A and B can complete a task in 12 days. However, A had to leave a few days before the task was complete and hence it took 16 days in all to complete the task. If A alone could complete the work in 21 days, how many days before the work getting over did A leave ?

Hint: Efficiency is defined as the work done in one day. So, to find the number of days in which B alone can do work we must first, find the efficiency of B ( i.e. work done by B in one day ).

Complete step-by-step answer:
If we had to find the days in which B alone can do the work so by the formula
\[ \Rightarrow \dfrac{{{\text{days taken by }}(A + B){\text{ }} \times {\text{ days taken by }}A{\text{ alone}}}}{{{\text{days taken by }}A{\text{ alone - days taken by }}(A + B)}}\]
So, \[ \Rightarrow \dfrac{{12 \times 21}}{{21 - 12}} = \dfrac{{252}}{9}\]= 28 days
So, B alone can do the work in 28days.
Now let us assume that A worked for ‘x’ number of days.
A can do \[\dfrac{1}{{21}}\]work in a day and B can do \[\dfrac{1}{{28}}\]work in a day and as we know that B worked for 16 days and A worked of ‘x’ days ( by assumption ).
So, the work done by A + work done by B = total work ( i.e. 1 )
\[ \Rightarrow \dfrac{x}{{21}}\; + \;\dfrac{{16}}{{28}}\; = \;1\]
Now solving this equation
\[ \Rightarrow \dfrac{x}{{21}}\; = \;\;1 - \dfrac{{16}}{{28}}\]
Now taking L.C.M and solving it will give us the value of ‘x’
\[ \Rightarrow x = {\text{ }}\dfrac{{12}}{{28}}\,\, \times \;21\; = 9\]days
Now, A worked for 9 days and the total work was completed in 16 days.
So, A left the work ( 16 – 9 ) days before the completion of work.
Hence A leaves the work 7 days before.

Note: Whenever this type of problem is seen then we must firstly know the work done by all the participants ( here A and B ) in a day or in an hour and this is known as the efficiency of the particulars. And if we know the work done by them in one day then we will easily calculate the work done by them in a given time. So this will make it easy for us to find the result that we are supposed to get.