On Your Own Practice Calculate the simple interest due and the final balance for each investment. Interest \( = \) Principal \( \times \) Rate \( \times \) Time or \( \mathbf{I}=\mathbf{P} \times \mathbf{R} \times \mathbf{T} \) Example: If you invest \( \$ 11,500 \) in a CD for 10 years with a \( 4 \% \) interest rate, how much will you earn and what is the final balance? \[ \begin{array}{l} \text { and what is the final balance? } \\ I=11,500 \times 0.04 \times 10=\$ 4,600 \text {, Final Balance }=\$ 11,500+\$ 4,600=\$ 16,100 \end{array} \] 1. To buy a laptop, Chris borrowed \( \$ 2,000 \) for 3 years at an annual simple interest rate of \( 5.85 \% \). How much interest will he pay if he pays the entire loan off at the end of the third year? \[ I=2000 \times 0.0585 \times 3=351 \] What is the total amount that Chris will pay for the laptop? \[ 2,351 \] 2. Mari deposited \( \$ 1,500 \) into a retirement account when she was 18 . How much will Mari have in this account after 50 years at a yearly simple interest rate of \( 7.5 \% \) ? \[ I=1,500 \times 0.075 \times 50=5,250 \] 3. Alex invested \( \$ 4,000 \) in a bond at a yearly interest rate of \( 4.75 \% \). He earned \( \$ 427.50 \) in interest. How long was the money invested? \[ I=4,000 x \] 4. Mario borrowed \( \$ 9,000 \) for 10 years to make home improvements. If he repaid a total of \( \$ 20,000 \), what was the interest rate on the loan? 20,000 4,000 \[ \ln T=\$ 11,000 \] \( 11,000=9,000 \times r \times 10 \) \( \frac{11,000}{90,000}=\frac{90,000 \times r}{90,000} \) \( 0.12=r \quad 12 \% \)

Sure! Let's work through each of the practice problems step-by-step using the simple interest formula: \[ \mathbf{I} = \mathbf{P} \times \mathbf{R} \times \mathbf{T} \] Where: - \( \mathbf{I} \) = Interest - \( \mathbf{P} \) = Principal (initial amount) - \( \mathbf{R} \) = Annual interest rate (in decimal) - \( \mathbf{T} \) = Time (in years) --- ### **1. Chris's Loan for a Laptop** **Problem:** Chris borrowed \( \$2,000 \) for 3 years at an annual simple interest rate of \( 5.85\% \). How much interest will he pay, and what is the total amount he will pay for the laptop? **Solution:** 1. **Convert the interest rate to decimal:** \[ R = \frac{5.85}{100} = 0.0585 \] 2. **Calculate the interest:** \[ I = P \times R \times T = 2000 \times 0.0585 \times 3 = 351 \] 3. **Calculate the total amount to be paid:** \[ \text{Total Amount} = P + I = 2000 + 351 = 2351 \] **Answer:** - **Interest Paid:** \$351 - **Total Amount Paid:** \$2,351 --- ### **2. Mari's Retirement Account** **Problem:** Mari deposited \( \$1,500 \) into a retirement account at age 18. How much will Mari have in this account after 50 years at a yearly simple interest rate of \( 7.5\% \)? **Solution:** 1. **Convert the interest rate to decimal:** \[ R = \frac{7.5}{100} = 0.075 \] 2. **Calculate the interest:** \[ I = P \times R \times T = 1500 \times 0.075 \times 50 = 5,250 \] 3. **Calculate the final balance:** \[ \text{Final Balance} = P + I = 1500 + 5250 = 6,750 \] **Answer:** - **Final Balance After 50 Years:** \$6,750 --- ### **3. Alex's Investment in a Bond** **Problem:** Alex invested \( \$4,000 \) in a bond at a yearly interest rate of \( 4.75\% \). He earned \( \$427.50 \) in interest. How long was the money invested? **Solution:** 1. **Given:** \[ I = 427.50, \quad P = 4000, \quad R = 4.75\% = 0.0475 \] 2. **Use the simple interest formula to solve for \( T \):** \[ I = P \times R \times T \\ 427.50 = 4000 \times 0.0475 \times T \] 3. **Calculate \( T \):** \[ T = \frac{427.50}{4000 \times 0.0475} = \frac{427.50}{190} = 2.25 \text{ years} \] 4. **Convert \( 0.25 \) years to months:** \[ 0.25 \times 12 = 3 \text{ months} \] **Answer:** - **Time Invested:** 2 years and 3 months (or 2.25 years) --- ### **4. Mario's Loan for Home Improvements** **Problem:** Mario borrowed \( \$9,000 \) for 10 years to make home improvements. If he repaid a total of \( \$20,000 \), what was the interest rate on the loan? **Solution:** 1. **Calculate the total interest paid:** \[ \text{Total Repayment} = 20,000 \\ \text{Principal} = 9,000 \\ I = \text{Total Repayment} - P = 20,000 - 9,000 = 11,000 \] 2. **Use the simple interest formula to solve for \( R \):** \[ I = P \times R \times T \\ 11,000 = 9,000 \times R \times 10 \] 3. **Calculate \( R \):** \[ R = \frac{11,000}{9,000 \times 10} = \frac{11,000}{90,000} \approx 0.1222 \text{ or } 12.22\% \] *(Rounded to two decimal places.)* **Answer:** - **Interest Rate:** Approximately **12.22%** per annum --- ### **Summary of Answers** 1. **Chris:** - **Interest Paid:** \$351 - **Total Amount Paid:** \$2,351 2. **Mari:** - **Final Balance After 50 Years:** \$6,750 3. **Alex:** - **Time Invested:** 2 years and 3 months 4. **Mario:** - **Interest Rate:** Approximately **12.22%** If you have any more questions or need further clarification on any of these problems, feel free to ask!