Hello. I'm Professor Von Schmohawk and welcome to Why U.
In the previous chapter
we saw how Algebra can be used to solve various types of percentage problems
including "percentage change" and "profit and loss".
In this chapter we will see how to solve problems
involving mixtures of solutions of different concentrations.
But what exactly is a solution?
And what does it mean to say that a solution has a particular concentration?
A solution is a mixture of one substance called the "solute"
dissolved in another substance called the "solvent".
The strength or "concentration" of a solution
is often specified as the percentage volume of the solution which is made up of solute.
This is called the "volume percent concentration".
Volume percent is usually used to specify concentration
when the solution is made by mixing two liquids.
For example, in a solution of antifreeze in water
the solute is antifreeze
and the solvent is water.
So a "25 volume percent" solution of antifreeze
is made up of one-quarter volume of pure antifreeze
and three-quarters volume of water.
The volume percent concentration of a solution
is calculated by dividing the volume of solute
by the total volume of the solution
and then multiplying this fraction by 100 to get a percentage.
For example, if the volume of solute is 80 milliliters
and the total volume of the solution made up of solvent plus solute is 800 milliliters
then the fraction of the solution which is solute is "80 over 800"
or one-tenth.
Multiplying this fraction times 100
we get a volume percent concentration of solute of 10%.
Given this relationship between the volume percent concentration
the volume of solute
and volume of the solution
we can calculate any one of these three quantities if we know the values of the other two.
Now, let's say that we have two flasks of antifreeze solution.
One solution contains 10% antifreeze
and the other contains 50% antifreeze.
If we mix 60 milliliters of the 10% solution
with 20 milliliters of the 50% solution
what will be the percentage of antifreeze in the mixture?
Let's start by creating a table which shows the volume of each solution
and the volume of pure antifreeze solute contained in each solution.
We know that the volume of the 10% solution is 60 milliliters
and the volume of the 50% solution is 20 milliliters.
Since the volume of pure antifreeze in the 10% solution was 10% of 60 milliliters
then the volume of antifreeze in that solution was "0.1 times 60"
or 6 milliliters.
Likewise, the volume of pure antifreeze in the 50% solution was 50% of 20 milliliters
or "0.5 times 20"
or 10 milliliters.
Now we can calculate the volume percent of antifreeze in the mixture of the two solutions
by using our formula.
The volume of pure antifreeze solute in the mixture
is "6 milliliters plus 10 milliliters"
or 16 milliliters.
And the volume of the mixture of the two solutions
is "60 milliliters plus 20 milliliters"
or 80 milliliters.
So the fraction of solution which is antifreeze is "16 divided by 80"
or 0.2.
Multiplying this fraction by 100
we get a volume percent concentration of 20%.
So the percentage of antifreeze in the mixture is 20%.
In our next example we will see how using Algebra
it is possible to produce a mixture of a desired concentration
by combining calculated quantities of two mixtures with different concentrations.
A.V. Geekman has borrowed a leaf blower to perform some yard work.
The blower runs on fuel which is a mixture of gasoline and oil
where 5% of the mixture is oil.
Unfortunately, A.V.'s fuel can is empty
and there are no other containers of gasoline or oil available.
However, there are two fuel barrels nearby which contain mixtures of gasoline and oil.
The fuel in one barrel contains 12% oil
and the fuel in the other contains 2% oil.
But since the blower requires a 5% oil mixture
the oil in one barrel is too concentrated
and the oil in the other is too diluted.
A.V. wonders if fuel from the two barrels could be combined in some way
to create a 5% oil mixture.
Fortunately, using Algebra
it is possible to calculate the amount of fuel to mix from each barrel
in order to fill his 10 quart can with fuel containing 5% oil.
We will start by creating a table which shows the quantity of fuel to take from each barrel
and the amount of oil contained in each of these quantities of fuel.
If we let x represent the volume of fuel to take
from the 12% barrel
then the amount of fuel to take from the 2% barrel must be "10 - x"
since the two volumes must total 10 quarts.
The amount of oil contained in the fuel from the 12% barrel will be "12% of x"
or 0.12x.
Likewise, the amount of oil contained in the fuel from the 2% barrel will be 2% of "10 - x"
or 0.02 times "10 - x".
So for any particular value of x
the top two expressions in our table tell us the volume of fuel to mix from each barrel
and the bottom two expressions
tell us the volume of oil which will be present in those volumes of fuel.
We can now use our equation for volume percent concentration
to calculate the value of x.
The "solution" in this equation is the fuel mixture of gasoline and oil we wish to create.
And the "solute" is the oil which is dissolved in the gasoline.
We know that the volume percent concentration of oil in the mixture we wish to create
must be 5%.
And the total volume of this "mixture" or in other words, this "solution"
must be 10 quarts.
The volume of solute is the total volume of oil in the mixture
in other words the volume of oil from the 12% barrel
plus the volume of oil from the 2% barrel.
We now have an equation which we can solve to find the value of x.
Since we are solving for x
we want to rearrange this equation so that x is alone on the left side.
Let's start by swapping the two sides of the equation.
Multiplying the fraction by 100
is the same as multiplying its numerator by 100.
Since 100 is "10 times 10"
we can cancel one of the 10's in the numerator and the 10 in the denominator
leaving just the remaining terms in the numerator.
Now, using distributive property
we distribute the 0.02 multiplier to the two terms in parentheses.
We can then eliminate the inner parentheses.
Completing the multiplication in the middle term
0.02 times 10 is 0.2.
And combining like terms in the parentheses
0.12x minus 0.02x is 0.1x.
We can then distribute the 10 to the two terms in the remaining parentheses
and eliminate the remaining set of parentheses.
Completing the multiplication in the first term
10 times 0.1x is 1x
or just x.
And in the second term 10 times 0.2
is 2.
Subtracting 2 from both sides of the equation
allows us to cancel the positive and negative 2's on the left
leaving us with the equation x equals "5 - 2"
or "x equals 3".
We can now substitute this value for x into our table
to find the quantity of fuel to take from each barrel.
We see that the quantity to take from the 12% barrel is 3 quarts
and the quantity to take from the 2% barrel is "10 - 3"
or 7 quarts.
So by using Algebra to calculate the quantities of the two fuel mixtures to combine
A.V. can complete his yard work.
In this lecture we have seen how to solve problems
involving mixtures of solutions of different concentrations.
In the next lecture we will see how to solve problems
involving other types of mixtures.