(y+40) boys can do a piece of work in (z-20) days. (y-5)

Solution

(y+40) boys can do a piece of work in (z-20) days. Total work = (y+40)x(z-20)    Eq.(i) (y-5) boys can do the same piece of work in (z+20) days. Total work = (y-5)x(z+20)    Eq.(ii) If (y+30) boys can do the same piece of work in (z-12) days. Total work = (y+30)x(z-12)    Eq.(iii) So Eq.(i) = Eq.(ii) (y+40)x(z-20) = (y-5)x(z+20) yz-20y+40z-800 = yz+20y-5z-100 -20y+40z-800 = 20y-5z-100 20y+20y-40z-5z+800-100 = 0 40y-45z+700 = 0 40y-45z = -700 8y-9z = -140 Eq.(1) So Eq.(i) = Eq.(iii) (y+40)x(z-20) = (y+30)x(z-12) yz-20y+40z-800 = yz-12y+30z-360 -20y+12y+40z-30z-800+360 = 0 -8y+10z-440 = 0 -8y+10z = 440 Eq.(2) Add  Eq.(1) and  Eq.(2). 8y-9z-8y+10z = -140+440 z = 300 Put the value of ‘z’ in Eq.(1). 8y-9x300 = -140 8y-2700 = -140 8y = 2700-140 8y = 2560 Value of ‘y’ = 320