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Transcript:

A Portland Community College mathematics telecourse.

A course in arithmetic review.

Produced at Portland Community College.

This lesson is often a frustrating one

for not only the student but the instructor.

Because a student frequently wants cut and dried methods

for solving all problems of a business nature,

and it doesn't exist.

One must learn a great, huge pot of techniques

for many, many problems.

So that all that we can hope to do in this tape

is to give you an indication of a strategy

for approaching such problems.

First, we're beginning to realize

that percent [ % ] is used constantly in business.

Let's look at just a few of the business concepts

which use percent.

Markup,

discount,

profit,

interest rate,

taxes,

salary increases,

commissions.

And the list goes on and on.

Of course, this is the purpose of accounting,

bookkeeping and business courses.

The point we're making is you not only must know mathematics

and various techniques with the calculators and percents,

but you in fact must know each of these

if you wish to work a problem involving them.

So in this lesson we will simply look

at a few of the simpler ones and leave the more complicated ones

to their appropriate business course.

A great number of those problems can be restated

as either simple percent problems, proportions,

or simple algebraic statements

like those taught so far in this course.

We'll give but a small sample of them here.

Other examples will constantly be encountered

through your own experience in other courses.

To solve such problems, first be sure that the terminology

is well defined and understood.

If you don't know it, stop, look it up, or ask.

Be able to state how the different terms

are related to each other,

preferably by a mathematical relationship.

Again if you don't know, stop, look it up, or ask.

And finally, state your problem as simply as possible

using all pertinent information.

And at this point is where mathematics usually begins.

Let's start with the simple type.

The cost of a jacket is $24.30

The markup is 40 percent [ 40% ] of the cost.

What is the selling price?

Now it's important to realize here you're looking at this

from the point of the merchant, not the buyer.

So let's look at how we can dissect this problem.

Let's look at the words we're using.

And of course, we know that business courses

will be much more precise about these than we will,

and they must be.

Okay, first:

Cost.

That's what the merchant pays for the item.

We'll use the letter 'C' to stand for that [ cost ]

in our future statements.

Markup.

That's the amount which is added to the cost

in order to get the...

selling price,

which is what you and I would pay for it.

So for 'markup,' let's us use the letter 'M,'

and for 'selling price,' let's use the letter 'S.'

Remember at the beginning of this course,

we said that the letters simply stand for some number,

and the reason we have different letters

is to help us to remember what it's standing for.

Now, how are these three things related,

cost, markup and selling price?

I think you know that already, but if we didn't,

we would ask the business instructor,

and he would tell us

that selling price is simply the cost plus markup [ S = C + M ]

That's a simple enough statement,

but notice how we're using those math symbols to say it.

And that's going to be part of our clue.

Now let's begin to get statements about our problem.

Let's go back and get some information out of it.

First, they wish to know: the selling price.

So at least we've identified what we don't know,

the selling price.

Now let's see what else the problem is telling us.

Okay, the cost of the jacket, $24.30

Then let's state that mathematically.

There, the cost. [ C = 24.30 ]

And of course, that's what the merchant paid for it

from the wholesaler or whoever he buys it from.

Okay now, do they tell us anything about the markup?

Now notice here, the markup [ M ]

is [ = ]

40 percent .40

of [ x ] the cost [ C ]

which they told me was 24.30

So although they didn't tell me what the markup was,

they very definitely told me

how to figure it out. [ M = .40 x 24.30 ]

So let's do that in case we can use it.

And we have a markup of $9.72 [ M = 9.72 ]

So let's take that information [ M = 9.72 ]

back to our statement of the problem.

Now, notice that the mathematical statement of the problem,

once we fill in what we don't know [ S ]

and what we do know [ C + M ]

is now instructing us exactly what is to be done.

The selling price [ S = C + M ] is the sum of those two numbers.

So performing that addition, [ S = 24.30 + 9.72 ] we get –

so the selling price would be $34.02

if it were to meet all of our conditions.

Now naturally, at this point

the merchant would probably change this to 34 or 34.50

to get a more attractive number,

more even number rather than down to the pennies.

Now naturally, after you work many, many problems like this,

you won't go through such a tortured movement as we did.

You'll simply solve it.

So the strategy we're giving you is what you do

if you approach a problem that's relatively new to you.

Notice how the use of algebraic variables

helped to get away from the confusion of words.

And an algebraic sentence can be solved,

whereas a verbal sentence cannot, at least not so easily.

Let's try another one.

The regular price on a stereo unit is $3.40

During a sale, it is to be discounted 20%.

What will be its sale price?

Okay now, you probably have done these in other classes

and it's automatic to you.

But let's use this simple problem

to sort of tie into our strategy a bit more firmly.

