional Information on Compound Interest Difference

The difference between the compound interest of consecutive years increases as the amount on which the interest is calculated grows. For annual compounding at rate R, the interest in year \(n\) is \(A_{n-1} \times \frac{R}{100}\), and the interest in year \(n+1\) is \(A_{n} \times \frac{R}{100}\).

The difference is:

\(CI_{(n+1)\text{-th year}} - CI_{n\text{-th year}} = \left(A_n \times \frac{R}{100}\right) - \left(A_{n-1} \times \frac{R}{100}\right)\)

\(= (A_n - A_{n-1}) \times \frac{R}{100}\)

\(= CI_{n\text{-th year}} \times \frac{R}{100}\)

This shows that the difference in CI between two consecutive years is the interest earned on the CI of the earlier year. In this problem:

Difference = \(CI_{\text{3rd year}} - CI_{\text{2nd year}}\) = \(CI_{\text{2nd year}} \times \frac{R}{100}\)

Let's verify this: \(2784 \times \frac{16}{100} = 2784 \times 0.16 = 445.44\). This matches our calculated difference, confirming the relationship.