Question

A can do a piece of work in $10$ days, B in $12$ days and C in $15$ days. All begin together but A leaves the work after $2$ days and B leaves $3$ days before the work is finished. How long did the work last?
A. $7{\text{ days}}$
B. $2{\text{ days}}$
C. $3{\text{ days}}$
D. $8{\text{ days}}$

Answer 1

Hint:
Here we are given the time period when each one of $A, B, C$ can complete the work individually. So we can find one day's work each and then according to the question, we can find the $2{\text{ day}}$ work of A and we also know that C works for the last three days alone. So we can find this work also and now with the leftover work we can find the total days also.

Complete step by step solution:
We are given here that A can do a piece of work in $10$ days, B in $12$ days and C in $15$ days.
We know that if the person completes work in $m{\text{ days}}$ then his one day work$ = \dfrac{1}{m}$
So we know A completes work in $10{\text{ days}}$ individually
So A’s one day work$ = \dfrac{1}{{10}}$
Similarly B’s one day work$ = \dfrac{1}{{12}}$
C’s one day work$ = \dfrac{1}{{15}}$
So we can find one day work when all are working together$ = $one day work of $(A + B + C)$
$ = \left( {\dfrac{1}{{10}} + \dfrac{1}{{12}} + \dfrac{1}{{15}}} \right)$
LCM of $10,12,15{\text{ is 120}}$ so making the denominators common which is $120$
One day work when all work together$ = \dfrac{{12 + 10 + 8}}{{120}} = \dfrac{{30}}{{120}} = \dfrac{1}{4}$
We know that all of them worked for two days together.
So $(A + B + C)'{\text{s}}$ two day work$ = {\text{their one day work}} \times {\text{2}} = \dfrac{1}{4}(2) = \dfrac{1}{2}$$ - - - (1)$
We also know that B also leaves three days before the work ends. So it means that for the last $3{\text{ days}}$ only C has worked.
So C’s three day work$ = C'{\text{s one day work}} \times 3 = \dfrac{1}{{15}}(3) = \dfrac{1}{5}$$ - - - (2)$
So we can add equation (1) and (2) to get the total work that has been completed..
So we get the work done till now$\dfrac{1}{2} + \dfrac{1}{5} = \dfrac{{5 + 2}}{{10}} = \dfrac{7}{{10}}$
Now we get that $\dfrac{7}{{10}}$ of the work has been completed.
Hence the work left$ = 1 - \dfrac{7}{{10}} = \dfrac{{10 - 7}}{{10}} = \dfrac{3}{{10}}$
So we can say that this leftover work has been done by $B{\text{ and }}C$
Now $(B + C)'s$ one day work$ = \dfrac{1}{{12}} + \dfrac{1}{{15}} = \dfrac{{5 + 4}}{{60}} = \dfrac{9}{{60}} = \dfrac{3}{{20}}$
Work done by $B{\text{ and }}C = \dfrac{3}{{10}}$
So we know that ${\text{total work done}} = {\text{one day's work}} \times {\text{number of days taken}}$
Substituting the values we get:
$
  \dfrac{3}{{10}} = \dfrac{3}{{20}} \times {\text{number of days taken}} \\
  {\text{number of days}} = \dfrac{3}{{10}} \times \dfrac{{20}}{3} = 2{\text{ days}} \\
 $
Hence we can say that total days to complete$ = (2 + 2 + 3) = 7{\text{ days}}$

Hence option A is correct.

Note:
Here in these types of problems, we need to find the one day work of a person and then proceed in the proper way as shown. We must know that ${\text{total work done}} = {\text{one day's work}} \times {\text{number of days taken}}$