Answer:
Option C
Explanation:
I. Let $B$’s $1$ min. work $=\frac{1}{x}$. Then, $A$’s $1$ min. work $=\frac{3}{x}$.
$(A+B)$’s $1$ min. work $=\left(\frac{1}{x}+\frac{3}{x}\right)$ $=\frac{4}{x}$.
$(A+B)$’s $10$ min. work $=\left(\frac{4}{x}\times 10\right)$ $=\frac{40}{x}$.
$\therefore$ $\frac{40}{x}=\frac{2}{3}$
$\Leftrightarrow x=60$.
$\therefore$ $B$’s $1$ min. work $=\frac{1}{60}$.
$\frac{1}{60}$ part is filled by $B$ in $1$ min.
$\frac{1}{3}$ part is filled by $B$ in $\left(\frac{4}{x}\times 10\right)$ $=\frac{40}{x}$.
$\therefore$ $\frac{40}{x}=\frac{2}{3}$
$\Leftrightarrow x=60$.
$\therefore$ $B$’s $1$ min. work $=\frac{1}{60}$.
$\frac{1}{60}$ part is filled by $B$ in $1$ min.
$\frac{1}{3}$ part is filled by $B$ in $\left(60\times\frac{1}{3}\right)$ = 20 min.
II. $B$’s $1$ min. work $=\frac{1}{60}$.
$\frac{1}{60}$ part is filled by $B$ in $1$ min.
$\frac{1}{3}$ part is filled by $B$ in $\left(60\times\frac{1}{3}\right)$ = 20 min.
Hence, the correct answer is (C).