MIXTURE PROBLEMS
Problem: A
chemistry experiment calls for a 30% sulfuric acid solution. If the lab supply room has only 50% and 20%
sulfuric acid solutions on hand, how much of each should be mixed to
obtain 12 liters of a 30% solution?
Solution: Let
x be the number of liters of the 50% solution to be used. Let
y be the number of liters of the 20% solution to be used. Note
that x + y = 12 because we want
a total of 12 liters. We can
use this equation anytime we want to – for now for simplicity lets keep
using y. 12
liters which is 30% sulfuric acid contains (.3)(12) liters of sulfuric
acid. x
liters which is 50% sulfuric
acid contains (.5)x liters of sulfuric acid. y
liters which is 20% sulfuric acid contains (.2)y liters of sulfuric
acid. Note
we will mix x liters with y liters to obtain 12 liters. This
completes the analysis of the problem and we are now in a position to
model the mixture problem with an equation. (.3)(12)
= (.5)x + (.2)y . Now
replace y with 12 - x to obtain (.3)(12) = (.5)x + (.2)(12 – x) Now we need only solve this equation for x and then calculate yMultiply
both sides by 10 and do some of the multiplications and additions to
obtain 36
= 5x + 24 – 2x
Problem: How
many gallons of a 3% salt solution must be mixed with 50 gallons of
a 7% solution to obtain a 5% solution?
Solution: Let
x be the number of gallons of the 3% solution to be added to the 50
gallons of 7% solution. Note
that the result will be (x + 50) gallons. The
desired amount of salt in this (x + 50) gallons is 5% of (x + 50) or
(.05)(x + 50) The
amount of salt contributed by the two components of the mixture will
be: 3%
of x gallons or (.03)x and 7%
of 50 gallons or (.07)(50) The amount of salt contributed by the two components must equal the desired amount of salt in the final mixture. This
completes the analysis of the problem and we are now in a position to
model the mixture problem with an equation: (.05)(x + 50) = (.03)x + (.07)(50) Now
we simply solve this equation. Begin
by multiplying both sides of the equation by 100 to get rid of the decimals (.05)(x
+ 50) = (.03)x + (.07)(50) Û 5(x + 50) = 3x +(7)(50) Û 5x + 250 = 3x + 350 Û 2x = 100 Û x = 50 Problem: To
make low fat cottage cheese, milk containing 4% butterfat is mixed with
10 gallons of milk containing 1% butterfat to obtain a mixture containing
2% butterfat. How many gallons
of the richer milk is used.
Solution: Let
x be the number of gallons of 4% milk to be used. The final mixture
will then be (10 + x) gallons and we want it to contain 2% butterfat. The
amount of butterfat in the final solution must be 2% of (10 + x) gallons
or (.02)(10 + x) The
amount of butterfat contributed by the 10 gallons of 1% milk is (.01)(10) The
amount of butterfat contributed by the x gallons of 4% milk is (.04)(x) This
completes the analysis of the problem and we are now in a position to
model the mixture problem with an equation (.02)(10 + x) = (.01)(10) + (.04)(x) Now
we simply solve this equation. Begin
by multiplying both sides of the equation by 100 to get rid of the decimals (.02)(10
+ x) = (.01)(10) + (.04)(x) Û 2(10 + x) = 1(10) + 4x Û 20 + 2x = 10 + 4x Û 10 = 2x Û x = 5
Problem: A 100% concentrate is to be mixed with a mixture having a concentration
of 40% to obtain 55 gallons of a mixture with a concentration of 75%.
How much of the 100% concentrate will be needed?
Solution: Let
x be the amount of the 100% concentrate to be used. Then
55 – x is the amount of the 40% solution to be used. The
amount of concentrate contributed by the 100% solution is 100% of x
or simply x.
The
amount of concentrate contributed by the 40% solution is 40% of (55
– x) or simply (.4)(55 – x). The
total amount of concentrate is the sum of these two contributions, so
the total is x + (.4)(55 – x). However,
the total amount of concentrate in the final solution is required to
be 75% of 55 gallons or (.75)(55). This
completes the analysis of the problem and we are now in a position to
model the mixture problem with an equation x + (.4)(55 – x) = (.75)(55)
Now
we solve this equation. Begin
by multiplying both sides of the equation by 100 to get rid of the decimals. x
+ (.4)(55 – x) = (.75)(55) Problem: A grocer mixes peanuts that cost $2.49 per pound and walnuts that
cost $3.89 per pound to make 100 pounds of a mixture that costs $3.19
per pound. How much of each
kind of nut is put into the mixture?
Solution: Let
x be the amount of peanuts to be put into the mixture. Then
100 – x is the amount of walnuts put into the mixture. The
cost of the peanuts in the mixture is 2.49x The
cost of the walnuts in the mixture is 3.89(100 – x) The
total cost contributed by the peanuts and walnuts is 2.49x + 3.89(100
– x). However,
the total cost of the final mixture is required to be (3.19)(100) or
319. This
completes the analysis of the problem and we are now in a position to
model the mixture problem with an equation 2.49x + 3.89(100 – x)
= 319 Now
we solve this equation. Begin
by multiplying both sides of the equation by 100 to get rid of the decimals. 2.49x
+3.89(100 – x) = 319
So
50 pounds of peanuts and 50 pounds of walnuts must be used to obtain
100 pounds costing $3.19 per pound. Problem: A forester mixes gasoline and oil to make 2 gallons of mixture for
his two-cycle chain-saw engine. This
mixture is 32 parts gasoline and 1 part two-cycle oil. How much gasoline must be added to bring the
mixture to 40 parts gasoline and 1 part oil?
Solution: Let
x be the amount of gasoline to be added. The
total amount of mixture will then be x + 2 gallons. The
concept of 32 parts gasoline and 1 part oil in the original mixture,
quite naturally suggests that we should consider each gallon as being
divided into 33 equal parts. The
total amount of gasoline in the two gallons of the original mixture
is then 32 + 32 = 64 parts.
When
we add x gallons of gasoline to the mixture we will have
This
completes the analysis of the problem and we are now in a position to
model the mixture problem with an equation
as the mathematical model for this mixture problem. Now
we solve this equation.
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