A can finish a piece of work in a certain number of days. B takes 45% more number of days to finish the same work independently. They worked together for 58 days and then the remaining work was done by B alone in 29 days. In how many days could A have completed the work, had he worked alone?

118 days

Given,

B takes 45% more number of days to finish the same work independently.

B : A = 145 : 100 = 29 : 20   (Time)

Efficiency α \(\frac{1}{Time}\)

Therefore, B : A = 25 : 29  (Efficiency)

⇒ A + B worked for 58 days (W1)= (29 + 20) x 58 = 49 x 58 = 2842 units,   ..(Efficiency × Days = Total work)

⇒ Remaining work is completed by B alone in 29 days = \(\frac{W - W1}{20}\) = 29 days,

⇒ Therefore, W - W1 = 580 units,

⇒ Putting value of W1 here,

⇒ W - 2842 = 580,

⇒ W = 3422 units,

⇒ Time required by A to complete the work alone = \(\frac{3422}{29}\) = 118 days,   ..(\(\frac{Work}{Efficiency}\) = Time)