At their usual efficiency levels, A and B together finish a task in 12 days. If A had worked half as efficiently as she usually does, and B had worked thrice as efficiently as he usually does, the task would have been completed in 9 days. How many days would A take to finish the task if she works alone at her usual efficiency?

Correct Option: 2 (18)

Let the rate of work done per day by A and B at their usual efficiency levels be $a$ and $b$ respectively, and let the total work be 1 unit.

From the first case, A and B together complete the task in 12 days:
$12(a+b)=1$ --- (Equation 1)

From the second case, if A works at half efficiency ($\frac{a}{2}$) and B works at thrice efficiency ($3b$), they complete it in 9 days:
$9\left(\frac{a}{2}+3b\right)=1$ --- (Equation 2)

Equating both expressions for the total work:
$12(a+b)=9\left(\frac{a}{2}+3b\right)$

Divide both sides by 3:
$4(a+b)=3\left(\frac{a}{2}+3b\right)$
$4a+4b=1.5a+9b$
$2.5a=5b$
$a=2b$

Substitute $b=\frac{a}{2}$ in Equation 1:
$12\left(a+\frac{a}{2}\right)=1$
$12\left(\frac{3a}{2}\right)=1$
$18a=1$
$a=\frac{1}{18}$

Since A's daily rate is $\frac{1}{18}$ of the task, A working alone at her usual efficiency would take 18 days to finish the task.