First, can you identify the technical words

we should be able to define before we start?

There's the 'regular price.'

There's a 'discount.'

And there's a 'sale price.'

So our first point is to identify these [ words ]

then ask, What are they? and, How are they related?

That's not a math question.

That or those are business questions.

Now, in this easy problem,

we'll presume you know what those are already.

But for the 'regular price' let's use the letter 'R.'

The 'discount rate,' let's use small 'r.'

The 'discount' itself, let's use 'D.'

And the 'sale price,' let's use 'S.'

What letter you use is really up to you

or your employer or your teacher.

But now, what do we know about these?

Well, we know the regular price

is the discount plus the sale price. [ R = D + S ]

Or the sale price

is the regular price minus the discount. [ S = R - D ]

Whichever way you say it, you're saying the same thing.

But we also know that the discount

is so much percent of the regular price. [ D = r% of R ]

Okay now, let's fill in these relationships

as best we can from our words.

So back to our problem.

And we bring to it our two formulas

of the now known relationships.

So let's see what we can fill in from these.

The 'regular price,' okay, here's regular price [ S = 340 - D ]

And they told me that's $340,

but also I might need it over here. [ D = r% of 340 ]

So $340.

Now, what else do they tell us?

During the sale, it is discounted at 20 percent.

So they're telling me

the discount rate is 20 percent. [ D = 20% of 340 ]

What else do we know?

What will be the 'sale price?'

So they're telling us it's the sale price [ S ] we want to know.

Okay, let's stop and just look at this.

Once we have gone from the words to the mathematical statement,

if we've done a good job, we won't have to go back to that.

So let's see if we can just stay here and find that sale price.

Notice over here it says the sale price, which is what I want,

is 340 minus the discount, [ S = 340 - D ] but I don't know that.

But over here it says

the discount is 20 percent of this, [ D = 20% of 340 ]

but this is not quite fully mathematics.

20 percent [ 20% ] mathematically is two tenths [ .2 ]

'of' is 'times' 340 [ (.2)(340) ] is mathematics.

And this [ D=(.2)(340) ] says discount is simply

that (.2) times that (340) [ D=(.2)(340) ]

which is $68. [ D = 68 ]

But now that I know the discount, I can put it up here,

and now it's clearly telling me that selling price

is the difference of these two, [ S = 340 - 68 ]

which is $272. [ S = 272 ]

Now, if you understood these things

and if you have worked problems like this before,

you probably wouldn't do it this way.

We're trying to use this as a vehicle to find out

how we can attack problems we don't know

by going through, first, to be safe problems we might know.

So the general strategy once again was this:

Identify in the statement of your problem

the technical things being dealt with.

If necessary, define the terminology,

get some letters to stand for those things,

then state the relationships between them.

And these relationships are relationships you must already know,

you must read, or you must ask somebody.

Then finally, take those statements, those relationships,

back to the problem, fill in what you do know,

and usually at that point

the mathematics will direct you to its solution.

Now, as you can see, we haven't even started

to scratch the possibilities, but perhaps these few problems

will give you an indication about how you can better organize

your approach to such problems.

Please know that: being good at verbal, or application, problems

is simply a matter of experience.

And 'experience' means reading worked examples

from the textbook possibly,

watching somebody work examples,

and in this case, it's going to be me.

And above all else,

a willingness to stumble and experience frustrations.

So for the remainder of this lesson, I won't teach,

I won't show you how to do things

because in fact that's not possible.

Every problem is different.

I'll simply let you experience watching me work some examples.

And try to store up those examples,

either in memory or in your notes,

so that you can use them to interpret other problems

which might be the same

or similar to the ones that we're going to work right now.

So here we go.

Have your problem in front of you,

and have scratch paper so that you can make notes as you read.

Because remember, our first task is to ask the question:

What are they saying?

Okay, so we have a new car

that depreciates 15 percent [ 15%] during the first year.

Its original cost was $12,500

What is the value at the end of the first year?

Now, let's begin to talk to yourself.

And you'll find this is really a very simple problem.

We're going to depreciate 15 percent of what?

Well, of its 'value,' and that's right here [ $12,500 ]

So we're going to take 15 percent of this [ 15% of $12,500 ]

And that's the amount of depreciation.

That's fairly obvious in this case, isn't it?

It's a good, simple problem.

So we want 15 percent [ .15 ]

'of' means 'times' [ x .15 ]

$12,500 [ 12500 x .15 ]

So now this [ 12500 x .15 ] is simply telling me to multiply.

15 percent tells me what by,

it's the 'of' that tells me I'm going to multiply.

So I do this either by hand or by calculator,

and I get $1875

Now be careful not to quit at this point

because this is only the depreciation of that first year.

