A group of men completes a work in 10 days, but five of them are absent and so the rest do the work in 12 days. Find the original number of Men.

• In time and work problems the important point to be remebered is (number of MENDAYS is always constant.)i.e product of men into days always remains constant.
  if one variable among men or days changes the other varible changes in accordance to make the value of remains constant.
  
  let us assume X be the number of initial persons. so mendays = 10* X( as told above), now 5 are absent.i.e (X-5) are present.as i told men days should be always constant
  10* X= (X-5)*12
  
  10X= 12X-60
  
  2X=60 implies X=30.
• 12 years agoHelpfull: Yes(21) No(1)
• original no. of men is 30
• 12 years agoHelpfull: Yes(8) No(1)
• 30 men originally
• 12 years agoHelpfull: Yes(8) No(2)
• Let us assume x Mens are working there initially.
  so x men completes the job in 10 days. i.e. x --> 1/10 part of work per day.
  similarly if 5 of them were absent means they will finish it at 12 days.
  i.e. (x-5)=1/12
  Here one man work will be same for the both condition, therefore
  1/10x = 1/12(x-5)
  10x=12(x-5)
  2x=60 ==> x=30.
  therefore total number of men at the begining is 30.
• 12 years agoHelpfull: Yes(7) No(0)
• let the original men be x so x*10=(x-5)*12
  by solving the above we get x=30 men
• 12 years agoHelpfull: Yes(2) No(0)