Ayden deposits $\$ 290$ every month into an account earning a monthly interest rate of $0.7 \%$ . How much would he have in the account after $13$ 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 7

Compound interest

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Q. Ayden deposits

\$ 290

every month into an account earning a monthly interest rate of

0.7 \%

. How much would he have in the account after

13

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: Identify the variables from the problem. $\newline$ We are given: $\newline$ Monthly deposit $d$ = $\$290$ $\newline$ Monthly interest rate $i$ = $0.7\%$ or $0.007$ (as a decimal) $\newline$ Number of years = $13$ $\newline$ Number of periods $n$ = $13$ years $*$ $12$ months/year = $\$290$ $0$ months
Convert interest rate: Convert the interest rate to decimal form. $0.7\%$ as a decimal is $0.7 / 100 = 0.007$
Calculate periods: Calculate the number of periods. $\newline$ Since Ayden deposits monthly and we are looking at a $13$ -year period, we multiply the number of years by the number of months in a year. $\newline$ $n = 13 \text{ years} \times 12 \text{ months/year} = 156 \text{ months}$
Use formula for future value: Use the formula to calculate the future value of the account. $\newline$ $A = d \times \left(\frac{(1 + i)^{n} - 1}{i}\right)$ $\newline$ $A = \$(290) \times \left(\frac{(1 + 0.007)^{156} - 1}{0.007}\right)$
Calculate future value: Calculate the future value using the values from the previous steps. $\newline$ $A = (\$)290 \times \left(\frac{(1 + 0.007)^{156} - 1}{0.007}\right)$ $\newline$ $A = (\$)290 \times \left(\frac{(1.007)^{156} - 1}{0.007}\right)$
Perform calculations: Perform the calculations inside the parentheses first. $\newline$ $(1.007)^{156} - 1$
Calculate exponentiation: Calculate the exponentiation part of the formula. $\newline$ $(1.007)^{156} \approx 2.03009$ (rounded to five decimal places for simplicity)
Subtract from result: Subtract $1$ from the result of the exponentiation. $\newline$ $2.03009 - 1 \approx 1.03009$
Divide by interest rate: Divide the result by the interest rate. $\newline$ $1.03009 / 0.007 \approx 147.15571$
Multiply by deposit amount: Multiply the result by the monthly deposit amount. $\newline$ $A = (\$)290 \times 147.15571$ $\newline$ $A \approx (\$)42675.1559$
Round final answer: Round the final answer to the nearest dollar. $\newline$ $A \approx \$(42675)$ (to the nearest dollar)