1 / 15

Solving Mixture Problems

Solving Mixture Problems. One variable equations. The Problem. A chemist has 20 ml of a solution that is 60% acid. How much 15% acid solution must she add to get a solution that is 25% acid?. Identify key words.

gamma
Télécharger la présentation

Solving Mixture Problems

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Solving Mixture Problems One variable equations

  2. The Problem • A chemist has 20 ml of a solution that is 60% acid. How much 15% acid solution must she add to get a solution that is 25% acid?

  3. Identify key words • A chemist has 20 ml of a solution that is 60% acid. How much15% acid solution must she add to get a solution that is 25% acid? How much guides us to the question: How much 15% solution? To get guides us to where = will go.

  4. A chemist has 20 ml of a solution that is 60% acid. How much 15% acid solution must she add to get a solution that is 25% acid? • First, we draw a chart for mixtures:

  5. A chemist has 20 ml of a solution that is 60% acid. How much 15% acid solution must she add to get a solution that is 25% acid? • We fill in the key information. Convert % to decimal:

  6. A chemist has 20 ml of a solution that is 60% acid. How much 15% acid solution must she add to get a solution that is 25% acid? • The first column is added down. 20 + “how much” = “to get” • We have two unknowns! The question words are x, so now we have 20 + x = “to get” 20 + x = “to get” • Because we have an equation, 20 + x means the same amount as “to get.” Until you can use two variables, you must use the 20 + x in place of “to get”

  7. A chemist has 20 ml of a solution that is 60% acid. How much 15% acid solution must she add to get a solution that is 25% acid? • Now we can fill in our table for the amounts:

  8. A chemist has 20 ml of a solution that is 60% acid. How much 15% acid solution must she add to get a solution that is 25% acid? • Now we multiply each row across. It’s better to write the mixture as “rate * amount” even though the columns are in the opposite order. (remember, 3*4 is the same as 4*3).

  9. A chemist has 20 ml of a solution that is 60% acid. How much 15% acid solution must she add to get a solution that is 25% acid? • To get our final equation, we add down the Mixture column:

  10. A chemist has 20 ml of a solution that is 60% acid. How much 15% acid solution must she add to get a solution that is 25% acid? • We have our final equation! .60(20) + .15x = .25(20 + x) • So simplify and solve: .60(20) + .15x = .25(20 + x) 12 + .15x = .25(20 + x) 12 + .15x = 4 + .25x 12 -.10x = 4 -.10x = -8 X = 80

  11. A chemist has 20 ml of a solution that is 60% acid. How much 15% acid solution must she add to get a solution that is 25% acid? Now we know that x = 80. Did we answer the question? How much 15% acid solution must she add to get a solution that is 25% acid? Remember, “how much” was an amount measured in ml. She must add 80 ml.

  12. Other ways to determine amounts The amounts column tends to be the difficult one in mixture problems solved with only one variable. Here are the usual patterns: • Has 60 ml. + “how much” = “to get” 60 + x = (60 + x) • “How much” + added to 60 ml = “to get”x + 60 = (60 + x) The trickiest ones have the second unknown in the middle: • “How much” + “added to” = to get 80 ml x + (80 - x) = 80 Remember to watch your question. You may need to provide more than one answer!

  13. There are other types of mixture problems that have slight differences in their table set-ups: • Financial interest mixtures have no final rate, and may not have a final amount. “Interest earned” is the key phrase for the interest total cell. Translation sentences are important in determining amounts.

  14. Coins and mixed purchases (e.g. coffee and lattes) also don’t have a final rate. Each rate is the value of the item: (e.g. nickels are .05, a coffee may be 1.10). In each case a total value or cost is in the final mixture cell. Again, translation sentences are important in these.

  15. There are other mixture problems where a final rate is important. These are usually determining the best price for a mixture of types of nuts or types of flowers. In these cases, a final rate is provided, and the problem is calculated like the solution mixtures: Add down in the amount and value columns, multiply across on the items and final.

More Related