A container contains pure milk.if 20% is replaced by
water and then the same processs is repeated thrice
then what will b the % of milk remaining?

• 40.96%
  
  The replacement is performed four times instead three as mentioned in the above solution.
• 12 years agoHelpfull: Yes(20) No(6)
• 51.2 %
  
  If 100 litres of milk is initially there, then
  80 ltrs of milk is left after 1st replacement.
  80*0.8= 64 ltrs of milk is left after 2nd replacement.
  64*0.8 = 51.2 ltrs of milk is left afyer 3rd replacement.
  
  so 51.2% of milk is left in final mixture of 100 ltrs.
• 13 years agoHelpfull: Yes(19) No(10)
• There is a formula to solve such problems
  [(a-b)/a]^n
  where a=initial quqntity of liquid
  b=% of liquid replacd every time
  n=how many times this action is happening
  So,here a=100, b=20, n=3
  100-20=80
  80/100=0.8
  0.8^3=0.512
  0.512*100=51.2%
• 12 years agoHelpfull: Yes(19) No(2)
• Let there be 'X' litres of pure milk initially .
  
  Since , 20 % of it is replaced by water thrice . i.e. 20 % reduction in milk content each time .
  
  0.8 * 0.8 * 0.8 * X = 0.512X is the milk content present
  
  So , its 51.2 % of milk .
  
• 13 years agoHelpfull: Yes(5) No(4)
• at first milk =100%
  after replacing by water milk= 100-20=80%
  that means replacing of 20% water is present of 80% of milk
  now again it will replaced by 20% of water then milk=80*80=64%
  after that again it is replaced so then milk is =64*80=51.20%
• 13 years agoHelpfull: Yes(4) No(5)