Question 704628
-------------- 	percent	----------------	Amount									
Antifreeze solution	10	----------------			4	kg						
Antifreeze	100	----------------			x	kg						
Mixture	20	----------------	4	+	x	kg						
												
10*4+100x=20(4+	x)
40+100	x=80+20	x				
100x-20	x=-40+	80				
80	x	=	40									
/	80											
x	=	0.5		kg	Antifreeze	


100 grams solution contains 60 gms water
600=50(10+x)				
										600=500+50x
100=50x
x=2 gms of water has to be removed		
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