# $A$ 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 the $3$ days before the work is finished. How long did the work last?

$

{\text{A}}{\text{. 7 }}days \\

{\text{B}}{\text{. 2 }}days \\

{\text{C}}{\text{. 3 }}days \\

{\text{D}}{\text{. 8 }}days \\

$

Last updated date: 17th Mar 2023

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Answer

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Hint –For such types of questions, the approach should be to read the question carefully and put all your effort in understanding the question and write what is given as it will help us to analyse the question intricately. This will then help us to lead towards the desired answer.

Complete step-by-step answer:

According to question,

$A's$ one day’s work $ = \dfrac{1}{{10}}$

$B's$ one day’s work $ = \dfrac{1}{{12}}$

$C's$ one day’s work $ = \dfrac{1}{{15}}$

$A,B{\text{ and }}C's$ one day’s work $ = \dfrac{1}{{10}} + \dfrac{1}{{12}} + \dfrac{1}{{15}}$

$ \Rightarrow \dfrac{{\left( {6 + 5 + 4} \right)}}{{60}} = \dfrac{{15}}{{60}} = \dfrac{1}{4}$

$A,B{\text{ and }}C's$ two day’s work $ = \dfrac{1}{4} \times 2 = \dfrac{1}{2}$

But $B$ leaves \[3\] days before the work gets finished, so $C$ does the remaining work alone.

$C's$ $3$ day’s work $ = \dfrac{1}{{15}} \times 3 = \dfrac{1}{5}$

Then work done in $2 + 3$ days $ = \dfrac{1}{2} + \dfrac{1}{5} = \dfrac{7}{{10}}$

Work done by $B + C$ together $ = 1 - \dfrac{7}{{10}} = \dfrac{3}{{10}}$

$\left( {B + C} \right)'s$ one day work $ = \dfrac{1}{{12}} + \dfrac{1}{{15}} = \dfrac{9}{{60}} = \dfrac{3}{{20}}$

So number of days worked by $B$ and $C$ together $ = \dfrac{3}{{10}} \times \dfrac{{20}}{3} = 2$ days

So $A,B,C$ work for $2$ days,

$B,C$ work for $2$ days,

$C$ works for $3$ days.

Then total work done $ = 2 + 3 + 2 = 7$ days

$\therefore $ work will last for $7$ days.

Note – As you know, this is a work-time problem so there are many approaches to solve these types of problems. Here we first find the days, where work is done by all three i.e. $2$ and then days where $C$ works alone i.e. $3$. So, this equals $5$ days and now we have to find out the number of days in which remaining work is completed i.e. by $B{\text{ and }}C$ then we will add up all and get our required answer. Then match that sum of days with that one particular given option.

Complete step-by-step answer:

According to question,

$A's$ one day’s work $ = \dfrac{1}{{10}}$

$B's$ one day’s work $ = \dfrac{1}{{12}}$

$C's$ one day’s work $ = \dfrac{1}{{15}}$

$A,B{\text{ and }}C's$ one day’s work $ = \dfrac{1}{{10}} + \dfrac{1}{{12}} + \dfrac{1}{{15}}$

$ \Rightarrow \dfrac{{\left( {6 + 5 + 4} \right)}}{{60}} = \dfrac{{15}}{{60}} = \dfrac{1}{4}$

$A,B{\text{ and }}C's$ two day’s work $ = \dfrac{1}{4} \times 2 = \dfrac{1}{2}$

But $B$ leaves \[3\] days before the work gets finished, so $C$ does the remaining work alone.

$C's$ $3$ day’s work $ = \dfrac{1}{{15}} \times 3 = \dfrac{1}{5}$

Then work done in $2 + 3$ days $ = \dfrac{1}{2} + \dfrac{1}{5} = \dfrac{7}{{10}}$

Work done by $B + C$ together $ = 1 - \dfrac{7}{{10}} = \dfrac{3}{{10}}$

$\left( {B + C} \right)'s$ one day work $ = \dfrac{1}{{12}} + \dfrac{1}{{15}} = \dfrac{9}{{60}} = \dfrac{3}{{20}}$

So number of days worked by $B$ and $C$ together $ = \dfrac{3}{{10}} \times \dfrac{{20}}{3} = 2$ days

So $A,B,C$ work for $2$ days,

$B,C$ work for $2$ days,

$C$ works for $3$ days.

Then total work done $ = 2 + 3 + 2 = 7$ days

$\therefore $ work will last for $7$ days.

Note – As you know, this is a work-time problem so there are many approaches to solve these types of problems. Here we first find the days, where work is done by all three i.e. $2$ and then days where $C$ works alone i.e. $3$. So, this equals $5$ days and now we have to find out the number of days in which remaining work is completed i.e. by $B{\text{ and }}C$ then we will add up all and get our required answer. Then match that sum of days with that one particular given option.

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