Question

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?

Answer 1

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.