SOLUTION: How many liters of a 10% alcohol solution must be mixed with 70 liters of a 90% solution to get a 30% solution? Thanks in advance for your help!

Question 155666: How many liters of a 10% alcohol solution must be mixed with 70 liters of a 90% solution to get a 30% solution?
Thanks in advance for your help!
Found 2 solutions by jojo14344, Earlsdon:
Answer by jojo14344(1513) (Show Source): You can put this solution on YOUR website!
There are 2 unknowns in the problem: VOLUME OF ALCOHOL ---> ? VOLUME OF OUTPUT SOLUTION---> ?
We'll go to the 1st condition: 10% of Alcohol mixed with 90% of 70L of initial solution to produce 30% of . To show:

----------------------> eqn 1
Now, when we add the Volume of and , we get the volume of . To show:
--------------------------------> eqn 2
Continuing, ---------------------> eqn 3
Substitute eqn 3 in eqn 1:

----->
-------------------> volume of Alcohol
: For Volume of , go back eqn 2:

-------------------> volume of output solution
In doubt? go back eqn 1:

thank you
Jojo

Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Let x = the number of liters of 10% alcohol solution to be added to 70 liters of 90% alcohol solution to get (x+70) liters of 30% alcohol solution.
After changing the percentages to their decimal equivalents, we can express this as:
(0.1)x + (0.9)(70) = (x+70)(0.3) Simplify this and solve for x.
0.1x+63 = 0.3x+21 Subtract 0.1x from both sides.
63 = 0.2x+21 Subtract 21 from both sides.
42 = 0.2x Finally, divide both sides by 0.2
x = 210 liters.
You would need to mix 210 liters of a 10% alcohol solution to 70 liters of 90% alcohol solution to obtain 280 (210+70) liters of 30% alcohol solution.
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