Math Problem Statement

How much must be deposited today into the following account in order to have

$ 55 comma 000$55,000

in

55

years for a down payment on a​ house? Assume no additional deposits are made.

An account with monthly compounding and an APR of

88​%

Question content area bottom

Part 1

​$enter your response here

should be deposited today.

​(Do not round until the final answer. Then round to the nearest cent as​ needed.)

Solution

To solve this problem, we need to determine the present value PP that must be deposited today to grow to $55,000 in 55 years with monthly compounding and an APR of 8%. We use the following present value formula for compound interest:

P=A(1+rn)ntP = \frac{A}{(1 + \frac{r}{n})^{nt}}

Where:

  • A=55,000A = 55,000 is the amount we want in the future (the future value).
  • r=0.08r = 0.08 is the annual interest rate (APR = 8%).
  • n=12n = 12 is the number of times the interest is compounded per year (monthly compounding).
  • t=55t = 55 years is the time period.

Let's plug the values into the formula and calculate.

P=55,000(1+0.0812)12×55P = \frac{55,000}{(1 + \frac{0.08}{12})^{12 \times 55}}

I'll compute this now.The amount that must be deposited today is approximately $685.19.

Would you like more details, or do you have any other questions?

Here are 5 related questions:

  1. How would the result change if the APR were 6% instead of 8%?
  2. What is the effect of quarterly compounding rather than monthly compounding?
  3. How much interest would be earned over the 55 years in this scenario?
  4. What would be the future value if $685.19 were deposited today at 8% for only 30 years?
  5. How does simple interest compare to compound interest in long-term investments?

Tip: Compound interest grows faster than simple interest because interest is calculated on both the initial principal and the accumulated interest.

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Math Problem Analysis

Mathematical Concepts

Compound Interest
Present Value Calculation

Formulas

P = A / (1 + r/n)^(nt)

Theorems

Compound Interest Formula

Suitable Grade Level

Grade 10-12 (High School Level)