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Math 31 Activity # 13
“Solving Mixture Problems”
Your Name: ___________________
Solving Word Problems: The Cohort Strategy!
Step 1) Read the problem at least once carefully. Look for key words and
phrases. Determine the known and unknown quantities. Let x or another
variable to represent one of the unknown quantities in the problem.
Step 2) If necessary, using the same variable from Step 1, write an expression
using the to represent any other unknown quantities.
Step 3) Write a summary of the problem as an English statement. Then write an
equation based on your summary.
Step 4) Solve the equation.
Step 5) Check the solution. Ask yourself "Is my answer reasonable?"
Step 6) Write a sentence to state what was asked for in the problem, and be sure
to include units as part of the solution. (inches, square feet, gallons,
ounces, etc., for example).
Task 1) Percent Mixture Problems
Example 1:
How many gallons of a 15% sugar solution must be mixed with 5 gallons of a 40% sugar solution to make a 30% sugar solution?
Begin by creating a drawing of the situation and filling in the known information, as shown below.
Amount of Solution + ___ gallons =
+ =
Concentration (Percents) ___^ ____^ ____
Amount of Ingredient in Mixture + =
Now we must decide how to represent the unknown quantities. Since the question is asking us to find the number of gallons of 15% solution we will use to obtain the required result, it will be useful to have our variable represent the amount of 15% sugar solution we are using. Therefore:
Step 1) Let x = # of gallons of 15% sugar solution Step 2) x + 5 = # of gallons in the mixture
We place this variable and the variable expression in the appropriate place in our drawing and then perform all the mathematical operations which are applicable. See the drawing below for these results.
( )
( )
( )
15 %^10340 % 5 30 %^1035
+ = ^ +
+ = ^ +
+ = ^ +
Step 6) So, we will need to add 103 or 3 13 gallons of 15% sugar solution to 5 gallons of 40% sugar solution to produce a 30% sugar solution.
Exercise 1:
How many ounces of 50% alcohol solution must be mixed with 80 ounces of a 20% alcohol solution to make a 40% alcohol solution?
Begin by creating a drawing of the situation and filling in the known information, in the appropriate place in the drawing below.
Amount of Solution + =
Concentration (Percents)
Amount of Ingredient in Mixture + =
Step 1) Let _____ = # of ounces of 50% alcohol solution. Then
Step 2) ________= _______________________________
Now place this variable and the variable expression in the appropriate place in the drawing below.
Amount of Solution + =
Concentration (Percents)
Amount of Alcohol in Mixture + =
Step 3) The resulting equation is:_________________________________
Step 4) Solve:
Step 5) Check:
Step 6)
Example 2:
A chemist needs to mix an 18% acid solution with a 45% acid solution to obtain 12 liters of a 36% solution. How many liters of each of the acid solutions must be used?
Step 3) The resulting equation would be: 18% x + 45% 12 ( − x ) =36% 12( )
Step 4)
( ) ( ) ( ) ( )
x x x x x x x x x
4 li
Step 5) Check:
( ) ( ) ( ) ( )
432 432
Step 6) The chemist needs 4 liters of 18% acid solution and 8 liters of 45% acid solution.
Exercise 2: Find the number of liters of an 18% alcohol solution that must be added to a 10% alcohol solution to get 20 liters of a 15% alcohol solution.
Step 1) Let x =__________________________. Then
Step 2) __________=________________________________
Amount of Solution + =
Concentration (Percents)
Amount of Alcohol in Mixture + =
4 liters of 18% solution
From step 2 ( 12 − x )is the # of liters of the 45% solution so 12 − 4 = 8
Step 3) Equation: ____________________________
Step 4) Solve:
Step 5) Check:
Step 6)
Task 2) Value Mixture Problems
Other types of mixture problems are called value-mixture problems such as mixing 2 ingredients that have different unit cost into one single blend, mixing of items with different monetary values such as coins, stamps, tickets, etc.
a) Mixing two ingredients into a single blend.
unit cost of ingredient (price) × amount of ingredient = value of ingredient
b) Mixing coins that have different monetary values.
value of one coin × number of coin = total value
c) Mixing stamps that have different monetary values.
value of one stamp × number of stamps = total value
Example 3: How many pounds of chamomile tea that cost $9 per pound must be mixed with 8 pounds of orange tea that cost $6 per pound to make a mixture that costs $7.80?
Step 6) 12 pounds of chamomile tea that cost $9/lb is needed.
Example 4: A piggy bank has a mixture of coins in dimes and quarters. There are 3 more quarters than dimes with a total value of $4.95 or 495 cents. How many quarters and dimes are there in the piggy bank?
Begin by creating a drawing of the situation and filling in the known information, as shown below.
Number of coins? +? =
Value per coin 10 25 Total Value + = 495 Step 1) Let d = # dimes.
Step 2) d+7 = # of quarters (since there are 7 more quarters than dimes)
Number of coins d + d+3 = d+(d+3)
Value per coin 10 25 Total Value 10d + 25(d+3) = 495 Step 3) 10 d + 25 ( d + 3 ) = 495
Step 4)
( )
d
d
d
d d
d d
Step 5) check:
Step 6)
2) A coin purse contains 24 coins in nickels and quarters. The coins have a total value of $4.40. Find the number of nickels and the number of quarters in the bank.
3) How many kilograms of hard candy that cost $7.50 per kg must be mixed with 24 kg of jelly beans that cost $3.25 per kg to make a mixture that costs $4.50 per kg?
4) Beatrice wants 100 ml liters of 5% salt solution but she only has a 2% salt solution and 7% salt solution. She will need to mix a quantity of a 2% salt solution with a 7% salt solution. How many ml of each of the two solutions will she have to use?
5) A chemist needs 5 liters of a 50% salt solution. All she has available is a 20% salt solution and a 70% salt solution. How many liters of each of the two solutions should she mix to obtain her desired solution?
