The “unitary method” for discounts allows for the calculation to be done in one step rather than two (hence “unitary”). It also provides the foundation of an algebraic approach that will be built upon in a subsequent algebra course. As I’ve mentioned, students who I taught for both seventh grade math and eighth grade algebra have said that what was confusing for them in seventh grade regarding percent problems was much easier when they saw it again in algebra.
The unitary method is presented in two ways. The first way is direct translation to an algebraic equation as has been introduced and practiced in a previous lesson on percent problems. Some students find this easy and natural while others have difficulty with it. And, as was done before, I present a proportional method as well, to which most seventh graders gravitate. Other students will cling to the two-step method. Ultimately, in a future lesson when students must find the original amount when the discounted (or marked-up) amount is given, most students will necessarily use the proportional method. (And a select few will elect to use a direct algebraic translation approach.)
Warm-Ups. These are general review of what has been done previously. Problems 1 and 2 are similar to what was covered in previous lessons so should not present difficulty. Problem 3 asks for the amount by which the price is reduced, not the sales price. Problem 4 is a review of tape diagram or equation.
1. A bookshop gives a 20% discount during a sale. Tom bought two books which cost $40 and $15. How much did he have to pay altogether?
Answer: 65 – (65×0.2 ) = 65 – 13 = $52.
2. What is the discount rate on a bottle of shampoo that is reduced from $20 to $12? Answer: 8/20 = 0.4 or 40%
3. A $600 television is discounted 24%. What is the amount of the discount? Answer: 0.24 × 600 = $144.
4. Susan and Nancy received some money in the ratio of 2:5. Nancy received $36 more than Susan. How much money did Susan receive? (Use algebra or tape diagram to solve.) Answer: Let 2x = amt of Susan’s money, 5x = amt of Nancy’s money; 5x – 2x = 36; 3x = 36; x = 12. 2x = 24, 5x = 60.
5. John’s uncle says he will help John purchase a bike by paying 30% of the cost. If the bike costs $800, how much will John’s share of the cost be? Answer: John will pay 70% of the cost; 0.7 × 800 = $560.
Problem 5 is directly related to this lesson. One prompt I use for students who ask for help is “What if his uncle paid 50%; what would be John’s share?” Obviously 50%. Then, “What if he paid 60%, what’s John’s share?” They usually can see it will be 40% and that the uncle’s share is subtracted from 100 percent. In the problem, John’s share will be 70 percent, so a prompt would be “What is 70% of $800?” which should be enough information to allow them to solve the problem.
Finding Discounts Using the Unitary Method 1: Problem 5 of the Warm-Ups serves as a starting point for the discussion. They know from the problem that John’s uncle chipped in 30% and that John’s share is figured to be 100% - 30% or 70%.
“When a store discounts an item by 40%, it’s like they are saying ‘We’ll chip in 40%’. So what would be your share?” Students understand from the earlier Warm-Up problem that it is 60 percent.
“So let’s say there’s a $900 bike on sale 25% off. What is your share going to be?”
Hearing “75%” I continue: “How much are you going to pay? Can we write it as an equation?” I have them work on mini-whiteboards or in their notebooks and go around to see that they are doing it correctly and getting $675.”
“This is a different way of finding the discounted price. It’s called the Unitary Method, because instead of doing the calculation in two steps like we did yesterday—find the amount of the discount and then subtract—it’s done all in one. There’s no subtraction step.”
(Actually there is a subtraction step—the discount rate is subtracted from 100 percent. But in all the time I’ve been teaching this, no student has ever pointed that out.)
Worked Example:
A radio normally sells for $140. It is discounted 20%. What is the sales price?
“What percent are you going to pay?” Hearing 80% I state “This is now asking 80% of 140 is what number? How do we write that as an equation?” I have them write on their mini-whiteboards or notebook. It should be 0.8∙140 = x. (Typical mistakes are not changing the 80% to 0.8. The answer is $112.
If a student just calculates the answer rather than writing it in equation form as shown above, that’s acceptable. No point in insisting on something that seems redundant. If they do write it as an equation, no problem with that either. It’s a win-win situation.
Examples:
A $420 watch is discounted 15%. What is the sales price?
Equation: 0.85 ∙ 420 = $357
Mr. Hallmeyer pays 70 percent of the price for a necklace which normally sells for $400. What is the discount rate?
This one is not asking for the sales price, but rather the rate of discount.
Finding Discounts Using Unitary Method 2. Some students forget to subtract from 100 or find the method confusing. I mention that all we are doing is solving percent equations, and that like before, we can either do it by direct translation to algebra or we can use the proportion method.
The proportion method uses this formula which I have students write down:
Worked Example:
To show how it works I give an example: A $60 book is discounted 15%; find the sales price.”
“Let’s plug in the numbers to the formula. The discount rate is 15 percent, so we plug that in. Now work on the right hand side. What do we know and what are we trying to find out?” I want to hear we are trying to find the discounted price and that we know the original price.
“And if we don’t know what the value is, what do we use?”
Hearing x or other letters standing for the value of interest I then instruct them to plug numbers in and write the proportion in their notebooks keeping in mind they have to keep track of what numbers go where. I will go around checking to see what students have, offering help when needed. In general, most students get this right away, but some might place the original price in the numerator than in the denominator. The proportion should now look like this:
This simplifies to:
For something like this I remind students that they can reduce (simplify) the left hand side to make the cross-multiplication easier. Some do, and some don’t. There are still a good number of students who are not comfortable with simplifying fractions.
The “100 – discount rate” in the numerator forces students to do the subtraction that eludes some students.
Examples:
A $150 CD player is discounted 18%. What is the sales price?
A school had 2,000 students, but this year that number was reduced by 22 percent. What is the current number of students?
This problem may confuse some students because it is not a discounted price problem—but the structure and method is exactly the same. “Population is reduced by a 22 percent decrease. Set it up as if the population were money.” Answer: 1,560 students
A laptop sells for $1,750. It is discounted by 45%. By how much is the price reduced?
This can be solved simply; 0.45 × 1750 = $787.50. Before students dive into it, I will ask students what question does the problem want answered? Final price or amount of discount? It is asking “What number is 45% of 1750?” Students may solve it by multiplying as shown above, or they can solve by using a proportion using the “part/whole” construct (using “is/of” as necessary) presented in a previous lesson:
Homework. Homework problems consist of problems that involve discounts or other types of reductions. Students are free to solve using whatever method they find easier. Some problems will ask for the amount of a discount or reduction, in which case students simply multiply the percent reduction by the original amount. Mixing these problems is important to build up their recall of what they have learned previously. Otherwise, they may be using the same construct for problems that require a different one.