Question

$A$ and $B$ can do a piece of work in 30 days. While $B$ and $C$ can do the same work in 24 days and $C$ and $A$ in 20 days. They all work together for 10 days before $B$ and $C$ leave. How many days more will $A$ take to finish the work?

• A. 36 days
• B. 18 days
• C. 24 days
• D. 30 days

Answer 1

Answer: (B)

$(A+B)$'s one day's work $=\frac{1}{30}$

$(B+C)$'s one day's work $=\frac{1}{24}$..........$\left(1\right)$

$(C+A)$'s one day's work $=\frac{1}{20}$

$\therefore (A+B+C)$'s one day's work $=\frac{1}{2}\left[\frac{1}{30}+\frac{1}{24}+\frac{1}{20}\right]=\frac{1}{2}\times \left[\frac{20+25+30}{600}\right]=\frac{75}{1200}=\frac{1}{16}$...........$\left(2\right)$

$(A+B+C)$'s 10 day's work $=\frac{10}{16}=\frac{5}{8}$

Remaining work, $1-\frac{5}{8}=\frac{3}{8}$ (which is done by A)

From $\left(1\right)$ and $\left(2\right)$, $A$'s one day's work $=\frac{1}{16}-\frac{1}{24}=\frac{1}{48}$

$\therefore$ Remaining $\frac{3}{8}$ of the work is done by $A$ alone in $\frac{3}{8}\times 48=18$ days.