And they want to know its value at the end of the first year.

And of course, the value is what you get

after you have depreciated

or taken this [ $1875 ] off of the value.

So common sense now says I'll take its current value [ $12500 ]

the first year value, subtract the depreciation [ 12500 - 1875 ]

and what is left is its new value at the end of that year.

So at the end of the first year, by depreciation alone,

the car's value will be $10,625.

So see, as I talked to you,

as you work problems you will talk to yourself.

And you're always asking not: How do I do it?

but: What are they saying? and: What do they want?

And that should tell you what to do.

So be very, very free about talking to yourself

over and over and over until the problem becomes crystal clear.

Let's do another one.

Your labor union negotiated a new contract

which would increase salary by 5 ¼ percent.

What would be your new salary if you were earning $9.2-- $9.32?

Now is this as easy as the last one?

You're going to increase this [ $9.32 ] by 5 ¼ percent.

So that means the increase

is this [ 5 ¼% ] times this [ 9.32 ] [ 5 ¼% x 9.32 ]

But I don't want the increase,

I want to know what my new salary is.

Well, if we know the increase, we can just add it back on.

Or you could think this way.

I'm going to increase this [ $9.32 ] and of course,

this is always 100 percent of itself by another 5 ¼ percent.

So what we have here is that the new salary,

and let's call that 'n,'

is [ = ]

105 ¼ percent of the old salary [ n = 105 ¼% of O ]

Now if you see this, this will save you a little bit of work.

Because if we did it the first way,

where we take 5 ¼ percent [ 5 ¼% ] times this [ $9.32 ]

then we have to turn around and add it back on.

But here [ n = 105 ¼% of O ]

we'll just have one problem, and when we're done,

we will have the answer in one move.

So this has a new salary [ n ] and an old salary [ O ]

but the old salary I know.

It's the $9.32

So what do we have?

We have a simple sentence: is [ = ] of [ x ] and percent [% ]

And the only unknown is the one I want.

So this says the new salary is [ n = ]

And if we don't like that ¼ as a fraction,

we can change it to its decimal equivalent,

but the percent is still there. [ 105.25% ]

So now we can get away from percent

by moving it two places away [ 1.0525 ]

'Of' means 'times' 9.32 [ n =(1.0525)(9.32) ]

And since this is money,

we would know to round to the nearest penny.

And we simply do the indicated arithmetic.

This (1.0525) times this (9.32) [ n =(1.0525)(9.32) ]

and we're done. [ n =(1.0525)(9.32) ]

And doing that either by hand or by calculator,

we get $9.81 per hour.

This was fairly simple, as was the last one.

And that's the way most textbooks operate.

They'll take some fairly simple problems,

where it's pretty obvious what you're going to do

just by a reading, a simple reading,

and a clear understanding of what they're saying.

Then very slowly they begin to get more and more involved,

more words usually,

or else they begin to give you more and more pieces

that have to be put together.

Consider this next problem.

A stereo set is advertised to sell for $110.

Markup must be at least 33 1/3 percent of the cost.

What is the most the store can afford to pay for the stereo set?

Now first, let's begin to write down

what they're telling me in simpler mathematical expressions.

Okay, so we want the 'selling price' to be $110.

Okay, let's make a note.

'Selling price,' we'll call 'S' temporarily,

and they're telling me that S is $110 [ S = 110 ]

They say 'markup' must be 33 1/3 percent of the cost.

Okay, so they're saying 'markup' is [ M = ]

33 1/3 percent of cost [ M = 33 1/3% of C ]

Or the wholesale, if you would.

Okay, so we have just restated this in mathematics.

Now what else?

What is the most the store can afford to pay for the stereo set?

So 'cost' is what the store pays,

so it's the 'cost' [ C = ] that they want.

Now, if you see what to do, then do it.

If not, then you begin to ask yourself:

What do I know about these kinds of things?

Well, let's see, we know that

selling price is cost plus markup [ S = C + M ]

Okay, in this particular case, we know the selling price from up here.

So that will give me

110 is equal to cost plus markup [ 110 = C + M ]

But this says markup is to be 33 1/3 percent.

So let me try something here that's a little bit different from

and see if you can compare this

with something we've done a moment ago.

I left that [ C ] alone.

And markup, I'm going to replace by what this is saying it is,

is 33 1/3 percent of cost. [ 110 = C + 33 1/3% · C ]

And I'll replace the 'of' by 'times.'

Now perhaps by writing it this way, [ 110 = C + 33 1/3% · C ]

you can see more easily

that anything is 100 percent of itself. [ C = 100% C ]

So can you see that I have 100 percent of cost

plus 33 1/3 percent of cost, or if you would,

133 1/3 percent of cost. [ 110 = 133 1/3% C ]

Now, if you see this, we're just about done.

