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?

Let the number of days required for A alone to complete the work = 100a

$$=$$>  Number of days required for B alone to complete the work = 145a

Let the total work = W

Work done by A in 1 day = $$\frac{W}{100a}$$

Work done by B in 1 day = $$\frac{W}{145a}$$

Work done by A and B in 1 day = $$\frac{W}{100a}+\frac{W}{145a}=\frac{245W}{14500a}=\frac{49W}{2900a}$$

Work done by A and B in 58 days$$=58\times\frac{49W}{2900a}=\frac{49W}{50a}$$

Remaining work $$=W-\frac{49W}{50a}$$

Remaining work is completed by B in 29 days

$$=$$>  $$W-\frac{49W}{50a}=\frac{29W}{145a}$$

$$=$$>  $$W=\frac{29W}{145a}+\frac{49W}{50a}$$

$$=$$>  $$W=\frac{1450W+7105W}{145\times50\times a}$$

$$=$$>  $$W=\frac{8555W}{145\times50\times a}$$

$$=$$>  $$W=\frac{59W}{50a}$$

$$=$$>  $$50a=59$$

$$=$$>  $$100a=118$$

$$\therefore\ $$Number of days required for A alone to complete the work = 100a = 118 days

Hence, the correct answer is Option B