Question

maria deposits \\$50,000 into an account that pays 2\\% interest per year, compounded annually.
john deposits \\$50,000 into an account that also pays 2\\% per year. but it is simple interest.

find the interest maria and john earn during each of the first three years.
then decide who earns more interest for each year.
assume there are no withdrawals and no additional deposits.

\

$$\begin{tabular}{|c|c|c|c|} \\hline \\textbf{year} & \\textbf{interest maria earns (interest compounded annually)} & \\textbf{interest john earns (simple interest)} & \\textbf{who earns more interest?} \\\\ \\hline first & \\$ & \\$ & select \\\\ \\hline second & \\$ & \\$ & select \\\\ \\hline third & \\$ & \\$ & select \\\\ \\hline \\end{tabular}$$

Explanation:

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<reasoning>

Calculate John's simple interest for each year

\[

$$\begin{aligned} &P = 50000,\quad r = 0.02\\ &I_{\text{John, Year 1}} = 50000 \times 0.02 = 1000\\ &I_{\text{John, Year 2}} = 50000 \times 0.02 = 1000\\ &I_{\text{John, Year 3}} = 50000 \times 0.02 = 1000 \end{aligned}$$

\]

Calculate Maria's compound interest for each year

\[

$$\begin{aligned} &A_0 = 50000\\ &I_{\text{Maria, Year 1}} = 50000 \times 0.02 = 1000\\ &A_1 = 50000 + 1000 = 51000\\ &I_{\text{Maria, Year 2}} = 51000 \times 0.02 = 1020\\ &A_2 = 51000 + 1020 = 52020\\ &I_{\text{Maria, Year 3}} = 52020 \times 0.02 = 1040.40 \end{aligned}$$

\]

Compare interest earned each year

\[

$$\begin{aligned} &\text{Year 1: } 1000 = 1000 \implies \text{They earn the same amount}\\ &\text{Year 2: } 1020 > 1000 \implies \text{Maria earns more}\\ &\text{Year 3: } 1040.40 > 1000 \implies \text{Maria earns more} \end{aligned}$$

\]
</reasoning>

<answer>

Year	Interest Maria earns (Interest compounded annually)	Interest John earns (Simple interest)	Who earns more interest?
Second	\$1,020	\$1,000	Maria earns more
Third	\$1,040.40	\$1,000	Maria earns more

</answer>

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Answer:

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<reasoning>

Calculate John's simple interest for each year

\[

$$\begin{aligned} &P = 50000,\quad r = 0.02\\ &I_{\text{John, Year 1}} = 50000 \times 0.02 = 1000\\ &I_{\text{John, Year 2}} = 50000 \times 0.02 = 1000\\ &I_{\text{John, Year 3}} = 50000 \times 0.02 = 1000 \end{aligned}$$

\]

Calculate Maria's compound interest for each year

\[

$$\begin{aligned} &A_0 = 50000\\ &I_{\text{Maria, Year 1}} = 50000 \times 0.02 = 1000\\ &A_1 = 50000 + 1000 = 51000\\ &I_{\text{Maria, Year 2}} = 51000 \times 0.02 = 1020\\ &A_2 = 51000 + 1020 = 52020\\ &I_{\text{Maria, Year 3}} = 52020 \times 0.02 = 1040.40 \end{aligned}$$

\]

Compare interest earned each year

\[

$$\begin{aligned} &\text{Year 1: } 1000 = 1000 \implies \text{They earn the same amount}\\ &\text{Year 2: } 1020 > 1000 \implies \text{Maria earns more}\\ &\text{Year 3: } 1040.40 > 1000 \implies \text{Maria earns more} \end{aligned}$$

\]
</reasoning>

<answer>

Year	Interest Maria earns (Interest compounded annually)	Interest John earns (Simple interest)	Who earns more interest?
Second	\$1,020	\$1,000	Maria earns more
Third	\$1,040.40	\$1,000	Maria earns more

</answer>

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