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 when $B$ and $C$ leave. How many days more will $A$ take to finish the work?

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

Answer 1

The correct option is A $18$ days

$2(A+B+C)$'s $1$ day's work $=(\frac{1}{30}+\frac{1}{24}+\frac{1}{20})=\frac{15}{120}=\frac{1}{8}$.
Therefore, $(A+B+C)$'s $1$ day's work $=\frac{1}{2\times 8}=\frac{1}{6}$.
Work done by $A,B,C$ in $10$ days $=\frac{10}{16}=\frac{5}{8}$
Remaining work $=(1-\frac{5}{8})=\frac{3}{8}$.
$A$'s $1$ day's work $=(\frac{1}{16}-\frac{1}{24})=\frac{1}{48}$.
Now, $\frac{1}{8}$ work done by $A$ in $1$ day.
So, $\frac{3}{8}$ work will be done by $A$ in $(48\times \frac{3}{8})=18$ days.