a beaker contains 180 litres of alcohol. On day 1, 60 litres of alcohol is replaced with water. On 2nd and 3rd days 60 litres of the mixture in the beaker is replaced with water. What will be the quantity of alcohol in the beaker after 3rd day?

• ans =(160/3) lit
  1st time 180-60=120
  2nd time 120-60*120/180=80lit
  3rd time 80-60*80/180= 160/3 lit
• 10 years agoHelpfull: Yes(33) No(5)
• ====================Alcohol=============================Water
  0day> 180 0
  1day> 120(60 lit reduced) 60(60 lit added)
  2day> 80(2x i.e. 40 l reduced as 120=60) 100(that 40 added)
  3day> 160/3 200/3 (similarly taken out ratio of 80 and 100)
  ans=160/3 lit
• 10 years agoHelpfull: Yes(17) No(6)
• 180(1-60/180)^3=160/3lit
• 10 years agoHelpfull: Yes(16) No(3)
• Initially:: 180alc
  1st day:: 120alc+60wat
  2nd day:: (2/3 of 120)alc+(1/3 of 60)wat is removed and 60 wat is added.
  As a result by 2nd day ending:: 80alc + 40wat + 60wat = 80alc + 100wat
  3rd day:: (4/9 of alc and 5/9 of wat is rem) and 60wat is added.
  As a result by 3rd day ending:: 53 lit(approx)alc is present.
• 10 years agoHelpfull: Yes(10) No(5)
• 1st day- 120 al 60 water(after replacement)
  2nd day- 80 al 100 water(after replacement)
  3rd day- 160/3 al 380/3 water(after replacement)
  
  use the concept of ratio.....
  
  Ans 160/3...
  
• 10 years agoHelpfull: Yes(7) No(2)
• ans is 53.33 letter
  starting a=180 w=0
  1st day a=120 w=60
  2nd day a=80 w=100
  3rd day a=53.33 and w=126.67
• 10 years agoHelpfull: Yes(6) No(3)
• soln is total quantity*(1-fraction of wine withdrawn)^n
  180*(1-1/3)^3
• 10 years agoHelpfull: Yes(6) No(3)
• 160/3 answer
• 10 years agoHelpfull: Yes(5) No(4)
• 60 litr alcohal
• 10 years agoHelpfull: Yes(3) No(10)
• Initially, 180L of alcohol
  
  removing 60L from 180L = 180* X/100=60
  solving, X=33.33%(amount of prcentage to be removed)
  
  alcohol water total
  1st day- =180-60=120 +60 180
  
  2nd day- =120-(120*33.33/100) 60-(60*33/100)
  =80(approx) =40(approx) 180
  =40+60
  (adding 60L water)
  
  3rd day- =80-(80*33.33/100) =40-(40*33.33/100) 180
  ANS="53.33" =26.68+100
  (adding 60L water)
• 4 years agoHelpfull: Yes(2) No(0)
• it'll be 80 litres if on 2nd and 3rd day,total 60 litres of mixture is replaced with water.......
• 10 years agoHelpfull: Yes(1) No(7)
• amount of alcohol left = 160/3 Ans.
• 10 years agoHelpfull: Yes(1) No(3)
• 0th day 180 lt of alcohol,,
  1st day due to replacement 120 lt of alcohol, 60 lt of water
  2nd day:mix of 60 lt is to be replaced then { 120-60*(2/3):60-60*(1/3)+60}=80:100
  3rd day: now alcohol and water are in 4:5 ratio then {80-60*(4/9):100-60*(5/9)+60}=53.5:126.5
  so 53.5 lt of alcohol is present
• 7 years agoHelpfull: Yes(0) No(0)
• 53.3 litre
• 6 years agoHelpfull: Yes(0) No(0)
• For solving this problem you have to use a simple equation:
  
  Final Concentration=(Initial Concentration)*(1-Volume Replaced/Final Quantity)
  
  1st Day: FC=1*(1-60/180)=2/3
  2nd Day: FC=(2/3)*(1-60/180)=2/3*2/3
  3rd Day: FC=(2/3*2/3)*(1-60/180)=2/3*2/3*2/3=0.296
  
  NOTE:Here the final volume is always 180l because once one take 60l out of the mixture,it is getting replaced by same amount of water.
  
  So 0.296 of the entire volume is alcohol;
  ie; 0.296*180=53.33
• 3 years agoHelpfull: Yes(0) No(0)