plzz solve this problem guyzz.... n give absolute explaination... soon...

If x ml pure alcohol is added to 400 ml of a 15% solution and after adding its strength becomes 32%, then the value of x is
(A) 50 ml (B) 75 ml
(C) 200 ml (D) 100 ml

• 400 ml of a 15% solution has 60 ml pure alchohal and 340 ml of water.
  when x ml of pure alchohal is added, strength becomes 32%,
  hence
  (60+x)/(400+x) = 32/100= 8/25
  1500+25x = 3200+8x
  17x = 1700
  x= 100 ml ..... option D
• 13 years agoHelpfull: Yes(8) No(1)
• 100 ml.
  If 100 ml pure alcohol is added to 400 ml of a 15% solution and after adding its strength becomes 32%,
  because total alchohal in 500 ml sol = 100 + 0.15*400 = 160 ml
  hence conc= 100*160/500 = 32%
• 13 years agoHelpfull: Yes(1) No(0)
• 400ml 0f a solution contain 15%alcohol
  so,alcohol content=400*15/100=60
  we add x ml to it.
  therefore,(60+x)/(400+X)*100=32
  solving this x=100
  so, we have to add 100 ml of alcohol
• 13 years agoHelpfull: Yes(1) No(0)