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Julian is saving money and plans on making quarterly contributions into an account earning a quarterly interest rate of 
1.875%. If Julian would like to end up with 
$17,000 after 10 years, how much does he need to contribute to the account every quarter, to the nearest dollar? Use the following formula to determine your answer.

A=d(((1+i)^(n)-1)/(i))

A= the future value of the account after 
n periods

d= the amount invested at the end of each period

i= the interest rate per period

n= the number of periods
Answer:

Julian is saving money and plans on making quarterly contributions into an account earning a quarterly interest rate of 1.875% 1.875 \% . If Julian would like to end up with $17,000 \$ 17,000 after 1010 years, how much does he need to contribute to the account every quarter, to the nearest dollar? Use the following formula to determine your answer.\newlineA=d((1+i)n1i) A=d\left(\frac{(1+i)^{n}-1}{i}\right) \newlineA= A= the future value of the account after n n periods\newlined= d= the amount invested at the end of each period\newlinei= i= the interest rate per period\newlinen= n= the number of periods\newlineAnswer:

Full solution

Q. Julian is saving money and plans on making quarterly contributions into an account earning a quarterly interest rate of 1.875% 1.875 \% . If Julian would like to end up with $17,000 \$ 17,000 after 1010 years, how much does he need to contribute to the account every quarter, to the nearest dollar? Use the following formula to determine your answer.\newlineA=d((1+i)n1i) A=d\left(\frac{(1+i)^{n}-1}{i}\right) \newlineA= A= the future value of the account after n n periods\newlined= d= the amount invested at the end of each period\newlinei= i= the interest rate per period\newlinen= n= the number of periods\newlineAnswer:
  1. Given Values: We are given:\newlineFuture value of the account, A=$17,000A = \$17,000\newlineQuarterly interest rate, i=1.875%i = 1.875\% or 0.018750.01875 (as a decimal)\newlineNumber of years, t=10t = 10\newlineSince contributions are made quarterly, there are 44 periods per year.\newlineNumber of periods, n=4n = 4 periods/year 10* 10 years = 4040 periods\newlineWe need to find the amount invested at the end of each period, dd.\newlineUse the formula A=d((1+i)n1i)A = d\left(\frac{(1+i)^{n}-1}{i}\right) to solve for dd.
  2. Convert Interest Rate: First, convert the interest rate from a percentage to a decimal by dividing by 100100.\newlinei=1.875%100=0.01875i = \frac{1.875\%}{100} = 0.01875
  3. Calculate Number of Periods: Next, calculate the number of periods over 1010 years with quarterly contributions.\newlinen=4n = 4 periods/year 10* 10 years =40= 40 periods
  4. Plug Values into Formula: Now, plug the values of AA, ii, and nn into the formula to solve for dd.A=d((1+i)n1i)A = d\left(\frac{(1+i)^{n}-1}{i}\right)$17,000=d((1+0.01875)4010.01875)\$17,000 = d\left(\frac{(1+0.01875)^{40}-1}{0.01875}\right)
  5. Calculate Factor: Calculate the factor (1+i)n1i\frac{(1+i)^{n}-1}{i}.\newline (1+0.01875)4010.01875=(1.01875)4010.01875\frac{(1+0.01875)^{40}-1}{0.01875} = \frac{(1.01875)^{40}-1}{0.01875}\newline First, calculate (1.01875)40(1.01875)^{40}.\newline (1.01875)402.11356(1.01875)^{40} \approx 2.11356
  6. Subtract from Result: Subtract 11 from the result of (1.01875)40(1.01875)^{40}. 2.1135611.113562.11356 - 1 \approx 1.11356
  7. Divide by Interest Rate: Divide the result by ii (0.018750.01875).\newline1.11356/0.0187559.39041.11356 / 0.01875 \approx 59.3904
  8. Solve for d: Now, solve for d using the calculated factor.\newline17,000=d×59.390417,000 = d \times 59.3904\newlined = 17,00059.3904\frac{17,000}{59.3904}\newlined \approx 286.26286.26
  9. Round to Nearest Dollar: Since the question asks for the nearest dollar, round the result to the nearest whole number. \newlined$286d \approx \$286

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