SOLUTION: How many liters of a 20% solution of acid should be added to 10 liters of a 30% solution of acid to obtain a 25% solution?

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Question 1089288: How many liters of a 20% solution of acid should be added to 10 liters of a 30% solution of acid to obtain a 25% solution?
Found 2 solutions by addingup, Edwin McCravy:
Answer by addingup(3677) About Me  (Show Source):
You can put this solution on YOUR website!
.2x + .3(10) = .25(x + 10)
.2x + 3 = .25x + 2.5
-.05x = -.5
x = 10 liters of 20% solution

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
How many liters of a 20% solution of acid 

That's 20% of x liters of solution which will include 0.20x liters of pure acid. 


should be added to 10 liters of a 30% solution of acid 

That's 30% of 10 liters of solution which will include 0.30(10) or 3 liters 
of pure acid. 

to obtain a 25% solution.

That's 25% of x+10 liters of solution which will include 0.25(x+10) liters of pure acid. 

Therefore the equation is:

0.20x + 0.30(10) = 0.25(x+10)

The first step is to multiply every term by 100 to
remove the decimals:

20x + 30(10) = 25(x+10)

You can finish solving that.

Edwin