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When they work alone, $\text{B}$ needs $25\%$ more time to finish a job than $\text{A}$ does. They two finish the job in $13$ days in the following manner: $\text{A}$ works alone till half the job is done, then $\text{A}$ and $\text{B}$ work together for four days, and finally $\text{B}$ works alone to complete the remaining $5\%$ of the job. In how many days can $\text{B}$ alone finish the entire job?

- $20$
- $16$
- $22$
- $18$

1 vote

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 $: \text {A}$