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Mathematics 11 Online
OpenStudy (dynamiz77):

Torren has 25 ounces of 30% salt solution. How much salt should he add to make it a 40% solution?

OpenStudy (mathstudent55):

x amount of 30% salt solution contains 30% of x in salt. x amount of salt solution contains 0.3x salt. Ok?

OpenStudy (mathstudent55):

Pure salt is 100% salt, so y amount of salt is y amount of salt, since all of it is salt. Let's say we start with x amount of 30% salt solution; we add y amount of 100% salt until we end up with an amount of 40% salt solution.

OpenStudy (mathstudent55):

If you have x amount of the 30% salt solution and you add y amount of salt, what amount of the 40% solution will you have?

OpenStudy (mathstudent55):

No. The percentages don't add. This question is easier than you think. You have x ounces and you add to it y ounces. How many ounces do you now have?

OpenStudy (dynamiz77):

x + y = 100

OpenStudy (dynamiz77):

i know i'm overthinking this

OpenStudy (mathstudent55):

Forget the 100. x ounces plus y ounces = x + y ounces

OpenStudy (mathstudent55):

You start with a given amount of the 30% salt solution. We call that amount x. (We actually know the amount, and we'll use it soon.) Then you add an amount of pure salt. Since pure salt is 100% salt, it is more concentrated than 30% salt solution, so the more pure salt you add, the higher the concentration of the resulting solution will be. You add enough pure salt, we'll call the amount y, until you reach a concentration of 40% salt. That means, you add y amount of pure salt to x amount of 30% solution. We now have x + y of the new 40% solution. Do you understand it so far?

OpenStudy (dynamiz77):

i think so

OpenStudy (mathstudent55):

Good. Remember from above, x amount of a 30% salt solution has 0.3x pure salt in it. The resulting solution is in the amount of x + y. Since it is a 40% salt solution, the resulting solution contains 0.4(x + y) pure salt.

OpenStudy (mathstudent55):

Now we write an equation for the amount of salt. 0.3x + y = 0.4(x + y)

OpenStudy (dynamiz77):

"You add enough pure salt, we'll call the amount y, until you reach a concentration of 40% salt. That means, you add y amount of pure salt to x amount of 30% solution. We now have x + y of the new 40% solution."

OpenStudy (dynamiz77):

now its starting to make sense

OpenStudy (mathstudent55):

Now we go back to the statement of the problem, and we see that the original amount of 30% salt solution is 25 oz. We use our equation, and we substitute 25 for x. 0.3x + y = 0.4(x + y) 0.3(25) + y = 0.4(25 + y)

OpenStudy (mathstudent55):

Now we solve the equation for y, the amount of pure salt.

OpenStudy (dynamiz77):

ok - this is making SO much sense

OpenStudy (mathstudent55):

Multiply both sides by 10 to get rid of decimals. 3(25) + 10y = 4(25 + y) Multiply left side; distribute on right side. 75 + 10y = 100 + 4y Subtract 4y from both sides; subtract 75 from both sides. 6y = 25 \(y = \dfrac{25}{6} \) \(y = 4 \dfrac{1}{6} \)

OpenStudy (mathstudent55):

Great.

OpenStudy (dynamiz77):

thank you SO much for your help - i am embarrassed that i was SUPER stuck with this problem

OpenStudy (mathstudent55):

1. You are very welcome. 2. Don't be embarrassed. None of us knew anything until we were taught. 3. You are, in fact, quite smart. You had a question, and you asked it. To me, that is smart.

OpenStudy (dynamiz77):

I am deeply humbled and am grateful for your time and teaching and thanks for the encouragement :)

OpenStudy (mathstudent55):

You're welcome.

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