Intermediate Algebra (11th edition)

OBJECTIVE 7 Solve mixture problems.

Solving a Mixture Problem

A chemist must mix 8 L of a 40% acid solution with some 70% solution to get a

50% solution. How much of the 70% solution should be used?

Step 1 Readthe problem. The problem asks for the amount of 70% solution to be

used.

Step 2 Assign a variable.Let x= the number of liters of 70% solution to be used.

The information in the problem is illustrated in FIGURE 5and organized in

the table.

EXAMPLE 7

SECTION 2.3 Applications of Linear Equations 73

8 L Unknown

number of liters, x

(8 + x) L

From 70%

40% From 40%

After mixing

+=70% 50%

FIGURE 5

Sum must

equal

Number Percent Liters of

of Liters (as a decimal) Pure Acid

8 0.40

x 0.70 0.70x

8 +x 0.50 0.50 18 +x 2

0.40 182 =3.2

NOW TRY
EXERCISE 7
How many liters of a 20% acid
solution must be mixed with
5 L of a 30% acid solution to
get a 24% acid solution?

NOW TRY ANSWER

7. 7 12 L

The numbers in the last column of the table were found by multiplying the

strengths by the numbers of liters. The number of liters of pure acid in

the 40% solution plus the number of liters in the 70% solution must equal

the number of liters in the 50% solution.

Step 3 Write an equation.

Step 4 Solve.

Distributive property

Subtract 3.2 and 0.50x.

Divide by 0.20.

Step 5 State the answer.The chemist should use 4 L of the 70% solution.

Step 6 Check.8 L of 40% solution plus 4 L of 70% solution is

of acid.

Similarly, or 12 L of 50% solution has

of acid.

The total amount of pure acid is 6 L both before and after mixing, so the

answer checks. NOW TRY

121 0.50 2 = 6 L

8 + 4

81 0.40 2 + 41 0.70 2 =6 L

x= 4

0.20x=0.8

3.2+ 0.70x= 4 +0.50x

3.2+ 0.70x=0.50 18 +x 2

Remember that when pure water is added to a solution, water is 0% of the

chemical (acid, alcohol, etc.). Similarly, pure chemical is 100% chemical.

PROBLEM-SOLVING HINT