RP 5 Traditional Math: Proportions and Solving Proportional Relationships
Seventh Grade (Corrected for email subscribers)
To my subscribers: The e-mail version which went out yesterday had some typos and misplaced numbers in it. These were corrected on the web version. Sorry if it was hard to read. This is the corrected version.
So far, we have talked about ratios. We move now into proportions and their relationship to ratios. Students generally like working with proportions and it has been my aim in teaching seventh grade math to keep it that way. As I’ve mentioned, there are some complications when introducing concepts such as constant of proportionality, direct variation, and slope, and relating it all to unit rates. In keeping things straightforward, understandable and workable, I focus on the fundamental idea of proportions and how to work with them in solving problems.
Having had a solid foundation in proportions and solving them, students then find that proportions in the context of linear functions make more sense in an algebra course. In such a setting, students learn about functions, graphing, slope and most importantly independent and dependent variables. The concepts of constant of proportionality (which is called “constant of variation” in algebra courses) and direct variation make more sense. The equation y = kx follows as a consequence, in which y is the dependent variable, x the dependent variable and k the constant of proportionality or variation.
But Common Core standards push these concepts into the seventh grade when students, in my opinion, are not ready for them. Students find these concepts confusing, and working with them even more so. To address this, the two sections following this one cover slope, and direct variation in a procedural approach.
Warm-Ups. As usual, recent topics should be included as well as older ones. A problem with averages is included that is multi-step.
1. The lengths of 3 rods are in the ratio 1:3:4.If the total length is 96 cm, find the length of the longest rod. (Solve using algebra). Answer: Longest rod is 4x. x + 3x + 4x = 96; 8x = 96; x = 12, so 4x = 48 cm.
2. The average cost of 3 books is $4.50. The average cost of two of the books is $3.90. Find the cost of the third book. Answer: Find the total cost for the two books: 3.90 × 2 = 7.80. Find the total cost for the 3 books: 4.50 × 4 = 18. The cost of the third book is therefore $18.00 - $7.80 = $10.20.
3. If 4 gallons of gasoline cost $14.00, what is the cost of 10 gallons? (Use unit rate to solve). Answer: Unit rate: 14 ÷4 = $3.50/gal. $3.50/gal × 10 gal = $35.00
4. Simplify. -5x +3y – 2x – 4y Answer: -7x – y
5. Richard rowed a canoe 3 ½ miles in ½ hour. What is his speed in miles per hour? Answer: 3 ½ ÷ 3/8 = 7/2 × 8/3 = 28/5 = 5 3/5 hours, or 5 hours 36 minutes.
Problem 2 might prove difficult. Hints may include “Can we find total cost for the 2 books? What about 3 books?”
Problem 3 is a unit rate problem. Students may try to find a unit rate of gallons/dollar rather than dollars/gallon.
Problem 5 is also a unit rate problem, except that the numbers are fractions. A hint might be “What if he went 6 miles in 2 hours? How could we express that as a unit rate of miles/hr?” Students will generally find whole numbers easier to work with—it is then a matter of applying the same operation (division) to the fractions.
What a Proportion Is. I ask the class what the Surf City ratio is. If they have learned nothing else in this particular unit, they seem to grab onto this particular ratio. In response to my question they will call out “Two girls for every boy” to which I write: 2/1 on the board.
“Do 6 girls and 3 boys form the Surf City ratio?” To the chorus of “Yes’s” that erupt, I then ask, “How do you know?” Students have little difficulty showing that 6/3 is the same as 2/1.
“You mean they are equivalent,” I say. “Then how about ratios of girls to boys of 10/5, 8/4, 12/6?” Ratios that simplify to the same basic ratio like these are called “equivalent ratios”, which bring us to the definition of “proportion” which I write on the board:
A proportion is an equation stating that two ratios or rates are equivalent.
For example:
Taking a look at the first equation, I try something. “What is the common denominator of 8 and 4? Not the lowest—a common denominator.” Hearing 32, I say “Let’s multiply both sides of the equation by 32 and see what happens.”
I write 32/1 rather than 32 because students will stubbornly refuse to see that a whole number can be multiplied by a fraction without writing the whole number as a fraction with denominator of 1. (That, and the fact that I have a lot to cover and want to minimize people not following and tuning out.)
