Percent Mixture Problem #1
In this lesson, students learn how to solve percent mixture problems using one variable. Using a dog food example, the teacher explains how to mix a certain amount of 50% rice dog food with 400 pounds of 80% rice dog food to make a 75% rice dog food. By setting up an equation to find how much of the 50% dog food to use, students learn how to solve for the unknown variable and check their work to make sure the amount of rice in the final mixture is correct.
Explains using picture how to solve this percent mixture problem using one variable: How many pounds of dogfood that is 50% rice must be mixed with 400 pounds of dogfood that is 80% rice to make a dogfood that is 75% rice?
More free YouTube videos by Julie Harland are organized at http://yourmathgal.com
- How do you solve percent mixture problems with one variable?
- How many pounds of dog food that is 50% rice must be mixed with 400 pounds of dog food that is 80% rice to make a dog food that is 75% rice?
- How do you write and solve an equation for a percent mixture problem?
- How do you solve .50x + .80(400) = .75(x + 400)?
- How do you draw a picture of a percent mixture problem in Algebra?
- How can you solve an equation by first getting rid of all decimals?
- How can you check to make sure that your solution to a percent mixture word problem is correct?
-
Staff Review
These mixture problems can be a major thorn in the side of Algebra students from middle school through college. This video explains very clearly every step involved in writing out and solving an equation to figure out the answer to this problem.
