Question

A can do a piece of work in 14 days and B in 21 days. They begin together, but 3 days before the completion of the work, A leaves off. In how many days is the work complete?
(a) $10$ days
(b) $10\dfrac{1}{5}$ days
(c) $15$ days
(d) $7$ days

Answer 1

Hint: We will first find the fraction of work they do in one day, each alone and together. And, we will calculate the work done in the last 3 days. Then we will use this data to solve the question.

Complete step-by-step answer:

According to question,

A completes the work in 14 days if he works alone. Therefore, he must have completed ${{\left( \dfrac{1}{14} \right)}^{th}}$ of the total work in one day.

And, B completes the work in 21 days if he works alone. Therefore, he must have completed ${{\left( \dfrac{1}{21} \right)}^{th}}$ of the total work in one day.

Now, if they work together, then the fraction of total work done by them in one day must be$=\dfrac{1}{14}+\dfrac{1}{21}$

Taking LCM and then adding, we get,

$\begin{align}

  & =\dfrac{1\times 3}{14\times 3}+\dfrac{1\times 2}{21\times 2} \\

 & =\dfrac{3}{42}+\dfrac{2}{42} \\

 & =\dfrac{3+2}{42} \\

 & =\dfrac{5}{42} \\

\end{align}$

Now, B has worked alone in the last 3 days after A leaves, to complete the work.
Also, fraction of total work done by B in 3 days = $3\times \dfrac{1}{21}$

=$\dfrac{1}{7}$.

Therefore, remaining fraction of work, that A and B have completed together = $1-\dfrac{1}{7}$

Taking LCM and then subtracting,

 $\begin{align}

  & =\dfrac{1\times 7}{1\times 7}-\dfrac{1}{7} \\

 & =\dfrac{7-1}{7} \\

 & =\dfrac{6}{7} \\

\end{align}$

Here, we have,

Number of days taken by A and B together to complete \[{{\left( \dfrac{5}{42} \right)}^{th}}\] of the total work = 1 day.

Therefore, the number of days taken by A and B together to complete the whole work = $\dfrac{42}{5}$ days.

Hence, number of days taken by A and B together to complete \[{{\left( \dfrac{6}{7} \right)}^{th}}\] of the total work,

$=\dfrac{42}{5}\times \dfrac{6}{7}$ days

Simplifying, we get,

$=\dfrac{6\times 6}{5}$ days

$=\dfrac{36}{5}$ days.

Therefore, time taken to complete the whole work,

$=\dfrac{36}{5}$ days + $3$ days

Taking LCM,

$=\left( \dfrac{36}{5}+\dfrac{3\times 5}{1\times 5} \right)$ days

$=\dfrac{36+15}{5}$ days

$=\dfrac{51}{5}$ days.

$=10\dfrac{1}{5}$ days

Hence, the correct answer is option (b).

Note: In such types of questions, do not confuse fraction of work with number of works. And keep in mind that if we are considering a fraction of work, then whole work will be represented by 1.