Hello. I'm Professor Von Schmohawk and welcome to Why U.
In the previous chapter we saw how Algebra can be used
to solve problems involving mixtures of solutions.
In this chapter we will see how to solve problems involving other types of mixtures.
Adrienne Scienstein has taken a part-time job
managing her Uncle Monopolus' coffee shop.
The shop sells two different types of roasted coffee beans at two different price points.
Their very expensive Tasmanian roasted coffee
hand-picked by specially-trained wombats
sells for $31 per pound.
The shop also offers an inexpensive coffee
roasted at the White Sands Nuclear Testing Range
for $3 a pound.
Adrienne thinks that a moderately-priced coffee
might appeal to a broader group of customers.
She decides to mix the two types of coffee beans
to create a special blend costing $10 per pound.
To do this, Adrienne must calculate the quantity of each type of coffee bean needed
to create a pound of coffee costing $10.
As we saw in previous chapters
this problem can be organized by creating a table.
In the first row of the table we will list the price per pound of each coffee.
We know that the Tasmanian coffee costs $31 per pound
the White Sands nuclear roast costs $3 per pound
and the blend we wish to create must cost $10 per pound.
In the second row we list the weight of each coffee to mix
to create one pound of coffee blend which costs $10.
If we let x represent the weight of the Tasmanian roast
then the weight of nuclear roast must be "one minus x"
since the combined weight of Tasmanian roast plus nuclear roast must be one pound.
In the third row, we will list the total cost of each coffee
given the price per pound
and the weight.
The cost of the Tasmanian roast is $31 per pound times x pounds
or "31x" dollars.
Likewise, the cost of the Nuclear roast is $3 per pound times "1 minus x" pounds.
Since the combined cost of the two coffees in the blend must be 10 dollars
we can create an equation which says that the cost of the Tasmanian roast
plus the cost of the nuclear roast
equals $10.
We can now solve this equation for x.
We start by distributing the 3
to the two terms in parentheses.
We can then multiply 3 times 1
and eliminate the parentheses.
Combining like terms we subtract 31x minus 3x
to get 28x.
Subtracting 3 from both sides of the equation
allows us to cancel the positive and negative 3's on the left
leaving the equation "28x equals 10 minus 3"
or "28x equals 7".
We then divide both sides by 28
allowing us to cancel the 28's in the numerator and denominator of the fraction on the left.
This leaves us with the equation "x equals 7 over 28"
or "x equals one-quarter".
So to create one pound of a $10 per pound blend
Adrienne must mix one-quarter pound of Tasmanian roast
with "1 minus one-quarter"
or three-quarters pound of Nuclear roast.
We can check our calculations by multiplying the price per pound for each of the two coffees
times their calculated weights
and check to see if the total cost for a pound of blend comes to $10.
The cost of the Tasmanian roast in one pound of Adrienne's blend
is 31 dollars times one-quarter pound
and the cost of the Nuclear roast is 3 dollars times three-quarters pound.
Multiplying 31 times one-fourth
we get 31 fourths.
And 3 times three-fourths
is nine-fourths.
Adding these two fractions we get "31 plus 9" fourths
or 40 fourths
which is equal to 10.
Since the total cost of the calculated amounts of the two roasts is 10 dollars
our calculations appear to be correct.
Our next example illustrates how algebra can be used to solve problems
involving mixtures of items selling at different prices.
The Why U Entertainment Committee is planning a concert
of the world-famous bagpipe quartet
"Alasdair, Ian, Angus, and Clyde".
The concert will be performed at the Why University concert hall
which seats 1000 people.
Normally tickets sell for $50 a seat.
However, to be sure that the 1000 seat auditorium is completely filled
the committee has decided to offer a number of discounted tickets at $30 a ticket.
The total revenue collected from ticket sales must be at least $45000
in order to cover the cost of the performance.
The Entertainment Committee therefore needs to know
how many discounted tickets can be offered and still collect $45000
assuming that a total of 1000 tickets are sold.
Fortunately, using Algebra, we can calculate how many discounted tickets can be sold
and still collect $45000.
Once again, we can organize this problem by creating a table.
In the first row of the table we list the ticket prices
for discounted
and non-discounted tickets.
In the second row, we list the number of discounted and non-discounted tickets
which will be sold.
If we let x represent the number of discounted tickets
then the number of non-discounted tickets will be "1000 minus x"
since the total number of tickets is 1000.
In the third row, we list the revenue generated by each type of ticket
given the ticket price
and the number of tickets sold.
The revenue produced by the discounted tickets is 30 dollars times x tickets
or "30x" dollars.
Likewise, the revenue produced by the non-discounted tickets is 50 dollars times "1000 minus x" tickets.
Since the total revenue generated by both types of tickets must be $45000
we can create an equation which says that the revenue from discounted tickets
plus the revenue from non-discounted tickets
is equal to 45000.
We can now solve this equation for x.
We start by distributing 50
to the two terms in parentheses.
We then complete the arithmetic in the parentheses multiplying 50 times 1000
and eliminate the parentheses.
Combining like terms we subtract 30x minus 50x
to get negative 20x.
And subtracting 50000 from both sides of the equation
allows us to cancel the positive and negative 50000
leaving the negative 20x term alone on the left.
Completing the arithmetic on the right we subtract 45000 minus 50000
to get negative 5000.
We then eliminate the coefficient of x by dividing both sides by negative 20
allowing us to cancel the negative 20's
in the numerator and denominator of the fraction on the left.
Finally, dividing negative 5000 by negative 20
we get "x equals 250".
So 250 discounted tickets can be sold
and "1000 minus 250"
or 750 non-discounted tickets can be sold.
We can check our calculations by multiplying the ticket price for each type of ticket
times the number of tickets
and check to see if the total revenue comes to 45000 dollars.
The revenue from the discounted tickets is 30 dollars times 250 tickets
and the revenue from the non-discounted tickets is 50 dollars times 750 tickets.
Multiplying 30 times 250
we get 7500
and 50 times 750
is 37500.
Adding these two numbers
we get 45000.
Since the total revenue from the calculated quantities
of discounted plus non-discounted tickets is 45000 dollars
we know that our calculations are correct.
In the last several lectures we have seen how to create linear equations
to solve various types of real world problems.
In the next few lectures we will explore algebraic methods
which we can use to create parallel and perpendicular lines.