SOLUTION: Sarah wants to mix 400 liters of a solution that is 57.5% iodine. She has two solutions availiable. One is 20% and the other is 70% iodine. How many of each should Sarah use?
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-> SOLUTION: Sarah wants to mix 400 liters of a solution that is 57.5% iodine. She has two solutions availiable. One is 20% and the other is 70% iodine. How many of each should Sarah use?
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Question 146027: Sarah wants to mix 400 liters of a solution that is 57.5% iodine. She has two solutions availiable. One is 20% and the other is 70% iodine. How many of each should Sarah use? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=amount of 70% solution needed
Then 400-x=amount of 20% solution needed
Now we know that the amount of pure iodine in the 70% solution (0.70x) plus the amount of pure iodine in the 20% solution (0.20(400-x)) has to equal the amount of pure iodine in the final mixture (400*0.575). So our equation to solve:
0.70x+0.20(400-x)=400*0.575 get rid of parens (distributive law)
0.70x+80-0.20x=230 subtract 80 from each side
0.70x+80-80-0.20x=230-80 collect like terms
0.50x=150 divide each side by 0.5
x=300 liters-------------------------amount of 70% solution needed
400-x=400-300=100 liters----------------amount of 20% solution needed
CK
300*0.7+100*0.2=400*0.575
210+20=230
230=230
We can also work this problem using two unknowns:
Let x=amount of 70% solution needed
And let y=amount of 20% solution needed
Now we are told that:
x+y=400-------------------------------eq1
0.7x+0.2y=400*0.575---------------------------eq2
From eq1, substitute y=400-x into eq2 and we have the same equation as before
