Given that, when they work alone, $\text{B}$ needs $25 \%$ more time to finish a job than $\text{A}$.
Let $100x$ days taken by $A$ to finish the job.
Then, time taken by $\text{B}$ to finish the job $= 100x \times \frac{125}{100} = 125x$ days.
$\text{A}$ works alone till half the job is done.
$\Rightarrow \frac{100x}{2} = 50x$ days
Then, $ \text{A}$ and $\text{B}$ work together for $4$ days.
And finally $ \text {B}$ works alone to complete the remaining $5\%$ of the job.
$\Rightarrow 125x \times \frac {5}{100}= \frac {25x}{4}$ days
$\text A$ and $\text B$ finish the job in $13$ days in the following manner :
$ \Rightarrow 50x +4+ \frac {25x}{4}=13$
$ \Rightarrow 50x + \frac {25x}{4}=13-4$
$ \Rightarrow \frac {200x+25x}{4}=9$
$ \Rightarrow 225x=36$
$ \Rightarrow x= \frac {36}{225}$
$\therefore$ Number of days in which $\text{B}$ alone can finish the job $=125x = 125 \times \frac {36}{225}=20$ days.
$\textbf{Short Method:}$
$\begin{array}{ccc} & \textbf{A} & \textbf{B} \\ \text{Time:} & 100 & 125 \\ & 4 & 5 \\\hline \text{Efficiency:} & 4 & 5 \end{array}$
- A work alone $ = 50\%$
- B work alone $ = 5\%$
- A and B together work $ = 100\% – 50\%-5\% = 45\%$
Let $x$ days they take to finish the entire job.
- $4$ days $\longrightarrow 45\%$
- $x$ days $\longrightarrow 100\%$
$x = \frac{4 \times 100\%}{45\%} = \frac{80}{9}$ days
Total work $ = \frac{80}{9} \times9 = 80$ units.
Therefore, B alone can finish the job $ = \frac{80}{4} = 20$ days.
Correct Answer $: \text {A}$