Let's start with a warm up ratio problem.
Then we can tackle some harder word problems. So I have the
ratio 13/6 is equal to 5/x.
I don't like having this x in the denominator, so let's
multiply both sides of this equation by x.
So if I multiply both sides by x, what's going to happen?
On the right hand side, this x cancels out with that x.
And then the left hand side going to become 13 over 6x is
equal to-- you're just going to have a 5 there.
And then to solve for x, you just multiply both sides by
the inverse of 13/6.
6/13.
These, obviously, cancel out.
That's why I multiplied it by the inverse.
And you get x is equal to 5 times 6, which is 30/13.
Now one way that you might see this done-- it's kind of
skipping a step-- is called cross multiplying.
You look at a ratio like this, and you immediately say the
numerator on this side times the denominator on that side
is equal to the numerator on this side times the
denominator on that side.
Let me write that out.
So you might sometimes see people immediately go to-- let
me just rewrite the problem actually-- So that original
problem was 13/6 is equal to 5/x.
You might sometimes immediately see someone go to
13 times x is equal to 5 times 6.
And it might look like magic.
How does that work out?
Why does that make sense?
And really, all they're doing to get to this point is they
are simultaneously multiplying both sides of the equation by
both denominators.
Let me show you what I mean.
If I multiply both sides of this equation by 6 and an x,
what's going to happen?
If I multiply it by 6x times both sides of this equation--
And where did I get the 6?
From here.
Where did I get the x?
From there.
Both denominators.
What's going to happen?
On this side of the equation, the 6 is going to cancel out
with this denominator.
And on the right hand side of the equation, the x is going
to cancel with this denominator.
So you're going to be left with 13 times x is
equal to 5 times 6.
So nothing fancy there.
You're just multiplying by the denominators of both sides of
the equation.
And it looks like you're cross multiplying.
13x is equal to 5 times 6.
And then from here, of course, you divide both sides by 13.
You get x is equal to 30/13.
Now that we're all warmed up, let's tackle some actual word
problems.
So we have the highest mountain in
Canada is Mount Yukon.
It is 298/67 the size of then Ben Nevis.
Let's Y for Yukon is equal to 298/67 the size of-- let's say
N for Nevis.
That's what this in green tells us.
The highest peak in Scotland.
Mount Elbert in Colorado is the highest peak
in the Rocky Mountains.
Mount Elbert-- so we have this other information here-- Mount
Elbert is 220/67 the height of Ben Nevis.
So let's say, E for Elbert.
E is equal to 22/67 times Nevis.
Times the same Ben Nevis, right there.
And they're telling us more.
And, it is 44/48 the size of Mont Blanc.
So Elbert is equal to 44/48 the size of Mont Blanc.
Let's write B for Mount Blanc.
They also tell us Mont Blanc is 4,800 meters high.
Mont Blanc is 4,800.
meters high.
So B is equal to 4,800.
And they ask us, how high is Mount Yukon?
So we have to figure out Y.
So let's see if we can work backwards, and figure out all
the variables in between.
So let's start with this information here.
B is equal to 4,800.
E is equal to 44/48 times B.
So E-- so Elbert-- is equal to 44/48 times Mont Blanc, which
is 4,800 meters.
Now if you divide that by 48-- 4,800 divided by 48 is 100.
So Elbert is 44 times 100 meters high.
So it's equal to 4,400 meters.
Fair enough.
Now we can use this information and
substitute it over here.
We get Elbert, which is 4,400 meters high, is equal to
220/67 times Ben Nevis.
N for Nevis.
To solve for Nevis, we multiply both sides by the
inverse of this coefficient right here.
So we multiply both sides by 67/220.
So times 67/220.
The 67 cancels with that 67.
That 220 cancels with that 220.
And then you get-- let's see, if I take 4,400 divided by
220-- 440 divided by 220 is 2.
So this is going to be 20.
So 4,400 divided by 220 is just 20.
So you get Nevis is equal to-- I'll swap sides.
So Ben Nevis is equal to 67 times 20 meters.
And now that's what?
1,340 meters.
Is that is right?
Well, lets just leave it like that, because we could--
actually it looks like that's 67-- I'm going to leave Nevis
as 67 times 20 meters.
And substitute it right there.
So Yukon-- I'll just go down here, because I have more real
estate there-- Yukon is equal to 298 over 67 times the
height of Nevis.
Nevis is 67 times 20.
So times 67 times 20.
Well I can divide 67 by 67, and I get Yukon is
298 times 20 meters.
So Yukon is equal to 298 times 20.
And what is that equal to?
That is equal to-- Let's see that's 2 times 298
is going to be 396.
Oh sorry, it's going to be 596.
This is almost 300, so it should be close to 600.
This is 2 less than 300, so this should
be 4 less than 300.
And then, I have a 0 here.
So it's going to be 5,960 meters.
And we are done.
Let's do one more of these word problems. All right.
At a large high school, it is estimated that 2 out of every
3 students have a cell phone.
And 1 in 5 of all students have a cellphone that is one
year old or less.
All right.
So let's think about it.
Let's say that x is equal to the total number of students.
This first line, 2 out of 3 students have a cell phone, so
we could say that 2/3 x have cell phone.
That's what that green statement tells us.
And then that purple statement-- 1 in 5 of all
students have a cellphone that is one year old or less.
So 1/5 x have less than 1/5 year cell phone.
So they want to know, out of the students who own a cell
phone-- so it's out of this-- that's our denominator.
So let me write that down.
That is our denominator.
So out of the students who have a cellphone-- that's
right there-- they want to know what proportion owns a
phone that is more than one year old.
So how many students have a cell phone that is more than
one year old?
Well, we could take the total number that have a cellphone,
which is 2/3 x.
2/3 of all the students have a cell phone.
We can subtract out all of the students that have a new
cellphone-- a cell phone that is less than one year.
Remember they're saying more than one year here.
So we want to subtract out all the students with the new
cellphone, minus 1/5x, and you will then have the proportion
of students who have this right here.
This is right here.
This is, have greater than 1/5 year cell phone.
They have a phone, but it's more than 1/5 years old.
This is all of them that have a cellphone.
We subtract out the new ones.
So this is, essentially, all of the people who have an
older than one-year old cell phone.
So to solve this, we just subtract the fractions.
So this is just going to be, let's see, 2/3 is the same
thing as 10/15.
That's 2/3 minus 1/5.
The same thing as 3/15 x.
Which is equal to 10 minus 3 is 7/15 x.
Is the total proportion of students-- that's this
orange-- what proportion owns a phone that is more
than one year old?
It's 7/15 x.
That's an actual number.
So if you want to know, out of the students who own a cell
phone-- so out of the students who own a cell phone, right
there-- 2/3x, what proportion owns a phone that it is more
than one year old?
This is the number that own a cellphone that is more than
one year old.
And this whole value is the proportion out of the students
who have a cell phone.
Lucky for us, the x's cancel out.
And we are left with this is equal to 7/15 times the
inverse of the denominator.
You divide by 2/3.
That's the same thing as multiplying by 3/2.
And what does this equal to?
Divide by 3.
We are left with 7/10.
So of the students who own a cell phone, 7 out of 10 of the
students who own a cell phone, own a cell phone that is more
than one year old.
And we are done.