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

• suppose 100 amount of milk 20% replaced we r left with 80% then process repeated twice when replace 20% we left with 64 and again 20% replaced finally left with 51.2%
  80-20%of80=64
  64-20%of64=51.2%
• 12 years agoHelpfull: Yes(23) No(6)
• 64 % milk is left after repeating the process twice.
  
  Suppose total 100 ltr of pure milk is there.
  then 20 ltr milk is taken out anf 20 ltr water is added.
  
  now mixture has 80 ltr milk and 20 ltr water.
  if 20 ltr mixture is taken out , it has 16 ltr milk and 4 ltr water.
  remaining mixture has 64 ltr milk and 16 ltr water .
  when 20 ltr water is added, water becomes 36 ltr.
  so milk is 64 ltr in mixture of 100 ltr .
  so 64 % milk is there.
• 12 years agoHelpfull: Yes(20) No(14)
• Let pure milk was 100L. So,
  Water is replaced 20% in per process = 20% of 100 = 20L.
  Now, we use short-cut formula for it.
  Quantity of Milk reduced to,
  = X* [1 - (Y/X)]n
  = 100 * [1- (20/100)]3
  = (100* 64)/(125)
  = 51.2 L.
  
  Here,
  X = Initial quantity of milk.
  Y = Replaced water in per process.
  n = No. of process repeated.
  
  Note:
  The formula used in above problem is quite similar to depreciation formula or Compound interest formula.
  
  Alternatively,
  
  Let pure milk be 100 litres initially.
  After third operation, milk will be
  100==20%↓(- 20L)==> 80==20%↓(- 16L)==>64==20%↓(-12.8L)==>51.2 L.
• 7 years agoHelpfull: Yes(2) No(0)
• let x=100, y=20, n= no of replacement Use this formula x ( 1 + (y/x)^n ).
  100 ( 1 + (20/100)^3 ) = 64 Ans
• 7 years agoHelpfull: Yes(1) No(5)
• The short cut formula is : x * ( (1+(y/x))^3 )
  
  x is amount of liquid to be replaced
  y is the amount of liquid used for replacement
• 7 years agoHelpfull: Yes(0) No(2)