SOLUTION: The solution in a 7 liter radiator system is 35% antifreeze. How much of the solution must be drained and replaced by a 95% antifreeze solution to obtain a 50% antifreeze solution?
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Question 257585: The solution in a 7 liter radiator system is 35% antifreeze. How much of the solution must be drained and replaced by a 95% antifreeze solution to obtain a 50% antifreeze solution? Found 2 solutions by dabanfield, stanbon:Answer by dabanfield(803) (Show Source):
You can put this solution on YOUR website! The solution in a 7 liter radiator system is 35% antifreeze. How much of the solution must be drained and replaced by a 95% antifreeze solution to obtain a 50% antifreeze solution?
Let x be the amount to be replaced. Then we have:
.95*x + .35*(7-x) = .50*7 = 35
Solve for x.
You can put this solution on YOUR website! The solution in a 7 liter radiator system is 35% antifreeze. How much of the solution must be drained and replaced by a 95% antifreeze solution to obtain a 50% antifreeze solution?
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Equation:
anti - anti + anti = anti
0.35*7 - 0.35x + 0.95x = 0.50*7
2.45 +0.60x = 3.5
0.60x = 1.05
x = 1.75 liters (amount drained and replaced)
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Cheers,
Stan H.
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