A substance is 99% water. Some water eveporates, leaving a substance that is 98% water. How much of the water evaporated?

• 50/99 of the total water evaporated.
  
  Let evaporated water be x.
  (99-x)/(100-x) = 98/100
  Or, (100-x)/(99-x) = 100/98
  Applying dividendo,
  1/(99-x) = 2/98 = 1/49
  Or, 99-x = 49
  Or, x = 50.
  Since there were 99 units of water available, and 50 units of water evaporated, 50/99 of the total water evaporated.
• 13 years agoHelpfull: Yes(3) No(2)
• Let's say we start with W units of water, and S units of other stuff.
  We originally have 99% water, so
  
  W 99
  --- = ---
  W+S 100
  
  Now we want to reduce the water to some fraction, F, of the original
  amount. And we want to end up with 98% water:
  
  FW 98
  ---- = ---
  FW+S 100
  
  We can solve each of these equations for S:
  
  W 99
  --- = ---
  W+S 100
  
  
  100W = 99(W+S)
  
  100W = 99W + 99S
  
  W = 99S
  
  W/99 = S
  
  and
  
  FW 98
  ---- = ---
  FW+S 100
  
  100FW = 98(FW+S)
  
  100FW = 98FW + 98S
  
  2FW = 98S
  
  2FW/98 = S
  
  Two things equal to the same thing are equal to each other, so
  
  W/99 = 2FW/98
  
  1/99 = 2F/98
  
  98/(99*2) = F
  
  0.495 = F
  
  So 49.5% of the water remains, which means that 50.5% evaporated
• 12 years agoHelpfull: Yes(2) No(1)