SOLUTION: The radiator in a car is filled with a solution of 75 percent antifreeze and 25 percent water. The manufacturer of the antifreeze suggests that for summer driving, optimal cooling

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Question 682458: The radiator in a car is filled with a solution of 75 percent antifreeze and 25 percent water. The manufacturer of the antifreeze suggests that for summer driving, optimal cooling of the engine is obtained with only 50 percent antifreeze. If the capacity of the radiator is 3.3 liters, how much coolant (in liters) must be drained and replaced with pure water to reduce the antifreeze concentration to 50 percent?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The capacity of the radiator is 3.3 liters.
The radiator is filled with a solution of 75% antifreeze
which is +.75%2A3.3+=+2.475+ liters of antifreeze
Let +x+ = liters,of coolant to be drained and
replaced with water
+.75x+ liters of coolant is in the liters of the
+x+ liters drained
It started with +3.3+ liters and ended with +3.3+ liters
--------------------
+%28+2.475+-+.75x+%29+%2F+3.3+=+.5+
+2.475+-+.75x+=+.5%2A3.3+
+2.475+-+.75x+=+1.65+
+.75x+=+2.475+-+1.65+
+.75x+=+.825+
+x+=+1.1+
1.1 liters of the 75% solution must be drained and
replaced with pure water
check:
+3.3+-+1.1+=+2.2+ liters left after draining
+.75%2A2.2+=+1.65+ liters of antifreeze left
Now add +1.1+ liters of water
The concentration is
+1.65+%2F+3.3+=+.5+ or 50%
OK