Bo deposits $\$ 490$ every year into an account earning an annual interest rate of $5.3 \%$ compounded annually. How much would he have in the account after $3$ years, to the nearest dollar? Use the following formula to determine your answer. $\newline$ $A=d\left(\frac{(1+i)^{n}-1}{i}\right)$ $\newline$ $A=$ the future value of the account after $n$ periods $\newline$ $d=$ the amount invested at the end of each period $\newline$ $i=$ the interest rate per period $\newline$ $n=$ the number of periods $\newline$ Answer:

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Math Problems

Grade 8

Compound interest

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Q. Bo deposits

\$ 490

every year into an account earning an annual interest rate of

5.3 \%

compounded annually. How much would he have in the account after

3

years, to the nearest dollar? Use the following formula to determine your answer.

\newline

A=d\left(\frac{(1+i)^{n}-1}{i}\right)

\newline

A=

the future value of the account after

n

periods

\newline

d=

the amount invested at the end of each period

\newline

i=

the interest rate per period

\newline

n=

the number of periods

\newline

Answer:

Identify variables and amounts: Identify the variables from the problem to use in the formula. $\newline$ We have: $\newline$ $d = \$490$ (the amount invested at the end of each period) $\newline$ $i = 5.3\%$ or $0.053$ (the interest rate per period) $\newline$ $n = 3$ (the number of periods)
Convert interest rate to decimal: Convert the interest rate from a percentage to a decimal. $i = 5.3\% = \frac{5.3}{100} = 0.053$
Plug values into formula: Plug the values into the formula to calculate the future value of the account. $\newline$ $A = d \times \left(\frac{(1 + i)^n - 1}{i}\right)$ $\newline$ $A = 490 \times \left(\frac{(1 + 0.053)^3 - 1}{0.053}\right)$
Calculate value inside parentheses: Calculate the value inside the parentheses. $\newline$ $(1 + i)^n = (1 + 0.053)^3$ $\newline$ $(1 + i)^n = 1.053^3$ $\newline$ $(1 + i)^n \approx 1.053 \times 1.053 \times 1.053$ $\newline$ $(1 + i)^n \approx 1.162727$
Subtract one from result: Subtract $1$ from the result obtained in Step $4$ . $\newline$ $(1 + i)^n - 1 \approx 1.162727 - 1$ $\newline$ $(1 + i)^n - 1 \approx 0.162727$
Divide by interest rate: Divide the result from Step $5$ by the interest rate $i$ . $\frac{(1 + i)^n - 1}{i} \approx \frac{0.162727}{0.053}$ $\frac{(1 + i)^n - 1}{i} \approx 3.071075$
Multiply by amount deposited: Multiply the result from Step $6$ by the amount deposited at the end of each period $d$ . $A \approx 490 \times 3.071075$ $A \approx 1504.82675$
Round to nearest dollar: Round the result to the nearest dollar. $A \approx \$1505$