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 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? How much guides us to the question: How much 15% solution? To get guides us to where = will go.

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: AmountRateMixture First Solution Second Solution Final Solution

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: Amount(ml)Rate (%)Mixture First Solution Second Solution How much.15 Final Solution To get.25

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 “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”

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: Amount(ml)Rate %Mixture First Solution Second + Solution X.15 Final = Solution 20 + x.25

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). Amount(ml) *Rate % =Mixture First Solution 20 *.60 =.60(20) Second Solution X *.15 =.15x Final Solution (20 + x) *.25 =.25(20 + x)

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: Amount (ml)Rate %Mixture First Solution (20) + Second Solution X.15.15x = Final Solution (20 + x).25.25(20 + x)

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) x =.25(20 + x) x = x x = x = -8 X = 80

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.

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!

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. AmountRate %Interest First Investment Second Investment Final/Total May or may not haveTotal Interest earned:

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. NumberRateValue/Cost First Item Second Item Final/Total May or may not haveTotal value/cost:

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. AmountRateValue/Cost First Item Second Item Final/Total