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Hello, and welcome to Bay College's online lectures for
College Algebra.
In this video, we're going to look at Section 1.7, which
deals with application problems, which are otherwise
known as story problems.
Which may strike fear into the hearts of many,
but you're not alone.
A lot of us do struggle with story problems.
It essentially takes a lot of practice, and we can employ
some strategies that we're going to take a look at in
this video as we do some examples.
The first strategy is-- the most simple as possible.
We want to keep it simple.
When it comes to any story problem, we need to read it,
and we need to read it many times to make sure that we
understand it.
But each time we read it, we want to approach it with a
different strategy.
Excuse me.
The first time we read any story problem, we
need to know the words.
We have to make sure we understand the terminology
being used.
As an example, if the story problem says, in
an isosceles triangle.
Well, if I don't know what an isosceles triangle is or what
those terms mean, I can't go any further
in the story problem.
So you have to make sure you understand
the words being used.
So that's your first strategy when you
read the story problem.
The second time you read it, you ask yourself, what is the
given information?
What's useful that I can pull from this explanation?
Then we read it a third time, and we ask ourselves, what do
I need to find?
What is this problem giving me as my variable?
What am I going to assign an x to or any other variable I use
to denote what I need to find?
Once we've done that, we've read it three times, now it's
time to attempt to build the equation and solve it.
And honestly, this is probably the hardest part, to build a
workable equation.
Solving it's relatively simple.
At this point in algebra, our algebra skills are pretty
defined, I guess you could say.
Once we find a solution, we need to read the problem a
fourth time to say, does the solution I found make sense?
Does it have the appropriate units?
Does it answer the actual question that was asked?
So let's go ahead and do an example here.
This one deals with simple interest.
And hopefully, when we see the term "simple interest," it
conjures up this equation in our mind.
This may not always be given, so you have
to know this equation.
Interest equals principal times rate times time.
That's what it translates to.
Interest is the amount of money earned on some principal
P at some interest rate r, which is
usually given as a percent.
And hopefully, we know, oh, percents have to be converted
to numbers.
And t is the time that it's earning that interest,
generally given in years.
So let's read the problem.
And the first time we're going to read it, it's just make
sure we understand the terminology.
A sum of money is invested--
and I'm underlining that because we have to know what
that term is-- at 10%, and twice that amount at 12%.
So what I know from this just from reading it is we're going
to invest some money at 10%, and we're going to do twice
that same amount at 12%.
The yearly income was 2,890.
So I know the income is going to be what I get after some
amount of time that it's invested at 10%
and invested at 12%.
I'm going to get a total amount of interest here.
And because this says yearly, that's given information, and
we'll get to that when we read it.
I know what a year is.
OK?
So I know that's a unit of time.
And then how much was invested at each rate.
So that's the first time we read it.
I'm just making sure I understand invested, twice,
yearly, understand what the terms are.
Now, I'm going to read a second time and say, what is
given information?
A sum of money is invested at 10%, so I have something being
invested at 10%.
Twice that amount at 12%, so 12% is given information.
The yearly income was 2,890.
I'm going to use a table in this example, and in some of
the other examples to come, and I'm just going to put in
the given information.
I already have 10% here for one account,
12% for another account.
I know my interest rate, 0.1--
convert percents to numbers--
or 0.12.
So that's the rate.
And I know the time because it says yearly.
That's given information--
one year that they're in there.
Now, we have a total amount of interest.
The interest total in this column is 2,890.
So now I have to assess my table and say, well, what's
the missing information?
I need P, r, t, and I. Well, I have some r, and I have some
t, and I have some I. What I'm missing is P. So I have to go
back and read it the third time and say, what
do I need to find?
A sum of money is invested at 10%, and twice that amount is
invested at 12%.
Well, that sum of money is my principal.
So I'm going to say, some amount of money
is invested at 10%.
Twice that sum is invested at 12%.
So now my table is complete.
I can start building my equation.
Principal times rate times time is interest.
So P times 0.1 times 1 is 0.1P.
In my 12% account, I have 2P times 0.12 times 1 is my
interest, so 0.24P is my interest.
And the total is 2,890.
Well, how do we find a total?
Well, the total, we have to do addition, sum, we have to add.
So I'm going to go over here.
This is 0.1P plus 0.24P equals 2,890.
So we find that, essentially, just by saying this interest
and this interest is the yearly interest, the interest
gained in an entire year.
