After his freshman year in college, a former math student paid me a visit. Aaron (not his real name) indicated that his first year of college was a great experience. This was no surprise to me. I found Aaron to be a fairly perceptive individual who got along well with his high school classmates. Then Aaron said to me, “You know Mr. Johnson, after a year of college chemistry, I finally understand division.” He went on to explain how the concepts of division, fractions, decimals, ratios, rates, proportions and percent all fit together for him. In order to solve chemistry problems, Aaron was naturally motivated to learn and understand division and related concepts. Of course, I was surprised to hear Aaron make such a statement. He was an above average student who had attended a highly regarded local private school through junior high school. I then realized that if high school students like Aaron don’t really understand division, many of my students probably don’t really understand division either.
Based on numerous conversations with high school math teachers, in both public and private high schools, I have come to the conclusion that many high school students don’t really understand division and related concepts. It’s almost like a dirty little secret that high school math teachers don’t like to talk about in public. Many students have learned how to apply a set of rules to get the right answer, but they don’t really understand why a rule works or what the answer means. I see this all the time when I tutor developmental math students and above who are graduates of local area high schools.
Elementary, middle school, and high school teachers should NOT in any why interpret the above remarks to mean that they are not doing their job. In my view, teaching elementary and middle school students is far more difficult than teaching high school students. Of course, teaching high school math is not so easy either. If anyone thinks that they have a bullet proof method of teaching division and related concepts to a class of 30 typical middle school students, they should tell middle school teachers how to do it. I’m sure that middle school teachers would be ever grateful.
We constantly hear comments about how public education has deteriorated over the years. My now deceased father-in-law once complained to me that one of his office workers couldn’t figure out how many parts to ship to customers in proportion to the number of parts the customers ordered. (There were more parts ordered than parts in stock.). He said something like this, “Schools don’t teach kids the basics anymore. In my day, schools taught the fundamentals.” My father-in-law and many of his high school classmates learned basic math very well, but only 29% of the general population graduated from high school in the 1920’s. It’s my observation that today’s top students and athletes perform much better than the top students and athletes 40 or 50 years ago. As I see it, the perceived problem with modern education is due to the fact that public schools are attempting to reach a student population with a far greater range of abilities.
So what’s the purpose of this post? The purpose of this post is to share some of the methods that I have used in my classes to explain division and related concepts. Readers may find some of my examples and explanations to be ordinary, but other examples and explanations are different in that they are not found in textbooks. I invite readers to share some of their favorite methods that they use to explain division and related concepts to kids. I would love to do a post based on reader response. Readers should reply by sending an email to firstname.lastname@example.org. If your example has a graphic, please attach a file that contains the graphic. I will create the graphic for you if you give me a good description.
My first set of examples illustrates why the division rule for dividing two fractions gives us the correct answer. I assume that students already know how to apply the multiplication and division rules for fractions, and they know how to reduce a fraction. These examples reinforce the idea that dividing by a number is the same as multiplying by the reciprocal of the number.
- Diagram A below illustrates the solution of following problem: If we have 12 pounds of hamburger and a meat loaf recipe that calls for 2 pounds of ground beef, how many meat loafs can we make? Solution: 12/2 = 12*(1/2) = 6 meat loafs.
- Diagram B below illustrates the solution of following problem: If we have 12 pounds of hamburger, how many 1/4 pound hamburger patties can we make? Solution: 12/(1/4) =12*(4/1) = 12*4 = 48 hamburger patties.
- Diagram C below illustrates the solution of following problem: If we have 12 pounds of hamburger, how many 3/4 pound hamburger patties can we make? Solution: 12/(3/4) = 12*(4/3) = 48/3= 16 hamburger patties. Alternatively, 12 pounds of hamburgers divided into 16 patties = 12/16 = 3/4 pounds per patty.
When I introduce right triangle trigonometry to my students, I give them the sides of a right triangle, and then we find the trig ratios of the two acute angles α and β of the triangle. Students quickly notice that Sin(α) = Cos(β). I want my students to not only understand the definitions of trig ratios and understand that trig function values are ratios, but I also want them to see that trig function values are percentages in disguise. The diagram below shows a right triangle, accurately drawn to scale, with 150 and 750 acute angles. Knowing only the acute angles of the right triangle and a scientific calculator, we can deduce the facts below. When students study the diagram, they often say, “When you look at it this way, it makes more sense.”
- Cos(150) = b/c ≈ 0.966. This tells us the length of the leg adjacent to a 150 angle is always about 96.6% of the length of the hypotenuse.
