Q. From a container having pure milk ,20% is replaced by water and the process is repeated thrice. At the end of the third operation , the milk is
Option
a) 40% pure
b) 50% pure
c) 51.2% pure
d) none of these

• Ans is C.As in 1st operation- milk is 80% and water is 20%.
  2nd operation- milk is 64%(80-80*0.2) and water is 36%(20-20*0.2+20).
  3rd operation- milk is 51.2% and water is 48.8%.
• 10 years agoHelpfull: Yes(28) No(4)
• Let total quantity of original milk = 1000 gm
  milk after first operation = 80% of 1000 = 800 gm
  milk after second operation = 80% of 800 = 640 gm
  milk after third operation = 80% of 640 = 512 gm
  strength of final mixture = 51.2 %
• 10 years agoHelpfull: Yes(19) No(1)
• y=x(1-fraction atken out)^n
  here y= remaining liquid
  x=initial liquid
  if initial liquid(x) is 100
  fraction will be 20/100=1/5
  y=100(1-1/5)^3
  y=51.2%
• 10 years agoHelpfull: Yes(6) No(1)
• Milk : Water
  
  step 0:-) 100 : 0
  step 1:-) 80 : 20
  step 2:-) 64 : 32
  step 3:-) 51.2 : 48.8
  
• 10 years agoHelpfull: Yes(4) No(2)
• c: 51.2%
  solve by using successive discount method
  
  
• 10 years agoHelpfull: Yes(3) No(0)
• right answer is = 40.96 %.
  method 1) :q o pure milk left = 100(1-20/100)^4 = 100(0.8)^4 = 40.96 %.
  method 2) : q of milk left after each iteration=80% then 64 % then 51.2 % and then finally 40.96 %.................................
  
  
• 10 years agoHelpfull: Yes(2) No(3)
• It will result into 4 operations...as in question it is given that , after one operation it is repeated thrice, so final answer =40% pure.
• 10 years agoHelpfull: Yes(1) No(9)
• d none of these
• 9 years agoHelpfull: Yes(1) No(1)
• See every time 0.8 is milk left
  =>0.8^2=0.512
  
• 9 years agoHelpfull: Yes(1) No(0)
• in first operation we will have 80% pure milk,
  in second operation 80*20% =>80*0.2=> 16 now milk percentage is 80-16=64%,
  in third operation 64*20%= > 64*0.2 => 12.8 now milk percentage is 64-12.8=51.2%
• 9 years agoHelpfull: Yes(1) No(0)
• ans : 51.2 %
• 6 years agoHelpfull: Yes(1) No(0)
• its 40.96% pure
• 9 years agoHelpfull: Yes(0) No(1)
• it has a simple formula REMAINING MIXTUTRE=TOTAL QUANTITY*[(TOTAL-REPLACE QUANTITY)/TOTAL]^n
  where n is no.of operation
  so we will get 51.2%
• 9 years agoHelpfull: Yes(0) No(0)
• 51.2%
  let assume there is 100 lit pure milk in a container when 1st time 20% milk is replaced by water than there is 80 lit milk and 20 lit. water in container then in 2nd time there 80(100-20) % of 80 will be pure milk left in a container i.e 64 lit. and in 3rd time again 80% of 64 pure milk left in container i.e 51.2%.
• 8 years agoHelpfull: Yes(0) No(0)