SOLUTION: A 5 gallon radiator containing a mixture of water and antifreeze was supposed to contain a 50% antifreeze solution. When tested, it was found to have only 40% antifreeze. How much
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-> SOLUTION: A 5 gallon radiator containing a mixture of water and antifreeze was supposed to contain a 50% antifreeze solution. When tested, it was found to have only 40% antifreeze. How much
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Question 106238: A 5 gallon radiator containing a mixture of water and antifreeze was supposed to contain a 50% antifreeze solution. When tested, it was found to have only 40% antifreeze. How much must be drained out and replaced with pure antifreeze so that the radiator will contain the desired 50% antifreeze.
You can put this solution on YOUR website! Let x=amount that needs to be drained and replaced with pure antifreeze
Now we know that the amount of pure antifreeze left after we drain x amount out (0.40(5-x))plus the amount of pure antifreeze added (x) has to equal the amount of pure antifreeze in the final mixture (0.50*5). So our equation to solve is:
0.40(5-x)+x=0.50*5 get rid of parens and simplify
2-0.40x+x=2.5 subtract 2 from both sides
2-2-0.40x+x=2.5-2 collect like terms
0.60x=0.5 divide both sides by 0.60
x=0.833333-- gal----------amount that needs to be drained and replaced with pure antifreeze
CK
0.40(5-0.83333333333)+0.8333333333333=2.5
1.666666666666+0.8333333333333333333=2.5
~2.5=2.5
