Math Problem Statement

If you deposit $10,000 at 1.85% interest, compounded daily, what would your ending balance be after three years?

Solution

To calculate the ending balance of an investment with daily compounded interest, 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 ending balance (what we're solving for),
  • PP is the principal (initial deposit), which is $10,000,
  • rr is the annual interest rate (as a decimal), which is 1.85%, or 0.0185,
  • nn is the number of times interest is compounded per year (since it's compounded daily, n=365n = 365),
  • tt is the time in years, which is 3 years.

Now, let's plug in the values and calculate the ending balance.

A=10,000(1+0.0185365)365×3A = 10,000 \left( 1 + \frac{0.0185}{365} \right)^{365 \times 3}

I'll calculate this for you.The ending balance after 3 years, with a deposit of $10,000 at an interest rate of 1.85% compounded daily, would be approximately $10,570.68.

Do you want further details on this calculation or have any other questions?

Here are 5 related questions for deeper exploration:

  1. How would the balance change if the interest rate were 2.5% instead?
  2. What would the balance be after 5 years instead of 3 years?
  3. How does compounding frequency (daily vs. monthly) affect the final amount?
  4. What is the formula for simple interest, and how does it differ from compound interest?
  5. How would continuous compounding affect the balance?

Tip: Compounding more frequently increases the final balance, though the effect diminishes as the frequency increases.