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Algebra (all content)
Course: Algebra (all content)ย >ย Unit 2
Lesson 10: Linear equations word problemsLinear equation word problem: saline
Sal solves the following word problem: You have 50 ounces of a 25% saline solution. How many ounces of a 10% saline solution must you add to make a solution that is 15% saline? Created by Sal Khan and Monterey Institute for Technology and Education.
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These problems kill me. Could someone develop a similar problem but much simpler to help me understand? Thanks!(4 votes)
You have a really great rock collection. It has 40 rocks and 50% of the rocks can be classified as rare. In other words, you have 20 rare rocks.
Your younger brother comes along with his inferior rock collection, which only has 10% rare rocks. Your clumsy brother drops his rocks and now it's mixed in with your collection!
Since your collection has been "diluted" with your brother's collection, the combined collection is no longer 50% rare. In fact, it is only 42% rare. Can you figure out how many rocks were in your brother's collection?
42% rare means that the total number of rare rocks (20 of yours plus the rare rocks from your brother) divided by the total number of all rocks (40 of yours plus the rocks from your brother) will equal 0.42.
Since we don't know how many rocks were your brother's, make a variable x to stand for it. That means the number of rare rocks your brother had is 0.1x (10% of x).
Now you can write that 42% fraction as (20+0.1x) / (40+x). Solve for x and you'll be done!
(Answer: x is 10, meaning your brother's collection was 10 rocks, with 1 of them being rare)(13 votes)
I am completely confused, can someone please explain how to do this problem, because I do not get it(7 votes)
Where can you find exercises to practice these sort of problems?(4 votes)
if you google "eighth grade mixture problems, worksheets with about 5-10 problems will pop up.(6 votes)
how do i do this without a table? Xx(3 votes)
c = (c1q1+c2q2)/q1+q2, this equation works :D . Here C is the final solution c1=1st concentration, q1= amount of c1, c2=2nd concentration, q2= amount of c2.(5 votes)
How did he change x=100 to 10 ounces?(3 votes)- O.O Could someone please explain how Sal did it and (maybe) what his reasoning was?
(And whoever deleted this last time, please don't delete it again because I'm asking questions on all the videos I need help with, so I'm just copying & pasting everything so it takes up less time. Thanks.)(3 votes)- When I first approached this video I did not understand it either. I thought we would be making a table and after a few columns we would have our answer. I thought we would have a 5% salinity or some other variable, changing, until we got to our 15% salinity. However I was reading the problem incorrectly. The 10% is not changing. It is how many ounces of a 10%, the ounces being what is changing. So X relates to the unknown quantity of ounces you are adding of a 10% solution that will eventually dilute your original 25% down to 15%. Think of it that way. You have your starting point of 50 ounces, and then you add x amount of a 10% solution.
We got 12.5% salinity by multiplying 50(amount in ounces) times .25 or 25%. We now know when we multiply the amount of ounces times the percentage of a saline solution we get the amount of saline of the product. We can also look at the amount of 12.5% salinity (original starting point) plus amount you added which is the unknown X times the unchanging 10 percent to get your total amount of saline, X times 10% as we know our solution is 10% saline but not how much we added. see what I am saying the original amount is 50 ounces plus x or 12.5% saline solution plus X solution added times 10% or .1X. So then he used 50 + X = or is the same quantity as 12.5+ .1x. Now we want a 15% solution, how many ounces of that 10% solution did we add to get 15%?? So that is why he has .15 (50+x) which is 15 % saline times the original amount of 50 ounces plus the added amount X. But we can't figure out what is X from that alone. Since we know 12.5 +.1x is the total amount of saline, we can take the 50 +X times 15% = 12.5 + 01x. In this case the X equates to an amount of ounces of a 10% saline solution. It is hard to understand where he just came up with that 15%, We could have chose a different percentage and get that amount, the X would change its value with a different percentage added. But we really want to know how many ounces we added to get 15% we don't really care about any other percentage.(0 votes)
- How many liters of a 30% alcohol solution must be mixed with 60 liters of a 50% solution to get a 40% solution?(0 votes)
If x is the required no. of liters of the 30% alcohol solution, the problem translates as follows:
30% of x (amount of alcohol in x liters of 30% solution) + 50% of 60(amount of alcohol in 60 liters of 50% solution) = 40% of (x + 60) ( the amount of alcohol SHOULD BE 40% of the new volume which is x + 60)
.3x + 30 = .4(x + 60) = 3x + 300 = 4(x + 60)
x = 60 liters
Another way to think of it is
Solution concentration(%) = amount of alcohol/volume of solution
