Now that students are working with algebraic fractions they have a tool that expands the types of word problems that they now can solve. Students are not typically overcome with joy at this news, so I am careful with my phrasing when I introduce the new types of problems.
For this book, I am focusing mostly on distance/rate/time problems. But algebraic fractions allow students to also solve problems involving work, and mixtures. Typical problems of these categories include the following:
Work: One bulldozer clears land twice as fast as another. Together they clear a large tract in 1 ½ hours. How long would the larger bulldozer take?
Mixtures: To reduce 16 mL of a 25% solution of sulfuric acid to a 10% solution, how much distilled water needs to be added?
In this section, I describe lessons for round-trip problems, and wind and current problems. Each lesson is presented on a separate day. Previously, distance/rate problems were set up on the basis of “distance = distance”. Since Distance = rate ∙ time, or D = rt, then we can express time as t = D/r. Both types of problems are based on “time = time”.
Warm-Ups.
Problems 1 through 4 are problems students have worked on in this unit. Problem 5 is a distance/rate problem that is a distance = distance type problem.
Day 1
Round Trip Problems. I make the following announcement to the class. “So far in this class, you’ve been solving distance/rate problems that are in the form of distance = distance.” They look at me as if waiting for the other shoe to drop.
“Today we’ll be working on problems that are in the form of time = time. Before we get into that, someone tell me the formula we’ve been using for distance in these problems.”
Hearing “distance = rate times time”, I write “D = rt” on the board.
“If I wanted to express this equation in terms of t for time, how would I write it?”
They know this—or at least some students do:
“This is the formula for time: distance over rate. We can use this to solve problems involving round trips, which is what we’ll be doing today.”
The following problem is projected on the board:
A troop of scouts hiked to the scout cabin at the rate of 2 mph. They rode back to headquarters at the rate of 18 mph. If the round trip took 10 hours, how far is headquarters from the cabin?
“What is the total time for the trip?” I hear 10 hours. “What are we trying to find?”
We’re trying to find the distance from headquarters to the cabin.
“Do we know how long it takes to get to the cabin, and how long it takes to get back?” No.
“What we know is that the time to the cabin and the time to return totals 10 hrs. So let’s write that.”
Time from HQ to cabin + time from cabin to HW = 10
“Do we know the rates of speed going to the cabin and returning? What are they?”
I hear 2 mph going to the cabin and 18 mph returning.
“Aaaand, we know the formula for time. How do I represent the time going to the cabin at 2 mph?” I have them write in their notebooks, and I see some have it. I write on the board:
“And how do I represent the time returning to headquarters at 18 mph?” Just about everyone gets it now:
“Give me an equation.” I search for a correct equation; there are typically many.
“Now solve it. You’ve solved things like this before.
“How do I know this is correct?”
I’m told we plug it in to the original equation. And sure enough: 18/2 + 18/18 = 10
Exercises. I work with the class on the first and perhaps second problems; the exercises are taken from the homework assignment.
Homework. The problems are a combination of round trip problems as well as what they’ve solved before: opposite direction and same direction (catch-up) problems.
Day 2
Wind and Current Problems. After going through Warm-Ups and any homework problems which they want to see worked out, I project the following problem on the board.
Jim rows 9 miles downstream in the same time that he rows 3 miles upstream. The current flows at 6 mph. How fast does Jim row in still water?
The students have a deer-in-the-headlights type look as they read it. “I know what you’re thinking,” I say. “You’re thinking there’s no way you can do this problem.”
They tell me that I’m right.
“I don’t believe you,” I say. “And here’s why. You recall we did some round-trip problems. And they were based on ‘time = time’. So is this one. Let’s look.” I have someone read the first line of the problem.
“There are two distances stated here. One is a downstream distance and the other is an upstream distance. What are they?”
They tell me it’s 9 miles downstream and 3 miles upstream.
“Now downstream means going with the current. The water is flowing at what speed? Look at the problem.” They look at it and tell me 6 mph.
We then ascertain that the problem is asking for Jim’s speed in still water; that is, if there were no current at all, how fast would he be rowing? “So his speed in still water is x I think it’s safe to say. What would his speed be downstream? That is, with a 6 mph current.”
Some students will say x – 6 for reasons I don’t quite understand, but I ask them if they were rowing with the current, would they be going faster or slower. They suddenly see the light and now they say the speed is x + 6. And rowing upstream, they can see he would be going slower: x – 6.
“Let’s put all this information in a table,” I say, and draw a table underneath the problem. They are to copy this in their notebooks.
“Now the time it takes him to row 9 miles downstream is the same as 3 miles downstream. What does ‘the same as’ mean to you?” They answer “Equal”. “And how do we represent time?” Hearing “Distance/rate” I ask them to help me fill out the table.
“Now we write ‘time downstream = time upstream” as an equation.” I have them write it in their notebooks and solve.
To check it, we plug 12 into the original equation.
Examples. I use the homework problems as examples, and work through the first two with them. Then I provide guidance as they work on the rest.
1. The riverboat Memphis Belle sailing at the rate of 18 miles per hour in still water can go 63 miles down the river in the same time it takes it to go 45 miles up the river. What is the speed of the current in the river?
After we get to this point, I ask “Can 63 and 45 be divided by anything?” They tell me 9 goes into both.
“We can then divide both numerators by 9 first, to make the numbers easier to work with.” Students ask me if that’s legal.
“It’s the same thing as multiplying both sides by 1/9,” I say and we go ahead and do it.
2. In still water, Jim’s motorboat is 4 times as fast as the current in Pony River. He takes a 15 mile trip up the river and returns in 4 hours. Find the rate of the current.
Homework. Problems are similar to these with some variation. I work with students as they do the problems I assign 5 or 6 problems. Usually students finish them all within the class period.