George works 5 times faster than his son and hence comptles a job 40 days earlier than his son.find the time they would take to finish the job together
a)12.5 b)8.3 c)7.5 d)6

• No. of days is inversely proportional to Speed.
  
  Thus, (son days)/(dad days) = (dad speed)/(son speed) .....(i)
  
  Now given, (Dad speed)/(son speed) = 5/1 ..................(ii)
  
  Now let, son take 'x' days ................................(iii)
  Thus, dad takes (x-40) days ...............................(iv)
  
  From (i),(ii),(iii),(iv): x=50
  
  Thus, son takes 50 days and dad takes 10 days.
  
  Hence, son's 1 day work = 1/50
  And, dad's 1 day work = 1/10
  
  1/10 + 1/50 = 3/25
  
  Thus, Both takes 25/3 i.e. 8.3 days.....
  
  
• 12 years agoHelpfull: Yes(22) No(0)
• 8.3 is correct answer.
• 12 years agoHelpfull: Yes(8) No(2)
• Let work done by George be"G " and his son be "S"
  Then G=5S,
  G:S=5:1
  Then the times for completing the job are in the ratio 1:5
  then takin difference 5-1=4 parts means 40 days that means 1 part means 10 days
  Therefore the SON can complete the work in 5*10=50 days.
  Taking LCM of 10,50 as the total number of units of work
  Then George can perform 5units work per day and son can perform 1 unit work per day.Both together can perform 5+1=6 units work per day.Therfore both can complete the work in 50/6 days=8.333 days
• 6 years agoHelpfull: Yes(0) No(0)