SOLUTION: "Sam wants to mix a 5% acid solution with a 40% acid solution to make 10 oz. of a 25% acid solution. How much of the 40% acid solution will he need?" I am not sure how to set i

Question 615119: "Sam wants to mix a 5% acid solution with a 40% acid solution to make 10 oz. of a 25% acid solution. How much of the 40% acid solution will he need?"
I am not sure how to set it up in a concentration and amount chart but i pretty sure its not right because the answer is 5.47 (teacher told me the answer then said figure it out) But I don't know how to set it up. If someone can show me how to get to a equation like this it would be amazing! EX: 25000+100x=75(500 + x)I can solve it from there i just need to know how to get to that step
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
"Sam wants to mix a 5% acid solution with a 40% acid solution to make 10 oz. of a 25% acid solution. How much of the 40% acid solution will he need?"
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Do it like this:
if a 14% acid solution is mixed with a 19% acid solution to produce 45 gallons of a 15% acid solution how much (in gallons) of each solution should be used?
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f = 14%
n = 19%
----
f + n = 45 (total gallons)
14f + 19n = 15*45 (total acid)
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n = 45 - f
Sub for n
14f + 19*(45-f) = 675
14f + 855 - 19f = 675
-5f = -180
f = 36 gallons of 14%
n = 9 gallons of 19%
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