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A tank contains 50 gallons of a 40% solution of antifreeze. How much solution needs to be drained out and replaced with pure antifreeze to
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A tank contains 50 gallons of a 40% solution of antifreeze. How much solution needs to be drained out and replaced with pure antifreeze to
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Question 57579This question is from textbook Elementary and Intermediate Algebra
: To Whomever can assist... Thank you!
A tank contains 50 gallons of a 40% solution of antifreeze. How much solution needs to be drained out and replaced with pure antifreeze to obtain a 50% solution? This question is from textbook Elementary and Intermediate Algebra
You can put this solution on YOUR website! A tank contains 50 gallons of a 40% solution of antifreeze. How much solution needs to be drained out and replaced with pure antifreeze to obtain a 50% solution?
:
Let the unknown amount of solution be: x
You start out with: .40(50)
you take out: .40x
You put in: 1.00x
Then you get: .50(50)
The equation to solve is:
.40(50)-.40x+1.00x=.50(50)
20-.40x+1.00x=25
20+.60x=25
20-20+.60x=25-20
.60x=5
.60x/.60=5/.60
x=8.33 gal
Happy Calculating!!!
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