A and B can do a work in 9 days and 12 days respectively. If they work on alternate days starting with A, then in how many days will the work be completed?

The correct option is:
$10\frac{1}{4}\text{days}$

Here is the step-by-step derivation to show how this answer is obtained:

Let's define the total amount of work to be done. A convenient way is to find the Least Common Multiple (LCM) of the time taken by A and B to complete the work individually.
Time taken by A = 9 days
Time taken by B = 12 days
The LCM of 9 and 12 is 36. So, let us assume the total work is 36 units.

Now, let's find the daily work efficiency (units of work done per day) of A and B:
Efficiency of A = Total Work / Time taken by A = 36 / 9 = 4 units per day.
Efficiency of B = Total Work / Time taken by B = 36 / 12 = 3 units per day.

They work on alternate days starting with A.
On Day 1: A works and completes 4 units of work.
On Day 2: B works and completes 3 units of work.
So, in a cycle of 2 days (Day 1 + Day 2), the total work completed is:
$4+3=7\text{units}$

We want to find how many such 2-day cycles can fit into the total work of 36 units:
Number of complete cycles = 36 / 7 = 5 cycles (with some remainder).
In 5 complete cycles, the number of days elapsed is:
$5\times 2=10\text{days}$
Work completed in 10 days = 5 cycles * 7 units/cycle = 35 units.

Remaining work to be done:
$36-35=1\text{unit}$

On the 11th day, it is A's turn to work again. Since A can complete 4 units of work in a full day, the time taken by A to complete the remaining 1 unit of work is:
$\text{Time}=\frac{\text{Remaining work}}{\text{A's efficiency}}=\frac{1}{4}\text{day}$

Therefore, the total time required to complete the entire work is:
$10+\frac{1}{4}=10\frac{1}{4}\text{days}$