### Video Transcript

In this video, we’re going to discuss using properties of operations to solve word
problems. So we’ll start by talking about estimation and some properties of operations in math,
and then we will apply that to some word problems.

There are different forms of estimation. When
we have multiples of ten, the idea is we can move the decimal point. So if we have sixteen of
something and we wanna know what one-tenth of that is, we just divide the sixteen by ten. We
move the decimal point here; it would be here. We move that over one place and we get one point
six.

Rounding is another form of estimation and this makes it easier to do quick math. So say we
had a hundred and seventy-eight pencils and we wanted to use that number in a problem and-and
do it quickly without having to maybe multiply this number or divide this number.

What we might
do is round this up to one hundred eighty, so that’s an easier number to work with because it’s
a multiple of ten and it would make multiplication or division quicker. And finally we have
grouping and that’s putting similar numbers together.

So say you had the fractions one-half
one-half one-half and you needed to add them as part of a problem; you would group those
together because you know that that would be three and a half, or if you have numbers such as
ten twenty thirty, you would group those together, because as multiples of ten they’re easy to
add as well.

Now we’ll discuss some properties of operations. With decimals, remember to line up
the decimal point when you’re adding or subtracting, and when you’re multiplying and dividing,
you might need to move the decimal point in the solution to the problem. And we’ll see that in
an example later.

When working with percents, remember to convert to decimals by moving the
decimal point two places to the left. So if you have seventy-two percent, we take the decimal
point here, one two, and that equals zero point seven two.

When working with fractions, you can convert a fraction
to decimals by dividing the numerator by the denominator. And always remember to make sure that
your units are consistent and you’re using the correct units; if the problem is given to
you in yards and you’re asked for the solution in feet, you need to make sure you convert from
yards to feet in your solution.

Now let’s practice with some word problems. Aaron practices his
trumpet for a half hour on Monday and Tuesday and three-quarters of an hour on Thursday. If he
follows this pattern for twelve weeks, how many hours will he have practiced?

So this is a
two-step problem because we’re looking for how many hours he will practice in twelve weeks.
First we need to figure out how many hours he practices each week. So we add the times together.
We have a half hour on Monday plus a half hour on Tuesday plus three-fourths of an hour on
Thursday.

And remember we need to have a common denominator; here it’s four two and two the
common denominator would be four because we cannot add these without the d- the common denominator, so
one-half is really two-fourths plus two-fourths plus three-fourths is one and three-fourths
which also equals seven-fourths.

Now that we know how much time each week he practices, we need
to multiply that by twelve so we can figure out how much time he spends over the twelve weeks.
And because we’re doing multiplication, it’s easier to use the improper fraction to multiply, so
we have seven-fourths times twelve.

Four divides into twelve; this becomes one and
this becomes three and this equals twenty-one hours. That’s how much time he spends over twelve
weeks practicing his trumpet.

For our next problem, Samantha needs to buy some school supplies. A
notebook costs sixty cents, highlighters cost forty-five cents, and pens cost ten cents. If she
buys three notebooks, two highlighters, and twelve pens, how much did she spend?

So we can break
this up into three parts, calculating how much the notebooks cost, how much the highlighters
cost, and how much the pens cost, and then add those together for her total. So we’ll start with
the notebooks and since each notebook costs sixty cents, we multiply that by three.

So we set up
the problem zero point six zero times three. Three times zero is zero, three times six is eighteen, and
again three times zero is zero, but notice here we have a decimal point right there, and we’ve
gone over one two places, so we need to do the same thing in the solution for one two and we
put the decimal point here.

So this is one point eight zero or a dollar eighty. Now for the
highlighters, each one is forty-five cents. We have zero point four five times two. Two times
five is ten; we carry a one. Two times four is eight plus one is nine. Two times zero is zero. And
again we have the decimal point here; we’ve gone over two places, so we need to do the same
thing in the solution and put the decimal point here. And this is zero point nine zero or
ninety cents.

And lastly we need to calculate the cost of the pens. And we set up our problem
zero point one zero times twelve. Two times zero is zero, two times ~~two~~ [one] is two, and
two times zero is zero.

Remember when we multiply here, this is really multiplying by ten, so we
move over; one times zero is zero; this goes underneath the two. We could also put a zero here as
a placeholder. And one times one is one; one times zero is zero; then we add down zero plus zero
is zero; two plus zero is two; zero plus one is one.

And again we go back up here to the decimal
point and we- this is two places over, so and again in the solution we go over two places, and this
is one point two zero or a dollar and twenty cents. So now that we know how much the notebooks,
the highlighters, and the pens cost alone, we need to add them together to get the solution.

So
we have a dollar and eighty cents plus ninety cents plus a dollar twenty cents equals three
dollars and ninety cents. So that’s how much she has spent on three notebooks, two highlighters,
and twelve pens.

And our last problem, Jose has a job making eleven dollars an hour. After three
months, he gets a six percent raise. What is his new salary?

So first we need to determine what
the raise is, how much more money per hour is he getting. So here we have a percent and remember
when we have percents, we need to take the decimal point which is here and move it two places
to the left.

Now this is an empty space, but what we do is we put a zero there to fill the space.
So this equals- we can write this as zero point zero six, so what we will do is multiply eleven
by zero point zero six. And here I dropped the zero in front of the point zero six to make it
easier to multiply, so six times one is six and six times one is six.

When we multiply by zero,
that’s just the same as putting a zero over here. And again we go back to the problem; here’s
the decimal point, one two; we start here; we go one two; put the decimal point here. And this is
zero point six six, so that’s how much more money he is making per hour.

And now that we know
how much more he’s making per hour, we just add that to the eleven dollars he was making at the
start. So we have eleven plus zero point six six, and that equals eleven dollars and sixty-six
cents per hour. And that’s how much more he’s making after three months.

So you see these are
just a few examples of the different types of problems that you might need to do two parts to
solve the problem you do. Here we did one part where we found out how much more he was making;
then we had to add that to his original salary. We determined how much time Aaron practiced in
one week, so that we could figure out how much he practiced in twelve weeks. And when we were
buying school supplies, we had to figure out the cost of each item first and then add them
together.