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May 25, 2016
Posted by **masha** on Thursday, July 7, 2005 at 1:12pm.

I have a problem like this: JJ borrowed 21,000 dollars for a car, the finance company gave him an interest rate of 11.4% for 5 years how much will he pay over a 5 year period??

The problem here is that this is not simple interest....the amount he owes changes with each payment, so the interest on the outstanding balance changes also with each payment.

Assuming monthly payments, five years is sixty months. 11.4 percent is 11.4/12 for each month.

A/P = (i (1 + i) n ) / ((1 + i) n - 1)

Where A is the monthly amount of payment, n is the periods (60), i is the interest rate (11.4/12), P is the present value (21000).

A comes out to be 460.79 per month, so the total payments is 27,647.40.

See this explaination...

(Broken Link Removed)

and this calculator...

(Broken Link Removed)

Since your problem is not a simple interest problem but rather a compound interst problem, the following should helpyou determine what you seek.

What is the annual payment required to retire a debt of P dollars in n years if payments start at the end of the first year and bear I% interest compounded annually?

For this typical loan payment calculation, R = i/[1 - (1 +i)^(-n)] where R = the periodic payment, P = the amount borrowed, n = the number of payment periods, and i = I/100.

Example: What is the annual payment required to retire a loan of $10,000 over a period of 5 years at an annual interest rate of 8%? Here, P = 10,000, n = 5, and i = .08 resulting in

R = 10000(.08)/[1 - (1.08)^-5] = $2504.56 per year

The total amount paid back becomes 5(2504.56) = $12,522.82 meaning that the use of the money cost the borrower $2,522.82. It is worthy of note that most loans are paid on a monthly basis. The significance of this to the borrower is that he is paying the money back more often, thus reducing the outstanding balance more rapidly. The effect of this is to reduce the total amount paid for the use of the money. Here, P = 10,000, n = 60, and i = .006666 resulting in

R = 10000(.006666)/[1.006666)^-60] = $202.76 per month.

The total amount paid back becomes 202.76(60) = $12,165.60, a saving of $357.22 by paying monthly.

My apologies for leaving out the P term in the payment formula.

What is the periodic payment required to retire a debt of P dollars in n periods (months or years) if payments start at the end of the first period and bear I% interest compounded periodically?

For this typical loan payment calculation, R = Pi/[1 - (1 +i)^(-n)] where R = the periodic payment, P = the amount borrowed, n = the number of payment periods, and i = I/100.

Example: What is the annual payment required to retire a loan of $10,000 over a period of 5 years at an annual interest rate of 8%? Here, P = 10,000, n = 5, and i = .08 resulting in

R = 10000(.08)/[1 - (1.08)^-5] = $2504.56 per year

The total amount paid back becomes 5(2504.56) = $12,522.82 meaning that the use of the money cost the borrower $2,522.82. It is worthy of note that most loans are paid on a monthly basis. The significance of this to the borrower is that he is paying the money back more often, thus reducing the outstanding balance more rapidly. The effect of this is to reduce the total amount paid for the use of the money. Here, P = 10,000, n = 60, and i = .006666 resulting in

R = 10000(.006666)/[1.006666)^-60] = $202.76 per month.

The total amount paid back becomes 202.76(60) = $12,165.60, a saving of $357.22 by paying monthly.

Simple interest is defined as Principal x Interest Rate per unit time x Number of units of time or I = Pin where I is the interest on a principal P at an interest rate of i per period for n periods. The interest rate is expressed as a decimal. For example, the simple interest for one year on $100 at a 5% annual interest rate (per year) is $100 x .05 x 1 = $5. The interest on $100 at a 5% annual interest rate for two years is $100 x .05 x 2 = $10. The interest on $100 at a 6% annual interest rate for one year, but paid monthly, is $100 x .06/12 x 12 = 50 cents per month x 12 = $6 for the year.

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