And now this is my equation.
Now, the easy part, solving it.
When I solve it, I just combine like terms, 0.34P
equals 2,890.
And then I can divide by 0.34, 2,890/0.34 is 8,000.
Let's clean that up a little bit, so hopefully you can see
it in the video there, 8,000.
Well, we have an answer now.
P is 8,000.
8,000 what?
We need to assign units.
Units are very important in story problems.
Well, we're dealing with a sum of money, so I know
this 8,000 is money.
But what is this value?
This is the money that's invested at 10%, so I'm just
going to write 10%.
So we found this.
Did we answer the question?
Well, let's go back and read that fourth and final time.
A sum of money is invested at 10% and twice
that amount at 12%.
The yearly income was 2,890.
How much was invested at each rate?
I only found one of these values, the one at 10%.
I need to find each rate.
So we look at this and say, well, this is just twice P. If
p is this value--
actually, I think I did a little math here.
It's 8,500--
obviously, I worked this out, this is not something I would
do in my head--
at 12%.
So now we've answered the entire question.
All right?
We found that there's $8,500 invested at
10%, $17,000 at 12%.
Now, does this answer make sense?
Well, 10% of this plus 12% of that, 2,890?
Well, that makes sense, right?
It's a value less than either of these.
That just means the interest rate was relatively
small, 10% and 12%.
And I say relatively small because it would be great in
today's economy if we could get those
kind of interest rates.
All right.
So our answer makes sense.
We're going to move on to another application problem,
and this one deals with mixtures.
A coffee company wants a new flavor of cappuccino.
How many pounds of coffee beans worth $7 a pound should
be added to 14 pounds of beans worth $4 a pound to get a
mixture worth $5 a pound?
So the first time I read it is just to understand the words.
OK, I understand what pounds are, and I understand that
there's two types of beans being mixed together, and they
have different values.
So I have bean number 1 and bean number 2.
So again, I'm going to use a table.
And I know something about value.
If I know I have some cost, and I know that cost per item,
if I know the number of items, I will have the total cost.
Well, value per pound times pounds equals the total value
of whatever bean I'm working with.
So we do a little bit of assessing.
Do I understand?
Do I bring enough information to the story problem that I
can complete it?
So now I'm going to read it a second time and look for that
given information.
We're told that some coffee bean is worth $7 a pound.
OK.
Well, bean 1 is $7 a pound, its value for every pound.
And it's going to be added to 14 pounds of a bean, which is
our second bean.
I'm told something about pounds.
Well, I know the number of pounds there.
And that's worth $4 a pound.
Hey, this bean, number 2, is worth $4 a pound.
To get a mixture worth $5 a pound, well, that's given
information.
The mixture's value per pound is $5, $5 a
pound or $5 per pound.
So to complete our table, we have this column, we have this
column, and I know how to find a total.
It has to do with something that's just adding, right?
That's what total means, to do a sum.
I'm missing some pieces to my table.
Well, bean number 1, how many pounds?
Well, let's read it the third time and say, all right, we
have how many pounds of coffee beans worth $7?
$7, how many pounds-- that's where I'm going
to assign my variable--
should be added to 14 pounds of the beans?
Well, here's a key word that I should have underlined before.
So many pounds of this are being added to 14
pounds of this bean.
Well, if we're going to do that, this is 14
plus x or x plus 14.
The order really doesn't matter.
So now we have our table complete.
Well, let's find what bean number 1's total value is.
The cost times the amount is the total value.
The cost times the amount is the total value, and here it's
going to be 56.
It's just four times 14.
Here we have 5 times 14 plus x.
And I'm just going to distribute, and I'm going to
get 5 times x and 5 times 14, which is 70.
Now, to find the value of the mixture, well, I have a value
for my $7 beans, and I have a value for my $4 beans.
If I put them together, I'm going to have a
value for my mixture.
Essentially, this value plus this value equals this value.
This is my equation.
And I'm just going to write it up here, 7x plus 56
equals 5x plus 70.
And now we can solve this equation.
If I subtract 5x from both sides, I can subtract 56 from
both sides, and I'm going to get 2x equals 14.
And so now I can solve that-- x equals 7.
7 what?
Again, we need our units.
Well, this was my column for pounds. x represented the
number of pounds of the $7 bean, so 7 pounds of $7 beans.
