This is the last section of the unit on percents. Simple interest is an application of a mark-up: it is a mark-up rate applied repeatedly over time. Textbooks such as Dolciani’s “Pre-algebra: An Accelerated Course”, prior to presenting the topic in its own right, provides a sneak peak at simple interest by presenting problems that require a percent of increase to be calculated over some period of time, and then ask for the annual rate of increase.
This is the approach I use here, though I caution that most students in seventh grade are in a “get to the point” frame of mind, and tend not to notice and/or retain the connection between simple interest and mark-ups.
Simple interest is not used now; it is compound interest. It generally is taught in accelerated Math 7 and Math 8 which is the next chapter, and where I discuss it.
Warm Ups. These include percent problems that students can solve in any manner they wish; proportion, algebra, or bar models.
1. A painting is discounted 12%. The amount of the discount is $24. What is the original price of the painting? Answer: 12/100 = 24/x; 0.12/1 = 24/x; 0.12x =24; x = $200
2. A flip phone is discounted 15%. The amount of the discount is $48. What is the sales price of the flip phone?: Answer: Method 1: 15/85 = 48/x; 15x = 4080; x=$272
Method 2: Original price – discount = sales price: 15/100 = 48/x; 0.15/1 = 48; x =$320; $320 - $48 = $272
3. In a class of 40 students, 60% are girls, 50% of the girls and 25% of the boys were glasses. How many students wear glasses? Answer: 0.6∙40 = 24 girls. There are 40-24 = 16 boys. 50% of girls = 0.5∙24 = 12; 25% of boys = 0.25∙16 = 4. 16 students wear glasses.
4. Mr. Andrews puts $1,000 in the bank in 2015 and in 2020 the amount with simple interest was $1,500. Find the percent change of increase over the five year period. Answer: (1500-1000)/1000 = 500/1000 = 50% increase.
5. For the 5-year percent of change in Problem 5, find the percent increase on a yearly basis. Answer: 50% ÷ 5 = 10% annual increase
Problems 1 and 2 are similar except that Problem 1 asks for the original price and Problem 2 asks for the sales price. The latter was discussed in the previous lesson. Problems 4 and 5 act as a segue to this particular lesson.
The Wind-Up. Discussion of Problems 4 and 5 serve as the “wind-up” to how simple interest is calculated. Students have done percent change of increase enough times that they will have the answer:
Problem 5 asks what is the rate of increase per year. Discussion of this problem during the Warm-Ups should make it obvious that dividing by the number of years provides the answer.
I will then explore this with them “Let’s see if this checks out.”
I write on the board:
10% of 1,000 the first year = 0.1 ∙ 1,000 = $100
10% of 1,000 the second year = 0.1 ∙ 1,000 = $100
I continue this pattern. It quickly becomes evident that for each of the five years, the increase is $100 per year.
The Pitch. I explain a bit about interest; that getting a loan entails “paying for money” via the interest rate. I let them know that simple interest is not common any more but it provides a starting point for understanding how interest works.
“The interest on $1,000 at 10% per year is $100. We obtained that by multiplying the principal—the amount deposited in the bank—by 10% for each of five years. Is there a more efficient way to do this rather than adding $100 five times?”
Students will suggest multiplying by five, since it is quite obvious where I’m leading them.
I write the formula finding the simple interest for some amount of money for a given amount of time:
where, I = interest earned, P = principal, r = interest rate, and t = time
I explain that principal is the amount which is earning interest, the interest rate is expressed as a decimal in the formula, and the time is in years unless otherwise stated.
Worked Example:
Frank deposits $5,000 in the bank at 2% interest per year. How much money in interest does he receive at the end of 10 years?
I have them write the formula in their notebooks and then we fill it in.
“What is the principal?” $5,000
“What is the rate? Remember, it’s 2% as a decimal.” Here I will undoubtedly here someone say 0.2 and upon hearing that will say “No, 0.2 as a percent is 20%. You need to be careful here. Many problems dealing with interest rates are in single digits, so be prepared. What is 2% as a decimal?” 0.02
“And what is t ?” 10
Plugging everything in, we obtain:
“Now that’s the amount of interest earned. If the person initially had $5,000 in the bank, and earned $1,000 on it, what is his balance in the bank?”
