SOLUTION: a chemist has one solution that is 25% acid and 50% acid. how many liters of each should be mixed to get 10 Liters of a solution that has 40% acid?

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Question 209549: a chemist has one solution that is 25% acid and 50% acid. how many liters of each should be mixed to get 10 Liters of a solution that has 40% acid?
Found 2 solutions by checkley77, Earlsdon:
Answer by checkley77(12844) (Show Source):
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.5x+.25(10-x)=.4*10
.5x+2.5-.25x=4
.25x=4-2.5
.25x=1.5
x=1.5/.25
x=6 liters of the 40% solution is uses.
10-6=4 liters of the 25% solution is used.
Proof:
.5*6+.25*4=4
3+1=4
4=4

Answer by Earlsdon(6294) (Show Source):
You can put this solution on YOUR website!
Let x = the number of liters of 25% acid solution required and this is to be mixed with (10-x) liters of 50% acid solution to obtain 10 liters of 40% acid solution.
The key here is to set up an equation showing the quantities of acid.
The quantity of acid in the x liters of the 25% acid solution can be expressed by: 25%(x) and the quantity of acid in the (10-x) liters of the 50% acid solution can be expressed by: 50%(10-x). When mixed together, these will produce 10 liters of 40% acid solution which is 40%(10) acid. So, after converting the percentages to their decimal quivalents, we can write the equation:
0.25(x)+0.5(10-x) = 0.4(10) Simplify and solve for x.
0.25x+5-0.5x = 4 Combine the x-terms.
-0.25x+5 = 4 Subtract 5 from both sides.
-0.25x = -1 Divide both sides by -0.25
x = 4
The chemist willl need to mix 4 liters of 25% acid solution with 6 liters (from 10-x) of 50% acid solution to obtain 10 liters of 40% acid solution.