\text{A can do a work in 9 days and B in 12 days. They work on alternate days, beginning with B. In how many days will the work be completed?}

The correct option is:
$10\frac{1}{3}$

Let's break down the solution step-by-step to understand how this answer is obtained.

Step 1: Determine the total work and individual efficiencies
Let the total work be the Least Common Multiple (LCM) of the days taken by A and B.
A can complete the work in 9 days.
B can complete the work in 12 days.
The LCM of 9 and 12 is 36. So, let the total work be 36 units.
Now, we can find the daily work (efficiency) of A and B:
Efficiency of A = Total Work / Days taken by A = 36 / 9 = 4 units per day.
Efficiency of B = Total Work / Days taken by B = 36 / 12 = 3 units per day.

Step 2: Calculate work done in alternate day cycles
They work on alternate days, starting with B.
On Day 1, B works and completes 3 units.
On Day 2, A works and completes 4 units.
Therefore, in a 2-day cycle:
Work done in 2 days = Efficiency of B + Efficiency of A = 3 + 4 = 7 units.

Step 3: Calculate the number of complete cycles
We need to reach close to the total work of 36 units.
Number of 2-day cycles = 36 / 7 = 5 cycles (with some remainder).
In 5 cycles, the total number of days is:
5 cycles × 2 days/cycle = 10 days.
The work completed in 10 days is:
5 cycles × 7 units/cycle = 35 units.

Step 4: Calculate the remaining work and the time to complete it
Remaining work = Total work - Work completed = 36 - 35 = 1 unit.
Since 5 complete cycles are finished (each cycle ending with A), the next turn on the 11th day is B's turn.
B can do 3 units of work in a full day.
To complete the remaining 1 unit of work, B will take:
Time = Remaining work / Efficiency of B = 1/3 day.

Step 5: Total time taken
Total time = Time for 5 cycles + Time taken by B on the 11th day
Total time = 10 + 1/3 = 10 1/3 days.
Thus, the work will be completed in:
$10\frac{1}{3}$ days.