Having introduced ratios and a pictorial and algebraic method for solving problems with ratios, the topic of unit rates is now introduced. Unit rates serve as a bridge to proportions—what they are and solving problems with them. Usually unit rates are used for comparisons; i.e., which is the better deal, 4 pounds of oranges for $6, or 5 pounds for $8? The problems that are solved using unit rates also include those that are typically solved using more general proportion techniques. This may be confusing to students, at first, who may wonder whether some problems must be solved using unit rates, while others use the more general proportion approach. Both approaches are connected, and in my experience, it is best to show students how to use both, but to take time to ensure mastery of each.
As stated earlier there is a push via Common Core for students to connect unit rate to 1) the constant of proportionality, 2) the slope of graphs of proportional relationships, and 3) the equation form of proportional relationships (i.e., y = kx). I will say again, because it bears repeating, that my emphasis in teaching ratios and proportions is on the fundamentals of problem solving. I cover the other topics because it is required by Common Core, but do it in a highly procedural way, knowing that students will cover the topics more thoroughly in algebra.
Warm-Ups. This set contains a problem about averages that is the culmination of versions leading up to it.
1. The ratio of math books to novels in a library is 4:7. The total number of books is 2,200. How many math books are there? (Use an equation to solve)
Answer: 4x= number of math books; 7x= number of novels; 11x =2200; x =200. Math books: 800, Novels: 1,400
2. 6 ÷ ¾ = ? Answer: 6 × 4/3 = 8
3. (5/6)y + 2/3 = -4/9 Answer: (5/6)y = -10/9; y = -10/9 × 6/5; y = -4/3 or 1 1/3
4. Peggy averaged 75 over four tests in math. What does she need to get on the next test for her average to be 80. Answer: Total for average of 75 = 75×4 = 300. Total for average of 80 = 400. 400 – 300 = 100. She will need to get 100 on the next test.
5. Alicia paid $90 for 100 mangoes. She threw away 16 that were rotten and sold the rest at 3 for $4. How much money did she make? Answer: 100-16 = 84 good mangoes. 84 ÷ 3 = 28 groups of 3 mangoes. 28 × 4 = $112. She made $112-$90 =$22.
Students are likely to ask for help in solving Problem 4. Again its solution relies on finding the total value from which the average was derived. The total value for the average of five tests will differ from the total for the average of four tests by the latest test score.
For Problem 1 this is more of the same as previous lesson, though many if not most will rely on a tape diagram to derive the equation.
Rates are Ratios. I start this topic by asking students what a ratio is. There will be some stammering around, and mention of two numbers as in “You know, one number over another” and other vague definitions. Knowing that maybe one (if I’m lucky) or two students (if I’m extremely lucky) will actually look it up in their notes, rather than beat around the bush I remind them of the definition: A ratio is a comparison of two numbers by division.
Having established once more what a ratio is, I then state that “A rate is a specific type of ratio,” and then write the definition on the board:
A rate is a ratio of two quantities with different units.
I give an example of a ratio that is not a rate: The ratio of boys to girls in a classroom. The ratio of yellow to blue paint.
Examples of rates. “Suppose I do a count of how many boys pass by in a hallway over a 5 minute period. I count 35 boys in 5 minutes. That is a rate: 35 boys/5 minutes.”
A common example of rate is speed. “If a car travels 300 miles in 5 hours, we can express that as a rate: 300 mi/5 minutes. Another one is price: Oranges are selling for $4/2 lbs.”
Unit Rates. A rate with a denominator of 1 is called a unit rate. Using the above examples, we can express them as unit rates, and in most cases it provides a better sense of what the rate represents. “35 boys in 5 minutes is 7 boys/1 minute, which we say as ‘7 boys per minute’. 300 miles in 5 hours as a unit rate is how much?” I have them write this on mini-whiteboards: 60 mi/hr, commonly called 60 miles per hour.
I discuss how we find the unit rate. “If I have the rate of 35/5, how did I get the unit rate?” Someone might say that you divide the numerator and denominator by 5 which is correct, but it amounts to just dividing numerator by denominator. 300 miles in 5 hours, is 300 ÷ 5.
Unit rates can be used for comparison. If Store A sells apples at $5/2 lbs and Store B sells them for $10/5 lbs, we can find the better deal by finding the unit rate for both, which I ask the students to do. Store A’s price is $2.50/lb and Store B’s is $2/lb.
Examples.
A subway car goes 2.5 miles in 5 minutes. Find its speed in miles per minute.
Answer: 2.5/5 = 0.5 mi/minute
The international space station travels the earth at a speed of 14.4 miles for every 3 seconds. Find the unit rate; i.e., the speed in miles per second? Answer: 4.8 mi/sec
What’s the better deal: $15 for 8 lb oranges or $6/3 lb? Answer: $15/8 lb. It equals $1.88/lb vs $2/lb.
Find the unit rate for $11 for 4 boxes of cereal. Answer: $11/4 = $2.75/box
Find the mi/gal unit rate: 450 mi on 15 gallons of gas Answer: 30 mi/gal
Using Unit Rates to Solve Problems. “If I travel 450 miles in 5 hours, what is the unit rate?” I will ask. Hearing 90 mi/hr I will first warn that they shouldn’t be driving that fast, to which a student may respond that they know someone who drives at 100 mi/hr. So be prepared for diversions, and keep them on track. (There are downsides to “engagement”.)
“So using the unit rate, how can I find out how far they’ll drive at that rate in 7 hours?”
Most students will see that it is simple multiplication: 90×7=630 miles.
What is not obvious for a problem dealing with speed, is that they are also cancelling units when doing the problem. One where it is more obvious would be the following:
“Lexi painted 2 faces in 8 minutes at the fair. At this rate, how many faces can she paint in 40 minutes?”
“What do we do first to solve this?” Hearing “Find the unit rate” “And what is it? Express it as a decimal.”
I’m looking for 0.25 faces per minute as the answer but students will balk. Seeing a fraction of a face per minute is unusual, but I tell them that unit rates will sometimes be fractions—it doesn’t mean it’s wrong. “We solve it in the same manner as the first problem. We multiply 0.25/min by what?”
They will multiply by 40, to get 10 faces. I will then write on the board:
0.25 faces/min × 40 min. “What happens to the ‘minutes’ in the problem?”
“You may not be used to cancelling units; you’ve just done it with numbers. But you can do it with units as well.”
Upon cancelling “minutes” from numerator and denominator, we are left with 10 “faces”. I do one or two more so they get the feel of it.
“A bakery can make 195 donuts in 3 hours. At this rate, how many doughnuts can the bakery make in 8 hours?”
I have them work this in their notebooks as I walk around the room. The unit rate that I want to see in their notebooks is 65 donuts/hr. The calculation I want to see next is 65 donuts/hr × 8 hr = 520 donuts.
Homework. The problems for this section should be devoted to unit rate. As mentioned earlier, students will be learning to use a general proportion approach to solve problems. For future homework assignments it is a good idea to require that some problems be solved by unit rate, and others by the general proportion method—but not before students have mastery with both methods. This process is called “interleaving” and having done that, there is no need to obsess over the mix of unit rate and general proportion method. This will be discussed further in the next chapter. (The interleaving, not the obsessing.)