Q. Some men can do a work in 10 days. with 10 less than original no of men it takes 10 more days to complete the work. what is the original no of men?

• let the no. of men be x
  x men can do in 10 days,so one man can do in 10*x days
  now (x-10) men can do in (10+10=20 days),so one man can do in 20*(x-10)days
  10x=20(x-10)
  or x=2x-20
  or x=20 men
  
• 10 years agoHelpfull: Yes(34) No(1)
• m1 * d1 = m2 * d2
  10x=20(x-10)
  x=2x-20
  x=20
• 10 years agoHelpfull: Yes(12) No(1)
• Let total no of men are x.
  Now as x*10=(x-10)*20
  x=2x-20
  so x=20.
• 10 years agoHelpfull: Yes(4) No(0)
• total original worker=x;
  worker days unit
  x-10 20 1
  x 10 10x/20(x-10)
  
  10x/20(x-10)=1
  or,x=20
• 10 years agoHelpfull: Yes(2) No(0)
• (x * 10 ) = (x - 10 ) * 20
  
  10x =20x -200
  
  x = 20 men
• 10 years agoHelpfull: Yes(1) No(1)
• 'x' men can do a work in 10 days
  'x-10' men can do the same work in "10+10=20 days'
  here the work is same
  therfore: (X*10)=(X-10)*20
  X=20
• 9 years agoHelpfull: Yes(1) No(0)
• x=20. Days*Men=Days*Men hence, x*10=x-10*20 therefore x=20
• 7 years agoHelpfull: Yes(1) No(0)
• s*t=w
  (s*t)/w=1
  x*10=(x-10)*20
  x=20
• 10 years agoHelpfull: Yes(0) No(0)
• 15,12 this alternativ series
• 10 years agoHelpfull: Yes(0) No(1)
• Soln
  10x=20(x-10)
  x = 20
  So there were 20 mens in the begening.
  
• 10 years agoHelpfull: Yes(0) No(0)
• OG number of men 20
  
• 10 years agoHelpfull: Yes(0) No(0)
• No.of men No.of working days
  x 10
  (x-10) 20
  as members are reduced days are increased
  10x=20(x-10)
  by solving,x=20men
• 10 years agoHelpfull: Yes(0) No(0)
• we know, M1*D1=M2*D2
  here, M1=m; D1=10; M2=m-10; D2=20;
  SO, m=20
• 9 years agoHelpfull: Yes(0) No(0)
• x men 1 day work=10%
  x-10 men 1 day work=5%
  In 1 day x=20.
• 8 years agoHelpfull: Yes(0) No(0)
• 20
  
  x+10=20
  
• 7 years agoHelpfull: Yes(0) No(0)