Compounding - Compare Accounts (Level 2)
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Question
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Maya invested $19,000 in an account paying an interest rate of 8% compounded annually.
Jaxon invested $19,000 in an account paying an interest rate of 83% compounded monthly. To
the nearest hundredth of a year, how much longer would it take for Maya's money to triple than
for Jaxon's money to triple?
Answer Attempt 1 out of 2
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B
May 6
8:38 US



Answer :

To solve this problem, we want to compare the time it takes for both Maya's and Jaxon's investments to triple. To do this, we'll use the compound interest formula: \[ A = P(1 + \frac{r}{n})^{nt} \] Where: - \( A \) is the final amount of money after interest. - \( P \) is the principal amount (the initial amount of money). - \( r \) is the annual interest rate (in decimal form). - \( n \) is the number of times the interest is compounded per year. - \( t \) is the time the money is invested, in years. We need to rearrange the formula to solve for the time (\( t \)) it takes for the investment to triple. We want \( A = 3P \). The rearranged formula for \( t \) is: \[ t = \frac{\log(A/P)}{n \cdot \log(1 + r/n)} \] Let's calculate this for both Maya's and Jaxon's investments. For Maya's investment: Principal \( P = \$19,000 \) Annual interest rate \( r = 8\% = 0.08 \) (converted to a decimal) Compounded annually, so \( n = 1 \). We want the investment to triple, so we set \( A = 3P = 3 \times \$19,000 = \$57,000 \). Using the formula for \( t \): \[ t_{\text{Maya}} = \frac{\log(3)}{\log(1 + 0.08)}\] For Jaxon's investment: Principal \( P = \$19,000 \) Annual interest rate \( r = 83\% = 0.83 \) Compounded monthly, so \( n = 12 \). Again, we want the investment to triple, so \( A = 3P \). Using the formula for \( t \): \[ t_{\text{Jaxon}} = \frac{\log(3)}{12 \cdot \log(1 + \frac{0.83}{12})}\] To find the difference in time, we subtract the two values. Let's calculate these values. For Maya: \[ t_{\text{Maya}} = \frac{\log(3)}{\log(1 + 0.08)}\approx \frac{1.098612}{0.076961} \approx \frac{1.098612}{0.076961} \approx 14.27 \text{ years} \] For Jaxon: \[ t_{\text{Jaxon}} = \frac{\log(3)}{12 \cdot \log(1 + \frac{0.83}{12})} \approx \frac{1.098612}{12 \cdot 0.052456} \approx \frac{1.098612}{0.629472} \approx 1.74 \text{ years} \] Now we find the difference: \[ \text{Time difference} = t_{\text{Maya}} - t_{\text{Jaxon}} \approx 14.27 - 1.74 \approx 12.53 \text{ years} \] Therefore, to the nearest hundredth of a year, it would take approximately 12.53 years longer for Maya's money to triple than for Jaxon's money to triple.