PERCENTAGES QUESTION :)
Two beakers of equal volume
Beaker A contains 30% salt
Beaker B contains 10% salt
How many ounces of solution A should be added to solution B, so that the resulting solution is 20% salt?

- anonymous

- jamiebookeater

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- anonymous

so far, I let x be the amount of solution in Beaker A.
then, 30% of that is the amount of salt in A....same for B...

- mathstudent55

Let's try 10 liters.
10 liters of A has 3 litres salt.
10 liters of B has 1 litres salt.
You end up with 20 liters with 4 litres salt.
4 litres out of 20 litres is 20%

- anonymous

Okay, yep, I;m following so far....

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## More answers

- mathstudent55

Since the concentration you want is exactly the average of the two concentrations you start with, the same amounts of A and B are used.

- anonymous

Oki doki, that makes sense - there might be a problem with the worked solution – its talking about ounces, and I have no information about how many ounced the beakers started out with....

- mathstudent55

Use ounces where I used litres.

- anonymous

heres the part that dosnt make any sense to me in the worked solution: "the resulting solution will be 50 + x ounces and its concentration of salt will be 20%(50+x)....

- anonymous

im guessing instead of your 10L base, theyve used 50 ounces?

- mathstudent55

Is there any info missing from the question you posted?
Where does the number 50 come from in the solution?

- anonymous

thats exactly my problem...

- mathstudent55

Beaker A has x ounces of 30% salt solution.
Beaker B has x ounces of 10% salt solution.
(Both beakers have x amount because we are the beakers have equal volume.)

- anonymous

Yes. Corecct

- mathstudent55

x ounces of solution B has 0.1x ounces of salt.

- anonymous

Yep

- mathstudent55

Now we need to start adding solution A to solution B until we have a new solution that is 20% salt.

- mathstudent55

Let's say we add y ounces of solution A to solution B.

- anonymous

To do that, we need to find the volume in A containing a concentration of 10%?

- anonymous

Okay, yep

- mathstudent55

y ounces of solution A has 0.3y ounces of salt

- anonymous

correct

- mathstudent55

We will have a total of salt:
0.3y + 0.1x
Ok?

- anonymous

Yep, all good so far. Thanks for going through it with me.

- mathstudent55

Next, we will have a total amount of solution of:
x + y
Ok?

- anonymous

Yes.

- mathstudent55

We want the new solution to have a concentration of 20% salt.
20% = 0.2
That means
\(\dfrac{0.3y + 0.1x}{x + y } = 0.2\)

- anonymous

Okay, cool.

- mathstudent55

Now let's solve that equation and what we get.

- anonymous

x=y?

- mathstudent55

\(\dfrac{0.3y + 0.1x}{x + y } = 0.2\)
\(0.3y + 0.1x = 0.2(x + y)\)
\(0.3y + 0.1x = 0.2x + 0.2y\)
\(0.1y = 0.1x\)
\(y = x\)
Correct.

- mathstudent55

Since we chose x to be the amount of solution B and y to be the amount of solution A, we see the amounts must be the same no matter how much each beaker has.

- anonymous

Understood

- mathstudent55

Great.

- anonymous

So do you think that they made x=y=50 ounces?

- anonymous

in the solutions?

- mathstudent55

That's what I am thinking.

- anonymous

Yeah, okay. Well thanks for going through it with me in detail. Appreciate your work :)

- mathstudent55

Before the part of the solution you quoted above mentioning 50 + x, was there any mention of 50?

- mathstudent55

You're welcome.

- anonymous

their equation is exactly what you had, except they made 50=y for some reason:
10%50 + 30%x = 20%(50+x)
I understand the equation now, thank you.
And no. No mention anywhere of 50 :(

- anonymous

Also, just wondering if you know how to change the username on open study? i can find the change passwords under settings, but is it even possible to change usernames?

- mathstudent55

I guess they just assumed they would use 50 ounces of one solution, and then they found how many ounces of the other solution they needed. It turns out to be 50, of course, bec now we know it's the same amount of both solutions.

- mathstudent55

I don't know if you can change username in OS.

- anonymous

Yes, I reckon you're right on that one.
Okay then, thank you!

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