Great Lesson Ideas: Algebra 1 Mixture Problems with Marlo Warburton

[01:00:08;10]

Marlo: Hi. I'm Marlo Warburton. I teach Algebra 1, and this is how I teach mixture problems.

"And now...Mixture Problems. Mixture problems are perhaps the most dreaded type of problem in Algebra 1. People are afraid of these problems. Now, let me read one to you, and then you'll understand why people are scared. 'A chemist wants to mix some 70% saline solution with 8 liters of a 25% saline solution to create a solution that is 40% salt. How many liters of the 70% solution does she need?'"

Over the past five years, I've tried several different ways to teach mixture problems, and, and I just settled on this mixture picture, this common sense and this seesaw method.

"To introduce the big idea, I've brought in some water, and blue food coloring. This solution has a lot of drops of blue food coloring with water. I'm gonna call this my strong solution. This solution has a tiny bit of blue food coloring, and I'm gonna call this my...weak solution. So I have a strong solution, and I have a weak solution. I want you to silently think to yourself about what will happen if I mix, get it mixture problems? If I mix a strong solution with a weak solution. And, team talk."

Student: "It's gonna make it bluer than..."

Marlo: It's important to have that basic understanding that the concentration is gonna be in between, and I, I don't think that that's obvious to everyone. So, that's what I was going for first.

"Now I'm ready to mix. It's somewhere in between. So, this is our starting point."

Students take the information from the problem and put it into the mixture picture.

"Which saline solution is the weak saline solution?"

Student: "25"

Marlo: "The 25% saline solution."

In this case, 25 is the weak percent.

"And, which saline solution is the strong one?"

Student: "70%"

Marlo: Seventy is the strong percent.

"And, which is the in between?"

Student: "40%"

Marlo: "Yes? Easy enough?"

Eight liters is the amount of the weak solution, so I put that under the 25%.

"And the question is asking us?"

Students: "X liters"

Marlo: "Sure. X liters."

At this point, I show students that they can solve this problem by common sense alone.

"How far is the weak solution from what we want?"

Students: "15%"

Marlo: "And how far is the strong solution from what we want?"

Students: "30%"

Marlo: "What does this make you think?"

Student: "Twice as much?"

Marlo: So the mixture picture is a graphic organizer. It's a way for students to organize the information to solve problems.

Student: "Yeah, I knew something was..."

Marlo: "So, how many liters of the strong do we need?"

Student: "One"

Student: "Less than..."

Student: "Twice..."

Marlo: "Eight liters should be twice as much as this."

Students: "Four"

Marlo: "Four. Four liters."

But, the numbers aren't always that friendly, and a formula can be very handy.

"Here's the equation that you've been waititng for. Seesaw method."

So, I bring in the concept of a seesaw.

"Did you know that a big person can balance with a very small person, like an adult and a child can balance on a seesaw? How? How's that possible? Shout it out!"

Student: "It depends on how close the uh..bigger person is to the center."

Student: "Oh yeah!!"

Marlo: "Yes, it depends on how close they are to the...."

Student: "Center"

Students: "Fulcrum"

Marlo: "The fulcrum. OK."

And, I show students that on a seesaw, weight times distance equals weight times distance and try to make a connection between the seesaw and the mixture problem.

I, I did see more students comfortable with actually solving with the seesaw method.

Student: "'Cuz you have less of the strong, and more of the weak."

Marlo: "Yes!"

Student: "And then, the, the farther the weak is, the closer the strong has to be to the fulcrum."

Marlo: "Bingo! Say it again!"

Student: "The strong...ARRGH! Wait...the farther the weak is, the closer the strong has to be to the fulcrum."

Marlo: "Exactly. It turns out that weight times the distance to the fulcrum has to equal, that's the equation, this person's weight times this distance. So, weight is 8 liters, 8, times the distance, which is 15, has to equal the other weight, call it X, times the distance, which is 30. It doesn't get easier than that."

Eight times 15 equals X times 30. Solve for X, and it turns out, as we knew, we need 4 liters of the strong solution.

The best part about a lesson like today is seeing students go from being afraid and skeptical to being confident and excited.

## 31 Comments

Margie Ezell Aug 31, 2017 3:00pm

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