G works five time faster than his son and hence completes a job 40 days earlier than his son.Find the time they would take together to finish the job? [JABALPUR]
(a)20/3 days (b)25/3 days (c)10 days (d)35/4 days

• let son finishes the work in x days.
  father will finish in x/5 days.
  x/5+40=x
  =>x=50
  father's time taken=50/5=10 days.
  combined one day work will be1/50+1/10=6/50
  now time taken to do complete work=50/6 or 25/3.
  so answer is b)25/3 days
• 9 years agoHelpfull: Yes(16) No(0)
• Ratio of times taken by G and son = 1 : 5.
  
  The time difference is (5 - 1) 4 days while son take 5 days and G takes 1 day.
  
  If difference of time is 4 days, son takes 5 days.
  
  If difference of time is 40 days, son takes (5/4 *40) = 50 days.
  So, G takes 10 days to do the work.
  
  G's 1 day's work = 1/10
  son's 1 day's work = 1/50
  (G + son)'s 1 day's work= (1/10 + 1/50)= 6/50 = 3/25
  G and son together can do the work in 25/3 days.
  Answer B) 25/3 =days
  
• 9 years agoHelpfull: Yes(3) No(0)
• G :son
  speeds 5 :1
  time 1 :5 ( because speed is reciprocal of time)
  assume x : 5x
  As G works 5 faster than his son we get 5x - x=40days
  therefore x=10days
  5x=50days
  assume total work done = 50units which is divisible by both 10 and 50
  therefore G= 5u/day
  son = 1u/day
  on adding both they together work 6u/day
  As time is inversly proportional to speed we get
  t = w/s => 50/6 => 25/3 ( so answer is b )
  
• 9 years agoHelpfull: Yes(2) No(0)
• G=5S--------------------------1
  (let son=s)
  G=S+40----------------------2
  from 2 eqn
  S=10--------------------------3
  Then G=50------------------4
  TOGETHER WORK
  1/10+1/40=3/25=====>25/3
  
• 9 years agoHelpfull: Yes(2) No(0)
• 5x=40-x
  6x=40
  x=20/3
  
• 9 years agoHelpfull: Yes(0) No(7)