A certain quantity of milk is found to be adulterated to the extent of 10%. What proportion of the adulterated milk should be replaced with pure milk to take the purity level to 98%?
A. 31 B. 25 C. 58 D. 47

• Ans-80%
  No option is true
• 9 years agoHelpfull: Yes(10) No(2)
• let us assume that initially the amount of mixture is 100L
  now for 10% adulteration of mixture there will be 90% milk and 10% water:
  amount of milk( M)= 90L
  amount of water(W)=10L
  
  now let us assume that x litres mixture is replaced by pure milk
  so now,
  amount of milk taken out = 9x/10
  amount of water taken out=x/10
  
  now x litre pure milk will be added
  now final amount
  M=90-(9x/10)+x
  W=10-x/10
  
  so
  98=(90-(9x/10)+x)/100*100
  98=(90-(9x/10)+x)
  finally
  x=80
  
• 9 years agoHelpfull: Yes(9) No(0)
• 100 90
  /
  98
  /
  8 2
  =4 1
  
  100*4/5=80
• 9 years agoHelpfull: Yes(4) No(1)
• Let 100 Lt adultered milk is present so milk is 90 and water is 10 Lt
  
  Let x Lt of purecmilk is mixed to make it 98%.
  
  So
  90+x=(100+x)×.98
  =>x=40
  
• 9 years agoHelpfull: Yes(3) No(2)
• 25 is right ans
• 9 years agoHelpfull: Yes(3) No(2)
• 25:2.... 90+x/100=98/100
• 9 years agoHelpfull: Yes(2) No(11)
• can anyone tell the answer??
• 9 years agoHelpfull: Yes(0) No(1)