So let's follow this through, hoping that you did see it.

Now, if you wish, we can get rid of the percent

by moving a decimal point two places away.

So we have 1.33 1/3 times the C.

And the question is: What do we do about this 1/3?

Well, we could get rid of it by making it exact,

but we might recall that 1/3 is .333 forever.

So we can add enough decimal points

to give me the accuracy I know I'll want.

Now we have something times the variable

equal to something. [ 110 = 1.33333 · C ]

So we undo multiplying by dividing both sides

by the same quantity. [110/1.33333 = (1.33333 · C)/1.33333 ]

And this says cost is 110 divided by this [ C = 110/1.33333 ]

And since we have a lot of decimal points there,

it probably would be better to do this division by calculator.

And rounding off to the nearest penny,

we can see that that's $82.50 is the cost.

Now, it was this sentence and this fact up here

that told me the selling price is going to be the original cost,

which is 100 percent of itself of course,

increased by another 33 1/3 percent,

which tells me if I increase 100 percent by 33 1/3 percent,

I get 133 1/3%, but percent of the cost. [ 133 1/3% · C ]

Now, if you see this will help you quite a bit in future problems,

which will be very similar to this.

In fact, let's look at that problem once again

because the purpose of working these problems

is not to just get an answer,

it's to build a model for future problems similar to this.

So in solving this, we went through a lot of contortions

trying to find out just what are they doing.

So when you're done,

back off and see if you can see the whole thing

in a smooth train of thought

that you can transfer to another problem.

So in looking at this, you might be able to conclude this:

In all cases, selling price will be equal to cost plus markup.

This is always the case. [ S = C + M ]

And it is indeed always the case that anything

is 100 percent of itself. [ C = 100% of C ]

So this is just another way of saying cost.

And if it occurred that the markup

was 33 1/3 percent of the cost, [ M = 33 1/3% of C ]

then we have 100 percent of cost [ 100% of C ]

plus another 33 1/3 percent of cost, which now becomes

133 1/3 percent of cost. [ 100%C + 33 1/3%C = 133 1/3%C ]

This step [ S = 100%C + 33 1/3%C ] through a few more problems,

will be eliminated, as would this [ S = C + M ]

And you'll simply think that, gee whiz,

if I have 100 percent of cost and increase it by 33 1/3 percent,

[ 100%C + 33 1/3%C ] then it's obviously 133 1/3% of cost,

and that is the selling price. [ S = 133 1/3% of C ]

And perhaps you might recall that 133 1/3 percent is 1-- [ 100%=1 ]

and 33 1/3 percent is 1/3 of cost. [ 100% + 33 1/3% = 1 1/3 ]

And you might also recall that this [ 1 1/3 ] is 4/3 times cost.

And in this case, we don't have that long decimal, do we?

But the selling price we knew we wanted to be $110,

and we can solve this simple equation several ways.

One is to multiply this side by the reciprocal [ 3/4 · 4/3 x C ]

because any fraction times its reciprocal is 1 times the cost.

But then I have to multiply

the other side of the equation by the same amount [ 110 · 3/4 ]

and this is easier arithmetic than we had with the decimals.

Do you see that?

But always remember

that anything is the same thing as 100 percent of itself.

And sometimes you need to think in terms of percents first,

particularly when you're increasing or decreasing something

from what it originally was.

So if you can begin to think this way,

then you are beginning to think mathematically.

Can you apply that directly to this?

A discount house advertised

that they plan to sell all appliances at cost plus 10 percent.

Now right here you want to say

that my 'selling price' is going to be the cost,

which is 100 percent of itself, plus another 10 percent of it.

So in fact, the 'selling price'

is 110 percent of cost [ S = 110% of C ] Do you see that?

With practice you'll go from here directly to here

without all those many, many steps we had in the previous problem.

But in this case, Sue bought the set for $740,

which is the selling price. [ S = 740 ]

And we can find the cost.

So this is 740 is 1.1 of cost. [ 740 = 1.1 · C ]

Divide both sides by the multiplier, [ 1.1 ] the unknown,

do the arithmetic, [ 740/1.1 = C ]

and we get the cost, [ C = $672.73 ]

but that's not what they wanted, they wanted the profit.

Well, the profit, of course,

is the difference between the selling price and the cost.

So we follow through on that statement. [ S - C = profit ]

And we're done.

And of course you need to see many, many, many problems.

The three or four that we've done here are just barely enough

to point your head and mind in the direction we want to go.

But you need ten times this many to become fairly comfortable.

This is your host, Bob Finnell,

we'll see you at the next lesson and the next chapter.