Next we cross cancel:
This yields an important result. We notice first that the 6 in the numerator on the left hand side is multiplied by the 8 in the denominator on the right hand side, and vice versa for the 12 and the 4. This is called “cross multiplication”, which is accomplished by multiplying diagonally. If both sides are equal, which they are in this case, it tells us that the two ratios are equivalent and therefore proportional.
If we tried this with two ratios that are not proportional, like:
Cross multiplication would result in 27 and 20 on the two sides—definitely not equal.
The above is an informal way to show this formal statement:
If a/b = c/d then a ∙ d = b ∙ c
Using Cross Multiplication to Solve Problems. “The process we just described is actually very useful; we can use it to solve proportion problems.”
I will then give an example. “Green paint is made with 3 parts yellow and 2 parts blue. If we use 12 quarts of yellow how many quarts of blue must we use to make green paint?”
We write the proportion with the two ratios of yellow to blue that we’re interested in:
“To solve for x we cross multiply like we did before.” I draw arrows to show what it is we’re multiplying.
“We multiply 3 by x, and 12 by 2. How do I write 3 times x?”
I wait until I hear 3x, and then write it down. 12 times 2 they know.
“The result is the equation 3x = 24. Solve it.”
I wait for them to solve it and write the answer (8 in this case) on their mini-white boards.
“It’s easy to check. We just cross multiply: 3/2 = 12/8 gives us 24 = 24. Correct.”
Examples. Part to Part, Part to Whole and Rates. The above example is a part to part problem—yellow to blue. But if the problem were stated “The ratio of yellow to blue paint is 3:2. How much yellow paint is needed to make 20 gallons?”
Since the ratio of yellow to blue is 3:2 the ratio of yellow to total (green) is 3: (2+3) or 2:5. I’ll ask the students how the equation should be set up and have them write it on their mini-whiteboards. I’m looking to see: 2/5 = x/20. Initially it helps if they give themselves a guide and write yellow/green = 2/5 = x/20, to keep track of the order.
Solving it: 5x = 40, and x = 8 gallons of yellow paint.
A problem with a rate would be: “The cost of 5 oranges is $2. How much will 15 oranges cost?” This is neither a part-part nor part-whole, since the problem is stating a rate. Oranges/cost = 5/2 = 15/x; 5x = 30, x = $6.
Other examples
1. “The ratio of Jim’s weight to Eva’s weight is 5:4. If their total weight is 90 lbs how much does Eva weigh?” Answer: Part to whole would be Eva/total = 4/9 = x/90; 360=9x; x = 40 lbs.
Some students may see that the proportion 4/9 = x/90 can be solved without an equation. Namely since 90 is ten times 9, then x would be ten times 4, giving the same answer: 40.
Also, some students may see that the problem could be solved as was done for ratios in a previous lesson: 5x + 4x = 90; 9x = 90, x = 10, and 4x = 40.
2. “The ratio of boys to girls in a school choir is 4:3. If there are 16 boys, how many girls are there?” Answer: Part to part: girls/boys = 4/3 = 16/x. 4x = 48, x = 12. Ask if there are different solutions.
3. “John reads 8 books in 12 weeks. If he keeps reading books at this rate, how many books will he have read in 15 weeks?” Answer: This is a rate problem: books/weeks = 8/12 = x/15. It can be solved in this form, but it should be pointed out that 8/12 can be reduced to 2/3 so we have 2/3 = x/15, and 3x = 30, x = 10 books.
4. A car travels 300 miles in 5 hours. How far does it travel in 2 hours? Answer: Rate problem: distance/time = 300/5 = x/2. 300/5 reduces to 60/1. 60/1 = x/2, x = 120.
Homework. There is a variety of different problems that textbooks provide. For some of them, they can be solved using the proportion method, or by other methods, as seen for example 1.
The next lesson will focus on solving problems by using a unit rate and developing a formula. It is best to do this in a separate lesson rather than include it in this one. Otherwise there is too much information causing the dreaded cognitive overload, frustration, cries of despair and angry calls from parents.