SOLUTION: The radiator in a car is filled with a solution of 80% antifreeze and 20% water. The manufacturer of the antifreeze suggests that, for summer driving, optimal cooling of the engine
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Question 154118: The radiator in a car is filled with a solution of 80% antifreeze and 20% water. The manufacturer of the antifreeze suggests that, for summer driving, optimal cooling of the engine is obtained with only 40% antifreeze. If the capacity of the radiator is 3.8 L, how much coolant should be drained and replaced with water to reduce the antifreeze concentration to the recommended level?
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Let x=amount of coolant that must be drained and replaced with water to yield a 40% solution.
Now we know that the amount of pure antifreeze in the radiator after we drain x amount out (0.80(3.8-x)) plus the amount of pure antifreeze in the water that we add (0) has to equal the amount of pure antifreeze in the final solution (0.40(3.8)). So our equation to solve is:
0.80(3.8-x)=0.40*3.8 get rid of parens
3.04-0.80x=1.52 subtract 3.04 from each side
3.04-3.04-0.80x=1.52-3.04 collect like terms
-0.80x=-1.52 divide each side by -0.80
x=1.9 liters----------------amount that must be drained and replaced with pure water
CK
0.80(3.8-1.9)=0.40*3.8
0.80*1.9=0.40*3.8
1.52=1.52
