When they work alone, B needs 60% more time to finish a piece of work than A does. Both of them finish the work in the following manner: B works alone till one-fourth of the work is done, then A and B work together for 12 days, and finally, A works alone to complete the remaining 10% of the work. In how many days can A alone finish the entire work?

Let's denote the time A needs to finish the work alone as 'a' days.

Since B needs 60% more time to finish the work than A does, B would need 1.6a days to finish the work alone.

We know that work done is inversely proportional to time taken. So, the work done by A in one day is 1/a and the work done by B in one day is 1/(1.6a).

According to the problem, B works alone till one-fourth of the work is done. So, B works for (1/4)*(1.6a) = 0.4a days.

Then, A and B work together for 12 days. The work done by A and B in one day is 1/a + 1/(1.6a). So, the work done by A and B in 12 days is 12*(1/a + 1/(1.6a)).

Finally, A works alone to complete the remaining 10% of the work. So, A works for (1/10)*a = 0.1a days.

Adding up all these times, we get the total time taken to finish the work as 0.4a + 12 + 0.1a = 0.5a + 12.

But we know that the total time taken to finish the work should be 'a' days. So, we can set up the equation 0.5a + 12 = a and solve for 'a'.

Subtracting 0.5a from both sides, we get 12 = 0.5a.

Dividing both sides by 0.5, we get a = 24.

So, A alone can finish the entire work in 24 days.