Car Payments

aaron-math · January 21st, 2013, 01:22 PM

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)

Denis · January 21st, 2013, 09:15 PM

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"

aaron-math · January 22nd, 2013, 06:05 AM

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?

Denis · January 22nd, 2013, 07:47 AM

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:
http://www.google.ca/#hl=en&sugexp=les% ... 24&bih=571

aaron-math · January 22nd, 2013, 08:15 AM

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}= $378.20$

Denis · January 22nd, 2013, 08:51 AM

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.

aaron-math · January 22nd, 2013, 10:05 AM

awesome. I can't thank you enough. You really helped me understand this.

January 21st, 2013, 01:22 PM	#1
aaron-math 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)
$aaron-math is offline$

January 21st, 2013, 09:15 PM	#2
Denis Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038	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"
$Denis is offline$

January 22nd, 2013, 06:05 AM	#3
aaron-math 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?
$aaron-math is offline$

January 22nd, 2013, 08:15 AM	#5
aaron-math 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}= $378.20$
$aaron-math is offline$

January 22nd, 2013, 10:05 AM	#7
aaron-math 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.
$aaron-math is offline$

Thread Tools
$Show Printable Version$ Show Printable Version $Email this Page$ Email this Page
Display Modes
$Linear Mode$ Linear Mode $Hybrid Mode$ Switch to Hybrid Mode $Threaded Mode$ Switch to Threaded Mode

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
Loan outstanding to number of payments made	Russell368	Economics	3	July 7th, 2011 09:50 PM