For a certain principal lent at 10% per annum compound interest, the interest earned during the second year is Rs. 1200. What will be the compound interest during the fourth year on the same principal at the same rate?

Introduction / Context:
This problem asks you to connect the interest earned in one specific year under compound interest with the interest earned in another year. Because the principal effectively grows each year, the interest in later years is larger even though the percentage rate remains the same. The question is designed to test the concept that the interest in each successive year forms a geometric progression when the rate is fixed.

Given Data / Assumptions:

• Rate of interest r = 10% per annum.
• Interest during the second year, I2 = Rs. 1200.
• The same principal and rate continue over multiple years.
• We are asked to find the interest during the fourth year, denoted I4.
• Interest is compounded annually.

Concept / Approach:
Under compound interest, the interest in the nth year is given by P * r * (1 + r) ^ (n - 1), where P is the principal, r is the annual rate in decimal form, and n is the year number. Therefore, the second year interest I2 equals P * r * (1 + r) ^ 1, and the fourth year interest I4 equals P * r * (1 + r) ^ 3. We can first use the expression for I2 to obtain the principal P and then substitute this P into the expression for I4. Alternatively, we can find a direct relationship between I4 and I2 by noting that I4 = I2 * (1 + r) ^ 2.

Step-by-Step Solution:
Step 1: Convert the interest rate to decimal: r = 10% = 0.10. Step 2: The formula for interest in the second year is I2 = P * r * (1 + r) ^ 1. Step 3: Write I2 explicitly: I2 = P * 0.10 * 1.10 = 0.11 * P. Step 4: We are given I2 = 1200, so 1200 = 0.11 * P. Step 5: Solve for P: P = 1200 / 0.11 = 10909.09 approximately, but we will keep it symbolic where needed. Step 6: Interest in the fourth year is given by I4 = P * r * (1 + r) ^ 3. Step 7: Substitute r = 0.10 and simplify: I4 = P * 0.10 * (1.10) ^ 3. Step 8: Note that (1.10) ^ 3 = 1.331, so I4 = 0.10 * 1.331 * P = 0.1331 * P. Step 9: Since P = 1200 / 0.11, I4 = 0.1331 * (1200 / 0.11) = 1200 * (0.1331 / 0.11) = 1200 * 1.21 = 1452.

Verification / Alternative check:
An alternative method is to express I4 directly in terms of I2 without computing P explicitly. We have I2 = P * r * (1 + r) and I4 = P * r * (1 + r) ^ 3. Dividing I4 by I2 gives I4 / I2 = (1 + r) ^ 2. Therefore, I4 = I2 * (1 + r) ^ 2. With r = 0.10, (1 + r) ^ 2 = 1.21. Hence I4 = 1200 * 1.21 = 1452, which agrees with the earlier result and confirms the correctness of the calculation.

Why Other Options Are Wrong:
The option 1320 is smaller than the second year interest and would correspond to a decreasing interest sequence, which is not correct under compound interest. The option 1552 is higher than 1452 and would require a larger multiplier than 1.21, which would not match the 10% rate. The option 1420 is close but still lower than the true value and suggests that the multiplier has been approximated incorrectly. The option 1200 is just the second year interest itself and ignores the growth of the effective principal by the time we reach the fourth year.

Common Pitfalls:
A very common mistake is to assume that each year earns the same interest amount, as in simple interest, and therefore to write I4 equal to I2. Another pitfall is to misapply the power of (1 + r), for example using (1 + r) instead of (1 + r) ^ 2 when moving from year two to year four. Some students also mix up the formula for the amount and the formula for the year wise interest and end up computing total interest for four years instead of the interest specific to the fourth year. Accurate handling of geometric growth is essential in this sort of question.

Final Answer:
The compound interest earned during the fourth year at 10% per annum, given that the second year interest is Rs. 1200, is Rs. 1452, which matches option A.