Least complete years to exceed double at 20% CI: Find the minimum number of complete years for a sum to become more than twice itself at 20% per annum compounded annually.

Introduction / Context:This repeats the classic doubling-time logic under compounding. “More than doubled” means the amount strictly exceeds 2P, not merely reaches or approximates it. We therefore need the smallest integer n with (1.20)^n > 2.

Given Data / Assumptions:

• r = 20% per annum, annual compounding
• We require amount > 2P

Concept / Approach:Compute powers: 1.2^2 = 1.44, 1.2^3 = 1.728, 1.2^4 = 2.0736. The first power strictly above 2 is at n = 4.

Step-by-Step Solution:Check n = 3 → 1.728P < 2P (not enough)Check n = 4 → 2.0736P > 2P (succeeds)Hence, least complete years = 4

Verification / Alternative check:Log method: n = ln(2)/ln(1.2) ≈ 3.80 → need full years, so n = 4.

Why Other Options Are Wrong:2 or 3 years are insufficient; “Data inadequate” is incorrect because the rate is fully specified; 5 years is not the minimum.

Common Pitfalls:Answering 3 years based on SI thinking or misreading “more than doubled” as “at least doubled.”

Final Answer:4 years