A course in arithmetic review.

Produced at Portland Community College.

This lesson is often a frustrating one

for not only the student but the instructor.

Because a student frequently wants cut and dried methods

for solving all problems of a business nature,

and it doesn't exist.

One must learn a great, huge pot of techniques

for many, many problems.

So that all that we can hope to do in this tape

is to give you an indication of a strategy

for approaching such problems.

First, we're beginning to realize

that percent [ % ] is used constantly in business.

Let's look at just a few of the business concepts

which use percent.

Markup,

discount,

profit,

interest rate,

taxes,

salary increases,

commissions.

And the list goes on and on.

Of course, this is the purpose of accounting,

bookkeeping and business courses.

The point we're making is you not only must know mathematics

and various techniques with the calculators and percents,

but you in fact must know each of these

if you wish to work a problem involving them.

So in this lesson we will simply look

at a few of the simpler ones and leave the more complicated ones

to their appropriate business course.

A great number of those problems can be restated

as either simple percent problems, proportions,

or simple algebraic statements

like those taught so far in this course.

We'll give but a small sample of them here.

Other examples will constantly be encountered

through your own experience in other courses.

To solve such problems, first be sure that the terminology

is well defined and understood.

If you don't know it, stop, look it up, or ask.

Be able to state how the different terms

are related to each other,

preferably by a mathematical relationship.

Again if you don't know, stop, look it up, or ask.

And finally, state your problem as simply as possible

using all pertinent information.

And at this point is where mathematics usually begins.

Let's start with the simple type.

The cost of a jacket is $24.30

The markup is 40 percent [ 40% ] of the cost.

What is the selling price?

Now it's important to realize here you're looking at this

from the point of the merchant, not the buyer.

So let's look at how we can dissect this problem.

Let's look at the words we're using.

And of course, we know that business courses

will be much more precise about these than we will,

and they must be.

Okay, first:

Cost.

That's what the merchant pays for the item.

We'll use the letter 'C' to stand for that [ cost ]

in our future statements.

Markup.

That's the amount which is added to the cost

in order to get the...

selling price,

which is what you and I would pay for it.

So for 'markup,' let's us use the letter 'M,'

and for 'selling price,' let's use the letter 'S.'

Remember at the beginning of this course,

we said that the letters simply stand for some number,

and the reason we have different letters

is to help us to remember what it's standing for.

Now, how are these three things related,

cost, markup and selling price?

I think you know that already, but if we didn't,

we would ask the business instructor,

and he would tell us

that selling price is simply the cost plus markup [ S = C + M ]

That's a simple enough statement,

but notice how we're using those math symbols to say it.

And that's going to be part of our clue.

Now let's begin to get statements about our problem.

Let's go back and get some information out of it.

First, they wish to know: the selling price.

So at least we've identified what we don't know,

the selling price.

Now let's see what else the problem is telling us.

Okay, the cost of the jacket, $24.30

Then let's state that mathematically.

There, the cost. [ C = 24.30 ]

And of course, that's what the merchant paid for it

from the wholesaler or whoever he buys it from.

Okay now, do they tell us anything about the markup?

Now notice here, the markup [ M ]

is [ = ]

40 percent .40

of [ x ] the cost [ C ]

which they told me was 24.30

So although they didn't tell me what the markup was,

they very definitely told me

how to figure it out. [ M = .40 x 24.30 ]

So let's do that in case we can use it.

And we have a markup of $9.72 [ M = 9.72 ]

So let's take that information [ M = 9.72 ]

back to our statement of the problem.

Now, notice that the mathematical statement of the problem,

once we fill in what we don't know [ S ]

and what we do know [ C + M ]

is now instructing us exactly what is to be done.

The selling price [ S = C + M ] is the sum of those two numbers.

So performing that addition, [ S = 24.30 + 9.72 ] we get –

so the selling price would be $34.02

if it were to meet all of our conditions.

Now naturally, at this point

the merchant would probably change this to 34 or 34.50

to get a more attractive number,

more even number rather than down to the pennies.

Now naturally, after you work many, many problems like this,

you won't go through such a tortured movement as we did.

You'll simply solve it.

So the strategy we're giving you is what you do

if you approach a problem that's relatively new to you.

Notice how the use of algebraic variables

helped to get away from the confusion of words.

And an algebraic sentence can be solved,

whereas a verbal sentence cannot, at least not so easily.

Let's try another one.

The regular price on a stereo unit is $3.40

During a sale, it is to be discounted 20%.

What will be its sale price?

Okay now, you probably have done these in other classes

and it's automatic to you.

But let's use this simple problem

to sort of tie into our strategy a bit more firmly.

