SOLUTION: A chemist has a 30% acid solution and a 60% solution already prepared. How much of each of the two solutions should be mixed to form 300 mL of a 50% solution? Thank you in adv

Question 472969: A chemist has a 30% acid solution and a 60% solution already prepared. How
much of each of the two solutions should be mixed to form 300 mL of a 50%
solution?
Thank you in advance!
Found 2 solutions by stanbon, jim_thompson5910:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A chemist has a 30% acid solution and a 60% solution already prepared. How
much of each of the two solutions should be mixed to form 300 mL of a 50%
solution?
----------------------
Using 1-variable:
Equation:
acid + acid = acid
0.30x + 0.60(300-x) = 0.50*300
---------------
Multiply thru by 100 to get:
30x + 60*300 - 60x = 50*300
-30x = -10*300
x = 100mL (amt. of 30% solution needed)
----
300-x = 200mL (amt. of 60% solution needed)
===========================
Cheers,
Stan H.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Let x = amount of 30% solution, y = amount of 60% solution (both amounts in mL)

Since the chemist wants 300mL, this means that . Solve for y to get

Because "A chemist has a 30% acid solution and a 60% solution already prepared" and the chemist wants "to form 300 mL of a 50% solution", we know that . Next, multiply everything by 10 to get every number to be a whole number. So

Now plug in and solve for x

Distribute.

Combine like terms on the left side.

Subtract from both sides.

Combine like terms on the right side.

Divide both sides by to isolate .

Reduce.

Now go back to and use that to find the value of y.

So and

This means that 100 mL of 30% acid solution and 200 mL of 60% acid solution is needed to mix to get 300 mL of 50% acid solution.
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