§ 2.6

Percent and Mixture Problem Solving

Martin-Gay, Beginning and Intermediate Algebra, 4ed 22

Strategy for Problem Solving

General Strategy for Problem Solving1) UNDERSTAND the problem

• Read and reread the problem• Choose a variable to represent the unknown• Construct a drawing, whenever possible• Propose a solution and check

2) TRANSLATE the problem into an equation3) SOLVE the equation4) INTERPRET the result

• Check the proposed solution in problem• State your conclusion

Martin-Gay, Beginning and Intermediate Algebra, 4ed 33

Solving a Percent Equation

A percent problem has three different parts:

1. When we do not know the amount:n = 10% · 500

Any one of the three quantities may be unknown.

amount = percent · base

2. When we do not know the base:50 = 10% · n

3. When we do not know the percent:50 = n · 500

Martin-Gay, Beginning and Intermediate Algebra, 4ed 44

Solving a Percent Equation: Amount Unknown

amount = percent · base

What is 9% of 65?

n = 9% · 65

n = (0.09) (65)

n = 5.85

5.85 is 9% of 65

Martin-Gay, Beginning and Intermediate Algebra, 4ed 55

Solving a Percent Equation : Base Unknown

amount = percent · base

36 is 6% of what?

n = 6% ·36

36 = 0.06n36 0.06 =

0.06 0.06n

600 = n

36 is 6% of 600

Martin-Gay, Beginning and Intermediate Algebra, 4ed 66

Solving a Percent Equation : Percent Unknown

amount = percent · base

n = 144 ·24

24 is what percent of 144?

24 = 144n

24 144 = 144 144

n

0.16 = n216 % = 3

n 224 is 16 % of 1443

Martin-Gay, Beginning and Intermediate Algebra, 4ed 77

Solving Markup Problems

Mark is taking Peggy out to dinner. He has $66 to spend. If he wants to tip the server 20%, how much can he afford to spend on the meal?

Let n = the cost of the meal.

Cost of meal n + tip of 20% of the cost = $66

100% of n + 20% of n = $66120% of n = $66

1.2 66n 1.2 66 1.2 1.2n

55nMark and Peggy can spend up to $55 on the meal itself.

Example:

Martin-Gay, Beginning and Intermediate Algebra, 4ed 88

Solving Discount Problems

Julie bought a leather sofa that was on sale for 35% off the original price of $1200. What was the discount? How much did Julie pay for the sofa?

Julie paid $780 for the sofa.

Discount = discount rate list price

= 35% 1200= 420 The discount was $420.

Amount paid = list price – discount

= 1200 – 420

= 780

Example:

Martin-Gay, Beginning and Intermediate Algebra, 4ed 99

Solving Percent Increase Problems

The cost of a certain car increased from $16,000 last year to $17,280 this year. What was the percent of increase?

Amount of increase = original amount – new amount

The car’s cost increased by 8%.

amount of increase = original amount

Percent of increase

= 17,280 – 16,000 = 1280

amount of increasePercent of increase = original amount

1280= 16000

= 0.08

Example:

Martin-Gay, Beginning and Intermediate Algebra, 4ed 1010

Solving Percent Decrease Problems

Patrick weighed 285 pounds two years ago. After dieting, he reduced his weight to 171 pounds. What was the percent of decrease in his weight?

Amount of decrease = original amount – new amount

Patrick’s weight decreased by 40%.

amount of decrease = original amount

Percent of decrease

= 285 – 171 = 114

amount of decreasePercent of decrease = original amount

114= 285

= 0.4

Example:

Martin-Gay, Beginning and Intermediate Algebra, 4ed 1111

Solving Mixture Problems

The owner of a candy store is mixing candy worth $6 per pound with candy worth $8 per pound. She wants to obtain 144 pounds of candy worth $7.50 per pound. How much of each type of candy should she use in the mixture?

1.) UNDERSTAND

Let n = the number of pounds of candy costing $6 per pound.

Since the total needs to be 144 pounds, we can use 144 n for the candy costing $8 per pound.

Example:

Continued

Martin-Gay, Beginning and Intermediate Algebra, 4ed 1212

Solving Mixture Problems

Example continued

2.) TRANSLATE

Continued

Use a table to summarize the information.

Number of Pounds Price per Pound Value of Candy$6 candy n 6 6n$8 candy 144 n 8 8(144 n)$7.50 candy 144 7.50 144(7.50)

6n + 8(144 n) = 144(7.5)

# of pounds of $6 candy

# of pounds of $8 candy

# of pounds of

$7.50 candy

Martin-Gay, Beginning and Intermediate Algebra, 4ed 1313

Solving Mixture Problems

Example continued

Continued

3.) SOLVE

6n + 8(144 n) = 144(7.5)

6n + 1152 8n = 1080

1152 2n = 1080

2n = 72

Eliminate the parentheses.

Combine like terms.

Subtract 1152 from both sides.

n = 36 Divide both sides by 2.

She should use 36 pounds of the $6 per pound candy.

She should use 108 pounds of the $8 per pound candy.

(144 n) = 144 36 = 108

Martin-Gay, Beginning and Intermediate Algebra, 4ed 1414

Solving Mixture Problems

Example continued

4.) INTERPRET

Check: Will using 36 pounds of the $6 per pound candy and 108 pounds of the $8 per pound candy yield 144 pounds of candy costing $7.50 per pound?

State: She should use 36 pounds of the $6 per pound candy and 108 pounds of the $8 per pound candy.

6(36) + 8(108) = 144(7.5)?

216 + 864 = 1080 ?

1080 = 1080?