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?
Answer & explanation
Correct answer: option 2
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)