SOLUTION: The radiator in a car is filled with a solution of 65% antifreeze and 35% water. The manufacturer of the antifreeze suggests that for summer driving, optimal cooling of the engine

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Question 906375: The radiator in a car is filled with a solution of 65% antifreeze and 35% water. The manufacturer of the antifreeze suggests that for summer driving, optimal cooling of the engine is obtained with only 50% antifreeze. If the capacity of the radiator is 4.3 liters, how much coolant (in liters) must be drained and replaced with pure water to reduce the antifreeze concentration to 50%?
please please help me!
Found 2 solutions by richwmiller, stanbon:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
0.65*4.3-0.65x+0x=0.5(4.3)
2.795+-0.65x=2.15
-0.65x=2.15-2.795
-0.65x=-0.645
x=0.99230769 liters means almost a liter
check
0.65*3.30769231+0*0.99230769=0.5(4.3)
2.15+0.0=2.15
2.15=2.15
ok

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
The radiator in a car is filled with a solution of 65% antifreeze and 35% water. The manufacturer of the antifreeze suggests that for summer driving, optimal cooling of the engine is obtained with only 50% antifreeze. If the capacity of the radiator is 4.3 liters, how much coolant (in liters) must be drained and replaced with pure water to reduce the antifreeze concentration to 50%?
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anti - anti + anti = anti
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0.65*4.3 - 0.65x + 0.00x = 0.50*4.3
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65*4.3 - 65x + 0 = 50*4.3
65x = 15*4.3
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x = 0.23*4.3
x = 0.9923 liters (amt of liquid to drain and replace with pure water)
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Cheers,
Stan H.
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