First, can you identify the technical words

we should be able to define before we start?

There's the 'regular price.'

There's a 'discount.'

And there's a 'sale price.'

So our first point is to identify these [ words ]

then ask, What are they? and, How are they related?

That's not a math question.

That or those are business questions.

Now, in this easy problem,

we'll presume you know what those are already.

But for the 'regular price' let's use the letter 'R.'

The 'discount rate,' let's use small 'r.'

The 'discount' itself, let's use 'D.'

And the 'sale price,' let's use 'S.'

What letter you use is really up to you

or your employer or your teacher.

But now, what do we know about these?

Well, we know the regular price

is the discount plus the sale price. [ R = D + S ]

Or the sale price

is the regular price minus the discount. [ S = R - D ]

Whichever way you say it, you're saying the same thing.

But we also know that the discount

is so much percent of the regular price. [ D = r% of R ]

Okay now, let's fill in these relationships

as best we can from our words.

So back to our problem.

And we bring to it our two formulas

of the now known relationships.

So let's see what we can fill in from these.

The 'regular price,' okay, here's regular price [ S = 340 - D ]

And they told me that's $340,

but also I might need it over here. [ D = r% of 340 ]

So $340.

Now, what else do they tell us?

During the sale, it is discounted at 20 percent.

So they're telling me

the discount rate is 20 percent. [ D = 20% of 340 ]

What else do we know?

What will be the 'sale price?'

So they're telling us it's the sale price [ S ] we want to know.

Okay, let's stop and just look at this.

Once we have gone from the words to the mathematical statement,

if we've done a good job, we won't have to go back to that.

So let's see if we can just stay here and find that sale price.

Notice over here it says the sale price, which is what I want,

is 340 minus the discount, [ S = 340 - D ] but I don't know that.

But over here it says

the discount is 20 percent of this, [ D = 20% of 340 ]

but this is not quite fully mathematics.

20 percent [ 20% ] mathematically is two tenths [ .2 ]

'of' is 'times' 340 [ (.2)(340) ] is mathematics.

And this [ D=(.2)(340) ] says discount is simply

that (.2) times that (340) [ D=(.2)(340) ]

which is $68. [ D = 68 ]

But now that I know the discount, I can put it up here,

and now it's clearly telling me that selling price

is the difference of these two, [ S = 340 - 68 ]

which is $272. [ S = 272 ]

Now, if you understood these things

and if you have worked problems like this before,

you probably wouldn't do it this way.

We're trying to use this as a vehicle to find out

how we can attack problems we don't know

by going through, first, to be safe problems we might know.

So the general strategy once again was this:

Identify in the statement of your problem

the technical things being dealt with.

If necessary, define the terminology,

get some letters to stand for those things,

then state the relationships between them.

And these relationships are relationships you must already know,

you must read, or you must ask somebody.

Then finally, take those statements, those relationships,

back to the problem, fill in what you do know,

and usually at that point

the mathematics will direct you to its solution.

Now, as you can see, we haven't even started

to scratch the possibilities, but perhaps these few problems

will give you an indication about how you can better organize

your approach to such problems.

Please know that: being good at verbal, or application, problems

is simply a matter of experience.

And 'experience' means reading worked examples

from the textbook possibly,

watching somebody work examples,

and in this case, it's going to be me.

And above all else,

a willingness to stumble and experience frustrations.

So for the remainder of this lesson, I won't teach,

I won't show you how to do things

because in fact that's not possible.

Every problem is different.

I'll simply let you experience watching me work some examples.

And try to store up those examples,

either in memory or in your notes,

so that you can use them to interpret other problems

which might be the same

or similar to the ones that we're going to work right now.

So here we go.

Have your problem in front of you,

and have scratch paper so that you can make notes as you read.

Because remember, our first task is to ask the question:

What are they saying?

Okay, so we have a new car

that depreciates 15 percent [ 15%] during the first year.

Its original cost was $12,500

What is the value at the end of the first year?

Now, let's begin to talk to yourself.

And you'll find this is really a very simple problem.

We're going to depreciate 15 percent of what?

Well, of its 'value,' and that's right here [ $12,500 ]

So we're going to take 15 percent of this [ 15% of $12,500 ]

And that's the amount of depreciation.

That's fairly obvious in this case, isn't it?

It's a good, simple problem.

So we want 15 percent [ .15 ]

'of' means 'times' [ x .15 ]

$12,500 [ 12500 x .15 ]

So now this [ 12500 x .15 ] is simply telling me to multiply.

15 percent tells me what by,

it's the 'of' that tells me I'm going to multiply.

So I do this either by hand or by calculator,

and I get $1875

Now be careful not to quit at this point

because this is only the depreciation of that first year.

