Determine Equivalent Ratios - Scale Factor Between Terms
Lesson 7 of 12
Objective: SWBAT determine if quantities in a table are in a proportional relationship by finding a scale factor equivalent to 1 between the quantities
I'll start the lesson by discussing the essential question. So far we have focused on one method for determining if two ratios are equivalent. Today we will discover another useful method.
To illustrate it, I will post a simple rate on the board. The context of the problem may be how much a child earns for doing chores at the house. $12 for 3 hours of chores. $16 for 4 hours of chores. $10 for 2.5 hours. $28 for 7 hours. etc
Many students will see the unit rate of $4 per hour. That is fine, but I will emphasize more the idea of one term being 4 times that of the other.
We will then work through the first example. I will give students a chance to discuss each problem before we complete the example.
Guided Problem Solving
There are two more problems for students to solve. Here the point is for students to get more practice identifying a constant scale factor between terms. For GP1 the more obvious relationship is to multiply gas * 28 to find miles, but it is worth pointing out that miles could be multiplied by 1/28 or divided by 28 to find the gallons of gas used.
Opportunities are given for students to be able to prove how they can use today's technique to determine if values form a proportional relationship (MP3).
For the second problem, students may need to be prompted to find how many times greater 5 is than 2. Calculators will be available to make computations quick and to check for errors.
Again it is important to show errors between terms and the scale factor being used.
I will encourage my students to draw models to make sense of problem 1 (MP1) - especially the rate of 1/2 miles to 1/4 hours. A simple bar model will show that the value for miles is twice that for hours. A double number line would be fine too but this is more closely related to finding a scale factor between ratios as opposed to within the terms of a given ratio.
Students would also benefit from drawing a model for problem 3. It is very similar to a problem they solved the previous day. Question i and ii are designed to emphasize the relationship between terms. Once this is discovered the values in the table (part iii) will be relatively simple to fill in.
Before the exit ticket, we will summarize today's method for finding equivalent ratios - finding a common factor between terms of each ratio.
This exit ticket will be worth 3 points.
Part a is worth 2 points for a correct answer and a valid explanation.
Part b is worth 1 point.
A successful exit ticket today will be 3 out of 3 points earned.