On the TEAS test, you are going to encounter a lot of word problems. Here we have one example
of something you might see when taking the TEAS test. It says, "If Sara can read fifteen
pages in ten minutes, how long will it take her to read forty-five pages?" Here we have
a rate: fifteen pages in ten minutes. Which can be written as a ratio, or a fraction.
Fifteen pages in ten minutes. Or, you can write it in the reverse: ten minutes in fifteen
pages, or ten minutes to fifteen pages. Then we have this third piece of information: forty-five
pages. And you see that the units there, the pages, match the units, or one of the units
you have in your first ratio. That needs to be a clue to use a proportion, which is just
a ratio that's equal to a ratio. And what's really important about setting up proportions,
is that you're consistent. So if your first ratio is pages to minutes, then your second
ratio should also be pages to minutes, which tells us to put the forty-five pages on top
of our fraction in our numerator. So what were trying to find is, how many minutes is
it going to take to read those forty-five pages. There is more than one way to solve
a proportion. One of the easier ways, and it won't always work because the easier ways
don't always work or we wouldn't come up with the harder ways. One of the easier ways to
solve this would be: to look at either the numerator or the denominator, and figure out
what you'd have to do to get your other numerator. For example, what do you have to do to fifteen
pages to get forty-five pages? Well fifteen times three is forty-five. So you just do
the same thing in your denominator. Ten times three would be thirty minutes. So it's going
to take thirty minutes to read those forty-five pages. But like I said, that won't always
work. What will always work, is using cross-products. And that just means cross-multiplying. Fifteen
times "x" is "15x". And that is equal to ten times forty-five, which is four hundred fifty.
What you need to know about proportions is that, their cross products are always equal.
So now you just need to solve your equation, which means undoing what's been done to your
variable "x". So in this case, we need to divide both sides by fifteen. And "x", again
equals thirty. So now we know it's going to take thirty minutes to read those forty-five
pages. On the TEAS test, you're going to see a lot
of word problems. Here is one example of one type of word problem that you might see, "Restaurant
customers tip their server only eight percent for poor service. If their tip was three dollars
and seventy cents, how much was their bill?" So here we have a percent, the tip, or part,
that's part of a whole, and then what we're asked to find is the whole. So you can use
the percent equation, that part is equal to percent times whole to solve this problem.
Your part, in this case, is the tip. The tip they left was three dollars and seventy cents.
The percent is this eight percent, but we don't do calculations with percents. So you
have to either, change that percent into a decimal or into a fraction. Eight percent,
as a decimal, is eight hundredths. The way you go from a percent to a decimal is: just
to take the decimal in eight percent, which is at the back of the number, and move it
two places to the left and then filling that empty place with a zero. So for this percent,
we put our eight hundredths times the whole, and the whole is what we're trying to find.
So that is our variable "x". Now to solve for "x", you just need to undo multiplying
"x" by eight hundredths by dividing both sides by eight hundredths. Three dollars and seventy
cents divided by eight hundredths is forty six dollars and twenty-five cents, which means
that, the bill for their meal was forty six dollars and twenty-five cents. So there is
one example of a problem that you might see on the TEAS test.
On the TEAS test, you're going to see a lot of word problems. Here I have one example
of a word problem that you might encounter on the TEAS test. "If Leonard bought two packs
of batteries for 'x' amount of dollars, how many packs of batteries could he purchase
for five dollars at the same rate?" This is a key word right here, rate, because it tells
you to set up a ratio, or a proportion actually. So our first rate that we have is two packs
of batteries for "x" dollars. We can set that up as two packs for "x" dollars. And then
it says "at the same rate", which tells us we're going to set up a proportion. The key
with proportions is to be consistent. Looking at our first ratio, we have packs to dollars.
So our second ratio should also be packs to dollars. And the information they've given
us is the five dollars, which goes in your denominator with dollars. What we don't know
is how many packs of batteries. That's what we're trying to figure out, "...how many packs
of batteries could he purchase..." So I'm going to us "y", since we already used "x"
down here. What we have to keep in mind since we have two different variables is that: were
trying to solve for "y" this time, not for "x". Notice, in our multiple-choice answers,
that they all have that variable "x" in it. So we're trying to solve for "y" this time,
not "x", which is kind of weird. Just having two variables is kind of weird. Okay, so to
solve this proportion and to solve any proportion, we can use cross-products, which just means
cross-multiply. Two times five is ten, and that's equal to "x" times "y". Again, we have
to remember that we're trying to solve for "y", not for "x". "X" is actually part of
our answer. In order to get "y" alone, we need to undo multiplying by "x" by dividing
by "x". So, "y" or the number of packs of batteries that Leonard can buy, is ten divided
by "x". So there you have an example of a problem that you may encounter when you take
the TEAS test. The TEAS test has a lot of word problems on
it. Here is one example of a type of word problem that you could encounter when you
take the TEAS test. It says, "Choose the algebraic expression that best represents the following
situation: Jeral's score, 'j', was five points higher than half of Kara's test score 'k'".
