SOLUTION: Please help me solve this problem: The radiator in a car is filled with a solution of 70 per cent antifreeze and 30 per cent water. The manufacturer of the antifreeze suggests t

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Question 913981: Please help me solve this problem:
The radiator in a car is filled with a solution of 70 per cent antifreeze and 30 per cent water. The manufacturer of the antifreeze suggests that for summer driving, optimal cooling of the engine is obtained with only 50 per cent antifreeze. If the capacity of the raditor is 3.6 liters, how much coolant (in liters) must be drained and replaced with pure water to reduce the antifreeze concentration to 50 per cent?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +x+ = liters of coolant to be drained
+.7x+ = liters of antifreeze in +x+ liters
+.7%2A3.6+=+2.52+ = liters of antifreeze
originally in radiator
---------------
Note that +x+ liters drained off are
replaced with +x+ liters of water, so
radiator ends up with +3.6+ liters
---------------
+%28+2.52+-+.7x+%29+%2F+3.6+=+.5+
+2.52+-+.7x+=+.5%2A3.6+
+2.52+-+.7x+=+1.8+
+.7x+=+2.52+-+1.8+
+.7x+=+.72+
+x+=+1.029+
1.029 liters of coolant must be drained
off and replaces with water
--------------------------
check:
+.7%2A1.029+=+.72+ liters of coolant drained off
+.3%2A1.029+=+.3087+ liters of water drained off
--------------------
+.3%2A3.6+=+1.08+liters+of+water+to+start+with%0D%0A%7B%7B%7B+1.08+-+.3087+=+.7713+ liters of water remaining
+.7713+%2B+1.029+=+1.8+ liters of water after
re-filling with pure water
+3.6+-+1.8+=+1.8+ liters of antifreeze left
+1.8%2F3.6+=+.5+
OK