Working alone at their respective constant rates, A can complete a task 'a' days and B in "b" days. They take turns in doing the task with each working 2 days at a time. starts they finish the task in exactly 10 days. If B starts, they take half a day more. How long does it take to complete the task if they both work together?

As, A starts the work

So, A works for = 6 days

B works for = 4 days

that means = $\frac{6}{a}+\frac{4}{b}=1.......\left(i\right)$

$$\begin{gathered} If, B starts the work, they take 10\frac{1}{2} days to complete work. \\ So, B works for = 6 days \\ \therefore A works for = 4\frac{1}{2} days \\ That mean\text{'}s =\frac{4.5}{a}+\frac{6}{b}=1............\left(ii\right) \\ Multiplying \left(i\right) by 3 and \left(ii\right) by 2 and subtracting \left(ii\right) from \left(i\right). we get \end{gathered}$$

$$\begin{gathered} \frac{{\displaystyle \frac{18}{a}}+{\displaystyle \frac{12}{b}}=3 \\ {\displaystyle \frac{9}{a}}+{\displaystyle \frac{12}{b}}=2}{{\displaystyle \frac{18}{a}}-{\displaystyle \frac{4}{b}}=1} \\ =\frac{18-9}{a}=1 \\ =\frac{9}{a}=1 \\ \therefore a=9 \end{gathered}$$

putting the value of a in (i), we get

$$\begin{gathered} \frac{6}{a}+\frac{4}{b}=1 \\ =\frac{6}{9}+\frac{4}{b}=1 \\ =\frac{4}{b}=1-\frac{2}{3}=\frac{3-2}{3}=\frac{1}{3} \\ \therefore b=12 \end{gathered}$$

so, A can complete the work in = 9 days

& B can complete the work in = 12 days

$\therefore$They together can do the woek in 1 day = $\frac{1}{9}+\frac{1}{12}=\frac{4+3}{36}=\frac{7}{36}$

$$\begin{gathered} \frac{7}{36} part of work in done by them in = 1 day \\ \therefore 1 or complete work is done by them in = \frac{36}{7} day \end{gathered}$$