Question

A can do \[\dfrac{2}{3}\] of a certain work in \[16\] days and B can do \[\dfrac{1}{4}\] of the same work in \[3\] days. In how many days can both finish the work working together?

Answer 1

Hint: Let us first take the whole work as \[1\] unit.
We need to find first the work done by \[A\] and \[B\] in \[1\] day. Then we can calculate the work done by \[A\] and \[B\] in \[1\] day.
Then applying the unitary method on the work done by \[A\] and \[B\] in \[1\] day, we can find out the number of days they both need to finish the work, working together.
(The unitary method is a technique for solving a problem by first finding the value of a single unit and then finding the necessary value by multiplying the single unit value.)

Complete step by step answer:
It is given that, \[A\] can do \[\dfrac{2}{3}\] of a certain work in \[16\] days and \[B\] can do \[\dfrac{1}{4}\] of the same work in \[3\] days.
We need to find out the number of days they both need to finish the work, working together.
Let us first take the whole work as a \[1\] unit.
Now, \[A\] can be done in \[16\] days \[\dfrac{2}{3}\] units of a certain work.
In \[1\] day \[A\] can do:
\[\Rightarrow \dfrac{{\dfrac{2}{3}}}{{16}} = \dfrac{2}{3} \times \dfrac{1}{{16}} = \dfrac{1}{{24}}\] units of work.
\[B\] can do in \[3\] days, \[\dfrac{1}{4}\] units of the work.
In \[1\] day \[B\] can do:
\[\Rightarrow \dfrac{{\dfrac{1}{4}}}{3} = \dfrac{1}{4} \times \dfrac{1}{3} = \dfrac{1}{{12}}\] units of the work.
Now, \[A\] and $B$ together can do in \[1\] day:
\[\Rightarrow \dfrac{1}{{24}} + \dfrac{1}{{12}} = \dfrac{{1 + 2}}{{24}} = \dfrac{3}{{24}} = \dfrac{1}{8}\] unit of work.
Thus \[A\] and \[B\] together can do \[\dfrac{1}{8}\] unit of work in \[1\] day.
\[A\] and \[B\] together can do the whole work that is, \[1\]unit of work in \[\dfrac{1}{{\dfrac{1}{8}}} = 1 \times \dfrac{8}{1} = 8\] days.

Hence, \[A\] and \[B\] together can finish the work in \[8\] days.

Note:
We have solved the problem with the unitary method.
Now we need to know what is the unitary method?
In short, this method is used to find the value of a unit from the value of a multiple, and hence the value of a multiple.