Mr Mudd gave his children $2000 each to invest and their investment along with its yield/losses are shown below;
[tex]\begin{gathered} \text{ALBERT} \\ \text{ \$1000 earned 1.2\% annual interest compounded monthly.} \\ \text{This means;} \\ A=P(1+\frac{r}{m})^{m\times t} \\ A=1000(1+\frac{0.012}{12})^{12\times10} \\ A=1000(1+0.001)^{120} \\ A=1000(1.001)^{120} \\ A=1000\times1.127429248861108 \\ A=1127.4294\ldots \\ A=1127.43\text{ (to the nearest hundredth)} \\ \\ \text{ \$500 lost 2\% over the course of 10 years, that means} \\ A=500-(500\times0.02) \\ A=500-10 \\ A=490 \\ \\ \text{ \$500 grew compounded continuously at the rate of 0.8\% annually;} \\ A=Pe^{rt} \\ \text{Note that the variable e is a mathematical constant; } \\ A=500(e^{0.008\times10}) \\ A=500(e^{0.08}) \\ A=500\times1.083287\ldots \\ A=541.6435\ldots \\ A=541.64\text{ (to the nearest hundredth)} \\ \text{Therefore, Albert's total profit from his investment is;} \\ \text{Total}=1127.43+490+541.64 \\ \text{Total}=2159.07 \end{gathered}[/tex][tex]\begin{gathered} \text{MARIE} \\ \text{ \$1500 earned 1.4\% annual interest compounded quarterly.} \\ \text{This means;} \\ A=1500(1+\frac{0.014}{4})^{4\times10} \\ A=1500(1.0035)^{40} \\ A=1500\times1.149992671987681 \\ A=1724.98900\ldots \\ A=1724.99\text{ (to the nearest hundredth)} \\ \\ \text{ \$500 gained }4\text{ \% over the course of 10 years;} \\ A=500\times0.04 \\ A=520 \\ \text{Total}=1724.99+520 \\ \text{Total}=2244.99 \end{gathered}[/tex][tex]\begin{gathered} (HANS)\text{ \$2000 grew compounded continuously at the rate of 0.9\% annually} \\ A=Pe^{rt} \\ A=2000(e^{0.009\times10}) \\ A=2000(e^{0.09}) \\ A=2000\times1.09417428370521 \\ A=2188.3485\ldots \\ A=2188.35\text{ (to the nearest hundredth)} \end{gathered}[/tex][tex]\begin{gathered} \text{MAX} \\ \text{ \$1000 decreased in value exponentially at a rate of 0.5\% annually} \\ A=P(1-r)^t \\ A=1000(1-0.005)^{10} \\ A=1000(0.995)^{10} \\ A=1000\times0.951110130465772 \\ A=951.1101 \\ A=951.11\text{ (to the nearest tenth)} \\ \\ \text{ \$1000 earned 1.8\% annual interest compounded bi-annually (twice a year)} \\ A=1000(1+\frac{0.018}{2})^{2\times10} \\ A=1000(1+0.009)^{20} \\ A=1000(1.009)^{20} \\ A=1000\times1.196253784515613 \\ A=1196.2537 \\ A=1196.25\text{ (to the nearest tenth)} \\ \\ \text{Total}=951.11+1196.25 \\ \text{Total}=2147.36 \end{gathered}[/tex]Finally, the results show that Marie had a total of $2,244.99 at the end of the 10 years. Hence, Marie is $10,000 richer at the end of the competition.
