Question

A does 80% of work in 20 days. He then calls in B and they together finish the remaining work in 3 days. How long B alone would take to do the whole work?
$
  \left( A \right)23\,days \\
  \left( B \right)37\,days \\
  \left( C \right)37\dfrac{1}{2} \,days \\
  \left( D \right)40\,days \\
$

Answer 1

Hint: If a person can do work in n days then his one day work will be $\dfrac{1}{n}$. Therefore the number of days is equal to total work done by work done in one day.

Complete step-by-step answer:
The objective of the problem is to find how long B alone would take to do the whole work.
Given that , A can do $80\% $ of work in 20 days.
We can write work done by A in 20 days $ = \dfrac{{80}}{{100}} = \dfrac{8}{{10}} = \dfrac{4}{5}$
Now , we find the work done by A in one day .
Work done by A in one day
$
  A = \dfrac{4}{5} \times \dfrac{1}{{20}} \\
  = \dfrac{1}{{25}} \\
$
It is given that A calls B and they together finish the remaining work in days. Thus, the remaining work done by A and B together is $20\% $.
Therefore , work done by A and B together $A + B = 20\% = \dfrac{{20}}{{100}} = \dfrac{1}{5}$
Then work done by A and B in one day $\left( {A + B} \right) = \dfrac{1}{5} \times \dfrac{1}{3} = \dfrac{1}{{15}}$
The work done B in one day can be obtained by subtracting work done by A and B together in one day and work done by A in one day. We get ,
$ = \dfrac{1}{{15}} - \dfrac{1}{{25}}$
The LCM of 15 and 25 is 150 .
$
   = \dfrac{{10 - 6}}{{150}} \\
   = \dfrac{4}{{150}} = \dfrac{2}{{75}} \\
$
Thus, work done by B alone is given by $\dfrac{1}{{work\,done\,in\,one\,day}}$
$ = \dfrac{1}{{2/75}} = \dfrac{{75}}{2} = 37.5$
Therefore , B can do work alone in $37\dfrac{1}{2}$ days.
Thus, c is the correct option.

Note: The total amount of a complete work is always equal to one unless otherwise it is specified .
If a person does his work in 1/n days then after m days his work will be $m\left( {1/n} \right)$.