Question

If $A$ can do a piece of work in $25$ days and $B$ in $20$ days. They work together for $5$ days and then $A$ goes away. In how many days will $B$ finish the remaining work?
a)$17$ days
b)$11$ days
c) $10$ days
d) None of these

Answer 1

Hint: Here we will be converting a number of days to one day work. So, we will take one day's work in fractions. Then take LCM for those numbers then substitute the days and derive the values we will get the $B$’s working days.

Complete step-by-step answer:
$A$’s $1$ day’s work $ = \dfrac{1}{{25}}$
Similarly, $B$’s $1$ day’s work $ = \dfrac{1}{{20}}$
$A$’s and $B$’s together $1$ day’s work
\[\dfrac{1}{{25}} + \dfrac{1}{{20}}\]
We will be taking LCM for $25,20$,
Here LCM is $ = 100$ ,
$\dfrac{1}{{25}}$ is multiply dividend and divisor as $4$ ,
$\dfrac{1}{{20}}$ is multiply dividend and divisor as $5$ ,
We will get the answer,
$\dfrac{4}{{100}} + \dfrac{5}{{100}} = \dfrac{{4 + 5}}{{100}} = \dfrac{9}{{100}}$
Their $5$ days work together $ = 5 \times 1$ day’s work
$1 - \dfrac{{45}}{{100}}$ (Here $45$means $A$’s piece of work+$B$’s piece of work as$20 + 25 = 45$ )
$\dfrac{{100}}{{100}} - \dfrac{{45}}{{100}} = \dfrac{{100 - 45}}{{100}} = \dfrac{{55}}{{100}}$
Now, this remaining work is done by $B$.
Let $B$ take $x$ days to complete it.
$
  \dfrac{1}{{20}} \times x = \dfrac{{55}}{{100}} \\
   \Rightarrow x = \dfrac{{55}}{{100}} \times 20 \\
   \Rightarrow x = \dfrac{{55}}{5} \\
   \Rightarrow x = 11 \\
 $
Hence, $B$ will finish the remaining work in $11$ days.
So, here option $b$ is the correct answer.

Additional Information:
Here we will also find $A$’s remaining work. Same fraction method will be used for this kind of question. We will commonly be taking $100$ as a total number of days.


Note: Here we will be taking $100$. As the total number of days. Because it is an easy method for converting and substituting and multiplying values in given questions. Then LCM as the common method for here finding total number of days