Before we begin, it's important to remember that the formula for the future value in compound interest is:
[tex]A=P(1+\frac{r}{m})^{mt}[/tex]where r = annual rate in decimal form, P = principal, m = number of conversions in a year, and t = time in a year.
Also, the formula for simple interest is:
[tex]Is=\text{Prt}[/tex]Let's start with Alberto's earnings.
a. $1000 earned 1.2% annual interest compounded monthly.
P = 1,000
r = 0.012
m = 12 (monthly)
t = 10 years
[tex]\begin{gathered} A=1000(1+\frac{0.012}{12})^{12(10)} \\ A=1000(1.127429249) \\ A=1,127.43 \end{gathered}[/tex]Alberto's $1000 became $1, 127.43 after 10 years.
b. $500 lost 2% over the course of 10 years.
[tex]\begin{gathered} Is=500(0.02)(10) \\ Is=100 \end{gathered}[/tex]$500 of Alberto lost $100, therefore, it became $400 only.
c. $500 grew compounded continuously at a rate of 0.8% annually.
[tex]\begin{gathered} A=500(1+\frac{0.008}{1})^{1(10)} \\ A=500(1.082942308) \\ A=541.47 \end{gathered}[/tex]The $500 of Alberto became $541.47.
Therefore, at the end of 10 years, Alberto's money is $1, 127.43 + $400 + $541.47 = $2, 068.90. (Alberto)
Now, let's move to Marie's earnings.
a. $1500 earned 1.4% annual interest compounded quarterly.
P = 1500
r = 1.4% or 0.014
m = 4 (quarterly)
t = 10 years
[tex]\begin{gathered} A=1500(1+\frac{0.014}{4})^{4(10)} \\ A=1500(1.149992672) \\ A=1,724.99 \end{gathered}[/tex]The $1500 of Marie became $1,724.99 after 10 years.
b. $500 gained 4% over the course of 10 years.
[tex]\begin{gathered} Is=\text{Prt} \\ Is=500\times0.04\times10 \\ Is=200 \end{gathered}[/tex]The $500 of Marie gained $200, therefore, it became $700 after 10 years.
To summarize, Marie has $1,724.99 + $700 = $2,424.99 in total after 10 years.
Let's calculate for Hans.
a. $2000 grew compounded continuously at a rate of 0.9% annually.
[tex]\begin{gathered} A=2000(1+\frac{0.009}{1})^{1(10)} \\ A=2000(1.093733873) \\ A=2,187.47 \end{gathered}[/tex]Hans has $2, 187.47 after 10 years.
Lastly, let's see how much Max has.
a. $1000 decrease in value exponentially at a rate of 0.5% annually.
[tex]\begin{gathered} A=1000(1-0.005)^{10}_{}_{} \\ A=1000(0.9511101305) \\ A=951.11 \end{gathered}[/tex]The $1000 of Max became $951.11.
b. $1000 earned 1.8% annual interest compounded biannually (twice a year)
[tex]\begin{gathered} A=1000(1+\frac{0.018}{2})^{2(10)} \\ A=1000(1.196253785) \\ A=1,196.25 \end{gathered}[/tex]The other $1000 of Max became $1, 196.25.
To summarize, after 10 years, Max has $951.11. + $1, 196.25. = $2,147.36.
To summarize all, the children with their respective balance after 10 years is listed below:
Albert = $2, 068.90
Marie = $2,424.99
Hans = $2, 187.47
Max = $2,147.36
Marie has the most money and will be awarded $10,000.