If x is the required no. of liters of 30% solution then
1.The amount of alcohol = 30% of x + 50% of 60 liters
2.The volume of solution = x + 60
(1)/(2) = 40% = (0.3x + 30)/(x + 60) = 40/100
and x = 60(3 votes)
- An alloy is consisting of 40% copper and 60% nickel.In 120 kg alloy how much copper to be added so that the amount of copper will become 46% in that mixture?(2 votes)
what does x have to do with anything? and why .1x?(2 votes)- x = the amount of 10% saline solution we must add. O.1x is the amount of saline we have in the 10% saline solution.(1 vote)
find two consecutive odd integers whose sum is 111(2 votes)- Hint: use x+2+x+4...(1 vote)
Video transcript
We're told to make a
table and solve. So they tell us that we have
50 ounces of a 25% saline solution, a mixture
of water and salt. How many ounces of a 10% saline
solution must you add to make a new solution
that is 15% saline? So let's make this table that
they're talking about. Let's write amount
of solution. Let me write total amount of
solution or maybe I should say total volume of solution. And then the next column I'll
say percent saline. And then we can use this
information to figure out total amount of saline. And let's list it for each of
the two solutions that they talk about. We're starting with 50 ounces
of a 25% saline solution. So this is what we start with. Starting solution, we have
50 ounces of it, so I'll just write 50. We'll assume everything
is in ounces. It is 25% saline. So if we wanted to figure out
the total ounces of saline, we say, well, we have 50 ounces. Multiply that by 25% and we
have the total amount of saline in this solution. So 50 times 25%, that's the same
thing as 50 divided by 4, so that's 12.5 ounces
of saline in this 50 total ounces. It's 25% saline. Now, let's talk about what we're
going to add to it, so solution added. Now, they say how many ounces
of a 10% solution? So we don't even know how many
ounces we're going to add. That's what we have to
actually solve for. So let's call that x for the
amount of solution that we have to add. So we don't know how much we're
going to add, but we do know that it is a 10%
saline solution. And if we know what x is, we
know the total amount of saline is going to
be 10% of x. If we had 50 here, it
would be 10% of 50. If we had 10 here, it
would be 10% of 10. So the amount of saline we have
in this solution, in x ounces of this solution, is
going to be 0.1x, or 10% percent of x. That's what 10% of the solution
being saline means. Now, when we add it, what
do we end up with? So let me do this in
a different color. Resulting solution. Well, if we started with 50
ounces and we add x ounces, we're going to end up with
50 plus x ounces. That's our total volume of
the resulting solution. What percent saline is it? Well, our goal is to make a 15%
saline solution, so it has to be 15% saline. Now, what's the total amount
of saline in it? And this is kind of the main
box or cell in this table. There's two ways to get
to the total amount of saline in the solution. One, we could just multiply the
percent saline times the total value, so we
could do that. So let write that. It's 0.15 times 50 plus x. All I did, I multiplied
the percent saline times my total volume. That's one way to get the
total amount of saline. The other way to get the total
amount of saline is to add these two numbers. The 50-ounce solution had
12.5 ounces of saline. We added 0.1x ounces of saline,
so if we add these two numbers, it should also
be equal to the total amount of saline. So this has to be equal to the
sum of these two things. It has to be equal to
12.5 plus 0.1x. And just like that, because 15%
of this has to be the same thing as the sum of this, we
have one equation and one unknown, and we can solve
for x, which is what we need to solve for. How much solution do
we need to add? So let's just do that. So 0.15 times 50. Let's see, that's 7.5, if
I'm doing that right. Yeah, because 0.15 times
100 would be 15. So this is 7.5 plus 0.15x--
that's the left-hand side-- is equal to 12.5 plus 0.1x. Let me scroll to the
right a little bit. Now, we can subtract 7.5 from
both sides of this equation. Let me do it in a new color. So if we subtract 7.5 from both
sides of this equation, the left-hand side,
that cancels out. We're just left with 0.15x
is equal to-- 12.5 minus 7.5 is just 5. 5 plus 0.1x. Now, we could subtract
0.1x from both sides of this equation. Let me scroll down
a little bit. Let me subtract 0.1, or I could
say 0.10x from both sides of this equation. These are the same number. So they cancel out. The left-hand side, we
just end up with 0.05x is equal to 5. And now we just divide
both sides by 0.05. And we get x is equal to
5 divided by 0.05. That's the same thing as 5
divided by 1/20, or the same thing as 5 times 20. So we get x is equal to 100. So we're done! If you add 100 ounces--
so we figured out that x is equal to 100. If you add 100 ounces of 10%
saline solution, you will end up with 150 ounces of
15% saline solution. So if you add 100,
10% of that, we actually added 10 ounces. So you have 12.5 plus 10 ounces
is 22.5 ounces of saline in a solution of 150
ounces, which will be 15%. And we're done.