Math Problem Statement

How much would you need to deposit in an account now in order to have $6,000.00 in the account in 16 years? Assume the account earns 9.46% interest compounded daily (excluding leap years).

You need to deposit Incorrect$1320.96 now to have $6,000.00 in the account in 16 years.

Solution

To calculate the amount you need to deposit now (the present value) in order to have $6,000 in 16 years, given a 9.46% annual interest rate compounded daily, we can use the formula for compound interest:

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

Where:

  • AA is the future value ($6,000),
  • PP is the present value (the amount to deposit now),
  • rr is the annual interest rate (9.46%, or 0.0946 as a decimal),
  • nn is the number of times the interest is compounded per year (365 for daily compounding),
  • tt is the number of years (16 years).

Rearranging the formula to solve for PP:

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

Let’s plug in the values and calculate.You would need to deposit $1,320.96 now in order to have $6,000 in the account in 16 years, assuming a 9.46% interest rate compounded daily.

Would you like further details or have any questions?

Here are some related questions for deeper understanding:

  1. How does the compounding frequency affect the amount needed to deposit?
  2. What would the deposit be if the interest were compounded monthly instead of daily?
  3. How would the deposit change if the interest rate was 8% instead of 9.46%?
  4. How does the time period (16 years) influence the initial deposit amount?
  5. What would the deposit be if the target amount was $10,000 instead of $6,000?

Tip: Increasing the frequency of compounding (like from monthly to daily) typically lowers the amount you need to deposit upfront, thanks to interest compounding more frequently.

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Math Problem Analysis

Mathematical Concepts

Compound Interest
Exponential Growth

Formulas

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

Theorems

Compound Interest Formula

Suitable Grade Level

Grades 10-12