Work the mixture problem. See ExampleEXAMPLE Solving a Mixture ProblemA chemist needs to m...

Work the mixture problem. See Example

EXAMPLE Solving a Mixture Problem

A chemist needs to mix 20 L of a 40% acid solution with some 70% acid solution to obtain a mixture that is 50% acid. How many liters of the 70% acid solution should be used?

Step 1 Read the problem. Note the percent of each solution and of the mixture.

Step 2 Assign a variable.

Let x = the number of liters of 70% acid solution needed.

Recall from Example (a) that the amount of pure acid in this solution is the product of the percent of strength and the number of liters of solution, or

0.70x. Liters of pure acid in x liters of 70% solution

The amount of pure acid in the 20 L of 40% solution is

0.40(20) = 8. Liters of pure acid in the 40% solution

The new solution will contain (x + 20) liters of 50% solution. The amount of pure acid in this solution is

0.50(x + 20). Liters of pure acid in the 50% solution

FIGURE 15 illustrates this information, which is summarized in the table.

FIGURE 15

Liters of Solution	Rate (as a decimal)	Liters of Pure Acid
x	0.70	0.70x
20	0.40	0.40(20) = 8
x + 20	0.50	0.50 (x + 20)

Step 3 Write an equation. The number of liters of pure acid in the 70% solution added to the number of liters of pure acid in the 40% solution will equal the number of liters of pure acid in the final mixture.

Step 4 Solve the equation.

Step 5 State the answer. The chemist needs to use 10 L of 70% solution.

Step 6 Check. The answer checks, since

How many gallons of a 12% indicator solution must be mixed with a 20% indicator solution to get 10 gal of a 14% solution?