Question: Mara and Braylen are both age $75$. When Mara was $25$ years old, she began depositing $1200$ per year into a savings account. She made deposits for the first $10$ years, at which point she was forced to stop making deposits. However, she left her money in the account, where it continued to earn interest for the next $40$ years. Braylen didn't start saving until he was $47$ years old, but for the next $28$ years he made annual deposits of $1200$. Assume that both accounts earned an average annual return of $6\%$ (compounded once a year). Complete parts (a) through (d) below.
Part 1
a. How much money does Mara have in her account at age $75$? At age $75$, Mara has $162592.21$ in her account. (Round to the nearest cent as needed.)
1. **State the problem:** Calculate the amount of money Mara has in her account at age $75$ given she deposited $1200$ per year for $10$ years starting at age $25$, then left the money to grow for $40$ years at an annual interest rate of $6\%$ compounded yearly.
2. **Formula used:** We use the future value of an annuity formula for the deposits period and then compound the total for the remaining years.
- Future value of annuity (deposits period):
$$FV = P \times \frac{(1 + r)^n - 1}{r}$$
where $P = 1200$, $r = 0.06$, $n = 10$.
- Then compound this amount for $40$ years:
$$FV_{total} = FV \times (1 + r)^m$$
where $m = 40$.
3. **Calculate future value of deposits:**
$$FV = 1200 \times \frac{(1 + 0.06)^{10} - 1}{0.06} = 1200 \times \frac{(1.06)^{10} - 1}{0.06}$$
Calculate $(1.06)^{10}$:
$$ (1.06)^{10} = 1.790847 $$
So,
$$FV = 1200 \times \frac{1.790847 - 1}{0.06} = 1200 \times \frac{0.790847}{0.06} = 1200 \times 13.18078 = 15816.94$$
4. **Compound this amount for 40 years:**
$$FV_{total} = 15816.94 \times (1.06)^{40}$$
Calculate $(1.06)^{40}$:
$$ (1.06)^{40} = 10.2857 $$
So,
$$FV_{total} = 15816.94 \times 10.2857 = 162592.21$$
5. **Answer:** Mara has $162592.21$ in her account at age $75$.
This matches the given answer, confirming the calculations.