10-8 mixture problems standard 15.0: apply algebraic techniques to percent mixture problems....
DESCRIPTION
#1 Solving Mixture Problems Let x = The number of lbs of the first solution. Let 175 – x = The number of lbs of the second solution. A grocer wishes to mix some peanuts worth $0.90 per pound with some M&M’s worth $1.60 per pound to make 175 pounds of a mixture that is worth $1.30 per pound. How much of each should she use?TRANSCRIPT
10-810-8Mixture ProblemsMixture Problems
Standard 15.0: Apply algebraic Standard 15.0: Apply algebraic techniques to percent mixture techniques to percent mixture
problems.problems.
VocabularyVocabulary Problem-Solving GuidelinesProblem-Solving Guidelines
UNDERSTAND the problemUNDERSTAND the problem Develop and carry out a PLANDevelop and carry out a PLAN Find the ANSWER and CHECKFind the ANSWER and CHECK
#1 Solving Mixture Problems
poundper $1.60 Solution Second
lb.per $0.90 Solution First
Let x = The number of lbs of the first solution.
poundper $1.30 Solution Final
Let 175 – x = The number of lbs of the second solution.
A grocer wishes to mix some peanuts worth $0.90 per pound with some M&M’s worth $1.60 per pound to make 175 pounds of a mixture that is worth $1.30 per pound. How much of each should she use?
#1 Solving Mixture Problems
First Solution
Second Solution
Number of pounds
Price per pound
Amount of mixture
Final Solution
x
x17590.060.1
x90.0)175(6.1 x
5.2276.12809.0 xx5.527.0 x
lbs. 75x
175 30.1 )175(3.1
lbs. 100
acid 30% Solution Second
acid 60% Solution First
acid 50% Solution Final
#2 Solving Mixture Problems
Let x = amount of the first solution.
Let 750 – x = The amount of the second solution.
A chemist has one solution that is 60% acid and another that is 30% acid. How much of each solution is needed to make a 750 mL solution that is 50% acid?
#2 Solving Mixture Problems
First Solution
Second Solution
Amount of solution
Percent acid
Amount of acid
Final Solution
x
x750%60%30
x6.0)750(3.0 x
3753.02256.0 xx1503.0 xmL 500x
750 %50 )750(5.0
mL 502