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

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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) About Me  (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) About Me  (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 x%2By=300. Solve for y to get y=300-x


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 0.3x%2B0.6y=0.5%28300%29. Next, multiply everything by 10 to get every number to be a whole number. So


10%280.3x%29%2B10%280.6y%29=10%2A0.5%28300%29


3x%2B6y=1500


Now plug in y=300-x and solve for x

3x%2B6y=1500


3x%2B6%28300-x%29=1500


3x%2B1800-6x=1500 Distribute.


-3x%2B1800=1500 Combine like terms on the left side.


-3x=1500-1800 Subtract 1800 from both sides.


-3x=-300 Combine like terms on the right side.


x=%28-300%29%2F%28-3%29 Divide both sides by -3 to isolate x.


x=100 Reduce.


Now go back to y=300-x and use that to find the value of y.


y=300-x


y=300-100


y=200


So x=100 and y=200


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.