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### Course: Algebra (all content)>Unit 2

Lesson 10: Linear equations word problems

# Linear 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.

## Want to join the conversation?

• These problems kill me. Could someone develop a similar problem but much simpler to help me understand? Thanks!
• 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)
• I am completely confused, can someone please explain how to do this problem, because I do not get it
• Where can you find exercises to practice these sort of problems?
• how do i do this without a table? Xx
• 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.
• How did he change x=100 to 10 ounces?
• 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.)
• 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.
• How many liters of a 30% alcohol solution must be mixed with 60 liters of a 50% solution to get a 40% solution?
• 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
• 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?