- Sin(150) = a/c ≈ 0.259. This tells us the length of the leg opposite a 150 angle is always about 25.9% of the length of the hypotenuse.
- Tan(150) = a/b ≈ 0.268. This tells us the length of the leg opposite a 150 angle is always about 26.8% of the length of the leg adjacent to a 150 angle.
- Tan(750) = b/a ≈ 3.73. This tells us the length of the leg opposite a 750 angle is always about 373% of the length of the leg adjacent to a 750 angle.
In the next two problems, an intuitive approach is used to solve proportional parts problems. This would have made my father-in-law happy.
Problem 1: Find the three angles of a triangle if the angles are in a ratio of 3 : 4 : 5.
Solution: Refer to the diagram below. The problem boils down to dividing 1800 into a ratio of 3 : 4 : 5 parts. Because 3 + 4 + 5 = 12, we divide 180 into 12 equal parts with each part equal to 180/12 = 15. The smallest angle = 3 * 15 = 450. The next largest angle = 4* 15 = 600 and largest angle = 5 * 15 = 750. (I tell my students that I will only draw a diagram like the one below only one time because it takes too much time to draw the diagram. To my amazement, I once observed a student draw a diagram with well over 200 dots to solve a proportional parts homework problem. I’m not kidding.)
Problem 2: Customers A, B, C, D, and E have placed orders for 5, 7, 10, 20, and 40 widgets respectively. Since the company has only 60 widgets in stock, it was decided to immediately fill the orders in proportion to the number of widgets ordered, and ship the remaining widgets when they become available. How many widgets should be shipped to each customer?
Solution: Because 5 + 7 + 10 + 20 + 40 = 82, we divide 60 into 82 equal parts with each part equal to 60/82 = 0.732.
- Customer A: Ship 5 x 0.732 = 3.66 or 4 widgets
- Customer B: Ship 7 x 0.732 = 5.124 or 5 widgets
- Customer C: Ship 10 * 0.732 = 7.32 or 7 widgets
- Customer D: Ship 20 x 0.732 = 14.64 or 15 widgets
- Customer E: Ship the remaining 29 widgets
The classic “working together” problems are difficult for beginning algebra students to understand. Before I show students the standard algebraic technique for solving a working together problem, I show them how to use ratios to solve a working together problem. The problem below is a typical working together problem.
Problem: It would take homeowner Bob 7 days to roof his garage and professional roofer Clyde could roof Bob’s garage in 3 days. If Bob and Clyde cooperate with each other, how many days should it take them to roof Bob’s garage if they work together? Assume that a work day equals 8 hours. The text box below shows the solution of the problem. I suppose Bob and Clyde would work over time the second day so that they could finish the job in 2 days. (There is always someone who will make an initial guess of 5 days!)
Now let’s take a look at a couple of percent problems from a slightly different point of view.
Problem 1: A sofa that normally sells for $2,799.95 is on sale at 20% off. Local sales tax rate equals 9%. Find the sale price of the sofa and the cost of the sofa with sales tax. The solutions are given in the text box below.
Problem 2: Brad and his lovely wife Angelina dined at an upscale restaurant to celebrate their 20th wedding anniversary. The couple was fortunate to have a 15% discount coupon to help cover the $76.80 cost of their meal plus 9% sales tax. Since Brad was a high school math teacher, he thought that he had two options as to how he should apply the 15% discount. With option 1, he could take a 15% reduction of the cost of the meal plus sales tax. With option 2, he could first take a 15% discount on the cost of meal, and then pay sales tax on the discounted meal. With either option, Brad will leave a 20% tip. What option should Brad choose? (If Brad had stopped to really think about his options for a minute, he would have realized that it makes no difference what option he chooses.)
- Option 1: Total cost = 0.85(1.09 x 76.80) = $71.16
- Option 2: Total cost = 1.09(0.85 x 76.80) = $71.16
- Total cost with tip = 1.20 x 71.16 = $85.39 or $85.00
I will close this post with some sample exercises that I have used to give students practice drawing valid conclusions and hopefully promote their mathematical reasoning ability. Each exercise involves a conditional relationship which is true or false depending on the values of the variables in the conditional statement. From a list of about 25 possible conclusions, student are to select all conclusions that are necessarily true. Given the long list of possible conclusions, most students don’t find these exercises so easy on first exposure. Each of the text boxes below shows a conditional statement and all valid conclusions which are contained in a list of possible conclusions.