Answered

Pepe and Leo deposit money into their savings accounts at the end of each month. The table shows the account balances.

If their patterns of saving continue, and neither earns interest nor withdraws any of the money, how will the balances compare after a very long time?

\begin{tabular}{|c|c|c|}
\hline
Month & Leo & Pepe \\
\hline
1 & \[tex]$50 & \$[/tex]2 \\
\hline
2 & \[tex]$70 & \$[/tex]4 \\
\hline
3 & \[tex]$90 & \$[/tex]8 \\
\hline
4 & \[tex]$110 & \$[/tex]16 \\
\hline
5 & \[tex]$130 & \$[/tex]32 \\
\hline
\end{tabular}

A. Pepe's balance will be greater.
B. Leo's balance will be greater.
C. The balances will be the same.



Answer :

To determine how Pepe's and Leo's balances will compare after a very long time, let's first analyze the information given in the table and identify the patterns in their savings.

From the table:

[tex]\[ \begin{array}{|c|c|c|} \hline \text{Month} & \text{Leo} & \text{Pepe} \\ \hline 1 & \$50 & \$2 \\ \hline 2 & \$70 & \$4 \\ \hline 3 & \$90 & \$8 \\ \hline 4 & \$110 & \$16 \\ \hline 5 & \$130 & \$32 \\ \hline \end{array} \][/tex]

### Step-by-Step Analysis:

1. Identifying Leo's Pattern:

Leo's balance increases by a fixed amount each month. Let's denote his balance in the [tex]\(n\)[/tex]-th month as [tex]\(B_{\text{Leo}_n}\)[/tex].

- Month 1: [tex]\( \$50 \)[/tex]
- Month 2: [tex]\( \$70 \)[/tex]
- Month 3: [tex]\( \$90 \)[/tex]
- Month 4: [tex]\( \$110 \)[/tex]
- Month 5: [tex]\( \$130 \)[/tex]

His balance increases by \[tex]$20 each month. This indicates an arithmetic sequence where the balance can be expressed as: \[ B_{\text{Leo}_n} = 50 + 20(n - 1) \] 2. Identifying Pepe's Pattern: Pepe's balance doubles each month. Let's denote his balance in the \(n\)-th month as \(B_{\text{Pepe}_n}\). - Month 1: \( \$[/tex]2 \)
- Month 2: [tex]\( \$4 \)[/tex]
- Month 3: [tex]\( \$8 \)[/tex]
- Month 4: [tex]\( \$16 \)[/tex]
- Month 5: [tex]\( \$32 \)[/tex]

This is a geometric sequence where each term is twice the previous term. His balance can be expressed as:
[tex]\[ B_{\text{Pepe}_n} = 2 \cdot 2^{(n - 1)} \][/tex]

3. Comparing the Balances Over Time:

- For Leo, the balance increases linearly with a constant addition of \[tex]$20 each month. - For Pepe, the balance increases exponentially as it doubles every month. Even though Leo starts with a higher balance (\$[/tex]50) compared to Pepe (\$2), the exponential growth of Pepe's balance means that eventually, his balance will surpass Leo's balance.

4. Long-Term Behavior:

- Since exponential growth outpaces linear growth, after a sufficient number of months, Pepe's balance will become greater than Leo's balance. This is evident as [tex]\(2^{(n-1)}\)[/tex] grows much faster than the linear increment of 20 per month.

### Conclusion:

After a very long time, Pepe's balance will be greater than Leo's balance.

Therefore, the correct answer is:

A. Pepe's balance will be greater.