And they want to know its value at the end of the first year.

And of course, the value is what you get

after you have depreciated

or taken this [ $1875 ] off of the value.

So common sense now says I'll take its current value [ $12500 ]

the first year value, subtract the depreciation [ 12500 - 1875 ]

and what is left is its new value at the end of that year.

So at the end of the first year, by depreciation alone,

the car's value will be $10,625.

So see, as I talked to you,

as you work problems you will talk to yourself.

And you're always asking not: How do I do it?

but: What are they saying? and: What do they want?

And that should tell you what to do.

So be very, very free about talking to yourself

over and over and over until the problem becomes crystal clear.

Let's do another one.

Your labor union negotiated a new contract

which would increase salary by 5 ¼ percent.

What would be your new salary if you were earning $9.2-- $9.32?

Now is this as easy as the last one?

You're going to increase this [ $9.32 ] by 5 ¼ percent.

So that means the increase

is this [ 5 ¼% ] times this [ 9.32 ] [ 5 ¼% x 9.32 ]

But I don't want the increase,

I want to know what my new salary is.

Well, if we know the increase, we can just add it back on.

Or you could think this way.

I'm going to increase this [ $9.32 ] and of course,

this is always 100 percent of itself by another 5 ¼ percent.

So what we have here is that the new salary,

and let's call that 'n,'

is [ = ]

105 ¼ percent of the old salary [ n = 105 ¼% of O ]

Now if you see this, this will save you a little bit of work.

Because if we did it the first way,

where we take 5 ¼ percent [ 5 ¼% ] times this [ $9.32 ]

then we have to turn around and add it back on.

But here [ n = 105 ¼% of O ]

we'll just have one problem, and when we're done,

we will have the answer in one move.

So this has a new salary [ n ] and an old salary [ O ]

but the old salary I know.

It's the $9.32

So what do we have?

We have a simple sentence: is [ = ] of [ x ] and percent [% ]

And the only unknown is the one I want.

So this says the new salary is [ n = ]

And if we don't like that ¼ as a fraction,

we can change it to its decimal equivalent,

but the percent is still there. [ 105.25% ]

So now we can get away from percent

by moving it two places away [ 1.0525 ]

'Of' means 'times' 9.32 [ n =(1.0525)(9.32) ]

And since this is money,

we would know to round to the nearest penny.

And we simply do the indicated arithmetic.

This (1.0525) times this (9.32) [ n =(1.0525)(9.32) ]

and we're done. [ n =(1.0525)(9.32) ]

And doing that either by hand or by calculator,

we get $9.81 per hour.

This was fairly simple, as was the last one.

And that's the way most textbooks operate.

They'll take some fairly simple problems,

where it's pretty obvious what you're going to do

just by a reading, a simple reading,

and a clear understanding of what they're saying.

Then very slowly they begin to get more and more involved,

more words usually,

or else they begin to give you more and more pieces

that have to be put together.

Consider this next problem.

A stereo set is advertised to sell for $110.

Markup must be at least 33 1/3 percent of the cost.

What is the most the store can afford to pay for the stereo set?

Now first, let's begin to write down

what they're telling me in simpler mathematical expressions.

Okay, so we want the 'selling price' to be $110.

Okay, let's make a note.

'Selling price,' we'll call 'S' temporarily,

and they're telling me that S is $110 [ S = 110 ]

They say 'markup' must be 33 1/3 percent of the cost.

Okay, so they're saying 'markup' is [ M = ]

33 1/3 percent of cost [ M = 33 1/3% of C ]

Or the wholesale, if you would.

Okay, so we have just restated this in mathematics.

Now what else?

What is the most the store can afford to pay for the stereo set?

So 'cost' is what the store pays,

so it's the 'cost' [ C = ] that they want.

Now, if you see what to do, then do it.

If not, then you begin to ask yourself:

What do I know about these kinds of things?

Well, let's see, we know that

selling price is cost plus markup [ S = C + M ]

Okay, in this particular case, we know the selling price from up here.

So that will give me

110 is equal to cost plus markup [ 110 = C + M ]

But this says markup is to be 33 1/3 percent.

So let me try something here that's a little bit different from

and see if you can compare this

with something we've done a moment ago.

I left that [ C ] alone.

And markup, I'm going to replace by what this is saying it is,

is 33 1/3 percent of cost. [ 110 = C + 33 1/3% · C ]

And I'll replace the 'of' by 'times.'

Now perhaps by writing it this way, [ 110 = C + 33 1/3% · C ]

you can see more easily

that anything is 100 percent of itself. [ C = 100% C ]

So can you see that I have 100 percent of cost

plus 33 1/3 percent of cost, or if you would,

133 1/3 percent of cost. [ 110 = 133 1/3% C ]

Now, if you see this, we're just about done.

