At what annual rate of simple interest (in % per annum) will a sum of money double itself in 16 2/3 years?

Introduction:
This question tests the relationship between time and rate for doubling under simple interest. Unlike compound interest, doubling under simple interest is not exponential; it happens when the interest earned equals the principal (SI = P). Using the amount formula A = P * (1 + r*t/100), doubling means A = 2P, which directly gives r*t/100 = 1. The only careful part is converting the mixed fraction time 16 2/3 years into an improper fraction.

Given Data / Assumptions:

• Time t = 16 2/3 years
• Final amount is double the principal: A = 2P
• Simple interest amount relation: A = P * (1 + r*t/100)

Concept / Approach:
Set 2P = P * (1 + r*t/100) and cancel P. This yields r*t/100 = 1, so r = 100/t. Convert t = 16 2/3 into an improper fraction to compute r exactly.

Step-by-Step Solution:
t = 16 2/3 = (16*3 + 2)/3 = 50/3 years Doubling condition: 2P = P * (1 + r*t/100) Cancel P: 2 = 1 + r*t/100 r*t/100 = 1 r = 100 / t = 100 / (50/3) r = 100 * 3 / 50 = 6

Verification / Alternative check:
If r = 6% and t = 50/3, then r*t/100 = (6 * 50/3)/100 = (100)/100 = 1. So A = P(1+1)=2P, confirming doubling exactly.

Why Other Options Are Wrong:
5% would require 20 years to double under simple interest. 4% would require 25 years. 6 2/3% would double sooner than 16 2/3 years. Only 6% fits the exact time given.

Common Pitfalls:
Using compound interest doubling rules (like the rule of 72), mis-converting 16 2/3 years, or forgetting that doubling means SI equals principal (not “amount equals interest”).

Final Answer:
The required annual simple interest rate is 6% per annum.