Here is what a character named Bart has in mind: 1. He wants to buy a car. He ha

Question

Here is what a character named Bart has in mind:

1. He wants to buy a car. He has his eyes on a sleek convertible in metallic blue. The only problem is that he does not have loose cash in the amount of $37000 which is a ticketed price for the beauty.

2. He wants to take a loan. He will put down the 15% of the car’s price as his bank requires for the car loan.

3. His bank (The Second and Third Bank of the Midwest - STBM) offers two kinds of loans: a.

Option A: 5 years, 5% a year in interest, b. Option

B: 7 years, 6.5% a year in interest.

4. Create a table that computes these two options for Bart. Use appropriate formula.

5. Bart is itching to take a loan, to run to the dealer and to let the wind comb his curly hair while driving down his county road for everyone to see. But the idea of making monthly payments of more than $100 for 5 or 7 years does not sit well with him.

6. His Wife, Wilfrieda unexpectedly came to the rescue. She confessed that for the past 5 years she has been setting aside $85 a week and saved it at another bank that offers the best saving account annual interest rate - 4.5%. How much has Wilfrieda saved over the past 5 years? It has been exactly 5 years!

7. You need to help Bart figure the whole plan out: Will Bart’s down payment together with Wilfrieda’s savings be enough to take a loan for which the monthly payments for the loan in option A and B are less than $100.

8. Create a table that computes Wilfrida savings plan AND car loan options as if Wilfrieda’s savings were paid toward the loan on top of Bart’s downpayment of 15%.

Design your tables to your convenience; though clear enough for me to figure them out.

Explanation / Answer

Cost Of Car                37,000.0 Down Payment @15%                5,550.0 Net Loan Required                31,450.0 Formula for loan amortization = A= [i*P*(1+i)^n]/[(1+i)^n-1] Amt $ Option A A = periodical installment ? P=Loan amount = 31,450 i= interest rate per period = 0.4167% per month =5%pa n=total no of payments 60 months =5 years A= [0.004167*31450*1.004167^60]/[1.004167^60-1] A =593.50 So Monthly installment under   Option A = $             593.50 Option B A = periodical installment ? P=Loan amount = 31,450 i= interest rate per period = 0.5417% per month =6.5%pa n=total no of payments 84 months = 7years A= [0.005417*31450*1.005417^84]/[1.005417^84-1] A =467 So Monthly installment under   Option B = $             467.00 The value of Savings of Wilfrieda Formula for Future value of Annuity = FV= A*[(1+k)^n-1]/k A=85 per week n=5 years =5*52=260 weeks k =4.5% pa =0.0865% per week FV = 85*[1.000865^260-1]/0.000865 FV = 24770.50 So The Current Value of the deposit is $                                                         24,770.50 So The Option C is' Wilfrieda's Deposit Amt                24,770.5 Add Bart's downpayment                5,550.0 Total fund Available              30,320.5 Cost Of Car              37,000.0 Loan Required                  6,679.5 Monthly Installments Option A A = periodical installment ? P=Loan amount = 6,679.5 i= interest rate per period = 0.4167% per month =5%pa n=total no of payments 60 months =5 years A= [0.004167*6679.50*1.004167^60]/[1.004167^60-1] A =126.05 So Monthly installment under   Option A = $             126.05 Option B A = periodical installment ? P=Loan amount = 6,679.50 i= interest rate per period = 0.5417% per month =6.5%pa n=total no of payments 84 months = 7years A= [0.005417*6679.50*1.005417^84]/[1.005417^84-1] A =99.19 So Monthly installment under   Option B = $               99.19 Therefore Wilfrieda's savings, Bart's down payment and Option B of Bank Loan can solve the condition of   monthly installment < $100

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---