Question

A can do a certain work in 12 days, B in 18 days and C in 24 days. A and B work together for 3 days and then A
leaves the remaining work is done by B and C. How long did B and C take to complete the work?

Answer 1

Hint: If a person can do a piece of work in ‘x’ days then one day work of that person will be $\dfrac{1}{x}$. After that we
add the work of A and B then we will solve it.
Formula used: \[one\,\,day\,\, = \dfrac{{{\text{1 }}}}{{work}}\].

Complete step-by-step answer:
A can do a piece of work in 12 days,
$\therefore $ one day work of A =$\dfrac{1}{{12}}$
B can do a piece of work in 18 days,
$\therefore $ one day work of B =$\dfrac{1}{{18}}$
C can do a piece of work in 24 days,
$\therefore $ one day work of C =$\dfrac{1}{{24}}$
According to the question, A and B work together for 3 days and then A leaves.
Now, one day work of (A and B) $ = \dfrac{1}{{12}} + \dfrac{1}{{18}}$
Take LCM of 12 and 18 = 36
$
= \dfrac{{3 + 2}}{{36}} \\
= \dfrac{5}{{36}} \\
$
3 days work of (A and B) = $3 \times \dfrac{5}{{36}} = \dfrac{5}{{12}}$
Remaining work = $1 - \dfrac{5}{{12}} = \dfrac{7}{{12}}$
Now, ${\left( {\dfrac{7}{{12}}} \right)^{th}}$ of work is completed by B and C together.
one day work of (B and C) =$\dfrac{1}{{18}} + \dfrac{1}{{24}}$
Take LCM of 18 and 24 = 72
\[
= \dfrac{{4 + 3}}{{72}} \\
= \dfrac{7}{{72}} \\
\]
$ \Rightarrow $B and C can do ${\left( {\dfrac{7}{{72}}} \right)^{th}}$ of work in 1 day
Or $\dfrac{7}{{72}}\,\,of\,\,work = 1\,day$
1 work =$\dfrac{{72}}{7}days$
$\dfrac{7}{{12}}(remaining)work = \dfrac{{72}}{7} \times \dfrac{7}{{12}}days$
= 6 days
Thus, B and C will take another 6 days to complete the work.

Note: Treat the total work done as 1 and from this calculate the remaining work of the respective quantities(that is of A and B together and B and C together) and from this finally, you can calculate the time taken by B and C to complete the work.Do not calculate the values directly.