A finance company declares that, at a certain compound interest rate, a sum of money deposited by anyone will become 8 times in 3 years. If the same amount is deposited at the same compound rate of interest, then in how many years will it become 16 times?

### Concept & Fractional Cycle Conversion To find the time required for an arbitrary multiplier, first determine the base multiplication factor for a single year (or base period). $$\text{If } Amount = P(1+R/100)^T, \text{ then } Multiplier = (1+R/100)^T$$ ### Step-by-Step Solution 1. **Find the 1-year multiplier:** The sum becomes 8 times in 3 years. $P\left(1 + \frac{R}{100}\right)^3 = 8P$ $\left(1 + \frac{R}{100}\right)^3 = 8$ Taking the cube root of both sides: $1 + \frac{R}{100} = 2$ This means the money doubles (becomes 2 times) every 1 year. 2. **Apply to the new condition:** We want the sum to become 16 times itself. Express 16 as a power of the base 1-year multiplier (2). $16 = 2^4$ 3. **Calculate the time:** Since it takes 1 year to double, it will take $4 \times 1 = 4$ years to become $2^4$ times. ### Exam Strategy & Shortcut Recognize the powers of 2. $8 = 2^3 \rightarrow$ 3 years. Since the exponent perfectly matches the number of years, each power of 2 represents 1 year. $16 = 2^4 \rightarrow$ 4 years. ### Common Pitfall Many students try to solve for the exact rate $R = 100\%$ and then plug it back into the formula $P(1 + 100/100)^T = 16P \implies 2^T = 16$. This works here, but directly comparing powers of the multiplier is a more versatile technique that avoids complex fractional rates in harder problems. ### Final Answer Therefore, the correct answer is **4 years**.