So we could say of $7 beans.
Is that what it asked me to find?
I found an answer.
Let's go back and read the story problem a fourth time.
How many pounds of coffee beans worth $7 a
pound should be added?
And we could continue, but at this point we realize, oh, we
were looking for the number of pounds of the $7 bean, which I
found right here.
And I'm sorry about crowding that, but the amount of board
space I have is relatively limited here.
So $7 a pound-- we did answer the question.
Does it make sense?
Well, let's kind of review here for a moment.
This is an expensive coffee, and this is a relatively
inexpensive coffee.
And if we're mixing them together, well, the cost
should lie somewhere between $7 and $4.
Makes sense that it's $5 a pound.
Well, we should look at our weight as well.
The sum of this weight plus the sum of that weight
shouldn't be any crazy weight.
If I got something like 50 pounds, well, that would be
kind of ridiculous.
But I see 7 pounds, so I'm using half of the amount of
this for that coffee.
And this price is closer to this value than that, so it
makes sense that this would be a greater value than this one.
So just doing a little critical thinking, we can say,
you know what?
That is a reasonable answer.
All right.
Let's move on to the next example here.
This one's uniform motion.
And I wrote our equations that we need to bring to the table
when it comes to uniform motion just like I did with
the interest.
Interest equals principal times rate times time.
Well, this is distance equals rate times time.
Now, this equation, if you think about it, we work with
it all the time, daily.
If you drive to school or work or wherever you drive to, if
you do drive, you see this equation every time you read
your speedometer.
Your speedometer is measured in miles per hour.
That is a variation of this.
Rate equals miles, which is a distance, per hour, which is a
unit of time.
Essentially, if I solve this for r, I divide
both sides by d.
So we actually look at this equation every
single day if we drive.
There's other things we can do.
We can say time equals distance over rate, another
variation of that equation.
And sometimes, maybe other variables are involved.
Displacement, which is sometimes labeled as s, equals
v, which is our velocity, times time, which is t.
Notice time is pretty consistent no
matter where you go.
And we can write any variation of that equation.
It's just rearranging it algebraically.
So let's look at this example of an
application problem here.
It says a boat heads upstream on a river with a current of 3
miles per hour, which takes 5 hours.
The return trip takes 2 and 1/2 hours.
What is the speed of the boat?
So I've read it the first time, and I feel comfortable I
understand the words, but I just want to highlight for a
moment upstream.
Well, to go upstream, we know that water
flows downhill, right?
Water doesn't flow uphill.
We don't have rivers that go up, so it flows down.
So if we're going upstream, we're going against the flow
of the river, and that's an important concept.
All right.
It has a current of 3 miles per hour, which takes 5 hours.
The return trip--
this is another key.
Well, if we're coming back, we're going with the current.
And it makes sense that the trip takes a little shorter
because we're going with the current.
We're not fighting the current.
What is the speed of the boat?
All right.
The second time we read it, we're going to say, what's the
given information?
Well, if we're going upstream--
we're going to assume that the boat has a constant speed--
but it's going upstream, which is against the current, 3
miles an hour.
So I know I'm fighting the current.
If I'm going upstream, it's against the current.
And if I'm coming downstream, the return trip,
I'm with the current.
All right.
And we're given something about time.
We know 5 hours upstream and 2.5 hours downstream.
But what do we know about the distance?
Well, here's a good place where not just using a table,
but maybe we want to do an illustration.
I'm going from some point on the river to some other point
on the river, and that's in the most basic sense.
I'm no artist, so we're just going to do this.
I'm traveling from here, against the current, and then
I'm traveling from here back to where I started.
It's a round trip, right?
I head upstream, and then I come back downstream.
I return to where I started.
That's the key word, and that's why I underlined it.
So the distance here is the same.
The distance upstream equals the distance downstream.
So I'm just going to say those two are equal,
whatever they may be.
So what am I missing here?
Well, either distance or rate, and I was told something about
rate, miles per hour.
Rate equals distance over time.
And when I read it the third time, it asks me, what is the
speed of the boat?
So I'm going to call that my variable r because I'm looking
for a rate, or maybe I want to use velocity,
whatever your choice.
So the rate of the boat is r, but it's going against the
current upstream.
So whatever the boat is minus 3 because I'm
fighting that current.
It's slowing me down.
If I go downstream, I'm traveling with the current.