I like to say that we're taking English and translating it into "math" in this case. So,
if you can figure out what the key words mean, then this is, actually, pretty simple. So
we're just starting with "j". This word "was" is the past tense of "is". "Is" in math means
equals. So this word "was" is an equals sign. And of course, five is five. "Higher than"
tells you to add. It's just like more than or greater than, and all those words mean
"to add". Half, of course, means half. "Of" is another really important word. "Of" tells
you to multiply. So, you multiply whatever comes on either side of the word "of". So,
it's half times Kara's test score "k". So it's half "k". Another thing with higher than,
or greater than, or more than, is that word "than" tells you that, actually, you need
to switch the order. In stead of writing five plus half "k", you just switch these around.
So it says "j" equals half "k" plus five. But really, these say the same thing, since
when you're adding, it doesn't matter the order that you add two things in, you get
the same answer. That's the commutative property of addition. So, um, we see then that "a"
is our answer here. And there you have one example of a type of word problem that you
may encounter when you take the TEAS test. You're going to encounter a lot of word problems
on the TEAS test. So, let me walk you through how to solve one type of word problem you
may encounter. "Enrique weighs five pounds more than twice Brendon's weight. If their
total weight is two hundred twenty-five pounds, how much does Enrique weigh?" Even though
this is a short problem, there are, actually, two different equations we can write from
it. The first one starts with the first sentence. "Enrique weighs..." this is another way of
saying "Enrique is", and "is" in math is an equal sign. Since we don't know Enrique's
weight, we'll use a variable. In this case, an "e" makes sense. So, "Enrique is five pounds
more than...", and more than tells us to add, "...twice Brendon's weight." We also don't
know Brendon's weight. So, we need another variable, and in this case, a "b" would make
sense. The "twice" tells us to multiply times two. So, since it's twice Brendon's this is
actually going to be two times "b". "...more than...", besides telling us to add, also
tells us to switch around the order. So, instead of writing five plus "2b" we would write "2b"
plus five. Now, honestly, it doesn't really matter, five plus "2b", "2b" plus five, it
is the same thing. That's the commutative property of addition. That's the first sentence
and our first equation. Our second sentence has our second equation. "If their total weight
is two hundred twenty-five pounds, how much does Enrique weigh?" "Total" tells you to
add. And what are we adding? Enrique and Brendon's weight together. So, Enrique's weight plus
Brendon's weight is, and again that "is" tells us that it's an equal sign, two hundred twenty-five.
Okay so, how are we supposed to solve these two equations, that each have two different
variables? Well you have several options: you could graph both of these equations, which
I know nobody wants to do. You could use elimination, but really they're not set up for that. This
one is, but this one isn't because the "2b" isn't on the same side. Like this equation
you have both variables on the same side of the equal sign, but you don't in this one.
So, what they're really set up to is substitution. And, since that's the easiest way to solve
it, I think that makes the most sense. We're going to use substitution. We're going to
take this "2b plus five" and use it to replace the "e" in this equation. Since, what this
says is that "e" and "2b plus 5" are the same thing, then we can replace the "e" with this
expression "2b plus five". This equation then becomes: "2b plus 5", instead of the "e",
plus "b" equals two hundred twenty-five. Now we're able to actually solve this equation,
because it only has one variable. Starting with combining your like terms of "2b" plus
another "b" which is a total of "3b"s plus five equals two hundred twenty-five. Now we
need to subtract five from both sides. "3b" is equal to two hundred twenty, and then divide
both sides by three. So two hundred twenty divided by three is not a pretty number. Three
goes into twenty-two seven times, that'd be twenty-one. One left over, ten, three. So,
it's seventy-three and actually three-tenths repeating, or you could say seventy-three
and one third. Okay, but going back to the question, they didn't ask us how much Brendon
weighed, they asked us how much Enrique weighed. So, now that we know Brendon's weight, we
can use that to find Enrique's weight, by using this equation that already says, Enrique
what does he weigh? He weighs twice Brendon's weight, which we just found is seventy-three
and three-tenths repeating or seventy-three and one-third plus five. So, Enrique weighs
two times seventy-three and one-third, is one hundred forty-six and two-thirds plus
five. So, that is one hundred fifty-one and two-thirds, which is closest to the one hundred
fifty-two pounds. So, there you have an example of a type of word problem that you may encounter
when you take the TEAS test.