Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Christopher is saving money and plans on making monthly contributions into an account earning a monthly interest rate of 
0.5%. If Christopher would like to end up with 
$29,000 after 12 years, how much does he need to contribute to the account every month, 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:

Christopher is saving money and plans on making monthly contributions into an account earning a monthly interest rate of 0.5% 0.5 \% . If Christopher would like to end up with $29,000 \$ 29,000 after 1212 years, how much does he need to contribute to the account every month, 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. Christopher is saving money and plans on making monthly contributions into an account earning a monthly interest rate of 0.5% 0.5 \% . If Christopher would like to end up with $29,000 \$ 29,000 after 1212 years, how much does he need to contribute to the account every month, 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. Identify Given Values: Identify the given values from the problem.\newlineAA (future value of the account) = $29,000\$29,000\newlineii (interest rate per period) = 0.5%0.5\% per month = 0.0050.005 (as a decimal)\newlinenn (number of periods) = 1212 years * 1212 months/year = 144144 months\newlineNow we can use these values in the formula provided.
  2. Plug Values into Formula: Plug the values into the formula to solve for dd (the amount invested at the end of each period).A=d×((1+i)n1i)A = d \times \left(\frac{(1 + i)^{n} - 1}{i}\right)$29,000=d×((1+0.005)14410.005)\$29,000 = d \times \left(\frac{(1 + 0.005)^{144} - 1}{0.005}\right)We need to calculate the right side of the equation to find the value of dd.
  3. Calculate Compound Factor: Calculate the compound factor.\newline(1+i)n=(1+0.005)144(1 + i)^{n} = (1 + 0.005)^{144}\newlineCalculate the power using a calculator.\newline(1+0.005)1442.0398873(1 + 0.005)^{144} \approx 2.0398873
  4. Calculate Numerator: Calculate the numerator of the fraction.\newline((1+i)n1)=2.03988731((1 + i)^{n} - 1) = 2.0398873 - 1\newlineThe numerator is 1.0398873\approx 1.0398873
  5. Divide by Interest Rate: Divide the numerator by the interest rate ii.1.03988730.005=207.97746\frac{1.0398873}{0.005} = 207.97746
  6. Solve for d: Solve for d by dividing AA by the result from Step 55.\newline$29,000/207.97746139.456\$29,000 / 207.97746 \approx 139.456\newlineSince we need to find the monthly contribution to the nearest dollar, we round this to $139\$139.

More problems from Compound interest