if 50% of the 2:3 solution of milk and water is replaced with water.what concentration of the solution is reduced

• (check on 100 ltr)
  milk:water=2:3 =40:60
  take out 50 %(50 ltr means milk=20 ltr water=30 ltr) remainig
  milk:water=20:30
  add 50 ltr of water
  ratio
  milk:water=20:(30+50) =20:80
  1:4
  so initially milk= 40%
  after milk=20%
  so reduction in concentration= 20%
• 11 years agoHelpfull: Yes(35) No(1)
• suppose we have 100 ltrs of solution
  so milk is (2/5)*100 =40 ltrs
  and water is (3/5)*100=60 ltrs
  so concentration of solutin is 40/100=40%
  now 50 ltrs of solution is repaced by water
  we have 50 ltrs of orginal solution left in which
  milk is (40*50)/100=20 ltrs
  water is (60*50)/100=30 ltrs
  as we knw 50 ltrs of water is added...so in 100 ltrs of solutin we have only 20 ltrs of milk
  so concentration of new soution is 20/100=20%
  so reduced concentration is 40%-20%=20% ans
• 11 years agoHelpfull: Yes(9) No(0)
• not satisfied with ur answer khan so explain detail
  
• 11 years agoHelpfull: Yes(3) No(3)
• i dontknow
  
• 8 years agoHelpfull: Yes(0) No(1)
• lets assume 50% is 2 units of milk and 3 units of water. So,
  100% will be 4 units of milk and 6 units of water.
  Hence, total unit of solution is 10 units and 50% is replaced with water so,
  10 - 5 = 5 units of solution of water and milk.
  M W
  4 6 when full
  2 3 when half
  now its replaced by 5 units of water. Now, 2 units of milk and 3+5 units of water in the solution.
  hence 50% of milk is reduced. (From 4 units to 2 units)
• 6 years agoHelpfull: Yes(0) No(0)