Question 251464: Mixture A is 7.5% acid. To raise the concentration of acid to 10%, some pure acid will be added to the mixture. How many liters of mixture A and how many liters of pure acid are needed to end up with 200 liters of 10% solution.
Not sure if I started it out right, that's why I'm here!
x = pure acid
A = Mixture A
7.5 + x = (200)(.10) x = 12.5 liters
If that's right, I'm not sure where to go from there. If you can help me, that would be great! Thank you.
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! mixture A is 7.5% acid.
let x = amount of mixture A.
7.5% of x is equivalent to .075 * x (you have to divide percent by 100% to get proportion).
this means that .075 * x = amount of acid in mixture A.
let y = amount of mixture C.
then 1.0 * y = amount of acid in mixture B.
this is because 100% of mixture B is equivalent to 1.0 * mixture B.
mixture C is equal to 200 liters.
this means that:
x + y = 200
mixture C will contain 10% acid which means that mixture C will contain .10 * 200 = 20 liters of acid.
this means that:
.075 * x + y = 20
you have 2 equations that need to be solved simultaneously.
they are:
x + y = 200 (first equation)
.075*x + y = 20 (second equation)
you can solve for y in either equation and then solve for x in the other equation.
we'll solve for y in the first equation to get:
y = 200-x
we'll substitute for y in the second equation to get:
.075*x + (200-x) = 20
remove parentheses to get:
.075*x + 200 - x = 20
combine like terms to get:
-.925*x + 200 = 20
subtract 200 from both sides to get:
-.925*x = 20-100 = -180
divide both sides by -.925 to get:
x = -180 / -.925 = 194.5945946
use this value of x to solve for y in the first equation to get:
y = 200 - 194.5945946 = 5.405405405
use the values for x and y in the second equation to confirm that they are good.
the second equation is:
.075*x + y = 20
substituting for x and y in that equation, we get:
.075*194.5945946 + 5.405405405 = 20
this becomes 20 = 20 which is true confirming our values for x and y are good.
your answer is:
you need to add 5.405405405 liters of pure acid to make a 200 liter mixture of 10% acid.
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