Two pipes A and B can fill a tank in 36 min. and 4S min. respectively. A water pipe C can empty the tank in 30 min. First A and B are opened. After 7 minutes, C is opened. In how much time, the tank Is full ?

• tank left after 7 mins = 1-( 7*(1/36+1/45)) = (1-( 7*9/180)) = (1-7/20) tank = 13/20 tank
  when 3 pipes are opened, part of tank filled in 1 min = 1/36 +1/45 - 1/30 = 1/20-1/30 = 1/60
  
  hence 13/20 tank will be filled in 60*13/20 = 39 mins.
  
  so tank will be filled in total 7+39 = 46 mins.
• 13 years agoHelpfull: Yes(5) No(1)
• Part filled in 7 min. = 7*((1/36)+(1/45))=(7/20).
  
  Remaining part=(1-(7/20))=(13/20).
  
  Net part filled in 1min. when A,B and C are opened=(1/36)+(1/45)-(1/30)=(1/60).
  
  Now,(1/60) part is filled in one minute.
  
  (13/20) part is filled in (60*(13/20))=39 minutes.
• 11 years agoHelpfull: Yes(2) No(1)