George does 3/5th of a piece of work in 9 days. He then calls Paul. and they finish the work in 4 days. How long would Paul take to do the work by himself

28/32/30/35

• g's 9 days work=3/5
  g's 1 days work=3/5*9=1/15.
  (g+p)'s 4 days work=2/5
  (g+p)'s 1 days work=2/5*4=1/10.
  p's 1 days work=((g+p)'s -g's)1 days work
  =1/10-1/15=1/30
  so p takes 30 days to complete work
  
• 8 years agoHelpfull: Yes(15) No(0)
• 3/5---9
  1-------9*5/3=15
  remaining work --2/5
  let x be the work done by paul
  2/5=(1/x+1/15)*4(day)
  x=30(ans)
• 8 years agoHelpfull: Yes(2) No(0)
• George does 3/5th of work in 9 days so 1day work is (3/5)*(1/9)=1/5
  George and Paul does the remaining 2/5th work in 4 days ..so Paul and George one day work is=pauls 1 day work +George 1 day work=(2/5)*(1/4)=1/10
  
  I.e pauls one day work+(1/15)=(1/10)
  On simplification
  Pauls one day work=1/30
  So total work can be completed by Paul in 30 days
  And:30
• 8 years agoHelpfull: Yes(1) No(0)
• eqtn will be
  3/5+4(1/15+1/x)=1
• 8 years agoHelpfull: Yes(1) No(0)
• George can do 3/5 work in 9 days then he can do 2/5 remaining work in 6 days.....george and paul can do 2/5 work in 4 days ......take LCM of 6 and 4....its 12......efficiency of george and paul+george is 2 and 3 resp.....hence then 1 is efficency of paul is 1....if george can complete total work in 15 days with efficiency 2 then paul can complete same total work in 15*2=30 days
• 8 years agoHelpfull: Yes(0) No(0)
• as 3/5 unit of work is done by george in 9 days
  now 2/5 unit of work is done by both george and paul in 4 days.
  therefore 1 unit of work by both of them is done in 4*5/2 = 10 days.
  now using the euqtion
  1/15 + 1/x = 1/10
  => x = 30
  that is work done by paul alone in 30 days.
  
  as 3/5 unit of work is done in 9 days
  therefore 1 unit work is done 15 days.
• 8 years agoHelpfull: Yes(0) No(2)
• (9+4)/15 + 4/x = 1
  if we solve the equation we can get 30 days
• 7 years agoHelpfull: Yes(0) No(0)