A sum of money doubles itself in 6 years. Find the rate of simple interest per…

Concept: If a sum of money grows to n times itself under simple interest in T years, the total interest earned equals (n − 1) times the principal (since Amount = Principal + Interest). Using SI = (P × R × T) / 100, the rate is given by R = [(n − 1) × 100] / T.

1. Here the sum doubles, so n = 2 and the given time is T = 6 years.

2. The interest earned equals (n − 1) × P = (2 − 1) × P = P, i.e., the interest equals the principal itself.

3. Substitute SI = P and T = 6 into SI = (P × R × T)/100: P = (P × R × 6)/100.

4. Cancel P from both sides: R × 6 = 100, so R = 100/6 = 16.67% per annum, which matches the closest offered option, 16.66% (a rounded/truncated form of the same value).

Cross-check: at R ≈ 16.67% for 6 years, SI = (P × 16.67 × 6)/100 ≈ P, confirming the amount becomes P + P = 2P, i.e., exactly double — consistent with the given condition. Among the offered options, 16.66% is by far the nearest match to the exact value 100/6%, while the others are not.