Welcome to the Cypress College math review on mixture word problems
in this video we are going to talk about mixture word problems
to start off with let me ask you a question
what do you get when you add five gallons and 3 miles
how do you add miles and gallons together is it 8... no you can't do that
the units for each term must match they must be identical
for example if you are taking dollars in the first term you better have dollars in the second term and your answer will be in
dollars in the problems that we will be working with which are mixture problems were going to have something like this
we're going to have
the gallons of pure acid plus the gallons of pure acid in the second
solution to get the gallons of pure acid in the final solution
let's talk about the formula you would use for solution mixture problems let's say that in this bucket
I have a gallon but it's a mix part of its water and part of it is antifreeze in fact I know that it is 50% antifreeze
50% antifreeze what does that mean
how much antifreeze is in the bucket how much actual pure antifreeze is in the bucket
½ gal right
how did you get that
will let's see what you did you took 50%
of
1 gallon yes 50% of one gallon and what is of in mathematics
of is multiplication so we take 50% percent and we multiply it
times the amount so we take the
concentration times the amount
that gives us the formula that we are going to use in all the solution
mixture problems and what's really, really cool about mixture problems
is the formula is the same for every term on both sides of the equation even over on
the right hand side you also can have concentration times amount
let's talk next about how the percentages relate to each other let's say you had one concentration
that was 15% and another mixture that its concentration was 50%
and you have another mixture that was let's say 40%
which of these three is the final mixture now I'm assuming the we're adding them together were not like draining something
off the traditional type where you are adding two solutions together and getting a third solution
so let's say
could the mixture at the end of the problem be the 50% one
could you take something this 15% pure
and add in something that's 40% pure and end up with something that's 50% pure so take something is really diluted
mix it in with something that's way more concentrated and get
something that's more concentrated than either of the two no way that's impossible
could this be the mixture at the end could you take something that's 50% pure
add in something that's 40% pure and get an answer
that's way more diluted than either of the two mixtures you to put together that's impossible
of these three numbers the one that has to be the concentration for your
answer must be the one on the right there the 40% because it's numerically in between
it's the number that's numerically in between the other two mixtures
that will always be the mixture or solution that you have at the end your final solution
we talked about how the concentrations relate to each other now let's talk about how the amounts relate to each other
if I have 3 gallons of one mix
and add it to 2 gallons of another mix
I'm definitely going to get a mixture that is 5 gallons right
here's the formula that
we are working with concentration times amount
so we are going to focus on the amounts here
the way I've set it up the amounts are in the second set of parentheses
what if you were given one of the amounts on the left hand side let's say it's 5 gallons that goes in the first
in first term in the second parenthesis
if you're given that that's 5 gallons then
this would be X you don't know that amount right
now what do you do for the amount on the right hand side
well we know that they add up just like 2 gallons and three gallons added to the 5 gallons right so 5 gallons plus
x gallons is right five plus x so five plus x would be the amount on the right hand side
what if instead of being given one of the amounts on the left hand side you were given one any amounts on the right hand
the amount on the right hand side yeah so it's 12 gallons
you're given that the amount on the right hand side
then you're gonna use x for one of the amounts on the left hand side
and then you have two options
you can either use y for the other amount on the left hand side
and then have a second equation that says that x plus y is 12 solve that
equation for Y which would be 12 minus X and then you can replace y with 12 - x
I prefer to just set it up originally with 1 unknown so if I know that I have 12 gallons altogether in the final solution
one of the amounts on the left is x the other is 12 - x 12 is the total subtract the part
total minus part
Sonya has a 55% antifreeze solution and a 10% antifreeze solution
she wants 30 gallons of a 49% antifreeze solution how many gallons of each must be mixed to get the desired solution
so here's our generic formula
now let's fill in what we know
let's look first at the concentrations
we have 55% 10% and 49% which one's mathematically in between the other two
49 so that's gonna be your final solution so that's the one that goes over on the right and the other two
concentrations go on the left hand side
let's look next at the amount you're given 30 gallons of the 49% you put the 30 next to the 49%
ok so we're given the amount on the right if we're given the amount on the right
then one of them is the part that's the X and the other is whole minus part yeah total minus part so
30 - x
and here's the solution to that problem
a chemist has 3 ounces of a 9% alcohol solution
how many oz of a 17% alcohol solution must be added in order to get a 15% alcohol solution
here's our regular formula
now let's fill in what we know
let's look first at the concentrations
we have 9%
17% and 15% which one's mathematically in between the other two
15 oh so that's the concentration of the final mix
the other two can be put in either order
be sure to change 9% to a decimal the correct way 0.09
now let's look at the amounts we're given 3 ounces of the 9% so put 3 ounces next to the 9%
so we're given one of the amounts on the left
X is the other amount on the left remember these mathematically have to add up to
get the amount at the end so the amount of your final mix is three plus X
here's the solution to that problem
how many liters of pure acid must be mixed with a 25% acid solution to get 10 liters of a 40% acid solution
so this one is a little different we have pure acid what's the concentration of pure acid
