How much pure acid must be added to a 30% acid solution to make a 60L solution which is 50% acid?
This was the answer I got from Reiny.
amount of pure acid to be added -- x L
1(x) + .3(60-x) = .5(60)
10x + 3(60-x) = 5(60)
10x + 180 - 3x = 300
7x = 120
x = 120/7 or appr 17.14 L
I was just wondering how did you come up 10x on the second part? Did you multiply everything by 10? Why so? Why not 100?
Oh I think I got it now. Never mind. Thanks much for the help.
To solve the equation, we need to convert the percentages into decimal form. The equation given is:
1(x) + 0.3(60-x) = 0.5(60)
First, let's simplify the equation by multiplying the decimal coefficients:
x + 18 - 0.3x = 30
Now, let's combine like terms:
0.7x + 18 = 30
To get rid of the decimal, multiply through by 10:
7x + 180 = 300
Now, subtract 180 from both sides to isolate the variable:
7x = 120
Finally, divide both sides by 7 to solve for x:
x = 120/7 or approximately 17.14 L
So, the amount of pure acid to be added is approximately 17.14 liters.