SOLUTION: Adding 25 gallons of acid to a mixture, makes it 80% Acid solution, while adding 25 gallons of water makes it 60% Acid solution. What is the original acid percentage of the mixture

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Question 993323: Adding 25 gallons of acid to a mixture, makes it 80% Acid solution, while adding 25 gallons of water makes it 60% Acid solution. What is the original acid percentage of the mixture?
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
You are not given the amount of original acid mixture, so give this a variable for its volume.
v, the initial volume in gallons of the starting mixture of unknown percent concentration.

Let p be the percent concentration of your original acid mixture.
Additionally, you are forced to assume that your acid to add in pure form, the 100%, is a liquid.

ADDING 25 GALLONS
%2825%2A100%2Bv%2Ap%29%2F%28v%2B25%29=80

ADDING 25 GALLONS OF WATER
%2825%2A0%2Bv%2Ap%29%2F%28v%2B25%29=60

The system of equations slightly reworked is
system%28vp%2B2500=80%28v%2B25%29%2Cvp=60%28v%2B25%29%29

STEPS SOLVING THE SYSTEM
vp%2B2500=80v%2B8%2A25%2A10
vp%2B2500=80v%2B2000
vp%2B500=80v
vp-80v=-500
80v-vp=500
v%2880-p%29=500
highlight_green%28v=500%2F%2880-p%29%29
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Next use the simpler, dilution equation.
%28500%2F%2880-p%29%29p=60%28%28500%2F%2880-p%29%29%2B25%29
500p%2F%2880-p%29=30000%2F%2880-p%29%2B1500
5p%2F%2880-p%29=300%2F%2880-p%29%2B15
p%2F%2880-p%29=60%2F%2880-p%29%2B3
p=60%2B3%2880-p%29
p=60%2B240-3p
4p=60%2B240
4p=300
p=300%2F4
highlight%28highlight%28p=75%29%29