Question 844963: acid 'a' is 30% acid. acid 'b' is 70% acid. How much of each should be mixed to make 40 ml of a 45% acid?.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! x = the number of ml of solution A.
y = the number of ml of solution B.
.30x = the number of ml of acid in solution A.
.70y = the number of ml of acid in solution B.
the equations you need to solve are:
x + y = 40
.30x + .70y = .45 * 40
the first equations tells you the total ml of each solution.
the second equation tells you the ml of acid in each solution.
simplify these equations to get:
x + y = 40
.30x + .70y = 18
solve for y in the first equation to get y = 40-x
substitute 40-x for y in the second equation to get:
.30x + .70(40-x) = 18
simplify to get:
.30x + 28 - .7x = 18
combine like terms to get:
-.4x + 28 = 18
subtract 28 from both sides of this equation to get:
-.4x = 18 - 28
simplify to get:
-.4x = -10
divide both sides of this equation by -.4 to get:
x = -10 / -.4 which result in:
x = 25
since x + y = 40, then:
y = 15 because 25 + 15 = 40
you have:
x = 25ml
y = 15 ml
the amount of acid in 25 ml of solution A is equal to .30 * 25 = 7.5 ml.
the amount of acid in 15 ml of solution B is equal to .70 * 15 = 10.5 ml.
total amount of acid in 40 ml of the mixture that contains 25 ml of solution A and 15 ml of solution B is 7.5 + 10.5 = 18 ml.
18 / 40 is equal to .45 which is equal to 45%, so the numbers are good.
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