A man starts a piece of work. Starting from 2nd day onwards,everyday a new man joins. With every man joining the work that each man can do per day doubles.The work is completed in 5 days.
On which day they would have completed the work if the work each man can do per day remained constant.
Please Explain!!!

• Let day 1,man does x work.
  On day 2,there are 2 men=2x work.So, 4x work gets done.
  On day 3,there are 3 men=4x work. So, 12x work gets done.
  On day 4,there are 4 men=8x work. So, 32x work gets done.
  So,on day 5,there are 5 men=16x work. So, 80x work gets done.
  => x + 4x + 12x + 32x + 80x = 129x work.
  Given that work remains constant, then the progression is:
  x + 2x + 3x + 4x + 5x + .... + nx, on which n is the last day. We need that series to sum to 129.
  =>n(n+1)*x/2
  substitute th values in n
  n=14=>Sum=105x
  n=15=>Sum=120x
  n=16=>Sum=136x
  As we need 129x Total no days to cpmplete work will be 16days
• 11 years agoHelpfull: Yes(3) No(2)
• Ans= 31 days
  If 1st,2nd,3rd,4th & 5th day work is 1/x, 2/x, 4/x, 8/x, 16/x respectively
  then (1/x) + (2/x) + (4/x) + (8/x) + (16/x) =1
  31/x=1 or x=31
  If each day work of each man remains same i.e = 1/31
  So days to complete the work= 1/(1/31)= 31
• 11 years agoHelpfull: Yes(2) No(3)
• x+4x+12x+32x+80x= total 129 unit work in 5 days, so, 1+2+3+.....+16=136>129. so, 16 days will be needed.
• 10 years agoHelpfull: Yes(1) No(0)
• i think its 28 not sure...
  1st day work=1;
  2st day work=2+2;
  3rd day work=4+4+4;
  4th day work=8+8+8+8;
  5th day work=16+16+16+16+16;
  
  so,
  
  1+2+2+4+4+4+8+8+8+8+16+16+16+16+16=139
  so if their are 5 person so the
  
  (Total work)/(No. of workers)=139/5
  
  =27+(4/5)=28 Days
  
  M I Right, Please Make me correct if i am wrong.
• 10 years agoHelpfull: Yes(0) No(1)