Computing Simple and Compound Interest
Chapter 9: Computing Simple and Compound Interest
The division doesn’t come out evenly, so Delores will pay $123.18 each month for the first 47 months, and then she’ll pay $123.14 for her last payment. How did I figure the last payment? Well, if you multiply $123.18 by 47, you get
$5,789.46 in payments. That leaves $5,912.60 – $5,789.46 = $123.14 for the 48th payment. Sure, it’s only 4 cents, but Delores is a stickler for detail.
Solving for different values using the simple interest formula
The simple interest formula allows you to solve for more than just the amount of interest accumulated. You also can use the formula to find the value of any of the variables — if you have the other three. For instance, you can determine the interest rate from the amount of interest paid. See how in the following example.
Freddy agreed to make quarterly payments of $781.25 for 4 years on a loan of
$10,000. What simple interest rate is he paying?
You first need to determine the total amount of money being repaid. If Freddy is making 4 payments a year for 4 years, you know that he’s making 16 pay- ments of $781.25. So his total repayment is 16 ×$781.25 = $12,500. You also know that he borrowed $10,000, so the interest he’s paying is: $12,500 –
$10,000 = $2,500. The interest amount, $2,500, is the answer to the interest formula, I =Prt.So replace the Iwith $2,500, replace Pwith $10,000, replace t with 4, and then solve for r. Your math should look like this:
, ,
, ,
, ,
, , .
r r
r r
2 500 10 000 4 2 500 40 000 40 000
2 500
40 000 40 000 0 0625
=
=
=
=
^ ^h h
So, as you can see, Freddy is paying 6.25% interest on his loan.
Now try out this example, which asks you to use I= Prtto solve for the princi- pal amount. Say Hank was helping out his aunt with her finances. Auntie Em took out a loan for a new television set 2 years ago and agreed to pay $121.88 each month for 5 years at 121⁄2% interest. How much did the TV cost (before interest is added in)?
The repayment amount is the principal (cost of the TV) plus the interest on the loan. If she’s paying $121.88 per month, that’s 60 payments (5 years ×12 months). So the total she’ll pay back over the 5-year period is $121.88 ×60 =
$7,312.80.
The total, $7,312.80, represents the principal plus the interest (P+ I= $7,312.80).
You replace the Iin this equation with Prt(so you can solve for the cost of the
TV), which gives you P+ Prt= $7,312.80. Now factor out the P, fill in the interest rate and time, and then solve for P,like so:
$ , .
$ , .
$ , .
.
$ , .
.
$ , .
$ , . P Prt
P rt
P rt
7 312 80
1 7 312 80
1 7 312 80
1 0 125 5 7 312 80
1 625 7 312 80 4 500 18 + =
+ =
= +
= +
=
=
^
^ h
h
The principal comes out to be $4,500.18. That’s some jazzy TV set that Auntie Em bought.
Stepping it up a notch: Computing it all with one formula
The formula for simple interest, I = Prt, gives you the amount of interest earned (or to be paid) given an amount of money, an interest rate, and a period of time. You want the amount of interest as a separate number when you’re figuring your expenses or taxes as a part of doing business. But, as I show in the previous section, you frequently want the total amount of money available (or to be repaid) at the end of the time period. You add the principal to the interest to get the total. But guess what? There’s a better way! Here’s a formula that allows you to find the total amount all in one computation:
A= P(1 + rt)
where Ais the total amount obtained from adding the principal to the inter- est, Pis the principal, ris the annual rate written as a decimal, and tis the amount of time in years.
Try your hand at using this formula: Imagine that Casey loaned her business partner $20,000 at 7% simple interest for 3 years. How much will Casey be repaid by the end of the time period?
Using A= P(1 + rt), Casey gets A= $20,000(1 + 0.07 ×3) = $20,000(1.21) =
$24,200. From this total, you can determine the interest paid by subtracting the principal amount from the total: $24,200 – $20,000 = $4,200.
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Chapter 9: Computing Simple and Compound Interest
Taking time into account with simple interest
Loans don’t have to be for whole years. You can borrow money for part of a year or for multiple years plus a fraction of a year. If the time period is half of a year or a quarter of a year or a certain number of months, the computation is pretty clear. You make a fraction of the time and use it in place of the tin I= Prt.
When you get into a number of days, however, the math can get pretty inter- esting. For instance, when dealing with a number of days, the discussion then involves ordinaryor exactsimple interest. I explain each of the scenarios in the following sections.
