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:
Any assistance would be greatly appreciated. Found 2 solutions by Masaries7, firstname.lastname@example.org: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.
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)
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 = 9
Check this in the amt of acid equation
.5(6) + 1(9) = .8(9 + 6)
3 + 9 = .8(15)
12 = 12