Given that carbon-14 decays at a constant rate in such a way that it reduces to 50% in 5568 years, find the age of an old wooden piece in which the carbon is only 12.5% of the original.

### Concept & Logic This is a classic application of **Half-Life Decay**. The half-life is the time required for a quantity to reduce to half (50%) of its initial value. In radioactive decay, the amount remaining after $n$ half-lives is given by: $$\text{Remaining Amount} = \text{Initial Amount} \times \left(\frac{1}{2}\right)^n$$ ### Step-by-step Solution * **Given:** * Half-life ($t_{1/2}$) = 5568 years. * Remaining carbon-14 = 12.5% of the original. * **Deduction:** We need to find how many half-lives it takes to reach 12.5%. Let's track the decay sequentially starting from 100%: * After 1 half-life: $100\% \rightarrow 50\%$ * After 2 half-lives: $50\% \rightarrow 25\%$ * After 3 half-lives: $25\% \rightarrow 12.5\%$ * It takes exactly 3 half-lives to reduce the carbon to 12.5%. * **Calculation:** $$\text{Total Age} = \text{Number of half-lives} \times \text{Length of one half-life}$$ $$\text{Total Age} = 3 \times 5568$$ $$\text{Total Age} = 16704 \text{ years}$$ ### Exam Strategy & Shortcut Instead of stepping through percentages manually, convert the remaining percentage directly into a power of $\frac{1}{2}$: $$12.5\% = \frac{12.5}{100} = \frac{1}{8} = \left(\frac{1}{2}\right)^3$$ The exponent $3$ immediately tells you that $3$ half-lives have passed. Multiply $3 \times 5568 = 16704$. ### Common Pitfall A frequent arithmetic mistake occurs when multiplying $3 \times 5568$ under pressure. Instead of full multiplication, use the unit digit method to quickly eliminate options. $3 \times 8 = 24$, so the answer must end in 4. Looking at the options, only (c) 16704 ends in a 4. You don't even need to finish the multiplication! ### Final Answer **Therefore, the correct answer is 16704 years.**