We deposit 4000 euros in a bank account with a periodic comp

To solve this problem, we'll calculate the total amount of money in both the bank account and the safe-deposit box for different time periods and find when the total reaches 35,000 euros.

1. Bank Account Calculation:
2. Initial deposit in the bank: 4,000 euros.
3. Annual interest rate: 5%
4. Compounding frequency per year: 12 periods

The formula for compound interest is:
$$A=P{(1+\frac{r}{n})}^{nt}$$

where: - $A$ is the amount of money accumulated after n years, including interest. - $P$ is the principal amount (initial deposit). - $r$ is the annual interest rate (decimal). - $n$ is the number of times that interest is compounded per year. - $t$ is the time in years.

1. Safe-Deposit Box Calculation:
2. Deposit: 5,000 euros annually

3. No interest

4. Total Calculation for Various 't':

We'll calculate the values year by year until the sum of the two parts is 35,000 euros. Please note, the problem has constraints, and based on the scenario, try to find the correct whole-year solution:

Let's try manually for the range stated already between 1 and 6 (ignoring complex intermediate .x decimals which cannot naturally occur year-based without more detail):

• Year 1:

  • Bank: (4000 \left(1 + \frac{0.05}{12}\right)^{12 \times 1} \approx 4,204.02)
  • Safe-Deposit: $5,000$
  • Total: $4,204.02+5,000=9,204.02$

• Year 2:

  • Bank: (4000 \left(1 + \frac{0.05}{12}\right)^{12 \times 2} \approx 4,420.40)
  • Safe-Deposit Total: $10,000$
  • Total: $4,420.40+10,000=14,420.40$

• Year 3:

  • Bank: (4000 \left(1 + \frac{0.05}{12}\right)^{12 \times 3} \approx 4,649.58)
  • Safe-Deposit Total: $15,000$
  • Total: $4,649.58+15,000=19,649.58$

• Year 4:

  • Bank: (4000 \left(1 + \frac{0.05}{12}\right)^{12 \times 4} \approx 4,891.92)
  • Safe-Deposit Total: $20,000$
  • Total: $4,891.92+20,000=24,891.92$

• Year 5:

  • Bank: (4000 \left(1 + \frac{0.05}{12}\right)^{12 \times 5} \approx 5,147.90)
  • Safe-Deposit Total: $25,000$
  • Total: $5,147.90+25,000=30,147.90$

• Year 6:

  • Bank: (4000 \left(1 + \frac{0.05}{12}\right)^{12 \times 6} \approx 5,417.94)
  • Safe-Deposit Total: $30,000$
  • Total: $5,417.94+30,000=35,417.94$

This suggests that the total reaches beyond 35,000 just in 6 full years, so the expected base nearest to full years is around 5.9, which tells consideration pre-end of 6th year.

Therefore, aligning the provided options and calculation: The answer that fits closest to this context 4) t = .7** corresponds to a term within the advancement towards completion of year 6 though it reaches the sum intended. However, in context defined as final total status, the selection appears to base on an in-between numeric estimate mid late fifth starting prior sixth year production.

Please refine further if necessary, or if requires more specified breakdown in different calculations - including whole or fractional periods likewise.