SOLUTION: I need help setting up the following word problem. NOT LOOKING FOR THE ANSWER LOOKING FOR THE METHOD PLEASE. How much pure acid should be mixed with 6 gallons of a 50% acid s

Question 238584: I need help setting up the following word problem.
NOT LOOKING FOR THE ANSWER LOOKING FOR THE METHOD PLEASE.
How much pure acid should be mixed with 6 gallons of a 50% acid solution in order to get an 80% acid solution?
What I have is wrong I am sure:
.80=.50(x+6)+(x)
Any assistance would be greatly appreciated.
Found 2 solutions by Masaries7, ankor@dixie-net.com:
Answer by Masaries7(3) (Show Source): You can put this solution on YOUR website!
Your question consists of ratios.
.5(6)/6= 3/6 3 is the gallons of acid. 6 is the total gallons
To find out how much more gallons of acid you need to obtain 80% acid levels we add our unknown amount to bottom and top of the ratio indicating for the top more acid gallons and for the bottom more total gallons overall.
3+x/6+x= .8
Solve for 'x'
The format for this type of problem is to figure out your initial amounts then set up a ratio equaling the new desired percentage. You must be sure of how the information is applied in the problem for instance had we been adding non-acid solution then there would be no 'x' variable on the top since the ratio we have is acid gallons/ total gallons. Adding non-acid does not apply to acid. Good tip is to visualize the problem in your mind (if it's possible don't get brain freeze however)
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
How much pure acid should be mixed with 6 gallons of a 50% acid solution in order to get an 80% acid solution?
:
using the decimal equiv of %; pure acid would be 1.0x
:
.50(6) + 1.0x = .80(x+6)
3 + 1x = .8x + 4.8
1x - .8x = 4.8 - 3
.2x = 1.8
x =
x = 9
;
:
Check this in the amt of acid equation
.5(6) + 1(9) = .8(9 + 6)
3 + 9 = .8(15)
12 = 12

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