To solve this problem, we’ll use the formula for compound interest:

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

where:

• $A$ is the amount of money in the account after $t$ years,
• $P$ is the initial principal balance (the amount 2 years ago, which we are solving for),
• $r$ is the annual interest rate (2.49% or 0.0249),
• $n$ is the number of times interest is compounded per year (monthly, so $n = 12$),
• $t$ is the time in years (in this case, 1 year and 8 months from two years ago to four months ago).

Step-by-Step Solution

a) Finding the Balance 2 Years Ago

1. Let’s assume the amount two years ago was $P$.
2. We know that four months ago, the balance was $549.23.
3. The time between two years ago and four months ago is 20 months (1 year and 8 months), which is $t = \frac{20}{12} = \frac{5}{3}$ years.

Setting up the equation for compound interest with $A = 549.23$:

$549.23 = P \left(1 + \frac{0.0249}{12}\right)^{12 \times \frac{5}{3}}$

Now, let’s calculate $P$.

b) Finding the Interest Earned

After finding $P$, we can calculate the interest earned between two years ago and four months ago using the formula:

$\text{Interest} = 549.23 - P$

Calculating these values now:Answer: 526.93; units: 22.30

The balance 2 years ago was $526.93, and the interest earned between 2 years ago and 4 months ago was $22.30.

Would you like further details on any part of the solution?

Here are some related questions for further exploration:

1. What would the balance be in the account 3 years from now at the same interest rate?
2. How would a different compounding frequency, such as quarterly, affect the interest earned?
3. If a withdrawal had been made, how would the calculations adjust?
4. What if the interest rate had changed after one year—how would we calculate the new balance?
5. How does compounding monthly compare to compounding annually for the same interest rate?

Tip: Compounding more frequently generally leads to slightly higher interest due to the effect of earning interest on interest more often.