Minimum years to more-than-double at 20% compound interest: Find the least number of complete years for which an amount invested at 20% per annum, compounded annually, will exceed double the principal.

Introduction / Context:Doubling under compound interest depends on the growth factor (1 + r)^n. We seek the smallest integer n with (1 + r)^n > 2. This is a classic time-to-double check without logarithms by testing small n values.

Given Data / Assumptions:

• r = 20% per annum, compounded yearly
• Need the least integer n such that amount > 2P

Concept / Approach:Compute powers of 1.20: 1.20^3 and 1.20^4; identify the first to exceed 2. Exact doubling time is ln 2 / ln 1.2 ≈ 3.80, so the least complete year is the ceiling of this value.

Step-by-Step Solution:1.20^3 = 1.728 (less than 2)1.20^4 = 2.0736 (exceeds 2)Therefore, the least complete n is 4 years.

Verification / Alternative check:Using logs: n = ln(2)/ln(1.2) ≈ 0.6931/0.1823 ≈ 3.80 → round up to 4 full years for “more than double.”

Why Other Options Are Wrong:3 years yields 1.728P < 2P; 5 or 6 years are not the least; 2 years is far below doubling at this rate.

Common Pitfalls:Answering 3 years by comparing to simple interest or by stopping at “approximately doubled” instead of strictly exceeding 2P.

Final Answer:4 years