A, B and C can do $50\%$ of a work in $15$ days, $12$ days and $20$ days, respectively. They began the work together but A left $2$ days before the completion of the work. In how many days was the work completed?

If they can complete $50\%$ of the work in the given days, then the time required to complete $100\%$ of the work is double:
* Time taken by A = $15\times 2=30$ days
* Time taken by B = $12\times 2=24$ days
* Time taken by C = $20\times 2=40$ days

Let the total work be the $\text{LCM}$ of $30,24,40$:
$$\text{Total Work}=120\text{ units}$$

Now, compute individual efficiencies per day:
* Efficiency of A = $\frac{120}{30}=4\text{ units/day}$
* Efficiency of B = $\frac{120}{24}=5\text{ units/day}$
* Efficiency of C = $\frac{120}{40}=3\text{ units/day}$

Let the total time taken to complete the work be $x$ days.
* A worked for $(x-2)$ days.
* B and C worked for the entire $x$ days.

Set up the equation based on work done:
$$\text{Work done by A}+\text{Work done by B}+\text{Work done by C}=\text{Total Work}$$
$$4(x-2)+5(x)+3(x)=120$$
$$4x-8+5x+3x=120$$
$$12x-8=120$$
$$12x=128$$
$$x=\frac{128}{12}=\frac{32}{3}=10\frac{2}{3}\text{ days}$$