Question 364944: you want to mix a 30% solution with a 80% solution to get 40 ounces of a 40% solution. how much of each solution do you need?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! you have a 30% solution and you have an 80% solution.
you want to get 40 ounces of a 40% solution.
x = 30% solution
y = 80% solution
x + y = 40
.3x + .8y = .4 * 40
first equation tells you the total ounces required for the solution.
second equation tells you the total ounces required for the percent of the solution you are looking for.
solve these 2 equations simultaneously and you have your answer.
your 2 equations are:
x + y = 40
.3x + .8y = .4*40 = 16
multiply both sides of the first equation by .3 to get:
.3x + .3y = 12
.3x + .8y = 16
subtract the first equation from the second equation to get:
.5y = 4
divide both sides of the resulting equation by .5 to getg:
y = 8
since y = 8, then x must equal 32 because x + y = 40
substitute 32 for x and 8 for y in the second equation to get:
.3x + .8y = 16 becomes:
.3*32 + .8*8 = 16 which becomes:
9.6 + 6.4 = 16 which becomes:
16 = 16 which is true, confirming that the answer of x = 32 and y = 8 is good.
In order to get a 40 ounces of a 40% solution, you require 32 ounces of a 30% solution and 8 ounces of an 80% solution.
16 / 40 = 40% so you're good.
question was how much of each solution do you need?
answer is 32 ounces of 30% solution and 8 ounces of 80% solution.
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