A contractor has two teams of workers,team A | vedclass

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(B) Work done by Team $A$ in one day $= \frac{1}{12}$.Work done by Team $B$ in one day $= \frac{1}{36}$.Work done by Team $A$ in first $4$ days $= 4 \

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) Work done by Team $A$ in one day $= \frac{1}{12}$.Work done by Team $B$ in one day $= \frac{1}{36}$.Work done by Team $A$ in first $4$ days $= 4 	imes \frac{1}{12} = \frac{1}{3}$.Work done by Team $A$ and $B$ together in the next $2$ days $= 2 	imes (\frac{1}{12} + \frac{1}{36}) = 2 	imes (\frac{3+1}{36}) = 2 	imes \frac{4}{36} = \frac{2}{9}$.Total work done in $6$ days $= \frac{1}{3} + \frac{2}{9} = \frac{3+2}{9} = \frac{5}{9}$.Remaining work $= 1 - \frac{5}{9} = \frac{4}{9}$.Team $B$ doubles its efficiency,so new efficiency $= 2 	imes \frac{1}{36} = \frac{1}{18}$.Additional days required for Team $B$ to complete the remaining work $= \frac{4/9}{1/18} = \frac{4}{9} 	imes 18 = 8$ days.

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