Cook Security Systems has a $$ \$ 37,500 $$ line of credit, which charges an annual percentage rate of prime rate plus $$ 4 \% $$. The starting balance on October 1 was $$ \$ 9,300 $$. On October 4 they made a payment of $$ \$ 1,800 $$. On October 13 the business borrowed $$ \$ 2,900 $$, and on October 19 they borrowed $$ \$ 4,400 $$. If the current prime rate is $$ 6 \% $$, what is the new balance (in $$ \$ $$ )? (Round your answer to the nearest cent.)

Question

UpStudy AI · Accepted Answer

We first note that the line of credit charges an annual rate of  
$$
\text{APR} = \text{prime rate} + 4\% = 6\% + 4\% = 10\%,
$$
so the daily interest rate is  
$$
\text{daily rate} = \frac{0.10}{365}.
$$

The account’s timeline in October is as follows:

1. **October 1–3 (3 days):**  
   The starting balance is $$\$9\,300$$.

2. **October 4:**  
   A payment of $$\$1\,800$$ is made; the balance becomes  
   $$
   9300 - 1800 = 7500.
   $$  
   This new balance is used from October 4 through October 12 (9 days).

3. **October 13:**  
   The business borrows $$\$2\,900$$; the balance increases to  
   $$
   7500 + 2900 = 10400.
   $$  
   This balance is in effect from October 13 through October 18 (6 days).

4. **October 19:**  
   The business borrows $$\$4\,400$$; the balance becomes  
   $$
   10400 + 4400 = 14800.
   $$  
   This final balance is used from October 19 through October 31. Since October has 31 days and the transaction takes effect on the 19th, the interest is calculated over  
   $$
   31 - 19 + 1 = 13\text{ days}.
   $$

Next, we calculate the interest for each period.

- **Period 1 (October 1–3, 3 days):**  
  $$
  I_1 = 9300 \times \left(\frac{0.10}{365}\right) \times 3.
  $$
  
- **Period 2 (October 4–12, 9 days):**  
  $$
  I_2 = 7500 \times \left(\frac{0.10}{365}\right) \times 9.
  $$
  
- **Period 3 (October 13–18, 6 days):**  
  $$
  I_3 = 10400 \times \left(\frac{0.10}{365}\right) \times 6.
  $$
  
- **Period 4 (October 19–31, 13 days):**  
  $$
  I_4 = 14800 \times \left(\frac{0.10}{365}\right) \times 13.
  $$

We now compute each interest amount using the factor  
$$
r = \frac{0.10}{365} \approx 0.0002739726.
$$

1. **Period 1:**  
   $$
   I_1 \approx 9300 \times 0.0002739726 \times 3 \approx 2.5479452 \times 3 \approx 7.6438356.
   $$
   
2. **Period 2:**  
   $$
   I_2 \approx 7500 \times 0.0002739726 \times 9 \approx 2.0547945 \times 9 \approx 18.4927505.
   $$
   
3. **Period 3:**  
   $$
   I_3 \approx 10400 \times 0.0002739726 \times 6 \approx 2.8493151 \times 6 \approx 17.0958906.
   $$
   
4. **Period 4:**  
   $$
   I_4 \approx 14800 \times 0.0002739726 \times 13 \approx 4.0547945 \times 13 \approx 52.7123285.
   $$

Now, sum the interest over the entire month:
$$
I_{\text{total}} = I_1 + I_2 + I_3 + I_4 \approx 7.6438356 + 18.4927505 + 17.0958906 + 52.7123285 \approx 95.9448052.
$$

The final balance at the end of October is the balance after the last borrowing plus the accrued interest:
$$
\text{New Balance} = 14800 + 95.9448052 \approx 14895.94.
$$

Thus, the new balance is $$\$14\,895.94$$.
The new balance is \$14,895.94.

Upstudy AI Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Arithmetic Questions