§ 2.6
Percent and Mixture
Problem Solving
Strategy for Problem Solving
General Strategy for Problem Solving
1) 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
2
Solving a Percent Equation
A percent problem has three different parts:
amount = percent · base
Any one of the three quantities may be unknown.
1. When we do not know the amount:
n = 10% · 500
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
3
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
4
Solving a Percent Equation: Base Unknown
amount = percent · base
36 is 6% of what?
36 = 6% ·
n
36 = 0.06n
36
0.06n
=
0.06
0.06
600 = n
36 is 6% of 600
Martin-Gay, Beginning and Intermediate Algebra, 4ed
5
Solving a Percent Equation: Percent Unknown
amount = percent · base
24 is what percent of 144?
24 =
n
· 144
24 = 144n
24
144n
=
144
144
0.16 = n
2
16 % = n
3
2
24 is 16 % of 144
3
Martin-Gay, Beginning and Intermediate Algebra, 4ed
6
Solving Markup Problems
Example:
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
100% of n
+
+
tip of 20% of the cost
20% of n
120% of n
1.2n  66
1.2n
66

1.2
1.2
n  55
=
=
=
$66
$66
$66
Mark and Peggy can spend
up to $55 on the meal itself.
Martin-Gay, Beginning and Intermediate Algebra, 4ed
7
Solving Discount Problems
Example:
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?
Discount = discount rate  list price
= 35%  1200
The discount was $420.
= 420
Amount paid = list price – discount
= 1200 – 420
= 780
Julie paid $780 for the sofa.
Martin-Gay, Beginning and Intermediate Algebra, 4ed
8
Solving Percent Increase Problems
Percent of increase =
amount of increase
original amount
Example:
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
= 17,280 – 16,000 = 1280
amount of increase
original amount
1280 = 0.08
=
The car’s cost increased by 8%.
16000
Percent of increase =
Martin-Gay, Beginning and Intermediate Algebra, 4ed
9
Solving Percent Decrease Problems
amount of decrease
Percent of decrease =
original amount
Example:
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
= 285 – 171 = 114
amount of decrease
Percent of decrease =
original amount
114 = 0.4
=
285
Patrick’s weight
decreased by 40%.
Martin-Gay, Beginning and Intermediate Algebra, 4ed
10
Solving Mixture Problems
Example:
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.
Continued
Martin-Gay, Beginning and Intermediate Algebra, 4ed
11
Solving Mixture Problems
Example continued
2.) TRANSLATE
Use a table to summarize the information.
$6 candy
$8 candy
$7.50 candy
Number of Pounds
n
144  n
144
Price per Pound
6
8
7.50
Value of Candy
6n
8(144  n)
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
Continued
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Solving Mixture Problems
Example continued
3.) SOLVE
6n + 8(144  n) = 144(7.5)
6n + 1152  8n = 1080
1152  2n = 1080
2n = 72
n = 36
Eliminate the parentheses.
Combine like terms.
Subtract 1152 from both sides.
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
Continued
13
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?
?
6(36) + 8(108) = 144(7.5)
?
216 + 864 = 1080
?
1080 = 1080 
State: She should use 36 pounds of the $6 per pound candy and
108 pounds of the $8 per pound candy.
Martin-Gay, Beginning and Intermediate Algebra, 4ed
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