Carlos wants to invest his savings in a bank for 3 years. He has [tex]£18000[/tex] to invest and receives offers from two banks.

[tex]\[
\begin{tabular}{|c|}
\hline
\text{Bank of Peace} \\
\hline
3.2\% \text{ per year} \\
\text{Compound interest} \\
\hline
\end{tabular}
\][/tex]

Which bank should Carlos choose to get the most interest over the 3-year period? You must show all your working.

[tex]\[
\begin{tabular}{|c|}
\hline
\text{Bank of Trust} \\
\hline
3.6\% \text{ for } 1^{\text{st}} \text{ year} \\
1.4\% \text{ for each extra year} \\
\text{Compound interest} \\
\hline
\end{tabular}
\][/tex]

[tex]\[
\begin{align*}
1.014 \\
£18648 \times 1.014^2 = 19 \\
£19173.80 + £18648 = £37821.80 \\
£37821.80 - £18000 = \text{interest}
\end{align*}
\][/tex]

Which bank should Carlos choose?

Bank: Bank of Peace



Answer :

Let's solve the problem by calculating the total interest Carlos would earn from each bank over the 3-year period.

First, we will consider the offer from the Bank of Peace:

### Bank of Peace:
- Annual interest rate: 3.2% per year
- Compound interest calculation for 3 years

1. Convert the percentage rate into a decimal:

[tex]\( \text{Rate} = \frac{3.2}{100} = 0.032 \)[/tex]

2. Use the compound interest formula to calculate the future value:

[tex]\[ \text{Future Value} = P \left(1 + r\right)^t \][/tex]

Here, [tex]\( P = 18000 \)[/tex], [tex]\( r = 0.032 \)[/tex], and [tex]\( t = 3 \)[/tex].

3. Calculate:

[tex]\[ \text{Future Value} = 18000 \times \left(1 + 0.032\right)^3 \][/tex]

[tex]\[ \text{Future Value} = 18000 \times 1.032^3 \][/tex]

[tex]\[ \text{Future Value} = 18000 \times 1.099168 \][/tex]

[tex]\[ \text{Future Value} = 19783.885824 \][/tex]

So, the future value with the Bank of Peace is [tex]\( \$19783.885824 \)[/tex].

4. Calculate the interest earned:

[tex]\[ \text{Interest Earned} = \text{Future Value} - \text{Initial Amount} \][/tex]

[tex]\[ \text{Interest Earned} = 19783.885824 - 18000 \][/tex]

[tex]\[ \text{Interest Earned} = 1783.885824 \][/tex]

Now, let's consider the offer from the Bank of Trust:

### Bank of Trust:
- 3.6% interest for the 1st year
- 1.4% interest for the next 2 years
- Compound interest calculation

1. Convert the percentage rates into decimals:

[tex]\[ \text{Rate for 1st year} = \frac{3.6}{100} = 0.036 \][/tex]

[tex]\[ \text{Rate for 2nd and 3rd years} = \frac{1.4}{100} = 0.014 \][/tex]

2. Calculate the future value after each year.

- After the 1st year:

[tex]\[ \text{Amount after Year 1} = P \times \left(1 + \text{Rate for 1st year}\right) \][/tex]

[tex]\[ \text{Amount after Year 1} = 18000 \times \left(1 + 0.036\right) \][/tex]

[tex]\[ \text{Amount after Year 1} = 18000 \times 1.036 \][/tex]

[tex]\[ \text{Amount after Year 1} = 18664.8 \][/tex]

- After the 2nd year:

[tex]\[ \text{Amount after Year 2} = \text{Amount after Year 1} \times \left(1 + \text{Rate for extra years}\right) \][/tex]

[tex]\[ \text{Amount after Year 2} = 18664.8 \times \left(1 + 0.014\right) \][/tex]

[tex]\[ \text{Amount after Year 2} = 18664.8 \times 1.014 \][/tex]

[tex]\[ \text{Amount after Year 2} = 18926.092800000001 \][/tex]

- After the 3rd year:

[tex]\[ \text{Amount after Year 3} = \text{Amount after Year 2} \times \left(1 + \text{Rate for extra years}\right) \][/tex]

[tex]\[ \text{Amount after Year 3} = 18926.092800000001 \times \left(1 + 0.014\right) \][/tex]

[tex]\[ \text{Amount after Year 3} = 18926.092800000001 \times 1.014 \][/tex]

[tex]\[ \text{Amount after Year 3} = 19173.799008 \][/tex]

So, the future value with the Bank of Trust is [tex]\( \$19173.799008 \)[/tex].

3. Calculate the interest earned:

[tex]\[ \text{Interest Earned} = \text{Future Value} - \text{Initial Amount} \][/tex]

[tex]\[ \text{Interest Earned} = 19173.799008 - 18000 \][/tex]

[tex]\[ \text{Interest Earned} = 1173.799008 \][/tex]

### Conclusion
By comparing the interest earned from both banks over 3 years:

- Bank of Peace: [tex]\( \$1783.885824\)[/tex]
- Bank of Trust: [tex]\( \$1173.799008\)[/tex]

Carlos should choose the Bank of Peace to get the most interest over the 3-year period.