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Percent and Mixture Problem Solving. § 2.6. Strategy for Problem Solving. General Strategy for Problem Solving 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. - PowerPoint PPT Presentation
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§ 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 equation.3) SOLVE the equation.4) 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 0.06
n
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 = 144n24 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 = $66100% of n + 20% of n = $66
120% of n = $661.2 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 amountPercent of increase
= 17,280 – 16,000 = 1280
amount of increasePercent of increase = original amount1280= 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 amountPercent of decrease
= 285 – 171 = 114amount of decreasePercent of decrease = original amount114= 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 ProblemsExample continued
Continued
3.) SOLVE6n + 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.) INTERPRETCheck: 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?