Suppose A takes twice as much time as B and thrice as much time as C to complete a work. If all of them work together, they can finish the work in 2 days. How much time B and C working together to finish it?

Hint- If one person does a work in x days and another person does it in y days then together they can finish that work in \[\dfrac{{xy}}{{x + y}}\] days.
Let’s take work done by A be x
A takes twice as much as B Therefore B takes half of what time A takes.
$ \Rightarrow B = \dfrac{x}{2}$
A takes thrice as much as C. Therefore C takes one third of what time A takes.
$ \Rightarrow C = \dfrac{x}{3}$
When all of them work together, they can finish work in 2 days.
$
   \Rightarrow \dfrac{1}{A} + \dfrac{1}{B} + \dfrac{1}{C} = \dfrac{1}{2} \\
   \Rightarrow \dfrac{1}{x} + \dfrac{2}{x} + \dfrac{3}{x} = \dfrac{1}{2} \\
   \Rightarrow \dfrac{6}{x} = \dfrac{1}{2} \\
 $
Now, Cross multiply
$ \Rightarrow x = 12$
 Now, we calculate how much time taken by B to complete work.
$
  B = \dfrac{x}{2} = \dfrac{{12}}{2} \\
   \Rightarrow B = 6 \\
 $
B takes 6 days to complete work.
Now, we calculate how much time taken by C to complete work.
$
  C = \dfrac{x}{3} = \dfrac{{12}}{3} \\
   \Rightarrow C = 4 \\
$
C takes 4 days to complete work.
Now, we calculate how much time taken by B and C to work together.
 B does work in 6 days and C does it in 4 days.
 If B and C can work together
$
   \Rightarrow \dfrac{1}{B} + \dfrac{1}{C} \\
   \Rightarrow \dfrac{1}{6} + \dfrac{1}{4} \\
$
Take LCM
$
   \Rightarrow \dfrac{{2 + 3}}{{12}} \\
   \Rightarrow \dfrac{5}{{12}} \\
$
So, If B and C work together they will take $\dfrac{{12}}{5}$ days.

Note- Whenever we face such types of problems we use some important points. Like we calculate how much time taken by a single person to complete their work then we calculate how much time taken by persons when they work together.