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

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Question 507045: a chemist has one solution that is 25% acid and a second that is 50% acid. how many liters of each should be mixed to get 10 L of a solution that is 40% acid?
Answer by Earlsdon(6294) (Show Source):
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Let x = the required number of liters of the 25% acid solution, then (10-x) is the required number of liters of the 50% acid solution. These are mixed to obtain 10 liters (x+(10-x)) of 40% acid solution. Let's write the equation to solve for x after changing the percentages to their decimal equivalents.
0.25x+0.5(10-x) = 0.4(10) Simplify.
0.25x+5-0.5x = 4 Combine the x-terms.
-0.25x+5 = 4 Subtract 5 from both sides.
-0.25x = -1 Finally, divide both sides by -0.25
x = 4 and 10-x = 6, so...
The chemist will need to mix 4 liters of the 25% acid solution with 6 liters of the 50% acid solution to obtain 10 liters of 40% acid solution.