well it's all acid so it's 100% pure right so that would be 100%
so the concentrations were dealing with on this problem are 100%
25% and 40%
here we go
so which one is mathematically in between 40% that's the one that goes on the right
and put the other two in and 100% as a decimal what's that's one exactly
the other one is 0.25
now for the amount you're given 10 liters of the 40% solution so put the 10 liters over there with the 40
so you're given the amount on the right anytime you're given the amount on the right
one of the amounts on the left is X the other
one is: total minus part so 10 minus X right total is 10 minus the
part over here on the left hand side is X cause we're going to end up with 10 gallons
or liters altogether
here's the solution to that problem
how much water should be evaporated from
here we go with a different kind of problem
from
this is being taken away so it's not like two mixtures being added together
how much water should be the evaporated from 240 gallons of a 3% salt solution to produce a 5% salt solution
so here's our generic formula for this problem
notice the subtraction on this problem instead
let's put in what we know
alright now we have to be careful about what we are starting with and what we're ending with
because the subtraction makes things different
how much water should be evaporated from so this right here
is your water
that's what's being evaporated from
what is the concentration
of water how much
salt is in water well unless it's seawater it's 0% salt yes we're assuming this is just pure water so the concentration for that
would be 0% so 0% as a decimal is still just the number zero you can take 0%
change it to a decimal it's still just 0.00 it's just zero yes
240 gallons of the 3% salt solution so you're starting off with a 3% solution
and your ending up with a 5% now once again
the concentrations aren't going to be the same as we're used to it's not going to be
mathematically in between the other two because we're
taking away this problem is different you're taking something away your starting off with something that's 3% pure
you're gonna make it more concentrated by evaporating the water away and now
it's gonna be 5% so it does make sense now so let's look at the amounts
you're given 240 gallons of the 3%
we don't know how much water is taken off
now just like before we're gonna take a look at the amount on the left hand side
we start off with 240 gallons we take away X gallons which means we're left with 240
minus X gallons
and here's the solution to the problem
let's turn our attention now to value mixture problems
what's the formula that we would use for a value mixture problem let's say you go to the grocery store
and you're looking at hamburger
and the price on the package says $3.00 per pound
and you buy 2 pounds of it
how much did that cost you
$6.00
yeah that would cost you $6.00
because you multiply the price per pound
times the number of pounds to get
the price look what happens to the pounds yeah the units cancel right off the
units can cancel off just like numbers and variables can
so this is the formula that you are going to use for value mixture problems
dollars per pound
times pounds and once again
the formula will be the same for every term on both sides of the equation
it's really nice sometimes it won't be dollars it could be cents instead of pounds it could be oz but you get the idea
let's look at how the prices relate to each other
let's say you have
one of your prices you are dealing with is $5.00 a pound
another one is $7.00 a pound
and the third one is $4.00 per pound and this is a traditional problem where you're
mixing two ingredients together to get a third final product
which one is the price of your final product
could you take these two
and mix them together and end up with something at the end that's only $4.00 a pound so you take two things that are
more expensive than the third and come up with something that's cheaper
no that's not going to work is it
could you take two things that are less expensive
and end up with something that's more expensive than either the other two
that doesn't make any sense either
so once again the price of the mix is going to be mathematically in between
the price of the two ingredients that you put together so $5.00 a pound will be the price of your final mix
how many pounds of peanuts that sell for $1.80 per pound should be mixed
with 3 pounds of cashews that sell for $4.50 per pound
to get a mixture that sells for $2.61 per pound
here's our regular formula
now let's put in what we know
let's look first at the three different prices
which one's mathematically in between the other two
the $2.61 yes the $2.61 must then be
the price of the final mixture
put in the other two prices
and now let's look at the amounts
you're given 3 pounds of the cashews that that cost $4.50 per pound so put the three next to the 4.5
so you're given one of the amounts on the left
the other amount on the left is X and once again the two amounts on the left must
mathematically add up to get the amount on the right
and here's the solution to this problem
in a local supermarket hamburger sells for $3.50 per pound
and ground sirloin sells for $4.20 per pound
how many pounds of each should be mixed in order to obtain 30 pounds of a mixture that sells for $3.78 per pound
here's our regular formula
now let's put in what we know
let's look first at the prices
which one's mathematically in between the other two
the $3.78 so that's the one that goes on the right hand side
put the other two in
now let's look at the amounts which amount were you given 30 pounds
of the mixture at the end 30 pounds of the $3.78 mixture
so the 30 goes next to the $3.78
ah so you're given the amount on the right if you're given the amount on the right
then one of the amounts on the left is X which is the part
so the 30 is the
total
the X is part of that 30
what's left total minus part right total minus
part so the amount that goes next to 420 would be 30 minus X these two are interchangeable you can interchange the
30 minus X with the X it works out just the same exact way
here's the solution to that problem
here are some more problems for you to practice
on mixture problems so you'll get comfortable with them
I encourage you to pause the video do each of these problems
and then watch the rest of the video to see the
answers
here are the answers to the extra problems
I hope this video has helped you learn some more about mixture problems