SOLUTION: A tank contains 50 gallons of a 40% solution of anti-freeze. How much solution needs to be drained out and replaced with pure anti-freeze to obtain a 50% solution?
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Question 165519: A tank contains 50 gallons of a 40% solution of anti-freeze. How much solution needs to be drained out and replaced with pure anti-freeze to obtain a 50% solution? Answer by Mathtut(3670) (Show Source):
You can put this solution on YOUR website! Let x=amount that needs to be drained off and replaced with pure antifreeze
Now we know that the amount of pure antifreeze left after x amount is drained off (0.4(50-x)) plus the amount of pure antifreeze added(x) has to equal the amount of pure antifreeze in the final mixture (0.5(50)). So, our equation to solve is:
0.4(50-x) + x=0.50*50 get rid of parenthesis(distributive)
20-.4x+x=25 subtract 20 from each side
and combine like terms
.6x=5 divide each side by 0.6
x=8.33 qts-----------------------------------ans
CK
.4(50-8.33)+8.33=25
(50/3)+25/3=25
75/3=25
25=25
Hope this helps
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