My Math Forum Car Payments

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 January 21st, 2013, 01:22 PM #1 Senior Member   Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 Car Payments Hi, I am stuck on part b of this engineering economics problem. I think I got part a correct. Can anyone give me a hand? Kara borrows \$10,000 to purchase a car she must repay the loan in 48 equal end-of-period payments. Interest is calculated at 1.25% per month. Determine the following: a) What are the nominal annual rate and the effective annual rate? Nominal annual rate: $r= (i)(m) = (1.25%)(12) = 15%$ where i = periodic interest rate and m is the number of periods effective annual rate: $i_a= (1+i)^m -1= (1+0.15)^{12} - 1 = 0.1608 =16.08%$ b) What's the amount of Kara's monthly payment? (this is the part I need help with)
 January 21st, 2013, 09:15 PM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,419 Thanks: 1025 Re: Car Payments 10000 * .0125 / (1 - 1/1.0125^4 = 278.30748... Btw, the lender will tell you while flashing an even row of generous teeth that the rate is 15% annual compounded monthly, but will not add "which is really 16.08% annual"
 January 22nd, 2013, 06:05 AM #3 Senior Member   Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 Re: Car Payments Thank you for the help! Two questions. 1) Is there any chance that "annual" should be looked at as the term of the loan, and I should have used 48 periods instead of 12? 2) Where did you get that formula?
January 22nd, 2013, 07:47 AM   #4
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Re: Car Payments

Quote:
 Originally Posted by aaron-math 1) Is there any chance that "annual" should be looked at as the term of the loan, and I should have used 48 periods instead of 12? 2) Where did you get that formula?
1) No. Always use 12 to get effective rate

2) It's the standard "loan payment" formula
Go here:

 January 22nd, 2013, 08:15 AM #5 Senior Member   Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 Re: Car Payments Got it. I was confused by something the prof said. We were given the formula $F=P[1+i]^n$ Which gives $F=\frac {10,000[1+0.0125]^{48}}{48}= 378.20$
January 22nd, 2013, 08:51 AM   #6
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Re: Car Payments

Quote:
 Originally Posted by aaron-math Got it. I was confused by something the prof said. We were given the formula $F=P[1+i]^n$ Which gives $F=\frac {10,000[1+0.0125]^{48}}{48}= 378.20$
That's the formula for the future value of a deposit, like a 4year term deposit in a Bank.
10,000(1+0.0125)^48 = 18153.548...

Has NOTHING to do with loan payments.

 January 22nd, 2013, 10:05 AM #7 Senior Member   Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 Re: Car Payments awesome. I can't thank you enough. You really helped me understand this.

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