So it's going to make the boat go a little bit faster, so the
rate of the boat plus the speed of the
current pushing it along.
And now what we have is rate times distance equals time.
Well, if their distances are the same, their rate times
their times are the same.
So I can build my equation.
So I have this times that.
5 times r minus 3 is equal, my distance, and let me just
write it in here.
We've completed our table.
We could build that equation.
This equals 2.5 times r plus 3.
And if we do the distributive property and combine some like
terms, we're going to get 2.5r.
And I know I'm skipping steps here, but you can work this
out at home.
Let's see, we're going to have 15.
And this says 22.5.
And we could divide both sides by 2.5, and we're going to get
the rate of the boat, which is 9.
22.5 divided by 2.5 is 9.
9 what?
Well, we have to look at what we're dealing with.
We're dealing with a distance that's in hours, because this
was 5 hours and 2.5 hours, and a rate, which
was miles per hour.
So our unit is 9 miles per hour.
Distance over time.
So we have rate is 9 miles per hour.
Let's make sure that we've answered the question because
we do have a solution here.
So we could read through it a fourth time, but the key is
right here.
What is the speed of the boat?
r was my variable that represented the speed of the
boat. r is 9 miles per hour.
That is the speed of the boat.
And that answer makes sense.
We're going 3 times the speed of the current, so at least
we're going to get somewhere, right?
It has to be greater than that, otherwise we wouldn't be
going anywhere.
So it's a reasonable answer.
And if we think about it, yep, 9 miles per hour there, it's
going to take some time because we're really only
going 6 miles an hour.
And on the way back, we're going that much faster.
We were coming back at 12 miles an hour.
Well, 6 miles an hour, 12 miles an hour,
you think about it.
If this is half the rate, there should be twice the
time, and we could see it is.
So it does make sense.
It is a reasonable answer.
All right.
Let's move on to the last example.
And this one is going to be your quiz, essentially.
It says Pat, by himself, can paint 4 rooms in 10 hours.
If his partner Bob helps, they can do the same job together
in 6 hours.
If Bob works alone, how long would it take him
to paint the 4 rooms?
Before I let you loose on your own to solve this, I'm going
to give you a little bit of background about how to deal
with constant rate equations.
Constant rate equations basically say,
one item or one job--
we'll call it a job here--
is completed in some amount of time, some unit of time.
So one job for so many units of time.
So if you read this, we can say, Pat can paint
4 rooms in 10 hours.
If we think about it, 4 rooms is the job.
So we could say this is 4 rooms, and he does that in
some unit of time, 10 hours.
All right.
4 rooms per 10 hours.
If his partner Bob helps, they can do the same job together
in 6 hours.
If Bob works alone, how long would it take him
to paint the 4 rooms?
So we've read the problem.
We're pulling that given information.
Well, what do we know about together?
What does that tell us?
It tells us to add, so I know I'm going to add Pat's
constant rate and Bob's constant rate.
Well, that's my missing piece.
That's what I'm going to read the second time and say, here
I need to assign some variable.
He can do 4 rooms in x amount of hours.
Well, in here, it says they could do
the same room together.
So together, that's another thing that I'm pulling from
here the second time that I read it.
So I'm going to give you a little bit
of heads up on this.
So Bob, his rate, he gets the 4 rooms done in an unknown
amount of hours.
And the time together, if I put them together, the 4 rooms
get done in 6 hours.
Now, because of this word right here, that's what we can
do to build the equation--
together.
Patrick and Bob equal together because
they're working together.
Patrick plus Bob is the time it takes together.
Now, one thing to simplify this equation before you
actually put it together is to realize that the 4 rooms is
the one job.
Because in every step that we read this,
Pat's doing 4 rooms.
That's his one job.
Bob's going to help do that one job of painting 4 rooms.
And it asks, how long would it take Bob to paint the room, or
the 4 rooms, by himself?
How long does it take him to do that one job?
So really, we don't need 4 rooms.
We can substitute it with one job.
This will get done in 10 hours.
This will get done in x amount of hours.
If they work together, that one job gets done in 6 hours.
So build your equation, solve it, make sure your answer
makes sense.
Go back to the problem, read it the fourth time, and say,
does this work the way it should?
Does this make sense that Bob's time is this long?
OK?
Is it a reasonable answer?
This has been Section 1.7, Applications.
Thank you for watching.