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Mixture Problems Made Easy!!!Really!!

How many liters of a solution that is 20% alcohol should be combined with 10 liters of a solution that is 50%alcohol to create a solution that is 30% alcohol?

What are the important parts here?

Who would like to come and high light the important components to this problem while the rest of you do it in your notes?

Percent strength

Amount of solution

Amount of target ingredient

Original

solution

Added

solution

Mixture

This is the magic grid that will lead you to the equation!!. Once you have filled in the grid you are looking at the equation in the last column.

Percent strength

Amount of solution

Amount of target ingredient

Original Solution

Added

Mixture

.50

.20

.30

Fill the third column with the total amounts in each container. We will use x for the amount in the second container.

10

x

x + 10

Percent strength

Amount of solution

Amount of target ingredient

Original Solution

Added

mixture

.50

.20

.30

Fill the last column with the products of the first and secondcolumns. Focus on the row for the original. If you multiply the percent strength by the total amount, you will have the amount of alcohol in the last column.

10

x

x + 10

5

.20x

.30(x+10)

Percent strength

Amount of solution

Amount of target ingredient

Original Solution

Added

mixture

.50

.20

.30

The amount of alcohol in the original plus the amount of alcohol in the added equals the amount of alcohol in the final mixture. The equation is in the final column.

10

x

x + 10

5

.20x

.30(x+10)

5 + .20x = .30(x + 10). 10 (5 + .20x) = 10 (.30x + 3) 50 + 2x = 3x + 30 -2x -2x 50 = x + 30 -30 -30 20 L = x

It takes 20 liters of a 20% solution to dilute 10 liters of a 50% solution to a final solution of 30%.

So here is the final equation to solve which should be quite easy for all of us to solve!

Example 2. How many liters of a solution that is 30% antifreeze should be added to how many liters of a solution that is pure antifreeze to create 5 liters of a solution that is 50% antifreeze?Round your answer to the nearest tenth of a liter.

Notice a couple of differences between this Example 1 and Example 2. In the first place, what do you think we are going to use as the percentage strength for pure antifreeze? Think before you click.

If you guessed 5 – x, congratulations!

If you guessed 100%, pat yourself on the back. Secondly, if we let x be the number of liters of the 30% solution, what do you think we should use as the number of liters at 100%? Think before you click.

Hint: the total number of liters has to be 5.

Percent strength

Amount of solution

Amount of target ingredient

original

added

mixture

Try to fill in what you know.

We are going to let x be the number of liters of the 30% solution. We also know that the mixture will total 5 liters

X .30

5

Percent strength

Amount of solution

Amount of target ingredient

original

added

mixture

.

.30

1.00

.50

x

5

So let us see what we have here now?What are we going to use as an algebraic name for the number of liters of pure antifreeze? Since the total number ofliters has to be 5, subtract the number of liters at 30% and get:

5 - x

Percent strength

Amount of solution

Amount of target ingredient

original

added

mixture

Multiply percent strength by the amount of solution and get theamount of the target ingredient (antifreeze).

.30

1.00

.50

x

5

5 - x

.30x

1(5 – x)

.50(5)

The third column is your equation. Solve it before you click.

From the third column, the equation is:

.30x + 1.00(5 – x) = .5(5)Let us continue multiplying through by 10 to

get rid of those decimals(10) ( .30x + 1.00 (5 – x )) = (.5 (5) ) (10)

3x + 10(5 - x) = 5(5)3x + 50 -10x = 25

-7x + 50 = 25-50 -50-7x = -25X = 3.6

5 – x = 1.4 liters

We will introduce our final example with a little common sense. It is also pretty interesting.

How much water would you have to add to dilute 4 liters ofa 70% acid solution down to a 35% acid solution? Think beforeyou click.

Congratulate yourself if you guessed 4 liters. Another 4 liters would dilute the original 4 liters to half of its original strength.

See if you can work this problem on your own before you click to the solution. Guess first, then use the grid to solve algebraically.

Example 3. How much water would you have to add to dilute 4 liters of a 75% acid solution down to a 25% acid solution?

Percent strength

Amount of solution

Amount of target ingredient

Original .75 4 .75(4)

Added 0 x 0

Mixture .25 4 + x .25(4 + x)

The target ingredient will be acid. The second container will be the water. Did you use 0% for the strength of the added column?

.75(4) = .25(4 + x)(100) .75(4) = .25(4 + x) (100)

75 (4) = 25 (4 + x)300 = 100 + 25x

200 = 25x8 = x

It takes 8 liters of water to dilute 4 liters of 75% acid down to 25%

strength.

Now you may practice with your home work problems. Make sure to refer to your notes if something is not clear. Good luck and I know you will do fine.