Question 251464
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.