Question: Cecelia went on a hiking trip. The first day she walked $27$ kilometers. Each day since, she walked $\frac{2}{3}$ of what she walked the day before.
What is the total distance Cecelia has traveled by the end of the 5th day?
Round your final answer to the nearest kilometer.
1. **State the problem:** Cecelia walks $27$ km on the first day. Each subsequent day, she walks $\frac{2}{3}$ of the previous day's distance. We want the total distance walked by the end of day 5.
2. **Identify the type of problem:** This is a geometric series problem where the first term $a = 27$ and the common ratio $r = \frac{2}{3}$.
3. **Formula for the sum of the first $n$ terms of a geometric series:**
$$ S_n = a \frac{1-r^n}{1-r} $$
4. **Apply the formula for $n=5$:**
$$ S_5 = 27 \times \frac{1 - \left(\frac{2}{3}\right)^5}{1 - \frac{2}{3}} $$
5. **Calculate $\left(\frac{2}{3}\right)^5$:**
$$ \left(\frac{2}{3}\right)^5 = \frac{2^5}{3^5} = \frac{32}{243} $$
6. **Substitute back:**
$$ S_5 = 27 \times \frac{1 - \frac{32}{243}}{1 - \frac{2}{3}} = 27 \times \frac{\frac{243}{243} - \frac{32}{243}}{\frac{1}{3}} = 27 \times \frac{\frac{211}{243}}{\frac{1}{3}} $$
7. **Simplify the fraction:**
$$ \frac{\frac{211}{243}}{\frac{1}{3}} = \frac{211}{243} \times 3 = \frac{633}{243} $$
8. **Multiply by 27:**
$$ S_5 = 27 \times \frac{633}{243} $$
9. **Simplify $\frac{27}{243}$:**
$$ \frac{27}{243} = \frac{\cancel{27}^1}{\cancel{27}^9} = \frac{1}{9} $$
10. **Final multiplication:**
$$ S_5 = \frac{1}{9} \times 633 = 70.333... $$
11. **Round to nearest kilometer:**
$$ \boxed{70} \text{ kilometers} $$
**Answer:** Cecelia traveled approximately $70$ kilometers by the end of the 5th day.