(Simple Interest)
Simple interest word problems refer to applications in which money is invested in an account paying simple interest rather than compounded. The relationship between principal (P), interest rate (r), length of time the money is invested (t), and earned interest (I) is given by the following formula:
I = Prt My approach to solving a simple interest problem runs pretty close to how I approach a mixture or distance problem. As in these other problems, I will organize the information in the form of a chart.
Example 1: Mr. Simone deposits $8000 in one simple interest account and $2000 in a second simple interest account. The interest rate on the $8000 account is 2% more than the rate on the $2000 account. If the total yearly amount of interest on the two accounts is $578, find the interest rate on each account. Step 1: Identify which column you can complete from the information given in the problem (without having to use a variable) and fill it out. In this case, we know the amount invested in each account.
Step 2: Identify what you are looking for and define your variable. Here we are looking for the interest rate on each account. But since they are not the same interest rate we cannot let x represent them both. Arbitrarily I will choose the following:Let x represent the interest rate for the $2000 account.
Note: We want to express the interest rate as a decimal rather than a percentage. Therefore, instead of using 2% (the percent form) we use .02 (the decimal form). Prt = I
Step 4: The equation that will model this problem will now be drawn from the last column
Using the first example as a template, let's try a problem that is only slightly different.
Example 2: Ms. Parker deposits $8000 in one simple interest account and $2000 in a second simple interest account. The interest rate on the $8000 account is 2% more than the rate on the $2000 account. If the total yearly amount of interest earned on the $8000 account is $459 more than the interest earned on the $2000 account, find the interest rate on each account. Now this problem is exactly the like first example up until the time that we develop our equation that models the relationship between the two amounts of earned interest. Therefore, Steps 1 - 3 will be identical to Steps 1 - 3 from Example 1. For convenience, I'll repeat the last table here.
Step 4: Once again, the equation that will model this problem will now be drawn from the last column. The relationship between the two interest rates is given by the following:Interest on $8000 is $459 more than the interest earned on $2000
The last example has to do with what I will call a blended rate. That is, suppose that two or more sums of money are invested at different interest rates. What rate would all of the money have to earn (if invested together) to earn the same amount of interest as in the original scenerio?
Example 3: Suppose that you deposit $3000 at 7.5% (simple interest) and $5000 at 3.2% (simple interest). What is the blended rate? Step 1: Identify which column you can complete from the information given in the problem without having to use a variable and fill it out. In this case, we know the amount invested in each account. Since the amount of time does not change the effective rate, we will assume that the length of time is 1 year.
Therefore, $3000 + $5000 = $8000
Step 3: Since the money is only deposited for one year, Time = 1. Fill out the last column by using the first two columns and following the equation at the top of the chart Prt = I
Step 4: The equation that will model this problem will now be drawn from the last column
Remember that not all simple interest problems follow exactly the same pattern - that is, in fact, one of the qualities that make them both frustrating and challenging. However, since applying algebra to life situations is one of the reasons for the study of college algbera, time spent with applications is expected. The examples given here are just a starting point for your study of simple interest applications - as you continue to work problems you will discover ways in which you need to modify the table or may want to do away with the table entirely. Experiment and be creative.
© 2009 Jo Steig |