I am of course introducing P + I as the total balance, which would be $6,000. I compare this to calculating the tip you leave for a meal, with the total cost of the meal.
Frank deposits $5,000 in the bank at 2% interest per year. How much money in interest does he receive at the end of 6 months?
This one is a bit different. We want everything in units of years. “How much is 6 months in years?”
The answer is one-half or 0.5. The equation then becomes:
More Examples. (These are worked independently, as I check their work in their notebooks. I do allow calculators for these in the interest of time)
1. A deposit of $10,500 in the bank at 5.5% interest earns how much interest in 3 ½ years?
Answer: I = 10,500(0.055)(3.5) = $2,021.25 (Biggest mistake I’ve seen is expressing 5.5% as a decimal.)
2. $3,000 is deposited in a bank at 1.5% interest for 3 months. What is the total balance at the end of 3 months?
I = 3,000(0.015)(0.25) = $11.25; Total balance = P + I = 3,000 + 11.25 = $3,011.25
Finding an Annual Interest Rate. The formula can be used to find other parameters. In an algebra class, we would solve for the parameter of interest, but in seventh grade we have no covered literal equations in which a formula is solved for one of the variables: e.g., F = MA, solved for A becomes A = F/M.
Instead, an in-situ approach is used.
“We can use this formula to find an annual interest rate, or the time it should be kept in. For example, suppose I have $1,000 in an account. It earned $100 simple interest in 4 years. I want to know what the annual interest rate was.”
I write the formula I = Prt on the board. I then ask for the variables that we know and write them in:
“What was the interest earned? What was the principal? How long was it in the bank?”
As each variable is named I write it down, ending up with the following:
“Now we need to simplify this. The left hand side looks pretty good, but what can we do on the right hand side?”
I want them to multiply the 1,000 by 4, but in case there is a deer-in-the-headlights reaction, I’ll prompt them: “Can we multiply anything?” What I want, and what we eventually end up with is:
“We’ve solved equations like this before. How do we solve for r ?”
My mistake when I first started teaching this was to assume that students would by now have no problem seeing that 100/4000 is a fraction and can be expressed as a decimal. Seventh graders are novices at equations and so they need some guidance to get over the mistaken belief that “You can’t divide a number by a bigger number.”
What I have done to overcome this is just to say “I know it looks like you can’t divide 100 by 4,000 but you can. It becomes a fraction: 100/4000, which we can simplify. What does it simplify to?”
It simplifies to 1/40 of course. “How do we turn that in to a decimal?” This they should know, but I end up showing them to remind them of something that they’ve done and will be doing. They will ask if they can use their calculators but for now I want them to remember how to do this division.
1/40 = 0.025 or 2.5% interest per year
Worked Example:
Find annual rate of interest for a loan of $1,240 for 3 years 6 mos and interest earned is $651.
We work through this together:
I provide one or two more examples; they generally get into the rhythm of the algebraic steps.
Finding Time and Principal. Finding the amount of time or the principal when all other parameters are given is just a variation of the above. I start with a problem.
Interest rate = 2% P = $800; find how much time for the principal to earn $100 in interest
“We fill in the blanks just like we did before.” I have them do this interactively with me as I write in on the board and they write it in their notebooks.
“What if we didn’t know the principal in the above problem but we knew everything else? So I = 100, t = 6.25, r = 0.02. P = ? Help me plug things in.” We get:
I make sure they do the calculation at the end, rather than just say “Well we know it’s $800, it’s the same problem as before.”
I give maybe two or three more examples for them to do in their notebooks, before I get them started on homework.
Homework. The homework is straightforward—more of the same. Since students still get confused with the latter type of problems, and make mistakes in converting percents into decimals, I’ve found that doing this assignment in class is a critical one. That said, it may be beneficial to use some of the homework problems as examples.
Doing this accomplishes a number of things. It gives them needed practice in understanding the procedures and reasoning with these problems. And perhaps more importantly, it makes them feel like they’re getting something for nothing even though they’re doing the same amount of work.