At what least number of complete years will a sum of money, invested at 20% per annum compound interest (compounded annually), become more than double its original amount?

Introduction:
This compound interest question focuses on growth over multiple years and asks after how many complete years an investment will become more than double. It tests your understanding of how to use the compound interest multiplier (1 + r/100)^n and how to compare it with a target multiple like 2 times the principal.

Given Data / Assumptions:
Rate of compound interest r = 20% per annum. Interest is compounded annually. We want the smallest whole number of years n such that amount > 2 * principal.

Concept / Approach:
For compound interest with annual compounding, the amount after n years is: A = P * (1 + r/100)^n. We need A > 2P. Since P is common and positive, the condition becomes: (1 + r/100)^n > 2. We substitute r = 20 and test integer values of n until the inequality is satisfied.

Step-by-Step Solution:
Rate per year = 20%, so multiplier = 1 + 20/100 = 1.20. Compute powers of 1.20: For n = 1: (1.2)^1 = 1.20 (less than 2). For n = 2: (1.2)^2 = 1.44 (still less than 2). For n = 3: (1.2)^3 = 1.728 (still less than 2). For n = 4: (1.2)^4 = 1.728 * 1.2 = 2.0736 (now greater than 2). Therefore, after 4 years the amount becomes more than double its original value.

Verification / Alternative check:
To check, assume P = 100 for simplicity. Then after 4 years: A = 100 * (1.2)^4 ≈ 100 * 2.0736 = Rs. 207.36, which is clearly more than double 100. After 3 years, A = 100 * 1.728 = Rs. 172.80, which is still less than double, confirming that 4 complete years are needed.

Why Other Options Are Wrong:
2 years and 2.5 years do not provide enough growth because the multiplier is far below 2. 3 years gives only about 1.728 times the principal, which is still less than double. 5 years is not the least; it does give more than double but we already crossed 2 times at 4 years.

Common Pitfalls:
A typical mistake is to use simple interest logic and set 20 * n ≥ 100 to get n ≥ 5, which ignores compounding. Others may stop at 3 years by approximating incorrectly. Always use the compound interest multiplier and test successive integer years when the question asks for the least number of complete years.

Final Answer:
The least number of complete years required is 4 years.