Question 72
Had been one man less, then the number of days required to do a piece of work would have been one more. If the number of Man. Days required to complete the work is 56, how many workers were there?
Solution
Let the number of men be 'x' and number of days be 'y'
Total number of men days=xy
also xy=(x-1)(y+1)
xy=xy-1+x-y
x-y=1
xy=56
x-(56/x)=1
$$x^{2}-x-56$$=0
$$x^{2}-8x+7x-56$$=0
x(x-8)+7(x-8)=0
(x+7)(x-8)=0
x=8 and y=7
Therefore 8 men are needed
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