Calculating simple interest for parts of years
The simple interest formula (which I explain earlier in the chapter) uses the annual interest rate and tas the number of years. When you use a fraction of a year, you essentially reduce the interest rate. For example, to compute the interest on $1 for 1 year at 4% interest, your math looks like this: I =Prt =
$1(0.04)(1) = 0.04. Now, if you take that same $1 at 4% for half a year, you get I=Prt = $1(0.04)(0.5) = 0.02.
It doesn’t matter whether you take half the interest rate or use half a year;
either way you get the same number. You just choose the one that makes the computation easier.
Say Abigail is borrowing $6,000 for 3 months at 8% simple interest. How much does she repay at the end of 3 months?
Using I = Prt, you let the time be 1⁄4of a year (3 months ÷ 12 months = 1⁄4). So you get I= $6,000(0.08)(0.25) = $120. In this case, I replaced the 1⁄4with 0.25 to enter the numbers in the calculator. I also could have taken advantage of the fact that 8% divides evenly by 4 and written I= $6,000(0.02) = $120. In any case, Abigail owes $6,000 + $120 = $6,120 at the end of the 3-month loan period.
Try out another example that stirs things up a bit: Say that Gustav repaid a loan in 30 months by making bimonthly payments of $880. How much did Gustav borrow, if the simple interest rate was 4%?
Bimonthlymeans every other month, so a 30-month loan period has 15 pay- ments. So you multiply the payment amount by the number of payments to get the total payment amount: $880 ×15 = $13,200. The total of all the pay- ments includes both interest and principal, so use this formula to solve for the principal amount:
P(1 + rt) = total repaid
I show you how to use this formula in the earlier section, “Computing simple interest amounts the basic way.” There I also show you how the formula is derived.
Solving for the principal, you let 30 months be 21⁄2years, and then you plug in all the numbers. Your math should look like this:
$ ,
$ ,
. .
$ ,
.
$ ,
$ ,
P rt
P rt
1 13 200
1 13 200
1 0 04 2 5 13 200
1 1 13 200 12 000 + =
= +
= +
=
=
^
^ h
h
So you can see that Gustav borrowed $12,000 21⁄2years ago.
Distinguishing between ordinary and exact simple interest
To liven things up a bit, here in this section I introduce two ways of looking at computing interest: using ordinarysimple interest and using exactsimple interest. You don’t need to be concerned about distinguishing between ordi- naryand exactinterest unless you’re dealing with small periods of time — usually a number of days instead of a number of years.
When computing simple interest in terms of a number of years or parts of years (half a year or a quarter of a year), you use whole numbers or fractions in your calculations. Short-term transactions, on the other hand, may be mea- sured in days. In that case, you have to decide whether to use ordinary or exact interest. The difference between these two is that ordinary interestis calculated based on a 360-day year, and exact interestuses exactlythe number of days in the year — either 365 or 366, depending on whether it’s a leap year.
(See the nearby sidebar, “Don’t forget about the leap years!” for more infor- mation.) The following examples will help you understand how to use both ordinary and exact interest.
What’s the difference between the ordinary interest and the exact interest on a loan of $5,000 at 6% interest for 100 days if the year is 2008?
First, to compute ordinary interest, you divide the number of days by 360 (the number of days in an “ordinary” year), which gives you the fraction you use for the amount of time in the formula. In this case, you divide100 by 360.
Now you plug all your numbers into I= Prt, like this, to get your answer:
$ , . $ $ .
I 5 000 0 06 360
100 300
18
5 . 83 33
= ^ hc m= c m
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Chapter 9: Computing Simple and Compound Interest
Next, you compute the exact interest. To do so, you divide 100 by 366 (because 2008 is a leap year), and then you use that fraction for the amount of time. Now fill in your formula:
$ , . $ $ .
I 5 000 0 06 366
100 300
183
50 . 81 97
= ^ hc m= c m
The difference between the two interest methods is just over $1. The ordinary interest number is more convenient, but the number 360 in the denominator of the fraction doesmake the fraction value larger than with a denominator of 365 or 366. And the difference will become more significant as the amount bor- rowed increases.
Exact interest also is used when the starting and ending dates of a loan are given and you have to figure out the number of days involved. Most desk cal- endars include numbers on each date indicating which day of the year it is.
For example, in the year 2007, November 4 was the 308th day of the year.
I include the year because in 2008, November 4 is the 309th day of the year due to adding the leap year day.
Say, for example, that Timothy borrowed $10,000 on April 17 and paid it back, with exact simple interest of 51⁄4%, on November 22 in the year 2007. What interest was charged on the $10,000 loan?
First refer to a calendar and verify that April 17 was the 107th day in 2007 and November 22 was the 326th day. Find out the total number of days of the loan