How do you solve simple-interest word problems? (2024)

Examples

Purplemath

What is "interest"?

Interest is the price of money. If I borrow money from you, with a promise to pay you back later, that money is no longer available for you to use as you please. In return for the loss of your use of that money, I would pay you a certain amount (usually being a percentage of the amount that I borrowed), over and above the amount what I'd borrowed, in order to pay you for my use of your money.

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What is an "interest rate"?

An interest rate is a rate (usually expressed as a percentage) of the money borrowed, which is to be paid back over and above the borrowed amount. For instance, if I borrow $100 from you and pay you back $108 at the end of a year's time, then the interest rate "per annum" (that is, per year) is 8%.

Interest rates are almost always expressed in terms of years. If I borrowed the $100 from you and paid you back in six months, then I would owe you $104, because six months is half of a year.

What is "simple" interest?

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Simple interest is an amount of money that is a fixed percentage of the amount borrowed, and which is added to the borrowed amount once each interest-rate period. So if I'd borrowed $100 from you and didn't pay it back until two years later, I would owe you an extra $8 for the first year and another extra $8 for the second year, for a total of $116 that I need to pay you.

In pre-algebra or beginning algebra, "investment" word problems usually involve simple annual interest — as opposed to compounded interest. Simple interest is earned on the entire investment amount for a given period of time.

This differs from compounded interest, where simple interest is earned for a smaller amount of time (for instance, for one month). Then, over the next period of time, the interest is earned on the original investment *plus* on the interest that was earned on that first time period. Then, during the third time period, interest is earned on the initial investment amount plus the interest earned during the first two periods. And so forth.

What is an example of a simple-interest situation?

One example of a simple-interest situation is getting a loan for the car you're buying. Another example would be investing in a certificate of deposit (that is, a CD) at your bank. Mortgages and student loans can also be issued in terms of simple interest.

What is the formula for simple interest?

The formula for the simple interest I earned on the amount of an investment (that is, the "principle") P with an interest rate of r over a time period t is I=Prt.

In the simple-interest formula I=Prt, the variable I stands for the interest earned on the original investment, P stands for the amount of the original investment, r is the interest rate (expressed in decimal form), and t is the time (usually in terms of years).

For annual interest (that is, for interest that is stated in terms of a percentage per year), the time t must be stated in years. If they give you a time of, say, nine months, you must first convert this to years. Otherwise, you'll get the wrong answer.

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How do you solve simple-interest word problems? (2)

The time units, in all cases, must match the interest-rate units. For instance, if you got a loan from your friendly neighborhood loan shark, where the interest rate is monthly, rather than yearly, then your time must be measured in terms of months.

When working on investment word problems, you will want to substitute all given information into the I=Prt equation, and then solve for whatever is left.

  • You put $1000 into an investment yielding 6% annual interest; you left the money in for two years. How much interest do you get at the end of those two years?

The invested amount (that is, the principal) of my investment is P=$1000, the interest rate (expressed in decimal form) is r=0.06 per year, and the amount of time is t=2. Substituting these values into the simple-interest formula, I get:

I = (1000)(0.06)(2) = 120

I will get $120 in interest.

  • You invested $500 and received $650 after three years. What had been the interest rate?

For this exercise, I first need to find the amount of the interest. Since simple interest is added to the principal, and since the principal was P=$500, then the interest is I =$650−500 =$150. The time is t=3. Substituting all of these values into the simple-interest formula, I get:

150 = (500)(r)(3)

150 = 1500r

Of course, I need to remember to convert this decimal to a percentage.

I was getting 10% interest.

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When you have just one account, one simple situation, requiring one use of the simple-interest formula, it's pretty easy to set up and solve the exercises. The hard part comes when the exercises involve multiple investments or some other complication. But there is a trick to these that makes them fairly easy to handle. You use a simple table or grid.

  • You have $50,000 to invest, and two funds into which you'd like to place those monies. The You-Risk-It Fund (Fund Y) yields 14% interest. The Extra-Dull Fund (Fund X) yields 6% interest. You'd like to earn as much as possible but, because of college financial-aid implications, you don't think you can afford to earn more than $4,500 in interest income this year. How much should you put in each fund?

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The problem here comes from the fact that I'm splitting the $50,000 in principal into two smaller amounts. Here's how to handle this:

I will make a table. The top row has, as its entries, the variables in the simple-interest formula. The left-hand column labels the two funds — and, in case I need it, a "totals" row. (Not all exercises will need the "totals" row.)

I can't add rates — that's just a thing; rates can't be added — and each of the investments is for one year, so there are no "total" values for the interest rates or the years. Hence, the dashes, to remind me not to try to put anything in there.

I P r t
Fund X
Fund Y
total 4,500 50,000

I know the interest total that I'm aiming for, and I know the total amount that I'm investing, so I can enter "total" values for the "interest" and "investment" columns.

I know the interest rates and the time (namely, one year) for the two investments, so I can enter these values in the "rate" and "time" columns, in each fund's row.

Putting it all together, I get the following start to my set-up:

I P r t
Fund X ? ? 0.06 1
Fund Y ? ? 0.14 1
total 4,500 50,000

How do I fill in for those question marks? I'll start with the principal P. Let's say that I put x dollars into Fund X, and y dollars into Fund Y. Then x+y= 50,000.

But this doesn't help much, since I only know how to solve equations in one variable. However, I then notice that I can solve x+y =50,000 to get y =50,000−x.

THIS TECHNIQUE IS IMPORTANT! The amount left for me to put into Fund Y is (the total that I invested) less (what I've already put into Fund X), or 50,000−x.

Any time you have a total that is divided into two parts, you can designate one part as (one part), and the rest will be (the total) minus the (one part), because the second part is whatever is left, from the total, after (one part) is accounted for.

You will need this technique — this "how much is left" construction — in the future, so make sure you understand it now.

So now I have a variable for the Fund X part, and an expression for however much was left to go into the Fund Y part. I can add these to my table:

I P r t
Fund X ? x 0.06 1
Fund Y ? 50,000 − x 0.14 1
total 4,500 50,000

Now I will show you why I set up the table like this. By organizing the columns according to the interest formula, I can now multiply across (in this case, I will multiply the three right-hand columns to get expressions for I in the second column) to fill in the "interest" column.

I P r t
Fund X 0.06x x 0.06 1
Fund Y 0.14(50,000 − x) 50,000 − x 0.14 1
total 4,500 50,000

The interest from Fund X and the interest from Fund Y will add up to $4,500. As a result, I can add down the "interest" column, set the sum of the two interest expressions equal to the total interest, and solve the resulting equation for the value of the variable:

0.06x + 0.14(50,000 − x) = 4,500

0.06x + 7,000 − 0.14x = 4,500

7,000 − 0.08x = 4,500

−0.08x = −2,500

x = 31,250

The value of x stands for the amount invested in Fund X. So the amount that is left, from the total invested, is given by 50,000−31,250 =18,750. And this (being the amount that is left after I'd put money into Fund X) is the amount that is invested in Fund Y.

I should put $31,250 into Fund X, and $18,750 into Fund Y.

Note that the answer did not involve nice, neat values like $10,000 or $35,000. You should understand that this means that you cannot always expect to be able to use guess-n-check to find your answers. You really do need to know how to do these exercises.

URL: https://www.purplemath.com/modules/investmt.htm

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How do you solve simple-interest word problems? (2024)
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