So let's follow this through, hoping that you did see it.

Now, if you wish, we can get rid of the percent

by moving a decimal point two places away.

So we have 1.33 1/3 times the C.

And the question is: What do we do about this 1/3?

Well, we could get rid of it by making it exact,

but we might recall that 1/3 is .333 forever.

So we can add enough decimal points

to give me the accuracy I know I'll want.

Now we have something times the variable

equal to something. [ 110 = 1.33333 · C ]

So we undo multiplying by dividing both sides

by the same quantity. [110/1.33333 = (1.33333 · C)/1.33333 ]

And this says cost is 110 divided by this [ C = 110/1.33333 ]

And since we have a lot of decimal points there,

it probably would be better to do this division by calculator.

And rounding off to the nearest penny,

we can see that that's $82.50 is the cost.

Now, it was this sentence and this fact up here

that told me the selling price is going to be the original cost,

which is 100 percent of itself of course,

increased by another 33 1/3 percent,

which tells me if I increase 100 percent by 33 1/3 percent,

I get 133 1/3%, but percent of the cost. [ 133 1/3% · C ]

Now, if you see this will help you quite a bit in future problems,

which will be very similar to this.

In fact, let's look at that problem once again

because the purpose of working these problems

is not to just get an answer,

it's to build a model for future problems similar to this.

So in solving this, we went through a lot of contortions

trying to find out just what are they doing.

So when you're done,

back off and see if you can see the whole thing

in a smooth train of thought

that you can transfer to another problem.

So in looking at this, you might be able to conclude this:

In all cases, selling price will be equal to cost plus markup.

This is always the case. [ S = C + M ]

And it is indeed always the case that anything

is 100 percent of itself. [ C = 100% of C ]

So this is just another way of saying cost.

And if it occurred that the markup

was 33 1/3 percent of the cost, [ M = 33 1/3% of C ]

then we have 100 percent of cost [ 100% of C ]

plus another 33 1/3 percent of cost, which now becomes

133 1/3 percent of cost. [ 100%C + 33 1/3%C = 133 1/3%C ]

This step [ S = 100%C + 33 1/3%C ] through a few more problems,

will be eliminated, as would this [ S = C + M ]

And you'll simply think that, gee whiz,

if I have 100 percent of cost and increase it by 33 1/3 percent,

[ 100%C + 33 1/3%C ] then it's obviously 133 1/3% of cost,

and that is the selling price. [ S = 133 1/3% of C ]

And perhaps you might recall that 133 1/3 percent is 1-- [ 100%=1 ]

and 33 1/3 percent is 1/3 of cost. [ 100% + 33 1/3% = 1 1/3 ]

And you might also recall that this [ 1 1/3 ] is 4/3 times cost.

And in this case, we don't have that long decimal, do we?

But the selling price we knew we wanted to be $110,

and we can solve this simple equation several ways.

One is to multiply this side by the reciprocal [ 3/4 · 4/3 x C ]

because any fraction times its reciprocal is 1 times the cost.

But then I have to multiply

the other side of the equation by the same amount [ 110 · 3/4 ]

and this is easier arithmetic than we had with the decimals.

Do you see that?

But always remember

that anything is the same thing as 100 percent of itself.

And sometimes you need to think in terms of percents first,

particularly when you're increasing or decreasing something

from what it originally was.

So if you can begin to think this way,

then you are beginning to think mathematically.

Can you apply that directly to this?

A discount house advertised

that they plan to sell all appliances at cost plus 10 percent.

Now right here you want to say

that my 'selling price' is going to be the cost,

which is 100 percent of itself, plus another 10 percent of it.

So in fact, the 'selling price'

is 110 percent of cost [ S = 110% of C ] Do you see that?

With practice you'll go from here directly to here

without all those many, many steps we had in the previous problem.

But in this case, Sue bought the set for $740,

which is the selling price. [ S = 740 ]

And we can find the cost.

So this is 740 is 1.1 of cost. [ 740 = 1.1 · C ]

Divide both sides by the multiplier, [ 1.1 ] the unknown,

do the arithmetic, [ 740/1.1 = C ]

and we get the cost, [ C = $672.73 ]

but that's not what they wanted, they wanted the profit.

Well, the profit, of course,

is the difference between the selling price and the cost.

So we follow through on that statement. [ S - C = profit ]

And we're done.

And of course you need to see many, many, many problems.

The three or four that we've done here are just barely enough

to point your head and mind in the direction we want to go.

But you need ten times this many to become fairly comfortable.

This is your host, Bob Finnell,

we'll see you at the next lesson and the next chapter.