If 15 men or 24 women or 36 boys can do a piece of work in 12 days working 8 hours a day , how many men must be associated with 12 women and 6 boys to do another piece of work 9/4 times as great in 30 days working 6 hrs/ day?

• Here the following logic will be used:
  If M1 men can do W1 work in D1 days working H1 hours per day and M2 men can do W2 work in D2 days working H2 hours per day (where all men work at the same rate), then
  
  M1 D1 H1 / W1 = M2 D2 H2 / W2 .............(1)
  
  So,given 15M=24W=36B...................(2)
  where M,W,B are work done by one man, one woman and one boy respectively in one day.
  
  Also given W2=9/4W1 ....(3)
  
  for W2 let x Men(M) are required.
  So total Human-power(for W2) = xM + 12W + 6B
  now using relations from (2) Human-power(for W2) = xM + 15/2M + 15/6M = (10+x)M
  
  So replacing all the values in eqn (1) and (3), we get
  
  => 15M*12days*8Hrs/W1 = (10+x)M*30days*6Hrs/W2
  
  => 15M*12days*8Hrs/W1 = (10+x)M*30days*6Hrs/(9/4W1)
  
  => x = 8 Men
  
• 10 years agoHelpfull: Yes(45) No(0)
• w2=9/4*w1
  so
  1 man can do the w2's part in 1 day=(9/4)*(15*12*8)
  1 woman --------------------------=(9/4)*(24*12*8)
  1 child ---------------------------=(9/4)*(36*12*8)
  
  in one day x men,12 women, 6 boys do 1/180th part of w2 so
  
  1/180=x/((9/4)*(15*12*8))+12/(9/4)*(24*12*8)+6/(9/4)*(36*12*8)
  x=8
• 10 years agoHelpfull: Yes(12) No(0)
• 3.5.
  get the work done by each of their of the respective category.
  
  12*8=72 hr.
  so 1 hr work=1/72,1man work in 1 hr=1/72/15.
  likewise,1woman in 1hr=1/72/24.
  1boy in 1hr=1/72/36.
  
  so now they must work for 30*6=180hr.
  
  next,
  let the number of men to be associated with others be equal to x,
  therefore,
  x*180*(1/72/15)+12*(1/72/24)*180+6*(1/72/36)*180=9/4.
  x*0.166662+1.249992+0.416664=2.25,
  x=3.5 man.
• 10 years agoHelpfull: Yes(0) No(10)
• any short trick to solve this ques.
  for e.g. mrt/w
  
• 10 years agoHelpfull: Yes(0) No(0)