SOLUTION: 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.
I really don't
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I really don't
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Question 163917: 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.
I really don't know to do this. This is what I think.
100%=x 50 gallons=x 40% solution=X 50% solution=x
Let x= amount of pure antifreeze
x+0.4 (100)= .05(100+x)
x+4=50+0.5
x+4=.50(100)+0.4
I'm lost after this.
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 x = amt of 40% solution removed, and amt of pure antifreeze added:
:
.4(50 - x) + x = .5(50)
:
Solve this and you should have it.
