60 men complete work in 50 days. every 10 days 5 men left the work then how many days work will complete?

• 65 days
  
  1 man's 1 day work=1/(60*50)=1/3000
  Work completed by 60 men in first 10 days=10/50=1/5, so remaining work=4/5
  For next 10 days 55 men will work, so work completed by them=55*10/3000=11/60
  For next 10 days 50 men will work, so work completed by them=50*10/3000=10/60
  For next 10 days 45 men will work, so work completed by them=45*10/300=9/60
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  Here we observe that, with decrease of 5 men after every 10 days, work completed is of the order 11/60, 10/60, 9/60, and so on , so we can sum these work completed to the remaining work=4/5 or 48/60
  We have (11/60)+(10/60)+(9/60)+(8/60)+(7/60)+x/60=48/60
  As 'x' here is equal to 3, so this is the duration where work is completed by 30 men=3/60=1/20
  As 30 men's 1 day work=30/3000=1/100, so days required to complete the 1/20 of work=(1/20)/(1/100)=5
  So total days required to completion of whole work=10 days (60 men)+10 days(55 men)+10 days(50 men)+10 days(45 men)+10 days(40 men)+10 days(35 men)+5 days(30 men)= 65
• 8 years agoHelpfull: Yes(0) No(0)
• since 60 man complete the work in 50 days
  so 1 man can complete the work in 60*50=3000 days
  
  w=worker, d= no. of days
  w 9
  60*10=600
  55*10=550
  50*10=500
  45*10=450
  40*10=400
  35*10=350
  30*5=150
  
  total = 600+550+500+450+400+350+150=3000 ans
  total
• 8 years agoHelpfull: Yes(0) No(1)