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## College Algebra

### Course: College Algebra > Unit 8

Lesson 2: Modeling with rational expressions# Combining mixtures example

Figure out how much of two gasoline mixtures to combine to get gasoline with a certain concentration of ethanol. Created by Sal Khan.

## Want to join the conversation?

- Not sure why these problems always throw me for a loop but, in case you get confused, here's a shorthand version:
`where c = content / v = volume / c3 is the total content / v3 is v1+v2`

c1(v1) + c2(v2) = c3(v3)

This is just an abbreviated version of cancelling the denominator out and multiplying by the ethanol content`18(30) + 25(v) = 20(30 + v)`

540 + 25v = 20(30 + v)

25v-20v2 = 600-540

5v = 60

v=12(27 votes) - i don't understand this topic at all, i don't know if it needs any background information, if I need please tell me which lessons I should watch.(8 votes)
- I didn't know the formula for concentration!(5 votes)
- Might be helpful to use the c1v1 + c2v2 =c3v3 formula where

c1 = .18, v1 =30, c2 = .25, v2 is unknown, c3 = .20 and v3= (30 + v2).(4 votes) - Isn't this the acid-base titration equation?(3 votes)
- To verify the concentration, let's model the equation.
`target concentration = original ethanol volume + new ethanol volume / (original total volume + new total volume)`

Translate the equation into the given numbers, where v equals the new volume.`target concentration = 5.4 + 0.25v / (30 + v)`

target concentration = 5.4 + 0.25(12) / [30 + (12)]

The target concentration equals 0.20 which proves our new volume is correct.(2 votes) - Sal did 6-5.4 and got 0.05. Shouldn't the answer be 0.06?(1 vote)
- Sal has the equation: 6 = 5.4 + 0.05V

Notice, the 0.05V is already in the equation. He subtract 5.4 from both sides to create: 0.06 = 0.05V, (not 0.6 = 0.05). He then finishes solving for "v" by dividing both sides by 0.05 to find V = 12.

Hope this helps.(1 vote)

- I made short summary of how to solve this problem using Sal's method.(And expanded the parts he skipped)

concentration=volume of ethanol/ total volume

>**The tank originally has: 30L with 18% conc. ethnal**.

>Fuel station has: 25% conc gas [conc=vol eth./ total vol]

>We are asked to find how much we need to add to make it 20% conc

18%=vol. eth/30L vol. eth= 18%x 30L=5.4L

20%=New vol eth./ new total vol.**Let**`v= volume of gas we need to add (from the fuel station)`

x= volume of ethanol of the gas we need (from the fuel station`)`

conc=vol eth./ total vol

20%=5.4+x/ 30+v

conc of the gas (from the fuel station)= 25%=x /v

x=25%*v (substitute back

20%=5.4+25%v/ 30+v

0.2=5.4+0.25v/30+v

0.2(30+v)=5.4+0.25v

6+0.2v=5.4+0.25v

0.6=0.05v v= 0.6/0.05

v=12

volume of gas we need to add (from the fuel station) is 12 liters(L)

Hope this helps(1 vote) - If the tank only holds 30 liters, how do you put more than 30 liters (30+v) into it?(1 vote)
- 30L is the amount of gas originally in the tank, not its maximum capacity. If we meant to say the tank had a 30L capacity, we would not specify the makeup of the gasoline, since the tank would be able to hold 30L of any substance.(1 vote)

- The Wording of this problem makes it seem like the Tank has a maximum capacity of 30L. It has taken an embarrassing amount of time to realize that is not what the sentence is saying. At the same time I would argue it is poorly worded. The Solve seems impossible if given just the max capacity and not the actual volume of gasoline in the tank.(0 votes)

## Video transcript

- [Instructor] We're told
a partially filled tank holds 30 liters of gasoline with an 18% concentration of ethanol. A fuel station is selling gasoline with a 25% concentration of ethanol. What volume in liters of
the fuel station gasoline would we need to add to the tank to get gasoline with a 20%
concentration of ethanol? Pause this video and see
if you can figure this out. All right, now let's work
through this together. So let's, first of all,
just remind ourselves how concentration relates to total volume, to the volume of the ingredient. The way that we calculate concentration is that it is equal to the
volume of the ingredient, which is, in this case, it is ethanol, over the total volume, over total volume. Now this is already interesting because this first
sentence tells us a lot. It tells us our concentration. It tells us our total volume. And so if we know two parts of this, in theory, we could
figure out the third part. Let's try that out. We know we're dealing
with an 18% concentration. That's going to be equal to, they haven't told us our
volume of ingredient. We just know that the
ingredient is ethanol. Volume of ethanol over the total volume they
have told us, 30 liters. So if we multiplied
both sides by 30 liters, that's going to give us
the volume of ethanol, 'cause those two cancel. And what we get is 18% of 30. Let's see, 18 times 3 is 54, so this is going to be 5.4 liters is equal to our volume of ethanol. Not only will this, hopefully,
make it a little bit clearer, how these three relate, but this is also likely
to be useful information for the rest of the problem. But now let's go to where
we're trying to get to. We're trying to find a volume, in liters, of the fuel station gasoline we would need in order to have this concentration. So let's set v equal to that. And we're trying to get
a 20% concentration. So what we could write
is our 20% concentration is going to be equal to
our new volume of ethanol. Actually let me write that out. So it's gonna be new volume of ethanol divided by our new total volume, our new total volume. Now what's going to be
our new total volume? We're starting with 30 liters, and then we're adding v liters to it. So our new total volume is going to be the 30
liters we started with plus the v liters that we're adding. And what's our new volume of ethanol? Well it's going to be the
ethanol that we started with, the 30 liters times 18%,
which is 5.4 liters, 5.4 liters plus the volume
of ethanol we're adding. Well to figure out the volume
of ethanol we're adding, we just have to multiply
the volume we're adding times the concentration of that volume. So it's gonna be 25%, that's the concentration of
the gas that we're adding, times V, plus 0.25v. And now we have an
equation to solve for v. And the best thing that I can think to do is let's start by multiplying
both sides of this times 30 plus v. I'm also going to multiply this side times 30 plus v as well. These two characters cancel, and we are going to be left
with, on the left-hand side, this over here, 20% of that is going to be,
if I distribute the 20%, 20% of 30 is 6, and then it's going to be plus, I'll write it as a decimal, 0.2v. I'm just distributing the 20%
over this expression here. And then that is going to be equal to, on the right-hand side, I
just have the numerator here, because the 30 plus v
cancels with the 30 plus v. I have 5.4 plus 0.25v. And now, let's see. My v coefficient is larger on the right. So what I could do is
try to subtract the 0.2v from both sides, so I
isolate the v's on the right. So let me do that, so
minus 0.2v minus 0.2v, and then, actually I'll
just do one step at a time. So that's going to get
me, on the left-hand side, 6 is equal to 5.4 plus, if I subtract here, this is 0.05v. Now I could subtract 5.4 from both sides. And what I'm going to get is 6 is, or actually 0.6, I have to be careful, is going to be equal to 0.05v. And now to solve for v, I can just divide both sides by 0.05, 0.05. That's going to get me,
this is the same thing. as 60 divided by 5. It gets me that v is equal to 12 liters. And we are done. And if you want, you can
verify the new concentration. When I add 12 liters of this concentration to the 30 liters of that concentration, you can verify that I now
have a 20% concentration of ethanol.