A alone can do a piece of work in 10 days. B alone can do the same work in 12 days. They work on alternate days starting with A. In how many days will they complete $$\frac{3}{4}$$ of the total work?

Let's assume the total work is 60 units.
A alone can do a piece of work in 10 days.

The efficiency of A alone = $$\frac{60}{10}$$ =  6 units/day

B alone can do the same work in 12 days.

The efficiency of B alone = $$\frac{60}{12}$$ = 5 units/day

$$\frac{3}{4}$$ of the total work = $$\frac{3}{4}$$ of 60

= $$\frac{3}{4}\times60$$

= 45 units

So 45 units of work need to be done on alternate day.

They work on alternate days starting with A.

Work done by A on the first day = 6 units.

Work done by B on the second day = 5 units.

So the work done by A and B in these two days on alternate basis = 6+5 = 11 units.

By this way when working on alternate basis in 8 days, 44 units of work will be completed, and the remaining 1 unit of work will be done by A in $$\frac{1}{6}$$ days.

Hence the time taken by A and B working on alternate basis to do the 45 units of work = $$8+\frac{1}{6}$$

= $$\frac{48+1}{6}$$

= $$\frac{49}{6}$$ days