1. The problem statement, all variables and given/known data A full bowl cointains water, its mass is 395g.An iron screw is placed on the bowl and the mass is now 565g. This seemed simple, but I asked my teacher to clarify and he said that upon placing the screw on the water an equal volume of water is removed so the volume remains the same. Find the screw's volume. 2. Relevant equations Iron's density = 7,86g/cm^3. water's density = 1g/cm^3 screw volume + water volume = 395cm^3 3. The attempt at a solution water volume = 395g/1g/cm^3 Water volume =395cm^3 565g-395g =170g 7,86g/cm^3 - 1g/cm^3 = 6,86g/cm^3 (it's pretty hard for me to explain why I did this but basically if I don't minus the water's density the volume, the answer won't compute the water that was removed) 170g/6,86g/cm^3 24,71cm^3 Teacher said it was wrong because this made no sense/was wrong and that I was too complicated, the part was wrong for him was this one: 7,86/cm^3 - 1g/cm^3 = 6,86/cm^3 so the solution was: 170g/7,86cm/cm^3 21,something cm^3 I tried to proove that I was right and I did this x= water's volume y= screw's volume dx= water's density yx= screw's density x+y=395 x.dx+y.dy=565 dx=1 dy=7,86 x+7,86y =565 x+7,86y-6,86y = 565 -6,86y x+y = 565 - 6,86y 395 = 565 - 6,86y 395-395 = 565 -395 - 6,86y 0 = 170 - 6,8y 0+6,86y = 170 -6,86y +6,86y 6,86y = 170 6,86y/6,86 = 170/6,86 y=24,78134 Where did I go wrong.