Hello. I'm Professor Von Schmohawk and welcome to Why U.
In the previous chapter we introduced the concept of percentages
and explored the principles on which percentages are based.
In this chapter we will use Algebra to solve real-world problems involving percentages.
It has been reported that in Sweden 20% of all road accidents involve an elk.
Out of the total number of Swedish road accidents in January
if 20% of them involved elks
and there were 24 of these road accidents with elks
then how many total road accidents were there in Sweden that month?
To solve this problem, let's let x represent the total number of road accidents in January.
We know that the 24 accidents involving elks were 20% of this number.
And since the decimal equivalent of 20% is 0.2
we can write this quantity as "0.2x".
Since "0.2x" is the number of accidents involving elks
we can now create the equation "0.2x is equal to 24".
To solve this equation we eliminate the coefficient of x, 0.2
by dividing both sides of the equation by 0.2.
This allows us to cancel the coefficient
leaving us with the equation "x equals 24 divided by 0.2"
or "x equals 120".
So according to our calculations
there were a total of 120 road accidents in Sweden during the month of January.
If in February, the number of accidents increased by 25%
then how many accidents occurred in February?
Let's let x represent the number of accidents in February.
The increase in accidents was 25% of the number of accidents in January
or 25% of 120.
Since the decimal equivalent of 25% is 0.25
we can write this increase as "0.25 times 120".
We can now create an equation which says that x, the number of accidents in February
equals the number of accidents in January plus 25% of that number.
To solve this equation for x we simply complete the arithmetic.
"0.25 times 120" is 30.
And "120 plus 30" is 150.
So 150 accidents occurred during the month of February.
This change represents a 25% increase from January to February.
If in March, the number of accidents returns to the previous amount
what percentage decrease from the accidents in February does this represent?
Since the change from February to March
is the same size as the change from January to February
it might be tempting to say that this change will also be 25%.
However, this is not the case.
We want to find the percentage of accidents in February which did not occur in March.
If we call the decimal equivalent of that percentage "p"
then "p times 150" is the difference in accidents between February and March.
We can now write an equation which says that the number of accidents in March
equals the number of accidents in February
minus the change in accidents, "p times 150"
or "150p".
To solve this equation we would like p alone on the left side
so we start by adding "150p" to both sides of the equation
which allows us to cancel the "150p" term on the right.
We then subtract 120 from both sides
which allows us to cancel the 120 term on the left.
Completing the arithmetic on the right
150 minus 120 is 30.
Finally, dividing both sides by 150
allows us to cancel 150 in the numerator and denominator of the fraction on the left
leaving the equation "p equals 30 divided by 150"
or "p equals 0.2".
So the decrease in accidents from February to March is 0.2 times 150.
Since 0.2 is the decimal equivalent to 20%
this represents a decrease in accidents from February to March of 20%.
Even though the size of the difference from January to February
is the same as from February to March
this same value represents a larger percentage of 120
than it does of 150.
Therefore, this same quantity represents a 25% increase from January
but only a 20% decrease from February.
For our next example
we will solve a problem which involves percentages of profit and loss.
For her birthday, Adrian's Uncle Monopolus
gives Adrian $1000 to invest in the stock market.
Adrian decides to split her investment between two start-up companies.
The first company, Alaskan Swimwear
plans to introduce a new line of hi-fashion swimsuits.
Unfortunately, hoping to capitalize on an underdeveloped niche market
they introduce their swimwear only to stores located north of the artic circle
and the stock drops to 50% of its initial value.
The second company, Smello-phone
has developed a smartphone which not only transmits sound and pictures
but can also transmit smells.
It's stock triples to 300% of its initial value.
In spite of her bad investment in Alaskan Swimwear
Adrian's brilliant Smello-Phone investment more than makes up for her loss
and the total value of her $1000 stock portfolio
increases to $1500.
Knowing only this information
is it possible to determine how much money Adrian invested in each company?
When solving Algebra problems
it is sometimes helpful to create a table of known and unknown values.
Let's create a table for the initial and final values of each investment
and Adrian's total investment portfolio.
We know that initially her portfolio had a value of $1000
and that the portfolio's final value was $1500.
Although we don't know the initial value of either investment
we do know the final value of the Smello-Phone investment was 300% of its initial value
and the final value of the Alaskan Swimwear investment was 50% of its initial value.
If we let x represent the initial value of the Smello-Phone investment
then the initial value of the Alaskan Swimwear investment must be “1000 minus x”
since both investments totaled $1000.
The final Smello-Phone value is then 300% of x
or 3x
and the final Alaskan Swimwear value is 50% of "1000 minus x"
or 0.5 times "1000 minus x".
We can now create an equation which states that the final value of the Smello-Phone investment
plus the final value of the Alaskan Swimwear investment
is equal to the final value of Adrian’s total portfolio.
To simplify this equation, we can use the distributive property
to distribute the 0.5 multiplier to the two terms in parentheses.
And completing the multiplication
0.5 times 1000 is 500.
To simplify further, we can combine all the terms containing x.
To do this, we rearrange the equation so that all the x terms are on the left side
and the terms without x are on the right.
Subtracting 500 from both sides of the equation
allows us to cancel the positive and negative 500’s on the left
leaving only the terms containing x.
Combining these terms, 3x minus 0.5x is 2.5x.
Completing the arithmetic on the right, 1500 minus 500 is 1000.
The coefficient of x can then be eliminated by dividing both sides by 2.5.
This allows us to cancel the 2.5’s in the numerator and denominator on the left
leaving the equation “x equals 1000 divided by 2.5"
or “x equals 400”.
Now we can substitute this value for x into the expressions in our table
to find the size of each initial investment.
We see that Adrian's initial Smello-Phone investment was $400
and her initial investment in Alaskan Swimwear was 1000 minus 400
or $600.
So far, we have seen how Algebra can be used to solve various types of percentage problems
including "percentage change" and "profit and loss".
In the next lecture we will see how to solve problems
involving mixtures of solutions of different concentrations.