SOLUTION: a radiator contains 8 quarts of fluid, 40% of which is antifreeze. how much fluid should be drained and replaced with pure antifreeze so the new mixture is 60% antifreeze
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Question 345966: a radiator contains 8 quarts of fluid, 40% of which is antifreeze. how much fluid should be drained and replaced with pure antifreeze so the new mixture is 60% antifreeze Found 2 solutions by stanbon, haileytucki:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! a radiator contains 8 quarts of fluid, 40% of which is antifreeze. how much fluid should be drained and replaced with pure antifreeze so the new mixture is 60% antifreeze
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Equation:
active - active + active = active
0.40*8 - 0.40x + 1.00x = 0.60(8)
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Multiply thru by 100 to get:
40*8 - 40x + 100x = 60*8
60x = 20*8
x = (1/3)8
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x = 2 2/3 quarts (amt. of fluid to replace)
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Cheers,
Stan H.
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You can put this solution on YOUR website! We have to figure out what mixture of pure antifreeze must be mixed with 40% antifreeze to give a 60% soln
let x be the amount of 40% (.4) soln and 6-x be the amount of 100% (1) antifreeze.
In 6 quarts of 60% antifreeze is 3.6 quarts of pure antifreeze.
0.4x+6-x=3.6
Since 0.4x and -x are like terms, add -x to 0.4x to get -0.6x.
-0.6x+6=3.6
Since 6 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 6 from both sides.
-0.6x=-6+3.6
Add 3.6 to -6 to get -2.4.
-0.6x=-2.4
Divide each term in the equation by -0.6.
-(0.6x)/(-0.6)=-(2.4)/(-0.6)
Cancel the common factor of -0.6.
x=-(2.4)/(-0.6)
Simplify the right-hand side of the equation by simplifying each term.
x=4
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