Interest is payment for the use of money. There are many ways of charging interest - the usual way is to ask for the original loan back plus an additional percentage of that loan, after a certain period of time. For most loans the money is paid back in regular installments, but there are also loan agreements where all money due is to be returned in one lump sum payment.

Simple interest has that name for a good reason - it is interest paid under one of the simplest possible loan agreements between a lender and a borrower. A certain sum is loaned, called the principal. A rate of interest, expressed as a percentage of the principal over a certain time period, is agreed upon. The interest on the loan is computed based on the principal, interest rate, and length of time of the loan. Finally, the borrower must return to the lender the principal plus the interest according to some prearranged plan.

In our formulas we will refer to the principal as P, the interest rate as r, the interest as I, the time period of the loan as t, and the amount to be returned to the lender as A. To understand the formulas let us look at a specific example. Suppose that Josiah, eager to start a new business venture, asks to borrow \$600 from his good friend Zeb. The obliging Zeb agrees to a loan at a 10% annual interest rate, with the principal plus interest to be repaid in a lump sum payment at the end of 4 years. (Obviously a mistake on Zeb's part - never loan money to a friend!)

The principal is P = \$600, and the interest rate is r = 10% = .10. Since the rate is annual, Josiah owes Zeb 10% of the principal as interest for every year that he uses the money. The interest for each year will be

P · r = \$600 · 10% = \$600 · .10 = \$60  .

The time period is t = 4 years, and so the total interest that Josiah will owe Zeb is 4 times the yearly interest, or

I = P · r · t = \$600 · .10 · 4 = \$60 · 4 = \$240  .

Finally, at the end of the 4 years Josiah must pay back to Zeb the principal plus the interest, or an amount of

A = P + I = \$600 + \$240 = \$840  .

Now suppose that Zeb, a little suspicious of Josiah's frugality, does not trust him to save up the \$840 for return after 4 years, but instead insists that Josiah pay him the amount in equal monthly installments over the length of the loan. As a year has 12 months, the number N of Josiah's payments over 4 years will be

N = 12 · 4 = 48  ,

and the amount of each payment will be the quotient

p = A/N = \$840/48 = \$17.50  .

Sometimes the terminology add-on interest is used to refer to simple interest that is added on to the principal to arrive at the amount to be returned - as in the above example. This financing plan is common in loans for moderately priced consumer items, such as appliances, electronic goods, used cars, etc. In much larger loans, such as for housing or large business expenses, different financing methods are employed; we will discuss larger loans in a later section.

For handy reference we summarize our notation and formulas for simple interest:

 SIMPLE INTEREST SUMMARY P = principal (amount of loan) r = interest rate (expressed as a percentage over a time period) t = time of the loan (expressed as a number of time periods) I = interest = P · r · t A = amount to be paid back = P + I N = number of payments (when there is an installment plan) p = amount of each payment = A/N

The next several examples illustrate the use of these formulas.

example 1 Simon bought a new ukulele from Harry's Music Store for \$450, by making a 20% down payment and financing the remainder at an 8% annual simple interest rate over 18 months. How much are Simon's monthly payments?

Simon's down payment was

20% · \$450 = .20 · \$450 = \$90  ,

so he had to finance the remainder,

P = \$450 − \$90 = \$360  .

The yearly interest rate is r = 8% = .08, and the time of the loan is t = 18 months = 1.5 years. (We change the time to years because the interest rate is per year.) The amount of interest Simon will pay is

I = P · r · t = \$360 · .08 · 1.5 = \$43.20   ,

and the amount of money he will pay back over the life of the loan is the sum

A = P + I = \$360.00 + \$43.20 = \$403.20  .

The number of payments he will make is N = 18, and so his monthly payment will be

p = A/N = \$403.20/18 = \$22.40  .

example 2 Genevieve, much in need of a new stove, purchased one for \$500, paying 10% down and agreeing to 6 monthly payments of \$81 each. How much simple interest will Genevieve pay, and what is the rate of interest of the loan?

Genevieve's down payment was 10% of \$500, or \$50, and she had to finance the difference, \$500 − \$50 = \$450. She makes 6 payments of \$81 each, so the amount she pays back over the life of the loan is 6 · \$81 = \$486. Her interest is the difference between the amount she pays back and the amount financed, or

I = A − P = \$486 − \$450 = \$36  .

To determine the interest rate r, we use the simple interest formula with the values P = \$450, t = 6 months = 1/2 year, and I = \$36. We find that

 I = P · r · t  , r = I / (P · t) = \$36 / (\$450 · 1/2) = .16 = 16%  .

Genevieve must pay interest at the rate of 16% per year (but for only half a year).

example 3 Rebecca borrowed \$12 from Rachel at a simple interest rate of 2% per month, for an indefinite period of time. If she makes no payments, after how long will Rebecca owe Rachel \$15?

The principal is P = \$12, and the interest rate is r = 2% per month. (Since the rate is monthly, we will work in months in this problem.) Rebecca will owe Rachel \$15 when the interest, I, has grown enough that the amount owed, A, is \$15. At this time we have

 A = P + I  , I = A − P = \$15 − \$12 = \$3  .

To determine when the interest is = \$3, we compute

 I = P · r · t t = I / (P · r) = \$3 / (\$12 · .02) = 12.5  .

Rebecca will owe Rachel \$15 after 12½ months.

EXERCISES 4B

1. Doug bought a used bright-red convertible from Windward Ford for only \$6000. He made a 10% down payment, and financed the rest of the cost at an add-on annual interest rate of 12%, with monthly payments spread out over 3 years.
1. How much of the price did Doug have to finance?
2. How much interest will he pay over the life of the loan?
3. What are his monthly payments?

2. If Rebecca borrows \$20 from Rachel at the simple interest rate of 1.5% per month, and if Rebecca makes no payments, how much will she owe Rachel after three years and 4 months? A young couple from Nebraska borrowed \$3000 at simple add-on interest to finance their Hawaiian vacation. In order to pay off the loan, they will make monthly payments of \$150 over 2 years. How much will the couple pay back over the life of the loan? How much interest will they pay? What annual interest rate are they being charged? The same Nebraskan couple could have gotten financing for two and one half years, but they would have been charged a higher interest rate of 12.5% per year. How much interest would they have paid under this alternative financing plan? What would have been their monthly payment under this plan?

1. Bryan, urgently needing a new motorcycle to impress his lady friends, borrowed \$4000 from his father to buy one. His father agreed to an add-on annual interest rate of 5%, to be paid back in semiannual payments for 4 years.
1. Under this plan, how much interest would Bryan pay his father?
2. What would be the total amount that Bryan gives back to his father?
3. What would be the size of Bryan's payments?
4. Regrettably, Bryan lost his job the day after borrowing the money from his father, and he is unable to pay his father anything at all! If he cannot find a job, after how long will Bryan owe his father \$4500? Mr. and Mrs. Kwok bought a TV priced at \$900, but they must pay a sales tax of 4%. They made a 25% down payment on the purchase (including tax), and financed the remainder at 12% annual add-on interest for a period of 3 years. The payments are bimonthly (i.e., every 2 months). What is the size of these payments?