Asked by Irene Godinez on Jun 02, 2024

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Amanda purchased $25,000 vehicle through Mazda's Graduate Program. She would pay monthly payments over 5 years, at a rate of 2.8% compounded monthly. At the end of the third year, Amanda wished to pay off her loan outright. Determine the Balance on the loan at the end of year 3.

A) $10,421.30
B) $10,830.60
C) $11,054.60
D) $11,265.50
E) $11,480.60

Compounded Monthly

Pertains to the process where interest is calculated monthly and added to the initial principal amount of the investment or debt.

Monthly Payments

Regular payments made every month, often in the context of repaying a loan or lease.

  • Examine the structure of loan repayment plans, focusing on computing the outstanding loan amounts following a specified duration of repayments.
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Zybrea KnightJun 06, 2024
Final Answer :
A
Explanation :
To find the balance on the loan at the end of year 3, we first calculate the monthly payment and then use it to find the remaining balance. The loan amount is $25,000, the annual interest rate is 2.8%, compounded monthly, and the loan term is 5 years.1. Calculate the monthly interest rate: 2.8%/12=0.002333...2.8\% / 12 = 0.002333...2.8%/12=0.002333... 2. Calculate the number of monthly payments: 5×12=605 \times 12 = 605×12=60 3. Use the formula for the monthly payment on an amortizing loan: PMT=P×r(1+r)n(1+r)n−1PMT = P \times \frac{r(1+r)^n}{(1+r)^n - 1}PMT=P×(1+r)n1r(1+r)n Where: - PMTPMTPMT is the monthly payment - PPP is the principal loan amount ($25,000) - rrr is the monthly interest rate (0.002333...) - nnn is the total number of payments (60)Plugging in the numbers: PMT=25000×0.002333(1+0.002333)60(1+0.002333)60−1PMT = 25000 \times \frac{0.002333(1+0.002333)^{60}}{(1+0.002333)^{60} - 1}PMT=25000×(1+0.002333)6010.002333(1+0.002333)60PMT≈443.87PMT \approx 443.87PMT443.87 4. Calculate the total number of payments made by the end of year 3: 3×12=363 \times 12 = 363×12=36 5. Calculate the remaining balance using the formula for the balance of an amortizing loan after kkk payments: Bk=P×(1+r)n−(1+r)k(1+r)n−1B_k = P \times \frac{(1+r)^n - (1+r)^k}{(1+r)^n - 1}Bk=P×(1+r)n1(1+r)n(1+r)k Where: - BkB_kBk is the balance after kkk payments - kkk is the number of payments made (36)Plugging in the numbers: B36=25000×(1+0.002333)60−(1+0.002333)36(1+0.002333)60−1B_{36} = 25000 \times \frac{(1+0.002333)^{60} - (1+0.002333)^{36}}{(1+0.002333)^{60} - 1}B36=25000×(1+0.002333)601(1+0.002333)60(1+0.002333)36B36≈10421.30B_{36} \approx 10421.30B3610421.30 Therefore, the balance on the loan at the end of year 3 is approximately $10,421.30, which matches choice A.